Can calculus find the momentum of light?

[matphysik]

OK, let me try. If we start with \frac{v}{c}\leq\sqrt{1-\epsilon}, then we can get m_{0}\leq\sqrt{\epsilon}\leq\frac{1}{\gamma}. Am I on the right track?

NO. You are not.

 

[lugita15]

NO. You are not.

Can you give me a hint?

 

[matphysik]

Can you give me a hint?

I`ve been trying to. Over the years, reading proofs in math books has taught me all sorts of slight of hand `tricks`. You would be surprised. Rest assured that what i wrote there is correct.

 

[matphysik]

Can you give me a hint?

You haven`t had any trouble accepting the left hand sides of the two inequalities: "0<m₀" and "0<v/c". Let`s label the two zeros 1 & 2. So that, "0₁<m₀" and "0₂<v/c". Would you say that 0₁=0₂ ?

 

[lugita15]

You haven`t had any trouble accepting the left hand sides of the two inequalities: "0<m₀" and "0<v/c". Let`s label the two zeros 1 & 2. So that, "0₁<m₀" and "0₂<v/c". Would you say that 0₁=0₂ ?

First of all, the two zeroes are in different units, and units is one of the major issues I have with your proof. Also, depending on whether we want v to be velocity or speed, v may not need to be greater than zero.

 

[Dickfore]

Without intending to overrule other people's claims in this thread, I want to give my final exposition. Let us consider the asymptotic behavior of the momentum and energy relations around the point (v, m) = (c, 0). Because of the square root, this point is a branch point w.r.t. to the complex variable \beta = v/c, so care must be taken how we approach \beta_{0} = 1. We will consider the case:
<br /> \beta = 1 - \varepsilon, \; \varepsilon &gt; 0<br />
Similarly, we will assume the (rest) mass to be infinitesimally small:
<br /> m = M \, \mu, \; \mu &gt; 0<br />
Here, M is simply a unit of mass and \mu is a positive infinitesimal quantity.

The asymptotic behavior of momentum and energy is:
<br /> \begin{array}{l}<br /> p \sim \frac{M \, c}{\sqrt{2}} \frac{\mu}{\sqrt{\varepsilon}} \\<br /> <br /> E \sim \frac{M \, c^{2}}{\sqrt{2}} \frac{\mu}{\sqrt{\varepsilon}}<br /> \end{array}<br />

We immediately notice that, no matter how we change \varepsilon and \mu, the relation:
<br /> E \sim c \, p<br />
holds in the above limit.

 

[matphysik]

First of all, the two zeroes are in different units, and units is one of the major issues I have with your proof. Also, depending on whether we want v to be velocity or speed, v may not need to be greater than zero.

YES. `v` is a speed (as is `c`). And epsilon is a variable number. One may imagine the quantity `√(ε)` multiplied by a unit mass (in the same units as m₀).

 

[matphysik]

Without intending to overrule other people's claims in this thread, I want to give my final exposition. Let us consider the asymptotic behavior of the momentum and energy relations around the point (v, m) = (c, 0). Because of the square root, this point is a branch point w.r.t. to the complex variable \beta = v/c, so care must be taken how we approach \beta_{0} = 1. We will consider the case:
<br /> \beta = 1 - \varepsilon, \; \varepsilon &gt; 0<br />
Similarly, we will assume the (rest) mass to be infinitesimally small:
<br /> m = M \, \mu, \; \mu &gt; 0<br />
Here, M is simply a unit of mass and \mu is a positive infinitesimal quantity.

The asymptotic behavior of momentum and energy is:
<br /> \begin{array}{l}<br /> p \sim \frac{M \, c}{\sqrt{2}} \frac{\mu}{\sqrt{\varepsilon}} \\<br /> <br /> E \sim \frac{M \, c^{2}}{\sqrt{2}} \frac{\mu}{\sqrt{\varepsilon}}<br /> \end{array}<br />

We immediately notice that, no matter how we change \varepsilon and \mu, the relation:
<br /> E \sim c \, p<br />
holds in the above limit.

YES. That`s also correct

 

[lugita15]

Without intending to overrule other people's claims in this thread, I want to give my final exposition. Let us consider the asymptotic behavior of the momentum and energy relations around the point (v, m) = (c, 0). Because of the square root, this point is a branch point w.r.t. to the complex variable \beta = v/c, so care must be taken how we approach \beta_{0} = 1. We will consider the case:
<br /> \beta = 1 - \varepsilon, \; \varepsilon &gt; 0<br />
Similarly, we will assume the (rest) mass to be infinitesimally small:
<br /> m = M \, \mu, \; \mu &gt; 0<br />
Here, M is simply a unit of mass and \mu is a positive infinitesimal quantity.

The asymptotic behavior of momentum and energy is:
<br /> \begin{array}{l}<br /> p \sim \frac{M \, c}{\sqrt{2}} \frac{\mu}{\sqrt{\varepsilon}} \\<br /> <br /> E \sim \frac{M \, c^{2}}{\sqrt{2}} \frac{\mu}{\sqrt{\varepsilon}}<br /> \end{array}<br />

By asymptotic behavior do you mean Taylor expansion? If we just plug in your \epsilon and \mu, we get p=\frac{M\mu c(1-\epsilon)}{\sqrt{\epsilon(2-\epsilon)}} and E=\frac{M\mu c^{2}(1-\epsilon)}{\sqrt{\epsilon(2-\epsilon)}}. So then what do you do, do you take the Taylor expansions ofp(\epsilon,\mu) and E(\epsilon,\mu) about the point (\epsilon,\mu)=(0,0)? If that's what you're doing, the first term of each Taylor series is obviously undefined, so do you mean something else by "asymptotic behavior"?

 

[matphysik]

By asymptotic behavior do you mean Taylor expansion? If we just plug in your \epsilon and \mu, we get p=\frac{M\mu c(1-\epsilon)}{\sqrt{\epsilon(2-\epsilon)}} and E=\frac{M\mu c^{2}(1-\epsilon)}{\sqrt{\epsilon(2-\epsilon)}}. So then what do you do, do you take the Taylor expansions ofp(\epsilon,\mu) and E(\epsilon,\mu) about the point (\epsilon,\mu)=(0,0)? If that's what you're doing, the first term of each Taylor series is obviously undefined, so do you mean something else by "asymptotic behavior"?

If the Taylor expansion happens to be an asymptotic expansion.

Just discard the h.o.t. in epsilon in both numerator and denominator.

 

[Dickfore]

By asymptotic behavior do you mean Taylor expansion?

No. The function \varepsilon^{-\frac{1}{2}} has a singularity at \varepsilon = 0 and no Taylor expansion.

 

[lugita15]

No. The function \varepsilon^{-\frac{1}{2}} has a singularity at \varepsilon = 0 and no Taylor expansion.

So then what do you mean by asymptotic behavior ?

 

[Dickfore]

So then what do you mean by asymptotic behavior ?

I meant the following:

http://en.wikipedia.org/wiki/Asymptotic_expansions"

We keep the leading term in the expansion.

 

[bcrowell]

I didn't see anything wrong with pervect's #3:

We know that for all particles, E^2 - (pc)^2 = (mc^2)^2

So, we can write directly that E = sqrt(p^2 c^2 + m^2 c^4), and take the limit as m->0 rather than worry about v,( which becomes irrelevant), and find that in the limit as m->0, E= pc

I don't understand how this topic has generated another 110 posts after that.

Ctxyz's #11 also works.

PAllen said the same thing in #35 that pervect did in #3. Ctxyz raised an objection in #49, which PAllen addressed in #52. But even if you don't agree with me that ctxyz was wrong in #52-71, we still had a valid answer to the OP's question by #11.

 

[atyy]

Doesn't Pervect's method work only if p is not γmv?

If mass is defined via momentum and Newton's second law, how does it work?

In ctxyz's post #11, I don't understand why m approaches 0.

matphysik's approach seems to be the only one which addresses the question. However, in trying to get units right, I don't understand whether one uses an invariant reference mass, or a relativistic mass, and how it fits with Lorentz transformation.

PAllen's posts use a completely different definition of mass as the length of (E,p), and avoid the question.

 

[PAllen]

Doesn't Pervect's method work only if p is not γmv?

If mass is defined via momentum and Newton's second law, how does it work?

In ctxyz's post #11, I don't understand why m approaches 0.

matphysik's approach seems to be the only one which addresses the question. However, in trying to get units right, I don't understand whether one uses an invariant reference mass, or a relativistic mass, and how it fits with Lorentz transformation.

PAllen's posts use a completely different definition of mass as the length of (E,p), and avoid the question.

Dickefore's limiting argument should satisfy any skeptic of such a procedure. His is precise with units, and allows any limiting 'path', showing the same result.

 

[bcrowell]

Doesn't Pervect's method work only if p is not γmv?

Huh? Nobody is saying that p doesn't equal mγv.

In ctxyz's post #11, I don't understand why m approaches 0.

Because light has no rest mass.

PAllen's posts use a completely different definition of mass as the length of (E,p), and avoid the question.

It doesn't matter whether you take your initial assumption to be m²=E²-p², or p=mγv and E=mγ. If you prefer the latter, then it's two lines of algebra to prove the former. (This is all in units with c=1.)

matphysik's approach seems to be the only one which addresses the question.

I disagree. The question was answered in #3.

matphysik's approach seems to be the only one which addresses the question. However, in trying to get units right, I don't understand whether one uses an invariant reference mass, or a relativistic mass, and how it fits with Lorentz transformation.

I lost patience with matphysik's coyness about how s/he obtained results.

 

[atyy]

Huh? Nobody is saying that p doesn't equal mγv.


Because light has no rest mass.


It doesn't matter whether you take your initial assumption to be m²=E²-p², or p=mγv and E=mγ. If you prefer the latter, then it's two lines of algebra to prove the former. (This is all in units with c=1.)


I disagree. The question was answered in #3.

So p=mγv for light?

 

[bcrowell]

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.

 

[bcrowell]

So p=mγv for light?

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.

 

[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.