# D'Alembertian question again (sorry)

1. Mar 30, 2007

### neu

Sorry to cluter forum with previous thread. I cannot work out the letex code

http://www.cmmp.ucl.ac.uk/~drb/Teaching/PHAS3201_RelativisticTransformationsFull.pdf

in above pdf i do not understand the transition from eq 32 to 33.

That is, I don't understand the presence of the last i*psi/c term in the new 4-vector potential. SHouldn't it just be a= (A1,A2,A3,A4) to get eq29?

2. Mar 30, 2007

### Thrice

You understand 4 vectors? $\phi$ is the electric potential & A is the magnetic potential. They're trying to combine the two.

3. Mar 30, 2007

### nrqed

But there is no "A_4" in equation 29 so what do you mean?

No, their equation 33 contains both eqs 29 AND 30!! (with the four-vector $j_\mu$ defined in eq 27)

4. Mar 31, 2007

### neu

I understand the intention not the method. so I guess I don't understand 4-vectors!

my problem is this:

if square=(d2/dx1,d2/dx2,d2/dx3,-1/c^2 d2/dx4) (eqn 31)

& the combined vector & scalar potential 4-vector a:

a=(A1,A2,A3,i*psi/c) (eqn 32)

Then surely:

square*a=laplacian A - i/c^3 d^2 (scalar)/dt^2

but instead they have (correctly) square*a = mu*j = mu(J1,J2,J3,icp)

why? I cant see how that relates to eq 29 & 30?

5. Mar 31, 2007

### Thrice

As nrqed said, you're combining eqs 29 & 30 to form a 4 vector. J and A in eq 29 range as {1,2,3} & eq 30 becomes {4}.

Compare 29 and 30 with 33. Factor out $\mu_0$ from 30 & you have $-\rho c^2$ .. which is why you need to divide by c in 32 to make the 4D $j_\mu$ that you see in 27.

Last edited: Mar 31, 2007