Zwiebach Pg 143: Defining the Current Tensor & Equation 8.55

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Homework Statement


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?


Homework Equations





The Attempt at a Solution

 

Discussion

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ehrenfest said:
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?
Equation (8.55) does not completely define j, only the sum. For instance, you could add an amount to [itex]j_{01}[/itex] and add the same amount to [itex]j_{10}[/itex] and the sum [itex]\epsilon j[/itex] would remain the same. This is what he means by "the symmetric part would drop out."
 
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?
 
ehrenfest said:
I do not see why the antisymmetry j^alpha_{mu nu} allows you to do that since you are still summing over 2 indices?
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
[tex]j_{01} = \frac{1}{2}(j_{01} - j_{10}) = -\frac{1}{2}(X_{0}{\mathcal P}_{1} - X_{1}{\mathcal P}_{0})[/tex]
 
jimmysnyder said:
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
[tex]j_{01} = \frac{1}{2}(j_{01} - j_{10}) = -\frac{1}{2}(X_{0}{\mathcal P}_{1} - X_{1}{\mathcal P}_{0})[/tex]

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...
 
ehrenfest said:
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...
Zwiebach doesn't claim that every antisymmetric matrix is invertible and it isn't true.
 
You're right. He is claiming that if epsilon^{\mu \nu} is antisymmetric and M^{\mu \nu} is antisymmetric, then if [tex]\epsilon^{\mu \nu} M_{\mu \nu} = \epsilon^{\mu \nu} N_{\mu \nu}[/tex], [tex]M_{\mu \nu} = N_{\mu \nu}[/tex], right? How do you prove that?