Decelerating charged particle and energy conservation

[Jakub Supel]

Consider a charged particle moving with velocity v, having the energy 1/2 m v^2. Now we deccelerate the particle very quickly; so quickly that the radiated energy is greater than the kinetic energy (it can be arbitrarily large). Note also that energy obtained from decceleration is positive. Where is the mistake?

 

[Bystander]

Consider a charged particle moving with velocity v, having the energy 1/2 m v^2. Now we deccelerate the particle very quickly; so quickly that the radiated energy is greater than the kinetic energy (it can be arbitrarily large). Note also that energy obtained from decceleration is positive. Where is the mistake?

"It" refers to the "radiated" energy, or the "kinetic" energy?

 

[berkeman]

Consider a charged particle moving with velocity v, having the energy 1/2 m v^2. Now we deccelerate the particle very quickly; so quickly that the radiated energy is greater than the kinetic energy (it can be arbitrarily large). Note also that energy obtained from decceleration is positive. Where is the mistake?

Can you show us the math you used to calculate this? Also, remember that energy is power * time, so if you make the time arbitrarily short, you end up with high power for a short time, not high energy...

 

[Dale]

Consider a charged particle moving with velocity v, having the energy 1/2 m v^2. Now we deccelerate the particle very quickly; so quickly that the radiated energy is greater than the kinetic energy (it can be arbitrarily large). Note also that energy obtained from decceleration is positive. Where is the mistake?

Work and energy are frame dependent. In all frames the change in the KE plus the change in the field energy equals the external work done. You can indeed make the energy radiated arbitrarily large, but only by making the work done by the external force correspondingly large.

I also recommend working out the math on this. It is a quantitative question, so a quantitative answer is needed.

 

[Jakub Supel]

Ok, let me elaborate on my question. I really don't think the math is necessary as long as one understands all the physics behind the phenomenon, but I appreciate your skepticism.

1. A particle with mass $m$ (we can assume the electromagnetic mass is included here) and charge $q$ is moving with constant velocity $v \ll c$ along the x axis. The kinetic energy is $\frac{1}{2} m v^2$. Since the EM mass is included in $m$, the kinetic energy of the field is taken into account.

2. At $t=0$ we switch on a force $F$ acting opposite to the velocity and deccelerate the particle until its speed drops to $0$. Of course, due to energy loss to radiation, the acceleration $a \neq \frac{F}{m}$. Instead we have

$$ -F - \frac{q^2}{6 \pi \epsilon_0 c^3} \dot{a} = m a $$

Let $t_0 = \frac{q^2}{6 \pi \epsilon_0 m c^3}$. The boundary conditions are $a(0)=0$, $v(0)=v$, $v(T)=0$ for some $T$ which depends on $F$. The solution is

$$ a(t) = -\frac{F}{m} \left( 1- e^{-t/t_0} \right),$$
$$ F = \frac{mv}{T + t_0(e^{-T/t_0} -1 )}.$$

We use the Larmor formula to compute radiated energy:

$$ E_{rad} = \int_0^T m t_0 a(t)^2 dt.$$

For convenience, take $v=m=t_0 = 1$ - now $F = \frac{1}{T -1 + e^{-T}}$. It is easy to see that $F>0$ and $F \to \infty$ as $T \to 0$, which is intuitive. WolframAlpha says

$$ E_{rad} = \frac{T-\frac{3}{2} + 2e^{-T} - \frac{1}{2} e^{-2T}}{(T-1+e^{-T})^2}.$$

It goes to infinity as $T \to 0$. In particular, $\exists T>0 \ s.t. \ E_{rad} > \frac{1}{2} m v^2$. Moreover, there is an energy gathered from decceleration (negative work is done on the particle):

$$W = \int \vec{F} \cdot \vec{dx} < 0.$$

This is the maths. A simpler argument is this: radiated power is proportional to $a^2$. This means, contrary to berkeman's claim, that integral of radiated power over time does depend on total time. This total time of decceleration can be arbitrarily short - although due to Abraham-Lorentz force the acceleration is not constant if we apply constant force. For very large force, acceleration will instead grow linearly, but there is no limit on the speed of this growth. Let $a = -kt$. Then $\int a(t)^2 dt = \frac{1}{3} k^2 T^3 = \frac{1}{3} k^2 \left( \frac{2v}{k} \right)^{3/2} \sim \sqrt{k}$.

 

[Jakub Supel]

Spoiler

There is a very interesting problem related to this, and it is probably the source of all error:
http://www.physicspages.com/2015/02/12/radiation-reaction-the-abraham-lorentz-force/
http://www.physicspages.com/2015/02/13/radiation-reaction-energy-conservation-with-a-constant-external-force/
Note that $t_0 = \tau$ is very tiny in most cases.

 

[Dale]

Spoiler

There is a very interesting problem related to this, and it is probably the source of all error:
http://www.physicspages.com/2015/02/12/radiation-reaction-the-abraham-lorentz-force/
http://www.physicspages.com/2015/02/13/radiation-reaction-energy-conservation-with-a-constant-external-force/
Note that $t_0 = \tau$ is very tiny in most cases.

Yes, you discovered the key.

 

[houlahound]

Can someone explain the spoiler please, preferably the OP.

 

[Jakub Supel]

Shortly speaking, the question about energy is directly related to a serious problem with formulation of classical electrodynamics. It's the equation $F+\frac{q^2}{6πϵ_0 c^3}\dot{a}=ma$ which is problematic. For $F=0$, there is a solution $a(t)=a_0 e^{t / \tau}$, so if the equation is correct, then particle knows in advance that force will be acting on it and will accelerate accordingly (or it will accelerate after the force dissapears, but then its speed would quickly approach c). I think the energy is conserved when we take this effect into account, but people have been working on a modification of this equation, a modification that would preserve causality.

 

[Dale]

directly related to a serious problem with formulation of classical electrodynamics.

Good post except for this little bit. The problem is with classical point particles rather than with the formulation of classical electrodynamics. If you were to do this problem with a uniformly charged ball then you would find the radiation reaction force just fine.