# Can calculus find the momentum of light?

• lugita15
In summary, the relativistic momentum formula does not directly apply to massless particles such as photons. However, by taking the limit of a specific path in the m-v plane, namely a curve of constant energy, we can still calculate the momentum of a photon. There is no physical rationale for choosing this path, but it leads to the expected result that the momentum of a massless particle is equal to its energy divided by the speed of light. Other methods, such as using wave-particle duality, can also lead to this result.
lugita15
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.

An equation for a massive photon is found in Eq. 3 of http://arxiv.org/abs/0809.1003. When m is taken to zero, we get a massless photon.

Footnote 4 is interesting! "In the period 1862-1867 the Dane Ludwig Lorenz independently derived the “Maxwell equations” of 1865, but received relatively little credit for this work."

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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

pervect said:
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
Yes, I'm already aware of this simple proof. I'm just trying to see whether other method can achieve the same result.

lugita15 said:
Yes, I'm already aware of this simple proof. I'm just trying to see whether other method can achieve the same result.

I don't think so. It's really through wave-particle duality that light and massive particles can both have mass. The 0/0 argument is a great heuristic, but I think it remains that at best.

lugita15 said:
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.

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+).

pervect said:
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

v is very relevant, since light propagates with speed c (in vacuum). We must approach c from the left Recall that, v∈[0,c). Furthermore, in the expression E^2 - (pc)^2 = (m₀c^2)^2 (for a free relativistic particle), p is understood to be equal to m₀v / √(1 - v²/c²). We cannot simply let m₀→0+ for we would have, E^2=0.

@op:

because the energy and momentum for a free particle must be constant.

atyy said:
I don't think so. It's really through wave-particle duality that light and massive particles can both have mass. The 0/0 argument is a great heuristic, but I think it remains that at best.

No, it has nothing to do with wave-particle duality. It's a purely classical fact.

There is nothing nonrigorous about taking the limit of the expression $p=m\gamma v$. Nobody has proposed substitution as a way of evaluating that limit, so your remarks about 0/0 are a straw-man argument.

bcrowell said:
No, it has nothing to do with wave-particle duality. It's a purely classical fact.

There is nothing nonrigorous about taking the limit of the expression $p=m\gamma v$. Nobody has proposed substitution as a way of evaluating that limit, so your remarks about 0/0 are a straw-man argument.

How do you get the energy of a photon? (Or a massless classical particle?)

matphysik said:
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+).

By definition

$$E=\gamma m_0 c^2$$
$$p=\gamma m_0 v$$

so

$$\frac{E}{p}=\frac{c^2}{v}$$

When $$v->c$$ $$\frac{E}{p}->c$$ ?

ctxyz said:

By definition

$$E=\gamma m_0 c^2$$
$$p=\gamma m_0 v$$

so

$$\frac{E}{p}=\frac{c^2}{v}$$

When $$v->c$$ $$\frac{E}{p}->c$$ ?

By dividing the expressions for the energy and momentum, you eliminated the mass and essentially expressed the speed as a function of E and p. You can go back and express the mass as:
$$m = \frac{E}{c \gamma}$$
So, when taking the limit $v \rightarrow c^{-}$, you are actually approaching the point $(v, m) = (c, 0)$ along a particular path.

Dickfore said:
By dividing the expressions for the energy and momentum, you eliminated the mass and essentially expressed the speed as a function of E and p. You can go back and express the mass as:
$$m = \frac{E}{c \gamma}$$

You must mean $$m = \frac{E}{c^2 \gamma}$$, surely.

So, when taking the limit $v \rightarrow c^{-}$, you are actually approaching the point $(v, m) = (c, 0)$ along a particular path.

I am not so sure, it is clear from the definition of E and p that $$\frac{E}{p}$$ is not a function of $$m_0$$. As an aside, it is not a function of $$\gamma$$ either.

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atyy said:
How do you get the energy of a photon? (Or a massless classical particle?)

If the question is how you get the energy of a massless classical particle, I don't really understand the question. Are you asking how you get the energy if you know its momentum? E=p.

bcrowell said:
If the question is how you get the energy of a massless classical particle, I don't really understand the question. Are you asking how you get the energy if you know its momentum? E=p.

How do I calculate the energy or momentum of a massless classical particle? More generally, what is a massless classical particle?

How do I use Newton's second law on a massless classical particle? Could I, say, formulate a consistent theory, of a massless charged classical particle? Is there any consistent theory of massless classical particles interacting with massive classical particles or classical fields?

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OK, I guess we can have a consistent theory of point particles having definite position and momentum traveling at the speed of light. By definition, they interact by enforcement of energy and momentum conservation. But are these really related to massive particles? ie. are they things with m=0, or things for which the notion of mass does not apply - ie. it isn't really fair to use Newton's second law on them, or to define p=γmv.

ctxyz said:

By definition

$$E=\gamma m_0 c^2$$
$$p=\gamma m_0 v$$

so

$$\frac{E}{p}=\frac{c^2}{v}$$

When $$v->c$$ $$\frac{E}{p}->c$$ ?

Thank you.

You are using the expressions for a relativistic free MASSIVE particle. E=E(m₀,v), p=p(m₀,v) so that: E(m₀,v)/p(m₀,v)=c²/v and, E(m₀,v)/p(m₀,v)→E(m₀,c)/p(m₀,c)=c as v→c-. What about m₀>0?

matphysik said:
Thank you.

You are using the expressions for a relativistic free MASSIVE particle. E=E(m₀,v), p=p(m₀,v) so that: E(m₀,v)/p(m₀,v)=c²/v and, E(m₀,v)/p(m₀,v)→E(m₀,c)/p(m₀,c)=c as v→c-. What about m₀>0?

True, the starting point is the expressions for massive particles, yet the limit can be taken for the case when $$m_0$$ tends to 0. Why do my posts appear so funny, on multiple lines?

ctxyz said:

By definition

$$E=\gamma m_0 c^2$$
$$p=\gamma m_0 v$$

so

$$\frac{E}{p}=\frac{c^2}{v}$$

When $$v->c$$ $$\frac{E}{p}->c$$ ?

Hello.
Ive been thinking about your result carefully, and i havent found anything wrong with it. The quotient of two simple algebraic expressions with the rest mass canceling out. Excellent!

matphysik said:
Hello.
Ive been thinking about your result carefully, and i havent found anything wrong with it. The quotient of two simple algebraic expressions with the rest mass canceling out. Excellent!

thank you, we have TWO excellent methods, I really liked yours. Do you know why my posts are appearing so funny, I am not hitting any "return" key...

ctxyz said:
thank you, we have TWO excellent methods, I really liked yours. Do you know why my posts are appearing so funny, I am not hitting any "return" key...

They look fine to me. Probably youre using TeX, and this site cant handle it :)

ctxyz said:
Why do my posts appear so funny, on multiple lines?

If you want mathematics in line with your text, use itex and /itex instead of tex and /tex. Use tex and /tex when you mathematics to appear on a separate line.

George Jones said:
If you want mathematics in line with your text, use itex and /itex instead of tex and /tex. Use tex and /tex when you mathematics to appear on a separate line.
Thank you, this is what I was doing wrong. Test1: $p=\gamma m v$.Test2 : $$p=\gamma m v$$

You two are absolutely right, thank you!

Dickfore said:
By dividing the expressions for the energy and momentum, you eliminated the mass and essentially expressed the speed as a function of E and p. You can go back and express the mass as:
$$m = \frac{E}{c \gamma}$$
So, when taking the limit $v \rightarrow c^{-}$, you are actually approaching the point $(v, m) = (c, 0)$ along a particular path.
What particular path would that in the m-v plane would that be?

matphysik said:
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+).
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.

atyy said:
How do I calculate the energy or momentum of a massless classical particle? More generally, what is a massless classical particle?

How do I use Newton's second law on a massless classical particle? Could I, say, formulate a consistent theory, of a massless charged classical particle? Is there any consistent theory of massless classical particles interacting with massive classical particles or classical fields?

See my derivation:

https://www.physicsforums.com/showpost.php?p=3333233&postcount=52"

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lugita15 said:
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.

He gave an extremely clever proof that the $\gamma m_0 v ->c$ when $m_0->0$ and $v->c$ by using epsilon-delta in two dimensions.
By aplying the same algorithm to $E=\gamma m_0 c^2$ rewritten as $E=\gamma m_0 v \frac{c^2}{v}$ you get $E->c^2$ so, $E/p->c$.

lugita15 said:
What particular path would that in the m-v plane would that be?

The path determined by:

$$m_{0} = \frac{E}{c^{2}} \, \sqrt{1 - \frac{v^{2}}{c^{2}}}$$

Dickfore said:
The path determined by:

$$m_{0} = \frac{E}{c^{2}} \, \sqrt{1 - \frac{v^{2}}{c^{2}}}$$
. OK, so it's just a curve of constant energy again, in which case my question remains why that's the right path to take the limit over.

Dickfore said:
@op:

because the energy and momentum for a free particle must be constant.
But we're not talking about what happens to energy and momentum as we vary time, which is where conservation laws would matter. We're varying mass and speed, and moreover we're already doing it in a way that makes the momentum change. So there's no obvious reason why energy should remain constant.

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lugita15 said:
But we're not talking about what happens to energy and momentum as we vary time, which is where conservation laws would matter. We're varying mass and speed, and moreover we're already doing it in a way that makes the momentum change. So there's no obvious reason why energy should remain constant.

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.

Dickfore said:
$\gamma = 0 \Leftrightarrow v = c$

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

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atyy said:
OK, I guess we can have a consistent theory of point particles having definite position and momentum traveling at the speed of light. By definition, they interact by enforcement of energy and momentum conservation. But are these really related to massive particles? ie. are they things with m=0, or things for which the notion of mass does not apply - ie. it isn't really fair to use Newton's second law on them, or to define p=γmv.

OK, I am convinced by matphysik's and ctxyz's approaches to p=γmv. I notionally add a reference non-zero mass M to matphysik's limit to get the correct dimensions. I had never seen either of those before. Very nice - thanks!

I think we still cannot use Newton's second law, is that right? We have to go directly by energy-momentum conservation.

matphysik said:
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+).
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?

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

implies that a particle with only kinetic energy, no rest energy, must have momentum:

E/c

Similarly, it is easy to derive that:

sqrt( 1 - v^2/c^2) = 1 - KE / (KE + E0)

From which you can directly state that v=c must be true for a particle with nonzero KE but zero E0.

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