# Can calculus find the momentum of light?

1. Jul 8, 2011

### 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.

Thank You in Advance.

2. Jul 8, 2011

### atyy

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."

Last edited: Jul 8, 2011
3. Jul 8, 2011

### pervect

Staff Emeritus
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

4. Jul 8, 2011

### lugita15

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

5. Jul 8, 2011

### atyy

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.

6. Jul 9, 2011

### 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+).

7. Jul 9, 2011

### matphysik

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.

8. Jul 9, 2011

### Dickfore

@op:

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

9. Jul 9, 2011

### bcrowell

Staff Emeritus
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.

10. Jul 9, 2011

### atyy

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

11. Jul 9, 2011

### ctxyz

This is a very nice proof. What do you think about this alternative:

By definition

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

so

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

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

12. Jul 9, 2011

### Dickfore

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.

13. Jul 9, 2011

### ctxyz

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

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.

Last edited: Jul 9, 2011
14. Jul 9, 2011

### bcrowell

Staff Emeritus
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.

15. Jul 9, 2011

### atyy

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?

Last edited: Jul 9, 2011
16. Jul 9, 2011

### atyy

OK, I guess we can have a consistent theory of point particles having definite position and momentum travelling 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.

17. Jul 9, 2011

### matphysik

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?

18. Jul 9, 2011

### ctxyz

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?

19. Jul 9, 2011

### matphysik

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!

20. Jul 9, 2011

### ctxyz

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....