Undergrad Index placement on 4-potential

[dyn]

Hi.
I am working through some notes which use the following metric diag(1,-1,-1,-1).
They give the 4-potential as ( A^μ ) = ( 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

 

[Orodruin]

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 A_x ie. the x-component of the 3-vector A

 

[Orodruin]

When I wrote A₁ I meant A_x 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 A_x

 

[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.