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?
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3$\begingroup$ Why does this question has so many downvote. This question doesn't deserve that. $\endgroup$– Rounak SarkarDec 9, 2020 at 14:57
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3$\begingroup$ @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. $\endgroup$– JEBDec 9, 2020 at 16:16
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1$\begingroup$ Hi Dylan Rodrigues. Welcome to Phys.SE. What is your definition of $m$? $\endgroup$– Qmechanic ♦Dec 9, 2020 at 18:53
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3 Answers
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No you can't. You need to use $p = \gamma m v$ where $m$ is the rest mass.
For reference: https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation
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$\begingroup$ Adding a link socratic.org/questions/how-to-prove-m-m-0-sqrt-1-v-2-c-2 . $\endgroup$ Dec 9, 2020 at 13:29
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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 $v\to c$.
answered Dec 9, 2020 at 13:32
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$\begingroup$ 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. $\endgroup$– AndreaDec 9, 2020 at 19:23
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Yes, but m is the relativistic mass. In terms of the rest mass $m_0$, $m = m_0/\sqrt{1 - {v^2/c^2}}$.
answered Dec 9, 2020 at 13:33
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9$\begingroup$ "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. $\endgroup$ Dec 9, 2020 at 16:18
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$\begingroup$ 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 ) $\endgroup$– robphyDec 9, 2020 at 17:00