I Origin of Deformation Energy in Curved Spacetime?

[timmdeeg]

If a gravitational wave passes an elastic body it will be deformed. The source of the work done on it comes from the energy of the gravitational wave.

In curved spacetime e.g. an accelerated expanding FRW- or static Schwarzschild-spacetime an elastic body will be deformed too. What is the analogy to the energy a gravitational wave carries with it in this case?

Sure tidal forces are due to tidal acceleration but the latter concerns geodesics. I don‘t see that this answers the question regarding the source of the energy. Any help is appreciated.

 

[PeterDonis]

If a gravitational wave passes an elastic body it will be deformed. The source of the work done on it comes from the energy of the gravitational wave.

Yes (but see a caveat below). But pinning down exactly "where" the energy of the gravitational wave is stored is not possible; it's not an ordinary energy density and doesn't have an ordinary stress-energy tensor. This was one of the reasons it took a while for physicists to believe that gravitational waves could carry energy at all.

Note also that elastic deformation, strictly speaking, means the deformation is reversible. That can make it harder to see how any work is involved, because after the gravitational wave passes, the body's structure returns to what it was before and no net energy has been transferred. The most common example showing that gravitational waves carry energy is just showing that they can heat up an object as they pass through it. This process is irreversible, so it's easier to see that it must involve the gravitational wave doing work and therefore carrying energy.

In curved spacetime e.g. an accelerated expanding FRW- or static Schwarzschild-spacetime an elastic body will be deformed too.

It's important to be clear about exactly what scenario you are proposing. A body sitting at rest on the surface of the Earth will be deformed compared to what its state would be, given the same microstructure of the body (same number of atoms, same atomic arrangement, same chemical bonds, etc.), in free fall. That means that, in order to move the body from a state of free fall to the state of sitting at rest on the surface of the Earth, you have to do work on the body to change its shape (over and above any work required to just move it). The energy that ends up stored in the body's deformation comes from whatever did the work, which won't be "gravity" by itself (yes, you could just let the object fall and hit the Earth, but then the deformation it suffers wouldn't be reversible).

However, I think you might be thinking of a scenario such as, for example, a body moving in a changing tidal gravity field--for example, a body in an elliptical orbit around a planet, where the tidal deformation of the body varies along its orbit (because its distance from the planet, and therefore the tidal gravity across it, varies). This, again, will be reversible deformation (assuming the tidal gravity isn't strong enough to disrupt the body's structure), so no net energy will be transferred over a complete orbit (when the body returns to the same place in the orbit, its structure will be the same). For the details of what happens in between, the best response I know of is Sean Carroll's article on energy not being conserved in a curved spacetime:

http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

 

[timmdeeg]

However, I think you might be thinking of a scenario such as, for example, a body moving in a changing tidal gravity field--for example, a body in an elliptical orbit around a planet, where the tidal deformation of the body varies along its orbit (because its distance from the planet, and therefore the tidal gravity across it, varies). This, again, will be reversible deformation (assuming the tidal gravity isn't strong enough to disrupt the body's structure), so no net energy will be transferred over a complete orbit (when the body returns to the same place in the orbit, its structure will be the same). For the details of what happens in between, the best response I know of is Sean Carroll's article on energy not being conserved in a curved spacetime:
http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/

Thank you for this enlightening answer. The key seems to be reversibility. What about a spring being stretched in free fall towards a mass or in an accelerated expanding universe for some time? Will in this case (no whole cycle with no net energy) energy be transferred to the spring and if yes what is the source? Sean Caroll says "Energy isn’t conserved; it changes because spacetime does." Does this refer to the whole universe or is this statement also valid locally? In the latter case does this mean no work is done on the spring even though the stretching causes tension which is a form of energy? Energy from nothing?

 

[PeterDonis]

The key seems to be reversibility. What about a spring being stretched in free fall towards a mass or in an accelerated expanding universe for some time? Will in this case (no whole cycle with no net energy) energy be transferred to the spring

Yes.

if yes what is the source?

There isn't one, in the sense of being able to identify some energy density that gets reduced by the same amount that the energy density in the spring gets increased. Sean Carroll's answer in his article is basically "spacetime can exchange energy with matter", but the "energy stored in spacetime" can't be localized as an energy density.

Does this refer to the whole universe or is this statement also valid locally?

In the sense that the spring's energy density is localized, yes. But the "energy stored in spacetime" can't be localized in this way. So it's local in one way, but not local in another. Which of course clarifies everything.

does this mean no work is done on the spring

No. But identifying, locally, what energy density is being reduced to correspond with the energy density increase in the spring cannot be done. See above.

Energy from nothing?

Not in the sense in which energy conservation is valid in GR. But that sense is somewhat counterintuitive. The spacetime geometry along the spring's worldline is changing; so the spring's stress-energy tensor still has zero covariant divergence, which is what enforces energy conservation in GR. But the reason it has zero covariant divergence is not that stress-energy is "flowing" into the spring from somewhere else (which is what happens in a "normal" case of work being done--the work is being done by a rocket engine or a hydraulic press or something that has stress-energy stored in it); it's that the changing spacetime geometry offsets the increase in the spring's energy density in just the right way to keep the covariant divergence zero.

 

[timmdeeg]

There isn't one, in the sense of being able to identify some energy density that gets reduced by the same amount that the energy density in the spring gets increased. Sean Carroll's answer in his article is basically "spacetime can exchange energy with matter", but the "energy stored in spacetime" can't be localized as an energy density.

So one can't argue that in principle the energy increase in the spring goes to the expense of the accelerated expansion, meaning to a decrease of the second derivative of the scale factor, correct? Or to bring it to the point, would the Friedmann equations remain unchanged, if the universe was filled with springs compared to a universe with a mass density that equals that of the springs and the same vacuum energy density in both cases?*)

The spacetime geometry along the spring's worldline is changing; so the spring's stress-energy tensor still has zero covariant divergence, which is what enforces energy conservation in GR. But the reason it has zero covariant divergence is not that stress-energy is "flowing" into the spring from somewhere else (which is what happens in a "normal" case of work being done--the work is being done by a rocket engine or a hydraulic press or something that has stress-energy stored in it); it's that the changing spacetime geometry offsets the increase in the spring's energy density in just the right way to keep the covariant divergence zero.

This goes deeper. To my understanding covariance in GR means that the laws describing it are the same (perhaps better invariant) regardless the frame of reference of an observer. But searching the web for "zero covariant divergence" it turnes out to be highly technical. If you say "it's that the changing spacetime geometry offsets the increase in the spring's energy density in just the right way to keep the covariant divergence zero" it's not really technical but still hard. Could you please elaborate a little more on this?

EDIT This wording isn't good. The Friedmann equations themselves don't change. What I suspect is that the tension in the springs which is stress/energy reduces the accelerated expansion a bit. But if this view is correct, I think the work done on the springs is not compensated by less work done to expand the universe because the universe doesn't expand against an outside pressure (means that P in - PdV is zero).

 

[PeterDonis]

So one can't argue that in principle the energy increase in the spring goes to the expense of the accelerated expansion, meaning to a decrease of the second derivative of the scale factor, correct?

If the accelerated expansion is caused by a cosmological constant, no, one can't, because the cosmological constant is constant.

However, if the accelerated expansion were caused by something that wasn't constant, for example, a scalar field--as I understand it this is considered unlikely, but mathematically it's possible--then it could be possible for that field (or whatever it is) to interact with the stress-energy in the springs, such that the field's stored energy density would change.

Also, I should stress that I have not done any math for any of these examples. I have so far been implicitly assuming that the stress-energy in the spring was negligible as far as any effect on spacetime curvature is concerned; but if we're going to talk about a possible effect of the change in the spring's energy density on the expansion history of the universe, that assumption isn't really valid any more. So we would have to solve a more complicated problem where there is both dark energy (cosmological constant or something else that causes accelerated expansion) and the stress-energy in the spring present. If we do that, things could change. For one example of what could change, see next comment.

would the Friedmann equations remain unchanged, if the universe was filled with springs compared to a universe with a mass density that equals that of the springs and the same vacuum energy density in both cases?*)

No, because the Friedmann equations assume a perfect fluid stress-energy tensor, and if springs are present instead of a uniform mass density, the stress-energy tensor cannot take the form of a perfect fluid (because a perfect fluid requires the SET to be spatially isotropic, but the springs are stretched in a particular direction so their stress-energy is not isotropic).

To my understanding covariance in GR means that the laws describing it are the same (perhaps better invariant) regardless the frame of reference of an observer.

That's "general covariance" in the sense of being able to choose any coordinates you like. But "covariant divergence" means "the actual physical divergence of the tensor, adjusted so any effects of your choice of coordinates are removed". In other words, the covariant divergence being zero is a coordinate-invariant expression of the law that no stress-energy can be created or destroyed in any infinitesimal volume of spacetime.

"it's that the changing spacetime geometry offsets the increase in the spring's energy density in just the right way to keep the covariant divergence zero" it's not really technical but still hard. Could you please elaborate a little more on this?

I can't elaborate much without taking the time to do the math in more detail. But it's similar to the way the energy density of an expanding universe decreases with the expansion. Along any comoving worldline, the energy density in an expanding universe decreases with proper time. That decrease is not offset by anything: the energy density isn't being transferred anywhere else, it's just decreasing because the universe is expanding. That can make it seem like stress-energy is being destroyed. But all that's actually happening is that the spacetime geometry is changing in concert with the energy density, in just the right way to make the covariant divergence of the stress-energy tensor zero.

The Friedmann equations themselves don't change.

They do if we take the stress-energy in the springs into account. See above.

 

[timmdeeg]

I can't elaborate much without taking the time to do the math in more detail. But it's similar to the way the energy density of an expanding universe decreases with the expansion. Along any comoving worldline, the energy density in an expanding universe decreases with proper time. That decrease is not offset by anything: the energy density isn't being transferred anywhere else, it's just decreasing because the universe is expanding. That can make it seem like stress-energy is being destroyed. But all that's actually happening is that the spacetime geometry is changing in concert with the energy density, in just the right way to make the covariant divergence of the stress-energy tensor zero.

This helps a lot!

Thanks for new insights.

 

[timmdeeg]

No, because the Friedmann equations assume a perfect fluid stress-energy tensor, and if springs are present instead of a uniform mass density, the stress-energy tensor cannot take the form of a perfect fluid (because a perfect fluid requires the SET to be spatially isotropic, but the springs are stretched in a particular direction so their stress-energy is not isotropic).

Do I understand it correctly, the mass of the springs can be considered homogeneous on scales sufficiently large, but the shear stress they are producing can't? Even if their direction is random?

The following is just to understand how shear stress acts, please correct. If the springs aren't a good example let's consider the perfect fluid model with matter density $\rho$ (i) and with matter density $\rho$ plus shear stress exerted by the matter (ii). To realize that the matter shall have elastic properties. So, in (ii) the matter should be represented by the energy density component of the SET and the shear stress in the corresponding non-diagonal components. Now, looking at the term $(\rho + 3P)$ of the acceleration equation, I suspect that the shear stress which is a source or gravitational attraction contributes positively to $P$. So, to obtain the same value for $\ddot{a}$ for both cases $\rho$ in (ii) has to be reduced such that the increase of $P$ is just compensated.
If correct so far, can these two universes with the same accelerated expansion be distinguished though? I think so, because in the first Friedmann equation $\rho$ is different and $P$ doesn't contribute to $\dot{a}$.

 

[PeterDonis]

Do I understand it correctly, the mass of the springs can be considered homogeneous on scales sufficiently large, but the shear stress they are producing can't? Even if their direction is random?

If you're averaging on a large scale over springs with random directions, so their shear stress cancels out, then there is no difference between "springs" and "uniform mass density", so your original question about this model is meaningless; you're basically asking "if the model stays the same, does the model stay the same?".

If we take a step back and generalize your question to "does shear stress affect the dynamics differently than uniform mass density?", then the answer is "yes, of course, since it leads to a different stress-energy tensor". Then it's just a matter of figuring out whether that difference is significant in a particular case.

 

[timmdeeg]

Got it, thanks.