# Zwiebach page 143

1. Oct 21, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
On page 143, Zwiebach says that we can define the current tenser to be antisymmetric in mu and nu since it is multiplied by the antisymmetric matrix epsilon^{mu nu}--any symmetric part would drop out of the left hand-side.

But I thought it already was defined in equation 8.55? What does he mean that the symmetric part would drop out?

2. Relevant equations

3. The attempt at a solution

2. Oct 21, 2007

### Jimmy Snyder

Equation (8.55) does not completely define j, only the sum. For instance, you could add an amount to $j_{01}$ and add the same amount to $j_{10}$ and the sum $\epsilon j$ would remain the same. This is what he means by "the symmetric part would drop out."

3. Oct 21, 2007

### ehrenfest

That makes sense. Then below that he says that "the currents can be read directly from this equation because the factor multiplying epsilon^{mu nu} on the RHS is explicitely antisymmetric"?

I do not see why the antisymmetry j^alpha_{mu nu} allows you to do that since you are still summing over 2 indices?

4. Oct 22, 2007

### Jimmy Snyder

Before you assume that j is antisymmetric, you have some play in the values of j since you can add a symmetric part and equation (8.55) will still hold. However, once you assume that j is antisymmetric, that arbitrariness is gone and you can equate the antisymmetic factors on both sides for each index. For instance
$$j_{01} = \frac{1}{2}(j_{01} - j_{10}) = -\frac{1}{2}(X_{0}{\mathcal P}_{1} - X_{1}{\mathcal P}_{0})$$

5. Oct 22, 2007

### ehrenfest

Yes. I understand that is what Zwiebach is claiming; its just that the proof that every antisymmetric matrix is invertible isn't coming to me...

6. Oct 22, 2007

### Jimmy Snyder

Zwiebach doesn't claim that every antisymmetric matrix is invertible and it isn't true.

7. Oct 22, 2007

### ehrenfest

You're right. He is claiming that if epsilon^{\mu \nu} is antisymmetric and M^{\mu \nu} is antisymmetric, then if $$\epsilon^{\mu \nu} M_{\mu \nu} = \epsilon^{\mu \nu} N_{\mu \nu}$$, $$M_{\mu \nu} = N_{\mu \nu}$$, right? How do you prove that?

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