# Why c?

1. Apr 28, 2008

### neopolitan

I know this has been asked before but I would like to float a related question.

Why does the second postulate specifically refer to c?

This is the wording that I am referring to:
Two answers I have seen are:

1. The second postulate follows directly from the application of the first postulate to electromagnetism (in which case it seems you really only need the first postulate). The invariance of c is a consequence of this.

and

2. "If you are looking for proof, then I am afraid you will be disappointed. Postulates can not be proven, they are assumptions or statements made." (from https://www.physicsforums.com/showpost.php?p=1140259&postcount=2")

Another answer can be is that a more rigorous wording would be "light is always propagated in empty space with an invariant velocity which is independent of the state of motion of the emitting body".

The issue I have is not that the speed of light is invariant, but that it seems there are people who stop all further debate as to why the speed of light is invariant and why it is the particular speed it is (ie c) with answer 2 above.

So, what c in particular? And why does the speed of light in a vacuum have the value it has?

For a possible answer we could look at natural units. As discussed http://en.wikipedia.org/wiki/Planck_length" [Broken]:

This indicates that Planck units have some physical meaning. They are also interesting because Planck length/Planck time = c.

I wonder if it is possible to consider that there is a link between these measurements, at which scale time and space become discrete, "foamy" or as I prefer "granular", and the speed limitation placed on photons.

For example, if the smallest measurable time duration is one Planck time and the smallest measurable length is one Planck length and these measurement limitations are physical, rather than being limitations associated with our measuring regimes, then we are left with a limitation where a discrete particle (or wavicle) can either move one discrete length in one discrete time duration at speed c, or stay motionless. This would make c not only the maximum speed, but also the minimum speed - for discrete particles, ie quarks.

Now the actual location of individual quarks can't be pinned down, we can only assign each possible location a probability (and this is a physical thing again, not a limitation on our measuring devices). Get enough quarks together, enough to constitute a mass as we know it, and you have a probability cloud. The centre of the mass is somewhere in that probability cloud and if you look at it from a macro perspective, you can now point at where the mass "is", with the tacit understanding that this is an approximation. A mass which consists of a large number of discrete particles/wavicles, each of which is restricted to a speed of c but not restricted in direction, could in fact travel, as a statistical average, slower than c - even if each individual discrete particle/wavicle travels at c. In fact, the more of them you have, the more difficult it will be to get them to all travel in the same direction, and more energy will need to be applied to get them to travel in pretty much the same direction.

Is this at all valid?

Is it valid to think that the light speed limitation of the second postulate is due to quantum level foaminess?

cheers,

neopolitan

Last edited by a moderator: May 3, 2017
2. Apr 28, 2008

### lbrits

The Planck length/Planck time thing is a bit backwards. Those are constructed from c to begin. In any event, why the particular value of c? Well, in my books, c = 1, so... Alternatively, light, being massless, happens to travel at speed c, and massive things travel slowly enough that we measure them first and base our measurement systems on meters instead of lightseconds.

The fact that c is a limiting speed follows from the fact that the Lorenz group seems to be a symmetry group of nature, locally.

As for "Is it valid to think that the light speed limitation of the second postulate is due to quantum level foaminess?", your guess is as good as mine... but my guess would be "No" =) At the level of quantum foaminess, we probably don't even have spacetime anymore, so the question gets a bit uncertain.

I would argue that a good deal of effort is put into asking why nature has the particular symmetries that it has. I.e., what is the origin of the minus sign in the metric. It's just that these efforts have been fruitless so you don't hear about them.

3. Apr 28, 2008

### rbj

i think this goes along the same lines as this thread and this thread.

essentially the speed of light is the speed of all other instantaneous interaction (gravity, nuclear, as well as EM) of things separated by a vacuum. why it takes that value is really an historical question for why the units we use to measure length and time are what they are. the only salient physics is that this speed of propagation is real, finite, and positive. it could take on any physical value and the rest of reality would be scaled accordingly. i.e., i think that Planck Units are very important in viewing the scaling of things. if all of the dimensionless constants including the ratios of like-dimensioned quantities, specifically how the size of any thing (particle, etc.) compares to the Planck Length, or how any period of time compares to the Planck Time, or how the mass of anything compares to the Planck Mass, then it just doesn't matter what any person's measurement of c is. same for other dimensionful constants like G.

4. Apr 28, 2008

### Phrak

If light traveled at any other speed would there be a measurable difference?

5. Apr 28, 2008

### Staff: Mentor

Along with the previous posts, I think that the important question isn't why the invariant speed has the particular value that it does (because that depends on your units), but why is the invariant speed finite?

6. Apr 29, 2008

### rbj

i have an opinion of a layman (but was informed to me by some pretty heavyweight physicists) that i have already stated too many times.

it seems more credible that the speeds of propagation of the fundamental interactions is finite (and even identical to each other). it's a pretty wild universe that something going on in a galaxy 2 billion lightyears away would affect us at the same time (but with perhaps less magnitude) as it affects things in its own galaxy (as observed by a 3rd party that is equi-distant to both). i guess it's a pretty wild universe as it is.

7. Apr 29, 2008

### neopolitan

A point I was making, perhaps not sufficiently clearly, is that the speed of light is an upper limit. But one could say that light just goes as fast as it can, but this would be a misrepresentation, since light would also go as slow as it can.

At the foamy level of Planck units, there are two possible speeds, zero and c. A discrete particle/wavicle such as a photon has those two choices.

The motion of collection of particles/wavicles, such as ourselves, represents the summation of all the individual motions of constituent particles/wavicles. I think it would be fair to think that the constituent particles/wavicles could interact in manner analogous to (not precisely the same as) the individual molecules of a gas cloud.

So I am wondering if anyone got that point, that all of the elemental particles/wavicles from which we are made would all be moving at c.

Someone made comment on it being interesting that the speed of light is finite. Well, at macro level, the speed is demonstrably finite. But what about at the foamy quantum level? It seems to me that if you look at a grainy universe, then the speed at which a photon shifts from one "quantum box" to the next could effectively be infinite. You could (in effect) have a jerky sort of motion: instantaneous translation to adjacent quantum box, wait one Planck time, instantaneous translation to adjacent quantum box, repeat. Perhaps it is not valid to think of the the instantaneous translation as being equivalent to infinite speed (since it is a dividing by zero issue), but it seems pretty damn close.

cheers,

neopolitan

8. Apr 29, 2008

### Staff: Mentor

I guess then that I don't understand your question. The fact that c is finite has physical relevance. The actual value that it has is a question of your system of units, which is arbitrary.

9. Apr 29, 2008

### DrGreg

I am no expert in quantum theory, but I think you have the wrong picture here. Particles don't "jump from one quantum box from another". The Planck units simply refer to uncertainty in measurement. A measurement can take any value, it needn't be an integer number of Planck units. There is just an error bar on your measurement.

Don't picture a particle as a hard, solid ball, but as an out-of-focus blur that you can't measure very well. You can't be sure whether the blurriness is caused by its position or its speed.

Despite what Wikipedia says, I think that in some circumstances you might be able to measure a particle's position with an accuracy much better than a Planck length, but then you would measure its momentum very poorly. Or vice versa.

Finally, in quantum theory, "instantaneous" speed is inherently difficult to measure. If the speed isn't constant, you'd need to measure both a short distance and a short time accurately, which you can't. It's a lot easier to measure momentum. And there is no limit to momentum, unlike speed which is limited by c.

10. Apr 29, 2008

### neopolitan

Hi DrGreg,

Did you notice my out-of-focus blurred terminology?

"effectively", "(in effect)", "sort of", "particle/wavicle"

I also referred earlier to probability distributions.

But I am tired, so I won't be putting too much effort into a reply. I do wonder what possible meaning momentum has if you divorce it from speed. Are you saying the momentum of a specific known mass is unlimited? That seems odd to me, but I am in the throes of a cold, perhaps there is some subtlety that I am missing here.

cheers,

neopolitan

11. Apr 29, 2008

### rbj

the quote is accurate, Dale, but i don't see a question mark in it. whom are you addressing? (and BTW, i don't think we're disagreeing about anything either.)

12. Apr 29, 2008

### Staff: Mentor

Oops, my mistake. I read your response on a little hand-held device where I cannot see very many lines at a time, and I got confused thinking that your response was from neopolitan. The question I was refering to is the OP's Q.

I don't think we are disagreeing either.

13. Apr 29, 2008

### neopolitan

I said:

rbj said:

Does this indicate some agreement?

I also draw DrGreg's attention to the original post.

cheers,

neopolitan

PS About the momentum thing, the fundamental equations (for SR at least) contain an explicit reference to c. There is no mention of momentum, irrespective of how much easier it might be to measure.

Last edited: Apr 29, 2008
14. May 1, 2008

### DrGreg

Oops, sorry, I didn't read your earlier post in as much detail as I should.

One of the points I was making is that you seem to be suggesting consecutive measurements might be, say, 137, 138, 138, 139 Planck lengths at intervals of 1 Planck time; whereas I am saying they could be 137.23 ± 1, 137.82 ± 1, 138.51 ± 1, 139.12 ± 1, etc.

In quantum theory, the act of measurement actually modifies the thing you are measuring, so any second measurement of the same thing may differ from what it would have been had you not made the first measurement.

So if you were trying to measure speed, you have to measure two distances a short time apart, and the second measurement is distorted by the first. It is indeed possible to get an answer of c (in fact I'm not sure if you might even get a higher value) if you try to measure speed this way, but (as I understand it) that doesn't mean the particle is "really" travelling at that speed, whatever that means, it just means you've unavoidably made a pair of erroneous measurements.

Whereas momentum is something you can measure directly, either by colliding something else (large) into the particle and measuring the large thing's change of momentum, or, bearing in mind wave/particle duality, by measuring frequency which is proportional to momentum.

In SR, momentum is given not by $p = mv$, but $p = \gamma mv$, where $\gamma$ is the Lorentz factor $1 / \sqrt{1 - v^2 / c^2}$, and $m > 0$ is rest mass. As $v \rightarrow c$, $p \rightarrow \infty$. That's what I meant by saying "there is no limit to momentum".

So uncertainty in momentum does not translate directly into uncertainty in speed, partly because of the $\gamma$ factor, partly because there may also be uncertainty over mass.

(I should repeat my warning that I'm no expert in quantum theory, so I stand to be corrected if any of the above isn't quite right.)