# Can't get this product of Lorentz matrices right -- help please

1. Feb 25, 2015

Hey guys,

So consider the following product of matrices:
$(p_{1}^{\mu}\cdot p_{1}^{\prime\nu} -(p_{1}\cdot p_{1}')\eta^{\mu\nu}+p_{1}^{\nu}p_{1}^{\prime\mu})(p_{2\mu}p_{2\nu}'-(p_{2}\cdot p_{2}')\eta_{\mu\nu}+p_{2\nu}p_{2\mu}')$

where eta is the Minkowski metric.

I keep getting

$2(p_{1}\cdot p_{2})(p_{1}'\cdot p_{2}')+2(p_{1}\cdot p_{2}')(p_{1}'\cdot p_{2}) - 3(p_{1}\cdot p_{1}')(p_{2}\cdot p_{2}')$

But apparently its wrong; I'm meant to get just
$2(p_{1}\cdot p_{2})(p_{1}'\cdot p_{2}')+2(p_{1}\cdot p_{2}')(p_{1}'\cdot p_{2})$

Last edited by a moderator: Feb 25, 2015
2. Feb 25, 2015

### Orodruin

Staff Emeritus
It would be easier to see where you go wrong if you include your middle steps. The answer you are supposed to get seems correct.

3. Feb 25, 2015

Okay I'll write it out explicitly for you, please bear with me a moment.

4. Feb 25, 2015

Here it is...

Btw is $\eta^{\mu\nu}\eta_{\mu\nu}=1$ or $-1$ lol XD I've assumed its +1

5. Feb 25, 2015

### Orodruin

Staff Emeritus
It is neither ...
$$\eta_{\mu\nu} \eta^{\mu\nu} = \delta^\mu_\mu = \ldots$$

Edit: Your first expression in your attempt also does not match what you wrote in the OP. Only what you wrote in the attempt makes sense together with the presumtive answer so I am going to edit your OP to reflect this.

6. Feb 25, 2015

OMG so those terms vanish? :O

7. Feb 25, 2015

### Orodruin

Staff Emeritus
That would depend on what terms you are referring to. What did you get for the trace of the delta?

8. Feb 25, 2015

That's fine there is a typo...there is a cdot somewhere it shouldnt be in the first term.

But mu isnt = nu and there is only one term where the two etas are being contracted. If that term goes to 0 I get

9. Feb 25, 2015

### Orodruin

Staff Emeritus
Sorry, but it is not clear what you are doing with your etas. What did you get for $\eta^{\mu\nu}\eta_{\mu\nu}$ in the end? This is of crucial importance for the problem so you need to write these steps out. Neglecting the term where the etas contract you should get -4 in front of the term where the p1s are contracted with each other (and te p2s with each other).

10. Feb 25, 2015

I think it's easier if I just tell you a few terms. So in each line there is an example of a product of terms:

11. Feb 25, 2015

### Orodruin

Staff Emeritus
Yes, the question mark is the crucial point here. What is the trace of the delta tensor?

12. Feb 25, 2015

is it 4..?

13. Feb 25, 2015

Yes doctor that solves my problem. The terms now cancel if I have a factor of 4 in front of one of them due to the trace of the delta tensor in spacetime. You saved the day once more doctor, you should consider becoming a superhero :D thank you!

14. Feb 25, 2015

### Orodruin

Staff Emeritus
Just for closure: Yes, it is 4. In general it is equal to the total number of dimensions.