How does the concept of retarded potential apply to an infinitely long wire?

[QuantumJG]

Ok so we've been given a problem to solve where:

$$I(t) = q_{0} \delta (t)$$

Find $$A(t,x) = \int^{ \infty }_{- \infty } \dfrac{ I(t_{ret}, z')}{| x - x'|} dz'$$

All that I want is a hind because it was shown for the case that:

$$I(t) = \left\{\begin{array}{cc} 0 , t \le 0 \\ I_{0} , t > 0$$

 

[fzero]

You will probably want to review what each expression in that integral means, because you haven't defined anything consistently.

First of all, you've defined

$$I(t) = q_{0} \delta (t)$$

as a delta function, but then later say that

$$I(t) = \left\{\begin{array}{cc} 0 , t \le 0 \\ I_{0} , t > 0$$

which is consistent with a step function $$I(t) = I_0 \theta(t)$$, where the step function is defined as

$$\theta(t) = \left\{\begin{array}{cc} 0 , t \le 0 \\ 1 , t > 0 .$$

The delta function is the derivative of the step function $$\theta'(t) = \delta(t)$$, but it is still unclear which function you want here.

Now, in your integral you have $$I(t_{ret},z')$$, which you haven't defined. It's possible that

$$I(t_{ret},z') = I(t-\tfrac{|z-z'|}{c}),$$

but you should really review your notes to figure out the correct expressions.

 

[QuantumJG]

I apologize for the ambiguity.

The example we did in a lecture was the step function. The first part was a completely different question.

 

[QuantumJG]

I'll give more details.

The question involves an infinitely long wire where you're evaluating the vector potential at a point p which is an azimuthal distance ρ from the wire.

|x - x'| = $$\sqrt{(z')^2 + \rho ^2 }$$

My dilemma was that we did an example in class where the current was suddenly turned on and left on, in this case the current was quickly turned on and off and the whole retarded time part is confusing.

 

[fzero]

I'll give more details.

The question involves an infinitely long wire where you're evaluating the vector potential at a point p which is an azimuthal distance ρ from the wire.

|x - x'| = $$\sqrt{(z')^2 + \rho ^2 }$$

My dilemma was that we did an example in class where the current was suddenly turned on and left on, in this case the current was quickly turned on and off and the whole retarded time part is confusing.

Well the idea behind using the retarded potential is that it takes time for the current to reach more distant parts of the wire. Therefore distant parts of the wire do not contribute to the potential until later times. You should try to confirm the expression I suggested for the retarded current. If it's correct, then you should try to set your integral up again and try to see how the step function puts finite, time-dependent, bounds on the region of integration.