When a particle keeps "accelerating" near the speed of light, is the energy gained expressed in linearly gained impulse, or relativistic mass? [duplicate]

Question

When a particle keeps "accelerating" near the speed of light, is the energy gained expressed in linearly gained impulse, or relativistic mass?

Answer 1

The concept of relativistic mass is not used these days and it isn't taught to students. The idea of relativistic mass arose because the momentum of a relativistic particle is given by:

$$ p = \gamma m v $$

and if you call $\gamma m$ the relativistic mass $m_r$ this looks like the usual Newtonian equation:

$$ p = m_r v $$

The problem is that this conceals what is actually going on, and in any case it doesn't extend to the energy. For example the kinetic energy is not given by replacing $m$ by $m_r$ in the Newtonian expression for the kinetic energy:

$$ KE \ne \tfrac{1}{2} m_r v^2 $$

In fact the kinetic energy is:

$$ KE = (\gamma - 1)mc^2 $$

The total energy of a relativistic particle is given in terms of its momentum and rest mass by:

$$ E^2 = p^2c^2 + m^2c^4 $$

or this can also be written as:

$$ E = \gamma m c^2 $$

Kinetic energy of a single particle does not affect the spacetime curvature because it can be made to disappear by transforming to the rest frame of the particle. Compare this with thermal energy, i.e. the kinetic energies of an assemblage of gas molecules, that does affect spacetime curvature. Because the velocities of the gas molecules are in random directions you cannot make the kinetic energy of the assemblage disappear by transforming to its rest frame. As a result a fast moving particle cannot form a black hole.