The expression in square brackets whose time derivative you are taking is the relativistic expression for momentum. If you wish to think of it as product of two factors, think of it as the invariant $m$ (that used to be called rest mass' and is now simply called 'mass') multiplied by the kinematic factor $\frac{\vec v}{\sqrt{(1-v^2/c^2)}}$.

You are strongly advised not to think of it as the product of $\vec v$ and a factor $\frac{m}{\sqrt{(1-v^2/c^2)}}$, especially if you regard the latter factor as some sort of speed-dependent mass. What's wrong with doing this? (a) It doesn't help, for example substituting $\frac{m}{\sqrt{(1-v^2/c^2)}}$ for $m$ in the Newtonian formula for kinetic energy won't give you the right relativistic formula for kinetic energy. (b) $\frac{m}{\sqrt{(1-v^2/c^2)}}$ is actually much more closely related to energy than to mass.

But to understand the equation that you have quoted, simply think of the left hand side as the rate of change of momentum of the charged particle.

The expression in big brackets whose time derivative you are taking is the relativistic expression for momentum. If you wish to think of it as product of two factors, think of it as the invariant $m$ (that used to be called rest mass' and is now simply called 'mass') multiplied by the kinematic factor $\frac{\vec v}{\sqrt{(1-v^2/c^2)}}$.

You are strongly advised not to think of it as the product of $\vec v$ and a factor $\frac{m}{\sqrt{(1-v^2/c^2)}}$, especially if you regard the latter factor as some sort of speed-dependent mass. What's wrong with doing this? (a) It doesn't help, for example substituting $\frac{m}{\sqrt{(1-v^2/c^2)}}$ for $m$ in the Newtonian formula for kinetic energy won't give you the right relativistic formula for kinetic energy. (b) $\frac{m}{\sqrt{(1-v^2/c^2)}}$ is actually much more closely related to energy than to mass.

But to understand the equation that you have quoted, simply think of the left hand side as the rate of change of momentum of the charged particle.

The expression in square brackets whose time derivative you are taking is the relativistic expression for momentum. If you wish to think of it as product of two factors, think of it as the invariant $m$ (that used to be called rest mass' and is now simply called 'mass') multiplied by the kinematic factor $\frac{\vec v}{\sqrt{(1-v^2/c^2)}}$.

You are strongly advised not to think of it as the product of $\vec v$ and a factor $\frac{m}{\sqrt{(1-v^2/c^2)}}$, especially if you regard the latter factor as some sort of mass. What's wrong with doing this? (a) It doesn't help, for example substituting $\frac{m}{\sqrt{(1-v^2/c^2)}}$ for $m$ in the Newtonian formula for kinetic energy won't give you the right relativistic formula for kinetic energy. (b) $\frac{m}{\sqrt{(1-v^2/c^2)}}$ is actually much more closely related to energy than to mass.

But to understand the equation that you have quoted, simply think of the left hand side as the rate of change of momentum of the charged particle.

The expression in square brackets whose time derivative you are taking is the relativistic expression for momentum. If you wish to think of it as product of two factors, think of it as the invariant $m$ (that used to be called rest mass' and is now simply called 'mass') multiplied by the kinematic factor $\frac{\vec v}{\sqrt{(1-v^2/c^2)}}$.

You are strongly advised not to think of it as the product of $\vec v$ and a factor $\frac{m}{\sqrt{(1-v^2/c^2)}}$, especially if you regard the latter factor as some sort of speed-dependent mass. What's wrong with doing this? (a) It doesn't help, for example substituting $\frac{m}{\sqrt{(1-v^2/c^2)}}$ for $m$ in the Newtonian formula for kinetic energy won't give you the right relativistic formula for kinetic energy. (b) $\frac{m}{\sqrt{(1-v^2/c^2)}}$ is actually much more closely related to energy than to mass.

But to understand the equation that you have quoted, simply think of the left hand side as the rate of change of momentum of the charged particle.