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Retarded time calculations: Electrodynamics

  • #1
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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


[/B]
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.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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$$E \sim \int_{t_1}^{t_2} dt \ a(t_r)^2$$
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.
 
  • #3
479
12
Cheers, thanks for taking the time! I'll give the thread you mentioned a read.
 

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