# What is the physical meaning of retarded time?

Consider this figure

Now, when I measure a field produced by the charge $e$ at the point $\mathbf r$, at the time $t=t_1$, it means that the charge sent the signal field at the time $t=t_r$, where $t_1$ and $t_r$ are related by $$t_{r}=t_1-\frac{||\mathbf{r}-\mathbf{r}_{e}(t)|{}_{t=t_{r}}||}{c}$$ Now, my question is, how is it possible that we can take $t_r$ like a time variable? I mean, when we want to measure the velocity of the charge $\mathbf v_e$, I must derive $\mathbf r_e$ respect to $t_r$: $$\mathbf{v}_{e}=\frac{d\mathbf{r}_{e}}{dt_{r}}$$ but why? I mean, why not to derive respect to $t$? So, what is the physical meaning of $t_r$? Or, in other words, how can we interpret the time $t_r$? Are there actually two time-axes?

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This is the usual argument for explaining retarded time -

Consider a charge moving with a constant velocity along a straight line. If the charge suddenly comes to a halt, there will be a change in the electric field due to the acceleration. But this change in the electric field isn't communicated instantaneously through the whole universe, that's prohibited by special relativity and the finite travel time of light.

Therefore one has to infer that observers that are closer to the charge "see" the change in the field earlier than observers farther away.

What that means for an observer too far away to see the change is this - If you are calculating a current/some other property dependent on the velocity of the particle, you have information that is old. The particle has already stopped moving, but you just haven't seen it as yet. So to get the physically accurate picture in the frame of the particle, you have to use retarded time, because otherwise you will infer that the particle is still moving at a constant velocity now. But you don't possess that information. You only know that it was moving at time $t - \frac{r}{c}$ earlier, so all calculations about the state of the system must be carried out with respect to that time.

So to answer your questions : $t_r$ is a time variable because it still depends on t $$t_r = t - \frac{r}{c}$$ which is clearly dependent on $t$. It isn't that you have two time axes, you are just evaluating your system at a finite point in the past (i.e., $t_r \leq t$ always).

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But, if the particle is actually in my past, it doesn't mean that it has another time parameter, the time $t$ is unique. Here we're not using special relativity. –  Anuar Nov 24 '12 at 20:57
I'm confused... what do you mean by "it doesn't mean that it has another time parameter, the time t is unique." ? –  Kitchi Nov 25 '12 at 5:54
I think that there is only one time-parameter $t$, regardless if the object is in my past or in my present or in my future. Am I right? –  Anuar Nov 26 '12 at 1:51
Precisely. Which is why you have to take into account light propagation time when you are looking at a system some finite distance away from you. Since it's technically in your past, you can't assume that you are both in the same frame of reference. –  Kitchi Nov 26 '12 at 6:27
OK, I think I got it. Because we are detecting not the actual position of the particle but the "past" image, we must derive respect to the parameter that describes the motion of the particle: $t_r$, because that time is not the same that our time. This last is because the expression between $t$ and $t_r$, that relation is not "linear". It means that the "scale" of time, would not be the same for me that for the particle. –  Anuar Dec 1 '12 at 15:20
The idea is that it takes time for a signal to travel from a source to where it is being observed--so the field here and now doesn't depend on the charge distribution now, it depends on the value that the charge distribution had $t - \frac{\ell}{c}$ ago, since information cannot travel instantaneously.
Yes, I understand that. My question is, what is the mathematical or physical argument for taking $t_r$ like a temporal variable? I understand that $t_r$ is a particular value of the time $t$. –  Anuar Nov 24 '12 at 3:29
@Anuar: what I said above is the physical reason--signals take time to reach the place where the field is calculated. The mathematical reason is that if $\rho$ is time and space variant, then taking $V = \int \frac{\rho(t,{\vec r})}{r}$ is not a solution to Maxwell's equations but $V = \int\frac{\rho(t_{r},{\vec r})}{r}$ is. Maxwell's equations, after all, are Lorentz-invariant, which would not be true if ti were possible to communicate across spacelike separations. –  Jerry Schirmer Nov 24 '12 at 3:57