# I What constrains the metric tensor field in GR?

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1. Jul 31, 2017

Do the field equations themselves constrain the metric tensor? or do they just translate external constraints on the stress-energy tensor into constraints on the metric tensor?
another way to ask the question is, if I generated an arbitrary differentiable metric tensor field, would it translate through the field equations to some stress-energy tensor field (and Weyl tensor field) that are physically valid?

The reason I am confused is because, while the field equation implies a continuity constraint (the covariant divergence is zero) and that constraint provides several important conservation laws, this constraint comes from a modification of the Bianchi identity, which itself just represents a symmetry of the Riemann curvature tensor, which exists for any differentiable metric field. So I don't see how any metric tensor field could ever breach the continuity constraint.

The same goes for the equation that constrains the derivative of the Weyl tensor, it is based on the Bianchi identity, which should be a property of (rather than a constraint on) the Riemannian curvature tensor.

But the metric tensor must be constrained, otherwise I'm sure it would be easy to generate a metric tensor field that involved a mass appearing out of nowhere, so breaching the conservation of mass.

Many thanks

2. Jul 31, 2017

### vanhees71

This is like in any gauge theory: The gauge invariance of the field enforces constraints on the possible sources.

You know this from classical electrodynamics: From the Maxwell equations alone, i.e., not taking into account the equations of motion for the matter-degrees of freedom the electromagnetic current must be conserved, $\partial_{\mu} j^{\mu}=0$.

The same holds for the Einstein-Hilbert equations of the gravitational field,
$$G^{\mu \nu} = R^{\mu \nu}-\frac{1}{2} R g^{\mu \nu}=-\frac{8 \pi G}{c^4} T^{\mu \nu}.$$
The tensor on the left-hand side fulfills the Bianchi identity
$$\nabla_{\mu} G^{\mu \nu}=0$$
and thus the dynamics of the matter-degrees of freedom must necessarily fulfill the constraint
$$\nabla_{\mu} T^{\mu \nu}=0$$
for the energy-momentum tensor.

3. Jul 31, 2017

### robphy

Appropriate Causality conditions on the spacetime and Energy conditions on the stress-energy tensor are imposed to help capture the notion of "physically valid".

4. Jul 31, 2017

Hi Vanhees71. I understand the zero covariant divergence equations that you have given, but these are true from the Bianchi identity, and from my understanding this is just a symmetry of the Riemannian curvature tensor, not a constraint on the metric tensor field. I think that for any differentiable metric tensor field the Bianchi identity should hold for its corresponding Riemannian curvature tensor field. Am I wrong on this?

Hi robphy. I have read about the idea of globally hyperbolic being a condition that requires all time-like curves to extend the length of the space-time section of study. I can accept this as a constraint on the metric... but I think it is quite a 'loose' constraint in that it wouldn't be the main equation shaping the metric tensor field. Is your energy condition on the stress-energy tensor the condition that the determinant of the stress-energy tensor be non-negative? If so, then yes, I accept that this constrains the metric tensor field to some extent, but again, it is more of a loose contraint.

5. Aug 1, 2017

### vanhees71

No the Bianchi identity is, as its name suggests, an identity for the metric components. It's rather a constraint for the equations of motion of the matter degrees of freedom (generalized coordinates of point particles or fields like the em. field), i.e., the laws must be such that the covariant divergence of the energy-momentum tensor vanishes. It's an integrability condition for the Einstein equations.

6. Aug 1, 2017

Yes, so it constrains what the stress-energy tensor field can be, but it doesn't doesn't constrain the metric tensor itself.

But here is my problem. Let's define a metric tensor field that is flat for t<10 and is curved for t>11 to represent the existence of a mass. Plus a small smoothness transition between t=10 and t=11. Surely that metric tensor field represents a mass appearing from nowhere, and therefore a breach of the conservation of mass and so a breach of the Bianchi condition.

Sorry if this is missing something obvious.

7. Aug 1, 2017

### vanhees71

Then you get an energy-momentum tensor whose covariant divergence is not vanishing and thus cannot be on the right-hand side of the Einstein equations since this would be a self-contradictory equation, which cannot have a solution.

8. Aug 1, 2017

9. Aug 1, 2017

### RockyMarciano

In the specific case of general curved spacetimes of GR it is not quite the integrability condition one could expect in the absence of curvature(one can no longer find a 4-velocity $v^a$ with $v^a v_a=-1$ and $\nabla_{(a} v_{b)}=0$ ) in which case the conservation laws would really be global and there'd be integrability. It however guarantees the local conservation essential to any physical theory.

10. Aug 1, 2017

### RockyMarciano

As others have mentioned they don't constrain the spacetimes physically all that much, one must recurr to ad hoc conditions to limit the spacetimes considered physical.
The physicality demand for a spacetime might require more than just the continuity one. On the other hand the differential Bianchi identity(or second Bianchi identity) that you refer to is not an identity specific to the curvature of Riemannian metrics, it is more general, it already holds for the curvature of any affine connection without metrics involved or for curvature of connections on principle bundles so it is not constraining the metrics themselves but the manifold connection and differentiability(and therefore continuity).
This particular case would be incompatible witht the continuity constraint, it would violate the Bianchi identity and as vanhees said it would not fulfill the EFE and would not have a local conservation law.
See the book by Wald, page 456 for the relation of the second Bianchi identity and general covariance/diffeomorphism invariance as seen in the GR lagrangian.

11. Aug 1, 2017

Thanks for the helpful replies everyone! I appreciate it. So the zero covariant divergence of the Einstein tensor occurs regardless of the 'shape' of the metric tensor field.

If we have a coordinate chart t,x,y,z and an arbitrary (non-flat) symmetric metric tensor field over it, and scale this tensor field by max(0,t), then the stress-energy tensor for t<0 is by definition zero and the metric tensor for t>0 is by definition non-zero, so surely this implies an energy-density (and pressure) have appeared out of nowhere, violating the zero covariant divergence condition.

I think I will need to read this Wald book to understand what is wrong with my counter-example.

12. Aug 1, 2017

### Staff: Mentor

Then the metric tensor will vanish for $t < 0$, which violates the requirement that the metric must be locally Lorentzian--i.e., at any event it must be possible to find a coordinate transformation that puts the metric into the Minkowski form at that event.

What you are really talking about here is the Einstein tensor--that is what should vanish for $t < 0$ and not vanish for $t > 0$ in your example. So let's suppose that that is possible. Then the question becomes: what is the Einstein tensor at $t = 0$? Is it possible to smoothly connect the flat (vanishing Einstein tensor) and curved (non-vanishing Einstein tensor) regions through the spacelike hypersurface at $t = 0$? If so, what does that imply, physically, about that hypersurface?

I have not tried to construct such a solution, but my guess would be that it is not possible--that the Einstein tensor would not be well-defined on the $t = 0$ hypersurface.

13. Aug 1, 2017

Sorry, you're right, bad counter-example. If you add the lorentz tensor diag(-1,1,1,1) over the whole field you get what I meant.
i.e. for the metric tensor, Lorentz flat becomes gradually non-flat after t>0.

These replies have been helpful, they suggest that if I actually tried this counter-example it would (against my intuition) still have zero covariant divergence.
I will need to work through the maths of this, and convince myself.

robphy, those links are very useful, thanks!

Last edited: Aug 1, 2017
14. Aug 1, 2017

### Staff: Mentor

Either that or it would be impossible to construct a well-defined solution at all.

15. Aug 2, 2017

### martinbn

It is possible, but the resulting solution will not be vacuum (for $t>0$) and the stress energy tensor will violate the energy conditions.

16. Aug 2, 2017

### Staff: Mentor

On consideration, I think you're right, but let me try to formulate the question in more detail to make clear what I take to be the objection being implicitly raised by @TGlad :

Consider a small neighborhood of the origin, i.e., the point $(t, x, y, z) = (0, 0, 0, 0)$, and suppose we are evaluating the covariant divergence of the stress-energy tensor at that point. To do that, heuristically, we look at the boundary of the small neighborhood and add up all of the stress-energy crossing that boundary; the sum total should be zero. To make things simpler conceptually, instead of a spherical boundary, consider the boundary to be a hypercube, with two timelike faces (one past face for $t < 0$ and one future face for $t > 0$) and six spacelike faces, which can be imagined, heuristically, as the faces of an ordinary cube centered on the spatial origin at $t = 0$.

By hypothesis, the SET vanishes for $t < 0$ and does not vanish for $t > 0$. So there is no stress-energy at all crossing the $t < 0$ timelike face of the boundary, but there is stress-energy crossing the $t > 0$ timelike face in the outward direction. I think this is what @TGlad was imagining, and I think he was implicitly assuming that there was no other stress-energy crossing the boundary, which would mean the covariant divergence would not vanish.

However, it is still possible for the total sum to vanish, but for it to do so, there must be stress-energy crossing the spacelike faces in the inward direction, just sufficient to balance the stress-energy crossing outward in the future timelike direction. If, for simplicity, we take the SET to be diagonal (for example, if it is a perfect fluid), this stress-energy will appear as a pressure gradient on the little cube centered on the spatial origin that appears suddenly at $t = 0$. Note that this is not just pressure--it is a pressure gradient, i.e., the pressure must change across the faces of the cube. (I think that if you work out the signs, the pressure itself turns out to be negative, i.e., it's tension, and its magnitude must indeed be enough to violate the energy conditions.)

I think what I've just described is mathematically possible, but I'm not sure it's physically reasonable.

17. Aug 2, 2017

I think that has been helpful, let me express it more from the metric tensor perspective:

The claim is that any metric tensor field gμν(t,x,y,z) has zero covariant divergence of the Einstein and stress-energy tensor.
Counter-example attempt 1:
gμν(t,x,y,z) = diag(-1 + f(t,x,y,z)*max(0,t),1,1,1)
where f is a scalar valued function of the local coordinate chart t,x,y,z.

This does not cause the appearance of a non-zero stress-energy tensor at t>0 due to gauge invariance. i.e. the physics is invariant to g00 changing scale over the manifold. (Am I right?).
Counter-example attempt 2:
gμν(t,x,y,z) = diag(-1 + f(t,x,y,z)*max(0,t),1+ g(t,x,y,z)*max(0,t),1+ h(t,x,y,z)*max(0,t),1+ i(t,x,y,z)*max(0,t))

This does not cause the appearance of a non-zero stress-energy tensor at t>0 for the same reason. Due to gauge invariance.

Counter-example attempt 3:
gμν(t,x,y,z) = diag(-1,1,1,1), g01(t,x,y,z) = g10(t,x,y,z) = max(0,t)2

This doesn't just change an eigenvalue but also changes the first eigenvector. There should be the appearance of a non-zero stress-energy tensor after t=0.

However, as PeterDonis said, this may be explainable by 'sideways flux' to account for the positive t side temporal flux. As he said, this would still be zero covariant divergence, but would require a pressure that may violate certain energy conditions. In other words, it isn't the field equation (or the Bianchi identity) that constrains the metric tensor, it is energy conditions (and equations of state).

18. Aug 2, 2017

### Staff: Mentor

Zero covariant divergence of the Einstein tensor is a mathematical identity (called the Bianchi identity--or more precisely the contracted Bianchi identity, since a similar identity is also true of the Riemann tensor). It's a consequence of how the Einstein tensor is computed from the metric tensor. So there is no way to construct a metric tensor with an Einstein tensor that violates it.

Zero covariant divergence of the stress-energy tensor is a consequence of the Einstein Field Equation. So it will be satisfied by any stress-energy tensor which is a valid solution of the Einstein Field Equation. Obviously one could construct an arbitrary 2nd-rank tensor with nonzero covariant divergence; but then you will find that there is no possible metric tensor whose Einstein tensor will be equal to it.

For all of your possible counterexamples, you are looking at the metric tensor, not the Einstein tensor. You need to actually compute the Einstein tensor for your examples. (Note that you don't have to worry about gauge freedom, that just corresponds to a choice of coordinate chart, which you've already made. Just turn the crank and compute the Einstein tensor for each of your examples. Or use a package like Maxima to do it.)

Yes, in the sense that, while you can always write down an arbitrary metric tensor, compute its Einstein tensor, and call that the "stress-energy tensor", there is no guarantee that this will result in something that has a reasonable physical interpretation. The energy conditions, equations of state, etc., are attempts to make more precise what we mean by "a reasonable physical interpretation".

19. Aug 2, 2017

I know, I'm not claiming this is wrong, I just want to convince myself why it is impossible. Because my intuition is saying "make a metric tensor field that is Lorentz flat which then goes curvy and surely the stress-energy tensor will magically become nonzero after t>0".

what I'm saying in response to counterexample 1 is the same thing I think. If you replace coordinate t with -∫(-1+f(t,x,y,z)min(0,t))dt then I think the counter example 1 metric tensor field becomes the flat diag(-1,1,1,1), therefore there cannot be covariant divergence as the counterexample is isomorphic(?) to a Lorentz flat tensor field.

I think this exactly answers my original question, so thanks, it has been very helpful! really. (I added the bit in parentheses and the word field in a couple of places)

20. Aug 2, 2017

### Staff: Mentor

Then I think you should dig into how the Bianchi identity is proven, which will show you the properties of the Einstein tensor that guarantee that its covariant divergence must be zero. For example, MTW has a good discussion about visualizing the Bianchi identity in terms of the geometric fact that "the boundary of a boundary is zero".