Dynamic equation

Question

I am asked to give the dynamic equations of a charged particle moving on a potential, under relativistic considerations. Basically $\frac{dp}{dt}$, with $p$ the 4-momentum vector.

The question for the three space coordinates is simply $d\vec p =-q\nabla\Phi$, with $\Phi$ as the potential, that is: $$\frac{d(\gamma m\vec v)}{dt}=-q\nabla\Phi$$

For the first term of the four vector, the answer given is this, and I have no idea where this comes from: $$\frac{d(\gamma mc^2)}{dt}=q(\vec E\cdot\vec v)$$

With $\vec E$ as the electric field, $m$ as the mass and $\vec{v}$ as the velocity.

Where does this expression come from?

Is it even correct?

Isn't the first term of the four momentum vector just $\gamma mc$ instead of $\gamma mc^2 $

Answer 1

The first term in the four-momentum is $\gamma m c$ as you have said, but given $c$ is constant we can multiply both sides by $c$ and pull a $c$ into the derivative. This doesn't really matter anyway as it is not required for a derivation.

An easy place to see where the last equation comes from is the following. Special relativity gives us the relativistic energy relation $E = \gamma m c^2$ (mass + kinetic energy), where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$. The LHS of your equation is thus the rate of change of relativistic energy with respect to coordinate time, $t$. What is this? Well we know from the Lorentz force law that (in the absence of a magnetic field): $$\vec{F} = q\vec{E}~~~(= -q\vec{\nabla}\Phi)$$ At any point along the path travelled by this charged particle, we can find the instantaneous work done on the charged particle due to $\vec{E}$, in moving an infinitesimal distance tangent to the path, described by $d\vec{x}$. This is given by $$dW = \vec{F}\cdot d\vec{x} = q\vec{E}\cdot d\vec{x} $$ by energy conservation, this work done ($dW$) is the energy gained by the charged particle ($dE$). So we have $$\frac{d(\gamma m c^2)}{dt} = \frac{dE}{dt} = q\vec{E}\cdot \frac{d\vec{x}}{dt} = q\vec{E}\cdot \vec{v} $$ where $\vec{v}$ is the instantaneous velocity of the particle.

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Note that this assumes that all of the energy is going into the motion of the charge particle, it does not take into account radiated energy.

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