Validity of Birkhoff's theorem in flat Minkowski spacetime

  • #1
Volterra
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TL;DR Summary
I would like to know/discuss, if Birkhoffs theorem is valid in flat Minkowski spacetime
I have a theoretical question regarding the validity of Birkhoff's theorem in flat Minkowski spacetime (noting that this is a hypothetical scenario, as mass curves spacetime).

Common derivations of Birkhoff's theorem are based on the assumption that the Ricci tensor
R_ij = R_kij^k (Einstein summation convention)
must vanish to satisfy Einstein's vacuum solution.

In flat spacetime, all components of the Riemann curvature tensor trivially vanish. Consequently, all components of the Ricci tensor in flat spacetime also vanish, regardless of any potential time dependence.
From this, one might conclude that Birkhoff's theorem is not applicable or meaningful in flat spacetime.
Is this conclusion correct, or am I overlooking an important aspect? How is this viewed in the forum?
 
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  • #2
Birkhoff's theorem just says that the only spherically symmetric static vacuum spacetime is Schwarzschild. Flat spacetime satisfies Birkhoff's theorem, which is fine because Minkowski spacetime is Schwarzschild spacetime with ##M=0##.
 
  • #3
Volterra said:
Common derivations of Birkhoff's theorem are based on the assumption that the Ricci tensor
R_ij = R_kij^k (Einstein summation convention)
must vanish to satisfy Einstein's vacuum solution.
It's not an assumption, it's a definition. "Vacuum solution" is defined as the stress-energy tensor being zero, hence the Einstein tensor is zero by the Einstein Field Equation, hence the Ricci tensor is zero.

Volterra said:
In flat spacetime, all components of the Riemann curvature tensor trivially vanish. Consequently, all components of the Ricci tensor in flat spacetime also vanish
Yes, which means flat spacetime is a vacuum solution.

Volterra said:
regardless of any potential time dependence.
What does time dependence have to do with anything? No time dependence is assumed in the derivation of Birkhoff's Theorem.

The fact that there is a fourth Killing vector field (in addition to the three that come with spherical symmetry) in a spherically symmetric vacuum spacetime is a conclusion of the theorem, not an assumption. Flat spacetime satisfies that conclusion of the theorem. (Note that the theorem does not say that the fourth KVF must be timelike everywhere, so even its presence does not necessarily say anything about "time dependence".)

Volterra said:
From this, one might conclude that Birkhoff's theorem is not applicable or meaningful in flat spacetime.
I don't see why. See above.
 
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  • #4
Thank you very much for the answers

I would like to refine my questions:
  1. Is it correct that Birkhoff's approach is based on the definition R_ij = R^k_ijk := 0?
  2. Does Birkhoff derive the time independence of the radial component of the Schwarzschild tensor from the condition R_10 = 0, thereby establishing the static nature of the Schwarzschild metric?
  3. Is it accurate that in flat spacetime, the condition R_10 = 0 is always satisfied by definition, even for time-dependent radial components of the metric tensor? Would this imply the possibility of dynamics for this trivial solution (M=0) of the Schwarzschild metric?

Thank you for the support


@Ibix: I agree that Minkowski spacetime corresponds to the Schwarzschild solution for M=0.

@PeterDonis: I agree in the first and second topic.
- My concern relates to the time independence of the Schwarzschild metric and the extent to which this is justified by Birkhoff's theorem.
 
  • #5
No there are no dynamics in M=0 Schwarzschild, no matter how much you play with the metric (e.g. if you reparameterise to introduce time dependency). It will still be Ricci=0
 
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  • #6
Volterra said:
Is it correct that Birkhoff's approach is based on the definition R_ij = R^k_ijk := 0?
I already answered this in post #3.

Volterra said:
Does Birkhoff derive the time independence of the radial component of the Schwarzschild tensor
What is the "Schwarzschild tensor"?

"Time independence" (not quite an accurate term since, as I pointed out in post #3, the KVF in question does not have to be timelike) is shown for the entire metric by Birkhoff's theorem. It is not shown component by component, it is shown for the entire metric all at once.

Volterra said:
Is it accurate that in flat spacetime, the condition R_10 = 0 is always satisfied by definition, even for time-dependent radial components of the metric tensor?
Your question doesn't make sense. Flat spacetime is a vacuum solution, as I said in post #3, so the entire Ricci tensor vanishes by definition. Also, there are no time-dependent components of the metric tensor in flat spacetime, which is obvious just by looking at the line element.

Volterra said:
My concern relates to the time independence of the Schwarzschild metric and the extent to which this is justified by Birkhoff's theorem.
I already answered this in post #3, and again above.
 
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  • #7
Thank you very much again for your answers, for which I am very grateful.

PeterDonis said:
Your question doesn't make sense. Flat spacetime is a vacuum solution, as I said in post #3, so the entire Ricci tensor vanishes by definition. Also, there are no time-dependent components of the metric tensor in flat spacetime, which is obvious just by looking at the line element.
I also see my question as a hypothetical one.
I think that Birkhoff's theorem would not be applicable in a flat spacetime, since R_ij = 0 by definition of flat spacetime.
Is this correct?
(I am aware that this question is only conditionally physically relevant. The answer would still be very helpful for me.)
 
  • #8
why not? flat spacetime is vacuum and spherically symmetric, birkhoff is satisfied trivially
 
  • #9
Volterra said:
I also see my question as a hypothetical one.
I don't know why. There's nothing hypothetical about it.

Volterra said:
I think that Birkhoff's theorem would not be applicable in a flat spacetime, since R_ij = 0 by definition of flat spacetime.
Is this correct?
No. As you have already been told in this thread, more than once. And that's enough. This thread is now closed.
 
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