# Index placement on 4-potential

• I

## Main Question or Discussion Point

Hi.
I am working through some notes which use the following metric diag(1,-1,-1,-1).
They give the 4-potential as ( Aμ ) = ( V/c , A ) where V is the scalar potential and A is the vector potential. This should mean in components A0 = V/c and A1 = A1 and so on but with the metric given shouldn't A1 = - A1 ?
Thanks

Related Special and General Relativity News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Be careful not to confuse potential different meanings of $A_1$ and $A_1$. When you write $A_1$, do you mean the Cartesian components of the 3-vector $\vec A$ or the covariant components of the 4-vector $A$?

When I wrote A1 I meant Ax ie. the x-component of the 3-vector A

Orodruin
Staff Emeritus
Homework Helper
Gold Member
When I wrote A1 I meant Ax ie. the x-component of the 3-vector A
Then no. It is not necessary that $A^1 = - A_x$ as $A_x$ is not the same as $A_1$. The sign difference is between the covariant and contravariant spatial components of the 4-potential. As I said, do not confuse the components of a 3-vector with the covariant (or contravariant for that matter) components of a 4-vector. Typically, the generalisation of a 3-vector to a 4-vector will be such that the 3-vector components are the same as the covariant components of the 4-vector, but this may sometimes be subject to sign conventions and if the 3-vector is more naturally viewed as having covariant or contravariant components. In some cases, there is no 4-vector generalisation of the 3-vector at all, such as in the case of the electric and magnetic field where their components instead together constitute the components of the electromagnetic field tensor.

• dyn
Thanks. I asked the question because I don't understand the following question. Using E = -∇V - ∂tA the x-component is given as E1/c = -∂x(V/c) -(1/c)∂tA1 = ∂1A0 - ∂0A1
Is this equation correct ? If so , I don't understand the sign change on the 1st term and it seems to me it uses A1 for Ax

Using E = -∇V - ∂tA the x-component is given as E1/c = -∂x(V/c) -(1/c)∂tA1 = ∂1A0 - ∂0A1
Say $x^0=ct, V=A^0$,
$$E^1=-\frac{\partial V}{\partial x^1}-\frac{\partial A^1}{\partial x^0}$$
$$=-\frac{\partial A^0}{\partial x^1}-\frac{\partial A^1}{\partial x^0}$$
$$=-\frac{\partial A_0}{\partial x^1}+\frac{\partial A_1}{\partial x^0}$$
$$=\frac{\partial A_1}{\partial x^0}-\frac{\partial A_0}{\partial x^1}$$
$$=F_{10}=-F^{10}$$
where
$$F_{\mu\nu}=\frac{\partial A_\mu}{\partial x^\nu}-\frac{\partial A_\mu}{\partial x^\nu}$$

Similarly $B^1=F_{23}=F^{23}$
Actually E and B are not vectors but components of antisymmetric electromagnetic tensor F.

• dyn