Retarded time calculations: Electrodynamics

[WWCY]

Homework Statement

A positive charge $q$ is fired head-on at a distant positive charge $Q$ that is held stationary. It comes in at speed $v_0$ and comes to an instantaneous halt at distance $r_f$ away from Q. What is the amount of energy radiated due to acceleration in this time interval? Assume $v_0<<c$

Homework Equations

Larmor's formula for power incident over a constant $R$ surface over the charge, at time $t$.
$$P(t) = \frac{\mu_0 q^2 a(t_r)^2}{6 \pi c}$$
where $t_r = t - R/c$ is the retarded time

The Attempt at a Solution

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Brief sketch:

Acceleration is $a \sim 1/r^2$, but can also be written as $a(t_r) \sim 1/r(t_r)^2$, which means that retarded acceleration is a function of retarded position at time t (call it $r'(t)$), and so $a(t_r) \sim 1/r'(t)^2$. Energy collected across a constant $R$ surface from $t \in [t_1, t_2]$ is
$$E \sim \int_{t_1}^{t_2} dt \ a(t_r)^2$$
where integral variable $dt$ refers to the present time. But $a(t_r) = a(r'(t))$
$$E \sim \int_{r'(t_1)}^{r'(t_2)} dr' \ \frac{dt}{dr'} \ a(r')^2$$
I'm not sure how right I have been so far, but I start to get confused here with the new limits and change of variables.

I'm guessing that $r'(t_1) = \infty$ while $r'(t_2)$ is set at the distance of closest approach $r_f$, which seems to make sense; the power you receive at $t_1$ should be the power released from the retarded position at $t_1$. Ditto for $t_2$

As for $\frac{dt}{dr'}$, $\frac{dr'(t)}{dt}$ seems to be the rate of change of retarded position, which is the retarded velocity $v'(t)$, and its relation to $r'$can be obtained via the law of energy conservation. Finally,
$$E \sim \int dr' \ \frac{1}{v'(r')} \ a(r')^2$$

Does any of this make sense? Many thanks in advance.

 

[TSny]

$$E \sim \int_{t_1}^{t_2} dt \ a(t_r)^2$$
where integral variable $dt$ refers to the present time. But $a(t_r) = a(r'(t))$
...
...
...
Finally,
$$E \sim \int dr' \ \frac{1}{v'(r')} \ a(r')^2$$

Does any of this make sense? Many thanks in advance.

I think it is OK. I believe the integration is easier if you choose the integration variable to be the retarded velocity $v(t') \equiv v'$. Note $dE_{rad} = Pdt = P\frac{dt}{dv'}dv' = \frac{P}{a'}dv'$

But, if you want to stick with integrating with respect to $r'$, it's not too bad.

This problem came up a few years ago
https://www.physicsforums.com/threads/energy-radiated-from-a-charge-in-electric-field.771453/

However, in your problem you are calcuating the total energy that eventually crosses a fixed spherical surface of large radius R. The energy that crosses the surface at time t depends on what the charge was doing at the retarded time.

In that other thead, the calculation adds up the energy emitted during each infinitesimal time $dt$. It does not make use of retarded time or of a fixed surface at large distance. I think the two calculations yield the same result.

 

[WWCY]

Cheers, thanks for taking the time! I'll give the thread you mentioned a read.