Can calculus find the momentum of light?

[PAllen]

This part. I don't see you getting any momentum vector. I see you getting a scalar.

I don't get what you don't get. We started with an energy momentum 4-vector. We defined that its t component is energy and its (x,y,z) components give momentum. We defined its norm as rest energy = energy in frame where momentum is zero. The to relate rest energy to other variables we Lorentz transform this *vector* (m,0,0,0). Then we get the conventional formulas (one matrix multiply). I don't see what part is not completely obvious.

 

[matphysik]

Is the limit a Lorentz invariant? How do we Lorentz transform it?

Why should it be?

 

[matphysik]

But it seems that by tying the bounds of $m$ and $v/c$ to the same variable $\epsilon$, you are controlling the rates at which these two variables go to 0 and 1 respectively. But if you really wanted to take a double limit independent of the path in the m-v plane, shouldn't the bounds of $m$ and $v/c$ be expressed in terms of completely different variables?

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.

 

[matphysik]

Why is it ok for the limits to be Lorentz invariant? I thought the limits were supposed to be the energy of a massless particle?

The exercise was to prove the existence of a limit of the relativistic 3-momentum (p). We are not concerned as to whether p or its limit is Lorentz invariant.

 

[Dickfore]

This isn't right, $\gamma=\infty$ for v=c, so the Jacobian is not zero for v=c.

True. However, the Jacobian does not exist for $v = c$, so this line in the $(v, m)$-plane is still singular.

 

[atyy]

The exercise was to prove the existence of a limit of the relativistic 3-momentum (p). We are not concerned as to whether p or its limit is Lorentz invariant.

Shouldn't we get a different number in each frame, in order for the result to be interpreted as the momentum of a massless particle?

For any ε:0<ε<<1, let 0<m₀≤√(ε) and 0<v/c≤√(1-ε). Then,
p=mv= m₀v / √(1 - v²/c²)= m₀(v/c)c² / c√(1 - v²/c²)≤√(ε)√(1-ε)c/√(1-(1-ε))= c√(1-ε)→c, as ε→0+.
Hence, p=O(1) (ε→0+).

In 0<m₀≤√(ε) to make the units come out right, we have to multiply by a reference mass. If we choose the reference mass to be an invariant mass, then it seems we get Mc as a Lorentz invariant limit. If we choose γM as a reference mass, then it seems to Lorentz transform as I naively expect, but I don't know how the limit works then.

 

[matphysik]

How are we varying mass and velocity exactly?

Energy and momentum depend on velocity and mass. You can think of it as a mapping from the (v, m) -> (p, E) The Jacobian of this mapping is:

$$\frac{\partial(p, E)}{\partial(v, m)} = \left|\begin{array}{cc} m \, \gamma^{3} & \gamma \, v \\ m \, v \, \gamma^{3} & c^{2} \, \gamma \end{array}\right| = m \, c^{2} \, \gamma^{4} - m \, v^{2} \, \gamma^{4} = m \, c^{2} \, \gamma^{2}$$

The Jacobian is zero if $m = 0$ or $\gamma = 0 \Leftrightarrow v = c$ and these are singular paths of the mapping.

∂(p,E)/∂(v,m)=m₀c²γ² ⇔ m₀=0 and v∈[0,c) is nonsense! What exactly are you (`Dickfore`) trying to prove??

 

[matphysik]

Shouldn't we get a different number in each frame, in order for the result to be interpreted as the momentum of a massless particle?



In 0<m₀≤√(ε) to make the units come out right, we have to multiply by a reference mass. If we choose the reference mass to be an invariant mass, then it seems we get Mc as a Lorentz invariant limit. If we choose γM as a reference mass, then it seems to Lorentz transform as I naively expect, but I don't know how the limit works then.

It is tacitly understood that the variable quantity `√(ε)` has units of mass.

 

[atyy]

It is tacitly understood that the variable quantity `√(ε)` has units of mass.

But if that is the case, won't v/c also pick up units?

 

[matphysik]

True. However, the Jacobian does not exist for $v = c$, so this line in the $(v, m)$-plane is still singular.

The family of hyperbolae in the v-m plane that you are alluding to, based on E² - p²c²=m₀²c⁴ (c.f., x² - Ay²=B or, x²/B - y²/(B/A)=1) refer to a free MASSIVE relativistic particle. They themselves will not help! Perhaps an asymptote (?), but since p=mv in E² - p²c²=m₀²c⁴ then we would get (as i have already pointed out), E²=0 if YOU naively set m₀=0.

 

[atyy]

I've mostly avoided discussing PAllen's tack not because I disagree with the heuristics (I'm not sure about all the details, but it should be basically ok, as I agreed with back in post #16), but because the point of the OP was can we do it via a limit? But PAllen's approach suggests we can also frame the question:

When can we consider a null vector the limit of time-like vectors?

 

[Dickfore]

The family of hyperbolae in the v-m plane that you are alluding to, based on E² - p²c²=m₀²c⁴ (c.f., x² - Ay²=B or, x²/B - y²/(B/A)=1)

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

 

[matphysik]

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

p=mv.

 

[matphysik]

But if that is the case, won't v/c also pick up units?

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

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

 

[Dickfore]

p=mv.

This is different from what you stated in the quoted post.

Nice try, but it's still wrong. When we say $(v, m)$-plane, we mean that the x-axis is $v$ and the y-axis is the (rest) mass $m_{0}$, so, what you wrote is:

$$p = \gamma \, m_{0} \, v$$

or:

$$p^{2} ( 1 - \frac{v^{2}}{c^{2}}) = m^{2}_{0} \, v^{2}$$

This is a fourth order polynomial and, hence, not a hyperbola in the discussed plane.

 

[Dickfore]

∂(p,E)/∂(v,m)=m₀c²γ² ⇔ m₀=0 and v∈[0,c) is nonsense! What exactly are you (`Dickfore`) trying to prove??

The Jacobian must exist and be different in order that we are able to invert the mapping in a neighborhood of a certain point. I was simply stating that:
$$\begin{array}{l} p = \gamma \, m \, v \\ E = \gamma \, m \, c^{2}, \; \gamma = \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}} \end{array}$$
can be inverted:
$$\begin{array}{l} v = \frac{p \, c^{2}}{E} \\ m = \frac{\sqrt{E^{2} - (p \, c)^{2}}}{c^{2}} \end{array}$$
These mappings are defined for:
$$E^{2} - (p \, c)^{2} \ge 0 \wedge E \neq 0$$
In the point $(p, E) = (0, 0)$, the inverse mapping is not defined. For example, imagine taking the limit along a straight line $E = A \, p, \; |A| \ge c$. Then, we have:
$$\begin{array}{l} v = \frac{c^{2}}{A} \\ m = \frac{|p| \, \sqrt{A^{2} - c^{2}}}{c^{2}} \rightarrow 0, \; p \rightarrow 0 \end{array}$$
The mass tends to zero along any of these paths (which does not imply that it is zero yet), but the velocity might have different values.

 

[PAllen]

I've mostly avoided discussing PAllen's tack not because I disagree with the heuristics (I'm not sure about all the details, but it should be basically ok, as I agreed with back in post #16), but because the point of the OP was can we do it via a limit? But PAllen's approach suggests we can also frame the question:

When can we consider a null vector the limit of time-like vectors?

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.

 

[matphysik]

This is different from what you stated in the quoted post.

Nice try, but it's still wrong. When we say $(v, m)$-plane, we mean that the x-axis is $v$ and the y-axis is the (rest) mass $m_{0}$, so, what you wrote is:

$$p = \gamma \, m_{0} \, v$$

or:

$$p^{2} ( 1 - \frac{v^{2}}{c^{2}}) = m^{2}_{0} \, v^{2}$$

This is a fourth order polynomial and, hence, not a hyperbola in the discussed plane.

E²=m₀²c⁴ + p²c²=constant (K², say). In the v-m₀ plane we get, K²v² + m₀²c⁶ = K²c² or,
v²/c² + m₀²/(K²/c⁴) = 1, a family of ellipses (c.f., x²/A + y²/B = 1).

 

[matphysik]

The Jacobian must exist and be different in order that we are able to invert the mapping in a neighborhood of a certain point. I was simply stating that:
$$\begin{array}{l} p = \gamma \, m \, v \\ E = \gamma \, m \, c^{2}, \; \gamma = \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}} \end{array}$$
can be inverted:
$$\begin{array}{l} v = \frac{p \, c^{2}}{E} \\ m = \frac{\sqrt{E^{2} - (p \, c)^{2}}}{c^{2}} \end{array}$$
These mappings are defined for:
$$E^{2} - (p \, c)^{2} \ge 0 \wedge E \neq 0$$
In the point $(p, E) = (0, 0)$, the inverse mapping is not defined. For example, imagine taking the limit along a straight line $E = A \, p, \; |A| \ge c$. Then, we have:
$$\begin{array}{l} v = \frac{c^{2}}{A} \\ m = \frac{|p| \, \sqrt{A^{2} - c^{2}}}{c^{2}} \rightarrow 0, \; p \rightarrow 0 \end{array}$$
The mass tends to zero along any of these paths (which does not imply that it is zero yet), but the velocity might have different values.

I already know well about Jacobians, and the inverse mapping theorem. Why are you so interested in the point (p, E) = (0, 0) ?

 

[matphysik]

The relativistic momentum formula is $p = \frac{mv}{\sqrt{1-v/c^{2}}}$. Since photons are massless and travel the speed of light, this formula doesn't apply to them directly, but it seems logical that we could find the momentum of a photon by calculating $lim_{(m,v) \rightarrow (0^{+},c^{-})} \frac{mv}{\sqrt{1-v/c^{2}}}$. Unfortunately, this double limit does not exist, but we can still calculate the limit along specific "path" in the "m-v plane". 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? There doesn't seem to be any physical rationale; is there any situation where a physical object gets lighter and lighter as it approaches the speed of light, all the while keeping its total energy constant?

Any help would be greatly appreciated.Thank You in Advance.

You should learn to distinguish between the `rest mass` (m₀) and the `relativistic mass` (m), where m := m₀/√(1 - v²/c²).

 

[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.