# Units in Special Relativity

I have several small questions about units in special relativity:

1. I've read somewhere that $$c=1$$ can be interpreted as lightyears being out unit of length, and years being our unit of time. But that would yield to $$c=1$$ lightyear/year, instead of just a dimensionless $$1$$. Does this mean that this interpretation is wrong? Otherwise, if I have to assume that $$c$$ is a dimensionless $$1$$, I have the following questions...

2. First, I don't really get why we talk about a $$ct$$-axis instead of a $$t$$-axis. Is the reason for this that $$c$$ is constant, so instead of talking about a $$t$$-axis, we might as well 'scale it up' with $$c$$, by means of speech? And does it matter how we set $$c$$, with or without dimensions? I would guess no, because it doesn't matter if we're dealing with time or space, they seem to be the same, somehow...

3. I am familiar with the notion of spacetime, but I don't get why we didn't introduce a new unit for the coordinates of events... Spacetime is supposed to be a blend of time and space, so why don't we get rid of both notions in our calculations, and work with a new unit? Or could someone elaborate on a justification why we choose to write time in terms of length?

I'm posting these questions simultaneously, because they overlap a lot.

I do understand snippets, like:

In a Newtonian world, space and time are considered fundamentally different dimensions, that don't interact with each other (they don't have an influence on each other).

However, as we've seen in the Lorentz transformations, the position of a particle in a different reference frame, will also depend on its time in the original reference frame. What we see is that space and time are interwoven, and as they depend on each other, we can express one — in our case $$t$$ — in terms of the other. I can almost grasp this, but not entirely. Does anyone know an example of a classical situation, where the units on the $$x$$-axis and $$y$$-axis are dependent on each other, which makes it possible for us to write the variable on the $$y$$-asis as a quantity that belongs to the x-axis?

I've written a lot of my thoughts down. I'm not expecting all my questions to be answered, but I'm just hoping that the more I've written, the easier it will be for someone to pinpoint the main problem I'm having, and from there on I will try to continue.

• As I explain here, deriving the classical limit of special relativity while using c = 1 units is quite instructive. You see that you need to rescale the variables in a proper way and consider the right scaling limit. – Count Iblis Jan 9 '17 at 1:01
• This question is far too broad. It should be split into several. – Dale May 27 '19 at 0:39

Well I am going to come in here. I have read (and written) a lot around this subject and have pondered it for a long time. I think I can comment both on what I find makes sense, and also on what are the currents of thinking in the physics community.

1. Is it "$$c=1$$" or "$$c = 1$$ distance-unit per time-unit"?

This is a good question and you are right to ask it. There is no consensus on the answer! Some people think that space and time are so alike that it is just bizarre to have different units for them, as it would be to suggest we should measure distance north-south in, say, inches, and distance east-west in metres. So they would say that once we have understood this then the value is $$c=1$$ and it is dimensionless.

Other people prefer to say that time and space are not quite identical, because, for example, a timelike line cannot be confused for a spacelike line, nor turned into one by Lorentz transformation. Conservation laws, such as conservation of energy and momentum, make a highly insightful appeal to the difference between timelike and spacelike directions. So these people would say that it is useful to adopt units in which the speed of light evaluates to 1 distance-unit per time-unit, and it is not a dimensionless quantity.

The reason that there is not complete consensus is that there is no objective principle which asserts which is the correct or better choice. This is because the use of units in physics is always somewhat subject to human convention. When we say, of a mass for example, "it is 12 kg" then ultimately we are saying that the mass is 12 times larger than some other mass (one we have to define another way, or just resort to pointing to a physical object). I think the concept of physical dimensions is tremendously useful in analysis and in physical understanding, so I would not want to set it aside. For this reason I prefer to say that $$c$$ (and velocities more generally) are dimensional not dimensionless quantities. If we take light-seconds as our unit of distance, and seconds as our unit of time, then the value of $$c$$ is one distance-unit per time-unit.

1. It is convenient to scale the time axis up by a factor which makes the diagram convenient when speeds of order $$c$$ are involved. I don't mind whether it is called $$t$$-axis or $$ct$$-axis; that doesn't matter.

Note that although time and space are parts of a single whole called spacetime, it is not quite right to say they are the same as one another. The future and past lightcone for any event is an objective idea, which separates spacetime into regions timelike-separated from that event, and another region spacelike-separated from it (and the boundary between them which is null). But nor are time and space completely distinct as they are in Newtonian physics.

1. In special relativity it is convenient, and it makes a lot of sense, to label coordinates in spacetime with values that all have the same physical dimensions. One can choose a temporal value or a spatial one (or some other possibility); it doesn't matter. Spatial ones are the standard choice.

In might interest you to know that in general relativity the mathematical framework adopted at the outset handles issues like this automatically as one "turns the crank" of tensor analysis, and as a result it is quite common to use sets of coordinates which do not all have the same physical dimensions. One could use $$t,r,\theta,\phi$$ for example. The use of the metric ensures that factors of $$c$$ are taken care of.

Further remark on units

After an exchange which you can see in the comments, I decided to add another remark. As I have said, the use of units in physics is to some extent a matter of human convention, but clearly some people judge that this does not extend to this issue in relativity, and they want to argue that only one formulation is correct. Here are the two formulations side by side. In the left column is the point of view where $$c$$ (and speed more generally) is taken as dimensionless. This can be done by reducing the number of physical dimensions employed to do physics. In the right column is the point of view where speed is dimensional.

(ls = light-second, s = second, m = metre) $$\begin{array}{ll} c=1 & c = 1\; {\rm ls/s} \\ 3 \times 10^8 \,{\rm m} = 1 \,{\rm s} & 3 \times 10^8 \,{\rm m} = 1 \, {\rm ls} \\ 3 \times 10^8 \,{\rm m/s} = 1 & 3 \times 10^8 \,{\rm m/s} = 1 \,{\rm ls/s} \\ E^2 = m^2 + p^2 & E^2 = m^2 + p^2 \mbox{ and see * below} \end{array}$$ (*) In the case of an expression like this, if one thinks of $$c$$ as dimensional then it might be said to be an abuse of notation to miss out the physical dimensions when substituting the value "$$c=1$$ speed-unit" into any formula, but this is merely a matter of convention on what the formula means. In either case it has been taken as understood that when evaluating any given expression, units will be adopted in which $$c$$ has the numerical value of 1.

• I find this answer misleading. What would you say about a radian? It's sometimes useful to specify that you're using radians and even to carry "rad" as a "unit" for a reminder. But ultimately it is dimensionless (a ratio between two lengths). There's no getting around that. Velocity within relativity is the same. – Brick Jun 26 '19 at 15:46
• @Brick I agree a radian is dimensionless. It does not follow that velocity is dimensionless, because temporal and spatial development are distinguished in the dynamics. I can tell what a second is, without mistaking it for a light-second, by observing the evolution of an oscillator, for example. – Andrew Steane Jun 26 '19 at 16:47
• I'm giving you the benefit of the doubt since you seem to be a legit physicist, but I cannot make sense of your last comment. There's nothing in the dynamics relevant to definition of units other than that $c$ is constant, your true-but-not-clearly-relevant comment about light cones in your answer not withstanding. The radian (usually implicitly) assumes that you measure the radius and arclength in the same linear units. If you measured arclength in cm and radius in inches, you'd have an analogy of $c$ in all of your equations there too, but so what? Would like to understand your point. – Brick Jun 26 '19 at 17:03
• FWIW, it's not that bizarre to use different length units in different directions. Vertical movement in Earth's gravity field is different to horizontal movement, and we tend to use smaller units to express heights than those used for horizontal distances, especially when not using the metric system. Traditionally, navigators used nautical miles for horizontal distances, and fathoms to measure depth. Similarly, (non-metric) aviators measure altitude in feet, not miles. – PM 2Ring Jun 26 '19 at 23:46
• @PM2Ring Exactly! So, in some cases, meters are part of the measurement of a time-like quantity. They’re not fundamentally different from seconds. – Bob Jacobsen Jun 27 '19 at 14:34

I would answer this first from the mathematical perspective. Mathematically, there is a difference between a quantity with a unit, like

$$1\ \mathrm{m}$$

and just

$$1$$

. This notion was - to a primeval extent - understood all the way back to Euclid of Alexandria (ca. 300 BCE), the same one for whom Euclidean geometry is eponymous. He considered the notion of magnitude, which had different "kinds" - lengths, areas, and so forth, against ratio, which was between two magnitudes of the same kind. Today, we might say magnitudes are the first type of quantity, and ratios as "pure numbers" of the second type of quantity.

The crucial difference is that magnitudes don't function like ratios do: you can add two magnitudes of the same kind, e.g. $$1\ \mathrm{m} + 5\ \mathrm{m} = 6\ \mathrm{m}$$, but you cannot add two magnitudes of different kinds, e.g. $$1\ \mathrm{m} + 5\ \mathrm{m}^2$$ is nonsense. You can multiply two magnitudes of different kinds, or the same kind, though - but the magnitude will be of another kind, e.g. two lengths multiply to an area, which we nowadays would use to describe something like

$$(a\ \mathrm{m})(b\ \mathrm{m}) = ab\ \mathrm{m^2}$$

But on the other hand, you can indeed add and multiply ratios as much as you want, because they're all the same type.

From the modern perspective, we would define ratios to be elements of the "pure" real number system $$\mathbb{R}$$, and while Euclid took magnitudes as his primitive notion from which are derived ratios, we take effectively ratios to be primary, and derive magnitudes from them. In a sense, elements of $$\mathbb{R}$$ are things that "want" to change the amount of something, to make it larger or smaller, or more or less numerous, according to the "sizing amount" they embody - if you follow this far enough you get to endomorphism rings and, even further, to the categorical product notion (see, e.g. Qiaochu Yuan's discussion here: https://math.stackexchange.com/questions/56663/is-there-a-natural-way-to-extend-repeated-exponentiation-beyond-integers/56710#56710). In fact, this is embodied in mathematical terminology: it's why we call them "scalars" in vector algebra. And it also connects with how we use them in natural language - when I say "three apples", "three" is something which "modifies", or acts upon, "apple" to conceptually produce the notion of three of them. Moreover, this is what multiplication is "really" about (instead of "repeated addition"): multiplication is, given two scalars, finding a third which acts upon things as one followed by another.

Magnitudes are a more complicated beast - the way you'd build them would be have to be basically a set of sets of "tagged" or "typed" real numbers that are effectively 1D vector spaces in that within any given such, tagged numbers can be added together and also rescaled by "pure" scalar (i.e. ratios) reals with an additional division operation which produces a ratio, and then finally a "trans-typal" multiplication operation that operates on the disjoint union of all the sets in the family (i.e. the "pooling" of all tagged reals). I am not sure at all what this kind of structure such a thing is called or if it's been studied in the mathematical literature before.

Now, returning to your question of physics and special relativity, we can employ this formalism to resolve the conundrum as thus. Instead of concentrating on all equations, we'll just concentrate on the basic one that describes the Minkowskian geometry, i.e. its line element

$$d\tau^2 = dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

. Because we are talking physical quantities, these are magnitudes. The $$dx$$, etc. are NOT real numbers(*). They belong in the second kind of structure I just discussed. They have "units", and the typal multiplication by $$\frac{1}{c^2}$$ ends up converting units of (squared) space to units of (squared) time, so that the subtraction from $$dt^2$$ can proceed.

If, however, we take "$$c = 1$$", then technically there is a difference: for one, as written, that would be nonsense in the above as that would make $$c$$ a scalar ("ratio" in Euclidean terminology). As you note, we have to take $$c = 1\ \frac{\mbox{distance unit}}{\mbox{time unit}}$$, in order to stay within the confines of the magnitude system. However, if we do this, we can then pass through the natural isomorphism which "strips off the units", that is, which "pulls the tags from the tagged reals" that make up magnitudes, which we may denote as $$\mathrm{strip}(x)$$, and come to

$$\mathrm{strip}(d\tau^2) = \mathrm{strip}\left(dt^2 - \frac{1}{c^2}(dx^2 + dy^2 + dz^2)\right)$$

and then percolating all the isomorphism relations through,

$$[\mathrm{strip}(d\tau)]^2 = [\mathrm{strip}(dt)]^2 - \frac{1}{[\mathrm{strip}(c)]^2}([\mathrm{strip}(dx)]^2 + [\mathrm{strip}(dy)]^2 + [\mathrm{strip}(dz)]^2)$$

and since $$\mathrm{strip}(c)$$ is now a real number $$1$$ (take "$$\frac{\mbox{distance unit}}{\mbox{time unit}}$$" off), $$\frac{1}{[\mathrm{strip}(c)]^2} = 1$$ and thus if we abuse the notation on the $$dx$$, etc. values to go to their stripped counterparts, we get

$$d\tau^2 = dt^2 - (dx^2 + dy^2 + dz^2)$$

and it is in this sense that setting $$c = 1$$ makes things "unitless". If we did not do this, we could still pass through the $$\mathrm{strip}$$ isomorphism, but we'd end up with a factor $$\frac{1}{c^2}$$ which depended on the unit system we originally were coming from.

(*) Yes I'm calling a differential 1-form a real number - blah! I have to keep it simple somehow! Just pretend they're small, finite real numbers instead and the form is merely suggestive :)

• My compliments for the mathematical explanation. – Sebastiano Jul 15 '19 at 12:00

EDIT

Interestingly, there seems to be some confusion about the phrase "set $$c=1$$" even among the physicists here. It's a bit of sloppy language that some people seem to be taking literally. What it actually means is "choose any set of units in which the value of the speed of light will be 1." That's not a unique choice - It's a compatibility condition between the units that you use for time and that you use for distance.

One consequence of making this choice is that time and distance get measured in the same unit as each other, but you can find any number of such compatible sets of units. Even if you "start" with seconds and meters, you still have at least two obvious choices as you can measure time in meters ("$$t \rightarrow ct$$") or distance in seconds ("$$x \rightarrow x/c$$"). But again, any units would do. You can make this work with furlongs, fortnights, length of your own left foot as it is at noon today, number of George-Washington-lifetimes... (Clearly some of these are easier to standardize and nicer for science, but the point is there is a $$c=1$$ compatibility possible starting from any of them.)

On the other side of it, this is a condition that you can always choose to meet as a consequence of the fact that the speed of light in constant. If you make this choice, then by definition of the choice, speed is unitless. You could have made a different choice, of course, where speed had units, but if you're in the "set $$c=1$$" convention, it means you didn't do that.

What makes this convention useful on the surface, of course, is that all of the factors $$c$$ in various equations become unnecessary.

If you choose some contrived set of units where you have the speed of light is something like "1 tu/du", where tu and du are some time and length units of your construction, this is still a different convention. For example, in that case you still need to carry the factor of $$c$$ in $$E^2 = (pc)^2 + (mc^2)^2$$ to keep the equation dimensionally consistent, which, of course, defeats the purpose of adopting the convention. (If you carry the factors, this is not a "wrong" convention, but it's probably a useless one.)

The only way this can make sense - carrying "units" on speed and dropping the factors of $$c$$ - is if you treat it like a radian ("rad"), which is actually a dimensionless ratio of two lengths. If you want to carry the "rad" around as a reminder of where it came from, ok, but at the end of the day that doesn't change the fact that it's unitless. (In that case, things like "small angle approximation" depend on it being unitless in analogy to the relativistic energy above. The Taylor series of trig functions, in their "usual" form, only works for arguments with no units. With units, there'd be an additional conversion factor in each term.)

With the discovery of special relativity and the associated notion of spacetime, it becomes clear that two distinct sets of units aren't quite right. While $$c$$ is the speed of light and therefore usually conjures a physical motivation, for the purposes of your question it is also essentially a unit conversion factor. There are (approximately) $$3 \times 10^8$$ meters in every second just the same as there are 12 inches in 1 foot.
After that, the rest of your questions are just questions of convention. If you've set everything up to measure time in seconds and distance in meters (in some frame of your choosing), then you need a factor of $$c$$ to do the conversion. If you choose to measure your time in, say, meters rather than seconds, then no further factor is needed to keep consistency. Setting $$c=1$$ is just another way of saying that you're going to use the same units for space and time.