EM vector potential - covariant or contravariant?

[pellman]

Are potentials appearing in the Maxwell equations the components of a contravariant vector or a covariant vector?

Let us be specific. metric is (+,-,-,-) . Let us write the potentials which appear in the Maxwell equations as $$\Phi$$ and $$\vec{A}=(A_x,A_y,A_z)$$

Is it then the case that

$$A^{\mu}=(\Phi,A_x,A_y,A_z)$$
$$A_{\nu}=(\Phi,-A_x,-A_y,-A_z)$$

or

$$A^{\mu}=(\Phi,-A_x,-A_y,-A_z)$$
$$A_{\nu}=(\Phi,A_x,A_y,A_z)$$


?

 

[Ben Niehoff]

Yes, but the OP is asking about sign conventions.

A is naturally a 1-form, and the sign is chosen such that F = dA. In index notation, this is

$$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

Furthermore, the sign of F is chosen such that the Lorentz force law reads

$$\dot u_\mu = \frac{q}{m} F_{\mu\nu} u^{\nu}$$

and the 4-velocity u is naturally a (contravariant) 4-vector.

 

[Meir Achuz]

Are potentials appearing in the Maxwell equations the components of a contravariant vector or a covariant vector?

Let us be specific. metric is (+,-,-,-) . Let us write the potentials which appear in the Maxwell equations as $$\Phi$$ and $$\vec{A}=(A_x,A_y,A_z)$$

Is it then the case that

$$A^{\mu}=(\Phi,A_x,A_y,A_z)$$
$$A_{\nu}=(\Phi,-A_x,-A_y,-A_z)$$

or

$$A^{\mu}=(\Phi,-A A^\mu_x,-A_y,-A_z)$$
$$A_{\nu}=(\Phi,A_x,A_y,A_z)$$
?

The $$A^\mu$$ has all positive signs. Then the divergence of A is $$\partial_\mu A^\mu$$.

 

[Jerbearrrrrr]

What effect does switching the signature have, from the perspective of calculation?
Would it switch covariant with contravariant?

The fields derived should be invariant surely, since it's just two ways of representing the same physical thing?

 

[Meir Achuz]

Changing the metric does not change that
$$A^\mu=[\phi,{\vec A}]$$.
It would make $$A^\mu A_\mu={\vec A}^2-\phi^2$$
and $$p^\mu p_\mu=-m^2$$, which is why I don't like the
=-1,1,1,1 metric.

 

[pellman]

What effect does switching the signature have, from the perspective of calculation?
Would it switch covariant with contravariant?

No, it only changes the sign of the invariant $$A^{\mu}A_{\mu}$$

 

[pellman]

which is why I don't like the
=-1,1,1,1 metric.

Amen! $$p^\mu p_\mu=-m^2$$ is obscene.

 

[pellman]

The $$A^\mu$$ has all positive signs. Then the divergence of A is $$\partial_\mu A^\mu$$.

Thanks!

 

[Phrak]

One thing of which you should be aware. The magnetic potential coefficients taken alone are equally covariant or contravariant. The subspace is three dimensions of space, and the space³ metric is (+++), so we can raise and lower indices without concern; the signs are unaltered. Aⁱ=A_i. When this subspace metric becomes an entry in the two-by-two metric of space and time, we choose to assign coefficients of either -1 and 1 or 1 and -1 respective to choice.