# Energy reference frames

1. Jul 19, 2004

### Moose352

Do two observers in different reference frames necessarily agree on the energy of another object in a third reference frame?

2. Jul 19, 2004

### chroot

Staff Emeritus
Define "the energy of another object" for me. If you mean kinetic energy, certainly not.

- Warren

3. Jul 19, 2004

### Moose352

I mean energy as in total energy, including speed, and mass and everything.

4. Jul 19, 2004

### Garth

According to the theory of General Relativity energy-mass is not even conserved. Consider being in the freely falling frame of a spaceship falling directly towards the Earth's centre from a stationary point deep in space. In the space-ship's frame of reference the Earth starts stationary but then accelerates towards you. Its velocity rapidly increases and therefore its relativistic mass also increases, which is its total energy ( E = mc^2) of rest mass plus the mass equivalent of its kinetic energy. However, in GR there is no gravitational force, just the curvature of space-time. The space-ship and the Earth are converging because they are both freely falling along their geodesics, which converge because they are 'warped' by the curvature of space-time, that is, the gravitational field around the Earth. No work is being done on the space-ship or the Earth, there is no force to do any work, and yet the Earth's total energy steadily increases. GR is an example of what Emmy Noether called an 'improper energy theorem', in GR energy is not locally conserved.

5. Jul 19, 2004

### Staff: Mentor

energy is frame dependent

No, the total energy of an object is not an invariant. The total energy equals rest energy (due to the object's mass, an invariant) plus kinetic energy. Since the KE depends on velocity, it and the total energy are frame dependent quantities.

6. Jul 19, 2004

### Moose352

Thanks for the answers. It's pretty interesting that energy is not invariant (you can see it easily from the relativistic energy equation, but I thought my interpretation was wrong).

So, just to see if this works out, consider two people A and B who are moving apart from each other. A asks B for some energy (lets say 5 joules), so B does some work and sends over a 5 joule photon. But A obviously gets a lower energy photon, so how does this work? Does A actually see B do less work and vice versa?

7. Jul 20, 2004

### Garth

Moose 352 - There are two effects to consider, the normal doppler effect, for which light is treated as a wave, and the relativistic correction, for which light can be treated as a stream of particles. In your example; the first effect works like this, rather than sending one photon, B would actually send the energy E as a stream of radiation over a period of time, say t', with a certain power E/t'. Because the radiation is red shifted A would receive that radiation at a lower power E/[t'(1+v/c)] but because they would move apart by vt' during the transmission A would receive the stream for longer t'(1+v/c). The energy received would be the same as transmitted {E/[t'(1+v/c)]}{t'(1+v/c)}=E. However the second effect also has to be included; when the relativistic effect is taken into account the radiation is red shifted even more because in B's frame A has actually increased in mass; so when the energy of the photon is compared to the mass of the object receiving it, that energy as measured by A appears to be less again.
Doc Al - Although energy is obviously frame dependent, there is a question of what should be the case if the two observers are not moving relative to each other but are separated up a gravitational well, such as in the Pound experiment. The photon is observed to be red shifted, but no work is done on or by it, so why? GR says the photon has lost energy, Self Creation Cosmology (see my thread https://www.physicsforums.com/showthread.php?t=32713) suggests instead that it is the higher apparatus that has gained mass by being carried up the hill.

Last edited: Jul 20, 2004
8. Jul 21, 2004

### DW

Energy as with anything physical that is not an invariant or tensor is actually an incomplete description of the physical entity. For an analogy: Consider a can that that has had half of it sliced off from the top edge to the bottom opposite edge. There is an orientation for that can for which the shadow cast on the floor will be a circle, the shadow cast on one wall will be a rectangle, and the shadow cast on a third wall will be a triangle. What frame you use to describe something will be analogaous to what set of surfaces you choose to make the projections of the can on. If you were to rotate the walls like changine frames each shape changes which is like taking a velocity transformation to find that each coordinate of momentum and the energy changes even though the physical entity was left untouched. Energy is like a single shadow. It is not invariant because it is the incomplete description of the physical entity, the can. The can itself is frame invariant. The choice of coordinates does not change it, just how the observer describes it. With the complete set of shadows analogous to the tensor one can describe the can in a frame covariant way from which one can extrapolate the invariant physical entity that they are describing. Even though energy is variant, it is only one of the shadows and by itself incomplete. One needs the complete energy-momentum four-vector to comepletely describe this particular physical aspect an object and the length of that is invariant. It is the mass and as such does not depend on speed.

9. Jul 21, 2004

### Garth

Thank you for that analogy, I shall use it in my lectures if I may. However the question of gravitational red shift may be re-expressed in its terms. If the two observers are both fixed relative to the centre of mass of the system and to each other would they not both see the can from the same perspective? GR would have it that the can rotates when observed from first one and then the other position, but should this be so? Remember that if a particle is 'lifted' out of an atom against the nuclear and electro-magnetic forces then its rest mass increases to include the nuclear and electro-magnetic potential energies. However if you lift an apparatus against a gravitational force then its mass does not increase, according to GR, it is this inconsistency that I question. See my thread "Self Creation Cosmology - a new gravitational theory " https://www.physicsforums.com/showthread.php?t=32713&page=1&pp=15 .

10. Jul 21, 2004

### DW

When you lift the particle out of the atom the mass of the atom-particle "system" is what increases by the amount of work you did seperating them, the amount of energy you transferred to the system according to the center of momentum frame. The mass of the particle by itself
$$m = \frac{\sqrt{g_{\mu \nu}p^{\mu }p^{\nu }}}{c} = \frac{\sqrt{g_{\mu \nu}(P^{\mu } - \frac{q}{c}\phi ^{\mu })(P^{\nu } - \frac{q}{c}\phi ^{\nu })}}{c}$$ *
did not change. By this definition it is invariant to speed, gravitation, any transformation, guage, or potential to which it is exposed.
Also, just fixing the observers with respect to each other is not sufficient to say that they should find the energy or frequency the same because at different locations in curved spacetime they use different standards of time and space. In the analogy they are still projecting against different surfaces. The shadow representing energy can then still look different.

* $$P^{\mu } \equiv p^{\mu } + \frac{q}{c}\phi ^{\mu }$$
where $$\phi ^{\mu }/c$$ is the four-vector potential of the em field in this scenario.
http://www.geocities.com/zcphysicsms/chap3.htm

Last edited: Jul 21, 2004
11. Jul 21, 2004

### Garth

Is not the rest mass of a free particle greater than when bound?

12. Jul 21, 2004

### DW

Rest mass is a bad fraise because mass does not depend on speed. The mass of the system is greater whenever the center of momentum frame energy is greater because that is what system mass is. If you do work on the system in removing the particle then you have increased the system's mass, not the particle's. Some authors have miss-associated that change in mass with the particle when it actually belongs to the system so no as the definition of mass that I gave clearly shows, the mass of the particle does not depend on whether it is bound or free. Changing the potential does not change $$p^{\mu }$$ at all. It changes $$P^{\mu }$$. It is the length of $$p^{\mu }$$ that is the mass of the particle. It is $$P^{0}$$ that is the total energy including the potential. $$p^{0}$$ is only the relativistic energy not including the potential.

13. Jul 21, 2004

### Garth

You can always choose a convention in which the allocation of changes in mass and energy are hidden. The question is, "Is this is a useful convention?". In the fission of a uranium atom the reallocation of potential energies releases a hell of a lot of energy and some would say the total mass of the fragments (consisting of more tightly bound components) is less than that of the original atom; but if your convention includes this energy in with those fragments as a total system then it tells you that no energy has been released. However, whether this is a useful way to describe the exposion of an atomic bomb or not is debatable to say the least.

Last edited: Jul 22, 2004
14. Jul 21, 2004

### DW

The center of momentum frame energy or system mass does not change in the explotion. The total energy does not change. Energy is conserved. As you worded it, it is a "reallocation" of energy from potentials to kinetic. The sum of masses comprising a system can change, changing the sum of kinetic energies of the system. But, the center of momentum frame energy, or system mass, does not change for the system just as the translational kinetic energy "of the system" does not change in the explosion. As invariant, as center of momentum frame energy for a system, is the most useful way to define mass. Without it defined the way it is in modern relativity we never would have arrived at the Klein-Gordon equation modified to include a four-vector potential or the same for the Dirac equation. See equation 3.1.5 and problem 3.1.8 at
http://www.geocities.com/zcphysicsms/chap3.htm

15. Jul 22, 2004

### Garth

In my definition of mass,
the mass of a proton (in units of 10^-24 gm) is 1.672661
the mass of an electron is 0.000911
and the mass of a hydrogen atom is 1.67352
which is less than both combined. Hence my point that the mass of a particle is less when bound than when free. The energy of the system must therefore be negative, which may or may not be an appropriate way of defining it. Rather like my bank calling its card a Credit card when actually they mean a Debt card!
I fully understand what you are saying, the point is, "Is the standard convention really the most useful way to define mass?" As you will see from my first post on the thread mentioned above "Self Creation Cosmology - a new gravitational theory " https://www.physicsforums.com/showthread...13&page=1&pp=15 [Broken] it does leave questions to be answered such as its question 2. "According to the EEP a stationary electron on a laboratory bench is accelerating w.r.t. the local Lorentzian freely falling inertial frame of reference. According to Maxwell's theory of electromagnetism an accelerating electric charge, such as an electron, radiates. So why doesn't it? Or, if it is thought that such an electron actually does radiate, what is the source of such radiated energy? However, note that in the preferred Centre of Mass frame of reference the electron is not accelerating."
A question that is resolved in the (published) theory of Self Creation Cosmology. That theory, while formulated with manifestly covariant field equations, uses Mach's Principle to select a preferred frame of reference, the Centre of Mass frame, as a preferrred foliation of space-time in which energy IS locally conserved. Mass in that frame is defined to include potential energies, gravitational as well as electromagnetic and nuclear.

Last edited by a moderator: May 1, 2017
16. Jul 22, 2004

### DW

But your last sentence does not follow logically from the previous. What follows is that the "system" mass is less when the constituent particles are bound than when they are free. The mass of each constituent particle did not change. What changed was the center of momentum frame energy of the system due to the change in potential energy. At that point I needn't address the rest of the post. System mass is not the sum of masses of the constituent parts. That sum is an incomplete part of the total center of momentum frame energy. The system mass is the center of momentum frame energy. This includes interaction potentials and kinetic energies of each with respect to the center of momentum frame. That is what changed when work was done in binding of the system's particles. The system mass changed, in your scenario, not the mass of either constituent particle.

Last edited: Jul 22, 2004
17. Jul 22, 2004

### Garth

I do value your approach so do keep replying! When I said the energy of the system is negative I was referring to your convention, in my convention the difference between the free masses of the electron plus proton and the re-associated hydrogen atom can be explained by the energy radiated away when they combined.
From your lectures I see that we have a difference of approach that can be expressed as, "Your concept of mass is something that is defined, my concept of mass is something that is measured."
Hence my use of the term "rest mass", which I do not find a confusing term, it is that mass as measured by an observer in the co-moving frame of reference in which it is at rest. It equals the energy-momentum four-vector, which is invariant in all frames of reference in motion relative to that frame. The question is, nevertheless does mass vary with potential energy? The problem with your definition is that it is tautological and could blind you to any actual variation that might occur. Your convention is consistent with the Equivalence Principle, but as the theory derived from the EEP, GR, does seem to have one or two difficulties being integrated with quantum theory I am prepared to question both GR and the EEP on which it is based. See the first question on my first post on the thread "Self Creation Cosmology An alternative gravitational theory".

18. Jul 22, 2004

### DW

I don't have a problem with calling it as something measured, but even then it shouldn't be referred to as rest mass because you hardly ever measure a particle's mass while it is at rest! You first define it in relation to other quantities like force and acceleration for example and then for an example may put a known charge at a known velocity in a magnetic field which therefor imparts a known force and then measure the radius of the circle to calculate with the velocity an acceleration and from that data actually calulate the mass for its "measurement".

The mass of a particle does not vary with potential. Imagine what a mess that would be as you can always add an arbitrary constant to a potential. Take a close look at the difference between $$p^{\mu }$$ and $$P^{\mu }$$. The first is the energy-momentum without the inclusion of a potential. The mass is the length of the first. The second includes the potential. The mass is related to it by $$m = \frac{\sqrt{g_{\mu \nu}[P^{\mu } - (q/c)\phi ^{\mu }][P^{\nu } - (q/c)\phi ^{\nu }]}}{c}$$ where you can see the potential is subtracted back out. The only reason I mention $$P^{\mu}$$ is because it is actually that four vector that carries the energy and momentum opperators for relativistic quantum mechanics.

There is only a problem with developement of quantization of gravity itself, not with relativistic quantum electrodynamics which is what these equations are about.

19. Jul 22, 2004

### Garth

Agreed; it is the theory of gravitation, GR, that I am first questioning and then modifying.
There is no problem over the global conservation of energy, the question is how should the local conservation of energy be treated? Ignored, explained away, or included in some way?
To give an example to illustrate the question, consider a system consisting of a planet with a mountain, a person and a boulder. According to the Equivalence Principle (EEP), GR and your convention defining mass, the mass of the boulder is invariant. The person carries the boulder up the mountain. In doing so she expends some energy, however her mass remains constant as well as that of the boulder. Globally the total mass-energy of the system is conserved and all appears well - from a distance. However she would be most put out to think that all the energy she has expanded has "vanished into thin air" - surely, after descending to the ground again, according to Special Relativity she should have lost some weight? In my convention defining mass, with the local conservation of energy, this is indeed what happens, she loses mass and the boulder gains it, this energy is then released when the boulder falls off the cliff and reappears as kinetic energy. In this convention mass is defined m = m' exp(Phi) where m' is a constant and Phi is the dimensionless Newtonian gravitational potential.
Although the field equations of SCC are manifestly covariant, there are two conformal frames of measurement in the theory - it is an example of conformal relativity theory such as Hoyle & Narlikar's in the 1960's, though with the significant difference of deliberately including the local conservation of energy in one of its frames. The first conformal frame of SCC is the Einstein frame in which energy-momentum is conserved and the EEP holds, and which would be consistent with your definitions. The second is the Jordan Frame in which a preferred frame of reference, or foliation of space-time, is selected by Mach's Principle; in this Centroid, or Centre of Mass/Momentum, frame of reference energy is locally conserved and the EEP is replaced by the Principle of Mutual Interaction. Although the EEP is violated in the theory, experimentally the difference between GR and my theory on ordinary metal blocks, as in Eötvös type experiments, is only about one part in 10^-17, i.e. three orders of magnitude smaller than the sensitivity of recent experiments.
However the theory is about to be tested as the Gravity Probe B satellite is entering its science phase about now. SCC predicts a geodetic precession of 5/6 that of GR that is SCC predicts a geodetic precession of 5.5120 arcsecs/yr. We shall know shortly!

Last edited: Jul 23, 2004
20. Jul 25, 2004

### Garth

DW - your posts have reminded me of a good discussion starter, which I shall post as a new thread, "Einstein's Inconsistency?" I shall value you replies.

21. Jul 25, 2004

### pmb_phy

Not in general. There are some cases when two observers can agree though. E.g. let a free particle (i.e. a particle not subject to any forces) be at rest in the inertial frame S. Let the inertial frame S+ be moving +x direction with respect to S with speed v. Let the inertial frame S- be moving +x direction with respect to S with speed v. Then observers in S+ and S- will agree on the inertial energy (i.e. the sum of the kinetic energy and rest energy) of the particle which is at rest in S.

If the spacetime is stationary then there exists a coordinate system in which the components of the metric tensor are time independant, i.e.

$$g_{\alpha\beta,0} = \frac{dg_{\alpha\beta}}{dt} = 0$$

In such a coordinate system the energy of a particle in free-fall is constant. For proof please see
http://www.geocities.com/physics_world/gr/conserved_quantities.htm

There is always a gravitational force in GR. It's just not a 4-force, its an inertial force. Also, the presence of a gravitational force does not depend on the presence of spacetime curvature. You can have a gravitational force in the absence of spacetime curvature. For the definition of gravitational force in GR please see
http://www.geocities.com/physics_world/gr/grav_force.htm

For a closed system, energy is always conserved locally. The complete statement of that fact is given by

$$T^{0\alpha}{,\alpha} = 0$$

For the conditions under which this is true please see A first course in general relativity, Bernard F. Schutz, page 104. See the section labeled conservation of energy-momentum.
The gravitational field does do work. In fact one widely known example is that of gravitational redshift. This is caused by the gravitational field doing work on a photon as it climbs out of the gravitational field.
Different physicists use the term total energy to refer to different quantities. Some, such as myself, use the term to mean the sum of rest energy, kinetic energy and potential energy. I prefer to call the sum of rest energy and kinetic energy inertial energy or free particle energy (as J.D. Jackson calls it in Classical Electrodyanmics - Second Ed.) so as not to confuse it with total energy. Inertial energy is proportional to the time component of 4-momentum whereas total energy is proportional to the time component of canonical 4-momemtum.
The proper mass of the particle does not change. However the (relativistic) mass decreases.

The rest mass of the particle is the same. The mass of the system changes. For the reason why please see On the concept of mass in relativity at www.geocities.com/physics_world. See the section called "Why does E = mc2" which starts on page 52.

Pete

Last edited: Jul 25, 2004
22. Jul 25, 2004

### Garth

The Centre of Mass/Momentum frame.
The Equivalence Principle relies on the fact that to an observer in a freely-falling frame of reference there are no forces acting on a small enough scale - although tidal forces apply to any realistic volume.

It is energy-momentum that is conserved because of that equation, which is entirely different.

Really? the photon has no forces acting on it, just the curvature of space-time. See my thread 'Self Creation Cosmology - a new gravitational theory' https://www.physicsforums.com/showthread.php?t=32713

GR replaces gravitational force with curvature of space-time, as there is no force there is no such thing as potential energy - not according to GR and the EEP that is.

23. Jul 25, 2004

### pmb_phy

I was responding to your comment According to the theory of General Relativity energy-mass is not even conserved. That staetment is incorrect.

In the case of a finding the energy of a particle as measured in a free-fall frame near a spherical body such as the Earth then no, the energy of a particle, as measured from that frame, is not constant. However the energy of the particle is constant as measured in the frame of reference attached to the surface of the Earth.

So from your response I take it that you mean "In some cases, the energy of a particle in a gravitational field is not constant." That's true in Newtonian Gravity too.

The energy of a particle in a gravitational field is given in terms of the time component of the particle's 4-momentum as E = P0/c. That is constant when the components of the metric tensor are not functions of time.
Now you're speaking of an inertial frame. Yes. There are no gravitational forces in an inertial frame of reference within a restricted region. But you said there are no gravitational forces. Gravitational forces are inertial forces and as such they are frame dependant. They exist only in non-inertial frames such as a frame attached to the surface of the Earth.
The equation

$$T^{0\alpha}_{,\alpha} = 0$$

is a statement of the conservation of energy. The equation

$$T^{k\alpha}_{,\alpha} = 0$$

is a statement of the conservation of the kth component of momentum. The equation

$$T^{\alpha\beta}_{,\beta} = 0$$

is the statement of conservation of energy-momentum.
I already explained to you why that is incorrect. You're speaking of four-forces which don't have a relative existance. Gravity is a force and as such it can do work. Just because it can be transformed away it doesn't mean that it isn't a force. It only means that its an inertial force. Curvature has nothing at all to do with this.
You'd be hard pressed to find a GR text which didn't refer to the components of the metric tensor as gravitational potentials. In fact it was Eisntein himself who called them that. He even called them that in his Nobel Prize acceptance speech - http://www.nobel.se/physics/laureates/1921/einstein-lecture.pdf

If you were to give a Nobel Prize acceptance speech would you include things in it that you didn't think were meaningful or important?

Sorry I can't read your thread. I'm in too much pain to sit in front of this computer for the time it would take to read that and determine what it has to do with what you implied it did.

Besides, the relationship between force and potential is different in GR than it is in non-relativistic/non-sr mechanics In any case I was not addressing something you said in that comment nor did that comment have anything to do with GR.

Pete

Last edited: Jul 25, 2004
24. Jul 26, 2004

### Garth

To be consistent to GR and the EEP you cannot select a preferred frame of reference. Statements have to be true of all inertial frames, therefore my statement was a true statement consistent with the theory. To select a preferred frame is exactly what my theory of self creation does, using Mach's Principle to do so, as such it is an alternative theory to GR

Newtonian gravitational potential is brought into GR only to calibrate the theory so that in the Newtonian limit the two theories agree. Once so calibrated the Newtonian gravitational potential plays no further part, it is replaced by the components of the metric. The temptation always is not to be consistent and work in one theory or the other but to confuse the two. If you are consistent then the problems of GR are not masked but force consideration, if you are bothered to.

25. Jul 26, 2004

### pmb_phy

Anyone can, at any time they see fit, choose a particular frame of reference to describe nature. That's what the subscripts/superscripts refer to in the equations of relativity. In SR and GR there are no preferred frames of reference.
No. It wasn't. If you're refering to this theory of yours then that is not GR. I'm speaking of GR and that is all.
No problemo! We all make mitakes.
Energy is always conserved locally in GR. Why would you think otherwise? Where did you hear this?
Who was speaking about a Newtonian gravitational potential? I wasn't.
That is why everyone who works in GR refers to the components of the metric tensor as a set of ten gravitational potentials. In GR things are more complicated. Potential *energy* is another concept in GR but it is one which is related to the gravitatonal potential. Gravitational potential energy is energy of position. Change the position of a particle and you've changed the energy of the particle. That is, by definition, a change in potential energy. In GR the energy is not a nice linear sum of rest energy, kinetic energy and potential energy but potential energy is still there. When you use the weak field approximation for GR, its still GR but now you can express the total energy of the particle in terms of potential energy as well as rest and kinetic energy. I can derive this if you'd like but for now I'll simply quote the result from Gravitation and Spacetime - 2nd Ed. Ohanian and Ruffini which is on page 157 in Eq. 101 which reads (Ohanian uses the symbol m for proper mass and sets c = 1 and uses u for 4-velocity)

$$P_0 \approx \frac{m}{\sqrt{1-v^2} - \frac{1}{2}\frac{m}{(1-v^2)^{3/2}} \kappa h_{\mu\nu} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} + m\kappa h_{0\alpha}u^{\alpha}$$

The authors state
If you're speaking about your theory again then, as I mentioned, I'm physically incapable of sitting at the computer long enough to read it. Please state what claims are from your theory and what claims are from GR.

As for Einstein on force and energy, let's not forget what he stated in his own text The Meaning of Relativity on page 83
Pete