Pauli-Lubanski pseudovector

[andrey21]

Hi can anyone help me prove the result of W₂ of the Pauli-Lubanski pseudovector :
This is very new to me and I've read I must use terms such as J¹³ and P³

Where totally antisymmetric symbol is defined by:
$\epsilon$₁₂₃₄=1 and $\epsilon$₁₂₄₃=-1

 

[dextercioby]

Well, what have you tried ?? As per the guidelines, you must post your attempt.

 

[andrey21]

Ive tried the method mentioned in the post below:

https://www.physicsforums.com/showthread.php?t=245130

 

[dextercioby]

Ok, where did you fail ? Post or attach a scan of your work.

 

[andrey21]

Well here is my answer:

W₂= 0.5 M_$\mu\nu$ M^$\mu\nu$ P² + M_$\mu\rho$ M^$\nu\rho$ P^$\mu$ P_$\nu$

Where the Pauli-Lubanski pseudovector given was:

W_μ= - 0.5 $\epsilon$_{$\mu\nu\rho\sigma$} J^$\nu\rho$ P^$\sigma$

 

[dextercioby]

I get 6 terms when I expand

$$W_2 = -\frac{1}{2} \epsilon_{2\nu\sigma\rho} M^{\nu\sigma}P^{\rho} = -\frac{1}{2} \left( \epsilon_{2013} M^{01}P^{3} + \epsilon_{2031} M^{03}P^{1} + \mbox{4 other terms}\right)$$

I don't think you can regroup them the way you did.

 

[andrey21]

Ok I am a little confused, could you go into more detail regarding your expansion? I am struggling to see how you got those terms!

 

[dextercioby]

Well, the epsilon for a fixed mu (=2) can only take 3 values in 3! combinations. 0,1,3 and the other 5 combinations. That's why from the possible terms you have only 6 remaining.

 

[andrey21]

OK so the other four terms would be:

ε₂₃₀₁ M³⁰P¹
ε₂₃₁₀ M³¹P⁰
ε₂₁₃₀ M¹³P⁰
ε₂₁₀₃ M¹⁰P³

 

[dextercioby]

Yes. Now the epsilons are +/-1 and you can regroup alike terms based on antisymmetry of M.

 

[andrey21]

OK so the antisymmetry rule again is:

M_ab=M_-ab

 

[dextercioby]

M_ab = - M_ba you mean...

EDIT: Yes, exactly. ;)

 

[andrey21]

:rofl: Thats what I meant...

So M₃₁= - M₁₃ for example...

 

[andrey21]

So using that information and that:

ε₁₂₃₄=1 and ε₁₂₄₃=-1

W₂= -1/2 (-J¹⁰P³+ J³⁰P¹+J¹⁰P³ +J³¹P⁰-J³⁰P¹-J³¹P⁰)

=-1/2 (0)

 

[dextercioby]

No, no, no. It's 2 times minus for 3 terms in the bracket, one minus from epsilon, one minus from M, so the terms don't have opposite signs, but equal (choose all 3 with plus). So you have 3 times double contribution. The 2 can be then factored and canceled with the 2 in the denominator.

 

[andrey21]

I understand so you end up with:

W₂= -(J¹⁰P³+J³⁰P¹+J³¹P⁰)

Correct?

 

[dextercioby]

Looks ok. Half the permutations of 0,1,3.

EDIT for post below: Don't mention it.

 

[andrey21]

So I have the correct answer, that's great. Thanks for all your help