I would like to understand what are the degrees of freedom in GR. I have read a few previous posts already, but none of them really help me. Below, I will try to write down the entangled web of thoughts I have/know and hopefully someone we help me reach epiphany.

Let us consider the vacuum Einstein's field equation (EFE). It has 10 independent equations (naively I guess...). The metric itself has 10 independent components (naively?) so all seems good. But now, trouble starts...

1) First, due to general covariance, 4 components of the metric are really redundant. That is, if some $g_{\mu \nu}$ solve the EFE, then $g_{\mu \nu} \rightarrow g'_{\mu \nu}$ will also solve it. So, actually, the EFE under-determine the metric!? There is a first confusion here. To me, it is a bit like saying: here we have a system of 10 polynomial equations with 10 unknowns, but the system is nonetheless under-determined.
I mean, if we look at say some standard PDE's like the wave equation, for appropriate initial data, we have a unique solution.
Are we then saying that even given appropriator initial data, the EFE are still under-determined?? Even though the number of unknowns and equations match? Or are we strictly speaking about the physical "interpretation" of the solution? In the sense that, given a solution, say in cartesian coordinates, we could translate it into polar coordinates, which would not "really" change anything, only the individual numerical values, but the description of the solution is still the same? In this case, cant we say that the wave equation is then also under-determined since we could play that same game here as well?

2)Ok, now let us say I am fine with the above. We end up with 6 "really" independent components of the metric, for still 10 equations. But, now, somehow, 4 equation of EFE are actually redundant because of the bianchi identity. I dont see how this makes sense. The Bianchi identity is,well, an identity. So it always holds, and therefore in my eyes it cannot provide with any additional constraints. More directly, if we start if $G_{\mu \nu} = 0$, which ones should we remove? and how do we recover me? (which we should be able to since they are supposedly redundant). For example, if we choose to "forget" about the first one, i.e. $G_{00}=0$, how can we recover it? From the Bianchi, we have: $\nabla_t G^{00} + \nabla_i G^{i0} = 0$, but that is not enough (even with all the other EFE) to deduce $G_{00} = 0$? In fact, with the EFE, we would have $\nabla_t G^{00} = 0$, but we can't get any further?

3)Now, even at this stage, I dont see how we get to 2 degree of freedom. I usually hear like diffeomorphism invariance removes 4 dof (as above), but Bianchi removes another 4?? However, Bianchi "at best" removes redundant EFE and does not "touch" the metric like diffeomorphism invariance does??

4) Finally, this was for the vaccum equation. Now, if we add a non-trivial stress-energy tensor, things get worse, since this looks like one is "adding" even more unknowns to the EFE. Even with Equations of state, one typically introduces more unknowns (for example for perfect fluid, pressure and 4-velocity). How does this work out? I hear that you introduce the continuity equation $\nabla_\mu T^{\mu \nu}=0$, but this is a direct consequence of the EFE, and therefore are not additional independent equations! So we still have only 10 equations!?

Thank you if you managed to read all this and thanks in advance for your responses.

  • $\begingroup$ I think you need to look at the initial value problem in GR. Or you need someone to write a summary here. $\endgroup$
    – MBN
    Commented Oct 18, 2018 at 14:37
  • $\begingroup$ These are good questions, but I agree with @MBN. There's just too much here for me to answer in a reasonable timeframe. I suggest reading through Wald's chapter 10 to understand the initial-value formulation. Also, the Bianchi identities are consistent with the EFE (including the stress-energy tensor); they don't further restrict the solutions to the EFE. A simple way of thinking about this is that $G^{\mu \nu} = 0$ is a stronger condition on the metric than $\nabla_\mu G^{\mu \nu} = 0$, so no additional information is gained from the latter. $\endgroup$
    – Mike
    Commented Oct 18, 2018 at 14:45
  • $\begingroup$ Thanks Mike. I will have a look at Wald then. I will see if this clears things up. $\endgroup$
    – Patrick.B
    Commented Oct 18, 2018 at 15:00
  • $\begingroup$ More on DOF in GR. $\endgroup$
    – Qmechanic
    Commented Oct 18, 2018 at 16:13

1 Answer 1


This is cleaner in the ADM formulation than it is directly from the Einstein equation.

If you're not familiar with the ADM formulation, it is done by converting the Hilbert Lagrangian to a constrained (and vanishing) hamiltonian, and getting Hamilton's equations out of it.


If you do this, you will get four "gauge" degrees of freedom which derive from the coordinate system of the underlying spacetime. Associated with these degrees of freedom, you will get four constraints. What will be left over will be twelve first-order equations in the 3-metric and its first derivative. Since we usually talk about degrees of freedom in terms of second-ordere equations, and you can always convert a system of equations of the form:

$$\begin{align} {\dot q_{i}} &= f_{i}(q,p)\\ {\dot p_{i}} &= g_{i}(q,p) \end{align}$$

into a set of equations of the form:

$${\ddot q_{i}} = h_{i}(q, {\dot q})$$

Then, given a fixed choice for your coordinate system (which is equivalent to choosing a fixed gauge in E&M), you have six second-order equations for the 3-metric, and four constraints, leaving you with two local degrees of freedom.

  • $\begingroup$ I see. I know a bit about the ADM formulation, but your explanation is rather helpful. If I try to put a parallel with what wrote above, the 4 constraints you mention come from the 4 diffeomorphism yes? How about the Bianchi identity? Does it have an "equivalent" in the ADM formulation? $\endgroup$
    – Patrick.B
    Commented Oct 19, 2018 at 17:29
  • $\begingroup$ @Patrick.B: you can think of them as arising from the diffeomorphisims, but fundamentally, it's the fact that you have four free functions arising as coordinate choices in your Hamiltonian, so their variations create constraints. Your EOM can't depend on perturbations of gauge. The Bianchi identify is going to be obscured in a 3+1 formulation, though, to my knowledge at least. $\endgroup$ Commented Oct 19, 2018 at 19:59
  • $\begingroup$ As in, the constraints are the variations of the lapse function and the shift vector, which ultimately give you the $g_{t\mu}$ components of the 4-metric, and are governed by your choice of 3-surfaces and time component in the 3+1 formulation. $\endgroup$ Commented Oct 19, 2018 at 20:00
  • $\begingroup$ I agree with the answer but not sure I agree with the logic. In the ADM formalism, the four constraints relate the 12 phase space degrees of freedom (not the 6 configuration space degrees of freedom). So initially by your logic we have 12 phase-space degrees of freedom (6 from the 3-metric, and 6 from the conjugate momenta) and the constraints reduce this to 8 independent phase space degrees of freedom (correspond to 4 d.o.f. in configuration space). It is also the case that one can remove 3 d.o.f. by choice of spatial coordinates. $\endgroup$ Commented Sep 20, 2021 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.