# Homework Help: Pauli-Lubanski pseudovector

1. Nov 30, 2011

### 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

2. Nov 30, 2011

### dextercioby

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

3. Nov 30, 2011

### andrey21

4. Nov 30, 2011

### dextercioby

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

5. Nov 30, 2011

### andrey21

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:

Wμ= - 0.5 $\epsilon$$\mu\nu\rho\sigma$ J$\nu\rho$ P$\sigma$

6. Nov 30, 2011

### 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.

7. Nov 30, 2011

### andrey21

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

8. Nov 30, 2011

### 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.

9. Nov 30, 2011

### andrey21

OK so the other four terms would be:

ε2301 M30P1
ε2310 M31P0
ε2130 M13P0
ε2103 M10P3

10. Nov 30, 2011

### dextercioby

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

11. Nov 30, 2011

### andrey21

OK so the antisymmetry rule again is:

Mab=M-ab

12. Nov 30, 2011

### dextercioby

M_ab = - M_ba you mean...

EDIT: Yes, exactly. ;)

Last edited: Nov 30, 2011
13. Nov 30, 2011

### andrey21

:rofl: Thats what I meant...

So M31= - M13 for example...

14. Nov 30, 2011

### andrey21

So using that information and that:

ε1234=1 and ε1243=-1

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

=-1/2 (0)

15. Nov 30, 2011

### 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 cancelled with the 2 in the denominator.

16. Nov 30, 2011

### andrey21

I understand so you end up with:

W2= -(J10P3+J30P1+J31P0)

Correct?

17. Nov 30, 2011

### dextercioby

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

EDIT for post below: Don't mention it.

18. Nov 30, 2011

### andrey21

So I have the correct answer, thats great. Thanks for all your help