Is the speed of light in vacuum constant or does the math just happen to work out?

Question

My apologies if my question is really idiotic, but I ask sincerely because I want to learn. Based on this question and lots of other places on the web, this topic seems to be really confusing.

Let's say for argument's sake that light is a speeding car. We know that it is travelling at 30m/s because we have confirmed this experimentally and then defined it as a constant. We know that it has travelled 300m. So we calculate time as:

speed = distance / time
time = distance / speed
time = 300m / 30m/s
time = 10s

If we have a stopwatch and time this to actually be 12s for some fringe case around a huge gravity well or something, we would conclude that the speed was actually not 30m/s, but rather 25m/s for that case

speed = distance / time
speed = 300m / 25m/s
speed = 12s

We would think that the gravity well affected the speed. With the speed of light, we don't do that. Instead, we seem to have done the following to explain this discrepancy:

Since the speed and the distance is fixed, the time must be variable. The gravity well must have distorted time itself, so that the time observed from the stopwatch was faster relative to the time "observed" by the photons. This seems to be the gist of Special Relativity. The math still works and you can make very accurate predictions, but the underlying concept is very different from what we intuitively understand.

To further confuse things, the meter is defined as the distance travelled by light in a vacuum for a certain amount of time. By defining the meter like this, we can logically never get a discrepancy because one of the variables rely on there never being a discrepancy. I don't understand why we would ever change the definition of a meter to be like this because it just makes discussions on the topic way more abstract.

So, in the following equation:

speed = distance / time

If we assume time to be constant rather than speed (Which is much more intuitive), would the math still work? Am I misunderstanding the whole thing?

Answer 1

With respect to your question, the immediate thing you need to clarify is: constant with respect to what?

How SR answers that question

The speed of light is usually held to be constant with respect to reference frames. In other words, if we're both at the same place in outer space, but you're passing by me in your spaceship, then every photon in either of our coordinates is moving at the same speed $c$.

This causes a lot of counterintuitive effects, for example, it sets up a real-life Zeno paradox where nobody can travel faster than the speed of light. Right now our particle accelerators are strongly based on this assumption: we dump gobs and gobs of energy into a proton going around a circular track with a constant magnetic field; if the proton didn't "top out" at a certain speed then everything we know about electromagnetism says that the magnetic field would need to be increasing, not constant, as we dump energy into it: if there were no "cosmic speed limit" the protons should smash through the walls, but there is, so they stay on the track. It's become an everyday engineering assumption for us physicists; that's "just how it is."

Other principles of relativity are also important in the same way. So, we see these muons that are created by cosmic rays in the upper atmosphere. Muons are a sort of heavy radioactive electron; after a certain half-life they decay into an electron plus some light. Muons, like electrons and protons, also seem to top out at the speed of light. And if you work this out, half life times the speed of light, you get a "half-life distance" of 450 meters. So by the time they pass through the 10km or whatever of atmosphere, we're talking 20 half lives have passed, we should see $(1/2)^{20}$ of them or one in two million of the muons created, right? But that's not what we see. Muons created in the upper atmosphere are seen on Earth with much greater frequency, because due to their "length contraction", they measure the atmosphere as much thinner than we do: so they traverse the shorter distance in only a couple half-lifes, if that. Equivalently, you can say that we see them because we think their clocks are running slow by the same factor: they travel the full 10km but their effective half life is much longer. Either way we do the math, we get a much larger number. And on the surface we see something like ten thousand of them per square meter per minute.

So: given that everyone who is moving relative to me at this present place agrees on the speed of light, why not use it as a universal conversion factor from locally-elapsed-time into local-distance?

How gravity interacts with that

You ask about a gravity well, but really this is a very special case: in general relativity we physicists play a trick that's a lot like the trick of assuming that your office is a 2D flat space when it's secretly located on the big round sphere that is Earth: every point in on the sphere gets a little "neighborhood" where things are approximately flat and you can just use Cartesian coordinates. In general relativity, each point in spacetime gets a little neighborhood where light moves at the speed $c$ in all directions for all reference frames; it's just that the space has an overall curvature so that if you "step back" from spacetime to see the global picture, you can see, for example, photons failing to leave a black hole's event horizon. It's very similar to the strange phenomenon that, say, a Muslim in the US, which would generally regard Mecca as east-southeast of it (local coordinates), will actually pray to Mecca by the geodesic path, which there points north-northeast (global perspective).

So if you're just "in a gravity well", you don't see the associated effects; there the spacetime is "locally flat." It's just that as you get further and further away from your local vicinity you see larger and larger deviations from those flatness expectations.

Because of these deviations, however, stopwatches from outside the gravity well and inside the gravity well see each other as going out of sync. And that is the real problem which we have to face. You want time to be constant? What are you going to do about the known effect that clocks in orbit around the Earth tick faster than the same-mechanism clocks on the surface of the Earth?

It's not that you're wrong, but...

I'm not saying that you can't get these effects with a "time is objective" theory like Newtonian mechanics. It is in the nature of physical theories that they are Turing complete and can therefore model anything with a complex enough model. Since a harmonic oscillator with mass $m$ and spring constant $k$ has period $T = 2\pi \sqrt{m/k}$ you might for example model the mass as instead decreasing with distance from the center of the planet, $m = m_0 / [1 - 2 G M/(rc^2)]$, or maybe you model springs as getting stiffer as they leave a gravitational field, or so. You also need to increase mass with speed, $m = m_0 / \sqrt{1 - v^2/c^2}$, to get the muons to "internally oscillate" slower when they're flying through the atmosphere. You can totally make these sorts of corrections until your model matches experiment.

But we have a physical theory, called special relativity, which models all of these things pretty simply, so most physicists will use that and will teach that to you. That's really the gist of it. The models are simpler when you just pay up-front the theory cost of thinking relativistically rather than conventionally. It's because general relativity is phrased in this "locally relativistic" way, for example, that we've been able to understand why Mercury isn't tidally locked 1:1 to the Sun like the Moon is to the Earth, but instead has a 3:2 spin-orbit resonance. You can get it with some really complicated Newtonian model, but why not do it the easier way?

Even if it seems a little bizarre: special relativity stems from the aggregate sum of a really simple little effect: if you accelerate a small velocity $u$ relative to me, then we start to disagree on whether far-away clocks are in sync, so that a clock at distance $L$ in the direction of your motion, which you think is in sync with your clock here, will look to me "forward" or "behind" by an amount $u L / c^2$. In addition to being describable as an aggregate sum of these simple little "sync disagreements", we can prove that the mathematics is totally, 100% consistent. And, it suggests freebies like $E^2 = m^2 c^4 + p^2 c^2$ which unify the things we know about light waves $(m = 0, E = p c)$ with everything we know about matter (in some rest frame $p = 0$ and $E = m c^2$, with deviations from that rest frame $E \approx m c^2 + \frac 12 m v^2$). So physicists pay that little cost of "little accelerations create little sync disagreements which can add up to length contractions and time dilations", in order to make all of their models simpler in the long run.

Answer 2

Is the speed of light constant or does the math just happen to work out?

None of the above. It's a tautology. What happens is that instead of having just one car, you count 9192631770 cars passing you by. See the defiition of the second which involves microwaves passing you by. Then you declare that a second has elapsed. If those cars are going slower, your second is bigger. Then whatever it is, you use that second to define the metre as the length of the path travelled by light in vacuum during a time interval of 1/299,792,458th of a second.

Note that this means you will always say the speed of cars is always 299,792,458 m/s. See http://arxiv.org/abs/0705.4507.

Answer 3

Let's say for argument's sake that light is a speeding car.

That's quite a silly analogy, because a car is generally thought of as some thing which can be observed while it's driving along on the road (e.g. being observed by landmarks such as delineators along the road, or by other cars), and which may observe them in turn.

A similarity to light is merely, that the (driving) car, too, may be attributed a value of (average) speed with respect to suitable participants (such as a set of delineators), for its passage from one such delineator to the next, for instance.

We know that it has travelled 300 m. So we calculate time as: speed = distance / time

Here we should more correctly write

speed = distance / duration;

because duration is the measure of the time (of some particular participant, from some particular initial indication of that participant, until some particular final indication of that participant).

It is important to note:
We attribute a value of distance to a suitable pair of participants, such as delineators $A$ and $B$ only if they were and remained at rest to each other.
As a consequence, $A$ and $B$ can determine which indication of one had been simultaneous to any particular indication of the other; and they can agree on the relevant value of duration in the trial under consideration (presuming that the car ran from $A$ to $B$):

• $A$'s duration from $A$'s indication of the passage of the car until $B$'s indication simultaneous to $B$'s indication of the passage of the car is equal to

• $B$'s duration from $B$'s indication simultaneous to $A$'s indication of the passage of the car until $B$'s indication of the passage of the car.

Consequently, the (average) speed of the car with respect to delineators $A$ and $B$ can be unambiguously determined (by the formula you know already); and likewise the speed of light (or more precisely: the (average) signal front speed, a.k.a. the "speed of light in vacuum") with respect to delineators $A$ and $B$ could be unambiguously evaluated.

Another consequence of delineators $A$ and $B$ being and remaining at rest with respect to each other is that their ping durations between each other are equal and constant; thus characterizing $A$ and $B$ as a system. Accordingly we define the distance of $A$ and $B$ with respect to each other (which is relevant for the trial under consideration), in the first place, in terms of their characteristic ping duration; as

distance[ A, B ] := c/2 ping_duration_AB,

where, based on the subsequent evaluation of speed described above, the symbol c turns out to be identified as the signal front speed (speed of light in vacuum).

Now:

a huge gravity well or something

This seems to refer to a case in which the relevant participants (say "the top and the bottom of a tower, $J$ and $K$") are not (strictly) at rest with respect to each other, but merely rigid with respect to each other;
specificly, a case in which $J$'s ping duration (wrt. $K$) and $K$'s ping duration (wrt. $J$) are separately constant but not equal.

In fact, $J$'s constant ping duration wrt. $K$ (i.e. always from having indicated a signal until having observed $K$'s reflection of that signal) was larger than $K$'s constant ping duration wrt. $J$ (i.e. always from having indicated a signal until having observed $J$'s reflection of that signal); that's why $J$ was referred to above as "top", and $K$ as "bottom".

The consequence is:
We cannot attribute a (any one) value of distance to the system comprised of $J$ and $K$;
and (therefore) we cannot attribute a (any one) value of (average) speed, with respect to $J$ and $K$, for instance to a car that might have been "dropped from $J$ to $K$";
nor, likewise, could we attribute a (any one) value of (average) speed to the signal front of a signal which had been exchanged between $J$ and $K$.

(Of course we might resort to certain approximations, considering the limit in which the ratio between $J$'s ping duration wrt. $K$ and $K$'s ping duration wrt. $J$ approached the value $1$; but I believe that wasn't the point of your question.)