D'Alembertian question again (sorry)

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In summary, the conversation discusses the transition from equation 32 to 33 in a PDF document on relativistic transformations. The presence of the last term in the new 4-vector potential (i*psi/c) is questioned and compared to equations 29 and 30. It is pointed out that the new 4-vector potential combines equations 29 and 30 to form a 4-vector, and that this relates to equation 33 and the factor of \mu_0 in equation 30. The conversation also mentions the intention behind using 4-vectors for combining the electric and magnetic potentials.
  • #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
You understand 4 vectors? [itex]\phi[/itex] is the electric potential & A is the magnetic potential. They're trying to combine the two.
 
  • #3
neu said:
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
Thrice said:
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 can't see how that relates to eq 29 & 30?
 
  • #5
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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1. What is the D'Alembertian operator?

The D'Alembertian operator is a mathematical operator that is used in differential equations to represent the wave equation. It is denoted by □ and is defined as the sum of the second derivatives of a function with respect to all of its independent variables.

2. What is the purpose of the D'Alembertian operator?

The D'Alembertian operator is used to describe the propagation of waves in a given medium. It is also used in the study of quantum mechanics, general relativity, and other areas of physics.

3. How is the D'Alembertian operator used in differential equations?

In differential equations, the D'Alembertian operator is used to represent the wave equation, which is a second-order partial differential equation that describes how a wave propagates through a medium. It is used to solve problems related to wave motion, such as sound waves, light waves, and electromagnetic waves.

4. Who was the D'Alembertian operator named after?

The D'Alembertian operator was named after the French mathematician and physicist Jean le Rond d'Alembert, who first introduced it in the 18th century. He used it to study the propagation of sound waves in air and other media.

5. How is the D'Alembertian operator related to the Laplace operator?

The D'Alembertian operator is closely related to the Laplace operator, as it is defined as the sum of the second derivatives of a function. The Laplace operator is a special case of the D'Alembertian operator, where the function is independent of time. In other words, the Laplace operator is the time-independent version of the D'Alembertian operator.

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