Photons have no Mass

Factual Questions
1

So they’re not Catholic?

No seriously. This has always bugged the shit out of me.

In this thread, we have;

[QUOTESaltire]
A photon will sometimes turn into an electron-positron pair
[/QUOTE]

Which would seem to indicate that a photon is comparable to or composed of smaller particles (quarks/gluons) equivalent to two electron/positrons.

Then we have Chronos explaining how photons are essentially massless.

But the whole idea of laser/photonic propulsion would seem to indicate that, NO, they are not massless, since you are bouncing photons against an object in order to impart inertia, aka increase it’s speed. Infinite zero is still zero, ergo photons must have mass in order to be able to accomplish this.

[QUOTE=Stranger on a Train]
The total mass of the system doesn’t actually change; while their might be a tiny “mass deficit” in the original components due to the energy in the “massless” photon, this is mere bookkeeping. Once the photon interacts with (absorbed by) another electron the mass of the overall system is returned, hence why photons are considered “force carriers”. The invariant masses of the nucleons and electrons do not change because the particles themselves are unchanged.
[/QUOTE]

Which seems to indicate, as I understand it, that photons are not really massless, but are practically so, being little more than a packet of energy (force or charge). Which would be a lot less than how Saltire describes them.

Can I get you guys to explain this better for me? Thanks.

2

Mass and energy are equivalent. Mass is a type of energy – concentrated if you will. Photons, like those coming out of your light bulb, have no rest mass, but then they’re never at rest. If you know the energy of the photon then it’s mass is E/c^2. It has momentum as well.

Exchange particles (i.e, the photons that make electrically charged particles attract or repel) are just virtual photons though and (as I understand it) have no mass or energy, but I may be wrong about that.

3

Propulsion relies not on mass but momentum. A single photon has no mass but does have momentum (proportional to its energy.)

As for pair production: A lone photon cannot turn into an electron/positron pair. Depending on how you like it, you can say it’s because the single photon (a system with no mass) can’t turn into a system with a mass of at least 2m[sub]electron[/sub], or you can say that you can’t simultaneously conserve energy and momentum if you try to make it happen.

Pair production requires two photons, although in practice one is usually virtual. This situation entails the original photon moving through matter, with that photon interacting with a nucleus as depicted by this Feynman diagram (to the extent that Feynman diagrams are “really” what happens) :



g = gamma (photon)
e- = electron
e+ = positron
N = nucleus
(v) ==> virtual
* = electromagnetic "interaction"
          
             g         e-
          ~~~~~~~*----------
                 |
                 |e(v)
                 |     e+
                 *----------
                 }
                 }g(v)
                 }
          =======*==========
             N         N
          
          
            (1)  (2)  (3)
          
          -------time------>

(1) photon and nucleus exist
(2) virtual [electron/positron] and photon interact
(3) nucleus, electron, and positron remain
4

[QUOTE=Chimera]
Which would seem to indicate that a photon is comparable to or composed of smaller particles (quarks/gluons) equivalent to two electron/positrons.
[/quote]
Also, particle decays/productions/transmutations of these sorts rarely imply anything about substructure. Many a well-meaning high school science teacher has told his/her class that a neutron is made up of a proton and electron bound together, the evidence being that a neutron decays to a proton and an electron (and an antineutrino) and that the sum of the proton and neutron masses is red-herringly close to the mass of a neutron. In reality, a neutron is just a slightly different bag of quarks than a proton, and before the decay, there is no proton and no electron and no antineutrino. There’s is just a neutron.

Similarly for the photon and pair production. That an electron and positron turned up does not imply they were around before.

5

[QUOTE=Pasta]
Propulsion relies not on mass but momentum. A single photon has no mass but does have momentum (proportional to its energy.)
[/QUOTE]
If a photon has momentum, why can’t we say it has mass of p/c = m[sub]photon[/sub] ?

6

[QUOTE=Dr. Lao]
If a photon has momentum, why can’t we say it has mass of p/c = m[sub]photon[/sub] ?
[/QUOTE]

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

(where E is energy, p is momentum, m is mass, and c is the speed of light)

Note that this reduces to E = mc[sup]2[/sup] for p = 0.

Basically, the first term in the above equation for E[sup]2[/sup] gives the kinetic energy, and the second term gives the energy at rest. All of a photon’s energy is kinetic energy. It has no rest energy, and thus no (rest) mass. (Occasionally people talk about “relativistic mass”, but usually “mass” means “rest mass”.)

[Edit: See also my post below]

7

[QUOTE=Chimera]
But the whole idea of laser/photonic propulsion would seem to indicate that, NO, they are not massless, since you are bouncing photons against an object in order to impart inertia, aka increase it’s speed. Infinite zero is still zero, ergo photons must have mass in order to be able to accomplish this.
[/QUOTE]

They need to have momentum, but not mass necessarily.

You might be used to thinking of momentum as mass times velocity, but this is only true for non-relativistic particles (those moving much slower than the speed of light). Particles moving at or near light speed can have a momentum which is not proportionate to their mass.

Edit: This is a further answer to Dr. Lao’s question. In relativity, momentum is not just mass times velocity, so you can’t just divide momentum by velocity to get mass.

8

[QUOTE=tim314]
E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup]

Basically, the first term in the above equation for gives the kinetic energy, and the second term gives the energy at rest.
[/QUOTE]

To be clear (and this may be more detail than you want):

The rest energy is mc[sup]2[/sup]
The kinetic energy is K = E - mc[sup]2[/sup]

Thus,
E = K + mc[sup]2[/sup]
E[sup]2[/sup] = K[sup]2[/sup] + 2mc[sup]2[/sup]K + m[sup]2[/sup]c[sup]4[/sup]
p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup] = K[sup]2[/sup] + 2mc[sup]2[/sup]K + m[sup]2[/sup]c[sup]4[/sup]
p[sup]2[/sup]c[sup]2[/sup] = K[sup]2[/sup] + 2mc[sup]2[/sup]K
p[sup]2[/sup] = K[sup]2[/sup]/c[sup]2[/sup] + 2mK

If m = 0 (as is true for photons),
p[sup]2[/sup] = K[sup]2[/sup]/c[sup]2[/sup]
p = K/c = E/c

If m > 0 (as is true for particles with mass),
p[sup]2[/sup]/2m = K[sup]2[/sup]/2mc[sup]2[/sup] + K
p[sup]2[/sup]/2m = K(K/2mc[sup]2[/sup] + 1)

Furthermore, if the kinetic energy is much less than mc[sup]2[/sup] (as is true for particles traveling much slower than light speed), then we can take K/2mc[sup]2[/sup] = 0
Thus,
p[sup]2[/sup]/2m = K
This is exactly what we expect for the non-relativistic case. (Compare to K = (1/2)mv[sup]2[/sup] and p = mv).

9

Restating what others have said above, but with different equations:

Momentum isn’t mv, it’s mv/sqrt(1-v^2/c^2). For massive particles at low speeds, that’s pretty close to mv. But for massless particles, v=c, then v^2/c^2=1 and sqrt(1-1) = 0. So, a photon’s momentum is what happens when the universe divides by zero.

10

[QUOTE=Pasta]
As for pair production: A lone photon cannot turn into an electron/positron pair. Depending on how you like it, you can say it’s because the single photon (a system with no mass) can’t turn into a system with a mass of at least 2m[sub]electron[/sub], or you can say that you can’t simultaneously conserve energy and momentum if you try to make it happen.

Pair production requires two photons, although in practice one is usually virtual. This situation entails the original photon moving through matter, with that photon interacting with a nucleus as depicted by this Feynman diagram (to the extent that Feynman diagrams are “really” what happens) :



g = gamma (photon)
e- = electron
e+ = positron
N = nucleus
(v) ==> virtual
* = electromagnetic "interaction"
          
             g         e-
          ~~~~~~~*----------
                 |
                 |e(v)
                 |     e+
                 *----------
                 }
                 }g(v)
                 }
          =======*==========
             N         N
          
          
            (1)  (2)  (3)
          
          -------time------>

(1) photon and nucleus exist
(2) virtual [electron/positron] and photon interact
(3) nucleus, electron, and positron remain

[/QUOTE]

But couldn’t a virtual photon (like in the above diagram) turn into an electron-positron pair without leaving behind another photon? Virtual particles don’t have to satisfy E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup]. I would have thought that such a reaction would occur, just at a lower rate than one in which the virtual particle was closer to being “on shell” (i.e., closer to satisying E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup]).

Or am I misremembering how quantum field theory works?

11

It could, and you could probably re-arrange that Feynman diagram so that exactly that happens. But although virtual particles don’t have to be on mass shell, they do still have rules they have to follow, so you still won’t get electron-positron pairs just appearing out of nowhere.

Also, remember that a virtual particle isn’t actually what happens. What actually happens is a superposition of a whole bunch of things, each of which involves a different set of virtual particles.

12

The short answer is that photons are inherently a very weird phenomenon. They seem to violate a lot of what we instincually assume about the universe. They are: it’s just very odd. They are a lot of things simulatenously, and do things we don’t expect with pretty much anything else.

13

Thanks for the responses. I’ve been reading physics, quantum physics and the like for my entire adult life, and I am capable of grasping it (largely), but at the very base of it, I’m too damned lazy and impatient to handle the math involved.

So next question. A very large number of photons strike the solar sail, imparting their momentum to the spacecraft. What happens to the photons? I assume that they are mostly just reflected, but obviously, there must be an energy transfer involved - inertia at least. How does that affect the photons?

(I realize this isn’t really anything different than the photons bouncing off my wall, but still…)

14

[QUOTE=tim314]
But couldn’t a virtual photon (like in the above diagram) turn into an electron-positron pair without leaving behind another photon?
[/QUOTE]

Oops, I missed the fact that Pasta has his arrow of time pointing to the right.

I was thinking the arrow of time pointed up the page, meaning he had something like an N and anti-N annihilating to produce an electron-positron pair via a virtual photon. Then he seemed to be saying that one of the electrons had to emit a photon for some reason.

Actually, I think he was just saying you can’t have a single real photon produce a particle-antiparticle pair without interacting with something else.

15

[QUOTE=tim314]
Actually, I think he was just saying you can’t have a single real photon produce a particle-antiparticle pair without interacting with something else.
[/QUOTE]
Yep, that’s all.

It doesn’t affect them in any terribly counter-intuitive way, once you grant that they carry momentum. In fact, a classical view works fine. If the photon is absorbed, the spacecraft experiences a momentum change equal to the the photon’s previous momentum, and the photon’s energy becomes heat or whatnot. If the photon is reflected 180[sup]o[/sup], the spacecraft gains twice the photon’s initial momentum, and the photon doesn’t lose any energy. (It’s momentum changes direction.)

16

Does light exert gravitational force?

My guess would be no. But it would just be a guess.

Another way to ask the question is: is it the rest mass of a body that that determines gravity, or is it just its energy?

-FrL-

17

[QUOTE=Frylock]
Is it the rest mass of a body that that determines gravity, or is it just its energy?
[/QUOTE]
It’s the system’s stress-energy tensor, which involves a variety of aspects of the system.

18

You could design a solar sail to absorb the photons, or to reflect them. In the most efficient design, you make your sail mirrored, and point the face of the mirror in the direction opposite the acceleration you want. This will typically not actually be facing straight towards the Sun (because then you’re fighting gravity), but to keep things simple, let’s look at a one-dimensional case and neglect gravity.

The first thing we want to do is to pick a reference frame. There are three logical reference frames to choose: The frame where the sail is stationary before the photon hits it, the frame where the sail is stationary after the photon hits it, and the zero-momentum frame. The latter one is the simplest, so that’s the one I’ll look at first. Without loss of generality, I’ll assume the Sun is on the left, and the sail is accelerating to the right.

In this frame, the photon has a momentum, since a single photon always must have a momentum. Its momentum is to the right, and since it’s just a single photon, it’s a minuscule momentum. Since we said that the total momentum is zero in this reference frame, this means that the sail must be moving, ever so slowly, to the left, initially. Then, the photon hits the sail and bounces off. In order to conserve energy and momentum, we must then have the sail moving to the right at the same speed (and hence with the same energy and magnitude of momentum) as it had before, and the photon ends up moving left with the same speed (of course), and the same energy and magnitude of momentum, as it had before. A nice, symmetrical arrangement.

Now, if we want one of the other reference frames, we need only shift these results. If, for instance, we look at the reference frame where the sail starts off at rest, then we see that in that frame, of course, the sail starts off with no kinetic energy, and ends up with some. Meanwhile, if we look at the photon in that frame, when it’s incoming, it’s slightly blueshifted relative to the zero-momentum frame, and after it’s reflected, it’s slightly redshifted relative to the zero-momentum frame. So in the frame we’re looking at, the photon got a little redder when it reflected, and therefore lost a little bit of energy. If you work through the math, you’ll unsurprisingly find that the energy lost by the photon is exactly equal to the energy gained by the sail.

19

[QUOTE=Pasta]
It’s the system’s stress-energy tensor, which involves a variety of aspects of the system.
[/QUOTE]

What’s the stress-energy tensor of a beam of light? (And what does this imply for the question whether light exerts a gravitational force?)

-FrL-

20

It’s worth pointing out that you can explain light pressure entirely without photons. You can view it as an electromagnetic efect – the fields that make up the light wave induce eddy currents in the material impacted, which set up a magnetic field, which the EM field of the original light wave then interacts with to produce the resultant force. Any good book on EM waves (Jackson, Marion, Bekefi) will cover this in detail.