In what real sense does the mass of an object increase with its speed? When we learn that the mass of an object increases according to the equation,

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

We can think of the mass of an object as its resistance to being moved (inertia). But suppose we devised a contraption that increased its mass by the motion of its internal parts. Would the inertia of the contraption be greater? In the following, I devise the simplest possible contraption that takes advantage of this fact and ask what happens when an object collides with it.

Consider a box with a negligibly light frame in which two balls each of mass $\frac{m_0}{2}$ start in the middle and move in opposite directions at velocity $v$. Call this box the "balls box".

The balls box has mass $m = \frac{m_{0}}{\sqrt{1 - \frac{v^2}{c^2}}}$.

Consider a box with rest mass $m$ when another box of rest mass $m$ hits it moving at velocity $v$. In an elastic collision, the first mass with move with velocity $v$ and the second will come to rest.

Now replace the box with rest mass $m$ with the balls box. Because the balls inside it are moving, it has rest mass $m_0$ but relativistic mass $m$. The balls box is at rest when a box with mass $m$ hits it at velocity $v$ in the direction along which the balls move. What happens?

According to a naive analysis, after the collision there will be some jiggling as information is transmitted down the length of the balls box. Eventually, the moving box will come to rest and the balls box will move with velocity $v$ to the right. Is this right?

What happens when we turn the balls box perpendicular to the motion of the moving ball?


The rationale behind the naive analysis is that when the balls box is treated as a self-contained system—and when it's at rest—it has mass $m$. By that reasoning, it should behave the same as a box with rest mass $m$. The problem arises when we look inside the box after the collision. After information has propagated, one ball moves at $\frac{v - v}{\sqrt{1 + \frac{v^2}{c^2}}} = 0$. The other moves at $\frac{2v}{\sqrt{1 + \frac{v^2}{c^2}}}$. The box previously in motion is at rest. This leads to a momentum and energy different from the energy and momentum before the collision.

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    $\begingroup$ Related: physics.stackexchange.com/q/133376 and many other similar question on the site. I'm firmly in the camp that prefers to use only the invariant mass. $\endgroup$ – dmckee Jul 4 '16 at 22:00
  • $\begingroup$ The mass of the ball-box is not $m_0$. Really. The mass of a system is not necessarily the sum of the masses of the parts, and the system you have constructed is one where it the mass of the system is greater than the mass of the parts. In all seriousness, trying to reason about masses and velocities is the hard way to do this. Reason about four-momenta: they obey simple rules (the four-momenta of a system is the sum of those of the parts, in the absence of external work and impulse they are conserved, and so on). $\endgroup$ – dmckee Jul 5 '16 at 0:17
  • $\begingroup$ Sure—it's not $m_0$ it's $m$. $\endgroup$ – Will Jul 5 '16 at 0:18
  • $\begingroup$ four-monenta, energy, velocity, mass—frame it however you like. The question asks what will the system look like after the collision. $\endgroup$ – Will Jul 5 '16 at 0:20
  • $\begingroup$ The question is not very clear. I am unable to see what you want to ask. Title and question in body do not match. Please try being more clear. $\endgroup$ – Anubhav Goel Jul 5 '16 at 10:14

This answer written in the "modern" perspective.

The balls box has rest mass $m_0$

But it doesn't. The (rest or invariant) mass of the box is found in the usual way from the square or it's energy-momentum four vector $(mc^2)^2 = \mathbf{p}^2 = E^2 - (\vec{p}c)^2$.

The energy-momentum four vector of the box is found by adding the four vectors of it's three parts together, so (noting that all the momenta are co-linear) \begin{align*} (mc^2)^2 &= \mathbf{p}^2 \\ &= \left[\mathbf{p}_{b1} + \mathbf{p}_{b2} + \mathbf{p}_{frame}\right]^2 \\ &= \left[ (\gamma m_0 c^2/2,+\gamma m_0 \vec{v}c/2) + (\gamma m_0 c^2/2,-\gamma m_0 \vec{v}c/2) + (0,\vec{0}) \right]^2 \\ &= \left[ (\gamma m_0 c^2, \vec{0}) \right]^2 \\ &= \gamma m_0 c^2 \end{align*}

So the invariant mass of the box is $\gamma m_0$ in the first place.

In relativity, the mass of a system is not necessarily equal to the sum of masses of it's component parts. Indeed, that is where the energy of nuclear reactions comes from/goes to: changes in the mass of the system by putting it together or taking it apart.

  • $\begingroup$ Ok sure—I edited the question to account for this correction. But that doesn't answer the question. What happens after the collision? Is the naive answer correct? $\endgroup$ – Will Jul 4 '16 at 22:52
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    $\begingroup$ It does answer the question. A object or system has only one mass—the invariant mass—but the invariant mass of a system is not necessarily the sum of the invariant masses of the parts (as is the case in your ball box). The fact that the balls are in motion means that as a system the ball-box has mass greater than $m_0$. But each ball continues to have mass $m_0/2$. The thing that adds up is the energy-momentum four-vector. $\endgroup$ – dmckee Jul 4 '16 at 22:55
  • $\begingroup$ Ok sure, but what happens? Is the naive analysis correct? After the collision, will the ball box move with velocity $v$? $\endgroup$ – Will Jul 4 '16 at 22:57
  • $\begingroup$ The question is "According to a naive analysis...the balls box will move with velocity $v$ to the right. Is this right?" $\endgroup$ – Will Jul 4 '16 at 23:02
  • $\begingroup$ @Will You may find the whole thing easier to understand in terms of four-momentum vectors. One of the problems with the rest/relativistic mass language is that it makes it hard to be clear in situations that involve multiple object in motion in several direction. What were the pre-conditions to your question again? I'm lost. $\endgroup$ – dmckee Jul 4 '16 at 23:29

The naive analysis cannot be correct because then the velocity of one ball would be $0$ and the other would be $\frac{2v}{1 + \frac{v^2}{c^2}}$, which gives an energy and momentum different from the initial energy and momentum of the system. As to what actually happens, I'll leave that to other posters...


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