62
$\begingroup$

Photons have no mass but they can push things, as evidenced by laser propulsion.

Can photons push the source which is emitting them? If yes, will a more intense flashlight accelerate me more? Does the wavelength of the light matter? Is this practical for space propulsion? Doesn't it defy the law of momentum conservation?

Note: As John Rennie mentioned, all in all the wavelength doesn't matter, but for a more accurate answer regarding that, see the comments in DavePhD's answer .

Related Wikipedia articles: Ion thruster, Space propulsion

$\endgroup$
10
  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/74013/44126 $\endgroup$
    – rob
    Commented May 15, 2014 at 14:54
  • 3
    $\begingroup$ A rocket powered by a flashlight strong enough to take off from the surface of Earth. Would likely burn a hole through the planet before it reached orbit. $\endgroup$
    – Reactgular
    Commented May 15, 2014 at 16:11
  • 1
    $\begingroup$ The fact that the ions are moving slower than the speed of light relative to the rocket means that the ratio of energy to momentum is less. $\endgroup$
    – DavePhD
    Commented May 15, 2014 at 17:16
  • 4
    $\begingroup$ Related: what-if.xkcd.com/87 $\endgroup$ Commented May 15, 2014 at 17:50
  • 1
    $\begingroup$ @HelloWorld: to your bullet question: yes and you don't even need to worry about relativity to deduce that. In Newtonian mechanics (good enough for bullets), kinetic energy $E = \frac{mv^2}{2}$ while momentum $p = mv = \frac{2E}{v}$. So if the two bullets have the same non-zero kinetic energy then the slower (more massive) bullet has more momentum. $\endgroup$ Commented May 18, 2014 at 19:04

4 Answers 4

73
$\begingroup$

Can photons push the source which is emitting them?

Yes.

If yes, will a more intense flashlight accelerate me more?

Yes

Does the wavelength of the light matter?

No

Is this practical for space propulsion?

Probably not

Doesn't it defy the law of momentum conservation?

No

In fact that last question is the key one, because photons do carry momentum (even though they have no mass). Photons, like all particles obey the relativistic equation:

$$ E^2= p^2c^2 + m^2c^4 $$

where for a photon the mass, $m$, is zero. That means the momentum of the photon is given by:

$$ p = \frac{E}{c} = \frac{h\nu}{c} $$

where $\nu$ is the frequency of the light. Let's suppose you have a flashlight that emits light with a power $W$ and a frequency $\nu$. The number of photons per second is the total power divided by the energy of a single photon:

$$ n = \frac{W}{h\nu} $$

The momentum change per second is the numbr of photons multiplied by the momentum of a single photon:

$$ P/sec = \frac{W}{h\nu} p = \frac{W}{h\nu} \frac{h\nu}{c} = \frac{W}{c} $$

But the rate of change of momentum is just the force, so we end up with an equation for the force created by your flashlight:

$$ F = \frac{W}{c} $$

Now you can see why I've answered your questions above as I have. The force is proportional to the flashlight power, but the frequency $\nu$ cancels out so the frequency of the light doesn't matter. Momentum is conserved because it's the momentum carried by the photons that creates the force.

As for powering spaceships, your 1W flashlight creates a force of about $3 \times 10^{-9}$ N. You'd need a staggeringingly intense light source to power a rocket.

$\endgroup$
22
  • 1
    $\begingroup$ Hypothetically, how much force would you need to power a spaceship if it was done this way? $\endgroup$
    – John Odom
    Commented May 15, 2014 at 15:31
  • 9
    $\begingroup$ @JohnOdom: it depends entirely on how much thrust you want to generate. The only realistic scenario I've seen for using light as propulsion is a solar sail. In that case it's the light from the Sun generating the force, and although the force is very small it's continuous so over time even the very small acceleration can build up high speeds. $\endgroup$ Commented May 15, 2014 at 15:41
  • 4
    $\begingroup$ According to Wikipedia, an ion thruster typically consumes 1–7 kW of power and produces 20–250 millinewtons of thrust. To get the same amount of thrust from a photon source would require roughly 6–80 megawatts of power. That's not trivial, but it's not totally infeasible either; a modestly sized nuclear fission reactor and several thousand halogen lamps would do it. (Lamp efficiency doesn't matter because any wavelength will do; reflector efficiency does matter, though.) $\endgroup$
    – zwol
    Commented May 15, 2014 at 15:46
  • 4
    $\begingroup$ ... The thrust-to-weight ratio of such a contraption is obviously terrible compared to just about any other option, but it would never run out of reaction mass. $\endgroup$
    – zwol
    Commented May 15, 2014 at 15:49
  • 3
    $\begingroup$ @Zack I think that's the important point about "photon drive technologies" : if you can build a source whose energy density is near-infinite, then you're better off in the long run. $\endgroup$ Commented May 15, 2014 at 15:51
20
$\begingroup$

Can photons push the source which is emitting them?

Yes, photons have momentum and momentum must be conserved. The source is pushed in the opposite direction of the photons.

If yes, will a more intense flashlight accelerate me more?

Yes, more photons means greater momentum.

Does the wavelength of the light matter?

Yes, shorter wavelength photons have higher momentum. $p = h / \lambda $

Is this practical for space propulsion?

Possibly, see Prospective of Photon Propulsion for Interstellar Flight (or use Alternative download site for pre-print version ) The concept of photon recycling is considered, for a potential enhancement of thrust/power ratio by several orders of magnitude.

Doesn't it defy the law of momentum conservation?

No, photons have momentum in one direction, the source has momentum in the opposite direction, so momentum is conserved.

$\endgroup$
9
  • 6
    $\begingroup$ So which is it? @JohnRennie says wavelength doesnt matter, but yours says yes. $\endgroup$
    – crthompson
    Commented May 15, 2014 at 15:45
  • 1
    $\begingroup$ @paqogomez momentum per photon = (Planck's constant)/wavelength. $\endgroup$
    – DavePhD
    Commented May 15, 2014 at 15:51
  • 2
    $\begingroup$ @JohnRennie expressed his answer in terms of the power of the flashlight. That is the sensible way to approach it IMO, and in those terms the wavelength doesn't matter. But wavelength does appear in the analysis. It would be possible to ask a similar but different question for which wavelength does matter, so pay attention to the exact wording of a question. $\endgroup$
    – garyp
    Commented May 15, 2014 at 16:00
  • 9
    $\begingroup$ @C4stor The question "does the wavelength of the light matter" is somewhat vague, I'm saying it matters on a per photon basis, and John is saying it doesn't matter on a per total energy basis, so I don't think there is any disagreement. $\endgroup$
    – DavePhD
    Commented May 15, 2014 at 16:02
  • 1
    $\begingroup$ @HelloWorld yes, but it would be better to say: the same energy could be transformed to nX photons of nM wavelength, to yield the same total momentum, even though the momentum and energy per photon is different $\endgroup$
    – DavePhD
    Commented May 15, 2014 at 16:41
4
$\begingroup$

This does not directly answer your question, but this is related. If you are floating in space the photons that hit you are also exerting a force. When you float in space a large number of photons emitted by the sun will hit you. These photons exert a force, this mechanism is referred to as radiation pressure. This force is significant enough that you can actually control a spacecraft with it.

NASA is doing that with the Kepler space telescope. The space telescope lost one of its reaction wheels. Reaction wheels are used to alter a spacecraft's orientation. With the remaining reaction wheels, the orientation of the telescope cannot be controlled with the accuracy needed for scientific missions. NASA devised a way to make use of the radiation pressure for controlling the spacecraft's orientation.

$\endgroup$
4
  • 1
    $\begingroup$ Actually, two of the reaction wheels are broken. With 3 of the 4 working, it could do its mission; once the 2nd broke it stopped being useful as an extremely sensitive planet finder $\endgroup$
    – Kyle Kanos
    Commented May 15, 2014 at 19:48
  • 3
    $\begingroup$ @KyleKanos Actually there is an announcement today that the technique Dohn Joe describes has been funded and NASA will try to revive Kepler. $\endgroup$
    – rob
    Commented May 16, 2014 at 19:13
  • 1
    $\begingroup$ @rob: That was made known to the community back in February (the white paper was back in September, I think). It won't be as stable as the original mission, hence it isn't an extremely sensitive planet finder, just a sensitive planet finder (plus, it's going to be looking at things that aren't planets). $\endgroup$
    – Kyle Kanos
    Commented May 16, 2014 at 19:19
  • $\begingroup$ Interesting about Kepler, but they found a way to balance it using reflected solar radiation. What OP describes is radiation, not reflection. The Pioneer Anomaly is a better real-life example. $\endgroup$ Commented May 18, 2014 at 15:41
0
$\begingroup$

See Solar Sails http://en.wikipedia.org/wiki/Solar_sail.

As other people have pointed out, this is extremely inefficient energy-wise, but has the advantage of being purely passive - no need to carry an energy source, and few or no moving parts to fail.

$\endgroup$
1
  • $\begingroup$ Solar sails is about being pushed by an external source, like laser propulsion. Although related, this is not the same. $\endgroup$ Commented May 19, 2014 at 8:34

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