# Lorentz transformation of the "bilinear spinor matrixelement"

• I
Gold Member

## Main Question or Discussion Point

there is a physics problem where I couldn't understand what the solutions.
It is about the lorentz transformation of a bilinear spinor matrix element thing.

So the blue colored equation signs are the parts which I couldn't figure out how.
There must be some steps in between which were skipped on the solution.

For example, we know that the original matrixelement contains Γ12T. Why does it turn back to ΓμνT? Where does the Γ02T element come from?

And lastly, why did we have to transpose the Γ's in the end, so that the indices get to the top?

Thank you very much for your time :)

-Garlic

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George Jones
Staff Emeritus
Gold Member
For example, we know that the original matrixelement contains Γ12T. Why does it turn back to ΓμνT?
First question first. Suppose
$$A_{\mu \nu} =\bar{\psi} \Gamma^{T}_{\mu \nu} \psi.$$
In the primed frame, what is ##A'_{\alpha \beta}##?

Gold Member
First question first. Suppose
$$A_{\mu \nu} =\bar{\psi} \Gamma^{T}_{\mu \nu} \psi.$$
In the primed frame, what is ##A'_{\alpha \beta}##?
I suppose:
$$A'_{\alpha \beta}= \Lambda_{\alpha}^{\: \mu} \Lambda_{\beta}^{\: \nu} A'_{\mu \nu} = \Lambda_{ \alpha}^{\: \mu} \Lambda_{\beta}^{\: \nu} \bar{\psi} \Gamma^{T}_{\mu \nu} \psi$$

So M should transform like this:
$$M_{\alpha \beta} \rightarrow M'_{\alpha \beta}= \Lambda_{\alpha}^{\: 1} \Lambda_{ \beta}^{\: 2} \bar{\psi} \Gamma^{T}_{1 2} \psi$$
My intuition tells me the opposite of the solution...

Edit: I have an idea. ##\Gamma_{12}^T =\gamma^1 \gamma^2## so the indices on lorentz transformations should be inverted?

$$M_{\alpha \beta} \rightarrow M'_{\alpha \beta}= \Lambda_{1}^{\: \alpha} \Lambda_{ 2}^{\: \beta} \bar{\psi} \Gamma^{T}_{1 2} \psi$$
But still the \Gamma matrix has non specific indices.

Also I probably did the index ordering wrong. ##\Lambda^{ 1}_{\: \alpha}## or ##\Lambda_{ \alpha}^{\: 1}##

Last edited:
George Jones
Staff Emeritus
Gold Member
So M should transform like

Sorry, my hint wasn't very good. Try this. Write
$$M' =\bar{\psi'} \left(x'\right) \Gamma^{T}_{12} \psi' \left(x'\right).$$
Now write each ##\psi' \left(x'\right)## in terms of ##\psi \left(x\right)##.

Gold Member
Sorry, my hint wasn't very good. Try this. Write
$$M' =\bar{\psi'} \left(x'\right) \Gamma^{T}_{12} \psi' \left(x'\right).$$
Now write each ##\psi' \left(x'\right)## in terms of ##\psi \left(x\right)##.
Maybe like this?

$$M' = \bar{\psi} \: (x) S(\Lambda^{-1}) \Gamma^{T}_{12} S(\Lambda) \psi \: (x)$$

George Jones
Staff Emeritus
Gold Member
Maybe like this?

$$M' = \bar{\psi} \: (x) S(\Lambda^{-1}) \Gamma^{T}_{12} S(\Lambda) \psi \: (x)$$
Now use ##I = S\left(\Lambda\right) S\left(\Lambda^{-1}\right)## between the ##\gamma## s.

Gold Member
So I just figured it out. :)
Although the indices are upside down, but I have a feeling they are the same as in the solution.

Now I'm stuck at the second part:
what am I doing wrong here? Am I using a wrong lorentz matrix?

Gold Member
I understand it! I can't believe I did it!! I feel intelligent

Thank you so much for helping me!!!

vanhees71