The relativitic energy formula for an accelerated particle

I know a particle at rest has an energy $$E=mc^2$$ But moving particle (constant velocity) has an energy (including kinetic energy) $$E= \gamma mc^2$$ where the Lorentz factor is $$\gamma= \frac{1}{\sqrt{1-(\frac{v}{c})^2}}$$ But when a particle moves with acceleration I still not know to express the energy since $v$ ( relative speed of the reference frames) must be constant in Special Relativity. I guess we can still write the relativistic energy in the same form.

• But $v$ is the velocity of a particle measured in an (inertial) reference frame, not of the reference frame itself. So why must it be constant unless $v=c$? The classical kinetic energy formula works whether a body is accelerating or not, same here. – Conifold Mar 11 '17 at 0:05
• $v$ is velocity of particle not frame.Also special relativity can deal with accelerated frame. – Lapmid Mar 11 '17 at 4:54
• You may find this old Usenet Physics FAQ article of interest: The Relativistic Rocket. – PM 2Ring Mar 11 '17 at 11:34

You can use the formula you have stated i.e.

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

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

If you want to calculate energy of an accelerated particle any instant , then you just calculate its velocity at that instant and put in the equation.

Energy of an accelerated particle will be changing with time so you need to specify the time at which you want to calculate its energy.

There is nothing in SR that stops you from calculating energy or momentum or anything for any accelerated particle.

You should use $\alpha =\frac{1}{\sqrt{1-(\frac{u}{c})^2}}$ where $u$ is a velocity of particle not a frame. The correct formula is $$\boxed{E= \alpha mc^2}$$

• Sorry but I will downvote you answer since you are just calling $\alpha$ what the OP calls $\gamma$ without providing additional information. – ZeroTheHero Mar 11 '17 at 2:29
• @ZeroTheHero I think $u$ in $\alpha$ is a velocity of particle which not need to be constant, while $v$ in $\gamma$ is a velocity of the frame of reference which need to be a constant in SR. – Saksith Jaksri Mar 11 '17 at 3:19
• @SaksithJaksri there is nothing to prevent $\gamma$ from changing as the velocity changes. Indeed, see en.m.wikipedia.org/wiki/Hyperbolic_motion_(relativity) for a classic example. – ZeroTheHero Mar 11 '17 at 3:24

protected by Qmechanic♦Mar 11 '17 at 6:33

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