I have received this question

"Einstein's ideas on mass mean in essence the equation $F=ma$ can only be used if the relativistic mass of the object is known. Describe in your own words to what extent this statement is true."

for a homework, by my understanding the equation $F=ma$ can't be used at velocities close to the speed of light because mass at those speeds is increased. But I am not sure this answers the question.

Could someone help me with answering the question, mainly:

  1. Can $F=ma$ be used without knowing the relativistic mass?
  2. If yes, why/(how)/ if no, why/(how)?

Thanks in advance!

  • 4
    $\begingroup$ Mass does not increase with velocity. See, for example, Why does the (relativistic) mass of an object increase when its speed approaches that of light? $\endgroup$ – AccidentalFourierTransform May 19 '16 at 12:50
  • $\begingroup$ So it is the relativistic mass which is increased? Or is it only the energy and momentum which matters? @AccidentalFourierTransform $\endgroup$ – P.ython May 19 '16 at 12:55
  • 3
    $\begingroup$ Relativistic mass is an old and obsolete concept. It's like the caloric theory: they thought it was a good model, but it turn out not to be. For some reason some teachers still speak about relativistic mass, confusing their students. But now you know it: relativistic mass is wrong, and people should forget about it. On the other hand, as you correctly point out (nice!) it's energy and momentum what matters: the energy and momentum diverge as $v\to c$, but $m$ is finite and independent of $v$. $\endgroup$ – AccidentalFourierTransform May 19 '16 at 13:27
  • $\begingroup$ As you said, this is the concept we have been taught. I'm trying to answer the questions but everywere I look for the answer they give the same explanation as you did and now I don't know how to answer the question. If we use the concept of relativistc mass would it be possible to use F=ma without knowing it? Thank you for making me understand this a bit more because physics is not my strongest subject. @AccidentalFourierTransform $\endgroup$ – P.ython May 19 '16 at 13:36

It would be more correct to state Newton's second law as following:

$$\vec F = \frac{d \vec p}{dt}$$

This holds in relativistic mechanics too:

$$\vec F = \frac{d \vec p}{dt} = \frac{d (\gamma m \vec v)}{dt} = \gamma^3 m \vec a_{//}+\gamma m \vec a_{\perp}$$

Where $\vec a_{//}$ is the component of the acceleration which is parallel to $\vec v$ and $\vec a_{\perp}$ is the orthogonal component. So the form $\vec F = m \vec a $ doesn't hold in relativistic mechanics.

Moreover, as AccidentalFourierTransform pointed out, relativistic mass is an obsolete concept. The $m$ I used in the above formula is a constant. It is $\gamma$ that is a function of velocity:

$$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$$

In old times people used to define the "relativistic mass" as

$$ m_r =\gamma m = \frac{m}{\sqrt{1-(v/c)^2}}$$

But even if we wanted to use this concept, $\vec F = m_r \vec a$ would not hold. You can verify that starting from

$$\vec F = \frac{d \vec p}{dt} = \frac{d (\gamma m \vec v)}{dt}$$

Since $\gamma$ is a function of $\vec v$ and $\vec v$ is a function of time, you cannot take $\gamma$ and $m$ out of the time derivative.

So the answer is that $\vec F = m \vec a$ just doesn't hold in relativistic mechanics.

  • $\begingroup$ Wow! Thank you @valerio92! I'm studying physics at quite beginner-level, could you possibly, if it doesn't cause you too much inconvenience use words to explain why F=ma does not hold in relativistic mechanics? I'm reading the verification on the lower middle of your answer, could you explain further? $\endgroup$ – P.ython May 20 '16 at 8:45
  • $\begingroup$ I'll try to explain better. The point here is that classical momentum is $p=mv$, and $F=dp/dt$. Since $m$ is a constant you can take it out of the derivative and obtain $F=m dv/dt=m a$. But relativistic momentum is $p = \gamma(v(t)) m v(t)$. Even if you define $m_r=\gamma(v(t)) m$ you cannot take it out of the time derivative because of the presence of the time variable inside of $\gamma$. So you cannot do the trick you do in classical mechanics to obtain $F=ma$. I'm not really able to explain it using only words I guess... :-) $\endgroup$ – valerio May 20 '16 at 9:10
  • $\begingroup$ Also notice that Newton's original statement of the second law is: "The alteration of motion is ever proportional to the motive force impress'd; and is made in the direction of the right line in which that force is impress'd". The "alteration of motion" he talks about should be identified with $dp/dt$. So Newton stated his law in the correct form, valid also in relativity ($F=dp/dt$) and not in the only-classical form $F=ma$. $\endgroup$ – valerio May 20 '16 at 9:12

Skipping the normal caveats about the obsolescence of relativistic mass, $F = ma$ still doesn't work. The more proper expression for force is the rate of change of momentum. So, \begin{array}{ll} F &= \frac{dp}{dt} \\ &= \frac{d(\gamma m_0v)}{dt} \\ &= m_0\left(\gamma\frac{dv}{dt} + v\frac{d\gamma}{dt} \right)\\ &= m_0\left(\gamma a + v\frac{-3v/c^2}{(1-v^2/c^2)^{3/2}} \right)\\ \end{array} where $m_0$ is the rest mass.

If it helps, here's what Einstein had to say in later years:

It is not good to introduce the concept of the mass $M = m/\sqrt{1 - v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.

— Albert Einstein in letter to Lincoln Barnett, 19 June 1948 (quote from L. B. Okun (1989), p. 42)



No, because $$F = \frac{dp}{dt}$$ (p is momentum) $$\rightarrow F = \frac{d \ m \times v}{dt}$$ $$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$ cross-product rule $$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$ $$\rightarrow F = \frac{d\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}}{dt} + \frac{dv}{dt}$$ which gives the correct relation between Force, mass and acceleration.

  • $\begingroup$ Momentum isn't a cross product, and that's not how the derivative of a cross product works. $\endgroup$ – Mark H May 19 '16 at 13:57
  • $\begingroup$ @MarkH here, $\times$ means "times", not cross product (I guess), but you are right on the second part. $\endgroup$ – AccidentalFourierTransform May 19 '16 at 13:59
  • $\begingroup$ Can you please tell me the MathJax phrase for cross product also the method for differentiating a cross product. $\endgroup$ – Normal Boy May 19 '16 at 14:08
  • $\begingroup$ It looks like you already know the MathJax for the cross product ... you've (incorrectly) used it in your answer. But note also that the real cross product is not used in this presentation, so asking about the derivative of a cross product is off topic, and can easily be found by asking Google. Finally, neither the classical nor relativistic version of Newton's second law a are valid in open systems, so $dm/dt$ must be zero. $\endgroup$ – garyp May 19 '16 at 14:52
  • $\begingroup$ I already accepted my mistake now please comment somewhere else. $\endgroup$ – Normal Boy May 19 '16 at 15:43

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