Coordinate transformations on the Minkowski metric

In summary, the conversation discusses using the adjugate method to find the inverse of a given line element in order to identify a coordinate transformation that will reduce it to the standard Minkowski line element. The conversation also discusses the confusion that can arise when treating the line element as a matrix without considering its true nature as a (0,2) tensor. The suggested method for finding the coordinate transformation involves using the law of transformation for tensors and solving a set of simultaneous partial differential equations.
  • #1
amathie
2
0
Homework Statement
Take the line element ds^2= (a^2t^2−c^2)dt^2+ 2atdtdx+ dx^2+ dy^2+ dz^2, where a and c are constants. Calculate the components of the inverse metric. By identifying a suitable coordinate transformation, show that the line element can be reduced to the Minkowski line element.
Relevant Equations
Minkowski line element: ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
Matrix diagonalisation: S^-1AS=D
The line element given corresponds to the metric:

$$g = \begin{bmatrix}a^2t^2-c^2 & at & 0 & 0\\at & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}$$

Using the adjugate method: ##g^{-1}=\frac{1}{|g|}\tilde{g}## where ##\tilde{g}## is the adjugate of ##g##. This gives me:

$$g^{-1}=\begin{bmatrix}-\frac{1}{c^2} & \frac{at}{c^2} & 0 & 0\\\frac{at}{c^2} & -\frac{a^2t^2}{c^2}+1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}$$

The question now asks me to identify a coordinate transformation to reduce the line element given to the Minkowski line element.

I think what I need to do is find a matrix that operates on ##g## to give ##\begin{bmatrix}-1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}##.

I've tried to find the eigenvalues of ##g## by the standard method (i.e. finding the characteristic equation and attempting to solve it), but as far as I can see it just makes an unfactorisable mess.

I get ##(a^2t^2-c^2-\lambda)(1-\lambda)(1-\lambda)^2=a^2t^2(1-\lambda)^2##, which gives one degenerate eigenvalue of ##\lambda=1## with multiplicity...3? But I can't find the fourth.

I'm self-teaching this stuff so have no idea what I'm doing really. Any help very appreciated!
 
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  • #2
Well, finding the eigenvalues of g isn't really the best thing to do. This is because even though we represent ##g## by matrix, we have to keep in mind that ##g## is a (0,2) tensor. What does this mean?
It means that when we act with ##g## on some vector, we don't get another vector. Instead, we get a dual vector. Or in otherwords, we need to act with ##g## on two vectors to get a scalar(this action is called scalar product). So, since we don't get a vector when acting with ##g##, we can't talk about eigenvectors, since those don't exist. This is a confusion that can arise when we're not taking into account that this matrix is just a convenient representation of ##g##, but we must know how to act with it on vectors in order to make use of it.

Now on to your question about the transformation of ##g##. Law of transformation of ##g## under coordinate transformations, reads:

$$g'_{\mu\nu} = \frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}g_{\rho\sigma}$$

Here indices represent different components of ##g##, so in your case you can take ##g'## to be standard Minkowski metric, and take ##g## to be your own metric. Summation convention is in use, which means we're summing over indices that are repeating. I will give you example of the first component, so you get the gist of how to use this, in case you haven't used this before.

$$g'_{00} = \frac{\partial x^\rho}{\partial x'^0}\frac{\partial x^\sigma}{\partial x'^0}g_{\rho\sigma} = \frac{\partial x^0}{\partial x'^0}\frac{\partial x^0}{\partial x'^0}g_{00} +2\frac{\partial x^0}{\partial x'^0}\frac{\partial x^1}{\partial x'^0}g_{01} + \frac{\partial x^1}{\partial x'^0}\frac{\partial x^1}{\partial x'^0}g_{11} + \frac{\partial x^2}{\partial x'^0}\frac{\partial x^2}{\partial x'^0}g_{22} + \frac{\partial x^3}{\partial x'^0}\frac{\partial x^3}{\partial x'^0}g_{33} $$

Now you proceed the same thing for the other components that should be transformed, and from the system of equations that you get, you find what the coordinate transformation is. This is like the most straightforward(brute-force) method that you get, so in case you have no other intuition on how to do it, it should work. Either way, even when guessing the transformation, this is how you check if it is right or wrong.
 
  • #3
Thanks Antarres, that's helpful. And clears up some conceptual issues I was having about the metric.

I've followed what you've suggested, and have ended up with the following set of equations:

$$g'_{00}=-1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{t'}})^2+2at\frac{\partial{x}}{\partial{t'}}\frac{\partial{t}}{\partial{t'}}+(\frac{\partial{x}}{\partial{t'}})^2+(\frac{\partial{y}}{\partial{t'}})^2+(\frac{\partial{z}}{\partial{t'}})^2$$
$$g'_{11}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{x'}})^2+2at\frac{\partial{x}}{\partial{x'}}\frac{\partial{t}}{\partial{x'}}+(\frac{\partial{x}}{\partial{x'}})^2+(\frac{\partial{y}}{\partial{x'}})^2+(\frac{\partial{z}}{\partial{x'}})^2$$
$$g'_{22}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{y'}})^2+2at\frac{\partial{x}}{\partial{y'}}\frac{\partial{t}}{\partial{y'}}+(\frac{\partial{x}}{\partial{y'}})^2+(\frac{\partial{y}}{\partial{y'}})^2+(\frac{\partial{z}}{\partial{y'}})^2$$
$$g'_{33}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{z'}})^2+2at\frac{\partial{x}}{\partial{z'}}\frac{\partial{t}}{\partial{z'}}+(\frac{\partial{x}}{\partial{z'}})^2+(\frac{\partial{y}}{\partial{z'}})^2+(\frac{\partial{z}}{\partial{z'}})^2$$

Can anyone give me some tips on how I solve for ##t', x', y', z'##? I haven't encountered sets of simultaneous PDEs like this before.
 
  • #4
amathie said:
Thanks Antarres, that's helpful. And clears up some conceptual issues I was having about the metric.

I've followed what you've suggested, and have ended up with the following set of equations:

$$g'_{00}=-1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{t'}})^2+2at\frac{\partial{x}}{\partial{t'}}\frac{\partial{t}}{\partial{t'}}+(\frac{\partial{x}}{\partial{t'}})^2+(\frac{\partial{y}}{\partial{t'}})^2+(\frac{\partial{z}}{\partial{t'}})^2$$
$$g'_{11}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{x'}})^2+2at\frac{\partial{x}}{\partial{x'}}\frac{\partial{t}}{\partial{x'}}+(\frac{\partial{x}}{\partial{x'}})^2+(\frac{\partial{y}}{\partial{x'}})^2+(\frac{\partial{z}}{\partial{x'}})^2$$
$$g'_{22}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{y'}})^2+2at\frac{\partial{x}}{\partial{y'}}\frac{\partial{t}}{\partial{y'}}+(\frac{\partial{x}}{\partial{y'}})^2+(\frac{\partial{y}}{\partial{y'}})^2+(\frac{\partial{z}}{\partial{y'}})^2$$
$$g'_{33}=1=(a^2t^2-c^2)(\frac{\partial{t}}{\partial{z'}})^2+2at\frac{\partial{x}}{\partial{z'}}\frac{\partial{t}}{\partial{z'}}+(\frac{\partial{x}}{\partial{z'}})^2+(\frac{\partial{y}}{\partial{z'}})^2+(\frac{\partial{z}}{\partial{z'}})^2$$

Can anyone give me some tips on how I solve for ##t', x', y', z'##? I haven't encountered sets of simultaneous PDEs like this before.

The first thing to do is to let ##y' =y, z' = z##. That should simplify things.

You need an equation for ##g'_{01}## as well.
 

Related to Coordinate transformations on the Minkowski metric

1. What is a coordinate transformation on the Minkowski metric?

A coordinate transformation on the Minkowski metric refers to changing the coordinate system used to describe a spacetime in special relativity. This transformation allows for the measurement of distances and time intervals to remain consistent across different reference frames.

2. What is the purpose of coordinate transformations on the Minkowski metric?

The purpose of coordinate transformations on the Minkowski metric is to maintain the principles of special relativity, which state that the laws of physics should be the same for all observers moving at constant velocities. By transforming coordinates, we can reconcile the differences in measurements made by different observers.

3. How do you perform a coordinate transformation on the Minkowski metric?

To perform a coordinate transformation on the Minkowski metric, you can use the Lorentz transformation equations, which involve the speed of light and the relative velocity between two frames of reference. These equations allow you to convert coordinates and measurements made in one frame of reference to those in another.

4. Are there specific conditions that must be met for a coordinate transformation on the Minkowski metric to be valid?

Yes, there are certain conditions that must be met for a coordinate transformation on the Minkowski metric to be valid. These include both frames of reference being inertial (moving at a constant velocity), and the transformation must preserve the speed of light as a constant in all frames of reference.

5. Why is the Minkowski metric used for coordinate transformations in special relativity?

The Minkowski metric is used for coordinate transformations in special relativity because it is a mathematical representation of the spacetime interval, which is a fundamental concept in special relativity. This metric allows for the measurement of distances and time intervals to remain consistent across different frames of reference, making it essential for maintaining the principles of special relativity.

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