Index placement on 4-potential

  • I
  • Thread starter dyn
  • Start date
  • #1
dyn
535
23

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
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,661
6,439
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##?
 
  • #3
dyn
535
23
When I wrote A1 I meant Ax ie. the x-component of the 3-vector A
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,661
6,439
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.
 
  • Like
Reactions: dyn
  • #5
dyn
535
23
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
 
  • #6
1,225
75
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##,
[tex]E^1=-\frac{\partial V}{\partial x^1}-\frac{\partial A^1}{\partial x^0}[/tex]
[tex]=-\frac{\partial A^0}{\partial x^1}-\frac{\partial A^1}{\partial x^0}[/tex]
[tex]=-\frac{\partial A_0}{\partial x^1}+\frac{\partial A_1}{\partial x^0}[/tex]
[tex]=\frac{\partial A_1}{\partial x^0}-\frac{\partial A_0}{\partial x^1}[/tex]
[tex]=F_{10}=-F^{10}[/tex]
where
[tex]F_{\mu\nu}=\frac{\partial A_\mu}{\partial x^\nu}-\frac{\partial A_\mu}{\partial x^\nu}[/tex]

Similarly ##B^1=F_{23}=F^{23}##
Actually E and B are not vectors but components of antisymmetric electromagnetic tensor F.
 
  • Like
Reactions: dyn

Related Threads for: Index placement on 4-potential

  • Last Post
Replies
4
Views
657
Replies
4
Views
2K
Replies
9
Views
580
  • Last Post
Replies
2
Views
1K
Replies
4
Views
2K
Replies
23
Views
3K
Replies
8
Views
813
Replies
36
Views
1K
Top