Exploring Lorentz Transformation Matrix: A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}

[Petar Mali]

$$\alpha=\left(\begin{array}{cccc} \gamma& 0&0& -\beta\gamma\\ 0&1& 0 & 0\\ 0 & 0 & 1 & 0\\ -\beta\gamma & 0 & 0 & \gamma \end{array} \right)$$$$x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}$$

$$\alpha$$ is Lorrentz transformation matrix. Can I see something more about it? . It's symmetric. That is important.

$$Tr(\alpha)=2\gamma +2$$

Is this value of trace important for something?

We can also say for four vectors

$$A'^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\nu}}A^{\nu}$$

using this relation and

$$x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}$$

we get

$$A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}$$

So

$$A'^{1}=\alpha^{1}_{\nu}A^{\nu}=\alpha^{1}_{1}A^{1}+\alpha^{1}_{4}A^{4}=\gamma A^{1}-\beta\gamma A^4=\gamma(A^1-\beta A^4)$$

$$A'^{2}=\alpha^{2}_{\nu}A^{\nu}=A^2$$

$$A'^{3}=A^3$$

$$A'^{4}=\alpha^{4}_{\nu}A^{\nu}=\alpha^{4}_{1}A^{1}+\alpha^{4}_{4}A^{4}=-\beta\gamma A^{1}+\gamma A^4=\gamma(A^4-\beta A^1)$$

Look now in antisymmetric tensors

$$A^{\nu\mu}=-A^{\mu\nu}$$

$$A'^{12}=\alpha^1_{\rho}\alpha^2_{\sigma}A^{\rho\sigma}$$

Nonzero terms are terms in which $$\rho$$ takes values $$1,4$$ and $$\sigma$$ takes $$2$$.

I have a question for component

$$A'^{14}=\alpha^1_{\rho}\alpha^4_{\sigma}A^{\rho\sigma}=\alpha^1_{1}\alpha^4_{1}A^{11}+\alpha^1_{1}\alpha^4_{4}A^{44}+\alpha^1_{4}\alpha^4_{4}A^{44}+\alpha^1_{4}\alpha^4_{1}A^{41}=-\beta\gamma^2A^{11}+\gamma^2A^{14}-\beta^2\gamma^2A^{14}-\beta\gamma^2A^{44}$$

Can I say if I look antisymmetric tensors $$A^{11}=A^{44}=0$$?

Then I will get

$$A'^{14}=A^{14}$$

Thanks for your answers!

 

[bcrowell]

Interesting question about the trace.

The determinant does have a nice interpretation. The determinant equals one, and this means that the Lorentz transformation conserves volume in spacetime. This is plays a prominent role in my favorite derivation of the Lorentz transformation. There is a purely geometrical argument, based only on the symmetries of spacetime, that shows that the Lorentz transformation must conserve volume. If you then demand causality, you can prove that there is an invariant velocity c, with Galilean relativity corresponding to the special case where c approaches infinity. So in this approach the existence of an invariant velocity is a theorem rather than an assumption (as Einstein took it to be in 1905).

 

[Petar Mali]

Yes its interesting me to trace question. What is easiest way to diagonalize matrix

$$\alpha=\left(\begin{array}{cccc} \gamma& 0&0& -\beta\gamma\\ 0&1& 0 & 0\\ 0 & 0 & 1 & 0\\ -\beta\gamma & 0 & 0 & \gamma \end{array} \right)$$

?

When we diagonalize this matrix trace must stay the same so

$$Tr(\alpha)=2(\gamma+1)$$

and determinant which will be product of the numbers in main diagonal must be equal $$1$$ also.

 

[bcrowell]

To diagonalize it, you'll have an easier time if you just work in 1+1 dimensions. Then the diagonal form has to have the form diag(a,b). The trace and determinant give $a+b=2\gamma$ and ab=1. Solve for a and b.