Wald synchronous reference frame proof

[cianfa72]

TL;DR

About the Wald's proof of existence of synchronous reference frame in a finite region of any spacetime

Hi, on Wald's book on GR there is a claim at pag. 43 about the construction of synchronous reference frame (i.e. Gaussian coordinate chart) in a finite region of any spacetime. In particular he says: $$n^b\nabla_b (n_aX^a)=n_aX^b\nabla_b \, n^a$$Then he claims from Leibnitz rule the above equals to $$\frac 1 2 X^b\nabla_b (n^an_a)$$ Why the above is true? Thanks.

 

[martinbn]

$n^an_a$ is the inner product, which is symmetric. You have $\frac12 \nabla (V\cdot V) =\frac12(\nabla V \cdot V+ V\cdot\nabla V) = V\cdot\nabla V$

 

[Orodruin]

… assuming the connection is metric compatible, which is the case in GR as the connection is taken as the Levi-Civita connection.

 

[cianfa72]

which is the case in GR as the connection is taken as the Levi-Civita connection.

$$\frac 1 2 X^b\nabla_b (n^an_a) = \frac 1 2 n_aX^b\nabla_b \, n^a + \frac 1 2 n^aX^b\nabla_b (g_{ac}n^c)$$The second term on RHS is:$$\frac 1 2 n^an^c X^b\nabla_b \, g_{ac} + \frac 1 2 n^ag_{ac}X^b\nabla_b \, n^c$$ $\nabla_b \, g_{ac}=0$ by definition of Levi-Civita connection, hence $$\frac 1 2 n^aX^b\nabla_b (g_{ac}n^c)=\frac 1 2 n_cX^b\nabla_b \, n^c $$ Therefore we get$$\frac 1 2 X^b \nabla_b (n^an_a)=n_aX^b\nabla_b \, n^a$$Is the above correct ? Thanks.

 

[Orodruin]

Is the above correct ?

Looks ok on a cursory glance.

 

[cianfa72]

According to Wald, the possibility of constructing a synchronous reference frame in a finite region of any spacetime, basically amounts to the existence of a coordinate chart path such that $g_{00}=1$ and $g_{0\alpha}=0, \alpha =1,2,3$.

Therefore for a finite region of any spacetime there is a coordinate transformation that bring the metric tensor components in that form.

 

[cianfa72]

In the The Classic Theory of Field from Landau - Lifshitz, there is a claim at p. 240 (section 85) where they say there exists a linear transformation defined at each point in spacetime such that the metric tensor transforms in galilean (inertial) form.

They claim that such linear transformation brings the metric in galilean form and at the same time makes the Christoffel symbols vanish.

My question is: doesn't just bringing the metric tensor in galilean form at a point make automatically the Christoffel symbols vanish ? Thank you.

 

[Orodruin]

In the The Classic Theory of Field from Landau - Lifshitz, there is a claim at p. 240 (section 85) where they say there exists a linear transformation defined at each point in spacetime such that the metric tensor transforms in galilean (inertial) form.

They claim that such linear transformation brings the metric in galilean form and at the same time makes the Christoffel symbols vanish.

My question is: doesn't just bringing the metric tensor in galilean form at a point make automatically the Christoffel symbols vanish ? Thank you.

No. A function can be one without its derivatives being zero.

Note: What is being discussed is a coordinate system such that the metric takes the Minkowski form at one particular event. At other events it will generally not be the case.

 

[cianfa72]

Note: What is being discussed is a coordinate system such that the metric takes the Minkowski form at one particular event. At other events it will generally not be the case.

But, as far as I understand, they claim there is a particular linear transformation that brings the metric in the Minkowski form at one particular event and at the same time makes Christoffel symbols vanish at that particular event.

 

[martinbn]

But, as far as I understand, they claim there is a particular linear transformation that brings the metric in the Minkowski form at one particular event and at the same time makes Christoffel symbols vanish at that particular event.

Yes, but I doubt that they say linear transformation.

 

[Orodruin]

Yes, but I doubt that they say linear transformation.

That depends on what they say is linear. The coordinate transformation may not be, but the corresponding transformation of the metric components at a specific event certainly are.

 

[vanhees71]

In the The Classic Theory of Field from Landau - Lifshitz, there is a claim at p. 240 (section 85) where they say there exists a linear transformation defined at each point in spacetime such that the metric tensor transforms in galilean (inertial) form.

They claim that such linear transformation brings the metric in galilean form and at the same time makes the Christoffel symbols vanish.

My question is: doesn't just bringing the metric tensor in galilean form at a point make automatically the Christoffel symbols vanish ? Thank you.

What LL show is that at a given (regular) point of a general-relativistic spacetime you can always make all Christoffel symbols vanish at a given point. The coordinate transformation is bilinear rather than liner. Geometrically you choose a coordinate system, where the four coordinate lines are all geodesics at this one point.

 

[cianfa72]

Yes, but I doubt that they say linear transformation.

The transformation is the equation 85.18
$$x^{'i} = x^i + \frac 1 2 (\Gamma^i_{kl})_0 x^k x^l$$Yes you are right, the transformation itself is not linear.

 

[Orodruin]

The transformation is the equation 85.18
$$x^{'i} = x^i + \frac 1 2 (\Gamma^i_{kl})_0 x^k x^l$$Yes you are right, it is not linear.

Yes, that is clearly not linear.

 

[cianfa72]

What LL show is that at a given (regular) point of a general-relativistic spacetime you can always make all Christoffel symbols vanish at a given point.

So if you do a coordinate transformation that makes all Christoffel symbols vanish at a given point then automatically the metric is brought in Minkowski form at that point, right ?

 

[Orodruin]

So if you do a coordinate transformation that makes all Christoffel symbols vanish at a given point then automatically the metric is brought in Minkowski form at that point, right ?

No.

As a counter example, take any oblique coordinate system on Minkowski space.

 

[vanhees71]

Not necessarily. You can, of course, also find always local inertial coordinates, where at a point $X$ the metric gets Minkowskian $g_{\mu \nu}(X)=\eta_{\mu \nu}$ and the Christoffel symbols vanish (i.e., the 1st derivatives of the metric vanish). That's the mathematical (and imho the only clear) definition of the equivalence principle. The result is that you have a pseudo-Riemannian spacetime, i.e., a spacetime with a Lorentzian fundamental form and the connection given as the unique torsion-free connection, the Christoffel connection.

This is nicely explained in

S. Weinberg, Gravitation and Cosmology, Wiley (1972)

 

[PeterDonis]

In the The Classic Theory of Field from Landau - Lifshitz, there is a claim at p. 240 (section 85) where they say there exists a linear transformation defined at each point in spacetime such that the metric tensor transforms in galilean (inertial) form.

They claim that such linear transformation brings the metric in galilean form and at the same time makes the Christoffel symbols vanish.

Yes, but only at one point, not over a finite region. The transformation is different at each point. So this is not a way to construct synchronous coordinates. The name for the coordinate chart you get by this means when you do it at some particular point is "Riemann normal coordinates" centered on that point.

 

[cianfa72]

So this is not a way to construct synchronous coordinates.

Ok, so to get a synchronous coordinate chart in a finite region of any spacetime, we need to do a coordinate transformation such that the metric tensor components $g_{00}$ and $g_{0\alpha}$ become $g_{00}=1$ and $g_{0\alpha}=0$ in that region.

 

[PeterDonis]

Ok, so to get a synchronous coordinate chart in a finite region of any spacetime, we need to do a coordinate transformation such that the metric tensor components $g_{00}$ and $g_{0\alpha}$ become $g_{00}=1$ and $g_{0\alpha}=0$ in that region.

In all of the finite region, yes. And in general the Christoffel symbols will not all vanish in such a chart.

 

[cianfa72]

That means in all of the finite region there is a timelike geodesic congruence that is hypersurface orthogonal.

 

[PeterDonis]

That means in all of the finite region there is a timelike geodesic congruence that is hypersurface orthogonal.

Yes. But the curves in the congruence will not be integral curves of a Killing vector field.

 

[cianfa72]

But the curves in the congruence will not be integral curves of a Killing vector field.

I would say that -in general- they are not integral curves of a timelike KVF, yes. Then, in any specific case, they might or might not be integral curves of a timelike KVF.

 

[PeterDonis]

in any specific case, they might or might not be integral curves of a timelike KVF.

I believe we can make a much stronger statement than that: if we have a timelike KVF that is hypersurface orthogonal and is also a geodesic congruence, we must be in flat Minkowski spacetime.

 

[cianfa72]

if we have a timelike KVF that is hypersurface orthogonal and is also a geodesic congruence, we must be in flat Minkowski spacetime.

But does the above claim also apply to a finite region of spacetime ?

 

[PeterDonis]

But does the above claim also apply to a finite region of spacetime ?

That is what we are talking about in this series of posts: a finite region of spacetime.

 

[cianfa72]

That is what we are talking about in this series of posts: a finite region of spacetime.

So in that finite region the spacetime might be flat and not flat outside of it.

 

[PeterDonis]

So in that finite region the spacetime might be flat and not flat outside of it.

Yes. But then in the non-flat region outside, the kind of congruence we are discussing would not exist.