Antisymmetry of the electromagnetic field tensor

[shinobi20]

Homework Statement

Show that the electromagnetic field tensor is antisymmetric from the fact that $p_\mu p_\nu F^{\mu \nu} = 0$.

Relevant Equations

$p_\mu p_\nu F^{\mu \nu} = 0$
where $p_\mu$ is the 4-momentum and $F^{\mu \nu}$ is the EM field tensor

I am trying to answer exercise 5 but I am not sure I understand what the hint is implying, differentiate with respect to $p_\alpha$ and $p_\beta$, I have done this but nothing is clicking. Also, what is the relevance of the hint "the constraint $p^\alpha p_\alpha = m^2c^2$ can be ignored ..."? Please help me clarify what the author wants me to think or do in accordance to the hint.

 

[mitochan]

$p_\mu p_\nu$ is a symmetric tensor.

 

[Dr Transport]

try writing F^{\mu\nu} as a sum of symmetric and anti symmetric pieces then do the contraction with the p^\alpha to show that the components cancel.

or try writing F^{\mu\nu} using its definition in terms of A^\mu and \partial^\nu

 

[shinobi20]

try writing F^{\mu\nu} as a sum of symmetric and anti symmetric pieces then do the contraction with the p^\alpha to show that the components cancel.

or try writing F^{\mu\nu} using its definition in terms of A^\mu and \partial^\nu

I cannot figure out what you want me to do, but then I want to follow the hint given in OP.

My attempt, differentiating with respect to $p_\alpha$ and $p_\beta$,

$\delta^\mu_\alpha p_\nu F^{\mu\nu} + p_\mu \delta^\nu_\alpha F^{\mu\nu} = 0 \quad \rightarrow \quad p_\nu F^{\alpha\nu} + p_\mu F^{\mu\alpha} = 0$

$\delta^\mu_\beta p_\nu F^{\mu\nu} + p_\mu \delta^\nu_\beta F^{\mu\nu} = 0 \quad \rightarrow \quad p_\nu F^{\beta\nu} + p_\mu F^{\mu\beta} = 0$

Multiplying both sides of the first equation by $p^\mu$

$p^\mu p_\nu F^{\alpha\nu} + m^2 c^2 F^{\mu\alpha} = 0$

Multiplying both sides of the second equation by $p^\nu$

$m^2 c^2 F^{\beta\nu} + p^\nu p_\mu F^{\mu\beta} = 0$

From these last two equations, I think there are only a few steps before I get the answer, but I cannot see it.

**I may be wrong by multiplying $p^\nu$ in the second equation, maybe it should also be $p^\mu$, but the flow is as above.

 

[mitochan]

(Further to post #2)
p_\mu p_\nu F^{\mu \nu}=p_\mu p_\nu (S^{\mu \nu}+A^{\mu \nu})
where S is symmetric and A is anti-symmetric components of F.
Contract of symmetric tensor, here $p_\mu p_\nu$, and anti-symmetric tensor, here $A^{\mu \nu}$, is always zero.
p_\mu p_\nu A^{\mu \nu}=0
This and
p_\mu p_\nu F^{\mu \nu}=0
give
p_\mu p_\nu S^{\mu \nu}=0
S^{\mu \nu}=0
So $F^{\mu \nu}$ has only anti-symmetric components.

 

[vela]

I cannot figure out what you want me to do, but then I want to follow the hint given in OP.

My attempt, differentiating with respect to $p_\alpha$ and $p_\beta$,

$\delta^\mu_\alpha p_\nu F^{\mu\nu} + p_\mu \delta^\nu_\alpha F^{\mu\nu} = 0 \quad \rightarrow \quad p_\nu F^{\alpha\nu} + p_\mu F^{\mu\alpha} = 0$

Try differentiating this result with respect to $p_\beta$.

 

[shinobi20]

Try differentiating this result with respect to $p_\beta$.

I don't consider myself stupid, but there are some days where I think I am one. It is like having a snake in front of me but still don't notice it. The hint given in OP is very clear, you just restated what it said but in a different way.

What is your advice on this? Is this just a physics thing that can be improved within physics?

 

[vela]

I don't consider myself stupid, but there are some days where I think I am one. It is like having a snake in front of me but still don't notice it. The hint given in OP is very clear, you just restated what it said but in a different way.

What is your advice on this? Is this just a physics thing that can be improved within physics?

I think everyone has had moments like these, and then we feel like idiots for not seeing what's right in front of us. The best advice I can give is to be aware of the possibility of tunnel vision so you can break out of a rut sooner rather than later.