D'Alembertian question again (sorry)

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  • #1
neu
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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?
 
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  • #2
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You understand 4 vectors? [itex]\phi[/itex] is the electric potential & A is the magnetic potential. They're trying to combine the two.
 
  • #3
nrqed
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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?
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 [itex] j_\mu[/itex] defined in eq 27)
 
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  • #4
neu
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You understand 4 vectors? [itex]\phi[/itex] is the electric potential & A is the magnetic potential. They're trying to combine the two.

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
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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 [itex]\mu_0[/itex] from 30 & you have [itex]-\rho c^2[/itex] .. which is why you need to divide by c in 32 to make the 4D [itex]j_\mu[/itex] that you see in 27.
 
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