Is this a correct 4-vector identity?

[rwooduk]

In our particle physics lecture this term comes up often, it doesn't 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 don't 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.

 

[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)

 

[rwooduk]

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

its from this first page of our notes:

So doesn't this imply that

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

or am I getting things mixed up?

thanks for the reply

edit is also appears here:

 

[ChrisVer]

please check again my first post for your own mistakes in your OP

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...

 

[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

 

[rwooduk]

please check again my first post for your own mistakes in your OP

That makes sense, thanks.

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

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

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

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

 

[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

 

[ChrisVer]

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

 

[rwooduk]

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

They are two different objects

 

[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...

 

[ChrisVer]

You're right, I guess when it comes to it what I am asking is, so FμνF^{\mu \nu} has no relation to AμA^{\mu}?

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

 

[rwooduk]

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...

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

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

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

Thanks!

 

[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}.

 

[ChrisVer]

^corrected some signs problem with E...

 

[rwooduk]

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}.

Also useful, thank you!

 

[ChrisVer]

Also useful, thank you!

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.