Friar Bellows

Answerer: the article is talking about rest mass.

xianseeker: if a photon had a really tiny but nonzero mass (like 10E-90kg), it could get extremely close to c with very little initial energy, so close that we might not be able to measure how close.

I think everyone (except possibly xianseeker) is missing my point, but that's probably because I'm not explaining it well. Here's another example which might get my point across:

On Planet Zoop live the Zoopians. They are extremely weird creatures in that they "see" via neutrinos, rather than photons. As far as their instruments are concerned, neutrinos are massless particles, and while they've detected photons they don't know much about them. One day, a very smart Zoopian called Albert Zoopstein invents the special theory of relativity. It predicts the same results as our special theory of relativity: time dilation, length contraction, relativity of simultaneity, E=mc^2, twin "paradox" etc etc. But there is one difference: instead of using the speed of light postulate, they use the speed of neutrinos postulate, i.e. the speed of neutrinos is the same in all inertial reference frames. And an important result they find is that no massive particle can attain the speed of neutrinos, just as we've found that no massive particle can attain the speed of light.

Now what would happen if the physicists on Zoop discovered that neutrinos did have a tiny but nonzero rest mass? Would that spell trouble for special relativity? Maybe the Zoopians would think so, and bloop their moopicles in distress, but if we knew about them, we here on Earth would tell the Zoopians, "Relax, Zoop dudes! Just change your postulate by replacing 'speed of neutrino' with 'speed of light' and all will be well", and then we'd wallow in self-satisfaction. But what if the opposite occurred? What if we found out that photons have a tiny but nonzero rest mass? Would that spell trouble for special relativity? Then maybe the Zoopians would tell us, "Fizzle, Earth gloops! Just change your postulate by replacing 'speed of light' with 'speed of neutrino' and all will be well", and then they'd wooble in self-satispooption.

The problem with the Zoop physicists is the same as the problem with the Earth physicists. The use of "speed of light" (or "speed of neutrinos" for that matter) is just pedagogical shorthand for "speed of massless particles". It's not really important what that particle is, as long as it's massless. So the differences between Zoopian relativity and our relativity are superficial. As the Zoopians show, the special theory of relativity works just as well without any mention of light. In fact, I think it would work just as well even if there weren't any massless particles in the universe. We could then just use an idealised massless particle.

OK, that's enough from me. I'm sure I'm making an ass out of myself.

ex-xian

Oh (light bulb icon over head), I see what you're saying. Other that what I posted, I know of no other problem with massive protons. But I see what you mean about the photon having a very, very small mass.

wade-w

My apologies, Friar. I was responding to alek0's post, not the OP, and I see now that I was taking the thread off topic. Echidna's link is essentially the same argument as xianseeker's post, btw.

I see what you are saying. IIRC, the lorentz transformations preceded special relativity. I have always assumed that the assumption about c being a limiting velocity is a direct result of the problems discussed above. Off the top of my head there does not seem to be any mathematical objection to redefining c from the velocity of light in a vacuum to the velocity of a massless particle in a vacuum. Though I'd like to hear from a mathematical physicist on the subject!

Jesse

Yeah, I made a mistake when I said y -> infinity. But how are you going to apply L'Hospital's rule when the choice of what function you use to represent m0 approaching zero is arbitrary? With different functions you may get different answers about the limit, which was basically my point about the mass being "undefined" (in contrast, if m0 != zero, then no matter what function you use to approach that number, you'll find that the limit goes to infinity, which is why all particles travelling at c had better not have a nonzero rest mass).

wade-w

I may not understand exactly what you mean by choosing a function to represent m0. I may be wrong, but isn't m0 a constant? We agree on the result in the case of m0 != zero. If you are right, though, it only reinforces my original point: that the mass transformation equation is not appropriate here.

Jesse

wade-w:
I may not understand exactly what you mean by choosing a function to represent m0. I may be wrong, but isn't m0 a constant? We agree on the result in the case of m0 != zero.

Yeah, but remember that my original comment was disagreeing with your claim that the limit would be zero if m0 is zero--I said that although the limit would be zero if you keep m0 as a constant and have the denominator approach zero, one should not conclude that a particle with rest mass zero will still have zero mass when travelling at c, because if you have the quotient approach zero then the limit will not necessarily be zero anymore. The equation:



So, I guess I'm thinking of having parametrized functions m(u) and v(u), where m(u) approaches zero as u approaches infinity, and v(u) approaches c. In that case, the limit will depend on your choice of functions. In contrast, if you just have m0 be the constant zero, then the answer will be zero for every finite u, so the limit will be zero too. And if you have m(u) approach some nonzero number, the limit will be infinity regardless of what functions you use for m(u) and v(u).

wade-w:
If you are right, though, it only reinforces my original point: that the mass transformation equation is not appropriate here.

I basically agree, but the I think mass transformation does at least tell you something--that a particle with nonzero rest mass should not be able to travel at c without being obviously inconsistent with the equation, whereas there is no inconsistency with the equation if a particle with zero rest mass has a finite nonzero mass when travelling at c.

wade-w

Jesse:

This is how I arrive at my statement about the limit being zero if m0 = 0.

The vb code doesn't seem to like my whitespace... so consider all limits to be as v->c. Also, g' is the derivative dg/dv.

Let's let g = (1 - (v/c)^2)^1/2. Then we should be looking at
lim m0/g. If m0 = 0, then since lim m0 = 0 and lim g = 0 we have to apply L'Hopital's rule to determine the limit. We will always have 0 in the numerator, and lim g' = inf, so we get lim m0'/g' = 0. Thus lim m0/g = 0.

Jesse

Ok, so you're saying we can let g(v) be (1 - (v/c)^2)^1/2, so g'(v) will be equal to (1/2)*((1 - (v/c)^2)^-1/2)*(-2v/c^2), which will approach infinity as v approaches c. The problem is, again, that our choice of m0(v) is arbitrary...if m0'(c) could be guaranteed to be finite, then your reasoning would be correct that the limit m0'(v)/g'(v) would have to be 0 as v approaches c, which would mean the limit of m0(v)/g(v) is also 0, but there's no reason to assume m0(v) will have that property. For example, one perfectly good choice for m0(v) which approaches 0 as v-->c is m0(v) = (1 - (v/c)^2)^1/2, but in this case the limit of m0(v)/g(v) as v-->c will be 1.

wade-w

How can m0(v) = (1 - (v/c)^2)^1/2? As far as I can see, m0(v) is a constant! There are then two cases, one where the constant is non-zero, and one where it is zero. In other words, either the object in question has no rest mass, or it does have some rest mass.

In other words, the m0 in the equation is the specific value of m(v) at some particular v. Even though in absolute terms (if that even means anything in this context) the initial v != 0, it is zero with respect to the initial frame of reference. So we can consider the equation to read (using the g in my above post) as m(v) = m(0)/g(v). What I was saying in my first post is that if m(0) > 0, then as v->c, lim m(v) = inf, and if m(0) = 0 then as v->c, lim m(v) = 0.

Maybe we should take any further discussion of this to PM, as I think its not really what Friar was asking about, and we seem to be derailing this thread.

Jesse

wade-w:
How can m0(v) = (1 - (v/c)^2)^1/2? As far as I can see, m0(v) is a constant!

In the original equation, of course it is, but then in the original equation we were just considering the case of v=c rather than taking the limit as v approaches c. The whole point of our debate as I understood it was about what happens when you replace the constant m0=0 with a function m0(v) that approaches 0 as the denominator approaches 0. That was why I wrote this comment early on:

(note that I should have written x and y approach 0 rather than infinity, but otherwise I think this comment is correct)

Although both 1/0 and 0/0 are technically undefined, it makes sense to say that 1/0 must be infinite while 0/0 is "indeterminate" because it can be set equal to any finite number (just use the cross-product: 0 = 0 --> 0*1 = 0*2 --> 0/0 = 2/1, for example). What you have been arguing is basically that 0/0 = 0 because you can replace the denominator with a function that approaches 0, therefore the limit is 0. But one can equally well replace the quotient by a function that approaches 0 as well, and then the limit will depend on which function you choose for quotient. In this way you can use calculus to back up the notion that 0/0 is "indeterminate" rather than being equal to any particular finite number, although for any particular choice of f(x) and g(x) which approach 0, f(x)/g(x) may have a well-defined limit.

In contrast, if m0 is some nonzero constant k, then even if you replace m0=k by a function that approaches k as the denominator approaches 0, then the limit will approach infinity regardless of what function you choose. This backs up the notion that 1/0 can be thought of as having a value of "infinity" rather than being equal to every possible number like 0/0.

Anyway, I think we're in agreement about mathematical issues but just don't agree about the most "natural" way to answer the question of what the mass of a particle with m0=0 and v=c would be, if a specific answer had to be given instead of just saying that relativity doesn't cover that case.