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You'll have to flesh out how you think EP says this.
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.
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?
This is all in the Earth frame.The transition from to meaning energy of is the same in perihelion and aphelion.
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.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 aboutYou 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.
Quote from: Jaaanosik on 12/12/2021 14:40:44Where is the more work coming fromGravity.
Where is the more work coming from
Quote from: Halc on 12/12/2021 04:19:44You'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.
Is there any force increasing photon's potential from perihelion to aphelion?
...Quote from: Jaaanosik on 12/12/2021 04:54:45The space station is on the Earth surface full of atoms in state.A rocket brings space station into the elliptic orbit.This 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....
...QuoteA 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.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....
I hope you see the problem.
...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....
...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....
...Quote from: Jaaanosik on 12/12/2021 14:40:44This is all in the Earth frame.The transition from to meaning energy of is the same in perihelion and aphelion.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.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.
Quote from: Jaaanosik on 13/12/2021 02:04:17Is 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.
If there is no work done on atoms flying from perihelion to aphelion then why would the Earth measure more energy?
Are you sayin the inertia is doing the work?
Quote from: Jaaanosik on 13/12/2021 03:04:55If 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.Quote from: Jaaanosik on 13/12/2021 03:15:54Are you sayin the inertia is doing the work?It's the force of gravity that's doing it.
Is gravity 'pushing' from perihelion to aphelion and doing work?
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).
Kryptid, I'm going to have to disagree with some of the stuff posted.Quote from: Kryptid on 11/12/2021 17:58:25It'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.
It's not true to say that no force is acting on Voyager.
Quote from: Jaaanosik on 13/12/2021 02:49:12We 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).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.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....
Quote from: Jaaanosik on 13/12/2021 16:25:01Assuming 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.
Assuming we have a simple hydrogen atom, for simplicity
E0 is n=1 quantum number - ground state
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?
Because of the Equivalence Principle the hydrogen atom barycenter observer and the space station observer do not see any change.