Performing a Global Boost in General Relativity: A Brief Guide

[keithdow]

Let us assume that I have set up a Cartesian coordinate frame in a system that follows the rules of general relativity. From the origin it seems that translation and rotation of coordinates should be trivial. All the action is in the metric and it will follow along with the transformations. However is there a simple way to perform a global boost? That is suppose I have an object at point (t, x, y, z) moving with x-velocity Vx, how do I shift from the reference frame at the origin to the object's reference frame?

Thanks

 

[jfy4]

In general, the transformations of boosts, rotations, and translations on a arbitrary Einstein manifold do not exist, at least not in the sense of preserving the metric. The transformations that preserve the metric are called the "group of isometries on a manifold". This is the group of transformations that preserves the metric. However, demanding that rotation, translation, and boosts must be possible simply means that locally every space-time can be modeled by the Minkowski spacetime and hence coordinate transformations are done with the Lorentz group. But if you are looking for coordinate transformations for boosts for an arbitrary Einstein metric, such transformations usually don't exist; once again, in so far as that, after the transformation you are in a different metric. The group of transformations that can manipulate spacetime by boosting and twisting and translating and all that jazz is the group of diffeomorphisms, this leaves the manifold "un-torn".

I hope this helps.

 

[keithdow]

Thanks for the help! It looks like I will have to master Wald to understand the details. Do you have any other references you could recommend?

Also if I am interested regular plain vanilla manifolds with just black holes and nothing extraordinary, it seems this meets your criteria of locally being a Minkowski space. For example take the standard Schwarzschild solution. Transform it to Cartesian coordinates and then apply a boost. It seems that this procedure should work. That is the coordinates will transform correctly as well as the metric. Thanks again!

 

[jfy4]

A passive coordinate transformation will not change the physics of the manifold. For example, I can write the flat-space metric in polar coordinates, but it still represents flat-space. This is simply a passive change in coordinates.

There is also the geometry of spacetime vs the metric of spacetime. A geometry is an equivalence class of metrics under diffeomorphisms. For example, the Minkowski metric and the Euclidean metric are different, but represent the same geometry (flat).

The isometries of a manifold are given by the geometry, not, per se, by the metric. Although symmetries in the metric and the Lagrangian can shed light onto what the possible isometries of a geometry are.

Now, take for instance, the Schwarzschild metric. This has 4 isometries associated with a translation through time and three rotations around the x, y, z axes (spherically symmetric). These are the isometries for this geometry regardless of the coordinates chosen; because we are not asking "what are the symmetries for the language I choose to express the geometry in?", we are asking "what are the symmetries for the geometry regardless of the language I express it in?"

Does that make sense?

 

[WannabeNewton]

The group of transformations that can manipulate spacetime by boosting and twisting and translating and all that jazz is the group of diffeomorphisms, this leaves the manifold "un-torn".

I assume you mean the group of isometries? You can define a diffeomorphism $\phi :M \mapsto N$ that doesn't necessarily result in $\phi ^{*}\mathbf{g} = \mathbf{g}$ but the result is always true for a member of the group of isometries on M.

 

[pervect]

The really really short version is this. Envision a general manifold as being curved - perhaps something simple like a sphere, or something more complex. Then you can perform a boost in the tangent space, which would be a tangent plane to your sphere. The tangent space is perfectly flat, but if you don't travel too far, it doesn't matter.

The tangent spaces at points that are a large distance apart are all different, and in general you can't compare them directly.

If Wald is too much, try reading Baez's "the meaning of Einstein's equation"

http://math.ucr.edu/home/baez/einstein/

note the section where he talks about how it's mipossible to compare velocities (boosts) at different points.

 

[atyy]

However is there a simple way to perform a global boost? That is suppose I have an object at point (t, x, y, z) moving with x-velocity Vx, how do I shift from the reference frame at the origin to the object's reference frame?

Cartesian frames that cover spacetime don't exist in GR. An observer can construct locally an orthonormal frame, and transport it with him. The resulting coordinates around him are given by Fermi normal coordinates.
http://arxiv.org/abs/gr-qc/9904078
http://arxiv.org/abs/1102.0529, section 9

 

[jfy4]

I assume you mean the group of isometries? You can define a diffeomorphism $\phi :M \mapsto N$ that doesn't necessarily result in $\phi ^{*}\mathbf{g} = \mathbf{g}$ but the result is always true for a member of the group of isometries on M.

In the passage you quoted from me above I mean the group of diffeomorphisms. The group of diffeomorphisms are all the manipulations you can do to a manifold that don't tear it apart. I don't understand your point, what you wrote doesn't seem contradictory to what I wrote. Care to elaborate for me?

 

[WannabeNewton]

In the passage you quoted from me above I mean the group of diffeomorphisms. The group of diffeomorphisms are all the manipulations you can do to a manifold that don't tear it apart. I don't understand your point, what you wrote doesn't seem contradictory to what I wrote. Care to elaborate for me?

When I posted, your second post on isometries didn't show up so I assumed in your first post that you meant isometries wherever you put the more general term diffeomorphism. I don't know why the second post didn't show up. Unfortunate misunderstanding on my part, cheers.