# What exactly are we doing when we set $c=1$?

I understand the idea of swapping from unit systems, say from ms$^{-1}$ to kms$^{-1}$, but why can we just delete the units altogether?

My question is: what exactly are we doing when we say that $c=1$?

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 you might find this paper interesting. – Mark Mitchison Jan 21 at 23:31 See this answer to nearly the same question. – Eduardo Guerras Valera Feb 11 at 20:04

All we're doing is using a set of units where certain quantities happen to take convenient numerical values. For example, in the SI system we might measure lengths in meters and time intervals in seconds. In those units we have $c = 3 \times 10^8 \text{m}/\text{s}$. But you could just as well measure all your distances in terms of some new unit, let's call it a "Finglonger", that is equal to $2.5 \times 10^6 \text{m}$, and time intervals in a new unit, we'll call it the "Zoidberg", that is equal to $8.33 \times 10^{-3} \text{s}$. Then the speed of light in terms of your new units is $$c = 3 \times 10^{8} \text{m}/\text{s} = 1 \frac{\text{Finglonger}}{\text{Zoidberg}} ~.$$ The units are still there -- they haven't been "deleted" -- but we usually just make a mental note of the fact and don't bother writing them.

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One oddity here is that in feet per nano-second the speed of light is very nearly 1 anyway. – dmckee Jan 21 at 16:54

If you are getting used to 'natural' units I think its best to think of it like this: we are basically defining a new time variable $t' \equiv c t$ to work in. $t '$ has units of distance. We can always go back to the old time variable, and old unit system using $t = \frac{t'}{c}$.

We do this to keep things simple as possible. For example the line element:

$d \tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 = dt'^2- dx^2 - dy^2 - dz^2$

and the relativistic dispersion relation:

$E = \sqrt{p^2 c^2+m^2 c^4} = \sqrt{p'^2 +m'^2}$

are much simpler in these units. This may not seem like a great step forward, but when dealing with complicated equations, anything that simplifies is of great use.

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Some theoretical physicist like to do that just to avoid constants while calculating, they choose a system of units in which h=c=1 (and some more of them), so the get rid of a lot of stuff. The point is just to do that, they get rid of constants by making them equal to 1, at the end they will have to change again to a system more usable, mks, IS, or someone else.

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Thanks, but I'm not so much looking for 'why we do it', but a deeper meaning, if there is one. +1 anyway. – user12345 Jan 21 at 16:37
"at the end they will have to change again to a system more usable" There is no fundamental need to convert back. Essentially all of particle physics is done in $c=h=1$ units all the time. Publications use those units, results are tabulated in them and so on. We only ever convert back to more human friendly units when we want to talk to people who don't use those units. – dmckee Jan 21 at 16:54

While the more careful approach is indeed to say the units are still there, we just don't write them as such, I prefer to think of it as DJBunk suggests:

By using certain ("god-given") constants, we are able to express the concept of time in meters just as well as seconds: Rather than say "something takes 10s", you might say "It takes as long as it would take a beam of light to travel $\frac{10\textrm{s}}{c}$ meters." Call it light-meters, if you will. It's analogous to the way we express distance in units of time, also using $c$, when we talk about "light-years". A similar reasoning lets you eliminate other units by simply expressing them in "more basic" units. Of course, which set of units you use as "fundamental" is completely up to you.

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Not much, although in a given reference frame, velocity being equal to zero is (usually) a fixed point on the scale. So you still have two points, which is enough to define this scale.

When looking at the Lorentz Transformations for something like $SO(1,3)^{\uparrow}$ (the space that special relativity takes place in), you'll see the term $\beta$ (defined as $v/c$) equate to the magnitude of $v$, as c is treated as an identity element. The domain of $v$ is therefore placed between zero and $c$, which reveals the primary reason people may set $c=1$: to represent all $v$-s as a fraction of $c$

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OK, so we need two points to define a scale. But, is there any meaning in setting $c=1$? The motion of just deleting units confuses me. Or, I am overthinking it: $c=1$ is just temporary and no-one would ever try to interpret a result with $c=1$? +1 to you also. – user12345 Jan 21 at 16:41
By setting $c=1$, we still have a scale- it is simply in units of $c$. That is, I could say my velocity is 0.1 speeds of light. – B. Elliott Jan 21 at 20:17

One conceptual reason for setting $c = 1$ is to make certain symmetries more apparent. For example, consider the relativistic relationship $E^2 = (|\vec{p}|c)^2 + (mc^2)^2$ with quantities expressed in SI units, as shown. If we set $c=1$ it becomes $E^2 = |\vec{p}|^2 + m^2$, which indicates that energy, momentum, and mass can be put on equal footing. One is just an expression of the other two. This relationship isn't quite as obvious when factors of $c$ are strewn about. As another example, consider the Lorentz transformation. Setting $c=1$ shows the true symmetry between space and time.

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We don't actually delete the units. They are still there.

In the system of units where the numerical value of $c$ is 1, any velocity can be expressed in terms of $c$. Just like in SI units, the metre has a numerical value of 1 and every distance can be expressed as some amount of m. So you could, for example, say that you are travelling at a speed of $v=0.000001\,\mathrm{c}$. We often omit the units in calculations for our convenience, but we do this in the SI system as well.

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