A charge moving through a magnetic field experiences a perpendicular force due to the Lorentz force. But things can get messi. What if the charge is moving at a relativistic velocity? Two main effects take place. Let's find out!

Suppose a charge q is moving with an initial velocity v = vex inside a magnetic field B(r,t) = Bez. What is the force acting on the charge and what is the force F(r,t) acting on the charge and its radius while in flight? For the sake of simplicity we will split the problem in two tasks.

## Problem Statement

1. Consider a constant total velocity v and discard radiation from the charge. However, take the relativistic mass effect into account!

2. Bonus: Take the radiation of the electron on its circular path into account as well. Note: you may not be able to find an explicit expression for the time-dependent velocity! A general expression that might be solvable numerically is sufficient.

## Solution

As in the problem statement, we will first focus on the solution only with relativistic mass effect.

As in the dispersion relation of a magnetized plasma let us start with the Lorentz force:

\begin{eqnarray*} m_{q}\ddot{\mathbf{r}} & = & -q\left[\mathbf{E}\left(\mathbf{r},t\right)+\dot{\mathbf{r}}\times\mathbf{B}\left(\mathbf{r},t\right)\right] \end{eqnarray*}

In our case we may simply state that at $$t = 0$$ $$\dot{\mathbf{r}} = v\cdot \mathbf{e}_x$$ and $$\mathbf{E}\left(\mathbf{r},t\right)= 0$$.

Without loss of generality we may switch to a scalar representation of the problem. This simplifies our task with respect to the vector notation. And reminds us a bit of High School!

In this formulation the (initial) scalar Lorentz force is given by:

$F = q \cdot v \cdot B$

Depending on q, we would have a force in positive or negative y direction, i.e. perpendicular to the charge's motion. Therefore we have a centrifugal force with radius r:

$F_\mathrm{cent} = \frac{m v^2}{r}$

However, since the charge is moving at relativistic speeds, we need to take the relativistic mass effect into account. The relativistic mass m_r of the charge is given by:

$m_r = m_0 / \sqrt{1-v^2 / c^2 }$

where m0 is the rest mass of the charge and c is the speed of light. Now we can rewrite the centrifugal force using the relativistic mass:

$F_\mathrm{cent} = \frac{m_r v^2}{r}$

Since the Lorentz force and the centrifugal force balance each other, we can set them equal:

$F = q v \times B = \frac{m_r v^2}{r}$

Thus,

$qvB = \frac{m_r v^2}{r}\ \mathrm{thus}$

$r = \frac{m_r v}{qB}$

Substituting the expression for the relativistic mass, we get:

$r = \frac{m_0 v}{qB\sqrt{1-\frac{v^2}{c^2}}}\ .$

This is already the solution to the first part of the problem.

Now, let's address the bonus problem and take into account the radiation of the electron on its circular path. When an electron is accelerated (as it is when moving in a circular path), it radiates energy in the form of electromagnetic radiation. This effect is called synchrotron radiation, and it leads to a loss of energy and a decrease in the electron's speed.

Now, let's address the bonus problem and take into account the radiation of the electron on its circular path. When an electron is accelerated (as it is when moving in a circular path), it radiates energy in the form of electromagnetic radiation. This effect is called synchrotron radiation, and it leads to a loss of energy and a decrease in the electron's speed.

The power radiated by the electron due to synchrotron radiation is given by:

$P = \frac{2}{3} \frac{q^2 a^2}{4 \pi \epsilon_0 c^3}$

where a is the acceleration of the electron, and ϵ0​ is the vacuum permittivity. Since the electron is moving in a circular path, the acceleration is given by:

$a = \frac{v^2}{r}$

We can now substitute the expression for r that we found earlier:

$a = \frac{v^2}{\frac{\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} v}{qB}}$ Simplifying the expression for acceleration:

$a = \frac{qBv}{m_0} \sqrt{1 - \frac{v^2}{c^2}}$

Now, substituting the expression for acceleration into the formula for the power radiated due to synchrotron radiation:

$P = \frac{2}{3} \frac{q^2 \left(\frac{qBv}{m_0} \sqrt{1 - \frac{v^2}{c^2}}\right)^2}{4 \pi \epsilon_0 c^3}$

Simplifying the expression for the power radiated:

$P = \frac{q^4 B^2 v^2}{6 \pi \epsilon_0 m_0^2 c^3} \left(1 - \frac{v^2}{c^2}\right)$

Now we have the power radiated by the electron due to synchrotron radiation.

To fully account for the radiation effect, we need to consider the energy loss of the electron as it moves along the circular path. The energy loss will cause the electron's speed to decrease, which in turn will affect the radius of the path.
The energy loss per unit time due to synchrotron radiation can be equated to the power radiated:

$\frac{dE}{dt} = P$

The relativistic kinetic energy of the electron is given by:

$E = (\gamma - 1)m_0 c^2$

where γ is the Lorentz factor, defined as:

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$

Now, we can differentiate the relativistic kinetic energy with respect to time:

$\frac{dE}{dt} = \frac{d}{dt} \left[(\gamma - 1)m_0 c^2\right]$

To find the change in velocity as a function of time, we can differentiate the Lorentz factor with respect to time and then substitute it into the equation for the energy loss:

$\frac{d\gamma}{dt} = \frac{1}{2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}} \cdot \frac{-2v}{c^2} \cdot \frac{dv}{dt}$

Now, we can substitute the expression for the time derivative of the Lorentz factor into the equation for the energy loss:

$\frac{d}{dt} \left[(\gamma - 1)m_0 c^2\right] = m_0 c^2 \frac{d\gamma}{dt}$

Now we can equate this to the power radiated by the electron due to synchrotron radiation:

$m_0 c^2 \frac{d\gamma}{dt} = \frac{q^4 B^2 v^2}{6 \pi \epsilon_0 m_0^2 c^3} \left(1 - \frac{v^2}{c^2}\right)$

This equation can be solved for dv/dt to obtain the rate at which the electron's velocity changes due to the radiation. This will allow us to account for the effect of synchrotron radiation on the electron's motion.

First, let's isolate the term dv/dt:

$\frac{dv}{dt} = -\frac{q^4 B^2 v^3}{6 \pi \epsilon_0 m_0^3 c^5} \cdot \frac{1}{\left(1 - \frac{v^2}{c^2}\right)^{\frac{3}{2}}}$

This equation gives the rate of change of the electron's velocity as a function of its current velocity, considering the energy loss due to synchrotron radiation. To find the electron's velocity as a function of time, we can integrate this equation with respect to time:

$\int_{v_0}^v \frac{dv}{\left(1 - \frac{v^2}{c^2}\right)^{\frac{3}{2}}} = -\int_0^t \frac{q^4 B^2 v^3}{6 \pi \epsilon_0 m_0^3 c^5} dt$

where is the initial velocity of the electron. Solving this integral is quite complex and may require numerical methods for a general solution. However, once the integral is solved, it will provide the relationship between the electron's velocity and time, accounting for the energy loss due to synchrotron radiation.

With the updated velocity function, we can re-calculate the radius of the path using the previously found expression:

$r(t) = \frac{\frac{m_0}{\sqrt{1 - \frac{v(t)^2}{c^2}}} v(t)}{qB}$

By accounting for the energy loss due to synchrotron radiation, we can analyze the motion of a charge moving at relativistic speeds in a uniform magnetic field more accurately.

## Background: Synchrotoron Radiation and Relativistic Effects

Synchrotron radiation is a type of electromagnetic radiation emitted by charged particles, such as electrons, as they move in a curved path or are accelerated by an external magnetic field. This phenomenon plays a crucial role in various scientific and technological applications, ranging from particle accelerators to synchrotron light sources, which are used for materials research, medical imaging, and more.

Understanding synchrotron radiation and its underlying principles, particularly the relativistic effects, is essential for students studying physics for several reasons:

1. Bridging classical and modern physics: Synchrotron radiation is an excellent example of a phenomenon that brings together both classical electromagnetism and special relativity. Studying synchrotron radiation helps students appreciate the interplay between these two fundamental areas of physics and develop a deeper understanding of their connection.

2. Real-world applications: Synchrotron radiation has numerous practical applications, including advanced imaging techniques, structural biology, and materials science. By understanding the fundamental concepts related to synchrotron radiation, students can better appreciate the importance of this phenomenon in various scientific and technological fields.

3. Particle accelerators: Synchrotron radiation is a crucial aspect of particle accelerator design and operation. As particles in accelerators reach relativistic speeds, the effects of synchrotron radiation become more significant. Understanding these effects is essential for the development and operation of modern particle accelerators, such as the Large Hadron Collider (LHC) at CERN. Knowledge of synchrotron radiation and relativistic effects allows students to comprehend the challenges and limitations faced in particle accelerator technology.

4. Astrophysics and cosmology: Synchrotron radiation is also observed in various astrophysical settings, such as supernova remnants, active galactic nuclei, and pulsars. It provides valuable information about the properties of celestial objects and their surrounding environments. Studying synchrotron radiation and relativistic effects helps students understand the physical processes occurring in the universe on both small and large scales.

5. Developing critical thinking and problem-solving skills: Analyzing the motion of a charged particle under relativistic conditions and accounting for synchrotron radiation requires students to apply various concepts from electromagnetism, special relativity, and calculus. This process helps students develop critical thinking and problem-solving skills, which are essential for success in physics and other STEM fields.

Understanding synchrotron radiation and the associated relativistic effects is not only essential for students to grasp the interplay between classical and modern physics but also exposes them to real-world applications, astrophysics, and particle accelerator technology. Moreover, studying these phenomena helps students develop essential critical thinking

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