Post by: Jaaanosik on 11/12/2021 16:38:58
As the subject says: Does the Equivalence Principle violates the conservation of energy law?
Here is the thought experiment.
(https://i.imgur.com/JPVwbu4.png)
[5] Feynman-Leighton-Sands, THE Feynman LECTURES ON PHYSICS, The NEW MILLENNIUM Edition, VOLUME II: MAINLY ELECTROMAGNETISM AND MATTER, page 42-10, ISBN 978-0-465-07998-8, 1964, 2006, 2010.
Post by: Kryptid on 11/12/2021 17:58:25
Post by: Jaaanosik on 11/12/2021 18:22:16
After the separation there is no force acting on Voyager, only kinetic energy changing to potential energy.
The total energy change is zero.
The heights is growing, what is doing the work in the Eq 2?
is supposed to be the same in any heights. That's the rest frame of the atom. This is the core of the Equivalence Principle.
But also Feynman showed us how the photon falling from the higher heights will have more energy when absorbed on the Earth.
Does the photon care how was the height gained?
Post by: Bored chemist on 11/12/2021 19:38:37
After the separation there is no force acting on Voyager,Yes there is; gravity.
Post by: Kryptid on 11/12/2021 20:53:09
Einstein's happiest thought: "No force acting on 'free falling body'".
Voyager isn't a free-falling body. It's moving away from the Earth, not towards it. As such, it is moving up a gravitational potential. Some of its kinetic energy is constantly being changed into gravitational potential energy.
Post by: Jaaanosik on 11/12/2021 21:22:41
After the separation there is no force acting on Voyager,Yes there is; gravity.
3min 40sec of this video...
Post by: Jaaanosik on 11/12/2021 21:33:37
Would it be easier for you to imagine a space station on an elliptic orbit?Einstein's happiest thought: "No force acting on 'free falling body'".
Voyager isn't a free-falling body. It's moving away from the Earth, not towards it. As such, it is moving up a gravitational potential. Some of its kinetic energy is constantly being changed into gravitational potential energy.
Photon
emitted at perihelion and 
emitted at aphelion.The absorbed energy on the Earth is different because of different heights of perihelion and aphelion.
Yet, the Equivalence Principle says
is the same in perihelion and aphelion.The space station energy is constant; perihelion more kinetic energy, aphelion more potential energy, the sum is the same.
Were is the delta of the absorbed energy coming from when atoms in orbit are not changing total energy sum?
Post by: Kryptid on 12/12/2021 00:25:45
Post by: Jaaanosik on 12/12/2021 00:29:52
The energy didn't appear out of nowhere. The photon with the higher frequency simply removed a larger percentage of the total energy than the lower frequency photon did.Higher frequency where?
The emission is the same according to the Equivalence Principle, agreed?
Post by: Kryptid on 12/12/2021 00:34:43
Alternatively, it may depend on the reference frame that the measuring is done in. I'm sure Halc will chime in if I've got that wrong.
Post by: Jaaanosik on 12/12/2021 01:00:59
As far as I'm aware, the equivalence principle assumes a uniform gravitational field (which the Earth's is not).
Alternatively, it may depend on the reference frame that the measuring is done in. I'm sure Halc will chime in if I've got that wrong.
http://www.people.vcu.edu/~rgowdy/mod/r43/imp.htm#1
Quote
In a sufficiently small laboratory, there is no experiment done entirely inside the laboratory which can tell whether the laboratory is in free fall in a gravitational field or in uniform motion in empty space.
In other words there is no experiment inside the space station to determine if the space station is falling in perihelion or in aphelion.
That's the reason
is the same. This is the claim of the Equivalence Principle.Post by: Kryptid on 12/12/2021 01:09:36
Post by: Jaaanosik on 12/12/2021 01:14:34
For an observer inside the lab, that is true because the experimental equipment is in the same reference frame as the excited atom. So like I said, whether or not you measure a frequency difference depends on your reference frame relative to the atom.OK, so we established
is the same amount of energy emitted.The Eq 2 is true as well, according to Feynman.
How do we get more energy from the aphelion photon if no work is done between perihelion and aphelion?
Post by: Kryptid on 12/12/2021 01:17:46
Post by: Jaaanosik on 12/12/2021 01:27:47
In what reference frame are you asking the question?The
is measured on the Earth.
Post by: yor_on on 12/12/2021 02:32:20
well, with a caveat, it's about displacements isn't it? As soon as that photon leaves it change your reference frame, you're no longer in a same frame of reference.
=
In a free fall that is, locally indistinguishable from a absence of gravity. A photon has no acceleration and should be able to be considered a point particle excited at different locations in a so called 'photon field' in where your measurement will define its existence.
Then again, frames of reference is still very confusing to me, how to define them and at what scale they are valid. Einstein only used 'point particles' defining it as a I remember reading about it a long time ago. So that part about being or not being in a same frame of reference is the way I think about it.
You can look here, it also contain a link to a series of lectures (Free book) by Einstein in Princeton 1921.
https://physics.stackexchange.com/questions/61757/how-or-why-equivalence-principle-led-to-einstein-field-equations
=
Thinking some more about it. If we define it such as there is no blue or red shift, then you should be in the same frame of reference, and that is one definition of how a free fall differs from a acceleration, locally defined.
Post by: Jaaanosik on 12/12/2021 03:00:22
It's a pity that Pete no longer is amongst us. I'm not totally sure what you are asking about here. If you're asking if different reference frames will see different energies then that is true. Inside that space station, assuming the observer being in a exact same frame of reference with what he observes, there shouldn't be any possibility of measuring different energies depending on ones location in a gravitational potential / field though. As far as I know?Fair comments, thanks.
well, with a caveat, it's about displacements isn't it? As soon as that photon leaves it change your reference frame, you're no longer in a same frame of reference.
=
In a free fall that is, locally indistinguishable from a absence of gravity. A photon has no acceleration and should be able to be considered a point particle excited at different locations in a so called 'photon field' in where your measurement will define its existence.
Energy is frame dependent, the delta of energy is supposed to be the same in all frames though.
The space station frame does not see any energy change that is stored between
and 
when going from perihelion to aphelion.The Earth frame is supposed see the same, no change, because there is no work done, just kinetic energy changed to potential energy.
Nevertheless the Eq 2 is showing us there is supposed to be a change in the Earth frame.
Post by: yor_on on 12/12/2021 03:26:14
Post by: yor_on on 12/12/2021 03:30:09
=
When I think of that it's frame dragging that comes to mind :)
Relativity is constant head ache Jaaanosik but I know of nothing more interesting.
If I use earth as my frame of observation then I invoke GR, where frame dragging also is about GR (spin). If I use the local definition I'm staying in SR. Relative motion as such belongs to SR but it is still dynamically changing gravity (versus the earth observer). You might be able to define it as taking itself out over a whole universe, 'unchanging', maybe that is possible, I just don't know :) but it seems almost reasonable to do so.
Post by: Origin on 12/12/2021 04:04:30
How do we get more energy from the aphelion photon if no work is done between perihelion and aphelion?It looks like there should be a difference between them according to your equation;
. There is a difference in the amount of work since H is different for the perihelion and aphelion.
Post by: yor_on on 12/12/2021 04:09:56
https://profoundphysics.com/does-gravity-do-work/
Post by: Halc on 12/12/2021 04:19:44
Kryptid, I'm going to have to disagree with some of the stuff posted.
It's not true to say that no force is acting on Voyager.True under Newtonian physics, but not true under GR, and GR is what's under discussion if EQ is the subject at hand. Gravity is manifest as spacetime curvature, not as a force. Voyager always moves on a geodesic, and hence has no proper force acting on it. Yes, it will gain or lose speed in accordance with changes in gravitational potential, but that's only coordinate acceleration, not proper acceleration.
Voyager isn't a free-falling body. It's moving away from the Earth, not towards it.This is not even true under Newtonian physics. Both define free-fall as being subject only to gravity, not any other force. A rock falling up is as much in freefall as it is after it reaches its maximum altitude.
Would it be easier for you to imagine a space station on an elliptic orbit?EQ does not apply here. Measurements are being taken in different frames. Yes, the energy (of the photon) measured on Earth will be more that the same one measured up high. GR (not EP) says that a photon emitted at different altitudes by similar means will locally be measured to be the same energy, but you're not making a local measurement here.
Photon
The absorbed energy on the Earth is different because of different heights of perihelion and aphelion.
Quote
Yet, the Equivalence Principle saysYou'll have to flesh out how you think EP says this. It is worded sufficiently ambiguously that I can't say it's wrong.
As far as I'm aware, the equivalence principle assumes a uniform gravitational field (which the Earth's is not).It doesn't. It assumes locality is all. Tests are done in a small box. Given a non-local test (a long box), one can always distinguish a uniform field from acceleration. For one, an accelerated reference frame has no escape velocity, something I see referenced in the OP.
Quote
Alternatively, it may depend on the reference frame that the measuring is done in. I'm sure Halc will chime in if I've got that wrong.
So like I said, whether or not you measure a frequency difference depends on your reference frame relative to the atom.Yes. That's what I mean by these measurements being frame dependent.
OK, so we establishedNo frame specified, so the statement implies absolute energy, which is wrong.
measured where? Somewhere other than where the emission event occurs? That's a frame change.Energy is frame dependent, the delta of energy is supposed to be the same in all frames though.This is incorrect. An excited atom emitting a photon at high altitude results in a higher energy emission than a similar emission locally at sea level, both as measured at sea level.
You've specified Earth (sea level) as your frame in post 14.
Post by: Jaaanosik on 12/12/2021 04:28:23
... and this is a problem.How do we get more energy from the aphelion photon if no work is done between perihelion and aphelion?It looks like there should be a difference between them according to your equation;
We are getting work done without spending any energy, because total energy is constant.
Kinetic energy is changing to potential energy for the space station.
The Eq 2, 3, 4 show that the same photon absorbed from higher height has more energy.
Hence the question: Does the Equivalence Principle violate the conservation of energy law?
Post by: Jaaanosik on 12/12/2021 04:30:53
so we are on the same page, from Feynman for quotation purpose:
(https://i.imgur.com/sXXVh0C.png)
(https://i.imgur.com/N26vWwj.png)
(https://i.imgur.com/dkY5AID.png)
(https://i.imgur.com/A1e6Bqo.png)
Post by: yor_on on 12/12/2021 04:41:58
spelling
Post by: Kryptid on 12/12/2021 04:52:13
We are getting work done without spending any energy, because total energy is constant.
It's normal for "spent" energy to not disappear, so of course the total energy is constant (that's conservation of energy for you). If I pick a rock up off of the ground, I did work but the energy that I used to lift the rock didn't disappear. It was transformed into the gravitational potential energy of the rock. The total energy of the system (The rock, the Earth and me) remained the same.
Post by: Jaaanosik on 12/12/2021 04:54:45
...Would it be easier for you to imagine a space station on an elliptic orbit?EQ does not apply here. Measurements are being taken in different frames. Yes, the energy (of the photon) measured on Earth will be more that the same one measured up high. GR (not EP) says that a photon emitted at different altitudes by similar means will locally be measured to be the same energy, but you're not making a local measurement here.
Photon
The absorbed energy on the Earth is different because of different heights of perihelion and aphelion.QuoteYet, the Equivalence Principle saysYou'll have to flesh out how you think EP says this. It is worded sufficiently ambiguously that I can't say it's wrong.
...
The space station is on the Earth surface full of atoms in
state.A rocket brings space station into the elliptic orbit.
A photon is emitted from perihelion back to the Earth and absorbed.
A photon is emitted from aphelion back to the Earth and absorbed.
The space station frame and the Earth frame agree there is no work done on the space station and atoms on it.
The Eq 2 or Feynman's 42.10 says otherwise, photon energy of a photon that 'fell' from higher height has more energy.
Post by: yor_on on 12/12/2021 11:43:03
Post by: Jaaanosik on 12/12/2021 13:42:32
Let's see. I very much doubt that Feynman didn't agree with relativity. And I'm sure he had a answer for his statement at the end of this excerpt. and the only thing I can see that explains it is gravity. I get confused reading about all this 'work done', lifting up and down photons. It doesn't fit my thinking at all. It's a gravitational field if I get it right and if it is then being in a exact same frame of reference becomes very tricky. I said that Einstein used point particles or 'dust', and the equivalence principle, if I now remember correctly, is defined as only over a small enough region will the equivalence principle hold. and that one is directly related to what one mean by a frame of reference.Interesting, isn't it?
One way to look at the atoms is as photon 'containers', how the energy of the atom changes so does the photon's energy in it.
Feynman pointed out the equivalence of the accelerated rocket to the rocket standing on the Earth.
The Eq 42.10 is supposed to be universal and it came from the EP.
When the analysis is done in the Earth frame, does it violate the conservation of energy law?
Post by: Kryptid on 12/12/2021 14:12:48
When the analysis is done in the Earth frame, does it violate the conservation of energy law?
No. The amount of energy removed from the atom by the photon never exceeds the total amount of energy that the atom has.
Post by: yor_on on 12/12/2021 14:18:46
=
I think I can say one thing. Using red and blue shifts to define being in a acceleration, then stating that a photon 'blue shifts' in a relative motion is quite extraordinary. Unless we're discussing different frames of reference there. Frames of reference can seriously hurt ones mind wandering into it :)
You can in a black box scenario define if you're 'accelerating' or in a relative motion by using light and red and blue shifts of it. I know, you can use gravity , locally defined, for it too but that's not my main point here.
Post by: Origin on 12/12/2021 14:26:59
Does the Equivalence Principle violate the conservation of energy law?No. It doesn't seem that you know what the equivalency principle is.
Post by: Jaaanosik on 12/12/2021 14:40:44
This is all in the Earth frame.When the analysis is done in the Earth frame, does it violate the conservation of energy law?
No. The amount of energy removed from the atom by the photon never exceeds the total amount of energy that the atom has.
The transition from
to meaning energy of is the same in perihelion and aphelion.
because height 
Where is the more work coming from when no force was acting on the space station and atoms between perihelion and aphelion?
Post by: Jaaanosik on 12/12/2021 14:41:10
Please, explain then...Does the Equivalence Principle violate the conservation of energy law?No. It doesn't seem that you know what the equivalency principle is.
Post by: yor_on on 12/12/2021 14:48:27
I might be forced to do that now :)
Post by: yor_on on 12/12/2021 14:58:37
https://www.feynmanlectures.caltech.edu/II_42.html
Post by: yor_on on 12/12/2021 15:01:21
" Our ideas about relativity, quantum physics, and energy conservation all fit together only if Einstein’s predictions about clocks in a gravitational field are right. The frequency changes we are talking about are normally very small. For instance, for an altitude difference of 20 meters at the earth’s surface the frequency difference is only about two parts in 1015. However, just such a change has recently been found experimentally using the Mössbauer effect.5 Einstein was perfectly correct. "
And I agree Jaaanosik, it is interesting and questions like yours makes us need to think. It deepens the understanding.
Post by: yor_on on 12/12/2021 15:34:31
And it is conserved that intrinsic energy this photon consist of, under a propagation. Using a field it's also 'conserved' and you don't need a propagation. You just need the emanation to 'pop up' under some dimensional constraints, And looked at that way it can't be anything else than 'conserved' to me, inside that matrix where you find it to exist.
Actually thinking of it this way makes locality something of a 'golden standard' :)
A new one, and observer dependent.
Maybe you shouldn't use words as observer at a distance for this one, but I see it as a relation between gravity, the clock, and the observer. Two clocks, the observers defining the 'frequency' of the observed. And thinking of it from frames of reference the only way you can synchronize those clocks is by being, or starting, in a same frame of reference. Distance wise that comes down to no distance at all. Well, if you think of it as a practical experiment with clocks as the ones NIST presents. But frames of reference are a lot more tricky than that. You can have the same frame of reference as something a thousand light years away, without anything else needing to be in that frame in-between as I gather. And if you think of it quantum mechanically it gets even weirder, how do you define a clock synchronization then? That's about those scales I'm always wondering about
You can use decoherence for it possibly? Defining it as a 'clock' only ticks above some scale? Under that scale they're indeterminate, unless you make a measurement of them? Possibly, as a wild guess. Any measurement you do will use your wristwatch and that one is macroscopic.
Post by: Origin on 12/12/2021 16:44:27
Please, explain then...I see that I was correct, you do not know what the equivalency principle says.
Post by: Kryptid on 12/12/2021 17:43:11
Where is the more work coming from
Gravity.
Post by: Halc on 12/12/2021 19:50:14
You'll have to flesh out how you think EP says this.Your reply here makes no mention of the EP. In fact, I've not seen you reference it in any post, except to quote Feynman who references it in a different example than the one you think violates energy conservation.
The space station is on the Earth surface full of atoms inThis is very different than Feynman's scenario. You've introduced velocity and associated time dilation. Stick to the simple scenario with just moving the atom up the stairs and forget adding orbital motion to it.
A rocket brings space station into the elliptic orbit.
Quote
A photon is emitted from perihelion back to the Earth and absorbed.Nonsense, as Feynman points out. You carried a larger mass up and didn't bring it all back down with you. Mass/energy was lost by the space station since it was not a closed system.
A photon is emitted from aphelion back to the Earth and absorbed.
The space station frame and the Earth frame agree there is no work done on the space station and atoms on it.
Quote
The Eq 2 or Feynman's 42.10 says otherwise, photon energy of a photon that 'fell' from higher height has more energy.Right. And it having higher energy has nothing to do with the EP.
I disagree with the completeness of Feynman's equations (like using 'g' like a constant), but his 42.10 is accurate enough for what he is illustrating.
Feynman pointed out the equivalence of the accelerated rocket to the rocket standing on the Earth.That he did, but not in the photon example. It was the prior example.
Quote
The Eq 42.10 is supposed to be universal and it came from the EP.It isn't wrong, but it didn't come from the EP. One can illustrate the same thing with Newtonian mechanics where a rock is raised up, a chip is ejected, and the remaining rock brought back down. The chip will have more energy when it reaches Earth than the energy measured up at the top of the stairs.
Quote
When the analysis is done in the Earth frame, does it violate the conservation of energy law?Not at all. Earth measures more energy, exactly as illustrated by 42-10. It all nicely balances, and no EP was ever invoked.
This is all in the Earth frame.Wrong at least twice. First of all, energy is frame dependent, and you have introduced different inertial frames here plus different potentials. Energy of a photon is dependent on the inertial frame used, so the high speed orbiting emission might be lower energy due to doppler, velocity time dilation, or both.
The transition from
As for potential (which is what Feynman was illustrating), E1 and E0 are different here than they are there, as is their difference. And yet again, EP has nothing to do with any of this, especially if orbits are involved.
Post by: Jaaanosik on 13/12/2021 02:00:38
And then it gets more understandable to me. He likens 'photons' to 'clocks' in a gravitational well (field) and uses their frequency. That frequency represents a energy and the deeper into that well you are the more 'energy' it accumulates, relative some stationary observer at a distance. As I read it then. But you need frames of reference for it to be defined this way. Locally defined that 'photon', if we make it 'propagate', is the same entity everywhere, us 'being at rest with it', which we can't be. Looked at differently it's a sort of 'photon field' in where you as the 'far observer' observe it to consist of different frequencies / energies depending on where you place your measurement in that gravitational well. That makes it observer dependent to me.Is
And it is conserved that intrinsic energy this photon consist of, under a propagation. Using a field it's also 'conserved' and you don't need a propagation. You just need the emanation to 'pop up' under some dimensional constraints, And looked at that way it can't be anything else than 'conserved' to me, inside that matrix where you find it to exist.
Actually thinking of it this way makes locality something of a 'golden standard' :)
A new one, and observer dependent.
Maybe you shouldn't use words as observer at a distance for this one, but I see it as a relation between gravity, the clock, and the observer. Two clocks, the observers defining the 'frequency' of the observed. And thinking of it from frames of reference the only way you can synchronize those clocks is by being, or starting, in a same frame of reference. Distance wise that comes down to no distance at all. Well, if you think of it as a practical experiment with clocks as the ones NIST presents. But frames of reference are a lot more tricky than that. You can have the same frame of reference as something a thousand light years away, without anything else needing to be in that frame in-between as I gather. And if you think of it quantum mechanically it gets even weirder, how do you define a clock synchronization then? That's about those scales I'm always wondering about
You can use decoherence for it possibly? Defining it as a 'clock' only ticks above some scale? Under that scale they're indeterminate, unless you make a measurement of them? Possibly, as a wild guess. Any measurement you do will use your wristwatch and that one is macroscopic.
> 
?
Post by: Jaaanosik on 13/12/2021 02:04:17
Gravity because the aphelion photon falls longer, agreed.Where is the more work coming from
Gravity.
Is there any force increasing photon's potential from perihelion to aphelion?
Post by: Jaaanosik on 13/12/2021 02:24:19
Feynman says: "Using the principle of equivalence we can figure out how much the speed of a clock changes with height in a gravitational field."You'll have to flesh out how you think EP says this.Your reply here makes no mention of the EP. In fact, I've not seen you reference it in any post, except to quote Feynman who references it in a different example than the one you think violates energy conservation.
... couple of paragraphs above the photon energy analysis. The frequency is energy for photon.
Is Feynman's equation 42.10 universal or not?
That equation was obtain from the equivalence principle analysis, rocket accelerating and making equivalence to rocket standing on the Earth.
Post by: Kryptid on 13/12/2021 02:27:31
Is there any force increasing photon's potential from perihelion to aphelion?
Yes, it's still gravity. It's like when a roller coaster goes up a rise after its initial drop. The coaster isn't powered, but the kinetic energy it gained by falling down the first drop carries it up against the pull of gravity when it ascends the first rise. In like manner, the kinetic energy gained by the satellite as it "falls" from aphelion to perihelion carries it back up on the opposite side of the orbit from perihelion back up to aphelion again.
Post by: Jaaanosik on 13/12/2021 02:28:35
...The space station is on the Earth surface full of atoms inThis is very different than Feynman's scenario. You've introduced velocity and associated time dilation. Stick to the simple scenario with just moving the atom up the stairs and forget adding orbital motion to it.
A rocket brings space station into the elliptic orbit.
...
I am interested in the scenario I presented.
Post by: Jaaanosik on 13/12/2021 02:35:15
...Now, this is very good observation. That's where Feynman made mistake.QuoteA photon is emitted from perihelion back to the Earth and absorbed.Nonsense, as Feynman points out. You carried a larger mass up and didn't bring it all back down with you. Mass/energy was lost by the space station since it was not a closed system.
A photon is emitted from aphelion back to the Earth and absorbed.
The space station frame and the Earth frame agree there is no work done on the space station and atoms on it.
...
1. Bring photon up in an atom.
2. Emit the photon.
3. "Bring the atom back down faster than emitted photon.... and receive the photon"
I hope you see the problem. That is why an atom of the same type in
that stayed back receives the photon.We should clarify difference between closed and isolated systems.
Post by: Kryptid on 13/12/2021 02:39:59
I hope you see the problem.
Nope.
Post by: Jaaanosik on 13/12/2021 02:49:12
The EP establishes the idea that falling photon gains energy during its fall. That is universal.
...QuoteThe Eq 2 or Feynman's 42.10 says otherwise, photon energy of a photon that 'fell' from higher height has more energy.Right. And it having higher energy has nothing to do with the EP.
I disagree with the completeness of Feynman's equations (like using 'g' like a constant), but his 42.10 is accurate enough for what he is illustrating.
...
Agreed on the 'g' change over the heights, but that is not important. We can look at it as two steps.
Falling photon from perihelion gains energy
, falling photon from aphelion gains energy 
.
Delta H is heights above perihelion height up-to aphelion.
Do you agree the falling photon gains energy and that is the universality of 42.10 (when done accurately to account for g variance factoring the heights).
Post by: Jaaanosik on 13/12/2021 03:04:55
...If there is no work done on atoms flying from perihelion to aphelion then why would the Earth measure more energy?QuoteWhen the analysis is done in the Earth frame, does it violate the conservation of energy law?Not at all. Earth measures more energy, exactly as illustrated by 42-10. It all nicely balances, and no EP was ever invoked.
...
Post by: Jaaanosik on 13/12/2021 03:13:39
...This is all in the Earth frame.Wrong at least twice. First of all, energy is frame dependent, and you have introduced different inertial frames here plus different potentials. Energy of a photon is dependent on the inertial frame used, so the high speed orbiting emission might be lower energy due to doppler, velocity time dilation, or both.
The transition from
As for potential (which is what Feynman was illustrating), E1 and E0 are different here than they are there, as is their difference. And yet again, EP has nothing to do with any of this, especially if orbits are involved.
Now this is very interesting point.
We can imagine photon as an energy stored in elasticity of the electron-proton interaction.
Just one electron and one proton for simplicity.
The electric field is responsible for the interaction, do you agree?
Post by: Jaaanosik on 13/12/2021 03:15:54
Are you sayin the inertia is doing the work?Is there any force increasing photon's potential from perihelion to aphelion?
Yes, it's still gravity. It's like when a roller coaster goes up a rise after its initial drop. The coaster isn't powered, but the kinetic energy it gained by falling down the first drop carries it up against the pull of gravity when it ascends the first rise. In like manner, the kinetic energy gained by the satellite as it "falls" from aphelion to perihelion carries it back up on the opposite side of the orbit from perihelion back up to aphelion again.
Post by: Kryptid on 13/12/2021 03:22:15
If there is no work done on atoms flying from perihelion to aphelion then why would the Earth measure more energy?
Work is done. There is displacement when the atoms change altitude from perihelion to aphelion.
Are you sayin the inertia is doing the work?
It's the force of gravity that's doing it.
Post by: Jaaanosik on 13/12/2021 03:27:54
Is gravity 'pushing' from perihelion to aphelion and doing work?If there is no work done on atoms flying from perihelion to aphelion then why would the Earth measure more energy?
Work is done. There is displacement when the atoms change altitude from perihelion to aphelion.Are you sayin the inertia is doing the work?
It's the force of gravity that's doing it.
Rocket gave the space station initial kinetic energy.
The energy is not changing after that.
Post by: Bored chemist on 13/12/2021 08:48:22
Is gravity 'pushing' from perihelion to aphelion and doing work?No, it is pulling; and work is done against it, causing a reduction in the KE of the orbiting object.
Gravity seldom pushes anything.
Post by: yor_on on 13/12/2021 10:06:10
If you do have something as a hypothesis or theory then you should present it in 'New Theories'
Post by: Halc on 13/12/2021 12:46:39
We can look at it as two steps.First of all, you seem to mean from apogee to perigee, and I don't see how the photon is going to get from one to the other without hitting Earth that sits in between.
Falling photon from perihelion gains energy
Delta H is heights above perihelion height up-to aphelion.
Do you agree the falling photon gains energy and that is the universality of 42.10 (when done accurately to account for g variance factoring the heights).
42.10 is a subtraction, and subtractions of two values are not different here than there, but the values being subtracted may very well be.
Feynman is subtracting EL from approximated EH (low and high altitudes) despite not accurately computing at least EH. That inaccuracy is a problem with 42.8, not 42.10. Both EL from EH are dependent on where they're measured, and I think 'at sea level' is implied.
Yours is even more complicated. You have EL (Earth), EA (apogee), and EP (perigee). EL may be for instance larger or smaller than either of the other two depending on the choice of frame (both potential and velocity) in which they are computed. EP might even be the highest of the three.
But once you've chosen your frame and actually computed everything correctly, then yes, 42.10 applies (a simple subtraction of two of the values), so I suppose that makes it universal. It's 42.8 that is so very frame dependent, and 42.10 only works if all values are computed in the same frame, and even then, the answer will be different when computed in different frames, so in that sense, 42.10 is not universal, but frame dependent. It's computing the energy of a photon at a given location in space, and that energy is frame and observer dependent.
If there is no work done on atoms flying from perihelion to aphelion then why would the Earth measure more energy?No work is done since the energy of your small orbiting thing is nearly constant until it ejects energy. Even without emitting photons, it is not a closed system, so its energy isn't actually constant. The emitted photon has more energy (in some frames, including the Earth one) because of the higher potential at apogee and the lowest speed relative to Earth, both of which result in more energy emitted in the Earth frame. Some other frames might measure EA to be lower than EP. Much depends on not only frame of choice, but exactly how high and how eccentric your chosen orbit is, none of which is reflected in any equation I've seen.
Also, time to at somewhat eat my own words:
Kryptid, I'm going to have to disagree with some of the stuff posted.As I indicated, there's no proper force on Voyager, but that doesn't mean no force. Its momentum (like the momentum of say the moon) is always changing, and that implies a force, even if not a proper force. So I will un-disagree with that statement.It's not true to say that no force is acting on Voyager.True under Newtonian physics, but not true under GR, and GR is what's under discussion if EQ is the subject at hand. Gravity is manifest as spacetime curvature, not as a force. Voyager always moves on a geodesic, and hence has no proper force acting on it. Yes, it will gain or lose speed in accordance with changes in gravitational potential, but that's only coordinate acceleration, not proper acceleration.
Post by: Jaaanosik on 13/12/2021 16:25:01
If we draw a circle in the perihelion heights, equipotential line, it goes to the other side of the Earth as well.We can look at it as two steps.First of all, you seem to mean from apogee to perigee, and I don't see how the photon is going to get from one to the other without hitting Earth that sits in between.
Falling photon from perihelion gains energy
Delta H is heights above perihelion height up-to aphelion.
Do you agree the falling photon gains energy and that is the universality of 42.10 (when done accurately to account for g variance factoring the heights).
42.10 is a subtraction, and subtractions of two values are not different here than there, but the values being subtracted may very well be.
Feynman is subtracting EL from approximated EH (low and high altitudes) despite not accurately computing at least EH. That inaccuracy is a problem with 42.8, not 42.10. Both EL from EH are dependent on where they're measured, and I think 'at sea level' is implied.
...
The aphelion photon emitted has to reach this perihelion line from the aphelion, this is the delta energy and then continues the same heights fall from the perihelion height as the perihelion photon would.
Yes, the sea level is implied.
Assuming we have a simple hydrogen atom, for simplicity, E0 is n=1 quantum number - ground state, E1 quantum number n=2, ...
Before absorbing the photons on the Earth, is the E1-E0 energy emitted at perihelion observed from the Earth frame at perihelion (not the space station frame) the same as E1-E0 observed form the Earth frame at the aphelion?
Post by: Jaaanosik on 14/12/2021 14:45:49
Assuming we have a simple hydrogen atom, for simplicityYou threw away simplicity when you dragged orbiting atoms into the fray.
You seem to be ignoring replies (even the one pointing out why 'aphelion' is wrong), so I see little point in continuing the discussion. For the last time:QuoteE0 is n=1 quantum number - ground stateNo it isn't. "E0 is n=1" only works for a stationary atom in the presence of the observer. Your atom is being emitted elsewhere, at a different potential, and at some speed. Measurement of the photons emitted at perigee or apogee could be anything. Either them three might be greater or less than any of the other values, as measured by your observer. This wasn't true of Feynman's simple example with at atom moved to the top of the stairs and back.QuoteBefore absorbing the photons on the EarthRemember that the photon is not going to be absorbed by some stationary hydrogen atom in ground state. It's probably the wrong energy for that.Quoteis the E1-E0 energy emitted at perihelion observed from the Earth frame at perihelion (not the space station frame) the same as E1-E0 observed form the Earth frame at the aphelion?First of all, about the bold part: The Earth observer is present at neither perigee nor apogee, and cannot observe anything there. His observation are on the ground, hopefully stationary on a non-spinning planet, unless you want to add yet further complication to this mess.
Secondly, No. There's no telling if the photons emitted at apogee and perigee will be measured with less than or greater than E1-E0 (the energy coming from is own local excited atom). You've just not given enough information. I can come up with scenarios where the apogee photon is the highest or lowest energy of the three, and ditto with the perigee photon. This ambiguity wasn't present in Feynman's example.
As for the Earth being in the way of one of the emissions, we can assume a solid cold planet with a little tunnel so the photon can pass through unmolested. I'm not worried about that problem.
Thanks Halc,
if I understand you correctly you are saying that centroids of isolated systems are frame dependent.
That's where the problem is.
If we start with hydrogen atom in the ground state as an isolated system inside of the space station at aphelion.
The hydrogen atom isolated system has a barycenter and inertial observer at the barycenter.
The inertial observer on the space station would observe the same thing.
The space station inertial observer is not looking outside, just at the hydrogen atom.
Due to the electron-proton spin orbit interaction and motion in the Earth frame the Earth observer sees the same hydrogen atom at higher energy level at aphelion, the energy is frame dependent.
Now the space station continues falling towards the perihelion.
Because of the Equivalence Principle the hydrogen atom barycenter observer and the space station observer do not see any change.
The space station starts to increase its speed in the Earth frame and the Earth observer is going to see higher transverse drift caused by the electron-proton spin orbit interaction and the higher space station speed.
This is the cause that leads to electron climbing the energy levels.
The elastic spring between proton and electron is being loaded with more energy based on the Earth observer.
If the energy would be higher then the Earth observer would even see electron separation from the proton, but he rest frame, no change because of EP, right?
This is the problem, disagreement on the energy change.
The Earth frame sees the energy change in the hydrogen atom but the space station frame is not supposed to see any energy change between electron-proton interaction.
That's the reason I ask the question does the Equivalence Principle violate the conservation of energy law.
Post by: Origin on 14/12/2021 15:27:11
Because of the Equivalence Principle the hydrogen atom barycenter observer and the space station observer do not see any change.I don't see how that has anything to do with the equivalency principle.
Post by: Jaaanosik on 14/12/2021 15:35:57
Because the falling observer is not going to detect any motion without any signal from the outside even though the space station and the hydrogen atom in it is gaining speed in the Earth frame.Because of the Equivalence Principle the hydrogen atom barycenter observer and the space station observer do not see any change.I don't see how that has anything to do with the equivalency principle.
Post by: Origin on 14/12/2021 18:01:24
Because the falling observer is not going to detect any motion without any signal from the outside even though the space station and the hydrogen atom in it is gaining speed in the Earth frame.So? I still don't see how the equivalency principle violates the conservation of energy. Energy is not conserved for different frames of reference. You have not made the case that shows there is a problem,
Post by: Jaaanosik on 14/12/2021 18:51:35
The Earth observer might see separation of electron from proton when atom speed is high enough.Because the falling observer is not going to detect any motion without any signal from the outside even though the space station and the hydrogen atom in it is gaining speed in the Earth frame.So? I still don't see how the equivalency principle violates the conservation of energy. Energy is not conserved for different frames of reference. You have not made the case that shows there is a problem,
The space hydrogen plasma.
The Equivalence Principle implies the hydrogen barycentric inertial observer sees the electron as part of the atom.
The electron cannot separate when the atom is in a free fall.
The observers do not agree on the reality, on physics.
In Newtonian mechanics different inertial observers have to agree on the acceleration, change of energy.
Why this would not be the case in GR? Are we getting energy out of nothing?
Post by: Kryptid on 14/12/2021 21:43:00
The Earth observer might see separation of electron from proton when atom speed is high enough.
Not if the proton and electron are being accelerated at the same rate. If they are being accelerated at different rates, then the observer on the ship will be able to see that too. You can't contrive a scenario where the electron separates in one reference frame but not another. Relativity doesn't produce contradictory results like that.
Post by: Jaaanosik on 14/12/2021 22:16:58
Yes, it does.The Earth observer might see separation of electron from proton when atom speed is high enough.
Not if the proton and electron are being accelerated at the same rate. If they are being accelerated at different rates, then the observer on the ship will be able to see that too. You can't contrive a scenario where the electron separates in one reference frame but not another. Relativity doesn't produce contradictory results like that.
If the space station observer is going to see the separation then it means we have an experiment to determine the acceleration of the free fall without any signal from the outside. The Equivalence Principle is not good any more.
We cannot have it both ways.
This just means we preserve the conservation of energy law and the Equivalence Principle has to go.
Post by: Kryptid on 15/12/2021 00:58:13
If the space station observer is going to see the separation...
...then that means some other force was present that caused the electron to separate. An atom in freefall isn't going to do that by itself, regardless of the reference frame you are in.
Post by: Halc on 15/12/2021 01:40:32
You've now crossed the line between just asking questions and asserting something very wrong.You can't contrive a scenario where the electron separates in one reference frame but not another. Relativity doesn't produce contradictory results like that.Yes, it does.
Quote
If the space station observer is going to see the separation then it means we have an experiment to determine the acceleration of the free fall without any signal from the outside. The Equivalence Principle is not good any more.This part is right. Can't have it both ways. Your way is the one with the contradictions. The Galilean principle of relativity ('PoR', 400+ years old) is the right way, and it is being correctly applied by Kryptid. No hydrogen atom is going to be pulled apart in an environment of zero proper acceleration as described in your scenario.
We cannot have it both ways.
Quote
This just means we preserve the conservation of energy law and the Equivalence Principle has to go.EP has never been applied by any words of yours. Origin is right. You've displayed no knowledge of it. I think you are confusing it with the PoR.
Admittedly you've applied some of the findings of general relativity (clocks run faster at high altitude), and EP is a postulate of GR, so in a long roundabout way, any GR reference is technically something that follows from EP, but certainly not directly. But the PoR that you're denying dates centuries before relativity theory came along.
Post by: Jaaanosik on 15/12/2021 14:00:21
(https://i.imgur.com/aktm8bi.png)
(https://i.imgur.com/vDNof5F.png)
Einstein shows the transformation how electric field changes to magnetic field in moving frames.
If we look at Hall effect: https://en.wikipedia.org/wiki/Hall_effect
(https://i.imgur.com/FVSrumK.png)
We see how electrons undergo transverse drift.
Now we do the Hall effect experiment on the space station but we drop the magnet, we will not use it.
The Earth frame sees the increased speed when the station falls from aphelion to perihelion and growing change of the electric field to magnetic field on the space station.
The electrons will start to drift for the Earth observer.
The electrons are not supposed to drift for the space station observer, the electrons are supposed to move in the straight line.
Clear contradiction, who is right?
It appears to me the space station observer will see the drift as well.
We have an experiment to detect the free fall speed without any signal from the outside.
Post by: Jaaanosik on 15/12/2021 16:06:01
As per Einstein's paper above, the force is the magnetic field observed by the Earth frame even though it is not observed by the space station frame and the rest frame of the atom.If the space station observer is going to see the separation...
...then that means some other force was present that caused the electron to separate. An atom in freefall isn't going to do that by itself, regardless of the reference frame you are in.
Post by: Kryptid on 15/12/2021 20:51:09
As per Einstein's paper above, the force is the magnetic field observed by the Earth frame even though it is not observed by the space station frame and the rest frame of the atom.
If the magnet is moving relative to the electron, then the electron will be deflected in both frames. If they are not moving relative to each other, then no deflection will occur in either frame.
Post by: Jaaanosik on 15/12/2021 21:12:37
Initially the electron is moving in the increasing magnetic field in the Earth frame and there is no electric field change in the space station frame.As per Einstein's paper above, the force is the magnetic field observed by the Earth frame even though it is not observed by the space station frame and the rest frame of the atom.
If the magnet is moving relative to the electron, then the electron will be deflected in both frames. If they are not moving relative to each other, then no deflection will occur in either frame.
The space station does not have a cause for drifting but the Earth frame has the cause, hence the contradiction.
Post by: Kryptid on 16/12/2021 01:05:12
Initially the electron is moving in the increasing magnetic field in the Earth frame and there is no electric field change in the space station frame.
The space station does not have a cause for drifting but the Earth frame has the cause, hence the contradiction.
So is the electron moving relative to the magnet or not?
Post by: Jaaanosik on 16/12/2021 02:03:38
It is the change of the magnetic field that is causing the transverse drift.Initially the electron is moving in the increasing magnetic field in the Earth frame and there is no electric field change in the space station frame.
The space station does not have a cause for drifting but the Earth frame has the cause, hence the contradiction.
So is the electron moving relative to the magnet or not?
The Hall effect image shows that very well.
If the magnet was large in size, covering the full conducting plate with a uniform field then the drift would not happen.
So the relative motion between the magnet and the electron does not guarantee the drift, the field change does.
The field is changing for the space station so the drift is inevitable.
Post by: Jaaanosik on 16/12/2021 02:05:46
Thanks,Clear contradiction, who is right?If I am running numbers and run into an apparent contradiction with centuries old principles, I presume it is I (an admitted non-physicist) who made the mistake. But that's not your reaction I see.
It appears to me the space station observer will see the drift as well.
We have an experiment to detect the free fall speed without any signal from the outside.
Your assertion directly violates PoR. I'm moving the topic.
any comments on the presented hypothesis?
Where is the mistake then?
Post by: Kryptid on 16/12/2021 03:19:56
It is the change of the magnetic field that is causing the transverse drift.
The Hall effect image shows that very well.
If the magnet was large in size, covering the full conducting plate with a uniform field then the drift would not happen.
So the relative motion between the magnet and the electron does not guarantee the drift, the field change does.
The field is changing for the space station so the drift is inevitable.
You don't even have to invoke the equivalence principle to see that something is wrong with your analysis. If I was simply walking towards the setup, then the magnet and electrons are both moving in my reference frame even if they aren't moving relative to each other. Obviously, my simple act of walking towards them isn't going to make the electron be deflected by the magnet. The conclusion that I must come to is that, in a reference frame where the electron and magnet are seen as moving, the field of the magnet and the field of the electron must change in precisely a way that prevents a net force between them from being any different than in a stationary frame.
Post by: Jaaanosik on 16/12/2021 04:30:59
If I understand correctly, you are saying that varying magnetic field is not going to affect current in a conductor?It is the change of the magnetic field that is causing the transverse drift.
The Hall effect image shows that very well.
If the magnet was large in size, covering the full conducting plate with a uniform field then the drift would not happen.
So the relative motion between the magnet and the electron does not guarantee the drift, the field change does.
The field is changing for the space station so the drift is inevitable.
You don't even have to invoke the equivalence principle to see that something is wrong with your analysis. If I was simply walking towards the setup, then the magnet and electrons are both moving in my reference frame even if they aren't moving relative to each other. Obviously, my simple act of walking towards them isn't going to make the electron be deflected by the magnet. The conclusion that I must come to is that, in a reference frame where the electron and magnet are seen as moving, the field of the magnet and the field of the electron must change in precisely a way that prevents a net force between them from being any different than in a stationary frame.
Post by: Kryptid on 16/12/2021 05:17:14
If I understand correctly, you are saying that varying magnetic field is not going to affect current in a conductor?
What I'm saying is that different reference frames can't have contradictory physical consequences. Walking towards a current-carrying conductor in a magnetic field cannot change the current in a way that causes an event to occur in my reference frame that does not also occur in all other reference frames. Any changes I observe in the magnetic field must be exactly compensated for by changes in the current and conductor so that a stationary observer will agree with me about what happens as a consequence of the current and field.
Post by: puppypower on 16/12/2021 11:36:24
For an observer inside the lab, that is true because the experimental equipment is in the same reference frame as the excited atom. So like I said, whether or not you measure a frequency difference depends on your reference frame relative to the atom.OK, so we established
The Eq 2 is true as well, according to Feynman.
How do we get more energy from the aphelion photon if no work is done between perihelion and aphelion?
One thing to consider is the quantum nature of electron energy levels; E0 and E1 and their photons. If the gravitational energy difference implies the electron/emitted photon should have an energy equal to a theoretical place that would be in the gaps between allowable quanta, that electron transition cannot occur, even though it has this extra or less energy.
The second thing to consider is connected to energy conservation. An electron is an elementary particle with two attributes; negative charge and mass. One nonreduceable thing with two attributes implies two attributes acting as one elementary thing. This implies the mass and negative charge are connected interchangeable by an aspect of the unified force.
Due to the nature of quanta and quantum gaps, any extra energy difference that should imply a gap position that is not allowed, can be compensated for by a shift in the integrated pendulum of mass and negative charge. Energy conservation would need to shifts this toward the mass side. This pendulum should be able to shift back if the total energy makes it to the next allowable quanta.
Post by: Bored chemist on 16/12/2021 12:56:49
One thing to consider is the quantum nature of electron energy levels; E0 and E1 and their photons. If the gravitational energy difference implies the electron/emitted photon should have an energy equal to a theoretical place that would be in the gaps between allowable quanta, that electron transition cannot occur, even though it has this extra or less energy.You just restated this, but less clearly.
For instance, for an altitude difference of 20 meters at the earth’s surface the frequency difference is only about two parts in 1015. However, just such a change has recently been found experimentally using the Mössbauer effect. Einstein was perfectly correct. "And the rest of you post is wrong.
Post by: Jaaanosik on 16/12/2021 13:22:31
Einstein showed us how the electric field transforms to magnetic field.If I understand correctly, you are saying that varying magnetic field is not going to affect current in a conductor?
What I'm saying is that different reference frames can't have contradictory physical consequences. Walking towards a current-carrying conductor in a magnetic field cannot change the current in a way that causes an event to occur in my reference frame that does not also occur in all other reference frames. Any changes I observe in the magnetic field must be exactly compensated for by changes in the current and conductor so that a stationary observer will agree with me about what happens as a consequence of the current and field.
This is observed by the Earth frame, external to the space station, this will cause the drift in the Earth frame.
I agree with you that this change has to be observed by the space station frame as well.
The conclusion, the space station observer has an experiment to detect the freefall without any signal from the outside.
Post by: Origin on 16/12/2021 13:48:49
Einstein showed us how the electric field transforms to magnetic field.I think it was actually Maxwell that showed that in the 1860s.
Post by: Jaaanosik on 16/12/2021 14:25:23
The Lorentz transformation came 20 years later.Einstein showed us how the electric field transforms to magnetic field.I think it was actually Maxwell that showed that in the 1860s.
It appears to me Einstein was the first who did it between reference frames.
I guess if he was not then the relativity should belong to someone else.
Post by: Kryptid on 16/12/2021 14:58:23
The conclusion, the space station observer has an experiment to detect the freefall without any signal from the outside.
No, it doesn't...
Post by: Jaaanosik on 16/12/2021 15:02:56
Here is another take to what you are saying.If I understand correctly, you are saying that varying magnetic field is not going to affect current in a conductor?
What I'm saying is that different reference frames can't have contradictory physical consequences. Walking towards a current-carrying conductor in a magnetic field cannot change the current in a way that causes an event to occur in my reference frame that does not also occur in all other reference frames. Any changes I observe in the magnetic field must be exactly compensated for by changes in the current and conductor so that a stationary observer will agree with me about what happens as a consequence of the current and field.
From Relativistic Hall Effect paper: https://arxiv.org/pdf/1112.5618.pdf
(https://i.imgur.com/G8YLStS.png)
The centroids are frame dependent...
(https://i.imgur.com/6SUxgkx.png)
The second image shows 0.2c and 0.5c velocities and different centroid positions.
The Earth has an Earth Center Inertial (ECI) reference frame.
If the wheel is falling on the space station from aphelion to perihelion it is increasing the speed and the centroids are undergoing the transverse drift in the ECI.
The space station inertial frame does not see any change.
What is going on? How do we reconcile that?
Post by: Origin on 16/12/2021 15:51:52
The Lorentz transformation came 20 years later.Since Maxwell was talking about relative movement they are obviously in 2 different frames. Inertial frames date back to Galileo.
It appears to me Einstein was the first who did it between reference frames.
I guess if he was not then the relativity should belong to someone else.I guess you would be wrong.
Post by: Jaaanosik on 16/12/2021 16:59:51
:)The Lorentz transformation came 20 years later.Since Maxwell was talking about relative movement they are obviously in 2 different frames. Inertial frames date back to Galileo.
It appears to me Einstein was the first who did it between reference frames.I guess if he was not then the relativity should belong to someone else.I guess you would be wrong.
Entertaining...
How about to address the presented EP problems?
Post by: Kryptid on 16/12/2021 20:53:52
Post by: Jaaanosik on 16/12/2021 21:01:16
One way to answer what would happen would be to consider your space station to be the only object in the Universe. Now ask yourself, does an observer onboard that space station see the electron beam deflected by the magnet?If we are talking about Figure 1 from post #68 then yes.
Post by: Kryptid on 16/12/2021 21:18:50
If we are talking about Figure 1 from post #68 then yes.
What caused the deflection?
Post by: Origin on 16/12/2021 21:40:09
Entertaining...just trying to help you get your facts straight.
How about to address the presented EP problems?I don't really see anything having to do with the equivalency principle, the question about the magnetic field in one frame and not the other is not something I can't answer off the top of my head. I am being lazy and waiting for someone else to answer before I go looking up the answer.
Post by: Jaaanosik on 16/12/2021 21:42:55
Electrons moving through the varying magnetic field.If we are talking about Figure 1 from post #68 then yes.
What caused the deflection?
Post by: Kryptid on 16/12/2021 21:53:38
Electrons moving through the varying magnetic field.
What causes the magnetic field to vary?
Post by: Jaaanosik on 16/12/2021 22:05:06
The shape of the magnetic field is not uniform.Electrons moving through the varying magnetic field.
What causes the magnetic field to vary?
The field lines are not straight.
Post by: Kryptid on 16/12/2021 22:32:26
The shape of the magnetic field is not uniform.
The field lines are not straight.
Okay, so you seem to think that:
(1) The electron will be deflected when the space station is alone in the Universe with no sources of gravity or movement outside the ship, and
(2) The electron will also be deflected if the space station is in freefall above Earth.
So how exactly is the electron deflection experiment supposed to tell you anything about whether you are in free fall or not?
Post by: Jaaanosik on 16/12/2021 23:02:40
As I said in the post #68, we will not use magnet when testing for free fall.The shape of the magnetic field is not uniform.
The field lines are not straight.
Okay, so you seem to think that:
(1) The electron will be deflected when the space station is alone in the Universe with no sources of gravity or movement outside the ship, and
(2) The electron will also be deflected if the space station is in freefall above Earth.
So how exactly is the electron deflection experiment supposed to tell you anything about whether you are in free fall or not?
The varying magnetic field will be generated due to the accelerated electric field in free fall, as per Einstein's equations for the Earth frame.
The space station observer does not expect any transverse drift because there is no magnet locally and the observer is not aware of any freefall acceleration.
To make the experiment consistent we would not use any magnet alone in the Universe and there would not be any transverse drift.
This is the contradiction. Who is right?
Similar problem for the flywheel experiment.
Post by: Kryptid on 16/12/2021 23:05:09
As I said in the post #68, we will not use magnet when testing for free fall.
The varying magnetic field will be generated due to the accelerated electric field in free fall, as per Einstein's equations for the Earth frame.
If there is no magnet, then what is the electron's magnetic field supposed to interact with?
Post by: Jaaanosik on 16/12/2021 23:07:09
Check the Einstein's equations...As I said in the post #68, we will not use magnet when testing for free fall.
The varying magnetic field will be generated due to the accelerated electric field in free fall, as per Einstein's equations for the Earth frame.
If there is no magnet, then what is the electron's magnetic field supposed to interact with?
Post by: Kryptid on 16/12/2021 23:10:59
Check the Einstein's equations...
So what is causing it to deflect?
Post by: Jaaanosik on 16/12/2021 23:17:17
With increasing space station speed in the Earth frame the magnetic field is getting stronger for the constant electric field on the space station.Check the Einstein's equations...
So what is causing it to deflect?
This is where the varying magnetic field is coming from when there is no magnet for the experiment.
Post by: Kryptid on 16/12/2021 23:19:40
With increasing space station speed in the Earth frame the magnetic field is getting stronger for the constant electric field on the space station.
This is where the varying magnetic field is coming from when there is no magnet for the Earth frame.
I still don't think you get what I'm asking: what is the source of the magnetic field that is interacting with the electron's magnetic field? There has to be a second field or there will be nothing for the electron to interact with to be deflected.
Post by: Jaaanosik on 16/12/2021 23:32:29
Moving electrons generate magnetic field.With increasing space station speed in the Earth frame the magnetic field is getting stronger for the constant electric field on the space station.
This is where the varying magnetic field is coming from when there is no magnet for the Earth frame.
I still don't think you get what I'm asking: what is the source of the magnetic field that is interacting with the electron's magnetic field? There has to be a second field or there will be nothing for the electron to interact with to be deflected.
This field gets magnified for the Earth frame observer as mentioned above.
This field is supposed to be unchanging for the space station observer.
Post by: Kryptid on 16/12/2021 23:34:22
Post by: Jaaanosik on 16/12/2021 23:44:10
That didn't answer my question: where is the second magnetic field?Why would you need a second magnetic field when the varying magnetic field provides dynamic feedback loop?
Post by: Kryptid on 16/12/2021 23:45:35
Why would you need a second magnetic field when the varying magnetic field provides dynamic feedback loop?
There is nothing for the electron's magnetic field to push against. It can't push against itself. That would violate conservation of momentum.
Post by: Jaaanosik on 17/12/2021 00:04:24
The energy of the system is increasing, because velocity is increasing in the Einstein's equations.Why would you need a second magnetic field when the varying magnetic field provides dynamic feedback loop?
There is nothing for the electron's magnetic field to push against. It can't push against itself. That would violate conservation of momentum.
The equations show the field energy increase is not uniform, it is directional, then feedback loop, … dynamic system.
Plenty of varying magnetic field to cause havoc.
Post by: Kryptid on 17/12/2021 00:08:15
The energy of the system is increasing, because velocity is increasing in the Einstein's equations.
The equations show the field energy increase is not uniform, it is directional, then feedback loop, … dynamic system.
Plenty of varying magnetic field to cause havoc.
If the electron is being deflected to one side, then the electron's momentum is changing. In order for momentum to be conserved (and it absolutely must be), then something else has to be carrying momentum equal and opposite to that of the electron. Recall Newton's laws of motion. What is carrying the momentum that is canceling out the change in the electron's momentum?
Post by: Jaaanosik on 17/12/2021 02:21:32
Increasing varying magnetic field will cause more heat in the system.The energy of the system is increasing, because velocity is increasing in the Einstein's equations.
The equations show the field energy increase is not uniform, it is directional, then feedback loop, … dynamic system.
Plenty of varying magnetic field to cause havoc.
If the electron is being deflected to one side, then the electron's momentum is changing. In order for momentum to be conserved (and it absolutely must be), then something else has to be carrying momentum equal and opposite to that of the electron. Recall Newton's laws of motion. What is carrying the momentum that is canceling out the change in the electron's momentum?
More energy more heat.
Measuring the varying magnetic field itself will show delta for the increased freefall speed.
Even if we consider a simple wire with a constant electrical current in it; we would measure increased magnetic field around he wire itself.
Electrons are trapped in the wire, the field would increase outside.
Post by: Kryptid on 17/12/2021 16:02:52
What is carrying the momentum that is canceling out the change in the electron's momentum?
Post by: Jaaanosik on 17/12/2021 16:51:36
Electrons are in the conducting plate, wire, …What is carrying the momentum that is canceling out the change in the electron's momentum?
That's where the electron physical interactions happen.
Post by: Kryptid on 17/12/2021 20:41:52
Post by: Jaaanosik on 17/12/2021 21:21:02
If they are in the wire, then they can't be deflected because they are stuck in the wire.Your point is?
The energy in the magnetic field increases as per the Earth frame.
If the energy increase is not observed by space station observer it is a disagreement on physics.
If the energy increase is observed by the space station observer then we have an experiment to detect the freefall.
They can be deflected in the conducting plate.
The system would be gaining energy so there would be Eddy currents and therefore more heat as a response predicted by the Earth frame.
Even if we look at the mechanical analysis:
(https://i.imgur.com/G8YLStS.png)
(https://i.imgur.com/6SUxgkx.png)
The centroids are undergoing the transverse drift in the Earth frame and no changes in the space station frame.
This is the contradiction.
What is the way out?
Post by: Kryptid on 17/12/2021 21:56:03
Your point is?
That you were wrong when you said this:
The electrons will start to drift for the Earth observer.
The energy in the magnetic field increases as per the Earth frame.
And the energy in the electric field decreases by the same amount. The total energy is the same.
Post by: Jaaanosik on 17/12/2021 22:23:44
Energy will not decrease because the increase in magnetic field will generate more electric field, feedback loop.Your point is?
That you were wrong when you said this:The electrons will start to drift for the Earth observer.The energy in the magnetic field increases as per the Earth frame.
And the energy in the electric field decreases by the same amount. The total energy is the same.
That's the conclusion based on Einstein's equations.
The velocity is going up, more energy in the system observed by the Earth frame.
The same as the mechanical example.
Transverse drift grows in the Earth frame but nothing in the space station frame.
The same cause, velocity increase.
Post by: Kryptid on 17/12/2021 22:42:59
Energy will not decrease because the increase in magnetic field will generate more electric field, feedback loop.
That's the conclusion based on Einstein's equations.
How do Einstein's equations say that? Not that it matters, as energy is frame-dependent anyway.
Let's go back to my empty Universe example. What is causing the electrons to drift in that Universe? There is no Earth present here and no gravity source for any kind of free fall.
Post by: Jaaanosik on 17/12/2021 23:04:15
Energy will not decrease because the increase in magnetic field will generate more electric field, feedback loop.
That's the conclusion based on Einstein's equations.
How do Einstein's equations say that? Not that it matters, as energy is frame-dependent anyway.
Let's go back to my empty Universe example. What is causing the electrons to drift in that Universe? There is no Earth present here and no gravity source for any kind of free fall.
(https://i.imgur.com/exWbUfj.png)
Page 14 from https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf
The Lorentz factor
is increasing.X', Y', Z' (electric) goes up and L', M', N' (magnetic) goes up as well.
There is no magnet so nothing.
If the space station is alone, no drift for the electrons.
That's how the space station observer knows there is no freefall.
Post by: Kryptid on 17/12/2021 23:16:12
If the space station is alone, no drift for the electrons.
That's how the space station observer knows there is no freefall.
So now we add another observer. Someone in a different spaceship is headed towards the space station at 50% the speed of light. Does that somehow cause an observer on the space station to observe drift?
Post by: Jaaanosik on 17/12/2021 23:41:55
No drift, the space station is not accelerating,If the space station is alone, no drift for the electrons.
That's how the space station observer knows there is no freefall.
So now we add another observer. Someone in a different spaceship is headed towards the space station at 50% the speed of light. Does that somehow cause an observer on the space station to observe drift?
is not changing.A logical question is accelerating in regards to what frame?
It appears the marriage between SR and GR didn't go well.
Post by: Kryptid on 18/12/2021 00:00:46
No drift, the space station is not accelerating, is not changing.
Why does it matter whether the Lorentz factor is changing?
Post by: Jaaanosik on 18/12/2021 14:33:25
I misunderstood your question. I was focused on the change and changing fields.No drift, the space station is not accelerating, is not changing.
Why does it matter whether the Lorentz factor is changing?
If there is just one observer at 0.5c then here is the static drift:
(https://i.imgur.com/6SUxgkx.png)
Post by: Kryptid on 18/12/2021 15:20:38
If there is just one observer at 0.5c then here is the static drift:
I hope you see the problem now. The electrons can't magically start to drift for the observer on the space station just because there is someone outside in a different reference frame. There is no causal connection between the two.
Post by: Jaaanosik on 18/12/2021 15:50:30
This is the boundary problem.If there is just one observer at 0.5c then here is the static drift:
I hope you see the problem now. The electrons can't magically start to drift for the observer on the space station just because there is someone outside in a different reference frame. There is no causal connection between the two.
When the spaceships start in the same inertial frame then the transverse drift is increasing in the accelerated frame.
If this is not the case then Einstein's equations are wrong.
All the discussion about relativity reciprocity, … is the length contraction real?
How about Bell's Spaceship Paradox?
https://math.ucr.edu/home/baez/physics/Relativity/SR/BellSpaceships/spaceship_puzzle.html
There is a causal connection between the two and the relativity has a problem.
This cannot be resolved unless there is an agreement on a preferred frame.
Post by: Jaaanosik on 18/12/2021 16:57:26
(https://i.imgur.com/kJGT2Tw.png)
The textbooks say the length contraction is only on the circumference because of the rotation.
But if the ring is observed from outside frame in relative motion moving along X axis that is parallel to the axle of the ring then the Einstein's equations say there is increased force pulling in Y and Z direction trying to increase the ring diameter.
(https://i.imgur.com/Yle6X3n.png)
Is the centrifugal force real?
If there is an acceleration along the X direction then the centrifugal force has to increase as well according to Einstein even though the rotation is constant.
If one observer sees the increase and the other does not then we have a disagreement on physics.
Post by: Origin on 18/12/2021 18:37:57
If one observer sees the increase and the other does not then we have a disagreement on physics.Look up "Ehrenfest Paradox solution" and you will see it is not really a paradox.
Post by: Origin on 18/12/2021 18:41:29
How about Bell's Spaceship Paradox?Look up "Bell's Spaceship Paradox Solution" and you will find it is not really a paradox.
Post by: Bored chemist on 18/12/2021 20:03:18
If we have a look at the Ehrenfest Paradox.I did
https://en.wikipedia.org/wiki/Ehrenfest_paradox#Resolution_of_the_paradox
Perhaps you should have.
Post by: Kryptid on 18/12/2021 22:21:13
As another example, consider a situation where you have multiple spaceships approaching the space station from multiple different angles and speeds. If what you claimed about there being a causal connection were true, then the electron would have to deflect in multiple different directions and by different amounts. Obviously, this is not possible. Only one physical consequence can result. The result is that it doesn't deflect at all, for any observer.
Post by: Jaaanosik on 20/12/2021 13:58:25
There is no problem. As I've said before, you can't create contradictory physical consequences using relativity. If you think you have, then it is significantly more likely that you have misunderstood something instead.That's exactly what happens with the Doppler effect.
As another example, consider a situation where you have multiple spaceships approaching the space station from multiple different angles and speeds. If what you claimed about there being a causal connection were true, then the electron would have to deflect in multiple different directions and by different amounts. Obviously, this is not possible. Only one physical consequence can result. The result is that it doesn't deflect at all, for any observer.
(https://i.imgur.com/7Y7MH6N.png)
The question is what happens to the frequency of the wave packet, one quanta of energy.
(https://i.imgur.com/Lr72Gu6.png)
What happens to a hydrogen atom when relative speed is too high?
Are we going to get/observe tera watts of energy from one forward hydrogen emission when a relative motion is close to speed of light?
Post by: Kryptid on 20/12/2021 14:19:12
What happens to a hydrogen atom when relative speed is too high?
Too high for what?
Are we going to get/observe tera watts of energy from one forward hydrogen emission when a relative motion is close to speed of light?
Terawatts are a unit of power, not energy. If you are moving very fast, yes, the energy of a photon emitted by an excited hydrogen atom will indeed appear to be very high. Energy is frame-dependent. Nothing wrong with that.
Post by: puppypower on 20/12/2021 16:19:06
What happens to a hydrogen atom when relative speed is too high?
Too high for what?Are we going to get/observe tera watts of energy from one forward hydrogen emission when a relative motion is close to speed of light?
Terawatts are a unit of power, not energy. If you are moving very fast, yes, the energy of a photon emitted by an excited hydrogen atom will indeed appear to be very high. Energy is frame-dependent. Nothing wrong with that.
One complication that stems from the observation that energy is frame dependent is, different frames by doing different energy balances, means someone will violate energy conservation since they will not all agree. Everyone needs to do the same total energy balance for energy conservation to work. Frame dependency makes that a pipe dream. If some frames are too high and some are too low; frame dependent, then what is the correct energy for universal energy conservation? This has led to the need for dark energy and dark matter to add some extra energy to our frame to explain observational anomalies that came after standard theory was instituted; add a bandaid. Einstein warned of this but nobody understood the implications as well as the math. Conceptual modeling is more fundamental than math.
The speed of light is the same in all frames. This would be a way to have a common observational reference that is the same for all and would allow agreement across the board in terms of a universal energy balance; conceptually speaking.
Post by: Bored chemist on 20/12/2021 18:26:52
One complication that stems from the observation that energy is frame dependent is, different frames by doing different energy balances, means someone will violate energy conservation since they will not all agree.A cricket ball has energy, but the fly sitting on it, and the batsman do not agree about how much kinetic energy the ball has.
This is not a contradiction.
There is no science in pretending that it is.
Post by: Origin on 20/12/2021 18:46:29
One complication that stems from the observation that energy is frame dependent is, different frames by doing different energy balances, means someone will violate energy conservation since they will not all agree.Excuse me, but have you lost your mind? Of course they don't agree! The conservation of energy is frame dependent as you noted in the first half of your sentence! Did you forget you wrote that??
Post by: Kryptid on 20/12/2021 20:41:27
One complication that stems from the observation that energy is frame dependent is, different frames by doing different energy balances, means someone will violate energy conservation since they will not all agree.
Conservation of energy only works within a reference frame, not between different reference frames.
Post by: Jaaanosik on 20/12/2021 23:02:27
Yes, screwup on the UoM, I guess I had too much to drink yesterday. :)What happens to a hydrogen atom when relative speed is too high?
Too high for what?Are we going to get/observe tera watts of energy from one forward hydrogen emission when a relative motion is close to speed of light?
Terawatts are a unit of power, not energy. If you are moving very fast, yes, the energy of a photon emitted by an excited hydrogen atom will indeed appear to be very high. Energy is frame-dependent. Nothing wrong with that.
Lorentz factor so high that Doppler effect gives us wavelength shorter than Planck length?
Post by: Jaaanosik on 20/12/2021 23:19:18
There is no problem. As I've said before, you can't create contradictory physical consequences using relativity. If you think you have, then it is significantly more likely that you have misunderstood something instead.That's the reason we have quantum physics.
As another example, consider a situation where you have multiple spaceships approaching the space station from multiple different angles and speeds. If what you claimed about there being a causal connection were true, then the electron would have to deflect in multiple different directions and by different amounts. Obviously, this is not possible. Only one physical consequence can result. The result is that it doesn't deflect at all, for any observer.
(https://i.imgur.com/wZaOQ8e.png)
This is the relativity.
The top is the S observer inertial frame with rotating flywheel. Collision …, straight on the line for centroids.
The bottom is SIV observer inertial frame with non-rotating flywheel. Collision …, the centroids are not on the line, deflection is expected.
Different results for different observers, disagreement on physics.
Comment: The rotating flywheel is mounted inside a non-rotating sphere/cylinder so the collision is between non-rotating objects.
Post by: Origin on 20/12/2021 23:58:32
That's the reason we have quantum physics.Seriously? You think quantum physics fixes paradoxes in relativity?
Let me assure you it doesn't and by the way 'paradoxes' in relativity are not really paradoxes.
Post by: Kryptid on 21/12/2021 01:15:33
Lorentz factor so high that Doppler effect gives us wavelength shorter than Planck length?
I can think of two possible solutions here:
(1) The Planck length is reference frame dependent, or
(2) The Planck length does not represent the shortest possible distance in space.
This is the relativity.
The top is the S observer inertial frame with rotating flywheel. Collision …, straight on the line for centroids.
The bottom is SIV observer inertial frame with non-rotating flywheel. Collision …, the centroids are not on the line, deflection is expected.
Different results for different observers, disagreement on physics.
Comment: The rotating flywheel is mounted inside a non-rotating sphere/cylinder so the collision is between non-rotating objects.
It would be nice if you would stop jumping around using different examples instead of sticking to just one until we agree on a solution to it. That being said, please make a choice as to what we are to focus on.
Post by: Jaaanosik on 21/12/2021 03:43:09
Lorentz factor so high that Doppler effect gives us wavelength shorter than Planck length?
I can think of two possible solutions here:
(1) The Planck length is reference frame dependent, or
(2) The Planck length does not represent the shortest possible distance in space.This is the relativity.
The top is the S observer inertial frame with rotating flywheel. Collision …, straight on the line for centroids.
The bottom is SIV observer inertial frame with non-rotating flywheel. Collision …, the centroids are not on the line, deflection is expected.
Different results for different observers, disagreement on physics.
Comment: The rotating flywheel is mounted inside a non-rotating sphere/cylinder so the collision is between non-rotating objects.
It would be nice if you would stop jumping around using different examples instead of sticking to just one until we agree on a solution to it. That being said, please make a choice as to what we are to focus on.
From the Relativistic Hall effect paper, emphasis mine:
Quote
Introduction.— Hall effects represent a group of in-
triguing phenomena which appear from the interplay be-
tween rotation and linear motion of particles. These phe-
nomena are associated with a transverse drift of the par-
ticle in the direction orthogonal to both its angular mo-
mentum (AM) and external force. For instance, in classi-
cal and quantum Hall effects, electrons rotate in a mag-
netic field and drift orthogonally to the applied electric
field [1]. In the past decade, various spin Hall effects at-
tracted enormous attention in condensed-matter [2], op-
tical [3], and high-energy [4] systems. These effects arise
from a spin-orbit-type interaction between the intrinsic
AM of the particle and its external motion. Despite strik-
ing differences between the systems, the spin Hall effects
are intimately related to universal AM conservation laws
[2–5]. Noteworthily, the intrinsic AM of classical waves
or quantum particles can be associated with a circulating
internal current which can originate not only from spin
but also from quantum (optical) vortices [6, 7].
The Einstein's electromagnetic equations, Hall effect, mechanical linear and rotational motion,...
They all appear to be connected.
The flywheel analysis is very much linked to the electron Hall effect.
Both undergo the transverse drift.
To understand the mechanical example may help us to understand the Hall effect.
Post by: Kryptid on 21/12/2021 03:51:48
Post by: Jaaanosik on 21/12/2021 04:45:01
So what example do you want us to discuss? Be specific.As you pointed out
Quote
As another example, consider a situation where you have multiple spaceships approaching the space station from multiple different angles and speeds. If what you claimed about there being a causal connection were true, then the electron would have to deflect in multiple different directions and by different amounts. Obviously, this is not possible. Only one physical consequence can result. The result is that it doesn't deflect at all, for any observer.... different observers different directions...
The post #135 shows how this is a problem for the relativistic angular momentum as well.
Do we know electron position and direction of motion in hydrogen atom?
When a photon is released in hydrogen atom, do we know in what direction it flies out?
It appears the local hydrogen rest frame is useless to determine electron motion, we have to use statistics.
Why is that?
Why the local rest frame in the post #135 disagrees on physics, the direction of the collision, with the other reference frame?
Post by: Kryptid on 21/12/2021 04:49:51
Do we know electron position and direction of motion in hydrogen atom?
When a photon is released in hydrogen atom, do we know in what direction it flies out?
It appears the local hydrogen rest frame is useless to determine electron motion, we have to use statistics.
Why is that?
Okay, since we are now talking about this, I am going to focus on it. Don't bring anything up about electrons flowing in conductors or spinning disks until this is dealt with first.
Quote
Do we know electron position and direction of motion in hydrogen atom?
No, an electron is in a superposition of different states until it is observed.
Quote
When a photon is released in hydrogen atom, do we know in what direction it flies out?
Only after it comes out. I'm pretty sure you can't know for sure in advance, due to quantum uncertainty.
Quote
It appears the local hydrogen rest frame is useless to determine electron motion, we have to use statistics.
Why is that?
Because electrons aren't classical objects. They are quantum.
Post by: Jaaanosik on 21/12/2021 11:50:44
Hydrogen atom is the most fundamental element, it is a good object to study.Do we know electron position and direction of motion in hydrogen atom?
When a photon is released in hydrogen atom, do we know in what direction it flies out?
It appears the local hydrogen rest frame is useless to determine electron motion, we have to use statistics.
Why is that?
Okay, since we are now talking about this, I am going to focus on it. Don't bring anything up about electrons flowing in conductors or spinning disks until this is dealt with first.QuoteDo we know electron position and direction of motion in hydrogen atom?
No, an electron is in a superposition of different states until it is observed.QuoteWhen a photon is released in hydrogen atom, do we know in what direction it flies out?
Only after it comes out. I'm pretty sure you can't know for sure in advance, due to quantum uncertainty.QuoteIt appears the local hydrogen rest frame is useless to determine electron motion, we have to use statistics.
Why is that?
Because electrons aren't classical objects. They are quantum.
I am trying to point out that the quantum uncertainty is a consequence of not knowing how the hydrogen atoms move through a preferred frame.
There is famous Hydrogen line: https://en.wikipedia.org/wiki/Hydrogen_line
'I am a free hydrogen in an interstellar space and suddenly my electron flipped. I am like, whiskey tango foxtrot, why my electron flipped? What is the cause?'
It appears to me an interaction with a preferred frame happened, according to Einstein's equations.
What would be your take on it?
Post by: Origin on 21/12/2021 13:34:57
I am trying to point out that the quantum uncertainty is a consequence of not knowing how the hydrogen atoms move through a preferred frame.A preferred frame has nothing to do with it AFAIK.
It appears to me an interaction with a preferred frame happened, according to Einstein's equations.Please show how "Einstein's equations" show this.
Post by: Kryptid on 21/12/2021 15:01:20
I am a free hydrogen in an interstellar space and suddenly my electron flipped.
If that hydrogen atom is unobserved, I'm pretty sure the electron wouldn't have a defined spin to begin with. It doesn't make sense to say that it "flipped".
It appears to me an interaction with a preferred frame happened, according to Einstein's equations.
Relativity states that there are no preferred frames, so this is wrong.
Post by: Jaaanosik on 21/12/2021 15:07:42
If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame the electron has a cause to flip to higher state based on the Einstein's equations.I am trying to point out that the quantum uncertainty is a consequence of not knowing how the hydrogen atoms move through a preferred frame.A preferred frame has nothing to do with it AFAIK.It appears to me an interaction with a preferred frame happened, according to Einstein's equations.Please show how "Einstein's equations" show this.
This atom is prerequisite for the 21cm photon emission. When this atom encounters right conditions and slows down in the preferred frame then again based on the Einstein's equations the electron can go back to the ground state emitting the 21cm photon.
If this is not the case then, please, explain how do we get the 21cm photons.
Post by: Jaaanosik on 21/12/2021 15:11:07
Are you saying what we measure from space is not real?I am a free hydrogen in an interstellar space and suddenly my electron flipped.
If that hydrogen atom is unobserved, I'm pretty sure the electron wouldn't have a defined spin to begin with. It doesn't make sense to say that it "flipped".It appears to me an interaction with a preferred frame happened, according to Einstein's equations.
Relativity states that there are no preferred frames, so this is wrong.
If it is real, please, how does 21cm photon leave the Hydrogen atom?
If we see the line, the energy change, then the Hydrogen atom frame has to see some energy change as well, right?
Post by: Kryptid on 21/12/2021 15:13:26
If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame
There is no preferred frame.
the electron has a cause to flip to higher state based on the Einstein's equations.
Are you talking about flipping spin, or merely being raised to a higher energy state? Those two things are not the same. It's also not accurate. Electrons don't become excited just because an atom is moving quickly.
This atom is prerequisite for the 21cm photon emission. When this atom encounters right conditions and slows down in the preferred frame
There is no preferred frame.
If this is not the case then, please, explain how do we get the 21cm photons.
Because that's how much energy an electron has to absorb in a hydrogen atom in order to raise it to the nearest excited energy level.
Are you saying what we measure from space is not real?
No, I'm saying that your understanding is wrong.
If it is real, please, how does 21cm photon leave the Hydrogen atom?
The electron falls down to its base energy level and the photon is released as a consequence of conservation of energy.
Post by: Jaaanosik on 21/12/2021 16:36:08
Flipping the electron spin costs energy.If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame
There is no preferred frame.the electron has a cause to flip to higher state based on the Einstein's equations.
Are you talking about flipping spin, or merely being raised to a higher energy state? Those two things are not the same. It's also not accurate. Electrons don't become excited just because an atom is moving quickly.
Proton and electron opposite spin + energy = Proton and electron spin in the same direction.
Where is the exact quanta of energy coming from?
Quote
What is the cause of the same direction electron spin changing to opposite direction electron spin, lowering Hydrogen atom energy?This atom is prerequisite for the 21cm photon emission. When this atom encounters right conditions and slows down in the preferred frame
There is no preferred frame.If this is not the case then, please, explain how do we get the 21cm photons.
Because that's how much energy an electron has to absorb in a hydrogen atom in order to raise it to the nearest excited energy level.Are you saying what we measure from space is not real?
No, I'm saying that your understanding is wrong.If it is real, please, how does 21cm photon leave the Hydrogen atom?
The electron falls down to its base energy level and the photon is released as a consequence of conservation of energy.
In general, what is a cause of any Hydrogen atom emitting photon energy change?
When electron goes to lower energy levels?
Why would electron go to lower energy levels?
Post by: Origin on 21/12/2021 16:50:38
If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame the electron has a cause to flip to higher state based on the Einstein's equations.Please tell us what those equations are.
This atom is prerequisite for the 21cm photon emission. When this atom encounters right conditions and slows down in the preferred frame then again based on the Einstein's equations the electron can go back to the ground state emitting the 21cm photon.That makes no sense at all.
If this is not the case then, please, explain how do we get the 21cm photons.This simple explanation should help you understand that it has nothing to do with a preferred frame https://www.britannica.com/science/21-centimetre-radiation (https://www.britannica.com/science/21-centimetre-radiation)
Post by: Jaaanosik on 21/12/2021 17:01:41
Post #123If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame the electron has a cause to flip to higher state based on the Einstein's equations.Please tell us what those equations are.This atom is prerequisite for the 21cm photon emission. When this atom encounters right conditions and slows down in the preferred frame then again based on the Einstein's equations the electron can go back to the ground state emitting the 21cm photon.That makes no sense at all.If this is not the case then, please, explain how do we get the 21cm photons.This simple explanation should help you understand that it has nothing to do with a preferred frame https://www.britannica.com/science/21-centimetre-radiation (https://www.britannica.com/science/21-centimetre-radiation)
https://www.thenakedscientists.com/forum/index.php?topic=83660.msg663629#msg663629 (https://www.thenakedscientists.com/forum/index.php?topic=83660.msg663629#msg663629)
Post by: Origin on 21/12/2021 17:10:48
Post #123You seriously think those equations address this, "If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame the electron has a cause to flip to higher state"?
https://www.thenakedscientists.com/forum/index.php?topic=83660.msg663629#msg663629
You have no idea what you are talking about.
Post by: Jaaanosik on 21/12/2021 17:41:08
When electron goes to lower energy levels?Post #123You seriously think those equations address this, "If there is a ground state Hydrogen atom in a freefall then after gaining speed in the preferred frame the electron has a cause to flip to higher state"?
https://www.thenakedscientists.com/forum/index.php?topic=83660.msg663629#msg663629
You have no idea what you are talking about.
Why would electron go to lower energy levels?
Post by: Origin on 21/12/2021 18:26:54
When electron goes to lower energy levels?On average about 10 million years.
Why would electron go to lower energy levels?This is the most stable state of the atom. In other words higher energy states are unstable and so they tend to transition to a lower energy state. Same reason that a hot object will cool to room temperature.
Post by: Jaaanosik on 21/12/2021 18:53:44
Is that the only option for the transition?When electron goes to lower energy levels?On average about 10 million years.Why would electron go to lower energy levels?This is the most stable state of the atom. In other words higher energy states are unstable and so they tend to transition to a lower energy state. Same reason that a hot object will cool to room temperature.
Post by: Bored chemist on 21/12/2021 19:43:30
There are, in principle,an infinite number of possible transitions of a hydrogen atom.Is that the only option for the transition?When electron goes to lower energy levels?On average about 10 million years.Why would electron go to lower energy levels?This is the most stable state of the atom. In other words higher energy states are unstable and so they tend to transition to a lower energy state. Same reason that a hot object will cool to room temperature.
Do you have any clue what you are talking about?
Post by: Jaaanosik on 21/12/2021 20:24:35
Essentially there is only one type of transition, or cause for Hydrogen atom transition when electron goes from higher energy state to lower energy state.There are, in principle, an infinite number of possible transitions of a hydrogen atom.Is that the only option for the transition?When electron goes to lower energy levels?On average about 10 million years.Why would electron go to lower energy levels?This is the most stable state of the atom. In other words higher energy states are unstable and so they tend to transition to a lower energy state. Same reason that a hot object will cool to room temperature.
Do you have any clue what you are talking about?
The electron has to slow down.
One instance is a long decay as mentioned above.
The other instances are variations of "braking radiation".
The imaginary "elastic string" between proton and electron has to get shorter.
Slower motion through the preferred frame translates to smaller electro magnetic forces for the electron transverse drift.
That's a cause for the transition and we do not have to wait 10 million years.
Post by: Kryptid on 21/12/2021 21:05:02
Flipping the electron spin costs energy.
Proton and electron opposite spin + energy = Proton and electron spin in the same direction.
Where is the exact quanta of energy coming from?
I never said anything about a spontaneous change in spin.
Quote
What is the cause of the same direction electron spin changing to opposite direction electron spin, lowering Hydrogen atom energy?
Again, I never said anything about such a thing happening. You really need to be more clear about what questions you are asking. I was under the impression that you were talking about hydrogen atoms entering an excited state.
Essentially there is only one type of transition, or cause for Hydrogen atom transition when electron goes from higher energy state to lower energy state.
That is not the same thing as an electron changing spin. Please make up your mind as to what you are talking about.
Slower motion through the preferred frame
Please stop saying that. No such thing exists.
Post by: Bored chemist on 21/12/2021 21:16:15
That's a cause for the transition and we do not have to wait 10 million years.There are lots of possible transitions; some do take millions of years.
I see your answer to my question
Do you have any clue what you are talking about?is "no".
Post by: Bored chemist on 21/12/2021 21:22:22
The electron has to slow down.No
One of the best known transitions is a spin flip.
https://en.wikipedia.org/wiki/Hydrogen_line#Cause
It takes about ten million years.
Post by: Bored chemist on 21/12/2021 21:23:39
preferred frameIf you think there's a preferred frame, your ideas are probably beyond help.
Post by: Jaaanosik on 22/12/2021 01:51:15
Flipping the electron spin costs energy.
Proton and electron opposite spin + energy = Proton and electron spin in the same direction.
Where is the exact quanta of energy coming from?
I never said anything about a spontaneous change in spin.QuoteWhat is the cause of the same direction electron spin changing to opposite direction electron spin, lowering Hydrogen atom energy?
Again, I never said anything about such a thing happening. You really need to be more clear about what questions you are asking. I was under the impression that you were talking about hydrogen atoms entering an excited state.Essentially there is only one type of transition, or cause for Hydrogen atom transition when electron goes from higher energy state to lower energy state.
That is not the same thing as an electron changing spin. Please make up your mind as to what you are talking about.Slower motion through the preferred frame
Please stop saying that. No such thing exists.
When you said this:
Quote
the electron has a cause to flip to higher state based on the Einstein's equations.
Are you talking about flipping spin, or merely being raised to a higher energy state? Those two things are not the same. It's also not accurate. Electrons don't become excited just because an atom is moving quickly.
There is this Hydrogen 21cm line: https://en.wikipedia.org/wiki/Hydrogen_line
Flipping the electron spin from opposite direction to the same direction is going from lower state to higher energy state.
'Spontaneous change in spin' as you say, does not exist.
There has to be a cause for it. Electron has to be under right condition to flip. We do not know the cause so we call it spontaneous, that makes sense. :)
As it was mentioned, it can take 10 million years or it can happen in much much shorter time if the electron is going to slow down.
The uncertainty principle is the consequence of the fact two inertial observers do not agree on barycenter position of a Hydrogen atom. That is where the Special Relativity failed. It cannot be reconciled with the Quantum physics.
In order to understand the Hydrogen atom barycenter problem, it is easier to see it in this image.
The reason is simple, the angular momentum of the flywheel represents electron behavior very well.
Top is S inertial observer view and bottom is SIV inertial observer view.
The inertial observers do not agree how the collision is going to happen.
The hydrogen barycenter disagreement is the cause why we do not know where the electron and proton are and where the photon is going to be emitted.
(https://i.imgur.com/wZaOQ8e.png)
The proof the electron in Hydrogen atom is affected by the preferred frame is the Lamb shift that leads to the fine structure constant. That's my hypothesis.
https://en.wikipedia.org/wiki/Lamb_shift
https://en.wikipedia.org/wiki/Fine-structure_constant
Post by: Jaaanosik on 22/12/2021 01:56:58
https://en.wikipedia.org/wiki/Hydrogen_maserThe electron has to slow down.No
One of the best known transitions is a spin flip.
https://en.wikipedia.org/wiki/Hydrogen_line#Cause
It takes about ten million years.
Quote
A hydrogen maser, also known as hydrogen frequency standard, is a specific type of maser that uses the intrinsic properties of the hydrogen atom to serve as a precision frequency reference.
Both the proton and electron of a hydrogen atom have spins. The atom has a higher energy if both are spinning in the same direction, and a lower energy if they spin in opposite directions. The amount of energy needed to reverse the spin of the electron is equivalent to a photon at the frequency of 1,420,405,751.786 Hz,[1] which corresponds to the 21 cm line in hydrogen spectrum.
… and it can be done on demand. :)
Post by: Jaaanosik on 22/12/2021 01:59:04
Please, explain the barycenter problem in the image couple of posts above, then we can continue.preferred frameIf you think there's a preferred frame, your ideas are probably beyond help.
Post by: Kryptid on 22/12/2021 02:17:35
The uncertainty principle is the consequence of the fact two inertial observers do not agree on barycenter position of a Hydrogen atom.
That doesn't make any sense because an observer in the same reference frame as the hydrogen atom in question is still subject to the uncertainty principle when trying to do measurements of it. Keep in mind that the electron is not a classical object. It does not move around the proton in the same way that a planet moves around a star nor does it spin in the common usage of the term.
That is where the Special Relativity failed.
I'm afraid not. Special relativity has a history of fantastically passing the tests it has been put through.
The proof the electron in Hydrogen atom is affected by the preferred frame is the Lamb shift that leads to the fine structure constant. That's my hypothesis.
Then your hypothesis is very probably wrong, as the results of tests of relativity very strongly support the notion that no such preferred frame exists. The very Wikipedia article you cited already has an explanation for it:
Quote
Interaction between vacuum energy fluctuations and the hydrogen electron in these different orbitals is the cause of the Lamb shift, as was shown subsequent to its discovery.
Post by: Bored chemist on 22/12/2021 08:56:40
And it even happens in space, but very rarely.
… and it can be done on demand. :)
https://en.wikipedia.org/wiki/Astrophysical_maser
The natural life of the upper state is still ten million years or so.
But that doesn't stop you being wrong, because all the tests ever done on relativity gave the expected answers.
Post by: Jaaanosik on 22/12/2021 20:51:20
The special relativity fails to predict the same position of Hydrogen atom barycenter.The uncertainty principle is the consequence of the fact two inertial observers do not agree on barycenter position of a Hydrogen atom.
That doesn't make any sense because an observer in the same reference frame as the hydrogen atom in question is still subject to the uncertainty principle when trying to do measurements of it. Keep in mind that the electron is not a classical object. It does not move around the proton in the same way that a planet moves around a star nor does it spin in the common usage of the term.
...
This failure leaves us with statistics of quantum physics.
The special relativity fails to predict the same position of flywheel barycenter.
Yes, it appears the vacuum fluctuations frame is the preferred one. That's the fabric of space and time.
Post by: Kryptid on 22/12/2021 20:58:01
Post by: Jaaanosik on 22/12/2021 21:04:41
Can a photon be released without a cause?And it even happens in space, but very rarely.
… and it can be done on demand. :)
https://en.wikipedia.org/wiki/Astrophysical_maser
The natural life of the upper state is still ten million years or so.
But that doesn't stop you being wrong, because all the tests ever done on relativity gave the expected answers.
Do we know a reason other than "breaking radiation" (incl. cooling off) for photon to be released?
Post by: Jaaanosik on 22/12/2021 21:07:06
Vacuum fluctuations are not a reference frame. They already appear in all reference frames.Which one?
How many vacuum fluctuation reference frame are out there?
Do you have a comment on the barycenter of Hydrogen atom and the flywheel?
Post by: Kryptid on 22/12/2021 21:13:32
I don't see any reason why a hydrogen atom's barycenter would differ between frames because the electron is in a superposition without a defined location.
As far as the flywheel goes, if relativity predicts that the centroid differs in different frames, then it differs in different frames. Simple as that.
Post by: Jaaanosik on 23/12/2021 14:03:22
In theory, there are infinitely many possible reference frames. All of them contain vacuum fluctuations.The bold part is not so simple.
I don't see any reason why a hydrogen atom's barycenter would differ between frames because the electron is in a superposition without a defined location.
As far as the flywheel goes, if relativity predicts that the centroid differs in different frames, then it differs in different frames. Simple as that.
It proves the inertial observers do not agree on physics, they cannot predict the same outcome of collisions.
The relativity is intrinsically antagonistic, broken.
Post by: Kryptid on 23/12/2021 14:56:33
The bold part is not so simple.
It proves the inertial observers do not agree on physics, they cannot predict the same outcome of collisions.
The relativity is intrinsically antagonistic, broken.
I will admit that the solution to your flywheel example isn't apparent to me, but I'm sure one must exist. There have been numerous proposals for paradoxes in special relativity, but solutions to them have been found. I'm sure the same will be true of yours. The experimental evidential support for special relativity pretty much guarantees it.
Post by: Jaaanosik on 23/12/2021 15:26:17
There is not solution.The bold part is not so simple.
It proves the inertial observers do not agree on physics, they cannot predict the same outcome of collisions.
The relativity is intrinsically antagonistic, broken.
I will admit that the solution to your flywheel example isn't apparent to me, but I'm sure one must exist. There have been numerous proposals for paradoxes in special relativity, but solutions to them have been found. I'm sure the same will be true of yours. The experimental evidential support for special relativity pretty much guarantees it.
This is how it goes, if I am in the top frame, rest frame of the rotating flywheel and I observer the unexpected collision then I am going to call it spontaneous, there is no cause for it, I just have to break causality in physics and nothing to see here, let us move on. :)
Hmmm, that sounds familiar.
Edit:... and I agree with Einstein: "God does not play dice."
Post by: Kryptid on 23/12/2021 22:00:54
There is not solution.
There obviously has to be one. If you performed this experiment in real life, you would get a result. That result would have be agreed on by all people watching the experiment. To say otherwise makes no sense.
This is how it goes, if I am in the top frame, rest frame of the rotating flywheel and I observer the unexpected collision
The only way that collision would be unexpected would be if you weren't watching for it.
Quote
then I am going to call it spontaneous, there is no cause for it, I just have to break causality in physics
The cause for the collision was whatever set the flywheels in motion in the first place.
Quote
and nothing to see here, let us move on.
Indeed, there is nothing to see here. This is what is called a "non-sequitur".
Hmmm, that sounds familiar.
Please tell me what that sounds familiar to (and don't say quantum physics, because what you just said has nothing to do with that).
Post by: Jaaanosik on 23/12/2021 22:46:45
This cannot be denied, two inertial observers do not predict the same outcome of the collision - disagreement on physics.There is not solution.
There obviously has to be one. If you performed this experiment in real life, you would get a result. That result would have be agreed on by all people watching the experiment. To say otherwise makes no sense.This is how it goes, if I am in the top frame, rest frame of the rotating flywheel and I observer the unexpected collision
The only way that collision would be unexpected would be if you weren't watching for it.Quotethen I am going to call it spontaneous, there is no cause for it, I just have to break causality in physics
The cause for the collision was whatever set the flywheels in motion in the first place.Quoteand nothing to see here, let us move on.
Indeed, there is nothing to see here. This is what is called a "non-sequitur".Hmmm, that sounds familiar.
Please tell me what that sounds familiar to (and don't say quantum physics, because what you just said has nothing to do with that).
This is a problem.
Only one observer will be right in predicting how the flywheels end up after the collision.
Post by: Kryptid on 23/12/2021 22:52:58
This cannot be denied
I deny it. Seeming paradoxes in relativity fade away under a more scrutinizing eye. Maybe @Halc or someone with a better understanding of relativity than me can shed some light on it.
Post by: Jaaanosik on 23/12/2021 22:59:00
This is your best honest answer, I appreciate it.This cannot be denied
I deny it. Seeming paradoxes in relativity fade away under a more scrutinizing eye. Maybe @Halc or someone with a better understanding of relativity than me can shed some light on it.
I challenge any physicist to show what is wrong with this analysis.
Denying it, without backing it up, is not good enough.
Post by: Kryptid on 23/12/2021 23:00:16
Denying it, without backing it up
Special relativity is backed up by experiment.
Post by: Jaaanosik on 23/12/2021 23:03:57
Nobody did the analysis I showed here.Denying it, without backing it up
Special relativity is backed up by experiment.
There is no experiment related to the presented problem.
Post by: Kryptid on 23/12/2021 23:06:39
Nobody did the analysis I showed here.
They didn't have to. We know that different observers see different things from experiment. Length contraction and time dilation are both real effects.
Post by: Jaaanosik on 23/12/2021 23:08:41
This one is about the specific collision.Nobody did the analysis I showed here.
They didn't have to. We know that different observers see different things from experiment. Length contraction and time dilation are both real effects.
Post by: Kryptid on 23/12/2021 23:09:49
This one is about the specific collision.
I'm aware of that. It doesn't change what I said.
Post by: Jaaanosik on 23/12/2021 23:12:22
I suggest we leave for now, unless someone else has an opinion.This one is about the specific collision.
I'm aware of that. It doesn't change what I said.
Post by: Halc on 24/12/2021 15:13:09
I deny it. Seeming paradoxes in relativity fade away under a more scrutinizing eye. Maybe @Halc or someone with a better understanding of relativity than me can shed some light on it.All I see is a single-word description "Collision." This is not a defined process. If two objects collide at relativistic speeds, you get an explosion and lots of parts flying in every direction. You want a better description, you need a far better description of what goes on, such as mass X applies force F on the axis of the spinning wheel, changing its momentum.
As for the wheel, its energy, spin rate, center of gravity, center of energy, and momentum are all indeed frame dependent, but that doesn't change any objective measurement made by any observer.
As for the angular momentum of the wheel, under special relativity, the motion and momentum and change in motion is all described by a tensor, which is frame invariant and hence does not have contradictory consequences as predicted by different observers.
I have pretty much stopped watching this troll thread, which first of all just a pathetic attempt to deny relativity along the lines of "Here's a dozen things I clearly don't understand about relativity, therefore relativity is inconsistent".
Secondly, the equivalence principle has not one single time been applied in nearly 200 posts. I don't think Jano could state the EP without just looking it up on a website.
Post by: Kryptid on 24/12/2021 15:57:00
All I see is a single-word description "Collision." This is not a defined process. If two objects collide at relativistic speeds, you get an explosion and lots of parts flying in every direction. You want a better description, you need a far better description of what goes on, such as mass X applies force F on the axis of the spinning wheel, changing its momentum.
It's in Reply #161. The gist of it is that a spinning wheel collides with a non-spinning wheel (but both are inside of casings of some kind that prevent the wheels from directly touching. That is supposed to eliminate friction from the wheel's spin from causing a deflection in itself). In the frame of the spinning wheel, the center of mass of both wheels are aligned so no deflection is expected upon collision. They are expected to simply bounce off along the same line as the collision.
In the frame of the non-spinning wheel, length contraction makes the spinning wheel's center of mass appear to be biased towards one side. In this case, the center of masses of the two wheels are not aligned, so we expect some kind of deflection upon impact.
I know there must be a consistent solution, I'm just not sure how to obtain it.
Post by: Jaaanosik on 24/12/2021 17:34:17
Kryptid,All I see is a single-word description "Collision." This is not a defined process. If two objects collide at relativistic speeds, you get an explosion and lots of parts flying in every direction. You want a better description, you need a far better description of what goes on, such as mass X applies force F on the axis of the spinning wheel, changing its momentum.
It's in Reply #161. The gist of it is that a spinning wheel collides with a non-spinning wheel (but both are inside of casings of some kind that prevent the wheels from directly touching. That is supposed to eliminate friction from the wheel's spin from causing a deflection in itself). In the frame of the spinning wheel, the center of mass of both wheels are aligned so no deflection is expected upon collision. They are expected to simply bounce off along the same line as the collision.
In the frame of the non-spinning wheel, length contraction makes the spinning wheel's center of mass appear to be biased towards one side. In this case, the center of masses of the two wheels are not aligned, so we expect some kind of deflection upon impact.
I know there must be a consistent solution, I'm just not sure how to obtain it.
Thank you for reading the posts carefully, I appreciate that!
Yes, there is a casing, I am working on a new drawing.
The problem is even before collision, the casing and the flywheel have a barycenter between them from the SIV frame point of view that is not on the axle.
That is different from S frame point of view where the barycenter is on the axle.
The barycenter is where the inertial observer is located. When I have the new image, it is going to be more obvious.
Thanks again,
Jano
Post by: Jaaanosik on 24/12/2021 18:28:00
Post by: Halc on 24/12/2021 23:33:07
The gist of it is that a spinning wheel collides with a non-spinning wheel (but both are inside of casings of some kind that prevent the wheels from directly touching.OK, so some force is applied between the two objects like a spring attached at the axle that forces an elastic collision, with one spinning wheel in one case and not spinning in the other. That's going to involve an objective vertical displacement of both objects. Not sure why the 2nd object needs to have a wheel in a case. It might as well be a rock with the same proper mass as the case with the non-spinning wheel..
Quote
In the frame of the spinning wheelWell, both wheels are going to be accelerating, so you probably mean the inertial frame of each wheel before they begin accelerating.
Quote
the center of mass of both wheels are aligned so no deflection is expected upon collision.No, because the spinning wheel is accelerated, so its CoM is going to shift upward, deflecting the wheel downward. The 'collision' will apply a torque to it and a counter-torque to the system.
Post by: Kryptid on 25/12/2021 20:34:58
No, because the spinning wheel is accelerated, so its CoM is going to shift upward, deflecting the wheel downward. The 'collision' will apply a torque to it and a counter-torque to the system.
Hmm... okay, I think I understand now. The solution involves the fact that the collision makes both wheels accelerate.
Post by: Jaaanosik on 27/12/2021 18:55:59
No, because the spinning wheel is accelerated, so its CoM is going to shift upward, deflecting the wheel downward. The 'collision' will apply a torque to it and a counter-torque to the system.
Hmm... okay, I think I understand now. The solution involves the fact that the collision makes both wheels accelerate.
The casing on the left does not rotate.
If we look at the image then the top view from S frame does not expect any angular velocity change because all centroids are aligned on the same line.
The collision will change the angular velocity of the flywheel in the SIV frame view because centroids are not aligned.
(https://i.imgur.com/Wpp4cMG.png)
Post by: Kryptid on 27/12/2021 19:39:10
The casing on the left does not rotate.
No one said it did.
If we look at the image then the top view from S frame does not expect any angular velocity change because all centroids are aligned on the same line.
The acceleration happens upon collision. When that occurs, neither wheel is in an inertial frame, but is an accelerated frame instead.
Post by: Jaaanosik on 27/12/2021 20:27:30
Well, is angular momentum going to be conserved in SIV frame?The casing on the left does not rotate.
No one said it did.If we look at the image then the top view from S frame does not expect any angular velocity change because all centroids are aligned on the same line.
The acceleration happens upon collision. When that occurs, neither wheel is in an inertial frame, but is an accelerated frame instead.
Is angular momentum going to be conserved in S frame?
Post by: Kryptid on 27/12/2021 20:44:31
Well, is angular momentum going to be conserved in SIV frame?
Is angular momentum going to be conserved in S frame?
As Halc said:
As for the angular momentum of the wheel, under special relativity, the motion and momentum and change in motion is all described by a tensor, which is frame invariant and hence does not have contradictory consequences as predicted by different observers.
Post by: Jaaanosik on 27/12/2021 23:23:04
It still comes down to L=rxp as per the linked paper.Well, is angular momentum going to be conserved in SIV frame?
Is angular momentum going to be conserved in S frame?
As Halc said:As for the angular momentum of the wheel, under special relativity, the motion and momentum and change in motion is all described by a tensor, which is frame invariant and hence does not have contradictory consequences as predicted by different observers.
The SIV frame has ri>0 and the S frame has ri=0.
The end result is going to be
in SIV frame and 
S frame. (https://i.imgur.com/5Pae3VF.png)
Post by: puppypower on 28/12/2021 15:58:56
If we knew how much energy was initially used by the train/station system; gallons of fuel to move the train, we can calculate the total energy used and from this we can predict the correct frame to use, if energy conservation is important. If not we can perpetuate fantasies via unintended energy conservation violation based on assuming all frames are equal when dealing with energy conservation.
Post by: Origin on 28/12/2021 16:47:27
Relative frames of reference can violate energy conservation.No they can't, the conservation of energy only applies to one reference frame by definition. Why don't you learn some physics? Oh, that's right, you're not interested in learning.
Post by: Kryptid on 28/12/2021 21:27:44
Post by: Jaaanosik on 28/12/2021 23:59:56
Fuel fantasy speculation?Please, any comments on the post #194?
The quote from the paper states the invariance applies only to the Galilean relativity.
Special relativity is different, … as you can see in the paper.
Halc's comment is not accurate.
As Halc said:As for the angular momentum of the wheel, under special relativity, the motion and momentum and change in motion is all described by a tensor, which is frame invariant and hence does not have contradictory consequences as predicted by different observers.
Post by: puppypower on 30/12/2021 12:03:55
The universe is bleeding useable free energy via the 2nd law. For example, we cannot just assume a cyclic universe, since so much of its original useable free energy is now tied up as entropy, which has caused the available universal free energy to lower. A half full universal fuel tank cannot retrace all the steps of a full tank. The universe also cannot go on forever, with less fuel each day.
It is possible that dark energy is the new placeholder theory for this ever increasing conserved entropic energy, that is more of less off net limits to the universe. By being conserved it will have an impact of the rest of the universe, acting as the platform via its time based history of its ever increasing complexity. We will see it entangled but not easy to isolate as particles.
Since an increase in universal entropy will be endothermic; loss of free energy, one can argue that what we called the universal red shift could be partially due to the 2nd law. The red shift of all photons will also make less useable energy in universe. If assume all this energy is transferred to the motion implied by a Doppler shift than what is left over for the entropy increase of the second law? There are not many laws of physics, but theories are a dime a dozen. Laws come first, theory then gets the leftovers, not the other way around. You guys appear to have used the wrong frame of reference.
Relative frames of reference can violate universal energy conservation. It can also cause problems within some frames which cause them to violate energy conservation.
As an example, say we have a train moving past a train station. If we make this motion relative to the train or relative to the station, we will get two different energy balances, since the train weighs more than the small station; 1/2MV2. If you choose the wrong reference; station, your energy balance will not add up to reality. This can create uncertainty and fuel fantasy speculation. A universal energy balance needs to be the same for all references or else it is not universal.
If we knew how much energy was initially used by the train/station system; gallons of fuel to move the train, we can calculate the total energy used and from this we can predict the correct frame to use, if universal energy conservation is important. Most of the time we do not know the amount of fuel used; universal starting energy. As such, some frames can fool us to perpetuate fantasies via unintended universal energy conservation violation; station.
The solution is to find a reference that is the same for all. That reference is the speed of light reference, which is the same in all inertial frames.
Post by: Bored chemist on 30/12/2021 13:12:49
The solution is to find a reference that is the same for all. That reference is the speed of light reference, which is the same in all inertial frames.The SoL is the same in all reference frames, but it is not a reference frame.
Post by: Jaaanosik on 03/01/2022 20:01:15
If I take the posts #190 and #194, ask the question about the inertial frames disagreement on the angular velocity after the collision in the physics forum, would it be moved to new theories?
I hope it would not because there is nothing new, I just quoted the paper and I added my question.
Post by: maul on 03/01/2022 20:18:09
Post by: Kryptid on 03/01/2022 21:03:12
Off subject, but why did you have to lock the DebatePhysics thread?
I see you are new to these forums, but this is the kind of question that should be sent via PM to the moderators instead of making an off-topic post in an existing thread. In any case, I'll do what I can to address your concerns.
We have, unfortunately, had a history of members making very long threads that end up running around in circles with nothing ever being accomplished because the member who started the thread consistently refuses to learn about the most basic principles of physics. As an example, we had someone who insisted (among other things) that black holes can make new energy (and thus violate the first law of thermodynamics). No matter how many times it was explained to that member, we could not convince him otherwise and the thread ended up running for over a year. It was a waste of forum resources that shouldn't have happened. I could tell, based on DebatePhysics' responses in the thread and the fact that he has a YouTube channel filled to the bring with physics nonsense that allowing that thread to continue would have resulted in the exact same kind of thing happening.
It was only recently decided by the staff here that we should work to minimize scientific misinformation. You want to propose a new theory that attempts to fill in the knowledge gaps of current science? That's perfectly fine. At least by doing that, denialism isn't required. It's a different matter altogether to say that a well-tested idea that is widely accepted by the vast majority of the scientific community is wrong, provide a fallacious understanding of the material as "proof" that it is wrong, and then refuse to concede that their understanding is wrong even after everyone else present has tried to correct them. That is useless. It doesn't do anything but make people get frustrated. As such, we don't want that to happen here.
Post by: Colin2B on 03/01/2022 22:55:44
Happy New Year to all!Happy New Year To you too.
If I take the posts #190 and #194, ask the question about the inertial frames disagreement on the angular velocity after the collision in the physics forum, would it be moved to new theories?
Best keep it where it is, saves us the trouble of moving it here.
Post by: Jaaanosik on 04/01/2022 00:55:38
The most noble inclination of the Human Nature is the pursuit of the truth.Happy New Year to all!Happy New Year To you too.
If I take the posts #190 and #194, ask the question about the inertial frames disagreement on the angular velocity after the collision in the physics forum, would it be moved to new theories?
Best keep it where it is, saves us the trouble of moving it here.
I appreciate an honest feedback from the knowledgeable members.
Do you agree the pointed out contradiction is end result of our current physics understanding?
I did not add/change any math.
Post by: Halc on 04/01/2022 01:22:03
The most noble inclination of the Human Nature is the pursuit of the truth.Then why are you constantly posting a thread asserting self contradictory nonsense? Asking questions is fine, but asserting bad science is why this topic is here.
Quote
Do you agree the pointed out contradiction is end result of our current physics understanding?No. It's the end result of your lack of it. The interaction mentioned can only have one single objective result, which was mentioned above, but you ignored it since it doesn't agree with your assertions (based it seems on Newtonian intuition, which are just plain wrong in relativistic scenarios.)
Post 188:
OK, so some force is applied between the two objects like a spring attached at the axle that forces an elastic collision, with one spinning wheel in one case and not spinning in the other. That's going to involve an objective vertical displacement of both objects.
Everybody in any frame sees a vertical displacement due to the collision. No contradiction. Your logic asserting other wise (in the frame of the spinning wheel) is wrong since it is based on assumptions demonstrated centuries ago to be inconsistent.
Post by: Jaaanosik on 04/01/2022 03:32:41
Halc,The most noble inclination of the Human Nature is the pursuit of the truth.Then why are you constantly posting a thread asserting self contradictory nonsense? Asking questions is fine, but asserting crap is why this topic is here.QuoteDo you agree the pointed out contradiction is end result of our current physics understanding?No. It's the end result of your lack of it. The interaction mentioned can only have one single objective result, which was mentioned above, but you ignored it since it doesn't agree with your assertions (based it seems on Newtonian intuition, which are just plain wrong in relativistic scenarios.)
Post 188:OK, so some force is applied between the two objects like a spring attached at the axle that forces an elastic collision, with one spinning wheel in one case and not spinning in the other. That's going to involve an objective vertical displacement of both objects.
Everybody in any frame sees a vertical displacement due to the collision. No contradiction. Your logic asserting other wise (in the frame of the spinning wheel) is wrong since it is based on assumptions demonstrated centuries ago to be inconsistent.
your concern from the post #188 was addressed in my post #190:
(https://i.imgur.com/Wpp4cMG.png)
There is a non-rotating casing around the rotating flywheel. There is no cause for objective vertical displacement in the rotating flywheel frame S.
All centroids are on the straight line collision in the S frame.
Post by: puppypower on 04/01/2022 12:05:20
One way to maintain the equivalency principle is to not assume an secondary force but to assume an action and reaction based only on gravity. The exothermic action of gravity; lowering gravitational potential, will create an endothermic reaction; increase gravitational potential elsewhere; expansion, so both appear to be from one indistinguishable source.
All force creates an acceleration which is one part distance and two parts time; d/t/t. The concept of space-time only has the units of d-t. The concept of space-time lacks the second vector of time associated with force and acceleration. Therefore the second time vector is not easy to measure based on standard assumptions using the concept of space-time. The second time vector involves time apart from space. It cannot be a sine wave in the traditional sense since this is 1-D and not 2-D like space-time.
For example, in the action/reaction of forming stars, the materials will rotate, with rotation expressed as frequency; reciprocal of time, without distance units. This creates a centrifugal force vector opposite the gravity vector; action-reaction. In the case of spiral galaxies, the second time vector is not uniform based on mass densities.
Post by: Colin2B on 04/01/2022 12:46:36
I appreciate an honest feedback from the knowledgeable members.and you are getting it in this section of the forum
Post by: Origin on 04/01/2022 15:08:12
If the equivalency principle is correct why do we experimentally isolate the accelerated expansion of the universe from gravity, using what we assume to be experimental proof? We should not be able to distinguish dark energy from gravity in term of the expansion from any frame of reference.Warning: no actual physics is in this rambling word salad. Time is still not a vector...
One way to maintain the equivalency principle is to not assume an secondary force but to assume an action and reaction based only on gravity. The exothermic action of gravity; lowering gravitational potential, will create an endothermic reaction; increase gravitational potential elsewhere; expansion, so both appear to be from one indistinguishable source.
All force creates an acceleration which is one part distance and two parts time; d/t/t. The concept of space-time only has the units of d-t. The concept of space-time lacks the second vector of time associated with force and acceleration. Therefore the second time vector is not easy to measure based on standard assumptions using the concept of space-time. The second time vector involves time apart from space. It cannot be a sine wave in the traditional sense since this is 1-D and not 2-D like space-time.
For example, in the action/reaction of forming stars, the materials will rotate, with rotation expressed as frequency; reciprocal of time, without distance units. This creates a centrifugal force vector opposite the gravity vector; action-reaction. In the case of spiral galaxies, the second time vector is not uniform based on mass densities.
Post by: Jaaanosik on 04/01/2022 18:33:45
your concern from the post #188 was addressed in my post #190I did not express a concern in post 188. It was simply the objective answer for which you asked, and which you continue to deny apparently because denial (not learning) is your goal. I didn't reply to 190 since your goal of denial was successfully met.
If your analysis contradicts the 400 year old PoR (Principle of Relativity), then it contradicts all the theories based on it, including Newton's. It means either that the last 4 centuries of science based on the principle is wrong (highly unlikely not to have been noticed by now), or it is you (who demonstrably has trouble with even high-school physics) is wrong. So take a humble pill and ask where your analysis went awry, and lay off the troll assertions that you've actually uncovered some kind of flaw in physics. That's what I do when I get a contradictory result. I assume the problem is with me, a known amateur. It happens a lot, but I don't go around asserting that physics is wrong. I just found a mistake in my understanding, and if I can't work it out myself (as is sometimes the case), then I seek help from those better informed.
We all make mistakes, and my answers are also not always correct.
...
We are analyzing the relativistic flywheel.
(https://i.imgur.com/5Pae3VF.png)
Newton/Galileo worked with absolute time therefore they achieved AM invariance.
The relativistic flywheel is different, the AM varies under the Lorentz transformation.
Why do you keep bringing up 400 years old PoR?
The relativistic AM is frame dependent.
Why is it so hard to understand the change of relativistic AM is going to be frame dependent as well?
The relativistic energy change is also frame dependent as well, agreed?
Post by: Jaaanosik on 04/01/2022 18:42:36
What force is going to cause vertical displacement when the collision in x direction is observed from the S frame?QuoteThere is a non-rotating casing around the rotating flywheel.That changes absolutely nothing. It just makes it more obvious that the force on the spinning wheel is being applied to the axle and not elsewhere, as it always has been. The post 188 reply is still unchanged.
Post by: Halc on 06/01/2022 17:12:20
Why do you keep bringing up 400 years old PoR?Because you seem to be denying it in almost every post, despite a different (and unreferenced) principle in the thread title.
Quote
The relativistic AM is frame dependent.I don't recall referencing the AM when making my argument. In any frame, the change of system AM is zero, per conservation laws. Ditto with energy since you're describing an elastic collision with no energy lost to heat. But I did not reference any of these things nor did I quantify the motion of this incompletely specified scenario. All I said was that after the collision, the left wheel will end up with a downward velocity component balanced by an upward one of the right wheel. This must be observed in all frames, per PoR of which you are in denial.
Why is it so hard to understand the change of relativistic AM is going to be frame dependent as well?
What force is going to cause vertical displacement when the collision in x direction is observed from the S frame?l was really hoping you'd figure this out yourself, but it seems your goals prevent that.
First of all, observation has nothing to do with it. Observation is not a cause (except arguably in QM).
A force is required only if the vertical linear momentum is changed, which it isn't, at least not at first when the forces between the two is entirely along the X axis. So it moves down not due to a force/momentum change, but rather to a change in the internal mass of the object. If an airplane is stationary in space, pointed north, and one pumps fuel from the west wing tank to the east wing tank, the airplane as a whole will move west despite no external force being applied to it and despite it having zero momentum the whole time. Same sort of thing happens as the rotating wheel accelerates left: The internal mass moves up, so the axle/casing moves down to compensate, preserving the center of mass.
Now the two masses (the two wheels) are no aligned along the x axis, and the force between them has a vertical component, giving the wheels nonzero vertical momentum components, so the motion will not cease once the acceleration of the collision does.
Post by: Kryptid on 07/01/2022 00:57:30
Because you seem to be denying it in almost every post, despite a different (and unreferenced) principle in the thread title.
Based on that, would this thread be a candidate for being locked since it's basically denialism?
Post by: Jaaanosik on 13/01/2022 16:58:55
... All I said was that after the collision, the left wheel will end up with a downward velocity component balanced by an upward one of the right wheel...
This is not going to happen in the S frame.
There is nothing in the collision that would force the rotating wheel in the casing to move downward when the analysis is done from the S frame.
The forces transferred from the casing to the axle and rotating wheel are symmetrical. Nothing to move the flywheel down, just left/right after the collision.
(https://i.imgur.com/Wpp4cMG.png)
Post by: Halc on 13/01/2022 22:26:37
Why do you keep bringing up 400 years old PoR?Because you seem to be denying it in almost every post
This is not going to happen in the S frame.And there it is again, an assertion that something happens only in one abstract coordinate system and not another, a direct denial of PoR.
Based on that, would this thread be a candidate for being locked since it's basically denialism?I agree. This has turned into straight denialism, without even pretending to have noticed the explanations provided. Thread locked.
The entire prior post is just pure repetition of prior assertions. You've seeming not bothered to read the prior explanations at all, which explain exactly why vertical displacement must result. I have nothing to add that hasn't already been said.
Quote
There is nothing in the collision that would force the rotating wheel in the casing to move downward when the analysis is done from the S frame.Repetition of a wrong statement doesn't make it more correct.
There is no symmetry. The left wheel is rotating, the case around it is not.
Between the two sides, the one side has a rotating wheel and the other does not. You want symmetry, put a wheel rotating the other way in the right side. You'll still get a vertical displacement, but no change in vertical momentum component.