How can the speed of light be a dimensionless constant?
This is a quote from the book A first course in general relativity by Schutz:
What we shall now do is adopt a new unit for time, the meter. One meter of time is the time it takes light to travel one meter. The speed of light in these units is $$\begin{align*}\end{align*}$$
$$\begin{align*} c &= \frac{ \text{distance light travels in any given time interval}}{\text{the given time interval}}\\ &= \frac{ \text{1m}}{\text{the time it takes light to travel one meter}}\\ &= \frac{1m}{1m} = 1\\ \end{align*}$$
So if we consistently measure time in meters, then c is not merely 1, it is also dimensionless!
Either Schutz was on crack when he wrote this, or I'm a dope (highly likely) 'cos I can't get my head around this:
The space-time interval between different events at the same location measures time, and between different events at the same time measures space. So they're two completely different physial measurents: One is a time measurement using a clock, the other a space measurement using a ruler. In which case the units of $c$ should be $ms^{-1}$
Does Schutz correctly show how $c$ can be a dimensioness constant?
dimensional-analysis physical-constants
Larry HarsonLarry Harson
$\begingroup$ I personally don't think this is the best way to introduce this. For an alternative, equivalent question and answer(s), see: physics.stackexchange.com/questions/51791/… $\endgroup$ – DJBunk Feb 3 '13 at 17:59
$\begingroup$ @DJBunk that other question is about setting $c=1$. Excluding your answer, all of the others, including the accepted and most upvoted one, makes clear that the units are still there. $\endgroup$ – Larry Harson Feb 3 '13 at 18:38
$\begingroup$ If time is one dimension of space, then naïvely, distance/time is unitless. However, isn't time usually considered to be orthogonal to the other spacial dimensions, and thus throwing away this information probably assumes some kind of non-trivial coordinate transformation? $\endgroup$ – kojiro Feb 3 '13 at 22:41
$\begingroup$ @kojiro: Coordinate transformation does not happen within a coordinate system; it is the transform from one coordinate system to another. All four coordinates, in fact, should have the same dimension, in order for many geometric constructs, like metric tensor to make sense. $\endgroup$ – Siyuan Ren Feb 4 '13 at 2:29
$\begingroup$ @C.R. Forgive my terminology gaffe. What do you call it when you rotate the axes of a graph of a line until the slope is 1? That's what I meant. I was really just saying that you can set any v to 1 by manipulating the axes, whether v has the speed of light or some other speed. $\endgroup$ – kojiro Feb 4 '13 at 12:51
The infinitesimal length interval between two events in spacetime $ds$ is defined by
$$ds^2=c^2 dt^2 - dx^2 - dy^2 - dz^2$$
The creature is dimensionally consistent, because time is multiplied with a speed. You can think of $(t,x,y,z)$ as the four coordinates of spacetime $(x^0,x^1,x^2,x^3)$ and $c$ appears naturally in the equations. However, the usual approach is setting $x^0=ct$ and so all the four coordinates have units of length. The definition of infinitesimal interval is then
$$ds^2=(dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2$$
By doing that, $c$ vanishes from equations. For convenience, $x^0$ is named $t$ and, since it is actually proportional to time, it is called time or the time coordinate, but it is not really time. It is a distance. The original idea of Minkowski was even more bizarre, since he originally did $x^0=ict$, an imaginary (complex) distance.
The quantity that matters is a distance. It is proportional to time, but it is nevertheless a distance. The whole conceptual business is quickly (but not very enlightening) explained by saying "we work in units such that $c=1$" and the beginner is lost...
So, when you see $t$ you must remember that you are calling "time" to a quantity that is actually a distance. This trick affects also other physical quantities, so that you will be calling energy to something that is NOT energy, but energy divided by $c^2$. You will call speed to an adimensional quantity that must be multiplied by c to recover the "real" speed... It is not so unfamiliar to you: a supersonic plane may fly at "match 2.5", which really means 2.5 times the speed of sound.
Look at this list you may derive and check by yourself as an exercise:
Things that happen when you call "time" to the distance $ct$:
What you call LENGTH is still a length.
What you call TIME is a distance.
What you call MASS is still a mass.
What you call SPEED is something adimensional.
What you call ACCELERATION is an acceleration divided by $c^2$
What you call MOMENTUM is a momentum divided by $c$.
What you call ENERGY is energy divided by $c^2$
What you call ELECTRIC CURRENT is an electric current divided by $c$
By looking at this list, you now know how to "undo" the definitions so that, for instance, you have to multiply by $c^2$ when you want to recover the energy from the E that appears in the equations (e.g. the famous Einstein equation relating mass and energy is written $E=m$).
Don't worry, you will get used very soon. If you think this is bizarre, then look at what particle physicists do: They rename things so that not only the speed of light vanishes from the equations, but also the Planck constant and the electron charge, all at once. Don't ask me how to "undo" that...
EDIT: Some users keep asking in the comments so, for the ease of beguinners, here is how to reproduce that list:
Decompose every physical quantity into the basic dimensions length, mass, time. For instance:
Energy=work (dimensionally)=
[Force]x[length]=
[mass]x[acceleration]x[length]=
[mass]x[length]x[length]/([tim‌​e]x[tim‌​e])
Now, take every [time] factor you see and change it into [length]:
Energy=
[mass]x[length]x[length]/([tim‌​e]x[tim‌​e]) turns into
[mass]x[length]x[length]/([length]x[length]) =
[mass]
That is, by pretending that time is a length, we are pretending too that Energy is mass. And so on with the other quantities.
SECOND EDIT, About the metric signature:
One user was puzzled about the sign of the interval $ds^2$. This edit attempts to make this point clear:
Einstein defined in the 1921 Princeton lectures the interval as:
$$ds^2= + c^2 dt^2 - dx^2 - dy^2 - dz^2 $$ this sign convention is referred to as $(+---)$ as a shorthand notation. It is used by many authors, both in the field of Relativity and Particle Physics, like Weinberg, Peskin & Schröder or Zee.
However, there are too very good, canonical books in both fields with the opposite convention $(-+++)$, that is $$ds^2= - c^2 dt^2 + dx^2 + dy^2 + dz^2 $$
For instance Schutz himself, Wald (except in the spinors chapter), MTW or Srednicki. It is also the convention used in Wikipedia.
The first thing you have to look at, when consulting a book, is which convention is the author using, because usually there are sign differences in the same equation when based upon one convention or the other.
Eduardo Guerras ValeraEduardo Guerras Valera
$\begingroup$ Schutz is not wrong, please read the post. And his book is probably the best one to start studying GR seriously. $\endgroup$ – Eduardo Guerras Valera Feb 3 '13 at 19:15
$\begingroup$ @Neo, no my friend, it isn't. Google for "metric signature". Relativists prefer usually (-+++) as Einstein in his lectures, and particle physicists (+---) although it is not a general rule and it is only a matter of taste. Minkowski wanted (++++) and so he had to define $x^0=ict$, so that $(x^0)^2=-c^2 t^2$ $\endgroup$ – Eduardo Guerras Valera Feb 4 '13 at 19:21
$\begingroup$ @Neo,as I said in the last comment, please google for "metric signature," and then you will understand. Or post a new question linking to this one (perhaps "Why the metric in that post has what it seems to me the wrong sign?"), I am sure many users will answer. Good luck! $\endgroup$ – Eduardo Guerras Valera Feb 4 '13 at 21:04
$\begingroup$ Keeping track of the symbols c, G, Planck constant and so on is a pain in the neck, and it is unnecesary. It distracts you from the essentials and obscures the meaning of the equations by loading them unnecesarily... Paying attention to covariant/contravariant indexes and all that stuff demands a lot of atention and is very prone to error, and thus every convention that simplifies your life is a great thing. After you spend some afternoons re-doing a couple of proofs and derivations in GR, believe me you will completely agree with calling your own mother dimensionless if necessary... $\endgroup$ – Eduardo Guerras Valera Feb 7 '13 at 14:54
$\begingroup$ Dear @Larry Harson, I have erased the comment that didn't add much physical information and perhaps might result sort of annoying. But the rest of the comments seem fine to me, since they reinforce what has been said in the main answer to other users that might also have the same doubts as you exposed in your (now erased) comments. However, I think the answer is clear enought as it is and I am not going to add anything more. $\endgroup$ – Eduardo Guerras Valera Feb 9 '13 at 15:02
The other answer are correct, so I'm only hoping to give an analogy to show that the situation is actually not that unusual at all. It is eminently mundane. If the analogy is extremely laboured and obvious, well... that's kind of the point. (Also I'm a little bored.) :)
The laws of physics are, or seem to be to a great deal of accuracy, rotationally invariant. There are no preferred directions in the fundamental laws and you can use the group theory of the rotation group to establish a great many things about the character of physical laws, conserved quantities and so on.
Nevertheless preferred directions arise all the time in everyday experience. "Vertical" is very different from "horizontal," though we know that the distinction is only a local one that makes a great deal of sense in the vicinity of the ground, but not elsewhere in the universe.
Now consider this: sailors measure (or used to) horizontal and vertical distances in different units: nautical miles and fathoms resp. The distinction is entirely practical: the methods convenient for measuring horizontal distances on the sea
judging by landmarks
navigating by the stars
are quite different from those for vertical distances
trailing a rope
And anybody who has fallen overboard will know there is an important difference between separated from safety by a hundred metres vertical versus a hundred metres horizontal.
It is plausible to imagine a race of people who grew up on the water and knew nothing of life on land. Perhaps they would raise up physicists, and the physicists, using the tools available, would measure $x$ and $y$ in nautical miles and $z$ in fathoms, never thinking this was a strange thing to do. It would not necessarily be obvious that horizontal and vertical, nautical miles and fathoms where connected in any significant way. Eventually they would find, perhaps by chance, that a certain "constant of nature" $$ c = 1013\ \frac{\text{fathoms}}{\text{nautical mile}} $$ appeared in all their equations, such as in the invariant quantity: $$ \mathrm{d}s^2 \equiv \mathrm{d}x^2 + \mathrm{d}y^2 + \frac{\mathrm{d}z^2}{c^2} $$
These physicists would only discover the significance of this quantity when, by careful experiments, they determined that there are transformations which mix up all of the $x,y,z$ which are symmetries of the fundamental laws. They would soon understand that the special importance that $z$ once had in their laws was due to the fact they never noticed the symmetries that involve $z$ in a non-trivial way: the overwhelming influence of Earth's gravity hid the true underlying symmetries of the laws and made ignoring the symmetry a decent approximation. They would also come to the conclusion that, at least for the purposes of understanding the fundamental laws, measuring horizontal and vertical distances in different units is nonsense and $c$ is nothing but a nuisance factor. Their articles would often begin with "we use units where $c=1$," and this would confuse all of their students who grew up on boats.
To complete the analogy:
rotations -> Lorentz transformations
rotation group SO(3) -> Lorentz group SO(3,1)
$(x,y),z$ -> $(x,y,z),t$
clocks for measuring $t$, and metre sticks for $x,y,z$
preferred spatial direction due to Earth's gravity -> preferred time direction due to the close alignment of the 4-velocities of all everyday objects (why this alignment exists is an interesting dynamical question, analogous to the question of "why the Earth?")
Finally, if you made it this far: the only reason nobody made a clock which reads in metres is convention. Here is what one would look like:
Michael BrownMichael Brown
1) The best answer here might be a little of both. This kind of question gets at the notion of what time is. You can define a system of measuring time by using light. The time between events is then the distance that light would travel in the duration between those events. Then by definition, the speed of light is $1$ and dimensionless, as we measure time in meters and distance in meters, and light will naturally traverse the same distance in meters as the time we measure between its endpoints (in meters).
2) Of course, historically we did not do this. We took some fixed duration between events and made that a unit of reference--all other durations of time would be referenced to this interval. Hence, we have the second. Here, time and distance are incompatible quantities, and there is no obvious relationship between them. The speed of light takes on some value that must be measured with both our reference rulers and reference clocks.
3) You could, alternatively, still maintain that distinction and just measure distance in light-seconds. Distance and time would still be separate quantities, but you'd measure the speed of light as 1 light-second per second--still a dimensioned quantity, just with numerical value of 1 in such a system of units.
Often, systems (1) and (3) are taken as equivalent. There are many systems of units in (3) that give a numerical value of 1 for the speed of light, so we tend to be flexible and use whatever system is convenient at the time. In numerical relativity, for instance, we often measure time and distance in units of mass (accomplished by setting $G=1$) and leave the conversion to "real" quantities until later.
Does this make $c$ a dimensionless quantity? Not really. The ability to set $c=1$ reflects on the arbitrary nature of the measuring sticks and clocks that we've chosen. It also shows that one can, in principle, do everything with only rulers, using the physical laws of the universe to infer things about time intervals and masses. But since you need to know these laws in order to get a good, consistent set of measurements, it's a point of view that only really works in hindsight.
MuphridMuphrid
$\begingroup$ How far is it? It's about 20 minutes down the freeway. How long till we get there? It's about five miles away now. $\endgroup$ – Jerry Asher Feb 3 '13 at 20:52
Does Schutz correctly show how c can be a dimensionless constant?
The reason why c is dimensionless is that special relativity requires that the units of space and time be interchangeable. That in turn makes it convenient to express both using the same units, such as meters. Doing so causes the speed of light to become $\frac{1 m}{1 m}=1$, which is dimensionless.
That largely answers your question, but if you are interested in more details about why special relativity implies that space and time are interchangeable, this answer about the velocity of time has more details.
Great green gobs of gooey gumdrops, why did someone minus one me for that answer??
It's not an insult against Shultz's textbook if that's your concern. It's just a statement of the flat-out fact: Shultz's explanation is entirely tautological.
Look at it. In the very first sentence Schulz says: "What we shall now do is adopt a new unit for time, the meter." Now come on: If you decide in your very first sentence that you are going to measure time in meters instead of seconds, is it really much of a surprise that from that point forward any kind of velocity, which by definition is a length traveled (measured in length-like meters) in a span of time (measured in time-like meters) is going to end up dimensionless?
So, the text after his first sentence is just noise. What Schulz really should have done was provide some kind of physics-based explanation or intuition as to why he suddenly declared that it is perfectly OK to redefine seconds in terms of meters.
Historically, by far the most important reason why physicists since Einstein feel it is OK to measure time in meters is because special relativity tells us that space and time really are in some sense interchangeable. And while that statement seems "obvious" to sensibilities of modern technological cultures where even people completely uninterested in science are likely to know it, the idea of interchangeable space and time was not clear at all in the 1800s and before.
What Schulz should have said is something more like this:
Since we can represent time as a length-like axis, it is convenient to use the speed of light to create a single definition of the length that applies to both space and time.
An answer like that is more honest to the reader about the lack of profound thinking behind what is, more than anything else, just a handy units choice. Special relativity then provides some real justification for making that choice, but I think it's fair to say that even that is somewhat secondary.
After all, once you've decided to represent time as an actual length-like axis, why not measure it in meters? And once you've decide to measure time in meters, why not make the particularly important and unifying constant $c$ come out to be exactly 1, so that you don't have to keep plugging in long funny-looking decimal numbers into your calculations?
(Down to -2 now? Wow! And not a single explanation of why by anyone. Remarkable, that!)
Continuing with an important point: Once you've made the conscious choice to measure time in meters, a dimensionless velocity constant is more accurately interpreted as the tangent function of an angle, specifically the angle called $\alpha$ in much of the special relativity literature.
The reason why is not hard to see. Take a look at this figure:
While it sounds at times as though special relativity is all about hyperbolic spaces, when it comes to light cones, dimensionless velocities, and angles such as $\alpha$ that are just another way of expressing velocity, it's really all about traditional trigonometry and the sides of right triangles in thoroughly Euclidean spaces. The hyperbolic views come into play only through a deeper analysis of the implications of $c$ being invariant.
I have to point out a bit of conceptual dyslexia that seems entrenched in how SR and GR text represent the concept of dimensionless velocities.
Look at my earlier figure, where I showed how the SR angle $\alpha$ can be used to defined a dimensionless velocity as $tan(\alpha)$.
So, what kind of space is that figure and its right triangle $\triangle{abh}$ using?
Well, Euclidean of course, with a [++] signature. It's all just good old Pythagorean $h^2=a^2+b^2$, and that's exactly what makes the $tan(\alpha)$ definition of dimensionless velocities possible.
But think about that: Isn't time always supposed to have a different sign from space? That is, for the simplified 2D space of the figure you should have either a [-+] or [+-] signature for $t$ and $x$, but not the [++] (simple Pythagorean triangle) of the figure.
The reason it comes out [++] is that velocities are ratios, and ratios are most directly expressed by using two axes of a Euclidean space with a [++] signature.
So, if you like dimensionless velocities based on the $c$-inspired assumption that $1 m = 3.33564095 ns$, it's hard to avoid also the need for a Euclidean space with a [++] signature to express those dimensionless velocities as simple ratios. Once you allow that you get that nice right triangle with sides $a$, $b$, and $h$, angle $\alpha$, and the desired dimensionless velocity $v=tan(\alpha)$.
However, this deviation from a [-+] signature is also why no one ever bothers to give that poor, lonely hypotenuse $h$ in the figure its own name. It's not that you can't find some geometric use for $h$, it's just that $h$ is a Euclidean mix-up of $t$ and $x$. That's a bit of a bit of a no-no in the [-+...] hyperbolic world of SR. So, most descriptions use only the safer $tan(\alpha)$, which relies only on the space and time deltas $a$ and $b$.
A tendency to avoid getting too deep into [++...] Euclidean spaces is I suspect also why no single, isotropic definition of the meter (call it $m_s$) is used to explain the equivalence of $t$ and $m$, as the inquirer suggested in a comment. But it's still a handy idea. By using an isotropic $m_s$ unit of measure for all lengths in spacetime figures, the equivalences become trivial: $1 m_s = 1 m$ when looking at horizontal $xyz$ distances, and $1 m_s = 3.33564095 ns$ when looking at vertical $t$ distances. Such lengths are of course emphatically not Lorentz invariant, but then again, neither are velocities in general.
My view?
I would humbly suggest that it's probably better to be explicit rather than silent about such deviations. Dimensionless velocities and light cones are examples of SR constructs that use [++...] Euclidean spaces to display and define certain relationships, so why not just say so? As long as you stick to the perspective of a single frame $\phi_0$, ratios (e.g. $v$) and lengths (e.g. $h^2=t^2+x^2+y^2+z^2$) actually provide a visually simple and understandable way to express many of the of the geometric transformations that occur to objects when they are viewed from a single rest frame $\phi_0$.
Terry BollingerTerry Bollinger
$\begingroup$ Who's downvoting this answer? $\endgroup$ – Larry Harson Feb 13 '13 at 0:16
$\begingroup$ +1 But I would edit your answer saying that space and time can be both measured using space-time-interval units. Calling them space or time units, and then using these to measure the other is confusing. Other than that I think your answer is great and similar to my thoughts: 3-velocities have dimensions of m/s, 4-velocities are dimensionless. I'm close to accepting this as the correct answer. $\endgroup$ – Larry Harson Feb 13 '13 at 0:22
$\begingroup$ Larry Harson, if I understand you, yes: Calling both by a different name might... work better? Hmm. Dunno if I've seen that, but it would make sense because you can then make all three a matter of perspective: An sti, space-time-interval...? no, because "interval" has a very specific non-Euclidean (hyperbolic) meaning. An stm, spacetime meter? That works... So, from a 4D Euclidean view, everything is in stm units. But within 3D space, "riding the wave" of time so to speak, an stm parallel to time looks like 3.33564095 ns, while any stm perpendicular to time looks like 1 m. $\endgroup$ – Terry Bollinger Feb 13 '13 at 5:49
what that absolutely confusing terminology is trying to say, is that you can switch time units to metric units, and instead of seconds to measure time, you'll use, for instance, a light-meter (the time that it takes light to cover one meter of distance), and it is written in units of meters. If you define time between events operationally in terms of how much distance light will travel in the current frame, then the resulting number is in units of meters, and by definition not just $c$, but any velocity will be adimensional too (since time is now translated to distance units)
lurscherlurscher
c in a vacuum is unique because it is the only speed in the universe that does not differ for all observers, regardless of their inertial frames or acceleration. While it might seem that one can use Schutz's formula to make any arbitrary speed a dimensionless constant, it doesn't work unless the speed does not vary among observers, and c is the only speed with this property.
In a Euclidean and classical mechanical universe, you would be correct to say that space and time are two "completely different physical measurements". But in a relativistic context, space and time are a single continuum, a manifold. Regardless of the coordinates of an event in that continuum, and regardless of the event's path (or world line) through the continuum, the speed of light in a vacuum is constant.
Each event in the space-time continuum is viewed variously by various observers. The reference frames of all the observers around an event are represented by x,y,z,t coordinates corresponding to each observer. Since c is the same for all observers, regardless of their reference frames, c may be represented by 1 for the purpose of computing how an observer's reference frame intersects another observer's reference frame at the event. Representing c as a dimensionless constant makes it easier to reconcile the disagreements among observers, while still showing that the laws of physics are not contravened in each observer's reference frame.
$\begingroup$ There is potential for this post to become a full answer to the question. I can see where you're going with it. But it's easy to see how most readers would not get how this answers the question. The question doesn't ask if any speed can be made a dimensionless constant, it asks how $c$ can be dimensionless because the OP sees time and space as unique and separate dimensions. Expanding on how c being unique is important would turn this into an answer to the question. As it is, I'd say it is insufficient. $\endgroup$ – Jim Jul 22 '15 at 14:56
$\begingroup$ @Jimself: Excellent comment. I hope my edits help. $\endgroup$ – Ernie Jul 22 '15 at 17:56
$\begingroup$ Yes, much better now $\endgroup$ – Jim Jul 22 '15 at 18:17
protected by Qmechanic♦ Sep 29 '13 at 18:03
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