Why does light travel?

Factual Questions
21

Sorry, but this is false. To a ray of light, no time elapses; from its ‘point of view’ (which doesn’t really exist in SR), it indeed traverses any distance ‘instantaneously’ (scare quotes because a better way to phrase this would be that the elapsed time asymptotically tends to zero the closer you get to the speed of light).

A photon’s clock would be entirely stopped; saying that it does not move through time is a fair, and very common, way of expressing this.

The picture that has light at a 45° angle may be what’s misleading you here. The problem is that the time experienced by something in that diagram is the length of the curve it traces out; but in special relativity, this length isn’t described in the usual, Pythagorean way (as x[sup]2[/sup] + y[sup]2[/sup]), but rather, in the Lorentzian way, such that the ‘length’ of the path of light in that diagram is x[sup]2[/sup] - c[sup]2[/sup]t[sup]2[/sup] = 0.

22

But we are not talking about ‘light’s point of view’. We are talking about an observer’s point of view of the light.

The OP’s question was ‘Why does light travel?’ i.e. Why do we see light travelling at a certain speed.

23

An introduction to this way of thinking can be found here: https://www.relativity.li/en/epstein2/read/c0_en/c1_en

24

Historical question:

From, e.g. Wiki, we see the equation
. . . ε[sub]0[/sub]·μ[sub]0[/sub] = c[sup]-2[/sup]
where, IIUC the three constants ε[sub]0[/sub], μ[sub]0[/sub], c were all known before Maxwell’s result. Is that correct?

My question is: Had anyone before Maxwell noticed the above arithmetic relationship? If so, what did they think of it?

25

Great question (to which I don’t know the answer). The numbers might not have been known to a high enough precision at the time of Maxwell to make that relationship obvious (but again, I don’t know)

26

well, yes they do. Plural theories. So 1905 and 1915.
That is what spacetime is.
How would you describe spacetime?

27

Maxwell probably was the first to notice that relation. Others had dealt with similar subject matter before, but it was Maxwell who first put it all together. Faraday, for instance, certainly knew that electricity and magnetism were related, but I don’t think he had the notion yet that they were two different manifestations of the same fundamental phenomenon.

Besides which, at the time that Maxwell made that realization, I think that his computation from the electromagnetic constants probably was the most precise determination of the speed of light. The really precise direct measurements were done by Michelson (yes, that Michelson), who was after Maxwell.

28

From Wikipedia I see that:

Using observations of Jupiter’s moon Io, the speed of light was determined by Jean Baptiste Joseph Delambre in 1809, with an error of about 1% when expressed in Astronomical Units (a.u.). Jérôme Lalande’s 1771 estimate of km/a.u. was off by about 2%; so (with same-direction errors) IIUC an 1809 estimate of c was about 3% too high.

I don’t know what the error in ε[sub]0[/sub] was at that time, but even a match within 15% or so might be remarked, since the exponent is large.

29

This article The Long Road to Maxwell’s Equations - IEEE Spectrum notes that Maxwell didn’t use ε0 and μ0 in his original calculations (using two other related constants instead), but annoyingly, doesn’t provide other details.

30

The simplest assumption would be that he was using Coulomb’s constant (k[sub]e[/sub] = 1/(4piepsilon[sub]0[/sub]) ), and the corresponding constant for magnetism. But it could have been almost anything. It takes a long time after a scientific discovery for the notation to settle down to a simple, consistent standard, and sometimes the discovery of the simple notation is itself a major discovery.

31

Your objection was to the notion that “things without mass only travel in space”. But that’s (within the limits of the level of discourse we’re at here) an entirely correct way of framing things. And it does explain why photons travel at c for an observer, since what travels at c in one reference frame, travels at c in all of them.

Anyway, we can make what Francis Vaughan said somewhat more rigorous. For this, we need the notion of proper time τ: that is, the time that elapses along a particular particle’s trajectory. From Eq. 3 in that wiki article, we see that
c[sup]2[/sup]dτ[sup]2[/sup] = c[sup]2[/sup]dt[sup]2[/sup] - v[sup]2[/sup]dt[sup]2[/sup],
where dt is (an infinitesimal interval of) coordinate time (the time measured by an external observer’s clock), and v is the ordinary three-velocity of the particle. Provided everything’s sufficiently well behaved, which it is, we can just divide through by dt[sup]2[/sup] to obtain
c[sup]2/sup[sup]2[/sup] = c[sup]2[/sup] - v[sup]2[/sup].

Now, what’s that funny quantity c[sup]2/sup[sup]2[/sup] supposed to be? Well, the quantity cdτ is the interval along the timelike direction of the particle, so cdτ/dt corresponds to the speed with which the particle traverses this interval. It’s the ‘speed of its movement through time’ in this sense. Consequently, if we rearrange the above:
c[sup]2/sup[sup]2[/sup] + v[sup]2[/sup] = c[sup]2[/sup],
we find that the sum of the squares of a particle’s speed through space and its speed through time must equal the squared speed of light. If we imagine an object at rest, i. e. v = 0, we get that dτ = dt, i. e. its coordinate and proper time agree—it experiences no time dilation, as it should be. Consequently, cdτ/dt = c, i. e. it moves through time at light speed.

On the other hand, if we want to have v = c, the only option for doing so is for cdτ/dt to be zero—i. e. to have zero speed through time. That’s also rather intuitive, because if we divide the above equation through by c[sup]2[/sup], we get
(dτ/dt)[sup]2[/sup] = 1 - (v/c)[sup]2[/sup] = (1/γ)[sup]2[/sup],
with γ being the Lorentz factor from above. Now, if v gets close to c, γ grows indefinitely, and thus, dτ/dt must go to zero.

So that a photon moves at c through space because it does not move through time is a perfectly adequate way of putting this.

32

So…the question I am asking is wrong, but I am hoping someone can explain why. The units of velocity through space are m/s, such that it is possible to say that the speed of light is approx. 3x10*9 m/s.What, then are the units of velocity through time? Obviously not seconds/second. Clearly, my oversimplified groping towards understanding is missing something. There must be some relationship (that I am not getting) between the units of movement in space and the units of movement in time so that it s meaningful to say they are covariant depending on whether an object is at rest or moving. What is it? And how is “rest” determined in a universe without grid lines?

33

Why “obviously not”?

34

“Rest” is where you are at. But anyone moving relative to you will not agree on your “grid lines”. This is sort of the whole point.

35

Light in relativity does not experience any “proper time”. Nor is it at rest with respect to some reference frame. The speed of light becomes a conversion factor between distance and time, which you could set to 1 if convenient.

36

The relationship is given by the constancy of the speed of light. Thus, c acts as a conversion factor between space and time. Note that the ‘speed through time’ I defined above is given as cdτ/dt, which has the ‘proper’ units of velocity. But actually, it’s common to work in units where c = 1 (natural units), so that the question of a conversion never really comes up.

There’s no such thing as ‘absolute rest’, of course. ‘Rest’ is defined with respect to a certain point of view—a reference frame. Everything that’s at rest with respect to you, will ‘move through time’ at the same speed (i. e., c) as you, as shown by the fact that your clocks agree. Everything moving with respect to you will consequently age more slowly.

Now, the weird thing about this is that it’s symmetric: picture two spaceships meeting in the void between the stars. To the passengers of each, their spaceship (provided nothing is being accelerated) will be at rest, and the other’s will move. Consequently, each ship’s crew will maintain that the other ship’s clocks are ticking more slowly! Isn’t that blatantly inconsistent?

Well, it turns out, it’s not: in order to actually compare clocks, at least one ship must turn around, to meet the other. And to do so, it must change reference frames—and then, the symmetry is broken: the crew of the ship that’s turned around will have experienced less time between the two meetings. This is the famous twin ‘paradox’.

However, maybe the answer given by ftg above is actually more appropriate to this question: Maxwell’s equations govern the dynamics of electromagnetic fields; light is an electromagnetic wave; and the only wave-solutions of Maxwell’s equations propagate with a speed of c. In particular, there is no ‘stationary’ oscillating solution—hence, light must always move at the speed of light.

In fact, this was seen early on as a problem of the theory: an observer moving with a certain speed v in the same direction as a beam of light, in a Galilean relativistic universe, ought to see the beam of light moving at c - v; and indeed, moving at c, ought not to see it move at all. But electrodynamics simply doesn’t provide such cases—which was indeed part of what inspired Einstein to formulate his notion of relativity. In a sense, it had been believed that Maxwellian electrodynamics was appropriate only in the rest frame of an observer, but Einstein proposed that we consider it to hold in every rest frame—and special relativity basically falls out of that. (So in a way, the answer from SR and the answer from EM aren’t different at all!)

37

Probably because seconds are not the unit of measurement of velocity, which is “an object’s speed and direction of motion” and is measured by meters per second. If I read the cite right, my fuzzy brain gives Noel Prosequi’s answer as (meters per second) per second, the rate of velocity’s change over a given period.

38

I was reasoning by analogy with other measures. Metres per metre as a metric of something is useless because the answer is always 1. Kilograms per kilogram will always be 1. Measuring something by reference to itself seemed to be trivially unproductive to my mind. If you use the same units in the numerator and the denominator, they cancel out.

This is, of course, in the absence of odd circumstances such as measuring the rate of evaporation of, say, water, which (at least in principle if not reality) might evaporate in a way dependent upon its intitial mass, so one might say that a body of water loses .5kg/kg of original mass per hour. But that is not a metric of “pure” kg/kg; rather it is kg/kg-hour.

And I suppose it is meaningful to say two seconds of my time pass for every second of yours for some varying frames of reference. But that is not (intuitively) the sense in which my question arises. But as I write, I can imagine that moving through time at seconds per second might be meaningful if I am comparing my own proper time with some other frame of reference’s time. Is that what is happening? Does my question about what are the units of motion through time resolve itself in this way?

But of course, I might be completely wrong in the deeply weird world of relativity.

39

But does light really travel?

From our perspective it does.

But from a photon’s perspective the universe is infinitely thin and no time passes. To the photon crossing your room is the same as crossing the universe. Travel (as used here) is, after all, distance/time so if there is no time spent moving or distance moved then…is it traveling?

40

The speed of light is, as mentioned, the conversion factor that changes your measure of moving through time at (eg) 1 second per second - it could be less due to time dilation - and converts it to metres per second. Your total speed through space+time always adds up to the speed of light.

As for a photon’s perspective, you cannot catch up with it no matter how fast you go, so maybe there is no “photon’s perspective” in that sense.