Why doesn't a ray of light have enough momentum to make us fall?

Why can't light be so powerful that it has enough momentum to make us fall?

• Commented Nov 15, 2014 at 7:03
• @GodParticle: Please link to arxiv abstracts rather than PDFs, thus: arxiv.org/abs/0806.2442
– rob
Commented Nov 16, 2014 at 0:47
• Possible duplicate: it can
– rob
Commented Nov 16, 2014 at 0:50

Supposing we could shield ourselves with a perfectly nonabsorbing, reflective shield so that light would perfectly elastically bounce off us, thus preventing high power beams from incinerating us as Anna V's answer validly argues they would. Then the "fundamental" answer to your question is "because light has zero rest mass"; to explain further: the 4-momentum norm relationship:

$$\frac{E^2}{c^2} - p^2 = m^2 c^2\tag{1}$$

implies a linear relationship $$|p|=\frac{E}{c}$$ for things without rest mass. This means a great deal of energy even for modest momentums. So then, if we elastically reflect a pulse of light with energy $$E$$, the impulse it imparts upon us is $$2\,\frac{E}{c}$$. What impulse would tip us off our feet? This is not a precise figure, but in the worst case, a force of one tenth of our weight acting on the centre of our chest for one second might do it. So the impulse is of the order of $$50{\rm N\,s}$$.

This means that the amount of light we should need to throw us off balance is of the order of $$50\,c/2 = 7.5\times10^9{\rm J}$$. In other words, the whole of the $$5{\rm GW}$$ of output of Fukushima Daiichi power station (before its demise) and a bit more for one second.

Quite simply, there are very few imaginable light sources that have this kind of output. And, as Anna V says, practically speaking, even a much weaker light source would likely fry us.

Now let's look at what happens when there is rest mass. (1) becomes:

$$\Delta E=E-m\,c^2 = \frac{p^2}{2\,m} -\frac{p^4}{4\,m^3\,c^2} + \cdots\tag{2}$$

The quantity on the left is the amount of (kinetic) energy we need to bestow on our massive object to get it moving with momentum $$p$$ (when $$p$$ is subtituted into the right). For a thrown one kilogram object, if we set $$p=25{\rm N\,s}$$ (so that it imparts $$50{\rm N s}$$ when it elastically bounces off us) then $$\Delta E$$ is a mere $$313J$$, seven orders of magnitude smaller than the quantity of light needed. A mere mortal human like you or I could easily impart this amount of energy to a $$1{\rm kg}$$ medicine ball.

Implications for Mass of Photon

Can we conclude by the fact that photons have no mass ?

Strictly no. You can never do that: you can only put bounds on the photon mass by experimental results. The classic measurement is that of the sphere inside a hollow conductor, see the section "Experimental checks on photon mass" on the Photon Wikipedia page, and this currently puts the rest mass limit at $$m=10^{-14}{\rm eV}$$. By way of comparison, an electron has a rest mass of about $$550\,{\rm keV}$$, twenty orders of magnitude higher. There are other, even smaller experimental limits, but they must rule out mass making mechanisms (e.g. the Higgs mechanisms) at the outset - i.e. they make more potentially falsifiable assumptions than the hollow sphere.

Coming back to our "experiment" of knocking someone over. You could put a postulated small photon mass $$m_p$$ in my equation (1) and check what kind of photon mass you would need so that you could observe a thrust from light outside the statistical confidence intervals from actual experimental measurements. This could then give an experimental limit on the photon mass, but it would be tricky. You'd need to be careful here, because you would need to think carefully about what your experiment were actually doing and what you were actually measuring with your beam energies. There are several alternatives:

1. You are truly measuring the beam's total energy, in which case, somewhat counter intuitively, you would find you would need a higher energy to produce a given momentum than for massless light because you would be storing some of the beam energy in rest mass;

2. The beam energy measured is the light's kinetic energy alone, in which case you would see more impulse for a given energy.

This time, instead of (2), you would use the Taylor series:

$$E=p\,c \left(1 + \frac{m^2\,c^2}{2\,p^2} - \frac{m^4\,c^4}{4\,p^4}+ \cdots\right) = p\,c + \frac{m^2\,c^3}{2\,p} + \cdots$$

(i.e. you would assume the $$p^2\,c^2$$ term were the bigger one under the square root.

Let's suppose you saw a one newton second difference from the calculate $$p=50{\rm N\,s}$$ for a pulse of $$7.5\times 10^9{\rm J}$$. Then we'd have $$1\approx m^2\,c^3/50$$, whence $$m\approx 10^{-12}{\rm kg}$$. If your light were optical, say $$500{\rm nm}$$ wavelength, then $$7.5\times 10^9{\rm J}$$ is about $$2\times 10^{28}$$ photons. This yields a photon mass of $$10^{-40}{\rm kg}$$, about nine orders of magnitude less than that of the electron. If we could measure $$10^{-10}{\rm N\,s}$$, then you would achieve about the same sensitivity as the hollow sphere experiment for bounding the photon mass.

• Can we conclude by the fact that photons have no mass ? Commented Nov 15, 2014 at 16:10
• A little bit of color: the most powerful light sources I know are used for inertial laser confinement fusion. The Natiional Ignition Facility laser has a pulse of about 4 MJ (delivered in picoseconds, that's 500 TW of instantaneous power!), and Laser Mégajoule - still under construction. They are about 3 orders of magnitude too small. Commented Nov 15, 2014 at 20:32
• @Vinayak see my updates. Commented Nov 16, 2014 at 0:44

A ray of light is a geometrical line describing the propagation of an electromagnetic wave. The electromagnetic wave is composed of zillions of photons each with a tiny momentum. The momentum is not large enough to sense an impact, it is pico newtons even for a laser beam.

Lasers can have very high energy and momentum, but like knives, they cut soft tissue as they have a small area. A laser large enough to cover a body area would burn up the body.

In contrast to the order of pico Newton force of a pointer laser beam, a velocity of 10m/second to a mass of 1 k gives a 10kg-m/sec momentum, ~10 Newtons force. Pico is a factor of 10^-12, there are 13 orders of magnitude to give a mass of a kilogram by a laser beam in vacuum a velocity of ten meters per second, using a light beam.

• And I did not post an answer thinking that I know no way of estimating how much energy/momentum do N photons need to transfer to a human body to topple the person, and expecting somebody would post it. (We can neglect that the ray would cause localized damage if sufficiently energetic, as a first approximation.) Can that be done? (Just an energy/momentum-transfer estimate.) Commented Nov 15, 2014 at 5:20
• @New_new_newbie there would need to be a way to transfer momentum without transferring energy. We'd have to be extremely reflective to have any hope of this happening. Commented Nov 15, 2014 at 5:31
• @BrandonEnright: You are misinterpreting. I never said they are going to be exclusive. I only said let's get an estimate for either of them, whichever of the two is easier to estimate. But "momentum is not large enough" is a vague statement, unless you know what you are comparing against. i.e. you know how much is "large enough". Commented Nov 15, 2014 at 5:37
• @New_new_newbie If you read the link in the answer where I state pico newtons, you will see the calculation for a laser pointer Commented Nov 15, 2014 at 6:18
• @annav - No, that calculation is the easy part (energy or momentum residing in a photon of wavelength $\lambda$, from which you can get the total amount residing in N such photons). The difficult part is - estimating how much energy or momentum would be required to topple a person over. That's what my comment is referring to. Commented Nov 15, 2014 at 6:31

As I indicated in my comment on Rod's answer, some very powerful lasers do exist - and while their photons don't have much momentum, they do "pack a mean punch".

In fact, laser ablation (where a laser beam produces significant local heating and material is ejected at high speed) may just produce the phenomenon needed.

Let's take the example of the LLNL laser at the National Ignition Facility. In a few nano seconds, it dumps a few MJ. Rod's answer assumed "perfect reflection". I would like to create an absorbing layer, and see what kind of punch that can give.

For example, if the ablated material had a velocity of 1000 m/s, you could get a 50 Ns pulse with just 50 gram of material. Let's assume that $\frac12 kT$ of the energy of the ablated atoms is available to provide thrust in the right direction, then for carbon atoms and a velocity of 1000 m/s you would need a temperature of about 700 K. This is quite possible, as demonstrated in this paper

If you wanted to ablate less material, you would need a correspondingly higher temperature.

My point is - if you shot a real laser at a person who was wearing an appropriate body armor, it would not be terribly hard (and certainly within the range of available technology) to "zap" a bit of their armor off and push them over.

The key difference between this answer and Rod's is in the mass used: the photons have a really, really terribly relationship between energy and momentum - the moment you can bring some "real world" mass into play, it becomes possible.

• In your answer you're using the energy of the photons to heat material so it expands and pushes something over. I think this violates the spirit of the question which was asking about photon's momentum. Commented Nov 15, 2014 at 20:58
• @BrandonEnright - I agree with your first point, and disagree with the second. I wanted to give another perspective. "Light is powerful". It packs a punch. Yet it needs to "borrow" mass from somewhere else in order to push anything over. I think this is a nice addition to the overall discussion / understanding. I don't think that I am "violating the spirit of the question" which is asking about the power of light as well as the momentum. Power is not the same thing as momentum. I hoped to highlight that. PS thanks for the edit. Commented Nov 15, 2014 at 21:12
• @BrandonEnright ... and, in a sense, my answer studying the mass versus massless particles, is kind of like Floris's: if you can begin with something with mass, then you get an efficient momentum store "for free" and so my discussion could be in a sense a bit misleading. The total energy of a one kilogram medicine ball is of course stupendous. I think Floris's answer is a great, somewhat oblique point that the rest of us didn't think of: good science almost always lies in the subtleties. BTW +1 Floris! Commented Nov 15, 2014 at 23:44
• +1. This is why nuclear weapons are the weapon of choice when it comes to diverting an asteroid or comet from colliding with the Earth. The explosion of the bomb itself doesn't provide all that much impulse. It's the ablation that results from the huge gamma (and neutron) flux hitting the asteroid/comet and making it shed its outer layer on one side that provides a huge impulse that changes the impactor into a non-impactor. Commented Nov 16, 2014 at 7:23