# Relativity from a Flashlight

Here’s an experiment to try. It’s a thought experiment – it would be almost impossible to carry out in reality, though more delicate experiments roughly along these lines have been done.

You’re in one of the space shuttles, or the Discovery One, or your favorite fictional but realistic spacecraft. It has a hallway extending the length of the spacecraft from bow to stern. You stand at one end of the hallway with a laser pointer, and shine a brief pulse of light down the other end. Make that a very brief pulse. You want the physical length of the pulse as it flies down the hall to be short compared to the length of the hall. Since the speed of light is about 1 foot per nanosecond, a pulse on the order of a nanosecond or shorter will do the trick. This isn’t difficult, by the way, light pulses more than a million times briefer than that have been generated in the lab.

Now we know that light can transfer momentum. It can and does exert a slight force on whatever it hits. This is true even in the purely classical context of Maxwell’s equations. Take an introductory course on the theory of classical electromagnetism and you’ll see that the force exerted by a beam of light is equal to its intensity in watts divided by the speed of light: F = I/c. We’re just dealing with a discrete pulse acting over a short time, so it’s more convenient to think of this in terms of momentum. The change in momentum is just the force multiplied by the time it acts over. And “I” in the force equation is the power, and power multiplied by time is just the energy of the pulse. Shall we write this down?

When you shoot that light pulse down the hallway, like any other gun that change in momentum is just recoil. Not much in this case, but recoil nonetheless. Since you’re standing against the wall on one end of the hallway, that momentum is transferred to you and from you to the ship. The whole ship begins recoiling in the direction opposite that of the laser pulse. By a ludicrously small amount, but that amount is not zero. According to the conservation of momentum, the momentum Mv of the ship is equal and opposite to that of the light pulse:

Where M is the mass of the ship and v is its recoil velocity. But it won’t go recoiling forever. In a few nanoseconds to microseconds depending on the length of the ship, the light will hit the far wall and impart momentum equal and opposite to the original recoil. The ship is now stopped again.

How far did the ship get displaced by the recoil during the time of flight of the light pulse? Well, distance is velocity multiplied by time, d = vt. The time of flight is just the length of the ship hallway divided by the speed of light, t=L/c. Substitute those in and we’ve got:

Which is great except for one thing: it’s impossible. The law of conservation of momentum says the center of mass of a system can’t move unless acted on by external forces. Our system here is a perfectly isolated spacecraft, and yet it has moved a distance d with no external forces.

We said earlier that light had momentum, which is a start, so perhaps we could say that the momentum of the spacecraft plus light is unchanged. True, but once the light hits and is absorbed by the far wall, it’s gone. All we have is the spacecraft, displaced in the absence of external forces. Is conservation of momentum toast as a concept?

Maybe not just yet. We know light doesn’t have mass, but if the light pulse could turn its energy into mass when it hit the far wall, then maybe that mass displacement would compensate for the displacement of the ship – that would leave its center of mass unchanged. You can do the math to verify, but it intuitively stands to reason that the ratio of the mass m transferred by the light to the total mass of the ship M would equal the displacement d divided by the length of the ship L. So, since

We have:

Rearranging:

Well, that looks familiar. What it tells us is that given only the classical fact that light has momentum, the energy of light must be turned into mass when it’s absorbed, and in a quantity given by Einstein’s famous equation.

Now this is far from a formally valid derivation. Relativistically the momentum of the space ship is only Mv for small v, and the handwaving m/M = d/L only works for small m. A more solid argument would have to go into more detail. Still, it’s more than a little cool to see the most famous equation in relativity come dropping out of a purely classical argument.

1. #1 Kent
April 14, 2010

I just love posts like this. Whimsical physics that reminds me of Einstein’s thought experiments.

2. #2 NoAstronomer
April 14, 2010

Ouch! That took me right back to my 1st year at university. Nowadays that’s a pretty long trip for me.

3. #3 BlackGriffen
April 14, 2010

Not to mention that the speed of sound in the walls of the spacecraft is somehow faster than the speed of light…

4. #4 Matt Springer
April 14, 2010

Yes, that too. ðŸ˜‰

Though to be sure, things would end up working out in about the same way. Once the disturbances from the emission and absorption distribute themselves into some kind of equilibrium, the spacecraft center of mass will end up having shifted by the same distance d – at least until the mass/energy transferred thermalizes and distributes itself evenly too, drifting the ship back about halfway to its initial position.

But this is why it’s a thought experiment.

5. #5 Len Bonacci
April 14, 2010

I like it, but the logic is somewhat circular. You’re not starting from first principles — the force due to a beam of light is I/c because E = mc2…

6. #6 csrster
April 15, 2010

Len, I don’t think so. As Matt says, the force can be calculated from Maxwell’s equations without introducing any “m”.

7. #7 Len Bonacci
April 15, 2010

Ah, okay — my physics background is weak enough that I don’t have a good grasp of Maxwell’s equations (not much more than what I’ve read on t-shirts!). Back to the books.

8. #8 hello kitty lamp
April 15, 2010

This subject is something that Einstein covered when he studied black holes.

9. #9 Anonymous Coward
April 15, 2010

Fun! That’s a terrific thought experiment, and one I’ve never seen before. Did you get that from some other source, or did you invent it?

Although it’s fun, I’m not sure that this is any more of a “purely classical argument” than the usual development of relativity. Your starting point seems to be a combination of “Maxwell’s equations properly describe the properties of light (in all reference frames)” and “Momentum is conserved”.

The usual development of relativity usually relies on “Maxwell’s equations properly describe the properties of light in all reference frames” to get the time-and-space transformations, then applies “momentum is conserved” to get non-Newtonian forces and relativistic mas, then uses those non-newtonian force equations (and the equation for work) to get E=mc^2.

So your way is a much more direct to get to E=mc^2 without the intervening stuff, but I’m pretty sure the underlying assumptions are exactly the same.

10. #10 rat
April 19, 2010

ok, i’m fascinated by thought experiments although I’ve never taken a physics class. But, I have to ask a question.

– If I understand the thing right, in order to generate the beam of light some mass has to be given up within a battery to fuel the laser and then a certain amount more is given up due to the inefficiencies within the laser pointer . This mass would have to be subtracted from the mass of the spaceship as a whole.

– Furthermore, it seems that much of the energy of the pointer would be absorbed into the mass of the receiving surface as heat.

– It further seems that any energy in the reflected light would have to be would have to be subtracted from the equation as it would still be available to do ‘work’.

Where does that leave us? It seems that the momentum change within the whole system would very possibly be zero.

11. #11 rat
April 19, 2010

one small addendum to my thoughts. since we have loss mass in the ship as a whole, it would then seem that since the ship is lighter than before, the same energy given by the spaceships thrusters, would then propel the ship faster by a minute amount which might be mistaken for momentum provided by the laser. Any thoughts?

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