As I understand it light has no mass and can act as both a particle and a wave.
Lets say we have a giant flashlight in a vacuum. Does the light coming from that flashlight have any effect on the flashlight? Like in newtons third law? Will the flashlight move from the momentum of the photons alone? Or light propulsion, if you will. Or do photons being massless not apply to newtons third?
Yes, the flashlight moves. If Newtons third law wasn’t obeyed, momentum wouldn’t be conserved and there wouldn’t really be much point in talking about light as having momentum.
Yep.
Initially the system’s momentum p[sub]0[/sub] = 0. Turn on the light and you have the flashlight’s momentum p[sub]1[/sub] and photon’s mommentum p[sub]2[/sub] summing to zero.
p[sub]0[/sub]=p[sub]1[/sub]+p[sub]2[/sub]
p[sub]1[/sub]=-p[sub]2[/sub]
p[sub]1[/sub]=-p[sub]2[/sub]=-E[sub]2[/sub]/c
m[sub]1[/sub]v[sub]1[/sub]=-E[sub]2[/sub]/c
v[sub]1[/sub]=-E[sub]2[/sub]/(cm[sub]1[/sub])
So theoretically one could have a “light propelled vehicle” in a space enviornment?
Would it be feasible? Or would we need a light source thats to big?
I just realized what I was asking and thought of Solar Sail technology. It is somewhat similar no?
What is the momentum of a massless particle? Maybe photons are a special case but in Physics 101 I learned that momentum = mv.
That’s exactly what you’re talking about, although real solar sails may also pick up momentum from actual physical particles when they’re close enough to a sun.
Tricky, but I’ll try.
- E=mc[sup]2[/sup]
- p=m[sub]0[/sub]v/sqrt(1-v[sup]2[/sup]/c[sup]2[/sup]) - relativistic momemtum. Now take take 2 and multiply by c
2a. pc= m[sub]0[/sub]vc/sqrt(1-v[sup]2[/sup]/c[sup]2[/sup])
Now square both sides
2b.p[sup]2[/sup]c[sup]2[/sup] = m[sub]0[/sub][sup]2[/sup]v[sup]2[/sup]c[sup]2[/sup]/(1-v[sup]2[/sup]/c[sup]2[/sup])
Play with the equation and you wind up with
- p[sup]2[/sup]c[sup]2[/sup] = -m[sub]0[/sub][sup]2[/sup]c[sup]4[/sup] + (mc[sup]2[/sup])[sup]2[/sup]
Well we know that E=mc[sup]2[/sup] so E[sup]2[/sup]= (mc[sup]2[/sup])[sup]2[/sup]
- p[sup]2[/sup]c[sup]2[/sup] = -m[sub]0[/sub][sup]2[/sup]c[sup]4[/sup] +E[sup]2[/sup]
Now m[sub]0[/sub] is 0 for a massless particle and the equation collapses to
5. p[sup]2[/sup]c[sup]2[/sup] = E[sup]2[/sup] → p=E/c
wow, that’s a lot of {sup}{/sup}s in there. I think I kept everything straight
Both the force from solar wind and the force from light pressure fall off as 1/r^2, and for a star like the Sun, it works out that the light pressure dominates over the solar wind at all distances.
As for practicalities, a light-propulsion drive produces very little force (and hence very low acceleration), so it’d take you a very long time to get up to speed. But it does so very efficiently, such that for a given amount of fuel, you’ll eventually get up to a higher speed than with other propulsion methods. The upshot is that a light drive isn’t practical within the Solar System, but it would likely be the way to go for a trip to another star.
From what I remember of physics (really I only took the first 2 semesters) waves can carry momentum and energy.(Yes, even if nothing really moves. I forget why this is true. Chronos, any chance you can explain why waves can carry momentum?) Since light is a wave it also carries momentum. Actually since someone brought it up an apparent paradox in relativity due to light carrying momentum is how Einstein figured out E = MC^2 .(It’s explained on Wikipedia. Might I add the explaination for this is better than the one I saw Michio Kaku which pretty much amounted to E = MC^2 because Einstein said so.)
FYI: Japan recently launched their solar sail spacecraft and the sail is working.
Yes I watched the video they released, very exciting stuff!
I am still a bit confused. How is it that a massless photon can push something with mass? I have just begun my foray into quantum physics, so please excuse my ignorance on the MATTER 
Im probably thinking about it all wrong.
Before I answer this, how’s your understanding of calculus?
[quote=“moldybread, post:11, topic:549400”]
Because a photon has MOMENTUM.
Force isn’t very relevant in QM, but,nonetheless, it equals dp/dt, and the momentum (p) of a photon is equal to Planck’s constant divided by the particle’s wavelength.
So, when a photon interacts with a massive particle both of their momentums change and thus a force is imparted.
Hmm Calculus. On a scale of 1 through ten, I’d say about a 1 
I do however plan on teaching myself in the near future. you know, to expand the horizons and all.
I Think I have a good grasp of how it works now from the following quote.
That helped alot. This is all extremely fascinating stuff.
ps Thanks Fiddlesticks for that link, it also helped fight my ignorance on the matter;)
Is any of this connected to the fact that light travels at c only in a vacuum, but when traveling through a medium it goes slower?
No! Between pseudo absorptions and emissions light still travels at c at all times.
I say pseudo because in a transparent medium there are no true energy levels that allow the photons to be absorbed.
And E = hc/λ, so p = h/λ; hence, momentum is strictly dependent upon wavelength, and so the smaller the wavelenght (and higher the corresponding frequency) the more momentum you get.
Also handy in case you run into an aggressive alien species that doesn’t believe your craft to be armed.
Rocket propulsion efficiency is a combination of two factors: the thermodynamic efficiency, η[SUB]c[/SUB], of the engine/motor (i.e. how much energy it converts into propellant momentum) and the propulsive efficiency, η[SUB]p[/SUB], which is a measure of the effectiveness of the exhausted propellant upon vehicle impulse. Since the former is dependent upon the process used to generate the propellant, we’ll ignore it and just look at propulsive efficiency. η[SUB]p[/SUB] = [2*v/c]/[1 + (v/c)[SUP]2[/SUP]], with c being the exhaust velocity and v is the vehicle velocity. If you plot this out you’ll find that efficiency is maximized (actually unity) when v=c; in other words, when the vehicle is moving the same speed and opposite direction as the exhaust.
For chemical rockets, even though using very low molecular weight propellants like hydrogen, that means that the effective use of the internal combustion-heated rocket engine is limited to a few tens of thousands of feet per second, even if you could carry enough propellant to continue accelerating. This also explains why the use of high molecular weight liquid propellants like kerosene/RP-1 and solid propellant boosters are often selected for the lower stages of orbital boost vehicles; in addition to taking up less bulk and thus requiring less tankage (and in the case of cryogenics, insulation), they also provide higher average thrust for the firing time and higher density impulse, while taking only a very small hit on efficiency when the booster is moving at low speeds. Basically, they get you up to a useful speed faster. Once you’re going at a significant clip (and especially once you’ve been relieved of the bulk of your mass, which is mostly propellant and tankage), a smaller engine with a lower mass flow rate and a low molecular weight propellant is significantly more efficient, hence why many modern upper stages like Centaur, Delta upper stage, and Saturn S-II and S-IVB use liquid hydrogen and liquid oxygen as propellants. (Many kick motors and upper stages use solid propellants and storable fuels like hydrozine, MMH, and UDMH for safety and maintenance reasons despite the inefficiencies.)
For non-chemically-powered rockets (i.e. thermal fission and fusion) using a low molecular weight propellant makes sense as higher temperatures will provide both more thermodynamic efficiency and higher exhaust velocities. However, there is a point–and it is far before you can achieve anything like interstellar speeds–that you just can’t accelerate a reasonable amount of onboard propellant to a high enough exhaust speed to keep thrusting effectively. Although you’ll still get some incremental thrust, once your speed significantly exceeds exhaust velocity you’re done pushing. With light as a propellant, you’ll never exceed the exhaust velocity of c, and in fact, your rocket will become more efficient the faster you go…but it will take a long, long time to get up to that speed.
No, light travels at c in all mediums, but not always in a straight line (or without being absorbed), hence the aggregate reduction of speed in physical mediums.
Stranger
OK, then, momentum of massless particles. The first thing to be aware of is that momentum isn’t actually mv: That’s just an approximation, valid at low speeds (i.e., much less than the speed of light). The correct formula is momentum is mu, where u is a quantity called “proper velocity”, and is given by u = gamma*v, where gamma is the relativistic dilation factor. At low speeds, gamma is only slightly more than 1, so it’s a good approximation to use v instead, but as speed approaches c, gamma (and hence also u) goes to infinity.
OK, so apply this to a photon. So far as we can tell, photons are massless and move at c. So for a photon, mu is 0infinity. Which we don’t know how to evaluate. What to do? Well, we would expect that a massless particle would behave an awful lot like a particle which had a mass that was really, really tiny, and that the approximation would get better the more really, really tiny the mass is. In fact, we don’t have any actual proof that the photon is truly genuinely massless: For all we can tell, it might actually just have a really, really tiny mass. So we can look instead at how a particle with really, really tiny (but nonzero) mass would behave. Particles with nonzero mass don’t travel at c, either (though they can get very close to it, depending on their energy, so we also can’t be sure of telling the difference that way), so they still have finite energy and momentum. And if the mass is very, very small, then we find that the momentum of the particle is very close to its energy divided by c. In fact, as the particle gets smaller and smaller, this approximation gets better and better. So a thing that has mass so small it’s indistinguishable from 0 should also have a momentum that’s indistinguishable from E/c.