Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Degrees of Freedom in Linearized Gravity (and beyond)

  1. Mar 19, 2013 #1
    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

    \square x^\mu = 0 \Leftrightarrow \partial_\nu \bar{h}^{\mu\nu} = 0

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

    x^\mu \to x^\mu + \delta^\mu : \square \delta^\mu = 0

    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?
  2. jcsd
  3. Mar 22, 2013 #2


    User Avatar
    Science Advisor

    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.
  4. Mar 22, 2013 #3


    User Avatar
    Science Advisor

    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.
  5. Mar 22, 2013 #4


    User Avatar
    Science Advisor

    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 :)
  6. Mar 27, 2013 #5
    Hey guys, thanks, this seems to be the direction I'm looking for, but I still have some questions.

    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.

    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?
  7. Apr 5, 2013 #6
  8. Apr 7, 2013 #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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Similar Threads for Degrees Freedom Linearized
B Is the speed of light a degree of freedom?