Can calculus find the momentum of light?

[atyy]

Yes, that's a good way to put it.

[I don't consider my main approach heuristics, but instead a change of decisions about what you consider definition versus consequence. As long as one is willing to define energy and momentum first, in a general way, then derive specific formulas for m>0, there are neither heuristics nor limits involved. Remember, metric and/or Lorentz transform include nothing about kinematics. You have to add those somehow. The traditional way is to start with mass, and take hints from Newtonian mechanics. Clearly, it is just as valid to start with energy/momentum, define rest mass (norm), then derive the traditional momentum/energy formulas for the case of rest mass > 0. ]

At first I thought a vector limit (timelike to null) would be absurd. Now I am not so sure. Coordinate transforms can get a null vector arbitrarily close (0,0,0,0). The limit of the rest frame timelike vector (m,0,0,0) as m goes to zero also reaches (0,0,0,0). Not sure what you can make of this rigorously.

I meant I agree heuristically with your approach, not that your approach is heuristic, just that I hadn't thought clearly for myself what could or could not be derived. For example, just saying (E,p) is a four-vector isn't enough to specify it, since the 4-current is also a four-vector. The distinction comes, I think with the conservation law that is imposed, which maybe can be addressed by an action.

I think the answer is yes (well, it seems you can do it for http://arxiv.org/abs/0809.1003" ). Originally, thinking of the question of the p=γmv limit, I had said I don't know if the limit really exists in a physically sensible way, and we need new axioms to be able to say m=0, and I suggested (w,k) for a field as the new axioms, just from the knee-jerk of always writing E=hf for a photon. However, I had forgotten that one can also just as well take (E,p) without needing to define E=hf (so that we have particles of "pure energy" ).

So I guess the reformulation isn't that interesting. We should still stick to the original question of if we start from p=γmv as a definition, whether the limit can be taken sensibly, as implied in https://www.physicsforums.com/showthread.php?p=3393005#post3393005. Rindler's text says something that seems to support it, but I can't find any reference that gives the details.

 

[matphysik]

Could you show that E^{2} - c^{2} p^{2} = m^{2}_{0} \, c^{4} describes a hyperbola in the (v, m)-plane?

YES. You have a valid point. Sorry.

Correction: Replace "family of hyperbolae" with "family of ellipses".

 

[atyy]

In YOUR earlier post #76 there was no mention of `v/c`.

It is tacitly understood that the variable quantity `√(1-ε)` is dimensionless.

Then how can √ε have units of mass?

 

[matphysik]

Then how can √ε have units of mass?

Because it`s associated with `m₀`.

 

[matphysik]

I meant I agree heuristically with your approach, not that your approach is heuristic, just that I hadn't thought clearly for myself what could or could not be derived. For example, just saying (E,p) is a four-vector isn't enough to specify it, since the 4-current is also a four-vector. The distinction comes, I think with the conservation law that is imposed, which maybe can be addressed by an action.

I think the answer is yes (well, it seems you can do it for http://arxiv.org/abs/0809.1003" ). Originally, thinking of the question of the p=γmv limit, I had said I don't know if the limit really exists in a physically sensible way, and we need new axioms to be able to say m=0, and I suggested (w,k) for a field as the new axioms, just from the knee-jerk of always writing E=hf for a photon. However, I had forgotten that one can also just as well take (E,p) without needing to define E=hf (so that we have particles of "pure energy" ).

So I guess the reformulation isn't that interesting. We should still stick to the original question of if we start from p=γmv as a definition, whether the limit can be taken sensibly, as implied in https://www.physicsforums.com/showthread.php?p=3393005#post3393005. Rindler's text says something that seems to support it, but I can't find any reference that gives the details.

With no disrespect intended, and in Einstein`s defense:

How easy it is to introduce `different` axiomatic derivations of special relativity AFTER having examined Einstein`s original derivation (the most natural).

 

[lugita15]

YES. Exactly, "controlling the rates" so that m₀ and v/c approach their respective limits concurrently. This was my objective, since the action of one limit should not occur sooner (or, later) than the other, in the case of light.

But in order for the double limit to exist, shouldn't we get the same result regardless of the rates at which the variables approach their respective limits? Otherwise you're just calculating the limit along a particular path.

Or at best, a set of particular paths: specifically, only paths in the m-v plane for which the rates of convergence of the two variables satisfy the bounds you have specified.

 

[lugita15]

Because it`s associated with `m₀`.

I really don't understand this. If \sqrt{1-\epsilon} is dimensionless, then so is 1-\epsilon, and so is \epsilon, and thus so is \sqrt{\epsilon}.

 

[atyy]

@PAllen, actually, how does one specify the interaction of a massless classical particle (from the forums, it seems that the gluon is a massless charged particle - but is it a classical particle, rather than a field)? For example, could one have a massless classical electron (I think massless dirac fermions in graphene are fields, not classical particles)?

I 'm guessing that if it is possible to specify interacting relativistic massless classical particles, they'd interact via a field, since the Currie-Jordan-Sudarshan no-interaction theorem prevents direct interaction between relativistic particles in a Hamiltonian formalism. Feynman got around it for massive particles, I suspect because he used a Lagrangian formulation. However, I'd be happy to hear about specifying the dynamics of massless classical particles by any means, even Newton's second law (Dickfore mentioned something some posts back, but I haven't read it carefully).

As a starting point, maybe we can ask if http://arxiv.org/abs/0905.2391" method goes through for massless charged particles?

 

[PAllen]

@PAllen, actually, how does one specify the interaction of a massless classical particle (from the forums, it seems that the gluon is a massless particle - but is it a classical particle, rather than a field)? For example, could one have a massless classical electron (I think massless dirac fermions in graphene are fields, not classical particles)?

I 'm guessing that if it is possible to specify interacting relativistic massless classical particles, they'd interact via a field, since the Currie-Jordan-Sudarshan no-interaction theorem prevents direct interaction between relativistic particles in a Hamiltonian formalism. Feynman got around it for massive particles, I suspect because he used a Lagrangian formulation. However, I'd be happy to hear about specifying the dynamics of massless classical particles by any means, even Newton's second law (Dickfore mentioned something some posts back, but I haven't read it carefully).

As a starting point, maybe we can ask if http://arxiv.org/abs/0905.2391" method goes through for massless charged particles?

No, all that stuff just says there is no action at a distance. I'm speaking only kinematics - direct collisions between particles. The classical massless particles would have no connection to any classical field theory (any more that classical point particles do, except to respond to them). Then there are no issues at all. Gralla and Ward is only about massive test particles following geodesics in the limit (in GR). I can show, trivially, that massless particles must follow null geodesics.

Massless charged particles + Maxwell fields, I had not thought about. There might be real issues here, in which case I would simply say that a classical theory including E/M precludes massless charged particles. I'm thinking about issues in the EM force law, the fact that a null geodesic can't change direction. I don't know if there is a way to make massless charged particles classically consistent.

[EDIT: As far as I am concerned, the whole treatment of point particles in classical EM is a bit of a kluge. So, in short, my feeling is:

1) For kinematics (which I mentioned in a couple of posts was my focus), these issues are irrelevant. Both elastic and inelastic collisions are perfectly well treated with abstract point particles.

2) For dynamics, at first glance, it seems like it would be hard to use the classical point charge techniques for massless point charges. My reaction is simply to say that a purely classical theory precludes them.

]

 

[lugita15]

The bounds that i used were chosen for a reason. Have fun trying to figure it out

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?

 

[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.