Transformation law in curved space-time

[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.