# Einstein’s Thought Experiment of a Stationary Box and a Photon

1) First of all, let us consider a particle of light, also known as a photon. One of the interesting properties of photons is that they have momentum and yet have no mass. This was established in the 1850s by James Clerk Maxwell. However, if we recall our basic physics, we know that momentum is made up of two components: mass and velocity. How can a photon have momentum and yet not have a mass? Einstein’s great insight was that the energy of a photon must be equivalent to a quantity of mass and hence could be related to the momentum.

2) Einstein’s thought experiment runs as follows. First, imagine a stationary box floating in deep space. Inside the box, a photon is emitted and travels from the left towards the right. Since the momentum of the system must be conserved, the box must recoils to the left as the photon is emitted. At some later time, the photon collides with the other side of the box, transferring all of its momentum to the box. The total momentum of the system is conserved, so the impact causes the box to stop moving.

3) Unfortunately, there is a problem. Since no external forces are acting on this system, the centre of mass must stay in the same location. However, the box has moved. How can the movement of the box be reconciled with the centre of mass of the system remaining fixed?

4) Einstein resolved this apparent contradiction by proposing that there must be a ‘mass equivalent’ to the energy of the photon. In other words, the energy of the photon must be equivalent to a mass moving from left to right in the box. Furthermore, the mass must be large enough so that the system centre of mass remains stationary.

My questions: 1) I'm not able to grasp the concept of centre of mass in paragraph (3). 2) What's the center of mass of the system of the box and photon? 3) If no external forces are acting on the system, does the location of the center of mass remains the fixed? Then what does it mean by the location of center of mass being fixed if the box has moved? 4) What's the relation between the mass being large enough and the center of mass to remain stationary in paragraph 4?

1.) As the box moves to the left, the photon moves to the right, and their momenta is conserved. Since the masses are moving proportionally and opposite to one another, the center of mass of that system remains fixed.

2.) It is the same as the center of mass of a system consisting of a large gymnasium and a tennis ball inside the gymnasium, if that helps make it clearer. It's just that photons are very, very, very, (very) "small" - but the idea behind it is the same.

3.) Yes, it does mean that. The box has moved, but so has the photon, so the center of mass of the box-photon system has not moved, they've just shifted relative to one another.

4.) It means that the mass must be non-negligible, so that it is accounted for in calculating the center of mass of the system, so that 1.) is true.

I hope this helps answer your questions, but please follow up if anything is unclear.

• Thank you, I've somehow grasped the concept, and will completely grasp it with time. Well, if only the photon had moved, and the box had not moved so would the center of gravity have been fixed? Forget the conservation of momentum for some time. – Samama Fahim May 3 '13 at 15:03
• If we ignore conservation of momentum, then the center of mass for the system would not remain the same - for the photon moves but the box does not. Thus, the center of mass would move in the same direction as the photon. – NWard May 3 '13 at 15:09
• That's a good explanation. :) – Samama Fahim May 3 '13 at 15:16

If you got this from www.adamauton.com, you should be aware that there is an error in the argument. The author assigns mass $M$ to the box and mass $m$ to the ball with the combined mass being $M+m$. In this case, the center of the mass of the combined system, with the ball on right side of the box, is at $$\frac{M (x_1-\Delta x) + m (L-\Delta x)}{M+m}.$$ Equating this to the previous center of mass, and solving for $\Delta x$, you'd get $\Delta x = \frac{Lm}{M+m}$. In order to get $\Delta x = \frac{Lm}{M}$, you need to consider a ball with mass $m$ and a box with mass $M-m$. In this case the center of mass before and after the ball is thrown is at $\frac{(M-m) x_1 + m (0)}{M}$ and $\frac{(M-m) (x_1-\Delta x) + m (L-\Delta x)}{M}$, respectively. Equating these two and solving for $\Delta x$, you'd get the "correct" result.

Added: By the way, there are other problems with the overall argument on that page that are more subtle. But since that is not part of your question, I won't address them.

• just ask a new question, we are all fond of subtleties :) – user46925 Jan 22 '16 at 17:23
• @igael I probably will after my current question physics.stackexchange.com/questions/231025/… is answered. – mhp Jan 22 '16 at 18:16