I never quite understood the quantum answer (next to last post linked thread below) given in this linked thread about how a photon can transfer momentum or “pressure” if it has no a mass. The Wikipedia article repeats the pressure without mass assertion. Could someone help me out? What’s hitting what to make the “pressure” or transfer of momentum happen?
Well, the key is right in that quote: “no rest mass”. That means that if a photon were to ever stop (which it can’t), it would not have any mass.
However, thanks to you-know-who, we know that E=mc2, which means that (removing the c multiplier) energy (E) = mass (m). Therefore, the kinetic energy of its movement also gives it a “relativistic mass”.
And it’s the relativistic mass that transfers the momentum?
Hmm, this is interesting…
e=mc[sup]2[/sup]
Kinetic energy = 0.5mv[sup]2[/sup]
But for a photon, v = c
hence energy = 0.5mc[sup]2[/sup]
Hence 0.5mc[sup]2[/sup] = mc[sup]2[/sup]
Hence 0.5 = 1.
::wanders off, scratching head::
Only in Newtonian physics. In this calculation you would have to use the rest mass of the photon, which is 0, zo E = 0.
An easier way to do it is:
Momentum is mass * speed, so p = mc
so the Energy (mc[sup]2[/sup]) is: E = pc
For a photon: p = h/λ (h = Planck’s constant; λ = frequency)
so: E = hc/λ and since the frequency f is v/λ, and for a photon v = c we get f = c/λ, and thus:
E= hf
Spoilsport. 
Sorry!
Actually, that’s consistent with the following proof:
Given a, b such that a = b
ab = bb = b[sup]2[/sup]
ab-a[sup]2[/sup] = b[sup]2[/sup] - a[sup]2[/sup]
a(b - a) = (b + a)(b - a)
a = b + a
a = a + a
a = 2a
1 = 2
Hence 0.5 = 1
Should I be a spoilsport again and point out the fallacy in this part?
Nope, I’ll let somebody else do it. Unless the fallacy hasn’t been pointed out within the next four posts. Then I’ll do it!
How about this?
a=2a
a=0
Bingo! In the next step, he divides both sides of the equation by a, which has to be zero to satisfy a=2a. Dividing by zero is a big no-no!
There was another one which I liked, which began with
For any a, b let a + b = 2c
and ended up proving that a = b, consequently all numbers are the same. I can’t remember how that one goes, though.
OP? We don’t need no steenkin’ OP!!
Actually, the fallacy comes earlier than that:
He divided by b-a, but since a=b, b-a=0.
Hey, you’re right. So, in essence he’s dividing by zero twice. If two wrongs did make a right, the proof would be valid again.
Also, since three lefts make a right, that makes two wrongs=three lefts. In that case, if left is wrong, then 2=3.
quod erat demonstrandum
So… let me shift the question a bit -
If the original mass is nothing, how does going C make it something. How does zero at rest become more than zero through speed? I appreciate the physics equations, but how about seeing if you can explain this in big crayon style thought balloons without using relativity equations.
How about looking at it this way – imagine that the photon had mass, but that it was moving slower than light. Now accelerate it to the speed of light. It turns out that acceleating an object of any mass to lightspeed requires an infinite amount of energy, and the relativistic mass asymptotically approaches infinity as well, regardless of how small the mass you started off with was. Far any value of v/c you can reduce the product of relativistic mass times relativistic velocity by making the mass smaller. The only limit that gives you a finite momentum for speed approaching lightspeed is mass approaching zero.
Here’s something I’ve been wondering about since a high school physics class some decades ago…
Black objects are black because they absorb light, right? What happens to that light? I assume that black objects trap a large number of photons. If photons can’t stop, are they bouncing around inside a black object. Do black objects get heavier and heavier when you shine lights on them? What’s going on with the color black?
They get hotter. Photons get absorbed by something (even if you have highly reflecting internal facets, as with the “light pump” in Theodore Sturgeon’s “Microscopic God” , you have a huge number of bounces. Eventually, even with teeny absorption per bounce, it all gets absorbed). Some of the energy might get re-emitted as photons (of a different wavelength), but at least some and usually all of it ends up going into heating the object, so it gets hotter. The photons have momentum but no mass, so it doesn’t get heavier. Some of the heat re-emerges as infrared photons of vastly different wavelength. most of the heat ends up being conducted away by objects your black object is in contact with, or by the surrounding air.
Actually, all else being equal, a hot object does weigh ever so slightly more than a cold object. The photon that got absorbed was massless, but in general, a system of photons does have mass, and a system of objects moving in different directions (as, for instance, a hot object) has a component of mass due to the kinetic energies of the individual particles.