This thought occurred to me after I began reading about the EM drive, and I know there are a lot of theories out there on how that works/doesn't work, I'm wondering why this solution wouldn't make sense. My background is in CS/Math, and I've only taken a few semesters of (Newtonian) physics so if this isn't reasonable at all I apologize.

Some things I believe to be true:

  • Photons always move at the speed of light. They have energy and momentum.

  • When photons are reflected, they impart a small amount of their momentum/energy into the reflector and are red shifted.

  • The speed of light is orders of magnitude faster than the speed of a pressure wave (sound) through every solid material.

  • (unsure) Reflected photons always impart their momentum as some kind of proportion of their total energy (a small percentage as opposed to a static amount).

So, I will try to explain as best as I can. I could make some clarifying diagrams if that helps. Suppose you have two mirrors facing one another, and also attached to the same vehicle. You shoot a photon (assume an energy of 1) at mirror one. The photon imparts a very small amount of momentum on it and it begins moving away. The photon bounces off with, I don't know, .9 energy now, leaving .1 energy with the first mirror and red-shifting itself. Normally if it weren't light going so fast, the first mirror's momentum would be transferred to the second mirror via a pressure wave through the vehicle, causing the relative energy between mirror 2 and the photon to be the same as mirror 1 and the photon prior to their collision. However, because it's going so much faster than the pressure wave the two mirrors are in a separate frame of reference so when the photon hits mirror two, it imparts the same proportion of its total energy, reflecting with I guess .81 using the same numbers as above. This means that when the two pressure waves meet somewhere in the middle, they are unequal (.1 vs .09 with above). Reflecting a photon back and forth continuously, you get asymptotically close to the photon's momentum reflection proportion (if that makes sense), and your vehicle actually moves, without reaction mass and without breaking conservation of momentum/energy.

Put a different way, if you're a static observer between initially stationary mirrors floating in space and you shoot a photon at mirror 1, which bounces off and hits mirror 2, and then is pulled out of existence, will mirror 1 be traveling away from you faster than mirror 2? I would think yes, since the photon lost momentum onto mirror 1 prior to hitting mirror 2.

Is this wrong? Correct? Impractical? Thanks in advance.

Edit: To clarify how the first bit is related to the second bit. Take the second example and attach the mirrors. When the photon hits the first mirror it will create a ripple that slowly propagates through the material connecting them. So until this ripple travels through the entire vehicle, you could consider the first mirror to be moving away from you and the second mirror to be static, which is the difference in the frame of reference of the two mirrors. After its first collision, the photon will hit the second mirror creating a second (presumably smaller) ripple before the first ripple has hardly propagated at all. If these ripples were equal, the vehicle wouldn't move, but because they are unequal, it does.

  • $\begingroup$ The first part of your question is quite long and confusing. The second part ("Put a different way...") is clear - and the answer is "yes". Since the photon lost energy on the first bounce, it has less energy for the second bounce. The difference might be measurable in the wavelength shifting of the photon - but I can guarantee you cannot measure it on your "mirror". But maybe if you repeated this billions of times. Of course once a mirror has started moving after the first impact, the situation changes... I wonder if you can tighten up the question a bit. $\endgroup$
    – Floris
    Commented Nov 6, 2014 at 20:26
  • $\begingroup$ Just added some clarification on how the two are related. $\endgroup$ Commented Nov 6, 2014 at 20:38
  • $\begingroup$ if you consider the sum of exchanged momentum on one side minus on the other, you get something like (1 - k) where k is the amount of conserved energy (very very close to 1). Also, how do you give the initial energy and momentum to the photon? $\endgroup$
    – njzk2
    Commented Nov 6, 2014 at 21:33
  • $\begingroup$ I envsioned it as sum(x^k-x^(k+1)) for every other number from 1 to n. Where x is the percent imparted momentum (.1 above), k is the number of times it's been reflected and n is the number of times it is reflected before it is expelled. As for where the photons come from, they could be generated on or outside of the vehicle, it wouldn't matter in the above example. $\endgroup$ Commented Nov 6, 2014 at 21:38

3 Answers 3


This part is incorrect:

When photons are reflected, they impart a small amount of their momentum/energy into the reflector and are red shifted.

Momentum is always conserved. If in the frame of reference you choose, the photon accelerates and increases the momentum of the mirror, then the reflected photon will have less momentum (or less eneergy). In that case you are correct that the photon is red shifted.

But if the mirror is approaching the photon in your frame of reference, then the photon may slow and decrease the mirror's momentum. That means the reflected photon will have greater momentum (energy) and be blue shifted. So it is incorrect to assume there is some fixed percentage that is always transferred.

So there are two problems I see with the scenario.

  • As the tension in the system that holds the mirrors in place increases, the photon can no longer accelerate the mirrors. It will stop transferring momentum after a (small) finite amount.

  • The mirrors begin moving in opposite directions. Any transfer that does occur is cancelled out by transfer in the other direction. The vehicle as a whole does not move since it is attached to both mirrors.

It would be more efficient to simply shoot the light in one direction from the vehicle. Then all of the momentum is transferred and the vehicle will accelerate. The problem is that the rate of change is tiny.

I can accelerate myself fairly quickly on a bike with a few hundred watts of power, but the same power of light does nothing,

On earth we use energy to separate a (relatively) low mass object (your bike) from a high mass object (the earth). With constraints of conservation of momentum, almost all the energy goes into the low-mass object. (yay!)

In space by generating photons, the situation is reversed. Now your vehicle is the high mass object and almost all the energy is delivered into the photons. Your vehicle gets almost none of it. (boo!)

You would get better acceleration by dumping the energy against a high-mass exhaust instead of a photonic one. Unfortunately, that's not as fuel efficient (because you run out of fuel faster).

  • $\begingroup$ The second mirror will not be moving toward the photon to be blue shifted, so as a static observer in space, you still see a difference in momentum between the two still, I believe. So even though the mirrors are moving in opposite directions, they are not moving in opposite directions equally, so it won't be cancelled out. As far as shooting light out directly, that's the workaround I'm trying to come up with. I can accelerate myself fairly quickly on a bike with a few hundred watts of power, but the same power of light does nothing, so I assume it's reflected the same when generated. $\endgroup$ Commented Nov 6, 2014 at 22:45
  • $\begingroup$ Oh, if you only want two bounces, then yes there is a slight difference. The difference would be identical to the momentum difference between the light when first launched and when it escapes from the vehicle. It would be better to just do one bounce. Then you get all the momentum transfer. $\endgroup$
    – BowlOfRed
    Commented Nov 6, 2014 at 22:49

You didn't say how big is the mirror. Assume the photon is in the visible domain, and the mirrors are classical object, i.e. big. Such objects will remain insensitive to the linear momentum lost by the photon, because the recoil velocity of the mirror is practically zero. The reflection of the photon will be elastic, no energy imparted to the mirror, however, since the mirror moves away from the photon, the photon will be red-shifted. On the other hand, when hitting the 2nd mirror, which moves toward the photon, the photon will be blue-shifted back, i.e. return to its initial color.

But if the mirrors are quantum objects, very small, they will be kicked out from their places, or will bend, or will be damaged.

Best wishes, Sofia


Whether or not your idea produces thrust depends on at least one of the following statements being true:

  • The laser doesn't travel with the vehicle.
  • The photon eventually leaves the mirror structure.

Forget the structures holding the mirrors and consider each component as the photon interacts with it. In all cases, I assume each emission and reflection conserves momentum because this is extremely well-tested experimentally.

  1. When the laser emits the photon, the photon travels forward and the laser is pushed backwards.

    • Photon momentum = $p_0$
    • Laser momentum = $-p_0$
    • Momentum of laser and photon = $p_0 - p_0 = 0$.
  2. The photon hits the front mirror.

    • Photon momentum before collision = $p_0$
    • Photon momentum after collision = $p_1$ (which will be negative due to traveling backwards and with a smaller absolute value due to losing energy to the mirror)
    • First mirror momentum after collision = $p_0 - p_1$
    • Momentum of mirror and photon after collision = $p_1 + (p_0 - p_1) = p_0$ (same as the starting photon)
    • Momentum of mirror, photon and laser = $-p_0 + p_0 = 0$
  3. The photon hits the rear mirror.

    • Photon momentum before collision = $p_1$
    • Photon momentum after collision = $p_2$ (which is now positive due to forward motion and smaller in absolute value due to giving up energy to the mirror)
    • Second mirror momentum = $p_1 - p_2$
    • Momentum of second mirror and photon = $p_2 + (p_1 - p_2) = p_1$ (same as photon after first reflection).
  4. Results
    • (a) Total momentum of both mirrors = $(p_0 - p_1) + (p_1 - p_2) = p_0 - p_2$ (result will be positive)
    • (b) Momentum of both mirrors and photon = $(p_0 - p_2) + p_2 = p_0$ (same as initial photon)
    • (c) Momentum of both mirrors and laser = $(p_0 - p_2) - p_0 = -p_2$ (opposite momentum of the photon, that is, traveling backwards)
    • (d) Momentum of both mirrors, the laser, and the photon = $-p_2 + p_2 = 0$

I've ignored the issues with pressure waves because the only thing the vehicle structure does is get everything attached to it moving at the same velocity. It does nothing to the total momentum. Let's examine the outcomes (a. through d.).

a. The laser is not attached to the vehicle and the photon leaves the mirrors. The final momentum of the vehicle is forward since $p_2$ is less than $p_0$.

b. The laser is not attached and the photon never leaves the mirrors. The final momentum of the vehicle is forward and equal to the initial momentum of the incoming photon.

c. The laser is attached to the vehicle and the photon exits the vehicle. The final momentum is $-p_2$, that is, opposite of the photon momentum after two reflections. The vehicle will be heading backwards.

d. The laser is attached to the vehicle and the photon never leaves. The final momentum is zero, so the vehicle never moves.

In order to move the vehicle, either an outside force has to push it (the unattached laser in a. and b.) or something has to be thrown out the back (situation c.). a. and b. are externally-powered solar sails (a single mirror would do much better). c. is a photon rocket (which would be better with no mirrors and the laser just pointed out the back).


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.