Do Photons Have Mass? Confused Student Seeks Answers

[Tach]

Hello, I'm quite confused about something. Maybe you all can help :)

I'm not sure I understand why the science community seems to assume photons have no mass. I can understand how it wouldn't make sense to say a photon has mass by the mathematics that is used to describe the energy of a particle, because it has no rest mass, but if a positron can interact with an electron and annihilate both particles, creating photons, then if photons have no mass, we have just destroyed mass. And if photons result from 'destroying mass' then shouldn't we also be able to 'create mass' from photons?

And in relation to the above paragraph and following from the same train of thought, then since photons travel at a constant speed and can create mass, then wouldn't it be wise to assume a constant mass for photons? Since relativistic mass wouldn't apply because of non-changing speeds.

I'm quite confused. Or perhaps the whole concept of mass is flawed to begin with? Or am I not understanding the basics right?

 

[nicksauce]

Sure, you can create and destroy mass. The "law of conservation of mass" is a classical approximation that is not valid in quantum theory.

 

[ansgar]

according to relativity, mass is just a type of energy.

this has nothing to do with quantum

 

[Char. Limit]

It actually has to do with mass-energy equivalence, I believe. In the creation of a photon, mass is converted to energy (and multiplied by c^2 in the process), creating the familiar packet of energy we call a photon, and getting rid of the massive particle in the process.

 

[Tach]

Thanks for your responses. I guess I'm having a hard time accepting this since I can't think of all the energy of the world around me in terms of a 'unit of space' or 'unit of some physical size'. I don't know how to imagine photons if they don't occupy a space and thus have some kind of mass; although I associate mass with particles occupying a space.

So it's as if photons are just a 'magical particle' because they don't actually exist in space? Such an idea really bugs me and seems extremely counter-intuitive and that there is some kind explanation of how photons do 'occupy a space', but that's assuming a lot, I suppose. I guess it's possible that some things can exist without making any rational sense.

 

[pallidin]

That an electron, for example, can exist and yet have no definitive spatial representation is a problem. Welcome to the quantum world.

 

[Passionflower]

Thanks for your responses. I guess I'm having a hard time accepting this since I can't think of all the energy of the world around me in terms of a 'unit of space' or 'unit of some physical size'. I don't know how to imagine photons if they don't occupy a space and thus have some kind of mass; although I associate mass with particles occupying a space.

Two photons going in opposite directions certainly have mass.

 

[Naty1]

"So it's as if photons are just a 'magical particle' because they don't actually exist in space?"

Perhaps you could say that, but nobody understands exactly what ANY fundamental "particle" is...photons are not necessarily more 'magical' than, say, an electron, that "doesn't actually exist in space" either but is rather an "electron probability cloud". For "heaven's sake" why can't we precisely locate a simple electron?? Oh yes, Heisenberg uncertainty...

We have lots of mathematic behavior descriptions, but alas, no really good ones of fundamental origins of particles.


".. Such an idea really bugs me and seems extremely counter-intuitive and that there is some kind explanation of how photons do 'occupy a space', but that's assuming a lot, I suppose. I guess it's possible that some things can exist without making any rational sense..."


Richard Feynman said "Nobody understands quantum mechanics"...the science of the very small...and that's still true today. But then general relativity is hardly "intuitive" either.

Another way to think about all the above is that everything is a vibrating string...a resonance of energy and what you THINK you see as a 'solid' mass is 99.999999999% empty space...gravitons are also massless and move at lightspeed like photons...so don't pick on the poor photon...and maybe you won't like black holes either as their singularity is also a point of no volume...

We apparently have a LOT more to learn about the relationship between space, mass, energy and time.

 

[Phrak]

Two photons going in opposite directions certainly have mass.

This would make the PF physics FAQ fairly meaningless without a proper make-over. Where two photons can 'go in opposite directions' so can a single photon, so that anything other than an ideal planar wave endows regions of spacetime with nonzero intrinsic mass.

 

[Rasalhague]

Two photons going in opposite directions certainly have mass.

That's how I understand it. Mass (i.e. rest mass), energy (i.e. total energy) and momentum are each conserved in special relativity. In this case, the mass of the system of particles that annihilated is the same as the mass of the system of photons they produce, isn't it? I have trouble understanding the concepts of mass-energy equivalence and mass turning into energy or vice-versa, but, following previous discussions here, I think these viewpoints depend on implicitly shifting the reference of the word. For example, Taylor/Wheeler say in Spacetime Physics

"Does the invariance of rest mass mean that rest mass cannot change as a consequence of a collision? No. Rest mass often changes in an inelastic encounter. Example 1: Collision between two balls of putty--hotter and therefore very slightly more massive after the collision than before."

But if I've got this right, then the "change" only comes from the fact that they start out talking about the sum of the rest masses of the balls (not the rest mass of the system), then switch to talk about the rest mass of the combined ball which latter is equal to the mass of the system throughout.

But I suppose it just depends how the system is defined, for conservation of mass, as for conservation of energy and conservation of momentum.

 

[Rasalhague]

Two photons going in opposite directions certainly have mass.

No, photons have energy but do not have mass or mass-energy.

But if they're going in opposite directions, the vector sum of their 3-momenta will be zero, so the magnitude of the total 2-momentum is zero. This being the case, the rest mass of the system of two photons is just the sum of their energies.

 

[D H]

That's how I understand it. Mass (i.e. rest mass), energy (i.e. total energy) and momentum are each conserved in special relativity.

Rest mass is not a conserved quantity. Think of an annihilation event.

 

[Rasalhague]

Rest mass is not a conserved quantity. Think of an annihilation event.

Taylor/Wheeler, excercise 97, deals with a positron and electron annihilating to produce radiation: "By considering the center-of-momentum frame (the frame of reference in which the total momentum of the initial particles is equal to zero), show that it is necessary for at least two gamma rays (rather than one) to result from the annihilation." The argument they give is that 3-momentum is conserved, therefore if 3-momentum is initially zero, it must remain zero. The 3-momentum of one gamma ray is equal to its energy and not zero, therefore there must be two (in opposite directions, right?).

But if 3-momentum is conserved and energy is conserved, rest mass must also be conserved, since it's a function only of momentum and energy.

Presumably there's a different way to describe this event, defining--or redefining--the system in such a way that it doesn't conserve energy or rest mass. Could you explain it from this viewpoint; and would you see 3-momentum as being conserved? How would you answer the question about the number of rays emitted from this viewpoint?

 

[D H]

You cannot treat an ensemble as one. Just because the energy and momentum of one photon are directly related to one another does not mean there is any direct relationship between the total energy and total momentum of a set of more than one photon.

 

[Rasalhague]

You cannot treat an ensemble as one. Just because the energy and momentum of one photon are directly related to one another does not mean there is any direct relationship between the total energy and total momentum of a set of more than one photon.

Could you elaborate? If by "no direct relationship" you mean that energy of an ensemble of massless particles doesn't necessarily equal their total momentum, that agrees with what Passionflower wrote, as I understand it.

The message I get from Spacetime Physics is that the equation

$$m^2 = E^2 - p^2$$

applies to any system (c=1, inertial frame), whether composed of one or more particles. Is that right?

 

[Tach]

Ah, okay. I think I understand now. So I shouldn't think of mass as 'occupying space', but as a construct to describe a mathematically observed difference between differing objects due to gravitational forces.

So when I think of photons not occupying space, that is incorrect to say, in this sense. And in that regard I'm going to imagine photons as occupying space with all other matter then. That seems to be the only way to make sense of this for me.

 

[Dale]

Rest mass is not a conserved quantity. Think of an annihilation event.

Passionflower is correct. The invariant mass of a system is preserved even across an annihilation event. The invariant mass of a system of particles is the norm of the sum of the individual particle's four-momenta. This can be different from the sum of the norms of the individual particle's four-momenta, which is not conserved. When people speak about mass not being conserved they are almost always erroneously identifying the latter quantity with the system's mass.

 

[Count Iblis]

Yes, or they implicitely redefine the system. So, when an atom emits a photon, you can focus on the atom, instead of taking your system to be the atom-photon system.

 

[ansgar]

That's how I understand it. Mass (i.e. rest mass), energy (i.e. total energy) and momentum are each conserved in special relativity. I

No they are not, only the invariant mass, momentum and total energy is conserved. Besides this, energy and momentum are frame dependent.

 

[Rasalhague]

That's how I understand it. Mass (i.e. rest mass), energy (i.e. total energy) and momentum are each conserved in special relativity.

No they are not, only the invariant mass, momentum and total energy is conserved.

That's what I said; rest mass = (frame)-invariant mass. Did you mean "yes they are...are conserved" or "no they are not...are not conserved"?

Besides this, energy and momentum are frame dependent.

Yes.

 

[Rasalhague]

Yes, or they implicitely redefine the system. So, when an atom emits a photon, you can focus on the atom, instead of taking your system to be the atom-photon system.

This must be what's going on in the Taylor/Wheeler inelastic-collision example I quoted in #12. That implicit redefinition threw me at first because it seemed inconsistent in the context of a chapter that discusses the conservation of energy and 3-momentum. It seemed to be saying that rest mass, unlike these quantities, is not conserved. But if the system is redefined so that mass is not conserved, then presumably energy and 3-momentum can't both be conserved either.

 

[starthaus]

It seemed to be saying that rest mass, unlike these quantities, is not conserved.

True. Interestingly, due to the fact that total energy is conserved:

$$\Sigma(\gamma(v_i)*m_i*c^2)=constant$$

it follows that relativistic mass is conseved since the above can be written as:
$$c^2\Sigma(\gamma(v_i)*m_i)=constant$$ !

But if the system is redefined so that mass is not conserved, then presumably energy and 3-momentum can't both be conserved either.

Both energy and momentum are conserved:

$$\frac{d}{dt}\Sigma(\gamma(v_i)*m_i*\vec{v_i}=\Sigma \vec{F_i}$$

If the sum of the forces acting on the system is zero, then:$$\Sigma(\gamma(v_i)*m_i*\vec{v_i}=constant$$

 

[Phrak]

Passionflower is correct. The invariant mass of a system is preserved even across an annihilation event. The invariant mass of a system of particles is the norm of the sum of the individual particle's four-momenta. This can be different from the sum of the norms of the individual particle's four-momenta, which is not conserved. When people speak about mass not being conserved they are almost always erroneously identifying the latter quantity with the system's mass.

This is why I said the physics forums FAQ, found https://www.physicsforums.com/showthread.php?t=104715", needs a slight make-over. There are few scenarios to examine.

One is the electron-positron annihilation into two gamma particles. The wave functions are even upon interchange of the two particles. There's no way to distinguish one from the other. Each photon is equally likely to have momentum to the left as to the right until measured.

The second scenario is a confined photon such as found within a laser cavity, as we've talk about before, with an uncertainty in position in the order of the cavity length. Some might argue that the photon cannot be treated in isolation without considering the cavity walls, so...

number three is the case of self interference. If you like, the two slits of an interference experiment. Withing some given region of space the momentum contributions from each slit add vectorially. So that the energy-momentum equation is satisfied, the photon must have intrinsic mass. Only under the ideal condition of a planar propagating wave is the momentum constant and the photon massless.

The best I can discern, is that in general, most photons have mass, where to say the photon has zero intrinsic mass refers to an idealized state of the particle.

 

[Rasalhague]

True.

Are you agreeing that "rest mass, unlike these quantities, is not conserved" (and redefining the system, as Count Iblis puts it), or just agreeing that the quote seems to be saying this? If the former, how does this tally with

Both energy and momentum are conserved

How can rest mass--which is defined sqrt(E²-p²)--not be conserved if E and p are? The square root of the square of a constant minus the square of another constant is also a constant.

 

[Char. Limit]

I thought that for light, E=pc, and thus when c=1, then E=p and m^2=E^2-p^2 cancels to zero... am I wrong?

 

[ansgar]

That's what I said; rest mass = (frame)-invariant mass. Did you mean "yes they are...are conserved" or "no they are not...are not conserved"?



Yes.

rest mass is frame dependent.

 

[DrGreg]

The phrase "rest mass is conserved" is somewhat ambiguous, which is why there has been some confusion in this thread.

1. The invariant mass of the whole system, $\sqrt{[(\Sigma E)^2 - (\Sigma p)^2 ]}$ (in units where c=1), is conserved.
2. The sum of the individual particles' rest masses is not conserved.

"Conservation of rest mass" is often taken to relate to the second statement, and therefore isn't true.

In fact some authors avoid this confusion by using "rest mass" to refer only to individual particles. For a whole system they use the phrase "invariant mass" or "system mass".

 

[DrGreg]

Phrak, quantum mechanics is outside the scope of special and general relativity and is dealt with in quantum field theory (QFT) instead. The discussion of photons in this thread refers to "classical photons" that are assumed to have a definite energy and momentum. As I've never studied QFT, I've no idea how the concept of mass is treated in QFT, but it's probably best discussed in the the Quantum Physics forum of this site rather than here.

 

[sweet springs]

Hi. In reading the discussions above, I state here my understanding of mass and "invariant mass".

In system of one photon, E=|p|c so mass=0.
In system of two photons, E1=|p1|c so mass=0 and E2=|p2|c so mass=0.
(E1+E2)/c^2 in mass centered frame of reference where p1+p2=0, gives "invariant mass" of the system.
In system of three photons, E1=|p1|c so mass=0, E2=|p2|c so mass=0 and E3=|p3|c so mass=0.
(E1+E2+E3)/c^2 in mass centered frame of reference where p1+p2+p3=0, gives "invariant mass" of the system.
...
In system of N photons, E1=|p1|c so mass=0, E2=|p2|c so mass=0, E3=|p3|c so mass=0, ... and EN=|pN|c so mass=0.
(E1+E2+E3+...+EN)/c^2 in mass centered frame of reference where p1+p2+p3+...+pN=0, gives "invariant mass" of the system.

Similary for electrons with mass m,
In system of one electron, E=sqrt(p^2c^2+m^2c^4).
In system of two electrons, E1=sqrt(p1^2c^2+m^2c^4) and E2=sqrt(p2^2c^2+m^2c^4).
(E1+E2)/c^2 in mass centered frame of reference where p1+p2=0, gives "invariant mass" of the system.
In system of three electrons, E1=sqrt(p1^2c^2+m^2c^4), E2=sqrt(p2^2c^2+m^2c^4) and E3=sqrt(p3^2c^2+m^2c^4).
(E1+E2+E3)/c^2 in mass centered frame of reference where p1+p2+p3=0, gives "invariant mass" of the system.
...
In system of N electrons, E1=sqrt(p1^2c^2+m^2c^4), E2=sqrt(p2^2c^2+m^2c^4), E3=sqrt(p3^2c^2+m^2c^4),..., and EN=sqrt(pN^2c^2+m^2c^4).
(E1+E2+E3+...+EN)/c^2 in mass centered frame of reference where p1+p2+p3+...+pN=0, gives "invariant mass" of the system.

Similarly for the mixed system of electrons, positrons, ... and photons, (E1+E2+E3+...)/c^2 in mass centered frame of reference where p1+p2+p3+...=0, gives "invariant mass" of the system.

It seems that "invariant mass" is mass of a single particle as if the whole system were to be. I hope I am on the right track.
Regards.

 

[sweet springs]

Hi, let me check my learning on conservation before and after annihilation of a positron and an electron of each mass m.

Before annihilation, E1=sqrt(p1^2c^2+m^2c^4) and E2=sqrt(p2^2c^2+m^2c^4). Coulomb energy is disregarded by long distance between. After annihilation, Eph1=|P1|c and Eph2 + |P2|c.

Conservations of momentum and energy are p1+p2=P1+P2 and E1+E2=Eph1+Eph2. "Invariant mass" before annihilation is (E1+E2)/c^2 in mass centered frame of reference where p1+p2=0. "Invariant mass" after annihilation is (Eph1+Eph2)/c^2 in mass centered frame of reference where P1+P2=0. We easily find that the mass centered frame of reference and "invariant mass" is conserved before and after annihilation process.

Sum of mass before annihilation is 2m, that of after is 0. Sum of mass over particles are not conserved quantity of the system.

Regards.

 

[filegraphy]

I thought that the photon could not have mass mathematically because since it is traveling at the cosmological constant it would have infinite mass. So stated in general relativity a photon does not have the property of mass. That's what I thought. Am I correct or mistaken. I would appreciate all of the help I can get from others.

 

[Char. Limit]

As is apparent from this discussion, either no one knows or several people are wrong...

 

[ansgar]

As is apparent from this discussion, either no one knows or several people are wrong...

since SR steams from 2 basic postulates and is standard physics, the most probable thing is that several people here are wrong ;)

 

[Dale]

I thought that the photon could not have mass mathematically because since it is traveling at the cosmological constant it would have infinite mass. So stated in general relativity a photon does not have the property of mass. That's what I thought. Am I correct or mistaken. I would appreciate all of the help I can get from others.

The mass of a photon is 0. The mass of a system of multiple photons may be non-zero.

 

[Dale]

rest mass is frame dependent.

Rest mass is another name for the invariant mass, which is (as its name implies) frame-invariant.