About the Larmor power formula

[snoopies622]

I first asked this question with much less detail in the recent thread about the Abraham-Lorentz formula but it was sort of passed over, and since it was only tangentially related to the central issue in that thread anyway, I am creating this new one about it specifically. Perhaps it won't need much attention.

Suppose I have a particle with electric charge q that moves according to the equation

$$<br /> x = \frac {1}{2}a t ^2<br />$$

in which the trajectory doesn't begin at t=0 but at some time before that. I consider two points on the trajectory that are very close to each other (x₁, t₁) and (x₂, t₂) and I want to know what force is being applied to the particle in the interval between these two points to maintain the constant acceleration a. If we let

$$<br /> k=\frac {q^2}{6 \pi \epsilon_0 c^3}<br />$$

then according to the Larmor formula the power emitted by the particle is $$ka^2$$ and so the energy emitted in this interval is $$k a^2 (t_2 - t_1)$$. If I assume that energy is conserved and therefore the energy invested into the particle is the same as the energy emitted from it, and that

$$W = \int \bold {F} \cdot d \bold {s}$$

then the average force being applied to the particle in this interval is

$$<br /> ka^2 \big {(} \frac {t_2 - t_1}{x_2 - x_1} \big {)} = ka^2 \big {(} \frac {t_2 - t_1}{(a/2) (t_2 ^2 - t_1 ^2)} \big {)} = ka^2 \big {(} \frac {t_2 - t_1}{(a/2) (t_2 - t_1)(t_2 + t_1)} \big {)}<br />$$

which as $$t_1 \rightarrow t_2$$

$$<br /> = \frac {ka^2}{(a/2)(2t)} = \frac {ka^2}{at}<br />$$ which in this case $$<br /> = \frac {ka^2}{v}<br />$$.

This is just another way of saying that force = power / speed. Obviously, this cannot be correct for this trajectory when t=0. The thread about the Abraham-Lorentz formula pointed out limits to its applications, but I haven't seen any mention of limits to the application of the Larmor formula other than that it is non-relativistic, which doesn't explain why there should be a problem using it when v=0.

So what is the problem?

 

[Meir Achuz]

Your formula is reasonable beyond v=0, but you have shown that the initial acceleration cannot be constant. The initial acceleration must be proportional to t, but it can be constant after the charge gets moving fast enough. If you assume a constant external force, rather than a constant acceleration, it gets a bit complicated but gives reasonable results.
Incidentally, there is a relativistic Larmor formula.

 

[snoopies622]

If you assume a constant external force, rather than a constant acceleration, it gets a bit complicated but gives reasonable results.

Thanks, clem. Do you know where I can find that kind of experimental data? I would like to try to interpolate my own formula for the reaction force since there doesn't seem to be one that works in all cases.

Incidentally, note that in my imaginary experiment the trajectory does not begin at the vertex of the parabola but before it is reached, so that when v=0, da/dt =0 (there is no "jerk") and the problem is actually not occurring at the initial acceleration.

 

[Meir Achuz]

I doubt if there is any exptl data for your case since the radiation force is very small except for a short time near t=0.
Even if you start at negative t, you can't have constant a near t=0, since that woyuld require infinite external and radiative forces.
You have shown that by the simple expression F=ka/t if a is constant.

 

[fizzle]

If I assume that energy is conserved and therefore the energy invested into the particle is the same as the energy emitted from it, and that

$$W = \int \bold {F} \cdot d \bold {s}$$

At a minimum you need to account for the change in kinetic energy of the accelerating charge over the interval. Right now, the equation says that all the input energy is converted to radiation, so it would take an infinite amount of input energy to accelerate the charge.

 

[snoopies622]

I doubt if there is any exptl data for your case...

I didn't mean experimental data for the parabolic case, but for the case you mentioned, where the external force is constant.

..you need to account for the change in kinetic energy of the accelerating charge..

Oops, you're right. I should have said,

"and therefore the energy invested into the particle is the same as the electromagnetic energy emitted from it, plus the particle's additional kinetic energy."

 

[snoopies622]

If you assume a constant external force, rather than a constant acceleration, it gets a bit complicated but gives reasonable results.

What gives reasonable results? What formula (or formulae) do you have in mind?

 

[Meir Achuz]

Your formula is reasonable beyond v=0, but you have shown that the initial acceleration cannot be constant. The initial acceleration must be proportional to t, but it can be constant after the charge gets moving fast enough. If you assume a constant external force, rather than a constant acceleration, it gets a bit complicated but gives reasonable results.
Incidentally, there is a relativistic Larmor formula.

With a retarding Larmor radiation force, taken as F_r=ka^2/v, and an external force F, NIII is
F-ka^2/v=ma. If a were constant, this would require an infinite external force as v-->0. Thus a cannot be constant in the limit v-->0. The next simple case is F being constant. Then as v-->0, a^2/v=F/k, a differential equation which has the solution a=Ft/2k.
As v gets larger, the differential equation is nonlinear, which I called "a bit complicated".
However, for most values of v, F_r is much smaller than F, due to the e^2 in the Larmor formula. Then the reasonable approximation that a=F/m, should give reasonable results. For relativistic v, the relativistic Larmor formula should be used.

 

[snoopies622]

Ah. I thought you were using an unrelated formula that I didn't know about. Thanks for explaining that.