Can calculus find the momentum of light?

[matphysik]

That was not the contention. The contention is that you are using a consequence to a pair of more fundamental definitions. This consequence you are attempting to use cannot be derived when \gamma is undefined (infinite).

I agree with you.

 

[ctxyz]

Are you really unfamiliar with the concept of equivalent sets of axioms?

Please don't talk down to me.

Given a formal system, you can make many choices as to what are axioms and what are consequences.

Not in the case you are inadvertently trying to use.

\vec{p}=\gamma m_0 \vec{v}

andE=\gamma m_0c^2

are the axioms.

E^2-(pc)^2=(m_0c^2)^2

is a (trivial) consequence valid only for finite \gamma. What you are trying to do is incorrect.

 

[PAllen]

Please don't talk down to me.

Not in the case you are inadvertently trying to use.

\vec{p}=\gamma m_0 \vec{v}

andE=\gamma m_0c^2

are the axioms.

E^2-(pc)^2=(m_0c^2)^2

is a (trivial) consequence valid only for finite \gamma. What you are trying to do is incorrect.

(I use c=1 in all below).
The norm is defined by a metric. The metric (1,-1,-1,-1) applied to (E,p) produces sqrt(E^2 - p^2). No gamma at all. In this framework, gamma comes into play only in the derivation of the group of transforms that preserve the metric. We then define that mass is the norm of 4-momentum. Then we derive that if mass, so defined, is > 0 (special case), there exists a frame where (E,p) = (m,0). Then, from Lorentz transform (derived from metric), applied to this, we get exactly what you claim must be a definition instead as a consequence. When you arrogantly resist logic, I may talk down to you.

You are fixated on the false idea that there is only one way to classify what is axiom and what is derived.

All my other arguments (unique continuation of valid expressions) remain valid and unanswered, as well. You keep repeating that one historic way of looking at things is the only possible way.

 

[ctxyz]

(I use c=1 in all below).
The norm is defined by a metric. The metric (1,-1,-1,-1) applied to (E,p) produces sqrt(E^2 - p^2). No gamma at all.

True, up to this point. Now, from the above, derive E^2-(pc)^2=(m_0c^2)^2. Let's see you obtain the RHS.

When you arrogantly resist logic, I may talk down to you.

You are fixated on the false idea that there is only one way to classify what is axiom and what is derived.

And you are doing it again. The reason that I resist is that you are making incorrect claims.

 

[PAllen]

True, up to this point. Now, from the above, derive E^2-(pc)^2=(m_0c^2)^2. Let's see you obtain the RHS.

This was already covered. I take this as a definition mass, in the special case where the norm > 0.
Then the conventional formulas for m>0 follow from this definition plus the Lorentz transform, which can be derived from the metric. I covered all of this and it really seems you refuse to read.

[Edit: actually, I take this as the definition of mass in all cases]

And you are doing it again. The reason that I resist is that you are making incorrect claims.

Nope.

 

[ctxyz]

This was already covered. I take this as a definition mass, in the special case where the norm > 0.

OK, so how do you get from this axiom , expressed in scalars only to the formula that describes the momentum , as a vector? I am very curious to see your derivation.

 

[PAllen]

OK, so how do you get from this axiom , expressed in scalars only to the formula that describes the momentum , as a vector? I am very curious to see your derivation.

I already described the derivation. Please read post #58 again. To recap:

Given m defined as norm of 4-momentum, then (if and only if) m>0, there exists a frame where the vector takes the form (m,0,0,0). Apply Lorentz transform to this, and you get the conventional representation involving gamma. I get the Lorentz transform as consequence of the metric.

 

[ctxyz]

I already described the derivation. Please read post #58 again.

I read it. I don't see any math. Can you indulge me , please?

 

[PAllen]

I read it. I don't see any math. Can you indulge me , please?

Which part?

- That the Lorentz transform follows from the Minkowski metric is covered in many elementary books. I am not inclined to repeat it. The method is to solve for the transforms that preserve the form of the metric.

- That for a metric of signature (+,-,-,-), then if a vector has a positive norm, there is coordinate transform leading to (norm,0,0,0) is an elementary result of vector algebra that I am not going to repeat the derivation of.

- That given (m,0,0,0), if you apply Lorentz transform you get (m*gamma, -beta*gamma*m,0,0) is one matrix multiply?

 

[ctxyz]

Which part?



- That given (m,0,0,0), if you apply Lorentz transform you get (m*gamma, -beta*gamma*m,0,0) is one matrix multiply?

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

 

[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:

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

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:

<br /> p = \gamma \, m_{0} \, v<br />

or:

<br /> p^{2} ( 1 - \frac{v^{2}}{c^{2}}) = m^{2}_{0} \, v^{2}<br />

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:
<br /> \begin{array}{l}<br /> p = \gamma \, m \, v \\<br /> <br /> E = \gamma \, m \, c^{2}, \; \gamma = \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}<br /> \end{array}<br />
can be inverted:
<br /> \begin{array}{l}<br /> v = \frac{p \, c^{2}}{E} \\<br /> <br /> m = \frac{\sqrt{E^{2} - (p \, c)^{2}}}{c^{2}}<br /> \end{array}<br />
These mappings are defined for:
<br /> E^{2} - (p \, c)^{2} \ge 0 \wedge E \neq 0<br />
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:
<br /> \begin{array}{l}<br /> v = \frac{c^{2}}{A} \\<br /> <br /> m = \frac{|p| \, \sqrt{A^{2} - c^{2}}}{c^{2}} \rightarrow 0, \; p \rightarrow 0<br /> \end{array}<br />
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:

<br /> p = \gamma \, m_{0} \, v<br />

or:

<br /> p^{2} ( 1 - \frac{v^{2}}{c^{2}}) = m^{2}_{0} \, v^{2}<br />

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:
<br /> \begin{array}{l}<br /> p = \gamma \, m \, v \\<br /> <br /> E = \gamma \, m \, c^{2}, \; \gamma = \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}}<br /> \end{array}<br />
can be inverted:
<br /> \begin{array}{l}<br /> v = \frac{p \, c^{2}}{E} \\<br /> <br /> m = \frac{\sqrt{E^{2} - (p \, c)^{2}}}{c^{2}}<br /> \end{array}<br />
These mappings are defined for:
<br /> E^{2} - (p \, c)^{2} \ge 0 \wedge E \neq 0<br />
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:
<br /> \begin{array}{l}<br /> v = \frac{c^{2}}{A} \\<br /> <br /> m = \frac{|p| \, \sqrt{A^{2} - c^{2}}}{c^{2}} \rightarrow 0, \; p \rightarrow 0<br /> \end{array}<br />
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).