Can calculus find the momentum of light?

[atyy]

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+).

If I take the equalities m₀=√(ε) and v/c=√(1-ε), then I seem to get m₀ as a function of v. But shouldn't m₀ be independent of v?

 

[matphysik]

If I take the equalities m₀=√(ε) and v/c=√(1-ε), then I seem to get m₀ as a function of v. But shouldn't m₀ be independent of v?

The variable `ε` has been introduced as a means of proving that the limit of the relativistic momentum exists, and is bounded as both m₀→0+ and v/c→1- together.

 

[atyy]

The variable `ε` has been introduced as a means of proving that the limit of the relativistic momentum exists, and is bounded as both m₀→0+ and v/c→1- together.

But doesn't this seem contrary to the physical meaning of rest mass?

 

[matphysik]

But doesn't this seem contrary to the physical meaning of rest mass?

NO. Given a free relativistic test particle. We are letting both the rest mass `m₀` approach zero arbitrarily close from the right, and `v/c` approach unity arbitrarily close from the left.

 

[atyy]

NO. Given a free relativistic test particle. We are letting both the rest mass `m₀` approach zero arbitrarily close from the right, and `v/c` approach unity arbitrarily close from the left.

So the use of the inequality, rather than the equality, is key?

 

[matphysik]

So the use of the inequality, rather than the equality, is key?

Yes.

 

[matphysik]

Could you please explain or elaborate on what you wrote? Having taken Real Anaysis, I'm familiar with epsilon-delta proofs for both single and double limits, but I'm not too clear on the details of your proof.

What i wrote was not an ε-δ proof.

 

[matphysik]

As I mentioned before, I don't understand the details of this proof. This is what I would consider to be an epsilon delta proof in this situation: we would have to find a number $L$, such that for any $\epsilon >0$ there exists a $\delta >0$ such that $\left|\frac{mv}{\sqrt{1-v^{2}/c^{2}}} -L \right| < \epsilon$ whenever $m>0$, $v<c$, and $m^{2}+(v-c)^{2} < \delta^{2}$. So what are $L$ and $\delta$ in this case?

No. It`s not an ε-δ proof.

 

[matphysik]

You say that, "..this double limit does not exist..":

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+).

So once the existence of the limit has been proved we may write:

p= m₀v / √(1 - v²/c²)= m₀(v/c)c² / c√(1 - v²/c²) → E`/c=: p`, as both v→c- and m₀→0+. Where the prime indicates of light/photon(s). Therefore, for light the equation E²=m₀²c⁴ + p²c² reduces to E²=E`².

 

[atyy]

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+).

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

 

[ctxyz]

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

He proved that the limit is c, so it is invariant. See post 27 for a detailed explanation.

 

[atyy]

He proved that the limit is c, so it is invariant. See post 27 for a detailed explanation.

How can E and p be Lorentz invariant? Do E and p of a massless classical particle transform differently from the non-Lorentz invariant E and p of a photon?

 

[ctxyz]

How can E and p be Lorentz invariant?

Not E and p, but their limits when $v->c$ and $m_0->0$. It is the limits that are Lorentz invariant, as explained in post 27.

 

[ctxyz]

I think all of this is unnecessarily complicated. I see nothing invalid about the idea that 'if a particle without rest mass exists' you want it to follow as much of SR as possible. Thus relationships independent of rest mass allow you state, if such a thing exists, it must behave as follows. Similar to what is done with tachyons.

To me, it is completely sufficient to say:

E^2 = E0^2 + p^2 c^2

I don't think that you can do that (pervect tried to use the same argument eralier in the thread).

$E^2-(pc)^2=(m_0c^2)^2$ is a consequence of $E=\gamma m_0 c^2$ and $p=\gamma m_0 v$ that results when one does the calculation of the norm of the energy-momentum four-vector. Indeed:

$$(\gamma m_0 c^2)^2-(\gamma m_0 vc)^2=(m_0c^2)^2$$

The evaluation of the LHS requires that $\gamma$ is defined properly (i.e. is not infinite) .

 

[atyy]

Not E and p, but their limits when $v->c$ and $m_0->0$. It is the limits that are Lorentz invariant, as explained in post 27.

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?

 

[ctxyz]

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?

Because they are numbers.

 

[PAllen]

I don't think that you can do that (pervect tried to use the same argument eralier in the thread).

$E^2-(pc)^2=(m_0c^2)^2$ is a consequence of $E=\gamma m_0 c^2$ and $p=\gamma m_0 v$ that results when one does the calculation of the norm of the energy-momentum four-vector. Indeed:

$$(\gamma m_0 c^2)^2-(\gamma m_0 vc)^2=(m_0c^2)^2$$

The evaluation of the LHS requires that $\gamma$ is defined properly (i.e. is not infinite) .

I strongly disagree this is the only way of looking at it. You have a theory that covers particles m>0, v < c. You can derive many true expressions in this theory. Some are formally undefined m->0 or v->c; some are not. You extend the theory by using the ones that don't diverge. There could only be a problem if different choices lead to different extensions - then you have ambiguity. In the case of SR, so far as I know, *all* expressions you may derive that allow you to plug in m (or E0) =0 directly, produce the same result.

You insist that momentum (similarly energy) must only be a representation of a particular formula. I insist that momentum can be treated as an independent variable (and except in the most elementary classical formulations, it usually is treated as an independent variable both classically and quantum mechanically).

 

[ctxyz]

I strongly disagree this is the only way of looking at it. You have a theory that covers particles m>0, v < c. You can derive many true expressions in this theory. Some are formally undefined m->0 or v->c; some are not. You extend the theory by using the ones that don't diverge. There could only be a problem if different choices lead to different extensions - then you have ambiguity. In the case of SR, so far as I know, *all* expressions you may derive that allow you to plug in m (or E0) =0 directly, produce the same result.

You insist that momentum (similarly energy) must only be a representation of a particular formula. I insist that momentum can be treated as an independent variable (and except in the most elementary classical formulations, it usually is treated as an independent variable both classically and quantum mechanically).

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).

 

[PAllen]

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).

Yes it can. We treat the consequence (true for all m>0,v<c) as an independent relationship between kinematic variables to extend coverage of a theory. This is common, valid, thing to do. In effect, without worrying about limiting schemes, we derive that any other extension will lead to inconsistencies.

Yet another way to say this: you declare some particular formulas are fundamental. Nonsense. SR can be derived from numerous different starting points. There is no deep reason to treat the Lorentz transform as the the most fundamental relationship. I can choose a starting point from the Minkowski metric, with 4-vector definitions of kinematic quantities. Then I derive the Lorentz transform (from the metric), and the v<c formulas for kinematic variables as a special case. This is just as valid a formulation of SR. Then not only are there no limits, there is no 'extension' at all. The only problem here is if there were different, consistent ways of doing this. There aren't.

 

[lugita15]

NO. Given a free relativistic test particle. We are letting both the rest mass `m₀` approach zero arbitrarily close from the right, and `v/c` approach unity arbitrarily close from the left.

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?

 

[ctxyz]

Yet another way to say this: you declare some particular formulas are fundamental.

I don't declare anything, you can check any book and you'll find the norm of the energy-momentum four vector calculated from the definition of the four vector. Simply put, start with $\gamma(\vec{p},E/c)$ and calculate its norm. This is how you get $E^2-(pc)^2=(m_0c^2)^2$

Nonsense. SR can be derived from numerous different starting points. There is no deep reason to treat the Lorentz transform as the the most fundamental relationship. I can choose a starting point from the Minkowski metric, with 4-vector definitions of kinematic quantities. Then I derive the Lorentz transform (from the metric), and the v<c formulas for kinematic variables as a special case. This is just as valid a formulation of SR. Then not only are there no limits, there is no 'extension' at all. The only problem here is if there were different, consistent ways of doing this. There aren't.

Perfectly valid but not relevant to the argument about how the norm of the energy-momentum four-vector is calculated.

 

[atyy]

Because they are numbers.

So how do I see that I get a different number in another frame?

 

[PAllen]

I don't declare anything, you can check any book and you'll find the norm of the energy-momentum four vector calculated from the definition of the four vector. Simply put, start with $\gamma(\vec{p},E/c)$ and calculate its norm. This is how you get $E^2-(pc)^2=(m_0c^2)^2$
.

No, the metric is just (+1,-1,-1,-1) (in units with c=1, and timelike signature convention). The 4 momentum is (E,p). No gamma in sight. From the metric definition, the Lorentz transform can be derived as that which takes the metric to the same form. Given the requirement that 4-momentum is a vector, p is simply defined from E and the required transformation rules. Rest mass becomes a consequence rather than an assumption: the norm of the 4-momentum vector. In any given frame, gamma is not used to define anything.

This is a perfectly consistent axiomatic approach to SR. Not only is this valid, but to me, it is a much more satisfying approach to expressing SR in its most general form without limiting arguments.

 

[ctxyz]

No, the metric is just (+1,-1,-1,-1) (in units with c=1, and timelike signature convention). The 4 momentum is (E,p). No gamma in sight.

$$(\vec{p},E/c)=\gamma (m_0 \vec{v}, m_0c)$$

By definition.

 

[PAllen]

$$(\vec{p},E/c)=\gamma (m_0 \vec{v}, m_0c)$$

By definition.

Are you really unfamiliar with the concept of equivalent sets of axioms? Given a formal system, you can make many choices as to what are axioms and what are consequences. *Nothing* requires this definition to be used. Instead, it can be derived as a *consequence* of the axioms I proposed, for the *special case* of m>0, v< c. You've really never heard that Euclidean geometry can be built on several different axiom sets? That there are half a dozen different axiom systems used to formally derive SR?

Have you also never heard of e.g. analytic continuation? That is an example of the idea that a given formula is undefined outside of some range, but a *unique* continuation consistently (by some definition) extends it. Even without choosing what are axioms and what are consequences, the norm of momentum 4-vector argument is an analog of analytic continuation without need of taking a limit. Even within conventional axioms and definitions, there is a unique extension to m=0 that preserves all expressions that are well defined for m=0.

 

[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.

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

and$$E=\gamma m_0c^2$$

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.

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

and$$E=\gamma m_0c^2$$

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.