Homework Help: Maxwell's Equations - Tensor form 2

1. Jan 22, 2016

IAN 25

1. The problem statement, all variables and given/known data
The Lorenz gauge ∂Φ/∂t + ∇. A = 0 enables the Maxwell equations (in terms of potentials) to be written as two uncoupled equations;
2Φ/∂t2 - ∇2Φ = ρ 1 and

2A/∂t2 - ∇2A = j 2

The tensor version using the Lorenz gauge
is, i am told,
μ∂μ Aα = jα 3

expanded this is: ∂2Φ/∂t2 - ∇2A = jα 4

where Jα is the 4-current; ρ + J

2. Relevant equations

If one adds 1 and 2 one gets two terms not included in 4. Namely;

2A/∂t2 - ∇2Φ My query is, what happened to these terms when we go to 3 (or 4). Have I misunderstood the transition from 1 and 2 to the tensor forms?

3. The attempt at a solution

If one tries to cancel these terms, or equate their sum to zero, one sees that the first is a vector and the second a scalar. Secondly; ∂2A/∂t2 = ∂E/∂t which > 0 for a time varying field. And ∇2Φ = ∇. E = div E which also is not zero unless no charges are present, which they can be in 3.

Any enlightenment would be greatly appreciated.

2. Jan 22, 2016

Orodruin

Staff Emeritus
No, this is not correct. Expanded you get four different equations, one for each value of alpha.

3. Jan 25, 2016

IAN 25

Please could you tell me where can I find out how to do this? I understood from my texts that ∂μμ acting on a tensor was equivalent to ∂2 /∂t2 - ∇2 the d'Alambert operator. How do I get the other values of alpha?

4. Jan 25, 2016

Orodruin

Staff Emeritus
Yes, this is not what was incorrect. What was incorrect is that you let the different terms act on different components of the 4-potential.

5. Jan 25, 2016

IAN 25

Thanks but I'm still not sure what's wrong. I proved the EM tensor by writing all the terms out long hand i.e I did ∂αAβ - ∂βAα . By constructing two matrices and subtracting to get a resultant matrix from which I identified the elements of Fαβ the EM tensor. Should i take ∂μAα which I think results in a 16 element square matrix and then let ∂μ act on that to get the correct elements for jα ? I am not conversant with tensor algebra yet you see.

6. Jan 25, 2016

IAN 25

Can anyone actually show me what I am doing wrong here!

7. Jan 25, 2016

Orodruin

Staff Emeritus
So you wrote
Can you go step by step to show how you obtained this from
In particular, there is a free index in this equation. Which value did you put it to?

8. Jan 26, 2016

IAN 25

Hi Orodruin, actually this morning I did what I said above; first I expanded ∂μAν i.e. (∂t, ∂x,∂y,∂z )(Φ, Ax, Ay, Az) and got the 16 element tensor. Then I let ∂μ act on that tensor and got the 16 element tensor in 2nd derivatives; with all the time derivatives in the 1st column. The top line is ∂2tΦ - ∂2xΦ etc, i.e. ≡ ∂2Φ/∂t2 - ∇2Φ and in the remaining array if one assumes the terms in ∂2iAj = zero (i ≠ j) for the mixed spacial derivatives; then one obtains the 3 spacial vector components as you said I should. That is ∂2tAx - ∂2x, and the same for y and z.

Adding gives me the equation: ∂2A/∂t2 - ∇2A

If this is right
(and please say it is!) then i get back to the two equations 1 and 2 I started with and all is accounted for. NB I should have used superscripts for the spacial comps of A perhaps.

9. Jan 26, 2016

IAN 25

The last sentence in paragraph one above should read, ∂2tAx - ∂2Ax, and same for y and z components of A (spacial vector).

10. Jan 27, 2016

IAN 25

Sorry, line 2, I meant 4 element column tensor/matrix (in 61 terms - 4 on each row) in 2nd derivatives. Anyway I am satisfied that my workings are correct.

11. Jan 28, 2016

ChrisVer

the equation: $(Something~ done) X^a = Y^a$ is not 1 equation... it's as many equations as are the values of a (a is not a summed/null index)... say that a=0,1,2,3...then it's 4 equations, which once you expand read:
$(Something~ done) X^0 = Y^0$
$(Something~ done) X^1 = Y^1$
$(Something~ done) X^2 = Y^2$
$(Something~ done) X^3 = Y^3$
As for example the $\nabla^2 \vec{X} = \vec{Y}$ is 3 equations and not 1.

your 3->4 transition is incorrect...in fact you destroyed a free index which is an unhealthy thing to do everywhere but especially in relativity. And you can also notice that from the fact that your equation 4 adds a scalar ∂2Φ/∂t2 with a vector ∇2A. What would you say as a result to the $2t^2 + \begin{pmatrix}2t \\ \sin t \\ u(t) \end{pmatrix}$? and of course equates it to the components of a 4 vector (which comes from destroying the free index)

Last edited: Jan 28, 2016
12. Jan 28, 2016

IAN 25

I have now understood where I went wrong (actually I said that I know you can't equate a scalar to a vector). Multiplying out the tensors long hand I found my error. I don't really get what these free indices are yet.