Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I know that $m$ in $E=mc^{2}$ is the relativistic mass, but can $m$ in $F=ma$ can also be relativistic? If the answer is yes, then can you tell me whether this equation is valid $E=\frac{F}{a}c^{2}$? If not, can you tell why this is not valid?

Advance thanks for your help and please forgive me my english as it is my second language.

share|improve this question
add comment

3 Answers

up vote 3 down vote accepted

Relativistic force is defined as $$\vec F = \frac {d} {dt} (\gamma m_o \vec v) = \frac {m_o\gamma^3} {c^2}\vec a\cdot\vec v + \gamma m_o\vec a$$ Although generally different, this becomes the same as your expression when $\vec a$ is perpendicular to $\vec v$ giving $\vec a\cdot\vec v = 0$.

share|improve this answer
add comment

The mass in F=ma is relativistic mass if the force is perpendicular to the velocity. If the force is parallel to the velocity, the mass is neither the rest mass or the relativistic mass, but the so-called "longitudinal mass" (deprecated term used by Einstein).

share|improve this answer
    
+1 you're welcome to copy my expression to make this clear to the op and I'll delete my answer. If not, I'll modify my answer to include yours ;) –  John McVirgo Feb 29 '12 at 16:07
    
Modify your answer, I'll delete mine. –  Ron Maimon Feb 29 '12 at 16:37
    
I think our answers are sufficiently different and upvoted so I'll leave it. –  John McVirgo Feb 29 '12 at 23:17
add comment

Here, $m=$ relativistic mass; $m_0=$rest mass

Not really. $F=ma$ only applies for a system with constant (relativistic) mass. The true equation is $\vec{F}=\frac{d\vec{p}}{dt}$, where $\vec{p}=m\vec{v}$ is the momentum. Since acceleration $\implies$ increase in relativistic mass, $F=ma$ is pretty useless here.

Anyways, in relativity, we tend to talk in terms of momentum and force, not acceleration.

So the second equation isn't valid except in special cases.

A better equation to use in place of the second one is: $$E^2=p^2c^2+m_0^2c^4$$

share|improve this answer
    
IMO it's good practice to always write $m=\gamma m_0$. It gets rid of such 'classical mechanics' prejudices. –  Manishearth Feb 29 '12 at 17:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.