Wald synchronous reference frame proof

  • #1
cianfa72
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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.
 
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  • #2
##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##
 
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  • #3
… assuming the connection is metric compatible, which is the case in GR as the connection is taken as the Levi-Civita connection.
 
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  • #4
Orodruin said:
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.
 
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  • #5
cianfa72 said:
Is the above correct ?
Looks ok on a cursory glance.
 
  • #6
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.
 
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  • #7
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.
 
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  • #8
cianfa72 said:
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.
 
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  • #9
Orodruin said:
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.
 
  • #10
cianfa72 said:
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.
 
  • #11
martinbn said:
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.
 
  • #12
cianfa72 said:
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.
 
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  • #13
martinbn said:
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.
 
  • #14
cianfa72 said:
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.
 
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  • #15
vanhees71 said:
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 ?
 
  • #16
cianfa72 said:
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.
 
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  • #17
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)
 
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  • #18
cianfa72 said:
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.
 
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  • #19
PeterDonis said:
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.
 
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  • #20
cianfa72 said:
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.
 
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  • #21
That means in all of the finite region there is a timelike geodesic congruence that is hypersurface orthogonal.
 
  • #22
cianfa72 said:
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.
 
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  • #23
PeterDonis said:
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.
 
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  • #24
cianfa72 said:
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.
 
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  • #25
PeterDonis said:
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 ?
 
  • #26
cianfa72 said:
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.
 
  • #27
PeterDonis said:
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.
 
  • #28
cianfa72 said:
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.
 
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FAQ: Wald synchronous reference frame proof

What is the Wald synchronous reference frame?

The Wald synchronous reference frame is a mathematical framework used to simplify the analysis of three-phase electrical systems. It transforms the three-phase currents and voltages into a rotating reference frame, making it easier to study their dynamic behavior and control.

Why is the Wald synchronous reference frame used in electrical engineering?

The Wald synchronous reference frame is used because it simplifies the analysis and control of AC machines and power systems. By transforming the three-phase quantities into a rotating frame, it allows for easier implementation of control strategies and better understanding of system dynamics.

How does the transformation to the Wald synchronous reference frame work?

The transformation involves converting the three-phase quantities (usually voltages or currents) into two orthogonal components, typically referred to as the direct (d) and quadrature (q) axes. This is done using a mathematical transformation matrix that rotates the reference frame to align with the rotating magnetic field of the machine or system.

What are the advantages of using the Wald synchronous reference frame?

Using the Wald synchronous reference frame offers several advantages, including simplified control algorithms, easier implementation of vector control techniques, improved stability analysis, and enhanced performance of AC machines and power systems. It allows for decoupled control of torque and flux in electric machines, leading to more efficient and precise operation.

Can you provide a basic proof or derivation of the Wald synchronous reference frame transformation?

A basic proof involves starting with the three-phase quantities and applying a Clarke transformation to convert them into two stationary orthogonal components (alpha and beta axes). Then, a Park transformation is applied to rotate these components into the synchronous reference frame (d and q axes). The transformations are mathematically represented by matrices that involve trigonometric functions of the electrical angle, aligning the reference frame with the rotating magnetic field.

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