Thought experiment: Let's take a positively charged particle and place it in a vacuum with no electric field. It just sits there. Now, we instantaneously introduce an electric field. The moment before it begins to move we can calculate the potential energy of the particle. OK. Particle feels a Coulombic force and begins to accelerate at a constant rate. After some time, t, we can calculate the new potential energy of the particle. The difference in potential energy between t = 0 and t = t must be the kinetic energy of the particle at time t.

Here is the wrench:

According to my professors, accelerating particles release electromagnetic waves. I was taught that EM waves are "packets of energy." How come in all my EM courses...I never had to account for this loss of energy in the form of EM waves? And is there a way to know/calculate how much energy was lost?

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    $\begingroup$ You applied conservation of energy as if there were only potential energy and kinetic energy in the situations, but you then go on to note that there is also energy in radiation. That is, you've identified that the situation doesn't meet the pre-conditions for your application of energy conservation. $\endgroup$ – dmckee --- ex-moderator kitten Jun 4 '16 at 0:35

If we assume that velocities are well below the relativistic region then in the absence of radiation the equation of motion of the electron is simply:

$$ \frac{d^2x}{dt^2} = \frac{Eq}{m} \tag{1} $$

where $E$ is the field strength and $q$ and $m$ are the electron charge and mass.

As Lawrence says, the power emitted as radiation is given by the Larmour equation:

$$ P = \frac{q^2a^2}{6\pi\varepsilon_0c^3} = \frac{q^2}{6\pi\varepsilon_0c^3} \left(\frac{d^2x}{dt^2}\right)^2 \tag{2} $$

and since power is force times velocity this produces a force on the charge of $P/v$ and therefore a deceleration of $P/(vm)$. So the equation of motion including the radiation is:

$$ \frac{d^2x}{dt^2} = \frac{Eq}{m} - \frac{1}{m\frac{dx}{dt}}\frac{q^2}{6\pi\varepsilon_0c^3}\left(\frac{d^2x}{dt^2}\right)^2 \tag{3} $$


Feel free to attempt a solution, but for now let's just get a feel for the quantities involved. Let's take a field strength of $10^5$ V/m, which is about as high as we can go without needing to include relativistic corrections. In that case equation (1) gives us:

$$ a \approx 1.76 \times 10^{16} \,\text{m/s}$$

Feed this into equation (2) and we get:

$$ P \approx 1.76 \times 10^{-21} \,\text{W} $$

Remember that we started out saying we were using a field strength of $10^5$ volts per metre, so the potential energy change per metre is $Vq$ or about $1.6\times 10^{-14}$J. So we have a seven orders of magnitude difference between the energy gain of the electron per metre and the energy radiated per second. And that's we ignore the energy lost to radiation in most circumstances.

As a footnote, the Larmour radiation is most certainly not always negligable. After all radio broadcasts use exactly this phenomenon - the energy radiated per electron is small but fortunately there are a lot of electrons in a radio aerial. The Larmour radiation is also why the LEP accelerator couldn't be made much more powerful than 200GeV - the energy radiated due to the circular motion made it prohibitively expensive to run at any higher energies.

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  • $\begingroup$ @ally: the power is proportional to $a^2$ so the sign of $a$ makes no difference. $\endgroup$ – John Rennie Jun 4 '16 at 7:05
  • $\begingroup$ @ally: change of direction is just an acceleration. The Larmour radiation at the LEP accelerator was due to a change of direction rather than a change in the magnitude of the velocity. $\endgroup$ – John Rennie Jun 4 '16 at 7:20

The Larmor formula comes from the Poynting vector $$ \vec S = \frac{c}{4\pi}\vec E\times\vec B $$ The electric field in a relativistic content is derived with the Lienard-Weichert potential $$ \vec E = q\frac{\hat n - (\vec v/c)}{\gamma^2(1 - \vec v\cdot\hat n)^3r^2} + \frac{q}{c}\frac{\hat n \times((\hat n - (\vec v/c)))\times\vec v_t/c}{\gamma^2(1 - \vec v\cdot\hat n)^3r} $$ with $\vec v_t = d\vec v/dt$ and $\vec B = \hat n\times\vec E$. The cross product calculation takes a bit of work but the result is $$ \vec S = \frac{q^2}{4\pi c^3}\frac{(|\vec v_t|^2 - \hat n\cdot\vec v_t)\hat n }{r^2} $$ where of course $|\vec v_t|^2 = a^2$ The Poynting vector is the energy flux rate. This can then be derived to compute the power to give the Larmor formula $$ P = \frac{2}{3}\frac{e^2a^2}{c^3}. $$

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    $\begingroup$ This is some impressive math, but I have no idea what any of it means...I was hoping for a conceptual answer backed up with the math and its meaning. What is the Lamor formula and Poynting vector? What do they have to do with my question? $\endgroup$ – Nova Jun 4 '16 at 2:20
  • $\begingroup$ @Nova Better than Wikipedia and nice presentation: "Chap.3: Dynamics of the Electromagnetic Fields", especially Sec. 3.2.2, "Poynting’s Theorem: Energy Conservation" here: ocw.mit.edu/courses/nuclear-engineering/…. Then continue to "Chap.4: Radiation By Moving Charges", ocw.mit.edu/courses/nuclear-engineering/…. $\endgroup$ – udrv Jun 4 '16 at 3:42

How come in all my EM courses...I never had to account for this loss of energy in the form of EM waves?

In the simple examples shown in introductory courses, the loss of energy of charged particle moving in external field via radiation is neglected, because radiation of EM waves is a difficult topic and even if it is not neglected, it can be shown its effect on the particle motion is relatively small.

If you take more advanced course, such as relativistic electrodynamics or design of antennae/particle accelerators, loss of energy via radiation is discussed there.

And is there a way to know/calculate how much energy was lost?

Yes there is, if the charged body has non-zero dimensions (is not a point), the Poynting theorem can be interpreted as work-energy theorem and if the body is far from other bodies (so that effect of other bodies on the Poynting vector can be neglected), the Larmor formula can be derived. The Larmor formula says energy radiated out by accelerated charged body per second is

$$ P = \frac{2}{3}\frac{Kq^2}{c^3}a^2 $$

where $K = \frac{1}{4\pi\epsilon_0}$,$q$ is charge of the body and $a$ is magnitude of its acceleration.

When energy loss is evaluated for electrons in CRT (old big TV vacuum chamber) or electrons in a microscope, one gets very small number compared to total energy of the electrons, so the effect is neglected most of the time. Where it cannot be neglected is particle accelerators - the designed parameters of motion are such that energy lost from one bunch of particles (one bunch is a small cloud of billions of particles that are moving together) in one orbit is comparable to energy that the machine is able to supply.

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