Potentials of a charge moving with uniform velocity

In summary, the conversation discusses finding the scalar and vector potentials for a point charge moving with uniform velocity in a laboratory frame. Two methods are considered, one using 4-vector potentials and the other using Lienard Wiechert potentials. Both methods result in potentials that satisfy Lorentz gauge. The question is raised whether the two methods always yield the same result and if there are any conditions for this to be true.
  • #1
am_smbh
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Homework Statement


A point charge q, moving with uniform velocity [itex]\vec{v}[/itex] = v [itex]\hat{z}[/itex], in the laboratory frame called [itex]\cal{K}[/itex]. The charge is in the origin of the system at time [itex]t=0[/itex].


Homework Equations


Find scalar potential [itex]\Phi[/itex] and vector potential [itex]\vec{A}[/itex] both in the frame in which the charge is at rest [itex]\cal{K'}[/itex] and the laboratory frame [itex]\cal{K}[/itex]. Verify the potentials to satisfy Lorentz gauge.


The Attempt at a Solution


I thought of two methods of proceeding:
  1. Transforming 4-vector-Potential [itex]A'^{\mu}[/itex]: I calculate the easy 4-potential in [itex]\cal{K'}[/itex] and I transformed it back into [itex]\cal{K}[/itex], transforming also the spacetime variables. The 4-vector-Potential [itex]A^{\mu}[/itex] that I found, satisfies Lorentz gauge [itex]\partial^\mu A_\mu = 0[/itex].
  2. Using Lienard Wiechert Potentials: I calculate the 4-potential in [itex]\cal{K'}[/itex] using the same method discussed above. In [itex]\cal{K}[/itex] I calculated the retarded Lienard Wiechert potentials, calculating the retarded time too that is needed to be substituted into the formula. This 4-potential satisfies Lorentz gauge too.

My question is: these two methods are pretty different as the first does not take into account any "retard" whereas the second is based on the assumption of "retard". Which is correct and why?

Thank you very much.
 
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  • #2
I just managed with more algebra to find that the two results were the same. now my question is: is that always true? is it necessary a uniform motion or whatelse?

Thanks.
 

What is the potential of a charge moving with uniform velocity?

The potential of a charge moving with uniform velocity is zero, since the electric potential is defined as the work done per unit charge in moving a positive test charge from infinity to a specific point in an electric field. In the case of a charge moving with uniform velocity, no work is done as the charge is already in motion and does not need to be moved from one point to another.

Does the potential of a charge change when it is in motion?

No, the potential of a charge remains constant regardless of its motion. This is because potential is a property of a point in an electric field and is not affected by the motion of a charge passing through that point.

What is the relationship between potential and kinetic energy of a charge in motion?

The potential energy of a charge in motion is converted into kinetic energy, as the charge moves through an electric field. This is known as the principle of conservation of energy, which states that energy cannot be created or destroyed, only transformed from one form to another.

Does the direction of motion of a charge affect its potential?

No, the direction of motion of a charge does not affect its potential. The potential at a point is determined by the properties of the electric field at that point, and is independent of the direction of motion of a charge passing through it.

Can a charge have a non-zero potential while moving with uniform velocity?

No, a charge cannot have a non-zero potential while moving with uniform velocity. This is because, as mentioned earlier, potential is the work done per unit charge in moving a positive test charge from infinity to a specific point in an electric field. If the charge is already in motion and does not need to be moved, no work is done and therefore the potential is zero.

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