Maxwell's tensorial equations.

[MathematicalPhysicist]

I wanto show that:
$$\partial_{\mu} T^{\mu \nu}=0$$ for
$$T^{\mu\nu} = F^{\mu \rho}\eta_{\rho\sigma}F^{\sigma \nu}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma})$$
by using Maxwell's equations.
Here are my steps (it's not for HW, I am auditing this course):
$$\partial_{\mu} T^{\mu \nu} = F^{\mu \rho}_{,\mu} \eta_{\rho \sigma} F^{\sigma \nu} + F^{\mu \rho} \eta_{\rho \sigma} F^{\sigma \nu}_{,\mu} + \frac{1}{4} \eta^{\mu \nu} (F_{\rho \sigma , \mu} F^{\rho \sigma}+F_{\rho \sigma} F^{\rho \sigma}_{,\mu})$$

I can't see the forrest from the trees, can someone hint me how to simplify this?

Thanks.

 

[LAHLH]

You need to use the Maxwell equations written in terms of F, i.e. $$\nabla_a F^{ab} =0$$ (kills your first term for e.g.) and $$\nabla_{[a}F_{bc]}=0$$

and remember that F is antisymmetric itself, which simplifies the second equation somewhat to $$\partial_a F_{bc}+\partial_b F_{ca}+\partial_c F_{ab} =0$$. You also have met compatibility and can relabel dummies and use the symmetry and antisymmetry of F, this should do the trick.

I think you will find you need to factor out an F after this, and then massage the rest into a form where you can use the second Maxwell relation by using symmetries and relabelling dummies, and then by the second equation you will get identically zero.

 

[MathematicalPhysicist]

So I get now:
$$F^{\mu \rho}_{,\mu} \eta_{\rho \sigma} F^{\sigma \nu} + F^{\mu \rho} \eta_{\rho \sigma} F^{\sigma \nu}_{,\mu} + \frac{1}{4} \eta^{\mu \nu} (F_{\rho \sigma , \mu} F^{\rho \sigma}+F_{\rho \sigma} F^{\rho \sigma}_{,\mu})= F^{\mu}_{\sigma , \mu} F^{\sigma \nu} + F^{\mu \rho}F^{\nu}_{\rho , \mu} -\frac{1}{4}[ F_{\rho \sigma , \nu} F^{\rho \sigma} +F_{\rho \sigma}F^{\rho \sigma}_{,\nu}]$$

How to procceed here? what is the compatibiliy condition here?

Thanks.

 

[homology]

Have you thought about working out $$F^{\mu\rho}_{,\mu}$$ in terms of the 4-potential. You'll get two terms and there will be one free index and two dummy ones. You can swap the dummy indices but when you do that for F you'll pick up a minus sign and the terms will add giving you one term. Then you can argue its zero because its the product of an antisymmetric part and a symmetric part.

I tend to have to write things out because I'm still new to this stuff and so I can't do it all just using F yet. Just a thought.

 

[LAHLH]

I'm not sure exactly what you've done here. Starting with (where I used Maxwell 1 to rid the first term): $$\eta_{cd}F^{ac} \ F^{bd}_{,a} - \frac{1}{4} \eta^{ab} (F_{cd , a} F^{cd}+F_{cd} F^{cd}_{,a})$$

$$\eta_{cd}F^{ac} \ F^{bd}_{,a} - \frac{1}{2} \eta^{ab} F^{cd} F_{cd , a}$$

antisym of F:

$$\eta_{cd}F^{ac} \ F^{bd}_{,a} + \frac{1}{2} \eta^{ab} F^{dc} F_{cd , a}$$

swap dummies a<->d on second term:

$$\eta_{cd}F^{ac} \ F^{bd}_{,a} + \frac{1}{2} \eta^{db} F^{ac} F_{ca , d}$$

now pull out the factor of F, and I use metric sym:

$$F^{ac}\left[\eta_{cd} F^{bd}_{,a} + \frac{1}{2} \eta^{bd}F_{ca , d}\right]$$

Because your in flat space the metric is constant so can be brought inside the partial derivs to raise/lower indices at will (in a curved space where your commas go to semicolons one would use metric compatability $$g_{ab;c}=0$$ at this stage):

$$F^{ac}\left[\eta_{cd}\eta^{be}\eta^{df} F_{ef,a} + \frac{1}{2} \eta^{bd}F_{ca , d}\right]$$

Now use $$\eta_{cd}\eta^{df}=\delta^{f}_{c}$$

$$F^{ac}\left[\eta^{be} F_{ec,a} + \frac{1}{2} \eta^{bd}F_{ca , d}\right]$$

Relabel the dummie e <->d in first term, then pull out a factor of the metric too:

$$F^{ac}\eta^{bd}\left[ F_{dc,a} + \frac{1}{2} F_{ca , d}\right]$$

Now this can be written as:

$$F^{ac}\eta^{bd}\left[\frac{1}{2} F_{dc,a} +\frac{1}{2} F_{dc,a}+ \frac{1}{2} F_{ca , d}\right]$$

All you have to do now is use the antisymmetry on F and symmetry on the metric, and relabel a few indices to manipulate the above in a form where the indices are correct to use Maxwell 2 : $$\partial_a F_{bc}+\partial_b F_{ca}+\partial_c F_{ab} =0$$ which will then be identically zero. (Note indices must end up cyclic)

 

[MathematicalPhysicist]

THanks.

As for the last eq.
$$F^{ac}\eta^{bd}\left[\frac{1}{2} F_{dc,a} +\frac{1}{2} F_{dc,a}+ \frac{1}{2} F_{ca , d}\right]$$
if I plug:
$$F_{dc,a}= F_{ad,c}$$
then I am done right?

 

[LAHLH]

So the first term is:

$$F^{ac}\eta^{bd} F_{dc,a}$$

antisym of F:$$-F^{ac}\eta^{bd} F_{cd,a}$$

now swap a<->c dummies:

$$-F^{ca}\eta^{bd} F_{ad,c}$$

antisym on first F:$$F^{ac}\eta^{bd} F_{ad,c}$$

so yep, you can do that (and the factor outside remains the same), so now it vanishes via second Maxwell.

 

[MathematicalPhysicist]

Thanks again.

Algebriac trickery... :-)