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This question already has an answer here:

What's so special about the speed of light?

Why do many equations in physics include the speed of light in vacuum $c$?

Why do so many thing depend upon it?

Why can't it be the speed of sound?

Or as a matter of fact any other variable?

For example, in mass-energy equivalence $E=mc^2$, why the speed of light $c$?

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marked as duplicate by Danu, Ben Crowell, JamalS, Kyle Kanos, David Z Oct 28 '14 at 1:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/80365/50583 $\endgroup$ – ACuriousMind Oct 27 '14 at 20:50
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    $\begingroup$ It's a duplicate down to the title of the question! $\endgroup$ – David Hammen Oct 27 '14 at 20:57
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    $\begingroup$ I actually think it is not a duplicate, despite the identical title. $\endgroup$ – user10851 Oct 27 '14 at 20:58
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I'd like to add to Chris White's excellent answer and summarise things thus:

$c$ is a number that parameterises the family of all possible linear transformations that follows from Galileo's relativity with the assumption of absolute time relaxed.

If you do Galileo's relativity with the assumption of an absolute time (that two relatively moving inertial observers will measure the same time interval between two given events), you can derive the intuitive vector addition of velocities rule to transform between relatively moving inertial frames.

Indeed, with an assumption of absolute time, this intuitive addition rule is the unique linear transformation rule in keeping with Galileo's notion of relativity, most wonderfully, poetically (and scientifically rock-solid-accurately) described in his allegory of Salviati's Ship.

However, if we relax the assumption of absolute time, and this was Einstein's bold and uniquely Einstein step, then derive all the possible linear transformations consistent with Galileo's postulates with the relaxed time assumption, you find that there are now a whole family of possible linear transformations, each characterised by a parameter $c$ with dimensions of speed. Galilean relativity does indeed belong to this family. It is the family member with $c=\infty$ (if we think of the parameter $c$ living on a compactified real line).

The deeply characteristic thing about the value $c$ for any member of the family of transformation laws is that if anything travels at this speed, then it will be observed to travel at this speed $c$ in all inertial frames.

It then follows that for any "universe", i.e. set of observers which could in theory possibly compare experimental results, then the particular value of $c$ characterising Galileo-consistent transformation laws must be unqiue for that "universe".

So now, it becomes a wholly experimental question:

What value of $c$ characterises transformations between inertial frames in our universe? Is it even finite?

So now, experimentally, we must look for something - anything- whose speed is the measured to be the same for all inertial observers. If we find anything like this, then we know that a finite $c$ characterises transformations between inertial observers in our universe, and its speed is the special parameter $c$.

You know of course that we did indeed find something with a speed that transforms between inertial observers in this very special way: it was the famous Michelson-Morley experiment, and light was that thing. But it turns out that any massless (i.e. with rest mass of nought) thing must be always observed to travel with this speed relative to any inertial observer. So it is also the speed of, for example, the massless graviton, if indeed this particle turns out to be real. For many years neutrinos were thought to be massless and indeed we observe them to travel at $c$ to within most experimental error bounds. We know that they must have mass indirectly through the observation of Flavour Oscillation. But we don't yet have a definite measurement for their nonzero rest mass.

This value of $c$ also appears, as discussed in Chris White's answer in the generalised $m^2 c^4 = E^2-p^2 c^2$, thus answering your last question. An intuitive way of getting to the $E=m\,c^2$ special case is to use the method that Einstein used in his 1905 paper (the one AFTER the big one):

A. Einstein, "Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?", Ann. der Phys. 18:639,1905 English translation "Does the Inertia of a Body Depend upon its Energy-Content?" is here


More on the Lorentz Group

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 $T(v_{1,3}) = T(v_{2,3}) \circ T(v_{1,2})$ 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 BIG 1905 paper "Zur Elektrodynamik bewegter Körper", Ann. der Phys. 18:891,1905 (english translation "On the Electrodynamics of Moving Bodies" is here).

Things get a bit more complicated with three-dimensional velocities: there is no Lie group of three dimensional boosts. This is a mathematical fact, not a physical one, so it's not simply a question that no-one has yet demonstrated such a group. It can be proven to be impossible. So when we make our bus problem three dimensional, we get a group that includes both boosts and rotations. The composition of two three dimensional boosts does not in general give a boost, but a boost composed with a rotation. This rotation begets the phenomenon of Thomas Precession (see wiki page of same name). This can also be thought of in terms of the "polar decomposition" that allows any square matrix to be written as a product of an orthogonal and upper triangular matrix (itself a special case of the Iwasawa decomposition of Lie groups).

The smallest Lie group containing three dimensional boosts is the famous Lorentz group, which I discuss in Example 1.10 "General Classical Groups and The Proper Lorentz Group $SO^+(1,3)$" on the page "Some Examples of Connected Lie Groups" on my website. There are also the Wikipedia references on the Lorentz group Chris White gave you; also see the Wikipedia page "Derivations Of The Lorentz Transformations" to see whether there is an explanation there that grabs you.

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  • $\begingroup$ Thanx for the answer, the link to the original papers really helped! $\endgroup$ – Investor Oct 28 '14 at 1:49
  • $\begingroup$ @Investor You're welcome. Thanks for the opportunity to bang on about Galileo! He's the boss-man of physics, IMO! Not that I don't wonder at Einstein, but I think history a few hundred years hence will look back and say relativity was Galileo's gig, and locality was Einstein's. Both people had the boldness to challenge deeply entrenched ways of thinking. $\endgroup$ – WetSavannaAnimal Oct 28 '14 at 4:09
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Long ago, English speaking sailors measured horizontal distances in units of nautical miles but depths in units of fathoms. The distance in fathoms to some point $z$ fathoms deep and $x$ nautical miles along the surface is $d^2 = 1012.68591^2 x^2 + z^2$. There's nothing physical to that factor of 1012.68591. It's solely a result of using inconsistent units for horizontal distance and depth.

A similar concept applies to the speed of light. In one sense, the only thing special about the speed of light is that it shows that the units we use are inconsistent. Just as horizontal distance on the ocean and depth into the ocean are different aspects of the same thing (distance), the special theory of relativity shows that passage of time and distance are different aspects of the same thing.

To illustrate, very short-lived muons that form in the upper atmosphere "see" the Earth as rushing toward them at a relativistic speed. From the perspective of those muons, length contraction appears to foreshorten the distance to the surface of the Earth. Some muons live to make it to the surface because of this length-contracted height of the atmosphere. From the perspective of an observer on the Earth's surface, it's the muons that are moving at relativistic speeds, and hence their half-life is time dilated. Which is right? Both are. Time and distance are different aspects of the same thing.

A consistent set of units would give the speed of light a numerical value of one. Using natural units, the proper distance between two points in space-time is given by $$ds^2 = dx^2 + dy^2 + dz^2 - dt^2$$ The speed of light needs to be taken into account in a system in which time and spatial distance have different units: $$ds^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2$$ The key difference between the above formulae and the Pythagorean formula we typically use to calculate distance is that time is subtracted rather than added. This sign change reflects that while the geometry of three dimensional space is Euclidean, the geometry of space-time is Lorentzian.

The relation between mass, momentum, and energy is simple and natural in a system where the speed of light is one: $$E^2 = p^2 + m^2$$ This nicely shows that momentum, mass, and energy are different aspects of the same thing. When you insist on using a non-unitary value for the speed of light the above becomes the ungainly $$E^2 = p^2c^2 + (mc^2)^2$$ Note that in the case $p=0$ (i.e., a stationary object) the above reduces to $E = mc^2$.


Update

In another sense, whether we give the speed of light units of 1, 299792458 m/s, or 1.8026175×1012 furlongs per fortnight, there is something very special about the speed of light. One of the key assumptions across all of physics is that space-time is a manifold, a differential structure that locally looks Newtonian. The geometric structure of a space-time manifold dictates that there will be one speed that is the same to all observers. Newtonian mechanics results if this universal speed is infinite. A Lorentzian space-time results if this universal speed is finite. That the speed of light appears to be this universal speed that is the same to all observers gives credence to the conclusion that the universe is not Newtonian. This was been tested many, many times, starting with the Michelson-Morley experiment, before the powers-that-be who decide what a second, meter, etc. are decided that the speed of light should be made a defined constant.

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The quantity $c$ is a very fundamental constant related to space and time. It is largely independent of the existence of matter or light or any other substance. It is probably more ontologically appropriate (if not pedagogically appropriate) to define it as the number that makes Lorentz transformations between reference frames work.

The quantity $c$ would retain this special place even if nothing actually traveled at that speed. It just so happens, though, that light in vacuum does travel at exactly $c$, and so we call $c$ "the speed of light" out of convenience.

If you expected a speed of sound to be special, at the very least you would have to justify why some particular material's sound speed is more important than others. After all, the speed of sound in air is very different from the speed of sound in a solid block of metal.

Moreover, due to the fundamental nature of $c$ mentioned in the first paragraph, this "speed" is entirely independent of who is measuring it. This property is not shared by any other velocity you can think of: If some object is moving at velocity $v \neq c$ relative to you, then if you start moving all of a sudden the object will appear to move at velocity $v' \neq v$. However, all observers, no matter how they're moving relative to one another, will always agree on the value of $c$.

As for the mass-energy equivalence, this is really fully formulated as $$ (mc^2)^2 = E^2 - (pc)^2, $$ where $p$ is the magnitude of the object's 3-momentum. This equation shows that there is some invariant quantity (the left-hand side) that you get by subtracting the squares of two things (a scalar and a magnitude of a 3-vector). This comes from the fact that in spacetime, the proper time $\Delta \tau$ between events with separation in time $\Delta t$ and separation in space $\Delta \vec{x}$ is given by $$ (c \Delta \tau)^2 = (c \Delta t)^2 - (\Delta \vec{x})^2. $$ The $c$ that relates space and time to an invariant interval is the same one that relates momentum and energy to an invariant mass.

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  • $\begingroup$ Thank you! Also a very complete answer, much appreciated. $\endgroup$ – Investor Oct 27 '14 at 21:05
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    $\begingroup$ Fantastic answer. "if not pedagogically appropriate": I'm not a teacher, but somehow I think it would be pedagogically appropriate to introduce in this way. I am so often asked by deeply mystified teenagers "what on Earth does the speed of light have to do with it?". I recall being nearly driven mad myself as a teenager by this question. You'd be more qualified to talk about this (do you teach?) and I'm guessing that your experience is that the more abstract approach doesn't fit with everybody - is this what you find? $\endgroup$ – WetSavannaAnimal Oct 28 '14 at 1:14
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    $\begingroup$ @WetSavannaAnimal I haven't completely settled on my opinion, perhaps because I've only taught at the pre- or post-SR level. I do love the abstract approach like in your answer, especially since I think the main alternative of teaching relativity via "paradoxes" is counterproductive (and often just a form of showing off to the unlearned - "Look how weird physics is, but I understand it!"). On the other hand, there are all those useful thought experiments that rely on bouncing light back and forth between mirrors. $\endgroup$ – user10851 Oct 28 '14 at 2:47
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This is a very complicated question and a complete answer would be very deep and very, very long.

Much of it stems from the fact that the speed of light is independent of the frame of reference from which it is being observed. In this sense, it is one of few "universal constants" - that is, quantities which do not depend on the observer. Since it has units of distance over time, it acts as a universal conversion factor between the two quantities. The speed of light is constant regardless of the observer, so it is the only reasonable way to convert between quantities of distance and time. It makes sense, then, that it would show up in problems that convert between these quantities.

In the example of $E = mc^2$ that you gave, for example: energy has units of $\frac{\text{mass}*\text{distance}^2}{\text{time}^2}$. If energy is to be related to mass, there is only one reasonable way to do it -- $c^2$ is constant for all observers and has units of $\frac{\text{distance}^2}{\text{time}^2}$, so the units only match up if we have $E = m c^2.$

This is very much an incomplete argument, but it gives you an idea of why the speed of light shows up so much in so many equations - because it is a universal constant for all observers, so any equation which should hold true regardless of the observer may reasonably contain the speed of light.

Another way to look at it is that $c$ isn't really the "speed of light", it's just a speed limit for the universe. The universe is built in such a way that nothing can move faster than some predetermined speed $c$. Since light has no mass, it makes sense that it would move through a vacuum at the fastest possible speed. This is why other things that propagate without mass (e.g. gravitational waves) also travel at this maximum speed $c$. So the maximum speed limit for the universe is a really important quantity - something we would expect to show up all the time.

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