Electromagnetic Wave have momentum

[iNow]

Pete,

 

Speaking of irrelevant comments, can you please explain to all of us how your tangent has anything whatsoever to do with the OP?

 

 

 

Well while reading a chapter on EM waves I got confused about EM have momentum.

 

This really little bit confusing because momentum can only be of them who has a mass and in EM waves there nothing like mass or there isn't any thing that can have mass.

 

So can anybody clear out my problem either I may wrong.

[Severian]

It hasn't stalled. Severian has probably chosen not to pursue it any further. Did you consider that perhaps he took my advice and looked the proof up in Rindler's text?

 

Sorry - I have just been a bit busy (with work and then family things over the weekend) so didn't get around to responding.

 

The reason that I asked about the box, it because it is crucial to the discussion. If you simply have a collection of particles then the sum of their four-momenta is also a four-vector. Since the four-vector's inner product with itself is a constant, you can define that as being the total mass of the collection of particles with no problem, and there is no ambiguity. If you claim there is a problem with this, then you are claiming that the Lorentz symmetry is broken.

 

However, you cannot define an energy density in this way, since not only do the "relativistic masses" change with a Lorentz transformation, but the volume of the space does too. Therefore, you need a rank 2 tensor (the stress-energy tensor) to describe the energy density. Although this is not a four vector, it transforms as a 2nd rank tensor under the Lorentz transformational.

 

Clearly, it is fundamental whether you are regarding the box as a closed system, such as a "rigid" body (however we choose to define that!), in which case you need to calculate the energy of the entire body, not just the density. Or is the box a part of something more?

 

Your claims seem to indicate the former - that you are calculating the energy and momentum of the entire object. But in that case, why do you have forces on the box. Where are they coming from and what is their point? What is pushing back against the force. And as I pointed out, if you do the sensible thing, and set [math]F^\prime_x=0[/math] then you do return to the usual energy and momentum satisfying [math]E^2-p^2c^2=m^2c^4[/math].

 

On the other hand, your calculation implies that you are only considering an element of a larger object. Indeed, Eq.(46.2) in Rindler (I had to walk to the Library to check this) is for this case. Rindler is very explicit in this (look at the discussion starting on page 146 about internal and external forces). Then you shouldn't be surprised that the energy and momentum don't transform as a four-vector since you need a 2nd rank tensor.

 

In either case, your "proof" must retain its inverted commas. For an isolated box, your forces should be zero and you end up proving my assertion yourelf(!). For an element of a larger fluid, you are not proving what you claim to prove, and your proof is inappropriate.

[swansont]

This discussion reminded me of the Abraham-Minkowski controversy — what happens to the momentum of a photon when it enters a medium. p=E/c, but you can't naively substitute c/n for the speed in the material and use that equation; there were competing derivations that said the momentum went up and went down, depending on the assumptions. Turns out that momentum of the photon is not well-defined in a medium — there are interactions, and those interactions must be accounted for. (I had suspected that a similar lesson would apply here to the mass issue as well) If the momentum does indeed increase, the surface of a bulk material should recoil backward when the photon enters it, but if the other approach is right, and the momentum decreases, the recoil is in the opposite direction. The latest experimental indication is that the latter is correct (Abraham's proposal), though one of the problems has been devising an experiment where the results are not confounded by other influences, since it's such a small effect.

 

Anyway, this is related to the OP and I had written this up a while back and there are a few links included.

[Sdogv1]

Well, I went through the obfuscations here and have a simple question to raise about the term "mass". Would anyone care to comment on what it might imply re: a mass frequency product?,. i.e. a mass cycle times a frequency cycle such that mass/cycle times cycles per second produces a "mass"? (mass to energy to mass "lifetime")

 

Comes up in photon "mass" re: http://greatians.com/physics/wave/photon.htm

[swansont]

Well, I went through the obfuscations here and have a simple question to raise about the term "mass". Would anyone care to comment on what it might imply re: a mass frequency product?,. i.e. a mass cycle times a frequency cycle such that mass/cycle times cycles per second produces a "mass"? (mass to energy to mass "lifetime")

 

A mass frequency product?

 

Comes up in photon "mass" re: http://greatians.com/physics/wave/photon.htm

 

That's a decidedly alternative view of physics

 

As an alternative view it should be brought up in the speculations forum, and not as part of a discussion in the physics section, as per rules 2.5 and 2.10

[pioneer]

Momentum and kinetic energy for a mass are related since both only require M and V, with a simple difference either MV or 1/2MV2. If we add energy to a mass we can increase its kinetic energy, and therefore increase its momentum.

 

I think one experiment was a light beam that shined on a mass foil within a vacuum. The light caused it to turn and therefore gave it momentum. But it also added energy or kinetic energy to create the V that is shared between kinetic energy and momentum. So technically there was a momentum transfer. One might also say a force impulse has momentum. We can do a mass experiment with iron and a magnetic pulse and transfer momentum. This would simulate a quantum of magnetic force containing momentum.

Edited January 10, 2009 by pioneer

[gre]

That fact photons have momentum does seem weird at first. My question is how do they have momentum? All objects that have momentum, must have at least one of the following, I believe.

 

1.) Mass and velocity .. (Photons don't)

 

2.) Electric field, and velocity .. (Photons do)

 

3.) Magnetic field, and velocity .. (Photons do)

 

 

Edit: Why not say any field with polarity and a velocity?

 

 

Why can't the momentum of photons be described at the transfer of an electro-magnetic field (with velocity)?

Edited January 16, 2009 by gre

[Mr Skeptic]

That fact photons have momentum does seem weird at first. My question is how do they have momentum?

 

Perhaps you could ask the opposite. How could they not have momentum? All waves have a momentum inversely proportional to their wavelength, and photons are no exception. Likewise, from Einstein's equation [math]E^2 = (m_0c^2)^2 + (pc)^2[/math], where p is momentum, how could a photon have energy but no mass, if it didn't have momentum?

 

Your trouble is using Newtonian mechanics at relativistic velocities. p = mv is an approximation for low velocities.

[gre]

I think I understand the wave nature of photons. But here's another thought.. Maybe photons do have mass in their own reference frame.

 

Since an electron's mass is converted into a photon instantaneously, since its acceleration is instant, exactly at what point does the electrons mass turn into a photon? And how is it proven photons don't have mass that we can't see.

[Klaynos]

Photons don't have a valid reference frame...

 

You can show that things moving at c cannot have mass. This is a result of relativity.

 

You can also measure the speed of light to a very high degree of accuracy.

[swansont]

And how is it proven photons don't have mass that we can't see.

 

Objects with mass behave a certain way, and massless objects behave a different way. photons behave the way massless objects behave.

[gre]

What are the requirements for a "valid" reference frame? Why can't entities moving at c have one? I thought a photon moving at c would be considered a "singularity" of a reference frame, but I guess not.

[npts2020]

Do conservation laws require photons to be emitted in pairs?

[timo]

Do conservation laws require photons to be emitted in pairs?

No, unless you put additional constraints on your question that lead to a yes. For example, in electron-positron annihilation conservation of energy and momentum prohibits e+ e- to react into a single photon. It does not forbid the reaction into three photons, though. The probability is just a little lower. I am even tempted to claim that there are no sensible constraints that leads to a yes - on the level of Feynman diagrams constraining to an even number of photons looks impossible, but cancellation of diagrams might do the trick.

The typical example of a single photon being emitted is when an excited atom decays to an energetically lower state emitting light. That's mostly single photons.

Edited January 16, 2009 by timo
Found a simple e+ e- -> 3 gamma kinematic

[Mr Skeptic]

I think I understand the wave nature of photons.

 

Then you should understand that, as waves, photons must have momentum.

 

But here's another thought.. Maybe photons do have mass in their own reference frame.

 

Photons aren't considered to have their own reference frame. This is because the equations for reference frames break down at c, and things go either to zero or undefined (plus or minus infinity). There might also be another reason that photons can't have their own reference frame.

 

If you really want to give photons mass, you can say they have zero rest mass but non-zero relativistic mass. Most people on the site hate relativistic mass, probably because it causes so much confusion and seems pretty much useless. Anyways, you can say photons have relativistic mass proportional to its energy [math]m_{relativistic} = E/c^2[/math]. Basically though, relativistic mass is just the energy given in units of mass.

[swansont]

What are the requirements for a "valid" reference frame? Why can't entities moving at c have one? I thought a photon moving at c would be considered a "singularity" of a reference frame, but I guess not.

 

You have to be able to describe physics in that frame, and the way we (mathematically) get from one to the other is with a Lorentz transformation. This breaks down at v=c, meaning you can't transform into the frame, or more problematically (AFAICT) you can't invert the math and do a transform back to your original frame.

[gre]

Why not just say, since space, time, and mass can't exist at c, then reference frames can't either. Would that be true?

[swansont]

Why not just say, since space, time, and mass can't exist at c, then reference frames can't either. Would that be true?

 

I don't know that it's meaningful to say that space doesn't exist at c. As for saying mass can't move at c, I think it's basically equivalent — we can't get into that frame of reference.

[gre]

Is mass the only requirement for frames of reference, or is it 3D space and mass?

[swansont]

Is mass the only requirement for frames of reference, or is it 3D space and mass?

 

There certainly are discussions on the topic, but since you can't eliminate one or the other to test, is there any way to test this? I think it's more philosophy than science. All you can do is make arguments either way.