1
$\begingroup$

In special relativity the partial derivative $\partial_\mu$ is a tensor. Now if some function $A^\mu$ was a tensor, then also the quantitiy $F^{\mu\nu}=\partial^\mu A^\nu - \partial^\nu A^\mu$ would be a tensor simply by composition.

Maxwell's equations can be broken down to a scalar potential $\phi$ and a vector potential $\vec{A}$. Now of course I can randomly construct a four potential $A^\mu=\left(\phi,\vec{A}\right)$ from these coincidentally 4 quantities which fits nicely in the 4-dimensional space-time formulation. But how do I know that $A^\mu$ is actually a tensor?! Is it just a phenomenological premise which cannot be proven? Given $A^\mu$ is a tensor and the relation of $F^{\mu\nu}$ to the electric and magnetic fields $\vec{E}$ and $\vec{B}$, I can obtain these fields in any inertial system by a simple Lorentz transformation of the entire thing $F^{\mu\nu}$.

But I need to start somewhere and know a priori that either $F^{\mu\nu}$ or $A^\mu$ is a tensor.

I hope that what I am asking is clear. So can it be proven, or is it just given by experiment?

$\endgroup$
1

3 Answers 3

4
$\begingroup$

First, note that if $A^\mu$ transforms like the components of a vector, then so does $A'^\mu = A^\mu + \partial^\mu \chi$ for any scalar field $\chi$ - therefore, the tensorial nature of the 4-potential is gauge-invariant.

Second, note that in the Lorenz gauge, we have $$\partial_\alpha \partial^\alpha A^\beta\equiv \square A^\beta = \mu_0 J^\beta$$ Since $\square \equiv \partial_\alpha\partial^\alpha$ transforms like a scalar operator under Lorentz transformations, it follows that $A^\beta$ shares the same transformation properties as $J^\beta$.

Lastly, we know $J^\beta$ transforms like a four-vector because the continuity equation $\partial_\beta J^\beta = 0$ holds in all reference frames. Letting primes denote transformed quantities,

$$\partial_\beta J^\beta- \partial'_\alpha J'^\alpha = \partial_\beta J^\beta - \left(\Lambda^{-1}\right)_\alpha^{\ \ \beta} \partial_\beta J'^\alpha = \partial_\beta \left(J^\beta - \left(\Lambda^{-1}\right)_\alpha^{\ \ \beta} J'^\alpha\right) = 0 $$

which implies that

$$ J'^\alpha = \Lambda^\alpha_{\ \ \beta} J^\beta + C^\alpha$$

for some divergence-free field $C^\alpha$. On physical grounds we can demand that if the 4-current vanishes in one frame then in vanishes in every frame (a Lorentz boost can't create charge or current density out of nowhere), implying that $C^\alpha = 0$ and that $J^\beta$ transforms like a 4-vector.

Therefore, $A^\beta$ transforms like a vector in the Lorenz gauge, and by extension, in every gauge.

$\endgroup$
1
  • $\begingroup$ I guess this is a good argument. Thanks $\endgroup$
    – Diger
    Oct 25, 2019 at 15:05
0
$\begingroup$

One way to prove it could be finding how the fields transform. Another way to prove that $A^{\mu}$ is a tensor is by trying to construct the theory of a massless spin-1 particle by using the Lagrangian formalism, since the Lagrangian should be lorentz invariant and gauge invariant you would eventually find that you need the same form of the Maxwell tensor $F_{\mu\nu}$ that you gave before and that it has to transform as a Lorentz tensor.

A pretty good derivation of the above is given in M.Schwartz Quantum field theory and the standard model, at the end electromagnetism is also a field theory (A classical one).

$\endgroup$
4
  • $\begingroup$ How do you know it has to be a massless spin-1 particle? $\endgroup$
    – user137661
    Oct 25, 2019 at 0:20
  • $\begingroup$ Since Maxwell equations describe waves moving at the speed of light the particle we want to describe needs to follow a massless dispertion relation in order to move at the speed of light and we need it to be spin-1 because it is the minimal spin that allows the theory to have two polarization states. (Three in the case of a massive spin-1 particle) $\endgroup$
    – J.Loz
    Oct 25, 2019 at 0:28
  • $\begingroup$ Sorry for my ignorance but, why do we need the theory to have two polarisation states? $\endgroup$
    – user137661
    Oct 25, 2019 at 0:49
  • 2
    $\begingroup$ Because Maxwell equations (no sources, free space) predict waves with two polarization states. :) $\endgroup$
    – J.Loz
    Oct 25, 2019 at 0:54
0
$\begingroup$

Well... $A^\mu$ is not really a tensor. The tensor form of that 4-potential is

$$ A = A^\mu \partial_\mu,\ A^\mu \in \mathbb{R} $$

So since you know that $\partial_\mu$ transform as a tensor, then $A$ is a tensor and you say that its components transform as a tensor. Let me explain this a little bit more:

Let's say I want to change coordinates $x \rightarrow y$, then

$$ \partial_\mu = \frac{\partial}{\partial x^\mu} \rightarrow \partial_\mu^{\ '} = \frac{\partial y^\nu}{\partial x^\mu}\frac{\partial}{\partial y^\nu} $$

Therefore, since $A^\mu$ is just a function, under this change

$$ A(x) = A^\mu(x)\partial_\mu \rightarrow A^{'}(y) = A^{'\ \mu}(y)\frac{\partial y^\nu}{\partial x^\mu}\frac{\partial}{\partial y^\nu}, \quad A^{' \mu}(y) = A^\mu(x(y)) $$

And now you can make an abuse of language and say that $A^\mu$ is a tensor because it transforms as

$$ A^\mu(x) \rightarrow A^{'\ \mu}(y)\frac{\partial y^\nu}{\partial x^\mu} $$

Now, how do you know that $A^\mu = (\phi , \vec{A})$ gives you the correct theory? As far as I know, that is due to a special relativity extension of classical lagrangian mechanincs. Classically you have that the interaction is given as

$$ S_{int} = \int dt\ L_{int} = -\int dt\ \phi $$

Since $S$, the action, must be invariant under Lorentz tranformation you suggest that $\phi$ is the zero-th component of some 4-vector $A^\mu = (\phi, \vec{0})$ so now you can write

$$ S_{int} = -\int dt\ A^0 = -\int dx_0\ A^0 $$

But this is not a Lorentz scalar, so you extend $A^\mu = (\phi, \vec{A})$ but without assuming that $\vec{A}$ is the usual magnetic vector potential. Therefore,

$$ S_{int} = -\int dt\ A^0 \xrightarrow{extension} -\int dx_\mu\ A^\mu = -\int dt\ \frac{dx_\mu}{dt}A^\mu \tag1$$

This is now a Lorent scalar (so invariant) since the index $\mu$ is dummy now. Let's see what $\vec{A}$ is by developing Eq. (1) (recall $x_0 = dt$):

$$ S_{int} = -\int dt\ \phi(dt/dt) - \vec{A}(d\vec{x}/dt) $$

The part on $\vec{A}$ is the usual interaction of the form current times magnetic potential if consider $\vec{A}$ like that and goes to lagrangian density formulation. By construction $A^\mu$ must be a 4-vector (tensor of rank 1) because if it wasn't that, $A^\mu dx_\mu$ wouldn't be a Lorentz scalar

$\endgroup$
5
  • $\begingroup$ I think that doesn't really address the issue. I can take any function $A^\mu$ and transform as you present to another inertial system. But how do I know that given $A^\mu$ in one inertial system and by that the physical quantities $\vec{E}$ and $\vec{B}$, that an observer in another system precisely sees these quantities $\vec{E}'$ and $\vec{B}'$ as given by the lorentz transformation. The point is, this must not be the case in reality. Saying this is just how a tensor transforms ignores whether this is actually physically the case in reality. $\endgroup$
    – Diger
    Oct 24, 2019 at 23:59
  • $\begingroup$ In essence; how do you know that $A=A^\mu \partial_\mu$ with the physical quantities $\phi$ and $\vec{A}$ can be represented as such. It could be possible that a more elaborate choice of basis vectors (e.g. $A=A^\mu b_\mu^\nu \partial_\nu$ with constants $b$) need to be used to get the correct transformation. $\endgroup$
    – Diger
    Oct 25, 2019 at 0:07
  • $\begingroup$ First, the fields don't transform by Lorentz, that's why you need the strenght tensor. Secondly, I'm going to edit the answer so check it out when I post it $\endgroup$
    – Vicky
    Oct 25, 2019 at 0:14
  • $\begingroup$ Why is $\int {\rm d}t A^0=\int {\rm d}x_\mu A^\mu$ if $\vec{A} \neq 0$? $\endgroup$
    – Diger
    Oct 25, 2019 at 22:47
  • $\begingroup$ It's not, a mistake using an equal instead of an arrow. Check edit $\endgroup$
    – Vicky
    Oct 25, 2019 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.