Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: jeffreyH on 13/02/2017 17:52:05
-
An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero. It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?
-
An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero.
I'm with you so far...
It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy.
I don't think this is right. The kinetic energy is still positive, even if the velocity is defined as negative (velocity is squared in the equation K.E. = 0.5*m*v2, so the sign of the velocity is inconsequential, only the magnitude matters here).
Where does the kinetic energy go? Is it conserved?
This problem is traditionally thought of as kinetic energy being converted into potential energy (gravitational potential energy in this case), which is then converted back to kinetic energy. Energy overall is conserved, but neither kinetic energy nor potential energy is conserved.
-
Basically, the equation Et = mv^2/2-GMm/r
Where
Et is the total energy of mass m with respect to M
M is the gravitating mass
r is the center to center distance between m and M
gives a constant value of Et for mass m in a free fall state (including orbits)
For a closed trajectory (circular or elliptical orbits), Et will be negative.
For a parabolic trajectory (v= escape velocity at r) Et will be 0. (Et is set at zero for mass m when it is at rest with respect to M at an infinite distance from M)
For a hyperbolic trajectory (v> escape velocity at r) Et will be positive.
-
The momentum of a body falling back towards the earth will be p = -m v. Do we distinguish which quantity 'owns' the minus sign? We can sensible assume that it is velocity since negative mass in this context makes no sense. If a force is attractive only it must remove energy. Potential energy balances the books but does it hide the underlying mechanism? The negative velocity squared does give a positive result but then information is lost since you cannot tell from the answer whether the kinetic energy was directed away from or towards the source. Since velocity, a vector, times mass, a scalar, is itself a vector momentum preserves directionality. Since energy is a scalar we cannot rely on it to tell us everything we need. This has to be the job of momentum.
So if the momentum is negative can we say the same about the energy? Do we need a concept of negative kinetic energy?
-
Basically, the equation Et = mv^2/2-GMm/r
Where
Et is the total energy of mass m with respect to M
M is the gravitating mass
r is the center to center distance between m and M
gives a constant value of Et for mass m in a free fall state (including orbits)
For a closed trajectory (circular or elliptical orbits), Et will be negative.
For a parabolic trajectory (v= escape velocity at r) Et will be 0. (Et is set at zero for mass m when it is at rest with respect to M at an infinite distance from M)
For a hyperbolic trajectory (v> escape velocity at r) Et will be positive.
That is just a Lagrangian. That was not my point.
-
Basically, the equation Et = mv^2/2-GMm/r
Where
Et is the total energy of mass m with respect to M
M is the gravitating mass
r is the center to center distance between m and M
gives a constant value of Et for mass m in a free fall state (including orbits)
For a closed trajectory (circular or elliptical orbits), Et will be negative.
For a parabolic trajectory (v= escape velocity at r) Et will be 0. (Et is set at zero for mass m when it is at rest with respect to M at an infinite distance from M)
For a hyperbolic trajectory (v> escape velocity at r) Et will be positive.
Thank you for your reply it has helped me to move in the right direction.
-
An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero.
So far, so good
It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?
And there's the mistake.As the body rises it loses kinetic energy and gains potential energy. As it descends it loses pe and gains ke.
Energy is conserved. Energy is the sum of ke and pe.
-
If a body gains potential energy at h from M, then how can the equivalence principle be upheld?
A caesium atom is thought to only 'appear' to have a higher frequency at h from M from the perspective of the lower reference frame.
It is supposedly the gravity potential that causes this to occur, but if one is in the reference frame with the clock in the higher gravity potential, there is no extra gravity potential energy apparent for the atom.
The equivalence principle states that the atom will be the equivalent in all reference frames.
So how can it be said that a body gains potential energy at h from m?
-
Begin with a test mass m on the surface of earth mass M and physicist Biff. The g-field from M is always on, i.e. m is accelerating downward but the ground gets in the way preventing any motion. Inserting a scale between the ground and m will easily prove this. To make the acceleration apparent m must be separated from the ground. The 3 masses define a closed system of energy. Biff tosses m vertically supplying the ke from his body. As m rises gravity imparts ke downward until m stops and reverses direction and returns to the ground. To label the acceleration of gravity as potential energy is misleading (what's new) since the g-field is a continuous dynamic process. The ke supplied by Biff returns to the ground with the accumulated ke from the g-field. If Biff picked up m and threw it back to the ground with the same effort, the result is obvious. A variation of the example would be to let m pass near the earth, gain momentum (ke) and continue into space. Since the g-field is produced by M, m has indirectly removed some energy/mass from M instead of returning it.
-
mathematically there is something decreasing ( speed of object) which decreases kinetic energy , and there is something increasing at the same time speed decreases which is height, at each small decrements in kinetic energy there is an equal increment in potential energy( conservation of energy ). you can add small change in g.
theoretically , the object does work against gravity - force-, it consumes this work done from its own kinetic energy, gradually , the object finally in fact LOSES its kinetic energy, but being moved again towards the opposite direction is an independent process, which is force(gravity ) acting on object and causing it to move , the object had lost its kinetic energy being given it again is another process.
-
As m rises gravity imparts ke downward until m stops and reverses direction and returns to the ground.
...and by what mechanism is gravity imparting ke downwards?
-
An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero.
At which point the mass of the body has increased. That external kinetic energy is now potential energy, which is mass-energy, which is actually internal kinetic energy. If some of that internal kinetic energy were released, what you'd have is a radiating body losing mass as described in Einstein's E=mc² paper (https://www.fourmilab.ch/etexts/einstein/E_mc2/www/).
It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?
That kinetic energy isn't negative, it's positive. When our body falls back down gravity converts some of the potential energy or mass-energy or internal kinetic energy into external kinetic energy. After the body hits the ground this external kinetic energy is dissipated, and you're left with a mass deficit. See the Wikipedia binding energy article (https://en.wikipedia.org/wiki/Binding_energy#Mass-energy_relation):
"Classically, a bound system is at a lower energy level than its unbound constituents. Its mass must be less than the total mass of its unbound constituents. For systems with low binding energies, this "lost" mass after binding may be fractionally small. For systems with high binding energies, however, the missing mass may be an easily measurable fraction. This missing mass may be lost during the process of binding as energy in the form of heat or light, with the removed energy corresponding to removed mass through Einstein's equation E = mc²".
Note that there is an effect on the large body, but whilst momentum is equal and opposite, kinetic energy is not. When you drop a brick there's a measureable change in the kinetic energy of the brick, but there's no measurable change in the kinetic energy of the Earth. It's so slight that we discount it.
-
...and by what mechanism is gravity imparting ke downwards?
By making light curve downwards, because of the wave nature of matter. Think about pair production, where we made an electron and a positron out of light. Also remember the Einstein-de Haas effect (https://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect#Description), which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". Then see Hans Ohanian’s 1984 paper what is spin? (http://aforrester.bol.ucla.edu/docs/Spin_Ohanian.pdf) He says this: “the means for filling the gap have been at hand since 1939, when Belinfante established that the spin could be regarded as due to a circulating flow of energy”. Check out the Poynting vector, then think of the electron as light going round and round, then simplify it further to light going round a square path. Remember gravity makes light curve downward, so what happens to that electron? This:
(https://i.stack.imgur.com/ATcJA.jpg)
The horizontals curve downwards a little, and the electron falls down. In doing so internal kinetic energy is converted into external kinetic energy. Because only the horizontals bend downwards, the deflection of light is twice the Newtonian deflection of matter. See Ned Wright's article (http://www.astro.ucla.edu/~wright/deflection-delay.html) for something on that. "Before Einstein developed the full theory of General Relativity he also predicted a deflection of 0.875 arcseconds in 1913, and asked astronomers to look for it. But World War I intervened, and during the war Einstein changed his prediction to 1.75 arcseconds, which is twice the Newtonian deflection".
-
Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.
Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass. He states that it is only kinetic energy that affects relativistic mass.
-
Ok. Negative kinetic energy requires an imaginary velocity, complex in other words. So that wasn't really an option. However the velocity itself can become negative due to choice of coordinates. Therefore negative momentum is also due to choice of coordinates. If we turn the tables and view the upward motion of an object to be due to a negative acceleration. Bear with me. Then Newtons laws are violated since a decreasing force should still cause a positive velocity. So gravity sucks big time. If I put a straw into a glass of milk and suck the amount of milk decreases and the force causing it, me, gains something from the milk. Potential milk anyone?
-
Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.
Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass. He states that it is only kinetic energy that affects relativistic mass.
You are certainly on fire at the moment. Keep it up!
-
Remember gravity makes light curve downward,
...obviously, but by what mechanism?
"Before Einstein developed the full theory of General Relativity he also predicted a deflection of 0.875 arcseconds in 1913, and asked astronomers to look for it. But World War I intervened, and during the war Einstein changed his prediction to 1.75 arcseconds, which is twice the Newtonian deflection".
Interestingly though John, 1.75 is the so far dimensionless gravitational coupling constant, negating the power 10 consideration...
https://en.m.wikipedia.org/wiki/Gravitational_coupling_constant (https://en.m.wikipedia.org/wiki/Gravitational_coupling_constant)
(Jeff, thanks for comment!)
-
Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.
You do. There is no magical mechanism by which the very real kinetic energy of a bullet fired upwards somehow disappears or zips across space to some other place. The bullet has that kinetic energy, and it stays with the bullet, as internal kinetic energy aka potential energy aka mass-energy. Conservation of energy applies. The mass deficit is not something I made up. A body at rest at a low elevation has less mass-energy than the same body at rest at the higher elevation.
Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass. He states that it is only kinetic energy that affects relativistic mass.
Relativistic mass is not rest mass.
...obviously, but by what mechanism?
It's akin to refraction. We don't call it gravitational lensing for nothing. Check out what Einstein said (http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22): "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". It's rather like sonar actually:
(https://i.stack.imgur.com/sNuHt.gif)
-
what Einstein said (http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22): "the curvature of light rays occurs only in spaces where the speed of light is spatially variable".
It can work just as equally if one says:
"The curvature of light rays occurs only in spaces where the speed of light is 'temporally' variable"
To far better and more sensible results!
-
Nope, I'm sorry John, but - according to the equivalence principle you cannot add that potential energy to the body because this means that the atom isn't the equivalent in each reference frame.
You do. There is no magical mechanism by which the very real kinetic energy of a bullet fired upwards somehow disappears or zips across space to some other place. The bullet has that kinetic energy, and it stays with the bullet, as internal kinetic energy aka potential energy aka mass-energy. Conservation of energy applies. The mass deficit is not something I made up. A body at rest at a low elevation has less mass-energy than the same body at rest at the higher elevation.
Also - Pmbphy has mentioned in another thread elsewhere that potential energy does not affect relativistic mass. He states that it is only kinetic energy that affects relativistic mass.
Relativistic mass is not rest mass.
...obviously, but by what mechanism?
It's akin to refraction. We don't call it gravitational lensing for nothing. Check out what Einstein said (http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22): "the curvature of light rays occurs only in spaces where the speed of light is spatially variable". It's rather like sonar actually:
(https://i.stack.imgur.com/sNuHt.gif)
Do you actually understand what potential energy actually is?
-
Potential energy should be at a maximum value at infinity. This is exactly where, mathematically, gravity ceases to operate. So that in the absence of gravity any mass should therefore be infinite since there have been infinite contributions of potential energy. Does this sound sensible to you John?
-
Oh look, Jeff's created a riddle!
Ok, hmmm... Well, I think the answer has got to be, that where there is no gravity, there is also no mass.
Does this give indication that potential energy is only applicable where there is mass in relation to Mass?
-
Since potential energy is the measure of kinetic energy required to 'get back' to (?) ...where you are calculating from...
... When gravity potential is set at infinity, in consideration that infinity potential energy is set at 0 gravity, 0 gravity being an absence of mass - can anyone tell me exactly 'what' it is from that reference frame that is mathematically being calculated as 'getting back'?
-
Since potential energy is the measure of kinetic energy required to 'get back' to (?) ...where you are calculating from...
... When gravity potential is set at infinity, in consideration that infinity potential energy is set at 0 gravity, 0 gravity being an absence of mass - can anyone tell me exactly 'what' it is from that reference frame that is mathematically being calculated as 'getting back'?
You can't have a reference frame at infinity. It can only be used in mathematical procedures to demonstrate something. Since it is infinitely far away you can never get there in a finite amount of time.
-
Which is exactly my point.
What is being demonstrated?
-
As m rises gravity imparts ke downward until m stops and reverses direction and returns to the ground.
...and by what mechanism is gravity imparting ke downwards?
The mass M tranfers energy into the surrounding space via an unknown process, producing the g-field.
Remove the mass and you remove the field.
-
Yes - and considering that the energy imparted by 'unknown process' to the g-field is greater near M than far from M, why is it thought that energy is greater in space at 0 gravity?
-
Yes - and considering that the energy imparted by 'unknown process' to the g-field is greater near M than far from M, why is it thought that energy is greater in space at 0 gravity?
Who thinks that? Energy anywhere depends on a source and specifically gravity depends on the distribution of mass.
-
Ideally in the over simplistic classical illustrations of momentum vectors (which represent motion of the fictitious center of mass, there is a large arrow for M and a small arrow for m, and initially they are parallel. If m (a rocket) departs M with mV, then M should react with Mv in the opposite direction to balance the 2 body system. In the 'real' world M does not react in that manner since the coordinated momenta of the exhaust disperses as heat.
If the strength of gravity decreases as the inverse square of distance then at large astronomical distances, it would approach zero. This is another ideal concept involving an isolated earth-object system. It is only necessary to leave the local sun system and approach another star system and it would dominate the gravitational effects. Selecting a random volume of 'empty' space a few 1000 ly distant, any contribution from Earth would be washed out by the g-noise from all the masses of the universe with a net effect of zero.
Is there really an image moving across your computer screen?
-
Yes - and considering that the energy imparted by 'unknown process' to the g-field is greater near M than far from M, why is it thought that energy is greater in space at 0 gravity?
Who thinks that? Energy anywhere depends on a source and specifically gravity depends on the distribution of mass.
Current physics thinks that via gravity potential energy which is set at infinity in a 0 gravity field.
Via the inverse square law it's virtually impossible to arrive at 0 gravity mathematically.
Yes - mass distribution would indeed have a bearing on what the g field in space, and outer space will be doing, and where there is less mass, the energy of the g field would be less, right?
-
Current physics thinks that via gravity potential energy which is set at infinity in a 0 gravity field.
Via the inverse square law it's virtually impossible to arrive at 0 gravity mathematically.
Yes - mass distribution would indeed have a bearing on what the g field in space, and outer space will be doing, and where there is less mass, the energy of the g field would be less, right?
Infinity has no value in any possible situation. It's a relation/state/property, literally 'without limit'. The gravitational relation is probably more complex than 1/d^2. If it is quantized, then it can reach a saturation level which avoids 'infinite' binding.
-
So - if one uses this infinity as a basis for ones mathematics, in relation to gravity = 0, for the weak field - then is it any surprise that where gravity is at the far greater value such as found in black holes, and the far weaker fields of individual particles, that these mathematics will result back to infinities?
-
So - if one uses this infinity as a basis for ones mathematics, in relation to gravity = 0, for the weak field - then is it any surprise that where gravity is at the far greater value such as found in black holes, and the far weaker fields of individual particles, that these mathematics will result back to infinities?
Not sure I understand what you're saying. In general "infinities" lead to illogical conclusions. An inverse square rule at zero distance resulting in an "infinite" force? Where would the energy come from? At the other end, a particle only needs to be at a large (but not 'infinite') distance to not have any significant effect on nearby particles. Water contracts as it cools except when within a degree of freezing, when it expands. The mind would prefer a simple linear behavior, but reality is more complex.
-
What I refer to is that potential energy and time dilation are set at infinity at 0 gravity, but time dilation and energy 'cease' before infinite gravity is reached.
-
From an engineering point of view, 2 bodies exert a force upon each other, if they have enough kinetic energy between them, they escape the attractive power of gravity. If however they do not have enough kinetic energy gravity becomes dominant and therefore they combine. Newtons inverse square law says gravitational attraction is inversley proportional to the square of the radius between them (i will stand to be corrected on that!) . Both bodies attract towards each other,to the gravitational centre . If the bodies had kinetic energy to begin with(one or both) the energywill cobine/counter the other, or have an effect upon to put it another way. If neither had initial energy neither will loose to or gain from the other, so initial energy cannot be required for gravitational attraction. The potential energy of each body is dependant upon the gravitational pull of the other, more potential energy for an apple attracted to earth than an apple to the moon When the bodies attract, the mass of the combined body determines the gravitational force. To seperate these bodies once more, will require energy on both bodies, ie does the apple move the earth or the earth move the apple? That means to conserve energy the problem is that 1) counteracting initial energies has to be accounted for 2)the gravitational centre of each body is not reached 3)Energy is required to seperate the bodies when none was removed initially from the bodies when individualy viewed
The initial energy if on collision can be accounted for by a big impact that kills the dinosaurs in a fireball, or if similar by orbiting around/combining in a centre of gravity in a combined directon , the path is altered. (if you can understand this with a bit o' leeway)
The kinetic gained by the attraction from the potential energy of the created gravitational field is concerved in the gravitational field, the bodies moving towards the centre of gravity The gravitational centres is not reached as both bodies wish to centre over it, it results in a bump, ie expended in heat light /molecular fracturingm . The kinetic energy of the attraction also transferred to a tragectory in the combined resultant mass. The potential energy to the gravitational centre is conserved and the kinetic energy conserved
The energy required to seperate the bodies is the same that was gained through the attraction plus the potential energy, you have to put energy in to overcome gravity. This is conserved by each body being freed from the combined specific field of gravity, the bodies tragectory are altered back to there original and the energy is conserved.
-
Potential energy should be at a maximum value at infinity. This is exactly where, mathematically, gravity ceases to operate. So that in the absence of gravity any mass should therefore be infinite since there have been infinite contributions of potential energy. Does this sound sensible to you John?
The maximal potential energy is attained infinitely far from the source of the gravity well, but the potential energy is always finite. Because the gravitational field falls off with the square of the radial distance, the integral of the change in potential energy converges as you rise from surface to infinite distance. It is actually very useful to treat this infinite distance limit as 0 potential energy, and then potential energy goes negative as the object is subjected to attractive fields (gravity or electrostatic).
-
Potential energy should be at a maximum value at infinity. This is exactly where, mathematically, gravity ceases to operate. So that in the absence of gravity any mass should therefore be infinite since there have been infinite contributions of potential energy. Does this sound sensible to you John?
The maximal potential energy is attained infinitely far from the source of the gravity well, but the potential energy is always finite. Because the gravitational field falls off with the square of the radial distance, the integral of the change in potential energy converges as you rise from surface to infinite distance. It is actually very useful to treat this infinite distance limit as 0 potential energy, and then potential energy goes negative as the object is subjected to attractive fields (gravity or electrostatic).
Yes of course with an infinite series you can get a finite limit. It still doesn't make John's assertion right.
-
ChiralSPO - I agree that it is useful to suggest that gravity potential energy is set at 0 in a 0 gravity field at infinite distance from M, but unless you have an m to apply gravity potential to, the statement is meaningless - unless applied to the reference frame of the g-field of space without mass. ie: 0 gravity
Taking m to an infinitely far distance from M, m has its own mass and gravity will not be 0...
Therefore m in a near 0 gravity field will have a very high gravity potential, but not infinitely high, and 0 gravity potential for m in relation to M would be at centre of body of M.
-
An object moving away from a large mass has an initial kinetic energy. Gravity removes this positive kinetic energy until it reaches a value of zero. It then continues to remove kinetic energy so that the object now travels towards the centre of the force with increasing negative kinetic energy. Where does the kinetic energy go? Is it conserved?
No. Kinetic energy is not conserved. Only total energy is conserved. In this case its the total mechanical energy that's conserved. I.e. the sum of potential and kinetic energy.
-
Pete - Then how can it be said that the frequency and therefore energy of a caesium atomic clock in a differing gravity potential is observer dependent?
-
Current physics thinks that via gravity potential energy which is set at infinity in a 0 gravity field.
Not as practised by current physicists. We say that gravitational potential is zero in deep space and negative at any point in a nonzero gravitational field.
I've quoted my old navigation instructor many times in this forum, but once more won't hurt. In the immortal words of Charlie Kovac "always start from where you are, then you won't get lost before you take off".
-
Pete - Then how can it be said that the frequency and therefore energy of a caesium atomic clock in a differing gravity potential is observer dependent?
There's nothing wrong with what I explained. I was describing the answer in terms of non-relativistic mechanics. You're speaking of general relativity. Atoms absorb and emit energy in the form of photons. In that sense they behave like ideal clocks. In GR the frequency measured by the observer of the clock depends on the difference of the gravitational potential where the clock is located and where the observer is located. What I explained above is sometimes explained (e.g. in Gravitation by Misner, Thorne and Wheeler as well as Nightingale and Foster) in terms of the energy of a photon, its energy being all kinetic. As the photon climbs out of the g-field its kinetic energy decreases and its potential energy increases, the sum being constant.
For the derivation see: http://www.newenglandphysics.org/physics_world/gr/grav_red_shift.htm
-
Current physics thinks that via gravity potential energy which is set at infinity in a 0 gravity field.
Not as practised by current physicists. We say that gravitational potential is zero in deep space and negative at any point in a nonzero gravitational field.
I've quoted my old navigation instructor many times in this forum, but once more won't hurt. In the immortal words of Charlie Kovac "always start from where you are, then you won't get lost before you take off".
There are certain problems used as exercises where one sets the zero at an arbitrary location since in those problems the g-field doesn't go to zero at infinity. E.g. a infinitely long rod.
-
Current physics thinks that via gravity potential energy which is set at infinity in a 0 gravity field.
Not as practised by current physicists. We say that gravitational potential is zero in deep space and negative at any point in a nonzero gravitational field.
I've quoted my old navigation instructor many times in this forum, but once more won't hurt. In the immortal words of Charlie Kovac "always start from where you are, then you won't get lost before you take off".
As one gets further away from M, gravity potential increases.
So why do current physicists say that gravity potential is zero in a deep space gravity field?
Zero for what?
For mass in that gravity field?
Or for the gravity field itself?
And if you start out with a negative value from any point in a non zero gravity field for m, this negative value will still escalate, negatively, as kinetic energy escalates positively.
Such a wonderful system, so beautiful and logical, oh - apart from the fact that:
This is the cosmological constant problem, the worst problem of fine-tuning in physics: there is no known natural way to derive the tiny cosmological constant used in cosmology from particle physics.
https://en.m.wikipedia.org/wiki/Cosmological_constant
So - never mind where you are, you can always try the other runway...
It's about where it is you will end up, as well as where you start from.
-
Pete - Then how can it be said that the frequency and therefore energy of a caesium atomic clock in a differing gravity potential is observer dependent?
There's nothing wrong with what I explained. I was describing the answer in terms of non-relativistic mechanics. You're speaking of general relativity. Atoms absorb and emit energy in the form of photons. In that sense they behave like ideal clocks. In GR the frequency measured by the observer of the clock depends on the difference of the gravitational potential where the clock is located and where the observer is located. What I explained above is sometimes explained (e.g. in Gravitation by Misner, Thorne and Wheeler as well as Nightingale and Foster) in terms of the energy of a photon, its energy being all kinetic. As the photon climbs out of the g-field its kinetic energy decreases and its potential energy increases, the sum being constant.
For the derivation see: http://www.newenglandphysics.org/physics_world/gr/grav_red_shift.htm
I did not say you hadn't described anything correctly.
Yes - if an atom emits a photon in a differing gravity potential, then as an observer observing the photon that has been emitted in the differing gravity potential, one will view the photon as it is when it has 'arrived in' the observation reference frame one is observing from, and not as it was when it was emitted in the differing reference frame.
So yes light is observer dependent...
But the cession atomic clock, although it uses light as part of the clocks apparatus, is making a reading of the light apparatus in its own frame of reference, and it is that reading that is sent to the computer.
If the clock is only ticking faster as an observer dependent phenomenon, (this can also be applied to SR time dilation), then the widely reported (NASA, NIST) and scientifically accepted (?) concept of a person ageing in keeping with their time dilated clock is wrong.
So, as a photon climbs out of a gravity well it's kinetic energy decreases, and it's potential energy increases?
When a caesium atom is ticking at elevation in a higher gravity potential, does it's potential energy also increase?
-
As one gets further away from M, gravity potential increases.
So why do current physicists say that gravity potential is zero in a deep space gravity field?
Because it starts out being negative when in the field and increases to zero outside the field. The potential for a spherical mass distribution outside the body is U = -GM/r
And if you start out with a negative value from any point in a non zero gravity field for m, this negative value will still escalate, negatively, as kinetic energy escalates positively.
Such a wonderful system, so beautiful and logical, oh - apart from the fact that:
Not trying to be rude whatsoever but it's not the theory that is flawed but your understanding of it. In this case things like the cosmological constant not being derived is no different than the fact that no theory in physics can be derived. They're all arrived at by things such as intuition.
This is the cosmological constant problem, the worst problem of fine-tuning in physics: there is no known natural way to derive the tiny cosmological constant used in cosmology from particle physics.
Fine tuning and the cosmological constant have little to do with each other. And you have no idea whether it can or can't be derived. All that can be said is that it hasn't been derived yet. And if it was the cosmological constant that drove inflation then it was very large at that time. That article was silly for saying that.
-
I did not say you hadn't described anything correctly.
I didn't say you did. I was merely saying that what I said was correct. However your responses indicate that what I said is in error and that's what I responded to.
Yes - if an atom emits a photon in a differing gravity potential, then as an observer observing the photon that has been emitted in the differing gravity potential, one will view the photon as it is when it has 'arrived in' the observation reference frame one is observing from, and not as it was when it was emitted in the differing reference frame.
Its not clear to me what you mean when you use the term "frame." That term is usually used only in SR, i.e. inertial frames with no spacetime curvature. one usually speaks in terms of coordinate systems, not frames.
If the clock is only ticking faster as an observer dependent phenomenon, (this can also be applied to SR time dilation), then the widely reported (NASA, NIST) and scientifically accepted (?) concept of a person ageing in keeping with their time dilated clock is wrong.
Not at all. I can't even see how you arrived at that conclusion.
So, as a photon climbs out of a gravity well it's kinetic energy decreases, and it's potential energy increases?
From these responses it appears to me that you didn't follow the derivation in the URL I posted. Is that correct? The reason I created those pages was to address these exact questions. I takes exact math to be precise and that's what I did on that page. I only assumed the reader knows basic algebra and physics.
If you'd prefer a textbook then read section 4.3 in A Short Course in General Relativity by James Foster, J. D. Nightingale.
You can download the text from - http://bookzz.org/book/681818/a5c827
Note: When I spoke of the kinetic energy of a photon/light it refers to the "intrinsic" energy of the light, i.e. as measured at the same location as the light. If one uses the frequency as reckoned from one location only then you should know that such a frequency doesn't change. Eg the frequency of light as it moves through a gravitational field does not change as reckoned by a Schwarzschild observer, i.e. an observer far away from the body.
When a caesium atom is ticking at elevation in a higher gravity potential, does it's potential energy also increase?
That depends on the nature of the field that you're talking about. E.g. if the field is uniform then the potential energy increases without bound as one goes higher and higher in the field.
-
As one gets further away from M, gravity potential increases.
So why do current physicists say that gravity potential is zero in a deep space gravity field?
Because it starts out being negative when in the field and increases to zero outside the field. The potential for a spherical mass distribution outside the body is U = -GM/r
And if you start out with a negative value from any point in a non zero gravity field for m, this negative value will still escalate, negatively, as kinetic energy escalates positively.
Such a wonderful system, so beautiful and logical, oh - apart from the fact that:
Not trying to be rude whatsoever but it's not the theory that is flawed but your understanding of it. In this case things like the cosmological constant not being derived is no different than the fact that no theory in physics can be derived. They're all arrived at by things such as intuition.
This is the cosmological constant problem, the worst problem of fine-tuning in physics: there is no known natural way to derive the tiny cosmological constant used in cosmology from particle physics.
Fine tuning and the cosmological constant have little to do with each other. And you have no idea whether it can or can't be derived. All that can be said is that it hasn't been derived yet. And if it was the cosmological constant that drove inflation then it was very large at that time. That article was silly for saying that.
Well Pete - In the case of arriving at a theory of everything, Wikapedia is not the only source by a long shot where I have read that the measured cosmological constant being many orders too small is a large problem that needs to be addressed.
And in stating that potential energy and kinetic energy escalate or diminish positively and negatively to each other to retain balance is repeating what I already said to Alan.
Leaving aside rudeness or not - to derive relativity one also needs to rely on unknowns that are necessary for this theories validity. One of these unknowns is dark energy and this directly relates to the cosmological constant, and this isn't really all that intuitive.
But my interest and question is:
Does the caesium atom in a higher gravity potential experience an increase in potential energy, such as the photon does?
-
I did not say you hadn't described anything correctly.
I didn't say you did. I was merely saying that what I said was correct. However your responses indicate that what I said is in error and that's what I responded to.
Yes - if an atom emits a photon in a differing gravity potential, then as an observer observing the photon that has been emitted in the differing gravity potential, one will view the photon as it is when it has 'arrived in' the observation reference frame one is observing from, and not as it was when it was emitted in the differing reference frame.
Its not clear to me what you mean when you use the term "frame." That term is usually used only in SR, i.e. inertial frames with no spacetime curvature. one usually speaks in terms of coordinate systems, not frames.
If the clock is only ticking faster as an observer dependent phenomenon, (this can also be applied to SR time dilation), then the widely reported (NASA, NIST) and scientifically accepted (?) concept of a person ageing in keeping with their time dilated clock is wrong.
Not at all. I can't even see how you arrived at that conclusion.
So, as a photon climbs out of a gravity well it's kinetic energy decreases, and it's potential energy increases?
From these responses it appears to me that you didn't follow the derivation in the URL I posted. Is that correct? The reason I created those pages was to address these exact questions. I takes exact math to be precise and that's what I did on that page. I only assumed the reader knows basic algebra and physics.
If you'd prefer a textbook then read section 4.3 in A Short Course in General Relativity by James Foster, J. D. Nightingale.
You can download the text from - http://bookzz.org/book/681818/a5c827
Note: When I spoke of the kinetic energy of a photon/light it refers to the "intrinsic" energy of the light, i.e. as measured at the same location as the light. If one uses the frequency as reckoned from one location only then you should know that such a frequency doesn't change. Eg the frequency of light as it moves through a gravitational field does not change as reckoned by a Schwarzschild observer, i.e. an observer far away from the body.
When a caesium atom is ticking at elevation in a higher gravity potential, does it's potential energy also increase?
That depends on the nature of the field that you're talking about. E.g. if the field is uniform then the potential energy increases without bound as one goes higher and higher in the field.
Which is why I corrected you, because I did not say that you ere in error. That was your interpretation.
*
ie: If an atom is omitting a photon at 22 metres top of Harvard bell tower, then when it is observed at bottom of tower it has changed frequency. The bottom of tower cannot observe the photon being emitted at top of tower, because it is not observable until it reaches the observer at bottom of tower.
Therefore light is observer dependent.
*
Because if the clock at elevation's frequency is observer dependent, then the caesium atom's frequency, when you place yourself with the clock will be the same as it was for the clock you were observing on the ground.
And if it's frequency is the same at elevation as it is on the ground, then the person with the elevated clock shall age no faster than the person on the ground.
*
I am well read in Relativity thank you...
You said in post 41:
As the photon climbs out of the g-field its kinetic energy decreases and its potential energy increases, the sum being constant.
I was directly quoting what you said, and didn't look at the link.
*
When you say uniform - do you mean unchanging. ie: not diminishing with the inverse square law with distance?
I am asking if the potential energy increases for the caesium atom when placed in elevation in a higher gravity potential than ground level earth.
-
And in stating that potential energy and kinetic energy escalate or diminish positively and negatively to each other to retain balance is repeating what I already said to Alan.
I don't read all of the posts in threads since it'd take too long to post the answer to the question asked and it serves no purpose to me.
- to derive relativity ...
As I explained above one does not "derive" SR or any other theory. That's not possible, at not using deductive logic. One takes intuitive guesses and hints from other laws to arrive at postulates which are then used to form postulates which are then used for construct theories. That's what's known as "inductive" logic. See:
http://www.iep.utm.edu/ded-ind/
A deductive argument is an argument that is intended by the arguer to be (deductively) valid, that is, to provide a guarantee of the truth of the conclusion provided that the argument's premises (assumptions) are true. This point can be expressed also by saying that, in a deductive argument, the premises are intended to provide such strong support for the conclusion that, if the premises are true, then it would be impossible for the conclusion to be false. An argument in which the premises do succeed in guaranteeing the conclusion is called a (deductively) valid argument. If a valid argument has true premises, then the argument is said to be sound.
...
An inductive argument is an argument that is intended by the arguer merely to establish or increase the probability of its conclusion. In an inductive argument, the premises are intended only to be so strong that, if they were true, then it would be unlikely that the conclusion is false. There is no standard term for a successful inductive argument. But its success or strength is a matter of degree, unlike with deductive arguments. A deductive argument is valid or else invalid.
Please keep this in mind when you again claim that theories such as SR are "derived." It a theory could actually be derived then that would be proof a theory. But that's not how science works. A very good friend of mine, Alan Guth (inventor of the inflationary model of the universe) did several interviews for me where he addressed several common misconceptions in physics. He did one concerning the common misconception Physics Is About Proving Things. See:
http://www.newenglandphysics.org/common_misconceptions/Alan_Guth_04.mp4
one also needs to rely on unknowns that are necessary for this theories validity. One of these unknowns is dark energy and this directly relates to the cosmological constant, and this isn't really all that intuitive.
I have no idea where you got the idea that ...one also needs to rely on unknowns that are necessary for this theories validity. Also, it's not necessarily true that dark energy relies on the cosmological constant. It's quite possible that dark energy is a result of negative pressure, i.e. stress.
Does the caesium atom in a higher gravity potential experience an increase in potential energy, such as the photon does?
Yes.
Note: Keep in mind that there is a difference between gravitational potential and gravitational potential energy.
It appears to me that not only that you didn't follow my derivation but that you have no intention of doing so. Perhaps you can't follow the algebra? I'll assume that's the case and no longer respond to your questions, especially since you keep repeating the same misconceptions over and over again regardless of my corrections to them. Good bye.
-
Which is why I corrected you, because I did not say that you ere in error. That was your interpretation.
So you claim, but you're wrong.
ie: If an atom is omitting a photon at 22 metres top of Harvard bell tower, then when it is observed at bottom of tower it has changed frequency. The bottom of tower cannot observe the photon being emitted at top of tower, because it is not observable until it reaches the observer at bottom of tower.
Did you see me say "as observed"? No. You didn't. I said "reckoned" because that's what it is. I.e. its how its calculated. Not observed. However I can also rightly say that its also possible to be observed because there's nothing wrong with having more than one observer. This is done all the time in SR when one speaks of observers in two frames measuring things like speed, energy, etc.
Therefore light is observer dependent.
I never said otherwise, did I???? :(
If you learned how to follow a derivation and/or reading the text I referred to then you'd learn this.
-
And in stating that potential energy and kinetic energy escalate or diminish positively and negatively to each other to retain balance is repeating what I already said to Alan.
I don't read all of the posts in threads since it'd take too long to post the answer to the question asked and it serves no purpose to me.
- to derive relativity ...
As I explained above one does not "derive" SR or any other theory. That's not possible, at not using deductive logic. One takes intuitive guesses and hints from other laws to arrive at postulates which are then used to form postulates which are then used for construct theories. That's what's known as "inductive" logic. See:
http://www.iep.utm.edu/ded-ind/
A deductive argument is an argument that is intended by the arguer to be (deductively) valid, that is, to provide a guarantee of the truth of the conclusion provided that the argument's premises (assumptions) are true. This point can be expressed also by saying that, in a deductive argument, the premises are intended to provide such strong support for the conclusion that, if the premises are true, then it would be impossible for the conclusion to be false. An argument in which the premises do succeed in guaranteeing the conclusion is called a (deductively) valid argument. If a valid argument has true premises, then the argument is said to be sound.
...
An inductive argument is an argument that is intended by the arguer merely to establish or increase the probability of its conclusion. In an inductive argument, the premises are intended only to be so strong that, if they were true, then it would be unlikely that the conclusion is false. There is no standard term for a successful inductive argument. But its success or strength is a matter of degree, unlike with deductive arguments. A deductive argument is valid or else invalid.
Please keep this in mind when you again claim that theories such as SR are "derived." It a theory could actually be derived then that would be proof a theory. But that's not how science works. A very good friend of mine, Alan Guth (inventor of the inflationary model of the universe) did several interviews for me where he addressed several common misconceptions in physics. He did one concerning the common misconception Physics Is About Proving Things. See:
http://www.newenglandphysics.org/common_misconceptions/Alan_Guth_04.mp4
one also needs to rely on unknowns that are necessary for this theories validity. One of these unknowns is dark energy and this directly relates to the cosmological constant, and this isn't really all that intuitive.
I have no idea where you got the idea that ...one also needs to rely on unknowns that are necessary for this theories validity. Also, it's not necessarily true that dark energy relies on the cosmological constant. It's quite possible that dark energy is a result of negative pressure, i.e. stress.
Does the caesium atom in a higher gravity potential experience an increase in potential energy, such as the photon does?
Yes.
Note: Keep in mind that there is a difference between gravitational potential and gravitational potential energy.
It appears to me that not only that you didn't follow my derivation but that you have no intention of doing so. Perhaps you can't follow the algebra? I'll assume that's the case and no longer respond to your questions, especially since you keep repeating the same misconceptions over and over again regardless of my corrections to them. Good bye.
You directly quoted this
Such a wonderful system, so beautiful and logical, oh - apart from
from the post I made to Alan in your post: 45, so you did indeed read my answer to Alan.
*
I think you are picking nits here Pete. But on the basis of a deductive argument, I see that it is deduced that Relativity requires unknowns to be a valid theory, that one of these unknowns is Dark Energy, and that Dark Energy 'can' be related to the cosmological constant.
*
If Dark Energy is negative stress, this still leaves particle physics in relation to relativity with a cosmological constant that is many orders too small.
*
Ah good, I see you have answered the question. Thank you! I know that your opinion is respected here and I shall quote you on that next time someone tells me that it doesn't.
*
Dear oh me - so touchy Pete. Really there is no need. You are quite correct, I have actually discussed with you on several occasions over the last few years that I am not much of a mathematician...
But the good news is Pete, that I'm getting a lot better at it than I was.
Good day to you... in Boston isn't it?
I will probably have a look at your maths, but it was the question that I was asking that was interesting me.
-
Which is why I corrected you, because I did not say that you ere in error. That was your interpretation.
So you claim, but you're wrong.
ie: If an atom is omitting a photon at 22 metres top of Harvard bell tower, then when it is observed at bottom of tower it has changed frequency. The bottom of tower cannot observe the photon being emitted at top of tower, because it is not observable until it reaches the observer at bottom of tower.
Did you see me say "as observed"? No. You didn't. I said "reckoned" because that's what it is. I.e. its how its calculated. Not observed. However I can also rightly say that its also possible to be observed because there's nothing wrong with having more than one observer. This is done all the time in SR when one speaks of observers in two frames measuring things like speed, energy, etc.
Therefore light is observer dependent.
I never said otherwise, did I???? :(
If you learned how to follow a derivation and/or reading the text I referred to then you'd learn this.
If it wasn't your interpretation then why did you mention it?
*
Clearly! Any relativity book will tell you that.
*
No you didn't. I was using the sentence to illustrate the difference between light and an atom.
*
I already know it and am the one who mentioned it first.
I'm sorry I just don't understand why you are so defensive. I only asked you a question.
Sorry, won't do it again...
P.S. Btw, you do realise that you 'are' using the word derivation yourself?
-
As one gets further away from M, gravity potential increases.
Yes
So why do current physicists say that gravity potential is zero in a deep space gravity field?
"deep space" is somewhere that there is no net gravitational field, so "deep space gravity field" is meaningless, and if the field is zero then the potential is also zero.
Zero for what?
For mass in that gravity field?
Or for the gravity field itself?
if the field is zero, the potential is zero and the potential energy of any mass at that point is zero.
And if you start out with a negative value from any point in a non zero gravity field for m, this negative value will still escalate, negatively, as kinetic energy escalates positively.
you've got it!
The logic is that the sum of kinetic and potential energy is always zero for an object in free fall, so we can calculate the orbit of anything if we know the starting parameters.
You can use this approach to model the banking crisis. Banks were supposed to invest deposits (potential) in loans (kinetic) in order to make a profit. But they started selling loans to each other for profit until the notional value of the loans far exceeded the deposits, which was OK until a depositor asked for his money back. Not surprising that accountants have borrowed the "black hole" metaphor, and they seem to turn up wherever you thought there was a deposit!