- #1

Petar Mali

- 290

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[tex]\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)[/tex][tex]x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}[/tex]

[tex]\alpha[/tex] is Lorrentz transformation matrix.

[tex]Tr(\alpha)=2\gamma +2[/tex]

We can also say for four vectors

[tex]A'^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\nu}}A^{\nu}[/tex]

using this relation and

[tex]x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}[/tex]

we get

[tex]A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}[/tex]

So

[tex]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)[/tex]

[tex]A'^{2}=\alpha^{2}_{\nu}A^{\nu}=A^2[/tex]

[tex]A'^{3}=A^3[/tex]

[tex]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)[/tex]

Look now in antisymmetric tensors

[tex]A^{\nu\mu}=-A^{\mu\nu}[/tex]

[tex]A'^{12}=\alpha^1_{\rho}\alpha^2_{\sigma}A^{\rho\sigma}[/tex]

Nonzero terms are terms in which [tex]\rho[/tex] takes values [tex]1,4[/tex] and [tex]\sigma[/tex] takes [tex]2[/tex].

I have a question for component

[tex]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}[/tex]

Then I will get

[tex]A'^{14}=A^{14}[/tex]

Thanks for your answers!

\gamma& 0&0& -\beta\gamma\\

0&1& 0 & 0\\

0 & 0 & 1 & 0\\

-\beta\gamma & 0 & 0 & \gamma \end{array} \right)[/tex][tex]x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}[/tex]

[tex]\alpha[/tex] is Lorrentz transformation matrix.

**Can I see something more about it?**. It's symmetric. That is important.[tex]Tr(\alpha)=2\gamma +2[/tex]

**Is this value of trace important for something?**We can also say for four vectors

[tex]A'^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\nu}}A^{\nu}[/tex]

using this relation and

[tex]x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}[/tex]

we get

[tex]A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}[/tex]

So

[tex]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)[/tex]

[tex]A'^{2}=\alpha^{2}_{\nu}A^{\nu}=A^2[/tex]

[tex]A'^{3}=A^3[/tex]

[tex]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)[/tex]

Look now in antisymmetric tensors

[tex]A^{\nu\mu}=-A^{\mu\nu}[/tex]

[tex]A'^{12}=\alpha^1_{\rho}\alpha^2_{\sigma}A^{\rho\sigma}[/tex]

Nonzero terms are terms in which [tex]\rho[/tex] takes values [tex]1,4[/tex] and [tex]\sigma[/tex] takes [tex]2[/tex].

I have a question for component

[tex]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}[/tex]

**Can I say if I look antisymmetric tensors [tex]A^{11}=A^{44}=0[/tex]?**Then I will get

[tex]A'^{14}=A^{14}[/tex]

Thanks for your answers!

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