# The reference frame of $c$

I don't have a lot of knowledge of special relativity and associated topics; some of the few things I know are that "all motion is relative" (that is, there is no 'stationary reference frame'), and the speed of light in vacuum ($$c \simeq 3 \cdot 10^8~\mathrm{m~s^{-1}}$$) is the absolute asymptotic speed limit (asymptotic meaning that you can never equal it, only get arbitrarily close). What escapes me is how those concepts work together - to my naive understanding, an object would never move in its own reference frame (and so would never reach $$c$$). What reference frame is $$c$$ measured against? (Is it measured against a reference frame?) Or am I looking at this the wrong way?

• $c$ is the speed that you measure for light propogating in free space. It is also the speed you buddy measures for light, even if you are moving at a large fraction of $c$ relative one another. $c$ is measured against every frame of reference for massive observers. – dmckee Oct 6 '13 at 7:21
• Yes, you are looking at it the wrong way, but insightfully so IMO! The relativity of motion and the constancy of the speed of light do appear to contradict each other, and that's why special relativity needed to be invented. – David H Oct 6 '13 at 7:25
• A very interesting, though non-intuitive fact: The speed of light is same in all reference frames. So no matter how fast you're going, you'll always see light moving at $c \approx 3\times 10^8 m/s$!!! – mikhailcazi Oct 6 '13 at 7:26
• Echoing what everyone has been saying: (i) Yes, a speed measurement is measured relative to a reference frame (RF) of a massive observer (how else would one perform a speed measurement?), and (ii), Yes, to measure the speed of light in vacuum, it doesn't matter which such RF that you use. The value should in every case be $c$. – Qmechanic Oct 6 '13 at 12:07
• And, a consequence of the invariance of the speed of light is that there is no reference frame with relative speed c, no "reference frame of c". An object moving with speed c in one RF, moves with speed c in all RFs or, put another way, there is no reference frame in which the object is at rest. – Alfred Centauri Oct 6 '13 at 13:09

## 3 Answers

Another way of thinking that might be helpful to you is to take heed that $c$ is not primarily the speed of light. It comes indirectly to mean the observed speed by any observer of any massless particle, and because, as far as we know, light is massless, it comes indirectly to mean the speed of light. But, in its most fundamental form, $c$ is only a parameter that happens to have the dimensions of speed. It doesn't primarily refer to a speed: here's how we go about defining it.

Think about the intuitive Galilean addition of velocities. The combination law is linear. So, assuming a linear combination law, there are some basic symmetries and characterisics of this everyday law you might like to think about. The following might look a bit daunting at first but it really is intuitive and we're not talking about at first anything that gainsays everyday Galilean relativity, so I'd urge you to think about applying these ideas to the simple problem where we have three frames: $F_1$, the street, $F_2$ a bus driving along the street and $F_3$ a person walking down the aisle of the moving bus. In the following, let us call the shift from one frame to another, uniformly relatively moving frame a boost:

1. (Linearity) If I transform from frame $F_1$ to a frame $F_2$ moving at a constant speed $v_{1,2}$ in some direction then my distance and time co-ordinates $(x, t)$ are transformed by some $2\times2$ matrix $T(v_{1,2})$, i.e. $X=\left(\begin{array}{c}x\\t\end{array}\right)\mapsto T(v_{1,2}) X$;
2. (Transitivity and Associativity): If I then transform to a third frame $F_3$, one moving at velocity $v_{2,3}$ in the same (original) direction relative to the transformed frame $F_2$ (using the matrix $T(v_{2,3})$, this has to be equivalent to a single transformation $T(v_{1,3})$ from the first to the third frame with some relative velocity $v_{1,3}$. Or, with our "boost" word: a boost combined with another boost in the same direction is still the same as a boost with some relative speed: transformations in the same direction do not change their character by dent of their being composed of boosts or indeed how (our of an infinite number of ways) they might be composed of boosts. If I walk at some speed along a bus itself moving along the road, then my motion should be describable as my moving along the road at some relative speed, forgetting about the bus;
3. (Symmetry of Description) In particular, if frame $F_3$ is moving relative to frame $F_2$ at velocity $-v$, then frames $F_1$ and $F_3$ have to be the same and $T(v) T(-v) = I$ (here $I$ = identity transformation - my running away from you at velocity $v$ should seem the same as your running away from me at the same speed in the opposite direction). This symmetry arises from a basic "homogeneity" (space and time are the "same" in some sense everywhere) and the Copernican notion that there is no special frame. Think carefully about these and you will see that the Galillean transformation fulfills all these intuitive symmetries.

Now for the killer question:

Do the conditions 1 through 3 fully define a Galilean transformation? Or, more mundanely, What is the most general form of the matrix $T(v)$ that fulfils conditions 1 through 3?

It turns out that, not only does the Galilean law $v_{1,2}+v_{2,3} = v_{1,3}$ fulfill all the above axioms, but there are a whole family of possible transformations, each parameterised by a parameter $c$, with the Galilean law being the transformation law we get as $c\to\infty$. Such laws are the Lorentz transformations. See the section "From group postulates" in the "Derivations of the Lorentz transformations" Wikipedia page. Notice how one has NOT assumed that $v_{1,2}+v_{2,3} = v_{1,3}$, aside from in the special case of when $v_{1,2} = -v_{2,3}$. It seems likely that Ignatowsky (see Wikipedia page) was one of the first to understand that one could derive relativity from these assumptions alone in 1911, although Einstein actually mentions the group structure of the Lorentz transformations in his famous 1905 paper "On the Electrodynamics of Moving Bodies".

So imagine we had carefully reviewed Galilean relativity as above but we didn't know anything about special relativity. This might well have been how science might have progressed in the late nineteenth century were it not for the Michelson-Morley experiment. We would now understand that our everyday Galilean looking laws might actually arise from a universe wherein we have this weird $c$ parameter that is not infinite but simply very big: this would still be consistent with our everyday addition of velocity laws with a big enough $c$. At this point, we'd only know the form of the Lorentz transformation and that there were a $c$ parameter (maybe infinite) with dimensions of velocity, so we'd like to come up with some experiment to measure whether our universe had a finite $c$ value. It would not be apparent straight away that this velocity parameter were the velocity of anything in particular or indeed whether it even could be the velocity of anything. But, now we say to ourselves, what if something were going at this velocity relative to us? A simple study of the Lorentz transformation would show us that:

1. The speed of this body $c$ would be measured to be the same in all inertial reference frames. Moreover, so as to enforce this invariance of $c$, there would be a peculiar addition rule for velocities not quite the same as the parallelogram rule;
2. No material object can go faster than $c$ and indeed something can travel at speed $c$ only if it has a rest mass of nought.

So now the Michelson Moreley experiment can be thought of not so much as validating relativity, but rather of showing that light, if made of particles, must be made of massless particles. The Michelson Morely experiment found something whose speed transforms precisely as foreseen by the general Lorentz transformation with a finite $c$, so it would then be a strong hunch (not a proof) that our universe indeed has a finite $c$ and that light is something that travels at this speed. In this context, a positive result of the Michelson Morley experiment (i.e. one showing a dependence of lightspeed on frame) could be thought of either as (i) detecting an aether (medium for light) but equally well (ii) it could be thought of saying that there is no aether but that the light particle has a small mass. Neither result would gainsay our newly found relativity laws.

Of course, many other experiments have since confirmed everything that a relativity grounded on a finite $c$ with $c$ set to the speed of light would foretell, so its quite reasonable to speak of $c$ as the speed of light in relativity. But I hope I have shown that this is not its primary meaning.

Footnote: Unfortunately these ideas don't quite work in more than one dimension. In one dimension, two boosts indeed compose to a boost, but a sequence of boosts in different directions in general compose to one boost together with a rotation. This rotation is called Thompson Precession. So we speak of the Lorentz group as the smallest group of all transformations that can be gotten from a sequence of rotations and boosts, but there is no multidimensional group of boosts, only the "one parameter" one dimensional group of boosts.

• I realize that the question may be more philosophical than physical (due to an inherent lack of possibilities for experimental proof), but do you feel that the same speed of light in vacuum would necessarily apply even in causally-separated regions? – Edouard Jul 17 '18 at 21:46
• I have to add that I'm making the above comment because of Terrell's well-accepted paper "Invisibility of the Lorentz Contraction", which seems to leave that contraction (implied by Special Relativity)asymptotically-imperceptible. – Edouard Jul 17 '18 at 21:57
• I think you meant Thomas, not Thompson precession. – Fizikus Dec 26 '18 at 7:36

It's one of the postulates of the special theory of relativity that the speed of light is $c=299,792,458\,{\rm m/s}$ in all inertial reference frames, regardless of the speed of the source and the speed of the observer.

This would conflict with other principles in Newtonian physics because one may always make light move faster or slower by adding the speed to the source or the observer. The relative speed would be $c\pm v$ where $v$ is the speed of the source or observer.

However, the simple addition of speeds is modified in special relativity due to its mixing of space and time. If two objects are moving against each other by speeds $u,v$ in a reference frame, their relative speed – the speed of the other object as perceived in the reference frame of either of the two objects – isn't $u+v$ but $$V_{\rm total} = \frac{u+v}{1+uv/c^2}$$ You may check that if one of the speeds is $c$, for example $v=c$, we get the relative speed $V_{\rm total}=c$ rather than $c+u$.

• Well SR doesn't postulate numeric value of $c$. – Ruslan Oct 6 '13 at 8:08
• Your opinions are usually helpful, s0, since the Wikipedia article "Variable speed of light" points out the fact that Einstein always believed in variations in c based on gravity (which doesn't figure into SR), I wonder if you'd care to comment on Unzicker's paper at arxiv.org/pdf/1503.06763.pdf. It appears to support such variations, and may not even require the causal separations I mentioned to the other answerer yesterday, although I'm sure the reference frames involved in variations of c would have to differ. – Edouard Jul 18 '18 at 15:45

The principle of relativity states that there is no preferred inertial frame. All frames are equivalent and that the motion between two frames is relative. You can choose any of the frames and call it at rest and the other in motion.

If you conduct an experiment on a platform of a station and also in a train moving with constant velocity, no experiment will be able to tell that the train is in motion. This is the principle of relativity.

As I've said above,

The speed of light is same in all reference frames. So no matter how fast you're going, you'll always see light moving at $c≈3×10^8m/s$!!!

Why is this so?
Maxwell's laws state that speed of an electromagnetic wave is:
$$c= \sqrt{\frac{1}{\mu_m\epsilon_m}}$$

where $\mu_m$ and $\epsilon_m$ are the permeability and permittivity of the medium $m$.

If Maxwell's Laws are applicable in all frames, light must travel with the same speed in all frames in same mediums. Experiments show that this is true.

How?

The faster you go, the more time slows down for you. And your 'length' kinda contracts (in the direction of your motion). Yes, it's weird.

The factor of time dilation and length contraction is same:

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

And:

$$\Delta t' = \gamma\Delta t$$ and $$L' = \frac{L}{\gamma}$$

(There's a long explanation for how)

So that means that for you to reach the speed of $c$, time would have to stop for you! And you'd have to contract to zero length. Which is $\text{(very very)}\times 10^{10}$ difficult (I'm not saying impossible because I don't know for sure if it is). That's why it's asymptotic, as you put it.

I hope you understood how it can be measured from every frame.

Even I've started off recently with relativity, but it's really interesting!

$P.S.$ Crediting a lot of this stuff from this physics textbook which I studied it from: "Concepts of Physics by H.C. Verma".