Transformation law in curved space-time

[jfy4]

Hi,

I am wondering what are the transformation laws (Lorentz transformations) for a general metric, if they exist.

 

[Dale]

Any diffeomorphism is allowed.

 

[jfy4]

Wait,

Are you saying that if I take the coordinate 4 vector, and put it through a diffeomorphism, that is an acceptable transformation? That is, realistic?

 

[atyy]

There are no position (t,x) 4-vectors in general relativity.

Coordinates (t,x) are just labels of points in a manifold.

There are velocity 4-vectors (in the tangent space at each spacetime point through which the particle's worldline goes),just as in special relativity.

 

[dextercioby]

There are no position (t,x) 4-vectors in general relativity.

Coordinates (t,x) are just labels of points in a manifold.

True, but changing a chart (U,x) (P in U) to a different one (U',x') (P in U') requires that the <labels> of that point transform in a specific way similar to 4-vector components in the tangent space at P.

 

[jfy4]

There are no position (t,x) 4-vectors in general relativity.

Coordinates (t,x) are just labels of points in a manifold.

There are velocity 4-vectors (in the tangent space at each spacetime point through which the particle's worldline goes),just as in special relativity.

Sure, fair enough. It's a little disconcerting the use of the phrase "spacetime point" though, when those points don't exist.


My OP is about how the lorentz transformations are valid in flat space-time. I was wondering if there is a transformation group or rules analogous to the lorentz transformations in flat space-time, except for a general metric.

 

[atyy]

In both special and general relativity, if you allow Christoffel symbols to be used in writing the laws of physics, then any transformation (more precisely, "diffeomorphism" as DaleSpam says above) will preserve the form of the laws.

In special relativity (flat spacetime), if Christoffel symbols are not allowed in writing the laws of physics, then Lorentz transformations will preserve the form of the laws.

In general relativity (curved spacetime), there is no way to write the laws of physics without Christoffel symbols.

 

[Dale]

In special relativity (flat spacetime), if Christoffel symbols are not allowed in writing the laws of physics, then Lorentz transformations will preserve the form of the laws.

In general relativity (curved spacetime), there is no way to write the laws of physics without Christoffel symbols.

That is a good way of putting it.

In flat spacetime the other thing that the Lorentz transform preserves is the form of the metric. I suppose for any given coordinate system on some spacetime you could look for the set of all diffeomorphisms which preserve the form of the metric.

 

[atyy]

That is a good way of putting it.

In flat spacetime the other thing that the Lorentz transform preserves is the form of the metric. I suppose for any given coordinate system on some spacetime you could look for the set of all diffeomorphisms which preserve the form of the metric.

In which case one would look for the Killing vectors of the spacetime.

The Killing vectors for the Schwarzschild spacetime are given in http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html

And those for Minkowski spacetime are given in http://mathworld.wolfram.com/KillingVectors.html .

 

[jfy4]

Thanks for you responses guys.

Is it possible for me to get an example of what your saying? Say, I want to boost in the schwarzschild metric the u_{1} component of velocity.

normally I would use a lorentz boost. How would I do it not for flat space?

 

[Dale]

I have limited experience with Killing vectors, but if the link atyy gave is correct then there are no Killing vectors corresponding to boosts in the Schwarzschild metric. This would imply that there is no boost which preserves the form of the metric for a Schwarzschild spacetime, which is not surprising since the spacetime is not homogenous.

 

[Mentz114]

Thanks for you responses guys.

Is it possible for me to get an example of what your saying? Say, I want to boost in the schwarzschild metric the u_{1} component of velocity.

normally I would use a lorentz boost. How would I do it not for flat space?

Instead of boosting the metric, try boosting the frame or coframe of a static observer. For instance, boosting a static observer in Schwarzschild spacetime by \beta=\sqrt{2m/r} will take you to the Painleve coords. I've got details somewhere because I've done it but no time to search now ...

A good example of a boost in Minkowski spacetime is here

http://en.wikipedia.org/wiki/Born_coordinates

where the frame-field is boosted to make a rotating frame.[edit: the sqrt is not showing properly ...]
\beta=\sqrt{2m/r}

 

[jfy4]

I really appreciate you guys bearing with me. Let me attempt to be more explicit.

In flat space-time, we can use a transformation to go between two different frames of reference by a linear transformation known as the Lorentz transformation, correct?

x_{\alpha}=\Lambda^{\beta}_{\alpha}x_{\beta}+a_{\alpha}

This will give us a transformation into a new reference frame. \Lambda is only good though, in its well known form, for flat space-time.

So now here is what I want to know, and I will once again try an example:

Lets say I have two different reference frames in say, the Scharwzschild geometry. I would like to transform between them. Is there an already known general way to transform between reference frames there, or must it be done in a case by case basis. Or is it not possible? Or something else...?

Thanks again.

 

[atyy]

I'm not sure exactly how, but I'd try something like this.

Look up the Killing vector at http://en.wikipedia.org/wiki/Schwarzschild_coordinates

For each Killing vector, find the integral curves.

eg. for sinVdW+cotWcosVdV,

the integral curve has coordinates t,r,V,W which are functions of a parameter L, and which are solutions of

dt/dL=0
dr/dL=0
dV/dL=cotWcosV
dW/dL=sinV

If you move each point along the integral curve that it's on by the same amount of the curve parameter, the form of the Schwarzschild solution should remain unchanged (assuming you started from the a coordinate system that is within the class of "preferred" coordinates).

An example of the type of computation is given in http://www.math.ku.edu/~lerner/GR09/LieDerivatives.pdf

 

[Mentz114]

Lets say I have two different reference frames in say, the Scharwzschild geometry. I would like to transform between them. Is there an already known general way to transform between reference frames there, or must it be done in a case by case basis. Or is it not possible? Or something else...?

Take two 'shell' observers in the Schwarzschild geometry. Suppose \lambda_a^A is the tetrad that transforms between the frame basis and the holonomic basis so that
<br /> \lambda_A^a \lambda_B^b g_{ab}=\eta_{AB}<br />

Now let \rho, \mu be \lambda evaluated at two points so we can write

<br /> \mu_A^a \mu_B^b g_{ab}^{(\mu)}=\eta_{AB}=\rho_A^a \rho_B^b g_{ab}^{(\rho)}<br />

from which

<br /> \mu_A^a \mu_B^b }=\left(g^{ab(\mu)}\right) \cdot \left( g_{ab}^{(\rho)}\right)\rho_A^a \rho_B^b<br />

(indexes don't have their usual significance here because the tetrads are matrices not tensors).

So it appears that the transformation that takes \rho\rho \rightarrow \mu\mu is

<br /> \left(g^{ab(\mu)}\right) \cdot \left( g_{ab}^{(\rho)}\right)<br />

This is meant to be the product of 2 matrices, giving a transformation matrix. I think that the square root of this matrix will transform one frame basis to the other.

That is expected for a comparison between static frames. It will be more interesting for two radially infalling frames.

[edit]
It comes down to a simple matrix equation
<br /> \mu=D \cdot \rho, \rightarrow \ D=\mu \cdot \rho^{-1}<br />

 

[Dale]

You have received two answers already, but I don't understand the question. Specifically, what do you mean by this?

Lets say I have two different reference frames in say, the Scharwzschild geometry.

What does it mean to have two different reference frames in the Schwarzschild geometry?

 

[Mentz114]

The transformation between different points the Gull-Painleve free-falling worldline is interesting.

We have
<br /> \mu=\left[ \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\\<br /> \,\sqrt{\frac{2M}{R_1}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; R_1 &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; \sin\left( \theta\right) \,R_1 \end{array} \right]<br />

and (obviously)

<br /> \rho=\left[ \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\\<br /> \,\sqrt{\frac{2M}{R_2}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; R_2 &amp; 0\\\<br /> 0 &amp; 0 &amp; 0 &amp; \sin\left( \theta\right) \,R_2 \end{array} \right]<br />

<br /> D=\mu \cdot \rho^{-1}=<br /> \left[ \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\\<br /> \,\sqrt{\frac{2M}{R1}}-\,\sqrt{\frac{2M}{R2}} &amp; 1 &amp; 0 &amp; 0\\\<br /> 0 &amp; 0 &amp; \frac{R1}{R2} &amp; 0\\\ 0 &amp; 0 &amp; 0 &amp; \frac{R1}{R2}<br /> \end{array} \right]<br />

Which seems to say that all the local times are the same and that Gallilean relativity is used for velocity addition.

This is born out (I think ) by the conclusions in the "River Model" paper arXiv:gr-qc/0411060v2 .

 

[jfy4]

You have received two answers already, but I don't understand the question. Specifically, what do you mean by this?What does it mean to have two different reference frames in the Schwarzschild geometry?

Perhaps my understanding of reference frames is wrong. But I meant reference frames in the same manner that they come across is a Relativity textbook. That is, just as they are in flat space, when one first learns about them in special relativity.

Except, I was wondering if there is a method to transform between frames in a geometry other than diag(-1,1,1,1), which is reserved for Lorentz transformations. And as you pointed out, It looks like there is. Is my question clear now?

 

[Mentz114]

Except, I was wondering if there is a method to transform between frames in a geometry other than diag(-1,1,1,1) , which is reserved for Lorentz transformations. And as you pointed out, It looks like there is.

Your question is clear to me. To find frames that correspond to physical observers in GR requires frame fields, and there is a way to transform between the frame bases .

 

[Dale]

Perhaps my understanding of reference frames is wrong. But I meant reference frames in the same manner that they come across is a Relativity textbook. That is, just as they are in flat space, when one first learns about them in special relativity.

Except, I was wondering if there is a method to transform between frames in a geometry other than diag(-1,1,1,1), which is reserved for Lorentz transformations. And as you pointed out, It looks like there is. Is my question clear now?

Are you talking about co-located observers or distant observers?

 

[Dale]

Your question is clear to me. To find frames that correspond to physical observers in GR requires frame fields, and there is a way to transform between the frame bases .

That is true if they are co-located or if the curvature is negligible. Is that true when they are not co-located and there is significant curvature?

 

[jfy4]

Your question is clear to me. To find frames that correspond to physical observers in GR requires frame fields, and there is a way to transform between the frame bases .

That is true if they are co-located or if the curvature is negligible. Is that true when they are not co-located and there is significant curvature?

This brings me confusion... if curvature is negligible and the observers are co-located, then what would be the point of using these "frame fields"? Regardless of curvature, locally curvature vanishes and space-time is flat, then one could use Lorentz transformations, correct? I am asking this question, or I thought I was, for the case that curvature is present. That is I am wondering if transformation laws exist for a general metric, not evaluated locally to approximate flat space-time.

Am I asking the question the wrong way? It seems this may be the case...

 

[Mentz114]

That is I am wondering if transformation laws exist for a general metric, not evaluated locally to approximate flat space-time.

In curved spacetime there's no globally valid transformation between frames as there is in flat spacetime.

To define a frame ( or observer, who has clocks and rulers ) in curved spacetime we must know something about the worldline, along which the frame is carried.

Although two observer may experience locally flat spacetime, there can still be differences in clock rate and ruler length between them. For instance the static observer example I gave earlier shows that the clock rates differ by the ratio of \sqrt{g_{00}} evaluated at the two points.

 

[jfy4]

In curved spacetime there's no globally valid transformation between frames as there is in flat spacetime.

To define a frame ( or observer, who has clocks and rulers ) in curved spacetime we must know something about the worldline, along which the frame is carried.

Although two observer may experience locally flat spacetime, there can still be differences in clock rate and ruler length between them. For instance the static observer example I gave earlier shows that the clock rates differ by the ratio of \sqrt{g_{00}} evaluated at the two points.

Ok, I hear you. This seems to make some sense. However, humor me for a bit... If it were possible,

it would need to satisfy these relations right?:

g_{ab}\frac{\partial x^{\prime a}}{\partial x^{c}}\frac{\partial x^{\prime b}}{\partial x^{d}}=g_{cd}

and transform accordingly:

x_a=T^{b}_{a}x_b+d_a

where T is the transform, and the set of partials are each a set of transforms.

 

[Mentz114]

Ok, I hear you. This seems to make some sense. However, humor me for a bit... If it were possible,

it would need to satisfy these relations right?:

g_{ab}\frac{\partial x^{\prime a}}{\partial x^{c}}\frac{\partial x^{\prime b}}{\partial x^{d}}=g_{cd}

and transform accordingly:

x_a=T^{b}_{a}x_b+d_a

where T is the transform, and the set of partials are each a set of transforms.

Yes, that looks right. I prefer primed indexes

<br /> g_{m^\prime n^\prime}=g_{mn}A^m_{m^\prime}A^n_{n^\prime}<br />

where A^m_{m^\prime}=\partial x^m /\partial x^{m^\prime } . The transformation is

<br /> x_{a^\prime}=T^{a}_{a^\prime}x_a+d_a<br />

This is a global coordinate transformation, which may change the form of the metric.

 

[atyy]

Ok, I hear you. This seems to make some sense. However, humor me for a bit... If it were possible,

it would need to satisfy these relations right?:

g_{ab}\frac{\partial x^{\prime a}}{\partial x^{c}}\frac{\partial x^{\prime b}}{\partial x^{d}}=g_{cd}

and transform accordingly:

x_a=T^{b}_{a}x_b+d_a

where T is the transform, and the set of partials are each a set of transforms.

When you define it this way, then any nice transformation ("diffeomorphism", as DaleSpam said right at the start) is ok, ie.

t'=f0(t,x,y,z)
x'=f1(t,x,y,z)
y'=f2(t,x,y,z)
z'=f3(t,x,y,z)

Basically, it depends on what you mean by "same form", whether the answer is "any diffeomorphism", "those generated by the Killing vectors", or "a field of Lorentz transformations acting on frame fields".

 

[Mentz114]

Basically, it depends on what you mean by "same form", whether the answer is "any diffeomorphism", "those generated by the Killing vectors", or "a field of Lorentz transformations acting on frame fields".

Yes, except that the transformation that connects the local frames along a geodesic is not always the LT. For instance the radially infalling frames in the GP chart are connected by a Gallilean transformation.

[edit] The transormation from the Schwarzschild coordinates ( holonomic frame) to the (local frame) G-P coords however is a boost by \beta=-\sqrt{2m/r}. So there are transformations from a holonomic frame to a local frame, and those between local frames along a curve. I'm not sure what the significance of this might be.

 

[jfy4]

Ok, After a day of researching (dont make fun if you think I accomplished little...)

this is what I have come up with.

I think we want this Lie group:

\mathcal{O}(1,3,\mathbb{F})

This group is the norm preserving linear transformations (matrices) whos entries come from a field \mathbb{F}. Then, we are interested in the set of these elements who are diffeomorphisms. Now this set is non-empty, cause it has the identity at least. Now I need to know if this whole group has elements who are all diffeomorphisms, or if it has some. If it has some, I need to know if this set forms a group.

what do you guys think?

 

[atyy]

You are allowed to make nonlinear changes of variables eg. between Rindler and Cartesian coordinates.

 

[jfy4]

You are allowed to make nonlinear changes of variables eg. between Rindler and Cartesian coordinates.

Doesn't this equation imply that transformations are linear?

g_{ab}\frac{\partial x^{\prime a}}{\partial x^{c}}\frac{\partial x^{\prime b}}{\partial x^{d}}=g_{cd}

 

[Mentz114]

Ok, After a day of researching (dont make fun if you think I accomplished little...)

this is what I have come up with.

I think we want this Lie group:

\mathcal{O}(1,3,\mathbb{F})

This group is the norm preserving linear transformations (matrices) whose entries come from a field \mathbb{F}. Then, we are interested in the set of these elements who are diffeomorphisms. Now this set is non-empty, cause it has the identity at least. Now I need to know if this whole group has elements who are all diffeomorphisms, or if it has some. If it has some, I need to know if this set forms a group.

what do you guys think?

I think that choice might be too restrictive. This article might be useful

http://en.wikipedia.org/wiki/Diffeomorphism

 

[jfy4]

I think that choice might be too restrictive. This article might be useful

http://en.wikipedia.org/wiki/Diffeomorphism

Here is my rational:

We know that when we need diffeomorphisms. So I looked for subgroups of the diffeomorphism group that could be used as tranformations. Then we need transformations that preserve the quadratic form, this is found in the orthogonal group. But it must work outside of just the flat space-time metric. So we need diffeomorphisms that preserve the norm and have inputs that are not just from \mathbb{R}. So I figured it must be extended to an (as of now) unknown field.

Does this line of thought make sense?

 

[Dale]

So we need diffeomorphisms that preserve the norm

All diffeomorphisms preserve the norm. That is kind of the point of using tensors.

 

[atyy]

Doesn't this equation imply that transformations are linear?

g_{ab}\frac{\partial x^{\prime a}}{\partial x^{c}}\frac{\partial x^{\prime b}}{\partial x^{d}}=g_{cd}

The equation will still work if each primed coordinate is a nonlinear function of all of the unprimed coordinates. Take a look at how they work out for Rindler and Cartesian coordinates in flat space. (There's a typo, I believe the a,b index on the left should be primed, and the primed xs go downstairs, but index gymnastics is hazardous and you should check this.)

BTW, in physics notation repeated upper and lower indices indicate that those should be summed from 0 to 3.

 

[jfy4]

All diffeomorphisms preserve the norm. That is kind of the point of using tensors.

Sorry I'm a little naive about some of the implications for diffeomorphisms. Last night I was reading about "The Classical Groups" and it occurred to me while reading that while the group is important and depends on what field it is taken over, the specific problem I'm trying to solve depends more on setting up a group/ group algebra around a general inner product.

I was more concerned with how the group was set up, which is important, but I need to go back a bit and make sure the group holds for any inner product (hence a general metric).

Q(Tx)=Q(x)

here Q is the norm of the vector and T is the transform.

 

[jfy4]

All diffeomorphisms preserve the norm. That is kind of the point of using tensors.

Is this true? The orthogonal group is a subgroup of the diffeomorphism group, so that group preserves the norm, but do all diffeomorphisms preserve the norm? it doesn't seem the converse is true. Tensors maintain the invariance of coordinates, not the norm right?

 

[atyy]

In a curved space, the metric does not act on "position" ie. coordinates. The metric acts on velocity.

 

[jfy4]

In a curved space, the metric does not act on "position" ie. coordinates. The metric acts on velocity.

In what context are you saying this? clearly this is a valid operation in GR, and a regular one:

ds^2=g_{ab}dx^{a}dx^{b}

This is the line element which is crucial in defining constant distance in space-time along with the notions of time-like, space-like, and null intervals. Here the metric is being used as the coefficients of the scalar product in a particular basis, and the product can certainly be between coordinates no?

 

[atyy]

In what context are you saying this? clearly this is a valid operation in GR, and a regular one:

ds^2=g_{ab}dx^{a}dx^{b}

This is the line element which is crucial in defining constant distance in space-time along with the notions of time-like, space-like, and null intervals. Here the metric is being used as the coefficients of the scalar product in a particular basis, and the product can certainly be between coordinates no?

The metric doesn't act directly on the coordinates to produce distance. Roughly speaking, it acts on the velocity vectors at each point to give their magnitude, and the distance is obtained by integrating the velocities over coordinates. It is the metric acting on the velocity vectors to give their magnitude that is coordinate-independent.

So the linear space is a tangent space at each point of the manifold. But different points have different linear spaces.

 

[jfy4]

Ok,

After a bit more reading here is what I have to report on discovering the group. We will take G\equiv g_{ab}. Now we are interested in the situation

B^{\top}GB=G

where B is our transformation. These transformations form a group, and it has a name and a notation!: the orthogonal group with the "metric ground form", \mathcal{O}_{g}(p,q). Now I quote a Lemma from Weyl's book:

"A non-exceptional transformation, B, [of the group mentioned above] may be written in the form

B=(E-T)(E+T)^{-1}

where T satisfies this condition

GT+T^{\top}G=0"

That is, T is anti-symmetric. I claim T is anti-symmetric since the indicies are raised and lowered by the metric, and E is the Identity.

Then our group of transformations are of the form, B.