When a body accelerates, it gains (relativistic) mass $m$ according to the relation $$m=\frac{m_0}{\sqrt{1-(v/c)^2}},$$ where $m_0$ is the (rest) mass. But after it stops is the gained (relativistic) mass still there?


3 Answers 3


First, to clarify: A body does not "gain mass" upon acceleration, it "gains mass" at high speeds. That is, whether or not the body's velocity is changing is not relevant, only its speed relative to the observer is important.

That being said, a body doesn't actually "gain mass" when it moves at a high velocity. The mass of a body is always the same, and is independent of the speed at which it is moving. This is the concept referred to as "rest mass."

What actually happens is that when a body accelerates its momentum increases according to the equation you posted. That is, if $p$ is a body's momentum and we put $\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}$, then the equations of relativity tell us that

$$ p = \gamma m_0 v = \frac{m_0 v}{\sqrt{1 - (v/c)^2}}, $$

where $m_0$ is the rest mass (the $m'$ you have in your equation) and $v$ is the velocity.

The tricky thing here is that in relativity, momentum works differently than it does at non-relativistic speeds. At low speeds (as $v$ goes to zero), we have $\gamma = 1$ and $p = m_0 v$, as we remember from non-relativistic mechanics. At high speeds, however, $\gamma$ can get arbitrarily large, and $p$ can be much much greater than $m_0 v.$

The idea of an object "gaining mass" as its speed increases comes from treating the momentum as though it behaved the same way as it does in the non-relativistic case. That is, let us pretend that the momentum were equal to $mv$ -- then we would have

$$ mv = \gamma m_0 v. \\ m = \gamma m_0 = \frac{m_0}{\sqrt{1 - (v/c)^2}}. $$

This is the equation you gave in your post.

In simple terms, the answer is this: An object does not gain mass as it increases its speed, it gains momentum. The way in which it gains momentum is very different from the way we expect based on our understanding of non-relativistic mechanics. However, from the standpoint of non-relativistic mechanics, this rapid change in momentum can be viewed as a change in mass given by $m = \gamma m_0.$ So an object does not gain mass when it moves at high speeds, but it acts as though it had a larger mass according to our standard rules of non-relativistic physics.

When it stops, it is no longer moving at high speeds, and we return to the equation $p = m_0 v.$ The mass never changed -- no "matter" was added to the body -- it simply acted as though it had a greater mass because it was moving at a high speed. When it stops moving it still has its rest mass $m_0$ because the only change it experienced was in effective mass.

  • 1
    $\begingroup$ It's largely a matter of definition--it did used to be common practice in relativity textbooks to define a "relativistic mass" equal to gamma times the rest mass, so if you specify that you're using that concept it's perfectly valid to say objects gain mass with velocity, and not just treat such talk as a proxy for talking about gain in momentum. But the concept of relativistic mass has fallen out of favor, probably for pedagogical reasons, and modern physicists do prefer to just talk about rest mass and relativistic momentum in place of relativistic mass. $\endgroup$
    – Hypnosifl
    Oct 27, 2014 at 15:01
  • 2
    $\begingroup$ Just a minor point about terminology. "Classical" in physics is used to mean not quantum-mechanical. The relativity we're talking about here is classical physics. $\endgroup$
    – user4552
    Oct 27, 2014 at 15:10
  • $\begingroup$ I'm not sure I agree with that - I hear the term "classical" used all the time to just mean "according to the previous model." I've heard plenty of physicists refer to various aspects of QM as "classical" because they don't fit in with modern QFT models. I understand how it could be confusing, though – I'll edit. $\endgroup$
    – user_35
    Oct 27, 2014 at 16:37
  • $\begingroup$ @user35736 to quote "whether or not the body's velocity is changing is not relevant, only its speed relative to the observer is important." does this imply that although the velocity of the body is constant, the 'mass gain' will occur at high speeds? $\endgroup$ Oct 29, 2014 at 12:56

I will offer a very short and simple answer.

The object does get more inertial mass,but no matter is added on it. The object at different speeds acts like it has different mass because it has more energy and bigger momentum. When the speed is small compared to that of light's the difference is negligible. Nevertheless when it's speed approaches the speed of light the differences cannot be overlooked.

Anyway the "new mass" of the body is always calculated as :

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

So when the object rests again the its mass is the same as in the beginning (rest mass m)


A good way of answering this question is to ask what would happen if any of the mass was still there.

Let's consider something made of wood, and we'll move it a bit (it helps if we move it fast: say we attach it to a flywheel and spin it up for a bit, and then spin it back down).

Now it's a bit heavier, right? So we'll chop off enough of it to make it the original mass again, and repeat.

Now we can burn the bit we chopped off to get some energy. And we can repeat this for ever.

Ah, but perhaps that extra mass comes from the work done on it by the machine we used to speed it up? Well, no: by clever design we can make the efficiency of the machine that does this as high as we like: the flywheel can have very good bearings, run in a vacuum and so on, and to spin it down we'll use it as a generator to charge up a big battery, which we then use later to spin it up again. Obviously there will be losses in the system but we can keep making them lower.

So this is a perpetual motion machine: it gives us free energy.

But such machines violate very basic principles of physics: specifically, Noether's theorem says such a thing would mean the laws of physics were not time-invariant.

So, just from symmetry, the answer must be no.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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