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There's a voltage difference of 1000 Volts between two points 2 meters apart. An electron starts at the point of lower potential and is left to travel alone in a straight line until it reaches the other point. Question: What speed does the electron have when it reaches the second point?

My concern is whether the retardation effects of the radiation of the electron are important here. I tried doing it using the relativistic formulae for the kinetic energy and got


where $m_e$ is the electron mass, $c$ the speed of light, $V$ the voltage difference between the two points, and $v$ the final speed of the electron.

This is not a homework exercise. I am just curious as to the effects of radiation.

Edit: What if the potential difference was 10 000 Volts in 2 meters? What if it was 1 000 000 Volts in 2 meters?

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In principle yes, there is some loss to Bremsstrahlung, but...half a kilovolt per meter over two meters is chump change. The electron is still mostly non-relativistic (around 2% of c) and the acceleration is quite modest (by accelerator physics standards). For comparison, high gradients are measured in MV/m. – dmckee Oct 15 '12 at 2:44
By the way, you'll find the math easier if you work in $c = 1$ units. Then $m_e \approx 511\text{ keV}$, and the above claim about the velocity becomes reasonably clear. – dmckee Oct 15 '12 at 2:47
@dmckee I added two additional cases. So at about a million Volts per meter is that the retardation effects of radiation start to be felt? – becko Oct 15 '12 at 3:36
up vote 1 down vote accepted

You need to distinguish between retardation and radiation.

Retardation usual refers to the effect of limited propogation seeds on interactions. That is we replace $$a_g = G\frac{M}{r^2}$$ with $$a_{g,\mathrm{ret}} = G\frac{M}{r(t - \frac{r}{c})^2} .$$

This only matters if

  1. The ration of velocities in the system to propogation speeds are significant, or
  2. The effect is constantly in one direction

in other cases (as in retarded Newtonian gravity in the solar system) it simply results in a constant correction to some effective parameter of the system (reduced mass in our example).

In your case, you have a static field which means that $\mathcal{E}(t) = \mathcal{E}$ and the problem reduced to original case. There is no effect due to retardation.

I'm taking "radiation" to mean the loss of energy by accelerating charges known as Bremsstrahlung.

Everything you really want to know is in the Wikipedia link. You propose a linear acceleration case, so we can use $$ P_{a\parallel v} = \frac{q^2 a^2 E^6}{6 \pi^2 \epsilon_0 m^6 c^{15}}$$ for the power lost to radiation. Here I have used $E = \gamma m c^2$ for the total energy of the electron (including it's mass).

As I noted above, the energies involved are such that the electron remains largely non-relativistic, so we'll stick to the Physics 101 for of the kinematic equations. The average velocity is about 1% of the speed of light, so the time to cover the distance is about 600 ns, and the acceleration is $a = \frac{0.02 * 3\times10^8\text{ m/s}}{6\times10^{-7}\text{ s}} = 10^{13}\text{ m/s}^2$.

Because the electron's energy only changes by about 2% over the whole range I'm not going to bother with a proper integration of the power, and instead just multiply (I make an error on order of 6% by doing so). The energy lost to radiation is about $$E_l = P_{a \parallel v} t = \frac{q^2 a^2 \left(1.01*mc^2\right)^6}{6 \pi^2 \epsilon_0 m^6 c^{15}} t = \left(1.01\right)^6 \frac{q^2 a^2}{6 \pi^2 \epsilon_0 c^3} t$$ and (after furiously checking the units) I just start plugging in

  • $q = 1.6 \times 10^{-16}\text{ C}$
  • $\epsilon_0 = 8.8 \times 10^{-12} \text{ m}^{-3} \text{ kg}^{-1} \text{ s}^2 \text{ C}^2$
  • $c = 3\times10^8\text{ m/s}$
  • $t = 6\times10^{-6}\text{ s}$

$$E_l \approx 1.2\times 10^{-28}\text{ J} = 7 \times 10^{-10}\text{ eV}$$.

So the losses are trivial.

This isn't surprising since a typical CRT runs at 10+ keV over a few centimeters and doesn't spew lots of x-rays around the room.

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