Why Is the Fourth Component Necessary in Relativistic Momentum?

[learningphysics]

I've been trying to see how relativity shows that the quantity gamma*m*c^2 (total energy) is conserved. I assumed that this would proceed from the conservation of momentum. So I researched momentum in relativity, and noticed that it has a time-component: gamma*m*c (which is E/c). So this partially answers my question. If four-momentum is conserved so is E.

But how and why does this fourth component get introduced? The spacelike components of momentum have been changed from m*v (classical physics) to gamma*m*v. This part I understand. But can someone show me how this fourth time component (gamma*m*c) is necessary from the first principles of relativity? Thanks a bunch!

 

[jcsd]

Clearly 3 vectors in general are not Lornetz invariant so if we wish to describe quantities in a Lornetz invariant manner we need four vectors, in any frame a four vector has a time component.

If we take an object, it's four vector velcotiy U is the vector tangent to it's worldline, it's four momentum is simply mU.

 

[learningphysics]

Hi. Thanks for replying. I still have some questions. The conservation of this fourth component is troubling me. My reasoning process is like this:

1. mv is not conserved in all frames. so we change the definition of momentum to gamma*m*v. This is conserved in all frames. So that's the first 3 components of momentum.

2. We need a fourth component to make the norm of the momentum lorentz invariant. This is y*m*c.

3. We guess that this fourth component of momentum is also conserved??

Is part 3 a new hypothesis? We can verify the gamma*m*v formula by studying elastic collisions from classical physics and seeing that this new space momentum is conserved. But this fourth time component is very different. Did Einstein just guess that it is conserved also?

Or does the fact that the norm of the momentum vector is Lorentz invariant and the first 3 components are conserved somehow prove that this fourth component is also conserved?

I think there is something fundamental to Lorentz invariance that I'm missing.

Thanks again!

 

[jcsd]

The conservation of the time component of four momentum is just the conservation of energy.

 

[learningphysics]

How can one prove the conservation of energy?

 

[selfAdjoint]

If you step back from momentum and consider position, you find that it has four components in Minkowski spacetime, three of space and one of time. The time one has an opposite sign to the space ones, and has a factor of c to make its units the same as the space components. The components of a four-vector like this are not Lorentz covariant, but the length is because it's the inner product of the vector with itself and the inner product is covariant. So now if you differentiate this position four-vector wrt proper time you get a velocity four-vector, and you multiple that by invariant mass and you get a momentum four-vector.

It time component is the kinetic energy of the particle, its space components span the spatial momentum of the particle mp, and its length (which is invariant, remember) is mc^2.

 

[robphy]

You probably meant to say that
"its time component is the relativistic energy of the particle"...
The relativistic kinetic energy (i.e., the non-rest-energy) is the relativistic energy minus the rest energy: $$K=E-mc^2=\gamma mc^2 - mc^2 =(\gamma-1)mc^2$$,
where $m$ is the rest mass.

 

[robphy]

How can one prove the conservation of energy?

If you accept that total relativistic momentum $\sum \gamma_{(i)} m_{(i)}v_{(i)}$ is conserved in (say) an elastic collision, then I would think that applying a velocity transformation would imply the conservation of relativistic energy.

 

[pervect]

I've been trying to see how relativity shows that the quantity gamma*m*c^2 (total energy) is conserved. I assumed that this would proceed from the conservation of momentum. So I researched momentum in relativity, and noticed that it has a time-component: gamma*m*c (which is E/c). So this partially answers my question. If four-momentum is conserved so is E.

But how and why does this fourth component get introduced? The spacelike components of momentum have been changed from m*v (classical physics) to gamma*m*v. This part I understand. But can someone show me how this fourth time component (gamma*m*c) is necessary from the first principles of relativity? Thanks a bunch!

You've already noticed that the momentum 4-vector is actually the energy-momentum 4-vector. So it should not be surprising to you that there are 4 quantities in the vector, one for the enregy, and 3 for the momentum.

But you may ask - why do energy and momentum mix in this manner? The answer is interesting, but it's unfortunately a bit advanced.

The relevant theorem is known as Noether's theorem. It points out that the existence of a conserved momentum is a consequence of the invariance of the laws of physics to spatial translation, plus the action principle.

THe existence of a consered energy is the result of the invariance of the laws of physics to time translation, plus the action principle.

Since space and time get intermixed by different observers, so do energy and momentum. This is ultimately "why" the energy-momentum 4-vector transforms in the same manner as the space-time vector.

There's a little more about Noether's theorem here:



I'll quote the points which I think are important from this article below:

* the invariance of physical systems with respect to translation (when simply stated, it is just that the laws of physics don't vary with location in space) translates into the law of conservation of linear momentum;
* invariance with respect to rotation gives law of conservation of angular momentum;
* invariance with respect to time gives the well known law of conservation of energy, et cetera.

In order to understand the argument fully, though, one has to be aware of the formulation of the laws of physics in terms of Hamilton's principle, the principle of least action. This may be mentioned in undergraduate physics courses, but it's usually not taught in full until the early graduate level.

The main point I want to make though, is that since space and time "intermix" because of relativity, so do energy and momentum, since energy is a time-translation symmetry and momentum is a space-translation symmetry.

 

[Garth]

The important point about using the Principle of Least Action is that it works for generalised coordinates. Hence the conserved quantity obtained from using that principle is that one appropriate for the manifold used. As energy-momentum or four-momentum is invariant in four dimensional space with a signature of -2 then the application of the principle of least action to such a manifold conserves energy-momentum of the field or the rest mass of a particle, rather than energy or three-momentum separately.

Garth

 

[pmb_phy]

I've been trying to see how relativity shows that the quantity gamma*m*c^2 (total energy) is conserved.

The term you speak of $E = mc^2$ (sometimes labeled T) is the inertial energy[/itex] of an isolated object (i.e. an object on which no force acts at all - a "free" object) and has the value

E = Rest energy + Kinetic Energy = E₀ + K

Total energy, W (sometimes labeled E) is a term which does not have a unique meaning in the relativity literature. Sometimes it refers to inertial energy, T, and sometimes it refers to T + V where V = potential energy of position in a field. The total inertial energy of a system is conserved if no force acts on the system and if the objects of the system do not interact. If they interact then you have to take into account the energy and momentum of fields through which they act. For a free object E is proportional to the time component of its 4-momentum. W is proportional to the time component of the canonical 4-momentum.

Consider a closed system of non-interacting particles. The mass, m, of a particle (not to be confused with proper mass m₀) is defined as the m such that the quantity m[sb]v[/sub] is conserved in all inertial frames of reference. The quantity m[sb]v[/sub] is then defined as the (three) "momentum" of the particle. This is true in both Newtonian physics and relativistic physics. It can then be shown that, if the particle is a tardyon (i.e. moves at speeds less than c) then the mass is related to the proper mass by $m = \gamma m_0$. For derivation please see - http://www.geocities.com/physics_world/sr/inertial_mass.htm

It can be shown that the mass of a tardyon transforms as

$m' = \gamma(m - v p_x /c^2)$

For derivatioin please see - http://www.geocities.com/physics_world/sr/energy_momentum_trans.htm

Thus m transforms in the exact same way that time does if momentum takes the role of position. Anything which transforms from one inertial frame to another in the same way as (ct, x, y, z) is called a Lorentz 4-vector. Therefore P = (mc, p) is a 4-vector called the 4-momentum (aka energy-momentum 4-vector).

By demanding that 3-momentum be conserved in all inertial frames of reference it can then be shown that this means that m is also a conserved quantity. The derivation is a bit long and I have not been able to create a new web page to post the derivation yet.

I assumed that this would proceed from the conservation of momentum. So I researched momentum in relativity, and noticed that it has a time-component: gamma*m*c (which is E/c). So this partially answers my question. If four-momentum is conserved so is E.

Yes.

But how and why does this fourth component get introduced?

See derivation in above URL.

1. mv is not conserved in all frames. so we change the definition of momentum to gamma*m*v. This is conserved in all frames. So that's the first 3 components of momentum.

If by "m" you mean "proper mass" (sometimes called "rest mass") then yes. That's true. But momentum is defined as the product of (inertial) mass times velocity. Inertial mass is defined such as to allow this.

2. We need a fourth component to make the norm of the momentum lorentz invariant. This is y*m*c.

Sounds good to me.

3. We guess that this fourth component of momentum is also conserved??

No! We do not guess it. It can be proved by deduction.

How can one prove the conservation of energy?

The principle of the conservation of energy cannot be proved. It is a postulate. However certain energy conservation relations can be proved. In this case it can be proved that E is conserved.

It time component is the kinetic energy of the particle, ..

The time component of a 4-momentum is not kinetic energy. The time component is the sum of rest energy and kinetic energy.

Note: Do not assume that $E = mc^2$ is valid in all cases and thus try to replace m with E. It won't work in general. In fact it won't work for objects which have a finite extent (like a rod). For a free particle in an inertial frame the relation $m/v = E/c^2$ is always valid. However if the body is an object (body with a finite size - not pointlike) rather than a particle then this relation is invalid. Missing this very important point has incorrctly led people to a paradox or two.

Take as an example a rod, or proper mass m₀ which lies parallel to the x-axis of an inertial frame S. Assume there are no forces acting on the rod and it is therefore stress/tension free. Consider the mass as observed from the inertial frame S' which is in standard configuration with S moving with speed v in the +x direction. The mass (as defined above -> m = p/v) in S' has the value $m = \gamma m_0$ as expected.

Now try this again but this time exert a force on each end of the rod such that the rod remains at rest in S. Assume that the object does not compress a noticible amount. The mass as measured in S is still m₀. The total force is zero but this is not a free object - there are forces acting on it. Now observe this from S'. The mass is now no longer equal to $m = \gamma m_0$. The mass is now a function of the stress acting on the body and therefore does not have the value $m = \gamma m_0$. The momentum of the rod is now no longer $p = \gamma m_0 v$.

I have not worked this particular example out on a web page. But I have worked out a similar example. See -- http://www.geocities.com/physics_world/sr/rd_paradox.htm

When I'm able to I will create a new page to explain why this is true.

Pete

 

[learningphysics]

Thanks to everyone that replied. It was all very helpful. I read Einstein's paper on mass and energy equivalence. He uses the situation of an elastic collision to derive formulas for momentum and kinetic energy. Then he uses the situation of an inelastic collision to derive the formula for rest energy.

 

[pmb_phy]

Thanks to everyone that replied. It was all very helpful. I read Einstein's paper on mass and energy equivalence. He uses the situation of an elastic collision to derive formulas for momentum and kinetic energy. Then he uses the situation of an inelastic collision to derive the formula for rest energy.

Okay. Last night I created a web page in order to provide this proof. I've been putting it off for some time but since there are so many people who don't understand that the conservation of P₀ can be proved through conservation of 3-momentum I thought it was about time I put this up.

Please see - http://www.geocities.com/physics_world/sr/mass_conservation.htm

(P⁰ is what you and some others have referred to as "energy" but what I and others refer to as "mass")

Pete

 

[Garth]

Okay. Last night I created a web page in order to provide this proof. I've been putting it off for some time but since there are so many people who don't understand that the conservation of P₀ can be proved through conservation of 3-momentum I thought it was about time I put this up.

Please see - http://www.geocities.com/physics_world/sr/mass_conservation.htm

(P₀ is what you and some others have referred to as "energy" but what I and others refer to as "mass")

Pete

Pete - on your web page you use P⁰ not P₀, which is it?
Also the real problem arises when the particles are freely falling in a general external gravitational field, then energy is not conserved even though no work is being done on or by them by that field.
Garth

 

[pmb_phy]

Pete - on your web page you use P⁰ not P₀, which is it?

I made an error typing. I typed "sub" when I meant to type "sup". It has been corrected.

Also the real problem arises ...

This thread seems to be about SR and inertial frames in flat spacetime. That's what I was addressing. I see nothing which indicates otherwise.

..when the particles are freely falling in a general external gravitational field, then energy is not conserved even though no work is being done on or by them by that field.
Garth

Incorrect

 

[Garth]

I made an error typing. I typed "sub" when I meant to type "sup". It has been corrected.

We all do that!

This thread seems to be about SR and inertial frames in flat spacetime. That's what I was addressing. I see nothing which indicates otherwise.

Indeed in SR the problem disappears, however it is always good to generalise concepts, in this case to non flat space-times.

Incorrect

You will find that most relativists disagree. The reason why I asked whether you meant P⁰ or P₀, is that it is P₀ that is conserved in the static gravitational field in the centre of mass frame in which the metric components are static, as you prove in your lecture notes. These are the same only in a Minkowski metric. I believe the reason there is confusion as to which one is to be used as energy is because in fact in general energy is not conserved in GR. (Though of course it is in SR)
To obtain a value for energy in a frame of reference U^a, the scalar product of four-momentum with U^a has to be taken,
E = - P_aU^a . [Minus sign to agree with most common sign conventions]

Garth

 

[learningphysics]

Thanks Pete! Your conservation of mass proof is exactly what I was looking for.

 

[learningphysics]

Thanks Pete! Your proof of conservation of mass was exactly what I was looking for.

 

[pmb_phy]

You will find that most relativists disagree.

I seriously doubt that.

Garth - I've explained all of this to before, several times in fact. But you're not listening to me. You have something in mind which is incorrect and you seem to be refusing to understand the correct concept. I.e. you're confusing the concept of the conservation of energy with the concpet of the constancy of the energy of a particle. These are two different things which are simply related to each other. But you constantly confuse them as being identical concepts.

The principle of energy conservation does not mean that the energy of a single particle remains constant, i.e. it doesn't change with time.
The principle of energy conservation refers to a sum of energies of a closed system. I.e. if the sum of the energies of a closed system is not a function of time then it is said that energy of the system is conserved.

Recall what you said ..

..energy is not conserved even though no work is being done on or by them by that field.

So what? Why did you say this? What was your point? Since when should work being done on a particle imply that the energy of the particle is constant? That has never been true in Newtonian physics so I can't fathom why you're always focusing all of your energy on it. Simply because work is being done on a particle is no guarantee that the total energy of the particle will remain constant. This is the same in Newtonian mechanics too. The only way that the energy of a single particle will remain constant is when the force is derivable from a potential energy function which is not an explicit function of time. When that is the case then E = K + V = constant. When the field is time dependant and V = V(t) then K + V is not a constant. This does not mean I'll win the Nobel prize for figuring out how to violate the principle of energy conservation since that principle only applies to closed systems.

Tell me something Garth - In classical (non-relativistic) electrodynamics the total energy of a charged particle in an EM field is not a constant in general. So why haven't you focused on that too instead of focusing exclusively on GR?

One a similar point so as to illustate what I'm saying - Consider the meaning of the phrase "momentum is conserved". Does this mean that a law of nature states that the momentum of a particle can never be changed? Of course it doesn't. Does it mean that the momentum of a charged particle in an EM field will remain constant? No. Of course not. It quite literally means that the total momentum of all particles and fields of a closed system is a constat.

The reason why I asked whether you meant P⁰ or P₀, is that it is P₀ that is conserved in the static gravitational field in the centre of mass frame in which the metric components are static, as you prove in your lecture notes. These are the same only in a Minkowski metric. I believe the reason there is confusion as to which one is to be used as energy is because in fact in general energy is not conserved in GR.

Who's confused?? And here once again your misusing the concept of energy conservation.

To obtain a value for energy in a frame of reference U^a, the scalar product of four-momentum with U^a has to be taken,
E = - P_aU^a .

That is incorrect. This is not what the notion that P₀ = constant refers to. E = - P_aU^a is the energy of a particle as measured by an observer whose 4-velocity is U. P₀ is the energy of a particle as measured by a (for lack of a better term) coordinate observer. E.g. in Schwarzschild coordinates this observer is referred to as the Schwarzschild observer aka "far away observer".

Before you continue with this please explain what you think "conservation of energy" means as it is pertains to Newtonian mechanics. Thanks.

Thanks Pete! Your proof of conservation of mass was exactly what I was looking for.

Your welcome. I wanted to get this up again before I had to leave the internet again.

I also created a new web page to explain why its incorrect to claim that m is defined as E/c<sup>2[/sub]. See

http://www.geocities.com/physics_world/sr/mass_momentum_density.htm

To see a worked out example of this please see
http://www.geocities.com/physics_world/sr/rd_paradox.htm

Notice that the mass of the struts is not proportional to the energy of the struts, i.e. In this case m does not equal E/c2 for the strut. This holds only for closed systems and thus does not hold in general. E.g. if you have a charged body in an electric field then m = E/c2 is invalid.

Pete</sup>

 

[Garth]

Pete - let me take this slowly:

. I.e. you're confusing the concept of the conservation of energy with the concpet of the constancy of the energy of a particle. These are two different things which are simply related to each other. But you constantly confuse them as being identical concepts.

The principle of energy conservation does not mean that the energy of a single particle remains constant, i.e. it doesn't change with time.
The principle of energy conservation refers to a sum of energies of a closed system. I.e. if the sum of the energies of a closed system is not a function of time then it is said that energy of the system is conserved.

A freely falling particle has no forces acting on it - according to GR - therefore it can be considered as a closed system. In which case the fact that its energy does not change means that energy is conserved and the opposite is true when its energy does change.

Tell me something Garth - In classical (non-relativistic) electrodynamics the total energy of a charged particle in an EM field is not a constant in general. So why haven't you focused on that too instead of focusing exclusively on GR?

This is a very interesting case to illustrate my point. In such a case the closed system consists of the total energy of the charged particle and the energy of the field, this is conserved. The energy of the charged particle varies because of the electromagnetic force acting on it from the field, work is done on the particle and its energy changes, however the system energy is conserved. However in the case of GR gravitation the energy of a freely falling and accelerating particle changes and no forces are acting (to first order ignoring tidal forces) as the Newtonian force is explained by space-time curvature. Energy is not conserved in this small closed system.

The four velocity U^a refers to the frame of reference of the observer, we do not disagree over this, if I have been ambiguous I apologise.

Energy is a slippery concept we have to be careful how we define it and how we measure it, however that being said the fact that GR does not in general or locally conserve energy is well accepted from Einstein onwards.

Noether's theorem showed how the conservation of energy required a symmetry under time translations. This is only possible in GR under certain special circumstances when there is a time-like Killing vector. Such as when the metric components in the observer's frame of reference are static.

As we know one of the basic principles of GR is that of the conservation of energy-momentum, but this is not a concatenation of the two principles of the conservation of energy and conservation of momentum. Energy and momentum are frame dependent whereas energy-momentum stands alone as a geometric object.


Garth

 

[Garth]

A second thought:
The question of whether energy is conserved in a closed system gives rise to a second question that is, "What is a closed system?"

This is a very interesting question to ask.

Normally, if we are thinking of energy conservation a closed system is one on which no forces are acting from outside the said system.
This issue in GR revolves around the understanding that, according to the equivalence principle, in the freely falling small enough region, Newtonian gravitational accelerations can be transformed away. Tidal forces cannot be so transformed, however to first order, and by considering a system small enough relative to the scale length of the gravitational field being considered such as a falling test particle, these can be ignored.

Hence we have a small closed system on which no forces are acting and yet which is being accelerated relative to an inertial observer situated at some other location in the field. She could be at the Centre of Mass or at an asymptotic null infinity (in which case the metric components may not be time dependent) , or anywhere in between.

So what comprises the closed system? One view would be to say the whole gravitational field is the closed system, and yet, I would argue, that goes against the spirit of the equivalence principle, because, by the EEP, forces that do work are not meant to be acting on the particle.

Garth

 

[pmb_phy]

A freely falling particle has no forces acting on it ...

That is incorrect. A particle in free-fall has zero 4-force acting on it. But 4-force is not the only kind of force that can do work.

- according to GR - therefore it can be considered as a closed system.

That is incorrect also. By definition[/b] a system is said to be closed if (1) there are no external forces acting on any part of it, whether those forces are 4-forces or inertial forces and (2) no part of the system interacts or is influenced by anything external to the system.

Note: Just in case you have this on your mind - Do not conclude a system is closed simply because the total energy of a system is constant. Also - do not assume that a system is closed if both energy and momentum is constant. A system is closed if and only if it does not interact with anything which is external to the system.

A particle in free-fall in a gravitational field has an inertial force impressed upon it and therefore its kinetic energy is not constant.

A system In which case the fact that its energy does not change means that energy is conserved and the opposite is true when its energy does change.

If the g-field is not static then the energy changes and hence the system of one particle is not a closed system, even though the 4-force is zero - the inertial force is non-zero.

This is a very interesting case to illustrate my point.

Actually it doesn't illustrate your point. From what I've gathered from your posts you have consistently implied that energy is not conserved in GR and your arguements have been based only on what parts of a system is doing and not of the system as a whole. In the past you spoke of the energy of a particle as observer from a system in free-fall. You focused only on the energy of the single particle and then you focused only on 4-forces, as if they were the only forces capable of doing work. You then implied that since the energy of that single particle was not constant then energy was not conserved in GR. That is an entirely incorrect arguement. In this example you're now changing your tune. You have now decided to focus, not only on the particle, but of that which is interacting with the particle. You're not addressing the fact that this example is nearly identical with your g-field exampe. I.e. in this case of particle in e-field, the constancy of the energy of the charged particle is a frame dependant concept. Consider a source charge whose mass is much greater than the test charge. Then in the rest frame, where the field of the source is static, the energy of the test charge is a constant of motion. In a frame moving with respect to the source charge the field of the source is time dependant and as such the energy of the test charge is not conserved.

That is exactly the problem that you were speaking of. You simply used gravity and you incorrectly assumed that inertial forces do no work.

.. (to first order ignoring tidal forces) as the Newtonian force is explained by space-time curvature. Energy is not conserved in this small closed system.

Invalid assumtion. The only way that a particle moving in a free-fall frame can have a non-constant energy is if you don't ignore tidal forces since its the time dependence of the tidal forces which ruins the constancy of the energy of the particle.

..fact that GR does not in general or locally conserve energy is well accepted from Einstein onwards.

That is incorrect. While energy may not be conserved in all cases in GR, you have yet to discuss those cases in which it may not be conserved and you've only discussed parts of a system and then claimed, from the non-constancy of the energy on that part of the system, that the consevation of energy principle is violated - that is incorrect.

Noether's theorem showed how the conservation of energy required a symmetry under time translations. This is only possible in GR under certain special circumstances when there is a time-like Killing vector. Such as when the metric components in the observer's frame of reference are static.

You're misusing Noether's theorem. You've changed from the laws of physics being invariant under time translations to a system being time dependant under time translations and then you've yet once again focued only on a single part of a system.

E.g. - Let's use your argument on EM. You've phrased that in terms using mathematical lingo rather than physical lingo - Let's use the physical lingo so that we can use an analogy from EM.

You claim that conservation of energy in GR requires a static field and since this is only possible in special cases it follows that energy is not conserved in GR.

Here is your argument applied to EM - Energy conservation in EM requires a static field and since this is only possible in special cases it follows that energy is not conserved in EM.

Do you see your mistake? In your GR argument you've once again incorrecly misused the idea of energy conservation and focused in on the notion that the energy of a single, non-isloated, particle is not constant. You've totally left out the fact that only the energy of the entire system, source of field , objects moving in field and field itself, that must have constant energy.

Before we go on please demonstrate that you fully understand that energy conservation does not refer to the energy of a single particle moving in a field. Please tell me that you understand that the concept of energy conservation applies the the system as a whole - source, field and particles in the field.

Please also tell me that you understand that inertial forces can do work as well as 4-forces. If yolu disagree then please provide proof.

Pete

 

[Garth]

Pete - My example of the freely falling particle, a test particle in Einstein's thought experiments, was used as it is in thought experiments about the equivalence principle in which it is in an inertial frame of reference and isolated from other non-gravitational influences.

As such, being a point mass there are no tidal forces, however if it were an extended body on which tidal forces act these would still not be the principle reason why energy conservation is violated.
There are two such problems; one is the local conservation of energy, the test particle case, and the global conservation of a complete field, which in general is not static.

The problem is both defining energy in the first place and then measuring it over an extended field at a distance from the observer.

As selfAdjoint posted in the thread "Sean Carroll Cosmology Primer" in the 'General Astronomy and Cosmology' Forum: (He was responding to my question of whether energy should be conserved on not)

Hmmm. Global energy conservation is a legacy to us from 19th century physics. It has no more warrant than that. When you say energy is conserved in "the natural world" you mean the world where h and v/c go to zero, which is the world of 19th century physics. Quantum theory and GR have each their modern statements of energy behavior, and in both cases AFAIK the 19th century limit gives energy conservation.

To me, to repine because modern physics doesn't match the views of Clausius and Maxwell sounds just like the people who can't accept Lorentz transformations or superposition because those ideas disagree with Newton and Kelvin.

Garth

 

[pmb_phy]

Garth - Regarding your comment that (paraphrasing) "everyone knows that the g-field does not work on falling particles". This may be your opinion but it is not one held by the GR community.

See "Gravitation," MTW page 187 (re - grav redshift)

The energy of the photon must decrease just as that of a particle does when it climbs out of a gravitational field. The photon energy at the top and bottom of its path through the gravitational field must therefore be related by ... The drop in energy because of the work done against gravitation implies a drop in frequency

See context since its quite clear that by "energy" he means in some places "kinetic energy" and in other places "Total energy = kinetic + potential"

The kinetic energy of a photon in an inertial frame in the absense of gravity is just E = hv

Pete

 

[Garth]

Garth - Regarding your comment that (paraphrasing) "everyone knows that the g-field does not work on falling particles". This may be your opinion but it is not one held by the GR community.

The energy of the photon must decrease just as that of a particle does when it climbs out of a gravitational field. The photon energy at the top and bottom of its path through the gravitational field must therefore be related by ... The drop in energy because of the work done against gravitation implies a drop in frequency

See "Gravitation," MTW page 187 (re - grav redshift)
See context since its quite clear that by "energy" he means in some places "kinetic energy" and in other places "Total energy = kinetic + potential"

The kinetic energy of a photon in an inertial frame in the absense of gravity is just E = hv

Pete

Thank you Pete - this is the heart of the problem.
You quoted MTW out of context, just above the passage you quote they say,

That a photon must be affected by a gravitational field Einstein (1911) showed from the law of conservation of energy, applied in the context of Newtonian gravitation...To avoid this contradiction of the principle of the conservation of energy, which can also be stated in purely classical terms, Einstein saw that the photon must suffer a red shift.

His argument was purely classical, the same classical physics that his theory of GR overthrew! The only advance over 19th century physics was his understanding that not only did a photon exist but also that its energy had a mass equivalent and therefore could acquire potential energy.

What MTW could have gone onto say but didn't, such authors tend to be silent on the issue although Weinberg was rather more transparent (pg 85), is that because GR conserves energy-momentum and not in general energy Einstein's argument is inconsistent within GR.

Weinberg says

However, I have insisted on including a non-relativistic emitter and absorber in the above calculation, because the concept of gravitational potential energy for a photon is otherwise without foundation.

The 'above calculation' examined "the total internal plus gravitational potential energy of the two pieces of apparatus" and equated them before and after the exchange of the photon. Again the concept of gravitational potential energy has slipped in - how?

The question raised is, "How does the gravitational field act on a freely falling particle or photon to change its energy?" The concept of the equivalence principle says such a particle is in an inertial frame of reference with no forces acting on it. (I am talking about a infinitesimal test particle)

It is on my conviction that herein lies a contradiction in GR that I based my theory SCC, GPB is testing them both, we shall know before long (well just over a year)!

Garth

BTW You may care to see how I derive gravitational red shift in my paper
'A New Self Creation Cosmology, a 'semi-metric' theory of gravitation'," http://www.kluweronline.com/oasis.htm/5092775, Astrophysics and Space Science 282: 683–730, (2002). section 1.3.1 The local conservation of energy, which you will have to pay for (or I could send you a preprint if interested) or my eprint 'The derivation of the coupling constant in the new Self Creation Cosmology', http://arxiv.org/abs/gr-qc/0302088 section 3.2 The Gravitational Red-Shift of Light which you will not have to pay for!

 

[Garth]

Furthermore: Wald gives a consistent derivation of gravitational red shift (page 136/7), and this is the method I use in the first part of my derivation; it is caused by a time dilation between the upper and lower levels due to the null geodesics diverging as they traverse curvature. However I continue on to show, that if energy is locally conserved (introducing a modification to GR), then this time dilation applies to the derivation of the mass of the apparatus as well as the photon. But there is a further increase of mass, due to the absorption of the energy required to lift the apparatus between the two levels:
m(r) =m₀exp[Phi(r)] where Phi(r) is the dimensionless Newtonian gravitational potential. This extra mass is the cause of the red shift. In SCC, incorporating the local conservation of energy, gravitational red shift is caused by the apparatus increasing in mass (potential energy) rather than the photon losing it.
Garth