# Proca Lagrangian with extra operator - find Laws of motion

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Maniac_XOX
TL;DR Summary
Im trying to find the laws of motion for a section of a lagrangian that would represent an extra operator to the proca theory for investigating additional properties of photons inside superconductors. Any help?
The euler lagrange equation I am using is:
$$\frac {\partial^\beta \partial L}{\partial(\partial^\beta A^\alpha) }= \frac {\partial L} {\partial A^\alpha}$$ Now the proca lagrangian i am using is $$L= -\frac {1}{16\pi} F_{\alpha\beta} F^{\alpha\beta} + \frac {\mu^2} {8\pi} A_\alpha A^\alpha - \frac {1} {c} J_\alpha A^\alpha$$ No problem in finding the laws of motion for this lagrangian, being $$\partial^\beta F_{\beta\alpha} + \mu^2 A_\alpha = \frac {4\pi}{c} J_\alpha$$ the extra operator that I have added to the proca lagrangian is $$-\beta A_\mu A^\mu (\partial_\rho A^\rho)$$ taken from https://arxiv.org/pdf/1402.6450.pdf, equation (3).
Using the euler lagrange equation, I found that:
(1) $$\partial^\beta \frac {\partial L}{\partial (\partial^\beta A^\alpha)}=-\partial^\beta(g_{\rho\sigma} \beta A_\mu A^\mu)$$ If i lower the index can this turn into ##-\partial^\beta(\beta A_\mu A_\mu)## ??
And also:
(2) $$\frac {\partial L}{\partial A^\alpha}=-2\beta A_\mu (\partial_\rho A^\rho)$$
Therefore
(3) $$\partial^\beta(\beta A_\mu A_\mu)=2\beta A_\mu (\partial_\rho A^\rho)$$ Is this okay?

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Homework Helper
$$\partial^\beta \frac {\partial L}{\partial (\partial^\beta A^\alpha)}=-\partial^\beta(g_{\rho\sigma} \beta A_\mu A^\mu)$$
Can you elaborate on how do you get this result?

Maniac_XOX
Maniac_XOX
Can you elaborate on how do you get this result?
$$\frac {\partial}{\partial(\partial^\beta A^\alpha)}[-\beta A_\mu A^\mu (\partial_\rho A^\rho)]$$ raise index to match the euler lagrange $$\frac {\partial}{\partial(\partial^\beta A^\alpha)}[-\beta A_\mu A^\mu g_{\rho\sigma}(\partial^\sigma A^\rho)]$$
Therefore since I am looking for the derivative in terms of ##\partial^\rho A^\rho## i thought the answer would be $$-\beta A_\mu A^\mu g_{\rho\sigma}$$

Homework Helper
Therefore since I am looking for the derivative in terms of ##\partial^\rho A^\rho## i thought the answer would be $$-\beta A_\mu A^\mu g_{\rho\sigma}$$
Not quite, what is the value of the derivative
$$\frac{\partial(\partial^\sigma A^\rho)}{\partial(\partial^\beta A^\alpha)}=?$$

Maniac_XOX
Not quite, what is the value of the derivative
$$\frac{\partial(\partial^\sigma A^\rho)}{\partial(\partial^\beta A^\alpha)}=?$$
Ohh, that's 0? I am actually a bit confused and i might not even know that

Homework Helper
No, it's also not zero. An even more simple example is to first compute the derivative
$$\frac{\partial A^\mu}{\partial A^\nu}$$
do you know how to compute this one?

Maniac_XOX
No, it's also not zero. An even more simple example is to first compute the derivative
$$\frac{\partial A^\mu}{\partial A^\nu}$$
do you know how to compute this one?
So I'm new to the use of indeces like this but I'm quite sure that expands to $$\frac {\partial A^t}{\partial t} + \frac {\partial A^x}{\partial x} + \frac {\partial A^y}{\partial y} + \frac {\partial A^z}{\partial z}$$

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Homework Helper
So I'm new to the use of indeces like this but I'm quite sure that expands to $$\frac {\partial A^t}{\partial t} + \frac {\partial A^x}{\partial x} + \frac {\partial A^y}{\partial y} + \frac {\partial A^z}{\partial z}$$
Ok, not quite. What you wrote is ##\partial_\mu A^\mu=\frac{\partial A^\mu}{\partial x^\mu}##. What I gave you is ##\frac{\partial A^\mu}{\partial A^\nu}##. Notice the main differences:
• What I gave you has no ##x##. So we are not taking derivatives with respect to the 4-position, but respect to the field itself.
• One has a repeated index, the other no
In Einstein sum convention each greek index can take 4 possible values ##(0,1,2,3)## (or ##(t,x,y,z)##), when the index is the same it is interpreted as being summed over the 4 values, as the first case. Since there is no other "free" index, we have only one expression.
The other one has no repeated index, so there no sum of any kind. On the other hand, it has two free indices. Therefore is a compact way to write a total of 4x4=16 expressions.

Maniac_XOX
Ok, not quite. What you wrote is ##\partial_\mu A^\mu=\frac{\partial A^\mu}{\partial x^\mu}##. What I gave you is ##\frac{\partial A^\mu}{\partial A^\nu}##. Notice the main differences:
• What I gave you has no ##x##. So we are not taking derivatives with respect to the 4-position, but respect to the field itself.
• One has a repeated index, the other no
In Einstein sum convention each greek index can take 4 possible values ##(0,1,2,3)## (or ##(t,x,y,z)##), when the index is the same it is interpreted as being summed over the 4 values, as the first case. Since there is no other "free" index, we have only one expression.
The other one has no repeated index, so there no sum of any kind. On the other hand, it has two free indices. Therefore is a compact way to write a total of 4x4=16 expressions.
Awesome, my lecturers should have explained it that way lol.
So i figured the answer to your first question is actually using the kroneker delta property.
$$\frac{\partial(\partial^\sigma A^\rho)}{\partial(\partial^\beta A^\alpha)}$$ turns out to be $$\delta^\alpha_\sigma \delta^\rho_\beta$$ ??

Homework Helper
Awesome, my lecturers should have explained it that way lol.
So i figured the answer to your first question is actually using the kroneker delta property.
$$\frac{\partial(\partial^\sigma A^\rho)}{\partial(\partial^\beta A^\alpha)}$$ turns out to be $$\delta^\alpha_\sigma \delta^\rho_\beta$$ ??
That much closer to the correct answer. However, it is not right yet. Notice that for example
$$\frac{\partial(\partial^1 A^0)}{\partial(\partial^1 A^0)}=1$$
As I hope is evident. However, your solution gives
$$\delta^0_1 \delta^0_1=0$$

Maniac_XOX
That much closer to the correct answer. However, it is not right yet. Notice that for example
$$\frac{\partial(\partial^1 A^0)}{\partial(\partial^1 A^0)}=1$$
As I hope is evident. However, your solution gives
$$\delta^0_1 \delta^0_1=0$$
Does it have to do with how i arranged the indeces? Because I'm confused since I used the same property when using proca lagrangian and it turned out alright

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Homework Helper
Does it have to do with how i arranged the indeces? Because I'm confused since I used the same property when using proca lagrangian and it turned out alright
Yes, the problem is with the indices. But anyway it is wrong, if you have used this property proving the proca EoM, then it is also wrong.

Maniac_XOX
Yes, the problem is with the indices. But anyway it is wrong, if you have used this property proving the proca EoM, then it is also wrong.
I disagree, if you have a look at Jackson's Classical Electrodynamics book (can download at https://physicsdir.wordpress.com/2018/11/02/classical-electrodynamics-solution/)
In section 12.8 he clearly uses that property to find the laws of motion. It is cut short but i have derived the entire process and i got the same result.. the laws of motion for the proca lagrangian i presented in my summary are def the correct ones since lecturers also approved.. but please, explain what's wrong with the indices? To elaborate, i used the identity $$\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\alpha_\mu \delta^\nu_\beta$$(which by the way i found from another forum in this site, and has helped another student with a similar purpose)
this is what has been working for me through the project

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Homework Helper
I don't know which edition you have, but the one I have there no such equation in section 12.8.
In any case, it is wrong at least with my understanding of the notation (and I'm not aware of any other notation that would make this correct, but if it is in Jackson...)
In any case, I don't care if it is in Jackon, or in the Bible. Compute the case ##\mu=\beta=1##, ##\nu=\alpha=0## as I did and tell if this equation holds...

Maniac_XOX
I don't know which edition you have, but the one I have there no such equation in section 12.8.
In any case, it is wrong at least with my understanding of the notation (and I'm not aware of any other notation that would make this correct, but if it is in Jackson...)
In any case, I don't care if it is in Jackon, or in the Bible. Compute the case ##\mu=\beta=1##, ##\nu=\alpha=0## as I did and tell if this equation holds...
I have attached the note from Jackson's book where it gives the technique for the notations i used.. I'm confused to why this doesn't work. Do you have a source where i can study those kronecker delta identities better maybe? I've been trying to google them but can't find any and it would help a lot

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Homework Helper
Seem perfectly fine, can you send a picture of where it says this:
$$\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\alpha_\mu \delta^\nu_\beta$$
Because this seems incompatible with what you just send

Maniac_XOX
Seem perfectly fine, can you send a picture of where it says this:

Because this seems incompatible with what you just send
Would it be instead ##\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\alpha_\beta \delta_\mu^\nu## ??
Because i remember that being the first case i tried out but it didn't give me the same indices arrangement as Jackson's result did

Homework Helper
I'm a little bit confused, does Jackson use the property ##\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\alpha_\mu \delta^\nu_\beta##? If it does, can you please send a picture of that part?
If it does not use that property, where do you get that property from? I'm telling you that property is wrong as far as my understanding goes. You have shown me a picture of Jackson's book where that property is not used and indeed everything seems fine. You have also linked a post where this property is never used and also, everything looks fine.
I have also shown you that this formula is clearly wrong unless you use very strange and weird conventions (and I'm not sure if such conventions even exist).

So please, can you prove this formula? Do you know of any textbook that uses such a formula? (and proves it, of course)

Would it be instead ##\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\alpha_\beta \delta_\mu^\nu## ??
Because i remember that being the first case i tried out but it didn't give me the same indices arrangement as Jackson's result did
Nop, it fails exactly the same way as the other. Try to plug ##\mu=\beta=1## and ##\alpha=\nu=0##

romsofia
Maniac_XOX
I'm a little bit confused, does Jackson use the property ##\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\alpha_\mu \delta^\nu_\beta##? If it does, can you please send a picture of that part?
If it does not use that property, where do you get that property from? I'm telling you that property is wrong as far as my understanding goes. You have shown me a picture of Jackson's book where that property is not used and indeed everything seems fine. You have also linked a post where this property is never used and also, everything looks fine.
I have also shown you that this formula is clearly wrong unless you use very strange and weird conventions (and I'm not sure if such conventions even exist).

So please, can you prove this formula? Do you know of any textbook that uses such a formula? (and proves it, of course)

Nop, it fails exactly the same way as the other. Try to plug ##\mu=\beta=1## and ##\alpha=\nu=0##
Jackson totally skipped the steps between to show them unfortunately. And if the formula in the post is correct but I'm still getting it wrong then I'm seriously lost as to how i got the same answer lol.
But by computing the guesses i made i understand why they're wrong since kronecker delta cannot have indices that are different in order to be equal to 1 and not 0. So is the answer $$\frac {\partial(\partial^\mu A^\nu)}{\partial(\partial^\beta A^\alpha)}=\delta^\beta_\mu \delta^\nu_\alpha$$??
A question that has also been burning me up is, why aren't the indices both superscripts instead of one superscript and one subscript since they were all high indices to start with?

Homework Helper
Ok, now the formula is correct... although maybe someone may complain about it.
This last formula is correct, you can indeed check for all the possible 256 equations and check that each one is satisfied. But in general, it is really useful to keep upper indices up and lower indices down. Notice how in the LHS ##\mu## is an upper index while in the RHS it is lowered. This is not strictly wrong in this equations, but be careful, because they could be considered wrong, or nonsensical depending on the conventions you use.

About your question on why we write one index superscript and other subscript is precisely for the reason I just told you, usually upper and lower indices are different, so it is better not to mix them. For the case of the Kronecker delta, it doesn't matter, you can write the indices in whatever position you prefer as long as the pairs are maintained.

Finally, in the LHS we don't have all upper indices. Yes it may seem that for example, ##\alpha## is an upper index (since you see something like ##A^\alpha##), but indeed you can prove that the derivative with respect to a contravariant vector is a covariant vector, therefore the ##\alpha## index is a covariant one and must be put down.

Maniac_XOX
Ok, now the formula is correct... although maybe someone may complain about it.
This last formula is correct, you can indeed check for all the possible 256 equations and check that each one is satisfied. But in general, it is really useful to keep upper indices up and lower indices down. Notice how in the LHS ##\mu## is an upper index while in the RHS it is lowered. This is not strictly wrong in this equations, but be careful, because they could be considered wrong, or nonsensical depending on the conventions you use.

About your question on why we write one index superscript and other subscript is precisely for the reason I just told you, usually upper and lower indices are different, so it is better not to mix them. For the case of the Kronecker delta, it doesn't matter, you can write the indices in whatever position you prefer as long as the pairs are maintained.

Finally, in the LHS we don't have all upper indices. Yes it may seem that for example, ##\alpha## is an upper index (since you see something like ##A^\alpha##), but indeed you can prove that the derivative with respect to a contravariant vector is a covariant vector, therefore the ##\alpha## index is a covariant one and must be put down.
Nice, so basically the arrangement of the indices has to remain the same as it started?
$$\partial^\beta \frac{\partial L}{\partial(\partial^\beta A^\alpha)}=\partial^\beta [-\beta A_\mu A^\mu \frac{\partial (\partial_\rho A^\rho)}{\partial(\partial^\beta A^\alpha)}]=$$
$$\partial^\beta(-\beta A_\mu A^\mu g_{\rho\sigma} \delta^\sigma_\beta \delta^\rho_\alpha)$$
Does this look correct now? if so, how's the right hand side of the euler-lagrange?

Homework Helper
Exact, now it looks correct, although you can simplify it more.
For the RHS you just need to use the product rule and find what is the value of the derivative ##\frac{\partial A^\mu}{\partial A^\nu}##

Maniac_XOX
Exact, now it looks correct, although you can simplify it more.
For the RHS you just need to use the product rule and find what is the value of the derivative ##\frac{\partial A^\mu}{\partial A^\nu}##
$$\frac{\partial A^\mu}{\partial A^\nu}$$
would be $$\delta^\mu_\nu$$for the same kronecker delta rules right?

Homework Helper
Yes, perfect!

Maniac_XOX
Yes, perfect!
So we know that
$$\partial^\beta \frac{\partial L}{\partial(\partial^\beta A^\alpha)}=\partial^\beta(-\beta A_\mu A^\mu g_{\rho\sigma} \delta^\sigma_\beta \delta^\rho_\alpha)$$

How does the right hand side look? $$\frac {\partial L}{\partial A^\alpha} = -\beta (\partial_\rho A^\rho) g_{\mu\gamma} (\delta_\alpha^\gamma A^\mu + A^\gamma \delta^\mu_\alpha)$$

Homework Helper
It looks fine to me, but you can still simplify it a little.

Maniac_XOX
It looks fine to me, but you can still simplify it a little.
awesome, how can I simplify though when it becomes $$\partial^\beta(A_\mu A^\mu g_{\rho\sigma} \delta^\sigma_\beta \delta^\rho_\alpha)=(\partial_\rho A^\rho) g_{\mu\gamma} (\delta_\alpha^\gamma A^\mu + A^\gamma \delta^\mu_\alpha)$$
When put inside the whole complete equation it becomes:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\sigma} \delta^\sigma_\beta \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) g_{\mu\gamma} (\delta_\alpha^\gamma A^\mu + A^\gamma \delta^\mu_\alpha) + \frac {4\pi}{c} J_\alpha$$
Can you help me simplify this?

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Homework Helper
Well, you can use the properties of the metric tensor and Kronecker delta, how would you simplify something like ##g_{\mu\nu}A^\nu## or ##\delta_{\nu}^\mu A^\nu##?

Maniac_XOX
Well, you can use the properties of the metric tensor and Kronecker delta, how would you simplify something like ##g_{\mu\nu}A^\nu## or ##\delta_{\nu}^\mu A^\nu##?
Wouldn't those become ##A_\mu## and ##A^\mu## respectively since i can sum over the repeated indices?

Homework Helper
Yes, so how you apply this on #28? Btw, there you forgot the ##-\beta##.

Maniac_XOX
i ended up using the property ##g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}## to come up with this solution:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\beta} \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) (g_{\mu\alpha} A^\mu + g_{\gamma\alpha} A^\gamma) + \frac {4\pi}{c} J_\alpha$$which finally becomes$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\alpha\beta}) + \mu^2 A_\alpha = (A_\alpha + A_\alpha) + \frac {4\pi}{c} J_\alpha$$
so then the right hand side becomes ##2A_\alpha + \frac {4\pi}{c} J_\alpha##

Homework Helper
i ended up using the property ##g_{\mu\gamma}\delta^\gamma_\alpha=g_{\mu\alpha}## to come up with this solution:
$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\rho\beta} \delta^\rho_\alpha) + \mu^2 A_\alpha =(\partial_\rho A^\rho) (g_{\mu\alpha} A^\mu + g_{\gamma\alpha} A^\gamma) + \frac {4\pi}{c} J_\alpha$$which finally becomes$$\partial^\beta F_{\beta\alpha} + \partial^\beta(A_\mu A^\mu g_{\alpha\beta}) + \mu^2 A_\alpha = (A_\alpha + A_\alpha) + \frac {4\pi}{c} J_\alpha$$
so then the right hand side becomes ##2A_\alpha + \frac {4\pi}{c} J_\alpha##
Apart from the ##-\beta## factor, you also lose a ##\partial_\rho A^\rho##. But yes essentially is correct. You can still simplify the RHS a little.

Maniac_XOX
Apart from the ##-\beta## factor, you also lose a ##\partial_\rho A^\rho##. But yes essentially is correct. You can still simplify the RHS a little.
Right i forgot those factors while focusing on the other stuff but yeah it's still there. Not sure how i can simplify more but a question i have involving the LHS is:
Does ##\partial^\beta(g_{\alpha\beta}A_\mu A^\mu)##
mean the same as $$\frac {\partial (g_{\alpha\beta}A_\mu A^\mu)}{\partial A^\beta}$$ ?

Homework Helper
No, the derivative is wrt the 4-position, not the ##A## field.