Massless particles

Ok, here's one for all you particle physicists out here.


So it's my current understanding that photons are massless bosons.

But, they exhibit certain properties such as radiation pressure, that a naive scientist (more accurately an electrical engineer getting into solid state physics) might interpret as exhibiting mass-like qualities.


Intuitively, since we have an assumed maximum velocity (speed of light) possible in a "vacuum" then it would seem that there would be an associated mass being transferred.


Say I am the sun, and I spew out radiant energy. At what point does this energy become mass, and if the answer is never, then how exactly does light cause radiation pressure, and "bend" around massive stars.

Originally posted by grad_student
how exactly does light ... "bend" around massive stars.

Interesting questions. I am not sure, but I believe that the reason for light 'bending' around
large gravitational sources is due to the curving of space itself, rather than gravitational force
being applied to the photons themselves.


editing for spelling.

[edit on 2005/8/23 by McGrude]

I think it has has something to do with in simple terms, light travelling in straight lines. More accurately on a line on space-time.

Think of it like a grid. Say, light must follow a grid line......if someone pushed there thumb onto the grid and caused a large indent (gravity). The light must still follow the grid lines and therefore is bent around. Does that help at all???

I dont think they have any kind of detectable "mass", but granted they do exibit some masslike properties. All energy effects its environment in some way, it would have to or it would be undetectable, so maybe this has something to do with it.

Also JUST INCASE ive misunderstood your question, could you possibly rephrase your question???

Originally posted by grad_student

Say I am the sun, and I spew out radiant energy. At what point does this energy become mass, and if the answer is never, then how exactly does light cause radiation pressure, and "bend" around massive stars.

Hello, I am an electrical engineering undergrad and I believe I can answer your question.

Light is an electromagnetic wave. Let's look at this at an atomic scale and consider the light hitting a single atom. That atom has electrons, so it has an electrical field, the electrons are moving which creates a magnetic field. The pulse of electromagnetic energy from the light pushes on the stationary electromagnetic field of the atom when it hits it, causing the "photon" of light to transfer some or nearly all of its energy to the mass it has struck.

As was stated above, the gravitational field of a star, black hole, etc will bend light because it is also bending space.

This website contains a lot of information about how antennas work and how an electron (which is loads smaller than a wavelength of light) can still emit and absorb photons. It'll help you understand the physics behind it all.

btw, what did you get your undergrad degree in? This is pretty basic, found in Physics 2 (electromagnetic and optical wave theory) for me.

[edit on 8/23/2005 by shbaz]

Originally posted by grad_student
But, they exhibit certain properties such as radiation pressure, that a naive scientist (more accurately an electrical engineer getting into solid state physics) might interpret as exhibiting mass-like qualities.

Energy, mass and momentum are related in relativity according to m2 = E2/c4 - p2/c2. Since light has no mass, it goes that E = pc. So light has no mass but it does have an impulse, momentum, that is very small however. Due to the law of conservation of momentum, if light is reflected of a flat surface according to the surface normal, there is a momentum transferred upon this reflected surface of 2p. If we know how many photons are falling onto a surface per unit of time, we can hence calculate the change in impulse p = mv per unit of time and hence the acceleration a=dv/dt=(dp/dt)/m and the force F = dp/dt on this surface. Dividing by the surface area A results in the radiation pressure P = F/A.

Since the energy of a photon is also given by E = hv with v the frequency, and E = pc, the impulse of a photon is given by p = hv/c. As said already, this is a very small value. Since impulse is classically calculated as p = mv and v=c for photons, we can calculate an equivalent mass of photons that is m = hv/c². This photon "mass" is however a purely mathematical concept.

Originally posted by Simon666

Energy, mass and momentum are related in relativity according to m2 = E2/c4 - p2/c2. Since light has no mass, it goes that E = pc. So light has no mass but it does have an impulse, momentum, that is very small however.

Since the energy of a photon is also given by E = hv with v the frequency, and E = pc, the impulse of a photon is given by p = hv/c. As said already, this is a very small value. Since impulse is classically calculated as p = mv and v=c for photons, we can calculate an equivalent mass of photons that is m = hv/c². This photon "mass" is however a purely mathematical concept.

Ok this makes more sense, you are saying there's actually an extra term in E=mc^2 that has to do with momentum. So that photons exhibit momentum and not mass.

This is more what I was digging for.

Now, let's consider a particle that can exhibit this ... this particle we will assume is constructed of some arrangement of quarks ... so this particle is made out of some "stuff" for lack of a better term. This stuff can have momentum, but no mass.

Quite intriguing it's not the definitions of math that confuse me ... it's just terribly bizarre intuitively to have "stuff" that has no mass! So it sounds from what you are saying that you simply accept this definition ...

Was looking for some real definitions, more like simon666's post ...

Originally posted by shbaz

This website contains a lot of information about how antennas work and how an electron (which is loads smaller than a wavelength of light) can still emit and absorb photons. It'll help you understand the physics behind it all.

[edit on 8/23/2005 by shbaz]

Nonetheless, the spot on Tesla's wireless power transceiver is of course interesting, and I've heard about it before. Yes I understand how antennas work in quite some gory detail!

I was in software engineering, worked in industry for 10 years, laid off, MS in control engineering, and some work with ultrafast electromagnetic pulse propagation in causal dielectrics.

I understand how electrons and photons interact, that's not the source of the confusion, the confusion comes out of several conversations I had with a cosmologist, and a friend of mine who works here:

www.perimeterinstitute.com...

Originally posted by Simon666

E2/c4 - p2/c2. Since light has no mass, it goes that E = pc. So light has no mass but it does have an impulse, momentum, that is very small however. Due to the law of conservation of momentum, if light is reflected of a flat surface according to the surface normal, there is a momentum transferred upon this reflected surface of 2p. If we know how many photons are falling onto a surface per unit of time, we can hence calculate the change in impulse p = mv per unit of time and hence the acceleration a=dv/dt=(dp/dt)/m and the force F = dp/dt on this surface. Dividing by the surface area A results in the radiation pressure P = F/A.
concept.

It's the mechanics of the momentum transfer that I am specifically curious about.

When you shoot a large ball at a solid plate with a large cannon, it moves the plate (transfers momentum to the plate). And when you shoot a plate with a photon, it also transfers momentum to this plate. This makes perfect sense.

However, at the microscopic level, the atoms of the large cannonball as they encounter the solid plate, interact by repulsive charge forces that transfer the momentum.

If the atom of the photon has zero mass, how exactly can it repulse the plate at the other end? Angular momentum due to the spin of the photon?

Well obviously photons are not atoms but you know what I mean.

Originally posted by grad_student
If the atom of the photon has zero mass, how exactly can it repulse the plate at the other end? Angular momentum due to the spin of the photon?

You're going to need to rephrase that..

doesn't make a lot of sense as-is. What is the plate at the "other end?"

[edit on 8/24/2005 by shbaz]

Originally posted by grad_student
Ok this makes more sense, you are saying there's actually an extra term in E=mc^2 that has to do with momentum.

It is not often known, but E does not exactly equal mc² indeed, that only gives you the energy of a mass in rest in your refererence frame, so with speed v and hence impulse p zero. Exactly how momentum gets transfered at microscopic level I wouldn't know. I do know that all things considered, momentum needs to be conserved so you can calculate the change in impulse of an atom or plate when a photon gets reflected at one.

[edit on 25-8-2005 by Simon666]

The way I see it is that photons are purely elements of angular momentum. Whilst they do not have mass they still have energy and therfore are effected by gravity. Did you know that recently optical physicists have discovered that light beams can carry orbital angular momentum as well as intrinsic spin angular momentum, I think that this has implications that the photon has structure of some kind.

Originally posted by McGrude

Interesting questions. I am not sure, but I believe that the reason for light 'bending' around
large gravitational sources is due to the curving of space itself, rather than gravitational force
being applied to the photons themselves.

[edit on 2005/8/23 by McGrude]

Thanks for the clarification! I have wondered this myself.