Mass of Light

[ACG52]

 

Quote

 

 

For photons, there is no frame in which p = 0.

 

IOW there is no frame in which a photon has zero mass.

 

No, p is not mass, p is momentum. There is no frame in which a photon has zero momentum. There is no frame in which a photon has a non-zero mass.

[swansont]

IOW there is no frame in which a photon has zero mass.

 

No, that's not what it means. In all inertial frames a photon has zero mass, because p = E/c

[elfmotat]

IOW there is no frame in which a photon has zero mass.

 

You're using a different definition of mass than everyone else. Your definition of mass (relativistic mass) is proportional to energy: E=m_rel. If you want to use relativistic mass, fine, but you need to clarify that that's what you're doing because it is not standard.

 

The mass everyone else is using is defined by E²-p²=m². We know that this quantity is zero for photons (or at least very very very close to zero) because their energy is directly proportional to their momentum. The fact that the quantity is called "rest mass" doesn't mean that particles need to be at rest for us to be able to measure it - it's just a name.

Edited February 9, 2013 by elfmotat

[michel123456]

You're using a different definition of mass than everyone else. Your definition of mass (relativistic mass) is proportional to energy: E=m_rel. If you want to use relativistic mass, fine, but you need to clarify that that's what you're doing because it is not standard.

 

The mass everyone else is using is defined by E²-p²=m². We know that this quantity is zero for photons (or at least very very very close to zero) because their energy is directly proportional to their momentum. The fact that the quantity is called "rest mass" doesn't mean that particles need to be at rest for us to be able to measure it - it's just a name.

 

That's the first reasonable post I have seen around for a while.

 

Rest mass is not just a name. It really means something for other particles. You are trying to elude the question.

 

Can a photon be at rest?

 

No. it is not physically possible.

 

Is it correct to use an equation to describe an impossible physical state? i don't think so. If you want to mathematically describe unicorns, you will obtain at best a fairy tale, nothing more. surely not "science".

 

Now, about observation: in which condition do we measure that the photon has zero mass?

Edited February 9, 2013 by michel123456

[D H]

Rest mass is not just a name.

 

Yes, it is just that, a name. If that name is too confusing, consider calling it invariant mass or intrinsic mass.

 

If that's still too confusing, just call it mass. There's no need for any qualifier.

 

 

Now, about observation: in which condition do we measure that the photon has zero mass?

 

You've already been told that. We observe energy and momentum.

[elfmotat]

E²-p²corresponds to the mass of a particle in its rest frame, sure. But that doesn't mean that the definition of E²-p² is "the mass of a particle in its rest frame." It is what it is: E²-p² is just a quantity. For photons this quantity is zero, and there's nothing more to it than that. Whether or not photons have a rest frame does not determine whether or not we can measure E²-p².

 

As for experimental tests of the relation E²-p²=0 for photons, it is trivial to check that the energy and momentum relations required by experiments such as the Compton Effect yield a rest mass of zero:

 

[math]\lambda f=c[/math]

[math]E=hf[/math]

[math]p=h\lambda^{-1}[/math]

 

[math]E^2-c^2p^2=h^2 ( f^2-c^2\lambda^{-2})=h^2 ( f^2-f^2)=0[/math]

 

There are numerous other reasons why the photon's mass must be zero, as outlined here.

Edited February 9, 2013 by elfmotat

[esbo]

Light does not have mass but it's energy can be converted to mass.

[ajb]

Yes or No, please.

 

Besides, if not in Minkowski space-time what is the difference?

 

besides#2, "all inertial frames" ; does that include the inertial frame in which a photon is at rest?

 

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(edit) forget the besides #1 & #2. A Yes or No will do. I really don't understand you problem. ACG52 has no problem answering a so simple question.

 

Can a photon be at rest?

There are no inertial frames in which the photon can be considered at rest. In all inertial frames the speed of light is c, but there are other frames you could use. In these, the speed of light is almost never going to be c. A common example here are light-cone coordinates. I think we have spoken about this before

 

http://www.scienceforums.net/topic/52070-is-there-such-thing-as-infinite-speed/?hl=%2Blight+%2Bcone+%2Bcoordinates+%2Bajb#entry568841

[michel123456]

There are no inertial frames in which the photon can be considered at rest. In all inertial frames the speed of light is c, but there are other frames you could use. In these, the speed of light is almost never going to be c. A common example here are light-cone coordinates. I think we have spoken about this before

Thank you again, but the wiki article about light-cone coordinates is not particularly enlightening.

 

 

There are no inertial frames in which the photon can be considered at rest

 

 

So the "rest mass" of a photon is something that does not exist. It is not "something" which value is zero. It is not even a limit that particles tend to approach. IIRC there is nothing that propulses a particle of low mass at any speed close to C.

[ajb]

Thank you again, but the wiki article about light-cone coordinates is not particularly enlightening.

I should write something up on my blog, eventually.

So the "rest mass" of a photon is something that does not exist. It is not "something" which value is zero.

That is the technically correct way to think of it. There are no inertial frames in which the photon is at rest, so any notions that would be deeply tied to a rest frame of the said particle, will not be well-defined.

It is not even a limit that particles tend to approach. IIRC there is nothing that proposes a particle of low mass at any speed close to C.

You should think of it as a limit of the speed as the particle's mass tends to zero, not actually at zero. This is a limit of a massive particle, which is a mathematical process and need not truly be physical. As the particle gets lighter and lighter, it can travel faster and faster. In the mathematical limit of the mass tending to zero you recover the top speed of c.

 

For example, neutrinos we know are very light and thus they can travel at speeds close to the speed of light without the input of huge amounts of energy.

[michel123456]

I should write something up on my blog, eventually.

That is the technically correct way to think of it. There are no inertial frames in which the photon is at rest, so any notions that would be deeply tied to a rest frame of the said particle, will not be well-defined.You should think of it as a limit of the speed as the particle's mass tends to zero, not actually at zero. This is a limit of a massive particle, which is a mathematical process and need not truly be physical. As the particle gets lighter and lighter, it can travel faster and faster. In the mathematical limit of the mass tending to zero you recover the top speed of c.

 

For example, neutrinos we know are very light and thus they can travel at speeds close to the speed of light without the input of huge amounts of energy.

They can, yes (it's hard to swallow but say yes).

They are not obliged to.

 

I mean if you reduce the mass of an object, you don't see it slowly going away at some velocity.

[ajb]

They can, yes (it's hard to swallow but say yes).

They are not obliged to.

Right, there is a dependency on the energy, via the mass-shell condition.

I mean if you reduce the mass of an object, you don't see it slowly going away at some velocity.

How would you do this?

 

For non-fundamental objects you would see a change in velocity due to conservation of momentum. For example a rocket.

[michel123456]

How would you do this?

 

Cutiing my sandwich in 2 parts.

There is no law of physics that say that my 2 sandwich pieces would be given the slightest velocity because they are of smaller mass than the original one. If you continue cutting in pieces smaller and smaller at my knowledge there is nothing in existing physics that would input any velocity because you reduced the mass of the object. Not even for a neutrino I suppose.

[ajb]

How would you do this?

 

Cutiing my sandwich in 2 parts.

There is no law of physics that say that my 2 sandwich pieces would be given the slightest velocity because they are of smaller mass than the original one. If you continue cutting in pieces smaller and smaller at my knowledge there is nothing in existing physics that would input any velocity because you reduced the mass of the object. Not even for a neutrino I suppose.

All I mean is that one should be careful with conservation of momentum. If some object in some frame is travelling at some velocity then part of it's mass disappears or really is expelled then the velocity must change to keep momentum conservation.

 

In the case of your sandwich, you simply cut the system in two rather than allow one half to expel the other. Both halves remain at rest relative to each other.

 

For a fundamental particle, the closest thing I can think of is a decay. The initial particle is at rest, but the two daughters could not be at rest as measured in the initial rest frame.

 

I don't know if this graphic helps.

 

It is a plot of the three momentum as a function of energy for several different masses. I have set the speed of light equal to one.

 

Some things to note.

 

• In the massless case the relation is simply linear.
• For the massive case we see that near E=M, small changes in the energy lead to big changes in the momentum.
• As energy increases changes in energy create smaller and smaller changes in momentum for massive particle. This approaches the massless case as energy gets higher and higher.
• As the mass gets smaller, the plot also approaches the massless case.

 

Essentially, for massive particles, the three momentum is the same as the speed, being quite slack here. Meaning, as the energy of the particle increases, it takes more and more energy to create a noticeable change in the velocity.

[michel123456]

All I mean is that one should be careful with conservation of momentum. If some object in some frame is travelling at some velocity then part of it's mass disappears or really is expelled then the velocity must change to keep momentum conservation.

 

In the case of your sandwich, you simply cut the system in two rather than allow one half to expel the other. Both halves remain at rest relative to each other.

 

For a fundamental particle, the closest thing I can think of is a decay. The initial particle is at rest, but the two daughters could not be at rest as measured in the initial rest frame.

 

I don't know if this graphic helps.

 

It is a plot of the three momentum as a function of energy for several different masses. I have set the speed of light equal to one.

 

Some things to note.

 

• In the massless case the relation is simply linear.
• For the massive case we see that near E=M, small changes in the energy lead to big changes in the momentum.
• As energy increases changes in energy create smaller and smaller changes in momentum for massive particle. This approaches the massless case as energy gets higher and higher.
• As the mass gets smaller, the plot also approaches the massless case.

 

Essentially, for massive particles, the three momentum is the same as the speed, being quite slack here. Meaning, as the energy of the particle increases, it takes more and more energy to create a noticeable change in the velocity.

(emphasis mine)

 

You must mean that the relation becomes linear.

 

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(edit)

 

As the mass gets smaller, the plot also approaches the massless case.

Yes but the diagram does not show that a small mass travels at close to C. Although a photon MUST travel at C.

 

 

_For massless photons, does the diagram go through zero? Is there a photon with E=0?

_It is half an half hyperbola, you have erased the other parts of the graph because they represent negative (unphysical) values. (I noticed the careful IPI in the equation)

_what about a graph with M on the abscissa and Energy on the ordinate?

Edited February 10, 2013 by michel123456

[derek w]

Question.Is there a threshold that when crossed energy converts to mass?Photons having not crossed that threshold?

[swansont]

In some instances, yes. A photon must have at least 1.02 MeV in order to form an electron-positron pair, because energy must be conserved. But just because a photon has this energy does not mandate that it will undergo this conversion. An interaction is required.

[ajb]

(emphasis mine)

 

You must mean that the relation becomes linear.

Right it approaches a linear relation. But at first you get out a lot of momentum for not much increase in energy and then that drops off. Meaning more energy is required to have the same increase in momentum until you reach the linear relation as a limit.

Yes but the diagram does not show that a small mass travels at close to C. Although a photon MUST travel at C.

True.

_For massless photons, does the diagram go through zero? Is there a photon with E=0?

You would interpret E=0 as no photons being present.

_It is half an half hyperbola, you have erased the other parts of the graph because they represent negative (unphysical) values. (I noticed the careful IPI in the equation)

_what about a graph with M on the abscissa and Energy on the ordinate?

I have plotted the absolute value of the three momenta and have kept energy strictly positive. Alo, I have only considered positive mass squared.

[michel123456]

Other comment on the graph:

The increase of energy is I suppose kinetic.

The graph of Mass 4 for example represents a single mass that we put in increasing motion (acceleration).

The graph of Mass-zero (photon), does it represent a single photon that we put in acceleration (how do we do that?), or the graph of multiple photons? Can we change the kinetic energy of a photon?

[ajb]

The increase of energy is I suppose kinetic.

All the graph really shows the relation between energy and momentum, neither of which is a Lorentz invariant. Only the mass, which is related to the energy and the momentum is invariant and so has some true meaning.

 

One could interpret the graph (backwards) as changing frames so that the particle's momentum is given and then reading off the energy.

The graph of Mass 4 for example represents a single mass that we put in increasing motion (acceleration).

Again, it really only shows how the energy and momentum are related.

The graph of Mass-zero (photon), does it represent a single photon that we put in acceleration (how do we do that?), or the graph of multiple photons? Can we change the kinetic energy of a photon?

For photons you get E=p. The energy and the momentum are (up to a c) the same thing. Changes in the momentum are exactly off set by changes in the energy.

[michel123456]

How come that I find everything so extraordinary and you find it so natural?

 

E=p for photons (not only, but for all massless particles)

In this diagram, you have not one single value for E, you have the whole range from zero to infinite.

For a specific mass (say mass4), each value of Energy corresponds to a single value of momentum.

Which is not true for m=0. Suddenly a single value of mass corresponds to all values of energy and all values of momentum.

So one could interpret the diagram stating that a photon can have a whole range of kinetic energy.

How do you change the kinetic energy of a photon?

 

On the other hand, P is frame dependent. So E (kinetic energy) is also frame dependent. But for photons, what does "frame dependent" mean, since all photons travel at the same velocity?

 

Am I the only one alive here?

[elfmotat]

But for photons, what does "frame dependent" mean, since all photons travel at the same velocity?

 

It means its frequency is Doppler-shifted.

[swansont]

Photon momentum depends on energy, which depends on frequency. You don't change the KE of a photon, per se, but you can create photons of different energies.

[michel123456]

Photon momentum depends on energy, which depends on frequency. You don't change the KE of a photon, per se, but you can create photons of different energies.

O.K.

How then do you change the energy of a particular photon?

 

It means its frequency is Doppler-shifted.

Do you mean that the doppler-shift formula matches the diagram (it is also a hyperbola) ?

[Przemyslaw.Gruchala]

O.K.

How then do you change the energy of a particular photon?

 

It must collide with other particle to give it part of energy. Which will be observed as acceleration of that particle, and changing frequency and direction of photon.