# Coordinate Transformation

1. Dec 12, 2008

### JustinLevy

Let's say I have a non-linear transformation for (ct,x,y,z) in one coordinate system to (cT,X,Y,Z) of another coordinate system.

Despite being nonlinear, I assume I can transform all four-vectors using the same non-linear transformation (correct?), but how in the world do I transform the component representation of tensors? Such as $$F_{\mu\nu}$$ (the electromagnetic field tensor) for example?

2. Dec 12, 2008

### Ben Niehoff

Can you give an example of a non-linear transformation?

3. Dec 12, 2008

### GDogg

What difference would it make whether it's linear or non-linear? $F_{\mu\nu}$ is a tensor and as such it follows the usual transformation rule for tensors:

$$\displaystyle{ F{'}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial X^{\alpha}} \frac{\partial x^{\nu}}{\partial X^{\beta}} F_{\mu\nu} }$$

Last edited: Dec 12, 2008
4. Dec 13, 2008

### JustinLevy

Sure. The most common example is Rindler Coordinates for an accelerated observer.
let (cT,X,Y,Z) be inertial coordinates, and (ct,x,y,z) be the coordinates for the accelerated frame with the spatial origin having a proper acceleration 'a', then:

$$\displaystyle{ ct = \frac{c^2}{a}\tanh^{-1}(cT/X)$$
$$x = \frac{-c^2}{a} + \sqrt{X^2 - (cT)^2}$$

$$y = Y$$
$$z = Z$$

the inverse is

$$cT = (c^2/a + x) \sinh(at/c)$$
$$X = (c^2/a + x) \cosh(at/c)$$

$$Y = y$$
$$Z = z$$

Hmm... I'm confused now. Because that would make $$x^\mu$$, which I thought I could consider a four-vector, transform differently than an actual four-vector. This is because the equation you wrote rewrites the transformation as linear at each point in spacetime.

Or maybe I'm just not understanding how to evaluate that when the transformation is non-linear.

Last edited: Dec 13, 2008
5. Dec 13, 2008

### atyy

$$x^\mu$$ is not a 4-vector, it's just the coordinates. A 4-vector would be a tangent vector at $$x^\mu$$.

6. Dec 13, 2008

### JustinLevy

Now I'm really confused.
Because that would make any four-vector or scalar derived from $$x^\mu$$ not an actual four-vector or scalar?
For instance proper time along a worldline, four-velocity, proper acceleration, four-momentum, etc.

7. Dec 13, 2008

### Fredrik

Staff Emeritus
Justin, check out #3 and #5 in this thread.

If you define X to be the map

$$p\mapsto(p,X_p)$$

where

$$X_p=x^\mu(p)\frac{\partial}{\partial x^\mu}\bigg|_p$$

then X is a vector field, and $x^\mu(p)$ are its components at p.

The matrix that appears in the transformation rules for tensors is the Jacobian of the coordinate change (or its inverse). When the coordinate change is linear, the matrix representing the change is equal to its Jacobian. (This is just the many variables version of the statement that if f(x)=kx, f'(x)=k).

Last edited: Dec 13, 2008
8. Dec 13, 2008

### GDogg

Like they said before me, $$x^\mu$$ are just the components of the four-vector. You need a basis to define the vector. The linearity comes from the fact that the new basis is a linear combination of the old basis.

So the components of four-vectors transform in the same way:

$$\displaystyle{ A{'}^{\alpha} = \frac{\partial X^{\alpha}}{\partial x^{\mu}} A^{\mu} }$$

And the transformation is valid for a general coordinate change

Last edited: Dec 13, 2008
9. Dec 13, 2008

### Fredrik

Staff Emeritus
A tensor can't be defined by specifying its components in one coordinate system. The definition must specify its components in all coordinate systems.

Let's take a closer look at the definition of 4-velocity, just as an example. The tangent vector T of the curve $C:(a,b)\rightarrow M$ at the point C(t) is defined by

$$T_{C(t)}f=(f\circ C)'(t)$$

(I wanted to call the tangent vector "\dot C" but the dot was barely visible). Let's find its components in the coordinate system x:

$$=(f\circ x^{-1}\circ x\circ C)'(t)=(f\circ x^{-1}),_\mu(x(C(t)))(x\circ C)^\mu'(t)=(x^\mu\circ C)'(t)\frac{\partial}{\partial x^\mu}\bigg|_{C(t)}f$$

Note that its components are the components of the "corresponding" curve in $\mathbb R^n$.

The 4-velocity is defined as the tangent vector normalized to unit length. If the t above is the proper time, then T is already normalized. The 4-velocity is a 4-vector, not because its components at parameter value t is $x^\mu(C(t))$, but because it's defined as the tangent vector that has those components.

Last edited: Dec 13, 2008
10. Dec 13, 2008

### atyy

In special relativity, you can choose coordinates that make Maxwell's equations nice and linear. These coordinates are position vectors with respect to an origin in spacetime. The four-velocity is a vector attached to a particular point that isn't the origin for the position vector. In flat spacetime, the position vectors with respect to an origin, and the tangent vectors at every point in spacetime have the same vector space structure, so it's often convenient to mix them up. In curved spacetime, the coordinates with respect to an origin don't form a "position vector space", so they are not vectors (unless you embed spacetime in a universe with more dimensions). However, the four-velocity, proper time (derived form the metric), are scalars, vectors, tensors that exist at each point in spacetime, and those are (multi)linear.

11. Dec 13, 2008

### JustinLevy

Okay, I still don't quite get this and so need to take some more time to think and let it settle.

So it sounds like:
For each spacetime point, there is an associated vector space 'attached' to that point. Apparrently I have always been fairly sloppy with the concept of the vectors, for I realized I need to "transport" them to the same point to compare them directly, but I didn't realize the full reason for this. So please bear with me here as I am learning.

So, given a coordinate system and transforming to a new one not only changes the label for a spacetime point, but also changes the basis in the vector space at that point? I'm getting kind of lost here. If we abstract stuff till we are considering a separate vector space to be attached to each point, why can't we freely choose the basis in the vector space separate from the basis used to provide the coordinate system. I'm clearly missing something important here about the connection between the two: coordinate system (giving labels to spacetime points) <--> coordinate representation in the vector space attached to a spacetime point (giving labels to each vector).

Undoubtedly you guys have provided the information already, but I'm still having trouble. I will reread what you wrote a couple more times and read other sources, but if you could offer more words of wisdom that would be great.

I thought that atyy was saying those AREN'T components of a four-vector. They transform using the non-linear transformation, and do not 'live' in the vector space attached to a spacetime point. No?

Are you guys all saying the same thing and I am just having trouble putting it together?

If you specify the components of a tensor in one coordinate system, doesn't that uniquely define the components in all other coordinate systems? I must be missing something here as well.

Those posts seem to say the metric can be derived just from the coordinate system. I am really confused here. I thought a metric was additional information you had to supply in addition to a manifold.

Maybe I've built up a horrible series of 'extensions' in my mind from what I thought things were in flat spacetime, and therefore really screwed things up. Maybe we should start further back to make sure I am understanding basic definitions correctly.

12. Dec 13, 2008

### atyy

You can use the coordinates {x,y} to make basis vectors for the tangent vector space at each point {d/dx,d/dy}. If you change coordinates to {u,v}, then you usually also change basis vectors to {d/du,d/dv}. Instead of using coordinates to form basis vectors, you can use arbitrary vectors at each point to form a basis, and these are called tetrads or vierbeins (in 4D), which apparently are related to your question about spin, torsion on the other thread (texts seem to mention vierbeins and spin, rather than torsion and spin), but to which I don't know the answer.

The reason you can use the coordinates to make basis vectors is that the coordinates form coordinate curves, and curves have tangent vectors. A curve is just a path on the manifold which is parameterised by a number, for example, a person carrying an ideal clock will have a path parametrized by proper time T. Then the tangent vector is the derivative operator d/dT, which acts on arbitrary scalar functions on the manifold. If you have a function f(x,y), then df(x,y)/dT=(dx/dT)(df/dx)+(dy/dT)(df/dy), which is why {d/dx,d/dy} are basis tangent vectors.

Last edited: Dec 13, 2008
13. Dec 13, 2008

### Fredrik

Staff Emeritus
You can of course choose another basis if you want. You can express any tangent vector at p as a linear combination of any n linearly independent tangent vectors at p, but the expression "the components of V in x" always refers to the numbers $V^\mu$ defined by

$$V=V^\mu\frac{\partial}{\partial x}\bigg|_p$$

Yes, it does, if you also say that you're defining a tensor. There's a difference between saying "$v^\mu$" and saying "the vector that has the components $v^\mu$ in the coordinate system x".

I think you're only having trouble with this because it's too simple. They aren't the components of a four-vector until someone defines a four-vector that has those components. If you just say that the position coordinates at time t of the particle with world line C are [itex]x^\mu(C(t))[/tex], then you haven't defined a vector.

Absolutely not. All I said about the metric is that its components in the coordinate system x are what you get when the metric acts on the basis vectors constructed from x. That stuff isn't relevant to your concerns in this thread. The reason I suggested that you take a look at those posts is that they define the tangent space (the first sentence in #3 can be taken as the definition), show how the basis vectors can be constructed from a coordinate system, and show how two such sets of basis vectors are related.