Is this a correct 4-vector identity?

1. Mar 7, 2015

rwooduk

In our particle physics lecture this term comes up often, it doesnt look right to me but the lecturer uses it so it must be:

${\partial }^{2}A^{\mu} = - {\partial }_{\mu}{\partial }^{\mu}A^{\mu}+ {\partial }_{\mu}{\partial }^{2}A^{\mu}$

I understand if you have:

$F^{\mu v} = {\partial }_{\mu}A^{v} - {\partial }_{v}A^{\mu}$

then

${\partial }_{\mu}F^{\mu v} = - {\partial }_{\mu}{\partial }^{\mu}A^{\mu}+ {\partial }_{\mu}{\partial }^{2}A^{\mu}$

but I dont understand how a term with a single indices can give the same? How would you derive the first equation? I've tried chain rule but there seems to be an extra delta in there. Any ideas would be really appreciated.

2. Mar 7, 2015

ChrisVer

That is wrong in many ways (eg. triple same indices)... are you sure it's $\partial^2 A^\mu$?

Also you write things with two indices up = 1 index up and 1 index down.

$F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$
So
$\partial_\mu F^{\mu \nu} = \partial_\mu \partial^\mu A^\nu - \partial_\mu \partial^\nu A^\mu$

Now $\partial_\mu \partial^\nu = \partial^\nu \partial_\mu$ and the first term $\partial_\mu \partial^\mu A^\nu= \partial^2 A^\nu$
So
$\partial_\mu F^{\mu \nu}= \partial^2 A^\nu - \partial^\nu (\partial \cdot A)$

Last edited: Mar 7, 2015
3. Mar 7, 2015

rwooduk

its from this first page of our notes:

So doesnt this imply that

${\partial }^{2}W^{\mu} = - {\partial }_{\mu}{\partial }^{\mu}W^{v}+ {\partial }^{2}W^{\mu}$

or am I getting things mixed up?

edit is also appears here:

4. Mar 7, 2015

ChrisVer

Then I don't know, did you extract the Equation of Motion for the field $A$ (or $W$) from the Lagrangian?

And, no, it doesn't....

5. Mar 7, 2015

ChrisVer

I think you only did that for the Photon field (in a general gauge/not Lorentz gauge) :
$- \partial^\mu \partial_\nu A^\nu + \partial^2 A^\mu = J^\mu$

And what you are told is that for the massive case, you only need to replace $\partial^2 \rightarrow \partial^2 + m^2c^2/\hbar^2$. So for a massive field $W$ the above becomes:

$-\partial^\mu \partial_\nu W^\nu + \Big( \partial^2 + \frac{m^2c^2}{\hbar^2} \Big) W^\mu = J^\mu$

6. Mar 7, 2015

rwooduk

That makes sense, thanks.
Sorry I have no idea what that means, last year we didnt have the best lecturer and the Lagrangian stuff wasnt taught very well. I just noticed that the term

$- {\partial }_{\mu}{\partial }^{\mu}A^{\mu}+ {\partial }_{\mu}{\partial }^{2}A^{\mu}$

appers quite often and trying to figure out its origin

Ok, so would there be another way to write $\partial^\mu \partial_\nu A^\nu + \partial^2 A^\mu$ i..e something more condensed? where did the left hand side come from?

thanks again

7. Mar 7, 2015

ChrisVer

I fixed the - sign for the first term ^_^
What do you mean "more condensed"?
The more condensed form would be to write the Maxwell equation $\partial_\mu F^{\mu \nu} = J^\nu$

8. Mar 7, 2015

ChrisVer

I think you posted the left-hand-side in your previous post #3

9. Mar 7, 2015

rwooduk

You're right, I guess when it comes to it what im asking is, so $F^{\mu \nu}$ has no relation to $A^{\mu}$?

They are two different objects

10. Mar 7, 2015

ChrisVer

Let me try to elaborate what he/she does:
you have the Maxwell equation : $\partial_\mu F^{\mu \nu} = J^\nu$
I've already written above what the $\partial F$ is equal to:
$- \partial^\nu (\partial_\mu A^\mu ) + \partial^2 A^\nu = J^\nu$
Then he applies the Lorentz condition gauge, which says that $\partial_\mu A^\mu = 0$ And the above eqaution gives:
$\partial^2 A^\nu = J^\nu$ , which for a source free (free maxwell equation as he writes it) is $\partial^2 A^\nu =0$. This is a set of four equations:
$\partial^2 A^0=0$
$\partial^2 A^1=0$
$\partial^2 A^2=0$
$\partial^2 A^3=0$
And that's why he says that the components follow a massless Klein Gordon equation ($\partial^2 \phi + m^2 \phi =0 \Rightarrow \partial^2 \phi =0$ for $c=\hbar=1$). To incorporate the massive case, he says just replace the $\partial^2$ with + the mass...

11. Mar 7, 2015

ChrisVer

$F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the relation.

12. Mar 7, 2015

rwooduk

Excellent this really helps, I see where I was getting confused they are two seperate things.

Still getting used to the notation, so your posts are appreciated!

Thanks!

13. Mar 7, 2015

ChrisVer

Also if $A^\mu$ or $F^{\mu \nu}$ are different, you should look at the relationships....
$A$ contains the electric potential $\Phi$ and the magnetic potential $\vec{A}$ as its components.
$F$ is a 4x4 antisymmetric matrix, which contains the electric field $\vec{E}$ and the magnetic field $\vec{B}$.
I guess you know how $E,B$ are connected to $\Phi,A$ from electrodynamics (?). That's also how you can see that $\partial_\mu F^{\mu \nu}$ reproduces your known maxwell equations in the form you came across them in an electrodynamics course :) ... Let's take for example:
$\partial_\mu F^{\mu 0}=\partial_0 F^{00}+\partial_i F^{i 0}$
Obviously because of antisymmetry $F^{00}=0$ and $F^{i0}= \partial^i A^0 - \partial^0 A^i =- \frac{\partial}{\partial x^i} \Phi - \frac{\partial}{\partial t } A^i = E^i$
So
$\partial_\mu F^{\mu 0}= \partial_i F^{i0}= \partial_i E^i = \vec{\nabla} \cdot \vec{E} = 0$
that's one of the free maxwell equations. You can get the other half of maxwell equations by taking the rest $\partial_\mu F^{\mu i}$.

Last edited: Mar 7, 2015
14. Mar 7, 2015

ChrisVer

^corrected some signs problem with E...

15. Mar 7, 2015

rwooduk

Also useful, thank you!

16. Mar 7, 2015

ChrisVer

You should do the same by yourself for the other three $\partial_\mu F^{\mu i}$ (i=1,2,3) and see which equations it reproduces.
You can either do that for 3 different i's, or if you feel comfortable with indices extract it in one line (by keeping i). The first approach won't give you something so known (except for if you have ever expanded the maxwell equations in components or if you combine the three results), the second will give you immediately a known Maxwell equation.

Last edited: Mar 7, 2015