Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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!

share|improve this question
    
I've read this question (v2) 3 times, and I must admit that I'm still unclear on what your confusion is. –  joshphysics Dec 12 '13 at 2:36
    
@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? –  anecdote Dec 12 '13 at 2:52
    
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. –  Prahar Dec 12 '13 at 3:23
    
@Prahar, Schwarzschild solution doesn't satisfy the initial condition $g_{\mu\nu}|_{t=0} = \eta_{\mu\nu} = (-1,1,1,1)$ –  anecdote Dec 12 '13 at 3:28
    
@josphysics, I have rephrased my question. Sorry, things are vague in my head and I can't put it clearly in words. –  anecdote Dec 12 '13 at 3:53

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
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? –  anecdote Dec 12 '13 at 15:00
    
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. –  user23660 Dec 12 '13 at 17:04
    
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 Dec 12 '13 at 17:12
    
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? –  anecdote Dec 12 '13 at 17:17
    
'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 Dec 12 '13 at 17:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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