# Can $p=mv$ be used for speeds close to $c$?

Can I use the formula $$p=mv$$ for a particle which is travelling at a speed which is very close to the speed of light?

• Why does this question has so many downvote. This question doesn't deserve that. Dec 9, 2020 at 14:57
• @RanjitKumarSarkar exactly, it is a clearly stated, and admits a yes or no answer. Just because the answer is obvious to non-beginners doesn't make it at poor question....even then it has a "yes" and a "no" answer, both of which are correct.
– JEB
Dec 9, 2020 at 16:16
• Hi Dylan Rodrigues. Welcome to Phys.SE. What is your definition of $m$? Dec 9, 2020 at 18:53

No you can't. You need to use $$p = \gamma m v$$ where $$m$$ is the rest mass.

No, you can't.

Instead you need to use the relativistic momentum $$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $$m$$ is the invariant mass (formerly also called rest mass).

You see, that this formula will result in $$p\to\infty$$ when $$v\to c$$.

• Also note that when $v^2/c^2$ is small, you can Taylor expand and get $p = mv(1+v^2/2c^2+\cdots)$ which is neat. Dec 9, 2020 at 19:23

Yes, but m is the relativistic mass. In terms of the rest mass $$m_0$$, $$m = m_0/\sqrt{1 - {v^2/c^2}}$$.

• "Relativistic mass" is a deprecated concept these days. We should not be encouraging students to you use it. It will not do them any favours come exam time. Dec 9, 2020 at 16:18
• It's easy to get in trouble, with, for example, relativistic kinetic energy ( physics.stackexchange.com/a/595075/148184 ) unless you are careful (physics.stackexchange.com/a/597577/148184 ) Dec 9, 2020 at 17:00