# Index placement on 4-potential

• I
dyn
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

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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##?

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

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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
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

sweet springs
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