6 independent Einstein field equations?

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Of course, the metric $\eta_{\mu\nu}$ is not a unique solution for Einstein vacuum equations compatible with your given initial data. And yes, we can interpret the alternatives as arising from coordinate functions.

Let us take the simplest of such function: redefine time by introducing new 'time' variable $\tau$ through a relation $t=f(\tau)$ (spacial coordinates we will keep intact). The metric in new coordinates $(\tau,x,y,z)$ would be $$ ds^2=(f'(\tau))^2 d\tau^2 - \delta_{ij}dx^i dx^j. $$ It is, obviously, a different metric. And by choosing the function $f$ satisfying some simple conditions ($f(0)=0$, $f'(0)=1$, $f''(0)=0$) this metric will be compatible with your initial data.

But at the same time it is equally obvious that this metric still corresponds to the same space-time - the Minkowski space-time (at least locally).

Addition. To make a solution of Einstein equations unique one can use coordinate conditions (which are analogous to gauge fixing conditions in EM theory). These work as constraints on metric imposed in addition to Einstein equations.

Also, if you are interested in initial data - time evolution formulation of general relativity, I recommend looking at the ADM formalism.

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Updated on December 13, 2020

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  • anecdote
    anecdote almost 3 years

    I can't understand the comment on page 409, Gravitation, by Misner, Thorne, Wheeler

    It follows that the ten components $G_{\alpha\beta} =8\pi T_{\alpha\beta}$ of the field equation must not determine completely and uniquely all ten components $g_{\mu\nu}$ of the metric.

    On the countrary, $G_{\alpha\beta} =8\pi T_{\alpha\beta}$ must place only six independent constraints on the ten $g_{\mu\nu}(\mathcal{P})$, leaving four arbitrary functions to be adjusted by man's specialization of the four coordinate functions $x^{\alpha}(\mathcal{P})$.

    I can't understand it. I think we can always solve the field equation with appropriate initial/boundary conditions to get unique $g_{\mu\nu}$. After all those are just second order differential equations. To be specific, let me try to construct a counter example, the vacuum Einstein equation, $$G_{\mu\nu}=0$$ If we apply the initial conditions $g_{\mu\nu}|_{t=0}=\eta_{\mu\nu}$ and ${\dot{g}_{\mu\nu}}|_{t=0} =0$, obviously the flat spacetime $g_{\mu\nu}=\eta_{\mu\nu}$ should be the solution. If the solution $g_{\mu\nu}$ is unique, what's the alternative solution?

    If there does exist an alternative solution, does it come from "specialization of the four coordinate functions"?

    Update: user23660 constructed an explicit alternative solution, which is $$ g_{00}=(f'(t))^2,\quad g_{ij}=-\delta_{ij} $$ with other components being zero.

    The function $f$ only need to satisfy $f'(0)=1,f''(0)=0$, that makes this metric compatible with the initial data; other than that, it's completely arbitrary! And we see that it does come from the coordinate transformation $t=f(\tau)$

    To get the solution to be $\eta_{\mu\nu}$, we need to put further constraints on the metric directly in this coordinate system, like $g_{00}=1,g_{0i}=0$.

    This redundant degrees of freedom(gauge) result from the contracted Bianchi identity, as explained in the following paragraph in MTW page 409, $$G^{\alpha\beta}{}_{;\beta}=0$$ is true automatically, and so the equation of motion of the matter fields $T^{\alpha\beta}{}_{;\beta}=0$ doesn't really put restrictions on the evolution of the metric. Therefore, there are only 6 independent equations!

    • joshphysics
      joshphysics almost 10 years
      I've read this question (v2) 3 times, and I must admit that I'm still unclear on what your confusion is.
    • anecdote
      anecdote almost 10 years
      @joshphysics, sorry maybe it's not properly worded. My question in simple terms is: Can be determine $g_{\mu\nu}$ uniquely by using Einstein equation? MTW states that we can't $\implies$ there must be an alternative solution to my specific example, what is it?
    • Prahar
      Prahar almost 10 years
      What about the Schwarzschild solution? Isn't that a solution to $G_{\mu\nu} = 0$? I'm sorry, I think that I, like @joshphysics am unable to understand your confusion.
    • anecdote
      anecdote almost 10 years
      @Prahar, Schwarzschild solution doesn't satisfy the initial condition $g_{\mu\nu}|_{t=0} = \eta_{\mu\nu} = (-1,1,1,1)$
    • anecdote
      anecdote almost 10 years
      @josphysics, I have rephrased my question. Sorry, things are vague in my head and I can't put it clearly in words.
    • MBN
      MBN almost 10 years
      Of course appropriate initial/boundary conditions will determine a unique solution. That is not what they are talking about. They are saying that the in some sense the ten equations are not independent, there are hidden relations, and in fact there are just six independent equations. I am of course oversimplifying things. The following two paragraphs as well as the preceding one do give some explanation. I would say don't worry about too much.
    • anecdote
      anecdote almost 10 years
      @MBN, the hidden relations are contracted Bianchi identity as pointed out in the next paragraph; but I don't see why it's related to coordinate transformation.
    • diffeomorphism
      diffeomorphism almost 10 years
      those 4 degrees of freedom are related to changes in the local frame coordinates. Remember that if you rotate your verbein from point to point you still have the same physical space, but the coordinates are labelled differently
  • anecdote
    anecdote almost 10 years
    thanks for you answer. You prove that ${\bf g}=g_{\mu\nu}dx^{\mu}dx^{\nu}$ is not unique. But if we restrict the coordinate system to be $(t,x,y,z)$, do you think the individual components solution $g_{\mu\nu}$ are unique or not?
  • user23660
    user23660 almost 10 years
    See addition. To restrict coordinate transform from my example to only produce $\eta_{\mu\nu}$ you can demand that the coordinates be synchronous. But these are conditions on the metric: $g_{00} = 1$, $g_{0i}=0$, not just coordinates.
  • anecdote
    anecdote almost 10 years
    That's exactly what am asking. So if I don't put restrictions on the metric(i.e. I don't specify the gauge), but use the $(t,x,y,z)$ coordinate to express my solution in the end, is this solution unique? The solution you gave in the $(t,x,y,z)$ coordinate is the same as $\eta_{\mu\nu}$ if expressed in $(t,x,y,z)$ coordinate system.
  • anecdote
    anecdote almost 10 years
    For the gauge condition, in E&M, I can change the vector potential $\vec{A}$ to $\vec{A}+\nabla \Lambda$ without spoiling the initial data and the equations: those are the redundant degrees of freedom, i.e. the gauge. But in this case, how can you change $g_{\mu\nu}$ to be a function $h_{\mu\nu}$ in the same coordinate system that solves the equation with the same initial conditions?
  • user23660
    user23660 almost 10 years
    'Specifying coordinates' without metric is just topology, so locally coordinates would always be a piece of $\mathbb{R}_4$, there are no restrictions on the metric, so no $\eta_{\mu\nu}$ is not unique in that case.
  • user23660
    user23660 almost 10 years
    Depending on the $\Lambda$, gauge transformation can change the initial data. So in GR, general coordinate transformations can make initial data invalid, but if we restrict them to keep them intact, they will produce other solutions. And of course, coordinate transformations will keep the same form of equations that is general covariance after all.
  • anecdote
    anecdote almost 10 years
    I agree that locally it's always $\mathbb{R}_4$, there is no restrictions on the metric locally, mathematically all those metrics in the matrix forms are connected by a congruent transformation. Note that my initial conditions do put restrictions on the global metric, that should make the evolution of the metric unique in this coordinate system. So what is the alternative solution in this coordinate system if not unique?
  • user23660
    user23660 almost 10 years