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4-Vector Problem.

Hi, I originally posted this in the maths forum but perhaps I should have posted in Physics to begin with, basically,

I have a Lagrangian:

Unparseable latex formula:

\mathcal{L}=\dfrac {-1} {16\pi }\left( \partial ^{\mu }A^{\nu}\right -\partial ^{\nu }A^{\mu })\left( \partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)+\dfrac {1} {8\pi }\left( \dfrac {mc} {h}\right) ^{2}A^{\nu }A_{\nu }



and I need to get to the result:

L(μAv)=14π(μAννAμ)\dfrac {\partial \mathcal{L}} {\partial \left( \partial _{\mu }Av\right) }=-\dfrac{1}{4\pi }\left(\partial^{\mu }A^{\nu }-\partial^{\nu }A^{\mu }\right)

The answer in the book that I'm working through does this by first finding the answer for all components involving 0A1\partial _{0}A_{1} and its contraction. I can follow this method fine and it gives the correct result for that choice of indices and then it just generalises it.

However, I'd much rather know how to do using the general indices the entire way through, as I need to know this kinda stuff for my general relativity course.

I've managed to get to:

(μAννAμ)(μAννAμ)=(μAν)2μAννAμνAμμAν(νAμ)2\left( \partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }\right) \left( \partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)=-\left( \partial _{\mu }A_{\nu }\right)^{2}-\partial ^{\mu}A^{\nu}\partial _{\nu}A_{\mu}-\partial^{\nu}A^{\mu}\partial _{\mu}A_{\nu}-(\partial _{\nu }A_{\mu})^{2}

Could someone show me what to do next? I haven't got a clue, I'm not very good at being able to manipulate the indices.

Thaanks! I'll give rep to whoever helps if that's any incentive :smile:
(edited 11 years ago)
Reply 1
Original post by You Failed
Hi, I originally posted this in the maths forum but perhaps I should have posted in Physics to begin with, basically,

I have a Lagrangian:

Unparseable latex formula:

\mathcal{L}=\dfrac {-1} {16\pi }\left( \partial ^{\mu }A^{\nu}\right -\partial ^{\nu }A^{\mu })\left( \partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)+\dfrac {1} {8\pi }\left( \dfrac {mc} {h}\right) ^{2}A^{\nu }A_{\nu }



and I need to get to the result:

L(μAv)=14π(μAννAμ)\dfrac {\partial \mathcal{L}} {\partial \left( \partial _{\mu }Av\right) }=-\dfrac{1}{4\pi }\left(\partial^{\mu }A^{\nu }-\partial^{\nu }A^{\mu }\right)

The answer in the book that I'm working through does this by first finding the answer for all components involving 0A1\partial _{0}A_{1} and its contraction. I can follow this method fine and it gives the correct result for that choice of indices and then it just generalises it.

However, I'd much rather know how to do using the general indices the entire way through, as I need to know this kinda stuff for my general relativity course.

I've managed to get to:

(μAννAμ)(μAννAμ)=(μAν)2μAννAμνAμμAν(νAμ)2\left( \partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }\right) \left( \partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\right)=-\left( \partial _{\mu }A_{\nu }\right)^{2}-\partial ^{\mu}A^{\nu}\partial _{\nu}A_{\mu}-\partial^{\nu}A^{\mu}\partial _{\mu}A_{\nu}-(\partial _{\nu }A_{\mu})^{2}

Could someone show me what to do next? I haven't got a clue, I'm not very good at being able to manipulate the indices.

Thaanks! I'll give rep to whoever helps if that's any incentive :smile:


Hmm I've worked through this but don't particularly fancy trying to write a pageful of index notation in Latex...

A few hints:

Firstly when you do the differentiation you need to change the indices ie if you have L in terms of mu and nu indices you'll want something like L(σAρ)\dfrac {\partial \mathcal{L}} {\partial \left( \partial _{\sigma }A_{\rho}\right) } to make sure the indices all match up.

Secondly you don't need to expand your original Lagrangian. You can just use the product rule as normal, and keeping in mind that (μAν)(σAρ)=δμσδρν \frac{\partial ( \partial_{\mu} A_{\nu})}{\partial ( \partial_{\sigma}A_{\rho})}= \delta^{\sigma}_{\mu} \delta^{\nu}_{\rho} and remembering that you can use the metric to raise and lower indices as needed.

I've attached a photo of my working, but not sure how clear it is...
(edited 11 years ago)
Reply 2
Original post by suneilr
Hmm I've worked through this but don't particularly fancy trying to write a pageful of index notation in Latex...

A few hints:

Firstly when you do the differentiation you need to change the indices ie if you have L in terms of mu and nu indices you'll want something like L(σAρ)\dfrac {\partial \mathcal{L}} {\partial \left( \partial _{\sigma }A_{\rho}\right) } to make sure the indices all match up.

Secondly you don't need to expand your original Lagrangian. You can just use the product rule as normal, and keeping in mind that (μAν)(σAρ)=δμσδρν \frac{\partial ( \partial_{\mu} A_{\nu})}{\partial ( \partial_{\sigma}A_{\rho})}= \delta^{\sigma}_{\mu} \delta^{\nu}_{\rho} and remembering that you can use the metric to raise and lower indices as needed.

I've attached a photo of my working, but not sure how clear it is...


Thaanks! Really appreciate you taking the time to do this!

Can I just ask why if you don't change the indices as you have done, why they won't match up?
(edited 11 years ago)
Reply 3
Original post by You Failed
Thaanks! Really appreciate you taking the time to do this!

Can I just ask why if you don't change the indices as you have done, why they won't match up?


It's because you'll end up with three of the same indices. The Lagrangian is a scalar which already contains a sum over the mu and nu indices, so adding in an extra mu and nu index doesn't make any sense.
Reply 4
Original post by suneilr
It's because you'll end up with three of the same indices. The Lagrangian is a scalar which already contains a sum over the mu and nu indices, so adding in an extra mu and nu index doesn't make any sense.


Ah yeah I see, cool, thanks :smile:
Reply 5
Original post by You Failed
Ah yeah I see, cool, thanks :smile:


No problem. It's good to see I still remember how to do these :smile:

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