Degrees of Freedom in Linearized Gravity (and beyond)

In summary, the conversation discusses the reasoning behind GR having only two polarizations. It is explained that the gauge conditions imposed in linearized terms reduce the degrees of freedom from 10 to 2. This concept is compared to electromagnetism, where one degree of freedom in the gauge transformation is "used twice", resulting in a reduction from 4 to 2 components. The conversation also touches on the restricted gauge transformation and how it further simplifies the equations on the mass shell. It is suggested to check Zwiebach's string theory book for a more detailed treatment on this topic.
  • #1
NanakiXIII
392
0
Dear all,

This is a continuation of my previous thread, but I figure it's cleaner to start afresh with this topic.

I'm trying to understand why GR has only two polarizations. I've only seen treatments of this in linearized terms, so I'll start there. The reasoning is often as follows (e.g. in Carroll's notes):

You can impose harmonic gauge conditions

[tex]
\square x^\mu = 0 \Leftrightarrow \partial_\nu \bar{h}^{\mu\nu} = 0
[/tex]

which costs you four degrees of freedom. However, since this does not completely specify your coordinates, you can still perform transformations of the form

[tex]
x^\mu \to x^\mu + \delta^\mu : \square \delta^\mu = 0
[/tex]

which tells you you lose an additional four degrees of freedom.

I find this highly suspect. The harmonic coordinate conditions do not uniquely determine your coordinates, so you've not yet specified anything, only constrained the set of coordinate systems you want to use. Then the second equation actually specifies a unique set of coordinates within this restricted set. It seems to me that only after doing both have you actually used up your four degrees of freedom.

What am I misunderstanding here?
 
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  • #2
You should look at what fixes your metric components. Fixing some of your metric components does not completely fix your coordinate choice, which you should check with an explicit calculation. A similar thing happens in e.g. U(1) gauge theories, so maybe it helps to compare. Hinterbilcher's notes on massive gravity can also be useful.
 
  • #3
I'm trying to understand why GR has only two polarizations.
Do you understand how it happens for electromagnetism? Namely, how one degree of freedom in the gauge transformation seems to be "used twice", reducing Aμ from four components to just two. It's exactly analogous to linearized gravity, but the algebra is simpler.

The equation of motion that comes out of the Lagrangian is not just the wave equation, it's ◻2A - ∇∇·A = 0. The gauge freedom is A → A + ∇λ. (Easier to Fourier transform everything and write them as k·k A - k k·A = 0 and A → A + k λ.)

We want to use the gauge freedom to satisfy the Lorenz condition k·A = 0, so we look at how k·A transforms: k·A → k·A + k·k λ. Sure enough, we can set k·A = 0 by choosing λ = - k·A/k·k. This works but only if k·k ≠ 0. However for k·k = 0 we can go back to the equation of motion and see that k·A was zero automatically. So we have completely satisfied the Lorenz gauge condition.

What about λ for the k·k = 0 case? ("on the mass shell") We still have not used that freedom. This is called the restricted gauge transformation. We can use it to further simplify A, although not in a Lorentz-invariant fashion. For example, we can set the time component A0 = 0.

All of this may already be familiar, but the gravitational case works exactly the same way. We use the gauge freedom hμν → hμν + ξμ,ν + ξν,μ to satisfy the Hilbert gauge condition. That reduces hμν from ten components to six. On the mass shell, k·k = 0, the gauge freedom is ineffective, but the condition is already satisfied from the Einstein equations. We then use restricted gauge transformations to reduce hμν to just two components on the mass shell.
 
  • #4
Btw, Zwiebach has a very detailed treatment about this in his string theory book, I believe chapter 10. He does basically every step very explicit, making it clear how to gauge away the 'gauge degrees of freedom' and which gauge transformations preserve the choice of gauge. That should answer your question :)
 
  • #5
Hey guys, thanks, this seems to be the direction I'm looking for, but I still have some questions.

Bill_K said:
Do you understand how it happens for electromagnetism? Namely, how one degree of freedom in the gauge transformation seems to be "used twice", reducing Aμ from four components to just two. It's exactly analogous to linearized gravity, but the algebra is simpler.

The equation of motion that comes out of the Lagrangian is not just the wave equation, it's ◻2A - ∇∇·A = 0. The gauge freedom is A → A + ∇λ. (Easier to Fourier transform everything and write them as k·k A - k k·A = 0 and A → A + k λ.)

We want to use the gauge freedom to satisfy the Lorenz condition k·A = 0, so we look at how k·A transforms: k·A → k·A + k·k λ. Sure enough, we can set k·A = 0 by choosing λ = - k·A/k·k. This works but only if k·k ≠ 0. However for k·k = 0 we can go back to the equation of motion and see that k·A was zero automatically. So we have completely satisfied the Lorenz gauge condition.

What about λ for the k·k = 0 case? ("on the mass shell") We still have not used that freedom. This is called the restricted gauge transformation. We can use it to further simplify A, although not in a Lorentz-invariant fashion. For example, we can set the time component A0 = 0.

All of this may already be familiar, but the gravitational case works exactly the same way. We use the gauge freedom hμν → hμν + ξμ,ν + ξν,μ to satisfy the Hilbert gauge condition. That reduces hμν from ten components to six. On the mass shell, k·k = 0, the gauge freedom is ineffective, but the condition is already satisfied from the Einstein equations. We then use restricted gauge transformations to reduce hμν to just two components on the mass shell.

I'm not sure I understand this. You want to impose a gauge condition, but looking at the equations of motion you find that (on the mass shell) this condition is already satisfied? It sounds to me like this gauge invariance is not even present then, given the equations of motion.

haushofer said:
Btw, Zwiebach has a very detailed treatment about this in his string theory book, I believe chapter 10. He does basically every step very explicit, making it clear how to gauge away the 'gauge degrees of freedom' and which gauge transformations preserve the choice of gauge. That should answer your question :)

I had a look at these notes and they look highly useful, but they leave me with the same question as did Bill's answer. He uses the gauge invariance to impose some condition, but then finds that the simplified equations of motion impose some additional constraint. Does this latter elimination really have anything to do with gauge invariance?
 
  • #6
Anyone?
 
  • #7
Actually, I've managed to clear up a false assumption I was hanging onto when thinking about all this. I believe I have an understanding now, but I have a follow-up question:

While the number of degrees of freedom in linearized gravity are abundantly described, I'm wondering whether the reasoning carries over to non-linear descriptions, e.g. full GR or the Post-Newtonian formalism. It is certainly suggested that they do, but does anyone have a reference that proves it?
 

1. What are degrees of freedom in linearized gravity?

Degrees of freedom in linearized gravity refer to the number of independent parameters needed to describe the gravitational field. In linearized gravity, these degrees of freedom correspond to the number of gravitational waves that can be produced by a given system.

2. How many degrees of freedom are there in linearized gravity?

In linearized gravity, there are two degrees of freedom, corresponding to the two possible polarizations of gravitational waves. This is in contrast to general relativity, where there are six degrees of freedom due to the non-linear effects of gravity.

3. What is the significance of degrees of freedom in linearized gravity?

The degrees of freedom in linearized gravity play a crucial role in understanding the behavior of gravitational waves and the effects of gravity on a small scale. They also help us to better understand the nature of gravity and its interactions with matter.

4. How do degrees of freedom in linearized gravity differ from those in other theories?

The number of degrees of freedom in linearized gravity depends on the specific theory being studied. In general relativity, there are six degrees of freedom, while in other theories such as string theory, the number of degrees of freedom may vary. However, the concept of degrees of freedom remains important in all theories as a way to understand the behavior of physical systems.

5. Can degrees of freedom in linearized gravity be observed?

Yes, degrees of freedom in linearized gravity can be observed through the detection of gravitational waves. By measuring the polarization of these waves, we can determine the number and nature of degrees of freedom in a given system. This can provide valuable insights into the behavior of gravity on a small scale and help us to test different theories of gravity.

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