Maxwell's Equations - Tensor form 2

[IAN 25]

Homework Statement

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 A^α = j^α 3

expanded this is: ∂₂Φ/∂t² - ∇²A = j^α 4

where J^α is the 4-current; ρ + J

Homework Equations

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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?

The Attempt at a Solution

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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.

 

[Orodruin]

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

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

 

[IAN 25]

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

 

[Orodruin]

I understood from my texts that ∂μ∂μ acting on a tensor was equivalent to ∂2 /∂t2 - ∇2 the d'Alambert operator.

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.

 

[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 ∂_αA_β - ∂_βA_α . 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 ∂^μA^α which I think results in a 16 element square matrix and then let ∂_μ act on that to get the correct elements for j^α ? I am not conversant with tensor algebra yet you see.

 

[IAN 25]

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

 

[Orodruin]

So you wrote

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

Can you go step by step to show how you obtained this from

∂μ∂μ Aα = jα 3

In particular, there is a free index in this equation. Which value did you put it to?

 

[IAN 25]

Hi Orodruin, actually this morning I did what I said above; first I expanded ∂^μA^ν 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 ∂²_tΦ - ∂²_xΦ etc, i.e. ≡ ∂₂Φ/∂t² - ∇²Φ and in the remaining array if one assumes the terms in ∂²_iA_j = zero (i ≠ j) for the mixed spatial derivatives; then one obtains the 3 spatial vector components as you said I should. That is ∂²_tA_x - ∂²_x, 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 spatial comps of A perhaps.

 

[IAN 25]

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

 

[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.

 

[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)

 

[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.