Atomic clocks in gravitational field

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  • #126
tom.stoer
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I can't imagine a situation where no definition is possible.
As said, non-stationary, or non-asymptotically flat spacetime

That sounds like another definition issue, and I think even a wrong one. When comparing energies one must use the same reference system
Yes, but in GR there is no global refence frame!

You have local frames which are not related a priori and which do not allow for a global definition of energy. In order to compare energies you have to define how to transport reference frames through spacetime (along null lines). But once yo have done that you get the redshift for free - w/o defining energy.
 
  • #127
WannabeNewton
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Consider for example a FLRW space-time with ##k = 0##. If we let ##\xi^{a}## be the tangent field of the congruence defined by the comoving observers then the spatial metric on a given isotropic and homogenous time-slice will be given by ##h_{ab}(\tau) = g_{ab} + \xi_{a}\xi_{b}## hence ##\xi^{b}\nabla_{c}h_{ab} = -\nabla_{c}\xi_{a}##. From this we find that ## \nabla_{\mu}\xi_{\nu}= \Gamma ^{\gamma}_{\mu \tau}h_{\gamma\nu} = \frac{\dot{a}}{a}h_{\mu \nu}## in the comoving coordinates hence ##\nabla_{a}\xi_{b} = \frac{\dot{a}}{a}h_{ab}## in general.

Now the frequency of a light signal with wave 4-vector ##k^{a}## as measured by an observer with 4-velocity ##\xi^{a}## is ##\omega = -k_{a}\xi^{a}## so ##\frac{\mathrm{d} \omega}{\mathrm{d} \lambda} = k^{b}\nabla_{b}\omega = -k^{a}k^{b}\nabla_{b}\xi_{a} = -\frac{\dot{a}}{a}k^{a}k^{b}\xi_{b}\xi_{a} = -\frac{\dot{a}}{a}\omega^{2}##. Then ##\frac{\mathrm{d} \omega}{\mathrm{d} \tau}\frac{d\tau}{d\lambda}= \omega \dot{\omega} = -\frac{\dot{a}}{a}\omega^{2}\Rightarrow \frac{\dot{\omega}}{\omega} = -\frac{\dot{a}}{a}## thus ##\frac{\omega_2}{\omega_1} = \frac{a_1}{a_2}## which is the usual cosmological redshift for the aforementioned space-time. Here all we have used is the definition of the frequency of a light signal at a given event as measured by an observer passing through that event and the parallel transport of this frequency along the associated null geodesic using the derivative operator ##\nabla_{a}##. There was no use of any killing symmetries nor any global energies (there is no meaningful notion of global energy for this space-time anyways). This was to basically give an example of what Tom said above.
 
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  • #128
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Yep that sums it up quite brilliantly. Check out ch 11 of Wald if you want more details. The problem is that there is no asymptotic flatness hence [..] you can't define a gravitational potential [..].
OK that remark made it immediately clear to me, thanks!

As far as I can see, the FAQ does not claim that it is impossible to define energies in GR; only that it's a can of worms. :tongue2:

For me, if no definition of gravitational potential is possible with a model, then that is an argument against that model. A quick search gave me a recent university paper about doing numerical calculations for asymptotic flatness, which suggests to me that this is still accepted as possible solution - http://eprints.ma.man.ac.uk/1535/01/covered/MIMS_ep2010_93.pdf
 
  • #129
WannabeNewton
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You can define various kinds of total global energy easily in asymptotically flat space-times (calculating and physically interpreting it is a whole 'nother story); some standard examples are Komar energy (only for stationary asymptotically flat space-times), Bondi energy, and ADM energy. The problem is with non-asymptotically flat space-times wherein you simply cannot define any meaningful notion of total global energy.

Gravitational potential is a purely Newtonian concept. There is no reason to expect that it has any meaning for arbitrary space-times in GR. Stationary space-times have a time translation symmetry which allows us to define a GR analogue of the Newtonian potential but this is a very special case.
 
  • #130
tom.stoer
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As far as I can see, the FAQ does not claim that it is impossible to define energies in GR; only that it's a can of worms.

For me, if no definition of gravitational potential is possible with a model, then that is an argument against that model.
An overview regarding non-local definitions of mass, energy etc. including open issues can be found here:
http://relativity.livingreviews.org/Articles/lrr-2009-4/ [Broken]
Quasi-Local Energy-Momentum and Angular Momentum in General Relativity
The present status of the quasi-local mass, energy-momentum and angular-momentum constructions in general relativity is reviewed. First, the general ideas, concepts, and strategies, as well as the necessary tools to construct and analyze the quasi-local quantities, are recalled. Then, the various specific constructions and their properties (both successes and deficiencies are discussed. Finally, some of the (actual and potential) applications of the quasi-local concepts and specific constructions are briefly mentioned.

Rejecting models w/o definition of gravitational potential means rejecting GR.
 
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  • #131
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An overview regarding non-local definitions of mass, energy etc. including open issues can be found here:
http://relativity.livingreviews.org/Articles/lrr-2009-4/ [Broken]
Quasi-Local Energy-Momentum and Angular Momentum in General Relativity
The present status of the quasi-local mass, energy-momentum and angular-momentum constructions in general relativity is reviewed. First, the general ideas, concepts, and strategies, as well as the necessary tools to construct and analyze the quasi-local quantities, are recalled. Then, the various specific constructions and their properties (both successes and deficiencies are discussed. Finally, some of the (actual and potential) applications of the quasi-local concepts and specific constructions are briefly mentioned.
Thanks for the link!

Rejecting models w/o definition of gravitational potential means rejecting GR.
Here you launch a similar strong opinion as fact like when this side discussion started; but instead of arguing about it, I simply take note that it does not match that of the FAQ nor of that article.
 
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  • #132
tom.stoer
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Here you launch a similar strong opinion as fact like when this side discussion started; but instead of arguing about it, I simply take note that it does not match that of the FAQ nor of that article.
Which FAQ?

It is a mathematical fact that energy cannot be defined as usual in GR. Potentials cannot be defined, either (this means only in special cases). Note that this has nothing to do with physics, it's due to the mathematical structure underlying GR. There's no choice.
 
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  • #133
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Which FAQ? [..]
The one I cited just before.

Hmm... a logical and equally strong follow-up of your claims would be that accepting GR means rejecting the principle of conservation of energy - one of the pillars of modern physics. Michael Weiss and John Baez don't go that far.
 
  • #134
tom.stoer
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The one I cited just before.
Sorry, can't find it.

Hmm... a logical and equally strong follow-up of your claims would be that accepting GR means rejecting the principle of conservation of energy - one of the pillars of modern physics.
Yes and no.

There's still a locally conserved energy-momentum tensor of all non-gravitationally d.o.f.

But instead of a conserved vector current

##\partial_a j^a = 0##

which is only available when a space-time symmetry encoded in a Killing vector field does exist

##j^a = T^{ab}\,\xi_b##

in general situations w/o space-time symmetry only a conserved tensor density exists

##(DT)^a = 0##

Due the the covariant derivative = the Levi-Cevita Connection the usual trick

##\partial_0 j^0 \;\to\; \partial_0 \int_V dV\,j^0 = \partial_0 Q = 0##

with

##\oint_{\partial V} d\vec{S}\,\vec{j} = 0##

does not work.

So when no symmetry is present, one cannot constructed a conserved vector current j, therefore one cannot construct a conserved charge Q b/c the volume integral

##\int_V dV\,T^{00}##

is neither meaningfull (no well-defined transformation property) nor conserved

##\partial_0 \int_V dV\,T^{00} \neq 0##

That means that there is a local conservation law, but there's no way to derive a globally conserved quantity simply due to mathematical reasons.

A simple fact is that a redshifted photon looses energy w/o transferring this energy to another system, i.e. the gravitational field. Locally the Einstein-Maxwell equations guarantuees energy-momentum conservation of the el.-mag. field. Globally energy is not conserved.
 
  • #135
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Sorry, can't find it.
Post # 121 : 'it depends on what you mean by "energy", and what you mean by "conserved" '

Yes and no.

[...]

That means that there is a local conservation law, but there's no way to derive a globally conserved quantity simply due to mathematical reasons.
That sounds very much like a "yes" ...

A simple fact is that a redshifted photon looses energy w/o transferring this energy to another system, i.e. the gravitational field. Locally the Einstein-Maxwell equations guarantuees energy-momentum conservation of the el.-mag. field. Globally energy is not conserved.
If you talk about basic gravitational redshift, that's simply wrong - see my posts #14 and #107 .
 
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  • #136
tom.stoer
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Post # 121 : 'it depends on what you mean by "energy", and what you mean by "conserved" '
OK. I checked John's articles. Very clear - as usual.

He says that there is no gravitational energy in general and he explains why globally defined energy doesn't exist. He writes "So the flux integral is not well-defined, and we have no analogue for Gauss's theorem". That's exactly my reasoning.

Of course you can play around with Theta, but there's one big issue (besides the fact that it's a pseudo-tensor): it has nothing in common with other definitions of energy-momentum density, there's no relation with Noether's Theorem, and there's no relation with a Hamiltonian as generator of time translations.

That sounds very much like a "yes" ...
I would say, global energy conservation - in the sense we are used to - does not exist. Playing around with Theta is a trick, but it has nearly nothing in common with what we usually call energy.

If you talk about basic gravitational redshift, that's simply wrong - see my posts #14 and #107 .
I don't see the relevance of your post. As I said a couple of times: in general (!) there is no potential energy, no globally defined energy, and no energy conservation. A photon in the CMB appears to have a different energy observed on earth than observed at emission, so energy is not conserved, but b/c there is no potential energy in expanding spacetime, there is nothing where the missing energy went to. The whole concept becomes meaningless, and I think we (including John) explained why. It does not make sense to rely on concepts from Newtonian mechanics valid in very special cases of GR, but not being able to generalize these concepts to full GR.
 
  • #137
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OK. I checked John's articles. Very clear - as usual.

He says that there is no gravitational energy in general and he explains why globally defined energy doesn't exist. He writes "So the flux integral is not well-defined, and we have no analogue for Gauss's theorem". That's exactly my reasoning. [..]
Yes it appears to be very difficult. Perhaps it is even impossible to do. Once more, their "it depends" is quite not the same as your "no"; and "is not well-defined" and "we have no", is quite different from "there is no" and "doesn't exist". Thus I noticed the difference in opinions.

I don't see the relevance of your post. [..] A photon in the CMB appears to have a different energy observed on earth than observed at emission, so energy is not conserved [..]
Of course features such as frequencies are measured differently with different reference standards; it is fundamentally erroneous to compute energies by mixing such reference systems - adding apples to oranges is a big No-No in physics. I also highlighted that point in post #77, and it turned out that the fact that energy is frame-dependent is just the point that PeterDonis tried to make too.

And although it is not the same problem as standard gravitational redshift, the threads that I linked bear out that same fact: a difference in measured frequency with different reference systems does not imply that "energy is not conserved".
 
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  • #138
tom.stoer
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ok; instead of saying that energy is not conserved in GR I should better say that in general energy cannot be defined in GR, so the question whether it is conserved becomes meaningless.
 
  • #139
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ok; instead of saying that energy is not conserved in GR I should better say that in general energy cannot be defined in GR, so the question whether it is conserved becomes meaningless.
That's more like it. :smile:
 

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