Can calculus find the momentum of light?

[atyy]

No if you mean to check the equality by direct substitution, because that produces an indeterminate form.
Yes in terms of a limit evaluated at constant E.
This is of course the reason that it's awkward to take p=mγv and E=mγ as fundamental, and preferable to take m²=E²-p² as a definition.

Yes, agree, completely. Especially, that in speaking about the mass of light, it's most convenient to take the length of (E,p) as the definition of mass, where (E,p) can be defined separately for the free Maxwell field (via Poynting) or the Einstein photon (via de Broglie). That definition also permits massless classical particles that interact by point collisions and energy-momentum conservation.

But anyway, we are back to the OP - why constant E? (Also, why does rest mass vary with speed?)

 

[bcrowell]

But anyway, we are back to the OP - why constant E?

A massless particle only has two properties that can be measured mechanically: p and E. There is therefore no possible interesting relation to be derived except one that connects p and E. If your goal is to find p in terms of E, then you are free to treat E as fixed, because in your final result, E is just going to be some number to be plugged in. You could choose to vary E simultaneously with m, but it would be pointless, because it would just complicate the calculation.

(Also, why does rest mass vary with speed?)

If you keep E fixed and vary m, then v has to vary as well. This is true even in nonrelativistic physics.

 

[atyy]

A massless particle only has two properties that can be measured mechanically: p and E. There is therefore no possible interesting relation to be derived except one that connects p and E. If your goal is to find p in terms of E, then you are free to treat E as fixed, because in your final result, E is just going to be some number to be plugged in. You could choose to vary E simultaneously with m, but it would be pointless, because it would just complicate the calculation.

Can't position and velocity be measured mechanically?

If you keep E fixed and vary m, then v has to vary as well. This is true even in nonrelativistic physics.

But how can m vary? Isn't constant m the definition of a particle?

 

[bcrowell]

Can't position and velocity be measured mechanically?

OK, but neither of them can be related to E or p. Position doesn't relate to E or p because a particle can have a given E and p regardless of where it's located in space. Velocity can't be related to E or p, because all massless particles have v=c, regardless of their energy or momentum.

But how can m vary? Isn't constant m the definition of a particle?

The limiting process isn't a physical process. We just want E and p to be continuous functions of m. For example, we used to think neutrinos were massless, but now we know that at least some flavors have mass. As long as the mass is small, we don't expect the nonzero mass to invalidate everything we ever inferred about neutrinos based on measurements.

 

[matphysik]

The following is to spell out the details of why pervect's #3 was a correct answer to the question, even if one prefers to take p=mγv and E=mγ as fundamental rather than taking m²=E²-p² as fundamental. (This is all in units with c=1.)

Starting from p=mγv and E=mγ, we have
$$E^2-p^2=m^2(\gamma^2-\gamma^2 v^2) = m^2 .$$
Substituting m=0 for light, we have E²=p², so |p|=E.

Hello. It is incorrect to substitute m₀=0 since then p=m₀γv=0, and hence E²=0.

 

[bcrowell]

Hello. It is incorrect to substitute m₀=0 since then p=m₀γv=0, and hence E²=0.

It would have been not so much incorrect as impossible to evaluate p=mγv by direct substitution, because it's an indeterminate form. No, direct substitution to evaluate p=mγv doesn't give zero, because for m=0, $\gamma=\infty$. That's why it's an indeterminate form.

But in any case, none of this has anything to do with the material you quoted, in which I did not evaluate p=mγv by direct substitution. What I evaluated by direct substitution was E²-p²=m², which is not an indeterminate form. (Maybe I shouldn't say "I." In the post that you quoted I was just elaborating on pervect's #3.)

 

[matphysik]

It would have been not so much incorrect as impossible to evaluate p=mγv by direct substitution, because it's an indeterminate form. No, direct substitution to evaluate p=mγv doesn't give zero, because for m=0, $\gamma=\infty$. That's why it's an indeterminate form.

But in any case, none of this has anything to do with the material you quoted, in which I did not evaluate p=mγv by direct substitution. What I evaluated by direct substitution was E²-p²=m², which is not an indeterminate form. (Maybe I shouldn't say "I." In the post that you quoted I was just elaborating on pervect's #3.)

Many textbooks do the same.

First: E=H=pv - L= γm₀v² + m₀c²/γ= γm₀[v² + c²(1-v²/c²)]= γm₀c²=mc² is derived, with L= -m₀c²√(1-v²/c²). Then, with p:=mv= m₀v / √(1 - v²/c²) we get, (1-ß²)E²= m₀²c⁴ OR, E²-ß²E²=m₀²c⁴ OR, E²=m₀²c⁴ + ß²E²=m₀²c⁴ + ß²m²c⁴=m₀²c⁴ + m²v²c²=m₀²c⁴ + p²c². Where ß≡v/c.

With `geometric` units (c=1): E²=m₀²+ p² or, E² - p²=m₀² which is the expression in your post.

 

[lugita15]

If your goal is to find p in terms of E, then you are free to treat E as fixed, because in your final result, E is just going to be some number to be plugged in. You could choose to vary E simultaneously with m, but it would be pointless, because it would just complicate the calculation.

Are you saying that you would have to do more work, but ultimately you'd still get the same answer?

 

[bcrowell]

Are you saying that you would have to do more work, but ultimately you'd still get the same answer?

Yes.

 

[atyy]

OK, but neither of them can be related to E or p. Position doesn't relate to E or p because a particle can have a given E and p regardless of where it's located in space. Velocity can't be related to E or p, because all massless particles have v=c, regardless of their energy or momentum.

Shouldn't the idea that velocity isn't related to E or p be derived, rather than assumed if we are starting from p=γmv? (In Newtonian mechanics, the energy of a massive particle in a conservative field can depend on position, but I don't think this can be carried over to special relativity?)

The limiting process isn't a physical process. We just want E and p to be continuous functions of m. For example, we used to think neutrinos were massless, but now we know that at least some flavors have mass. As long as the mass is small, we don't expect the nonzero mass to invalidate everything we ever inferred about neutrinos based on measurements.

For a photon and neutrino, aren't these due to wave-particle duality or at least field-particle duality? E and p are not defined by γm, but are instead defined in terms of fields, eg. the Proca action, then m can be taken to zero. I imagine one might be able to do something like this for a classical particle, but this would be like starting from E²=p²+m², with E and p defined by means other than γm.

If we choose a "curve of constant energy", namely $\frac{mc^{2}}{\sqrt{1-v/c^{2}}} = E$, then we find the that the limit along this path is $\frac{E}{c}$, as expected.

But what a priori reason do we have for choosing such a path, as opposed to any other path?

I wonder if it'd work to say that the path is chosen so that the limit and Lorentz transformation commute?

 

[bcrowell]

Shouldn't the idea that velocity isn't related to E or p be derived, rather than assumed if we are starting from p=γmv?

Pervect's calculation in #3 did not use v=c as an assumption. Since his result shows that p doesn't vanish, it immediately follows from his result that v=c.

(In Newtonian mechanics, the energy of a massive particle in a conservative field can depend on position, but I don't think this can be carried over to special relativity?)

We're not doing a calculation where there's an external potential.

For a photon and neutrino, aren't these due to wave-particle duality or at least field-particle duality? E and p are not defined by γm, but are instead defined in terms of fields, eg. the Proca action, then m can be taken to zero. I imagine one might be able to do something like this for a classical particle, but this would be like starting from E²=p²+m², with E and p defined by means other than γm.

No, none of this has anything to do with quantum mechanics. It's all purely classical.

 

[atyy]

No, none of this has anything to do with quantum mechanics. It's all purely classical.

How do you make light a particle without quantum mechanics?

 

[bcrowell]

How do you make light a particle without quantum mechanics?

You don't. If people (probably including me) have used the word "photon" in this thread, just translate it into "electromagnetic wave."

We discussed this in #9, 10, 14, and 15, and I guess didn't resolve it to your satisfaction. I simply don't understand what you're getting at. None of this has anything to do with waves or particles. The equations we're discussing apply to any physical phenomenon that carries mass-energy, whether it's a wave, a particle, a ray, or pixie dust.

 

[atyy]

You don't. If people (probably including me) have used the word "photon" in this thread, just translate it into "electromagnetic wave."

We discussed this in #9, 10, 14, and 15, and I guess didn't resolve it to your satisfaction. I simply don't understand what you're getting at. None of this has anything to do with waves or particles. The equations we're discussing apply to any physical phenomenon that carries mass-energy, whether it's a wave, a particle, a ray, or pixie dust.

Let's say we have a massive classical electromagnetic wave. In the massless limit, I should get p~EXB (and some integration). Is that limit achieved via p=γmv with m going to zero?

I guess it should look something like http://arxiv.org/abs/hep-th/0602190, Eq 62, which at least naively to me looks different from p=γmv, and m can be set to zero easily. (There is a gauge invariance problem with the Proca action, but http://arxiv.org/abs/0809.1003 seems to say it can be addressed.)

 

[bcrowell]

Let's say we have a massive classical electromagnetic wave. In the massless limit, I should get p~EXB (and some integration). Is that limit achieved via p=γmv with m going to zero?

Yes.

 

[atyy]

Yes.

Ok, thanks! So that justifies the constant E path easily then, since the (E,p) vector applies only to free electromagnetic fields.

No, none of this has anything to do with quantum mechanics. It's all purely classical.

All I mean is that everything goes through very cleanly for classical fields. I don't really know how to do it for classical point particles, and I need some way of saying that particles are all really fields anyway.

 

[matphysik]

So p=mγv for light?

This is the typical `method` that one usually finds in the best textbooks on SR:

Using the 4-momentum, Pᵅ:=m₀Uᵅ=m₀γ(c,v₁,v₂,v₃)=m(c,v): gᵤᵥPᵘPᵛ= -m₀²γ²c²+m₀²γ²v²=-m₀²γ²(c²-v²)=-m₀²γ²c²(1-v²/c²)=-m₀²c² <0 ⇒ P is `timelike when m₀≠0 . So if P is `lightlike`(m₀=0), gᵤᵥPᵘPᵛ=0. Where the LHS is (using m₀γ(c,v)=( m₀γc,m₀γv )=(m₀γc²/c,m₀γv)=(E/c,p)): -E² + c²p² = 0 or E=cp.

Do you see the flaw in this reasoning (?). Namely, we begin with Pᵅ:=m₀Uᵅ=m₀γ(c,v₁,v₂,v₃)=m(c,v), so that p=γm₀v.

 

[atyy]

This is the typical `method` that one usually finds in the best textbooks on SR:

Using the 4-momentum, Pᵅ:=m₀Uᵅ=m₀γ(c,v₁,v₂,v₃)=m(c,v): gᵤᵥPᵘPᵛ= -m₀²γ²c²+m₀²γ²v²=-m₀²γ²(c²-v²)=-m₀²γ²c²(1-v²/c²)=-m₀²c² <0 ⇒ P is `timelike when m₀≠0 . So if P is `lightlike`(m₀=0), gᵤᵥPᵘPᵛ=0. Where the LHS is (using m₀γ(c,v)=( m₀γc,m₀γv )=(m₀γc²/c,m₀γv)=(E/c,p)): -E² + c²p² = 0 or E=cp.

Do you see the flaw in this reasoning (?). Namely, we begin with Pᵅ:=m₀Uᵅ=m₀γ(c,v₁,v₂,v₃)=m(c,v), so that p=γm₀v.

From what I understand (going off bcrowell's hints), we begin http://arxiv.org/abs/hep-th/0602190" . These have (E,p) vectors only if they are free. If they are not free, then we need the energy-momentum tensor. Free fields have constant E. So we choose the constant E path in taking the massless light speed limit. (I still don't really understand the last sentence, it's much easier to substitute m=0 directly into the field action, which is a lot more like defining m as the length of the (E,p) vector).

 

[Dickfore]

It is my opinion that the question by the OP had been answered by many users. Any further discussion seems useless and is just going off in circles. May I suggest a mod locks this thread.

 

[atyy]

Yes.

A wave packet for a massive classical field has dispersion. Do I define v in p=γmv as dw/dk?

 

[matphysik]

A wave packet for a massive classical field has dispersion. Do I define v in p=γmv as dw/dk?

Dispersion, and apparent superluminal speed is off-topic.

 

[matphysik]

It is my opinion that the question by the OP had been answered by many users. Any further discussion seems useless and is just going off in circles. May I suggest a mod locks this thread.

You are right. But i think that this topic should be allowed to quiet down on its own. It doesn`t hurt anyone to discuss.

 

[atyy]

Yes.

Hmmm, I am really having trouble seeing this.

The equation for a massive free classical field is really like E²=p²+m². (Eq A)

The group velocity will be: v=dE/dp. (Eq B)

This gives: v²=p²/(p²+m²). (Eq c)

Rearranging: p=γmv. (Eq D)

If m=0, Eq B and C give v=1, which is correct. If we allow division by 0, Eq D gives an indeterminate form which is also correct, since p is not dependent on v for m=0. But that's fine since E=p by definition in Eq A. So I'm really not sure that the indeterminate form is wrong for a massless classical field, except for the division by 0 part. But if division by 0 is not permitted, then we can't even get to Eq D, let alone take its limit.

 

[bcrowell]

It is my opinion that the question by the OP had been answered by many users.

I agree. It was answered 141 posts ago, to be exact.

Any further discussion seems useless and is just going off in circles. May I suggest a mod locks this thread.

Done.