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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 $

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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.


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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  • $\begingroup$ Ok, thank you for your complete answer, you explained it perfectly $\endgroup$ – MyUserIsThis Jun 10 '13 at 20:01

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