Momentum of electromagnetic field

[neerajareen]

Hello, I am trying to prove that the momentum of an electromagnetic field is $$E \times B$$ by considering the conserved quantity due to the spatial translation of the Lagrangian.

$$L = - \frac{1}{4}\int {{F^{\mu v}}{F_{\mu v}}} {d^3}x$$

So far, I have calculated the canonical momentum.

$${\Pi _{{A_\mu }}} = \frac{{\partial {A_\mu }}}{{\partial t}} = \left( {\begin{array}{*{20}{c}}<br /> 0\\<br /> {{E_x}}\\<br /> {{E_y}}\\<br /> {{E_z}}<br /> \end{array}} \right)$$

I know that the conserved quantity is

$$Q = \int {{\Pi _{{A_\mu }}}} \delta {A_\mu }{d^3}x$$

But I am not sure how $$\delta {A_\mu }$$ is going to give me components of the curl.
Thank you very much.

 

[samalkhaiat]

The 3-momentum of the field is given by
$$P^{ j } = \int d^{ 3 }x \ T^{ 0 j } ,$$
where
$$T^{ \mu \nu } = \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ \mu } A_{ \sigma } ) } \partial^{ \nu } A_{ \sigma } - \eta^{ \mu \nu } \mathcal{ L } .$$
If you use the relation
$$\frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ \mu } A_{ \sigma } ) } = 4 F^{ \mu \sigma } ,$$
you find
$$P^{ j } = - \int d^{ 3 }x \ \eta_{ \mu \sigma }\ F^{ 0 \sigma }\ \partial^{ j } A^{ \mu } .$$
Now, add and subtract $\partial^{ \mu } A^{ j }$, integrate by parts, neglect surface term, and use the equation of motion $\partial_{ \sigma } F^{ 0 \sigma } = 0$. You will find
$$P^{ j } = \int d^{ 3 }x \ \eta_{ \mu \nu } F^{ 0 \mu } F^{ \nu j } = \int d^{ 3 }x \ \eta_{ i k } F^{ 0 i } F^{ k j } = \int d^{ 3 }x \left( \vec{ E } \times \vec{ H } \right)^{ j } .$$

Sam

 

[dextercioby]

It's important to know the difference between momentum and volumic momentum density. E × H is the momentum density of the free field.