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When an electron is stationary it has an electric field around it. The strength of that electric field (called the electrostatic field) is given by the equation

$E_{r} = \frac{ 1 }{ 4 \pi \varepsilon_{0} } \cdot \frac{ e }{ r^{2} } \text{ (Equ. 1) }$

where $E_{r}$ is the radial component of the electric field (which, in this case, is the only component), $e$ is the charge on the electron, $r$ is the radial distance from the electron, and $\varepsilon_{0}$ is the permittivity of free space $(8.854 \times 10^{-12} \text{ F/m or C}^{2} \text{/N/m}^{2})$.

An electron is surrounded by a radial electric field

If the electron is moving in a wire we have a current. A moving electron produces a magnetic field, but this magnetic field is constant because the electron’s motion has a constant velocity. In order for an electron to radiate Electromagnetic (EM) radiation, it needs to accelerate. Remember, acceleration means its velocity needs to change, which can be achieved either by changing the electron’s speed or its direction. Both types of change will produce an acceleration, and hence produce EM radiation.

In 1907 J.J. Thomson, who discovered the electron in 1897, presented an argument which helps us understand why an accelerated electron produces EM radiation.

Thomson’s argument of why an accelerated electron radiates

Let us suppose we have an electron which is initially at the origin of a coordinate system at time $t=0$. Thus the electric field lines radiate from the origin as shown in this diagram.

At time $t=0$ the electron is at the origin of the x-y coordinate system. We consider a sphere which is a distance $r(=ct)$ away from the centre.

We are now going to accelerate it in a time $\Delta t$ to the right, along the x-axis, and we shall see how this affects the electric field surrounding this electron. We are going to accelerate it to a small velocity $\Delta v$ which is much less than the speed of light $c$ (so that we do not have to consider relativistic effects). The acceleration is given by $a = \Delta v / \Delta t$.

We are going to consider the electric field lines at some distance $r$ from the electron. This distance is shown by the black dashed circle in the diagram above. Because of the finite propagation speed of EM radiation, the sphere can also be said to have a radius of $r=ct$. The radius of the sphere $r$ is chosen to be large enough so that the distance moved by the electron in the time $\Delta t$ is much less than the radius of this sphere. That is, $r \gg \Delta v \times \Delta t$.

At a time $t$ later, the electron has moved a distance $\Delta v \, t$ to the right. Because the black dashed sphere is large, with $r \gg ct$, the electric field lines outside of this sphere are indisturbed by the motion of the electron, as its change of position has not had time to propogate out to such distances. However, close to the electron the field lines are centred on the new position of the electron. These new field lines are represented in red in the figure below, with the black dashed lines representing the original field lines from when the electron was at the origin. The diagram also shows the original black dashed circle centred on the origin, but also a new solid red circle centred on the new position of the electron.

The electron is accelerated to the right along the x-axis. The black dot represents its position at time $t=0$, the red dot its position at time $t=t$. The black dashed circle is large enough so that its radius $r \gg ct$, where $c$ is the speed of light. Therefore the field lines outside of this black dashed circle are unaffected by the movement of the electron. The red solid lines represent the new field lines due to the electron’s new position, with the red solid circle being centred on the electron’s new position.

There is a transition region where the new field lines (the red ones) connect up with the old field lines (the black ones), in a shell. As the electron moves for a time $\Delta t$, the thickness of this transition region (a shell) is $\Delta t \, c$. We are going to consider the radial component and the tangential component (the tangential component is at right angles to the radial component) of the electric field in this shell. That is, the radial and the tangential components of the electric field line from point $A \text{ to point } C$ in the diagram below.

In the transition region the electric field line goes from point $A \text{ to point }C$. We are going to split this field line into its radial and tangential components $E_{r} \text{ and } E_{\theta}$.

As the electron has moved a distance $\Delta v \, t$, and we are considering a field line which is at an angle $\theta$ to the x-axis, the ratio of the tangential component of the electric field to the radial component is just given by the tangential and radial components of the triangle $ADC$, that is

$\frac{ E_{\theta} }{ E_{r} }= \frac{ \Delta v \, t \sin \theta }{ c \, \Delta t }$

But, the radial component $E_{r}$ is given in Equ. 1 above, so we can write

$E_{\theta} = E_{r} \cdot \frac{ \Delta v \, t \sin \theta }{ c \, \Delta t } = \frac{ 1 }{ 4 \pi \varepsilon_{0} } \cdot \frac{ e }{ r^{2} } \cdot \frac{ \Delta v \, t \sin \theta }{ c \, \Delta t }$

Re-ordering out terms we can write

$E_{\theta} = e \cdot \frac{ 1 }{ 4 \pi \varepsilon_{0} } \cdot \frac{ \Delta v }{ \Delta t } \cdot \sin \theta \cdot \frac{ 1 }{ cr^{2} } \cdot t =$

Remembering that $r=ct$ and the acceleration of the electron $a = \Delta v / \Delta t = \ddot{r}$ we can write

$\boxed{ E_{\theta} = e \cdot \frac{ 1 }{ 4 \pi \varepsilon_{0} } \cdot \frac{ \ddot{r} \sin \theta }{ c^{2} r } }$

So, this is the pulse of EM radiation which is produced by an accelerating electron. Notice that

1. it depends on the acceleration $\ddot{r}$ of the electron, the larger the acceleration the greater the EM pulse.
2. it is dependent on $1/r \text{ and not } 1/r^{2}$, and
3. it is tangential (at right angles) to the direction of motion of the electron.

10 Responses

1. Old puzzle (posted here before): If the equivalence principle holds, why doesn’t a charge in a gravitational field radiate?

• Yes, I avoided answering that didn’t I 😉

2. […] as synchrotron radiation) is something I have been planning to blog about for a while. I mentioned synchrotron radiation here when I derived the reason that accelerated electrons emit electromagnetic radiation. I will blog […]

3. does that mean that it radiates an EMW in every direction? If so they are of different wavelengths, yes?

• Yes, in every direction (but not isotropically). The wavelength is inversely proportional to the acceleration; greater acceleration means shorter wavelength.

• Doesn’t that mean it creates an infinite number of EMW as there are an infinite number of directions around this field? Also once the electron starts emitting it would lose energy which would cause a loss of energy thus deceleration leading to more emission. This implies that every electron that has undergone a change of velocity would imminently lose all its energy.

Thanks in advance, sorry if the questions are stupid 🙂