Imparting momentum to light

If I hit a cricket ball with a bat it doesn’t just bounce off, it changes direction as it takes momentum from the bat.

So, if I could swing a mirror, say, at some absure velocity like .99c, and hit some light with it, would it have any effect on the light different to what you would expect from a stationary mirror in the same position? Obviously the speed of the light would be the same, but what about the direction?

The direction will change, but so will the frequency of the light. If the mirror is moving towards the direction the light is coming from, the frequency will increase and the photons will have more energy. If the mirror is moving away from the light, the frequency and photon enrrgy will decrease.

Exactly, this is the principle used by the police to catch you speeding. The “light” they use is in the microwave portion of the spectrum.

With Lidar, actual light. Yes, the frequency change, the doppler effect. Oh, for light mv=0

True, but irrelevant, since mv is not the formula for momentum. It’s a pretty good approximation at low speeds, but of course light is the exact opposite of “at low speeds”, so the approximation becomes totally useless.

Better let the editor of the CRC handbook of Chemistry and Physics know.

Out of curiosity, do you really think you know more than the physicist who just corrected you?

eta added link

So it would just bounce of as if the mirror was stationary, in the same direction as if the mirror was stationary, but doppler shifted, then?

I don’t understand what the difficulty is. Light has momentum, it’s not particualrly new physics and it’s an experimental fact.

mv is the Newtonian definition of momentum, but as Chronos points out is only a low velocity approximation in relativity. For a particle travelling at c, momentum is not a function of velocity.

As Chronos has said, not applicable, since in the case of photons it’s not applicable. Photons do carry momentum. Those “Crookes Radiometer/Light Mills” were a 19th century atempt to directly observe this momentum. It didn’t work, mainly because they couldn’t get a good enough vacuum (and the rotation they observed had a different, and surprisingly complex, cause), but direct observation of light momentum was made in the 1930s. With the advent of laser beams, the momentum of light was significantly boosted by the greater intensity (and, for some effects, coherence), and people were able to levitate small balls and thin films. I’ve seen the (unintended) effect in my own lab. Such momentum transfer is behind the principle of Light Sails,

The momentum of a photon is prooportional to its wavenumber (and its frequency), and this changes with reflection from a moving reflector, so you certainly can cause reflected photons to have different momentum from he incident photons, even at speeds well below relativistic speeds.

At the risk of incurring the rath of the younger physicists on this board, this is one case where the “relativistic mass” helps, in my opinion. In this now disparaged way of talking (which goes against current conventions), the relativistic mass (m) of a photon is given by the famous formula E=mc^2, where c is the speed of light. The momentum is given by E^2 = p^2c^2 + m0c^2, where p is the momentum and m0 is the rest mass. The rest mass (which people now just call the mass) is zero for a photon, so for a photon E=pc=mc^2 and thus, p = mc. So for a zero rest mass particle, the momentum is still mv if we use the relativistic mass instead of the rest mass. As Chronos observed, this formula is invalid for a finite rest mass object at relativistic speeds and we have to use p = sqrt(E^2 - m0*c^2)/c.

First of all it’s

E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]

So if m = 0 then it’s p = E/c or for a photon E = hf so p = hf/c which is correct. The relativistic mass of a photon is zero.

Actually, are you thinking of adding light energy to a closed system where the photons bounce around and therefor have p = 0? If so then you could say that the mass of the system increases by E/c[sup]2[/sup]. But that’s no different than saying that a system of photons with a zero momentum frame has mass.

I’ve also read of people calling E/c[sup]2[/sup] the effective mass of a photon in order to explain why it follows a curved path near a gravitating mass. But in this case the photon is simply following a straight path in curved spacetime. However, I think this would only account for the time curvature and not the space curvature, which, for a photon, is equal to the time curvature. But Chronos would have to answer that.

Relativistic mass is just a function on energy so it makes sense for a photon’s relativistic mass to be E/c[sup]2[/sup].

Due to the equivalence principle, except for non-linear effects(usually negligble and hence ignored for photons), the path of a particle moving through space is a function of it’s velocity only so there’s no need to assign any particle (particualrly photons) a ‘mass’ in order to explain it’s path through space. For a test particle all you need consider is that it’s mass small enough that these non-linear effects can be safely ignored.

Yes, of course, I neglected to square the rest energy, and I fully agree that p=hf/c.

No, the relativistic mass m is E/c and for a photon p = mc. In fact, for any particle moving in the x direction at speed v, the x component of the four momentum is p = mv, where m is the relativistic mass. So p = mv is always true with that definition of m, even for zero mass particles.

P.S. I am not arguing to go back to using relativistic mass. It is a perfectly valid concept, but conventions change and I see the value of following Einstein’s 1948 advice on the subject.

Well, yes, because the definition of “relativistic mass” was specifically contrived to make that one exact equation work. It’s not really useful for anything else, though. Proper velocity, meanwhile, does just as good a job of making that one specific equation work, but is also useful in other contexts, and makes more conceptual sense to boot.

I think if you can reduce a problem to one (spatial) dimension then relativistic mass is useful and also it’s useful pedagogically to be aware of i.e. it’s useful to know that in special relativity a bodies inertia is a function of it’s mass AND it’s velocity.

On the other hand it’s got plenty goign against it, i.e. redundancy (because it’s directly proportional to the relativsitic energy of a body), frame-dependency (i.e it’s not Lorentz invariant), m[sub]relativistic[/sub] = E/c[sup]2[/sup] is only directly proportional to a bodies inertia when you only need consider one dimension.