# 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 '13 at 23:31
See this answer to nearly the same question. – Eduardo Guerras Valera Feb 11 '13 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 '13 at 16:54
@dmckee Don't say things like this loud! Harald Fritzsch claims that his accidental mentioning of the elementary charge being close to a natural number in some imperial units to a Texan senator caused a vote on Metrication to fail due to the claim that "imperial units were (more) god-given". Only in America m-/ – Tobias Kienzler Jun 11 '13 at 14:15
There are a few neat co-incidences in various units - pure co-incidences of course, but: the solar energy incident on Earth (after accounting for 30% reflexion by clouds, atmoshphere, ground) is about 1kg per second (i.e. $9\times10^{16}\mathrm{J\,s}^{-1}$, 1eV is roughly a 1um wavelength photon (actually 1240nm, but near enough for rough reckonnings on the back of an envelope) and so forth ... – WetSavannaAnimal aka Rod Vance Jul 6 '13 at 12:08

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:

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

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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In the first equation "$\mbox{d}\tau^2=c_0^2\mbox{d}t^2-\mbox{d}x^2-\mbox{d}y^2-\mbox{d}z^2$", you are using Natural units on the LHS but not on the RHS! Either that or my computer screen renders $\tau$ instead of $s$ : ) And you are using a $(+---)$ signature... which is a very bad metric signature... But still, +1... (The bad conventions may have been the reason for the anonymous downvote...) – centralcharge Jul 6 '13 at 9:52
@dimension10 - Thanks for the correction regarding $d \tau$ verse $ds$. As far as the metric signature goes, I am an unapologetic particle physicist. – DJBunk Jul 6 '13 at 15:52

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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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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another such symmetry is $(E,p,p,p)$ (where I didn't write the subscripts)... +1! – centralcharge Jul 6 '13 at 9:55

Some theoretical physicist like to do that just to avoid constants while calculating, they choose a system of units in which $\hbar=c_0=1$ (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 '13 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 '13 at 16:54

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 '13 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 '13 at 20:17
@user12345, B.Elliot Huh? This (as in the first sentence of this answer) is wrong. If you set the reference frame to be moving with the light beam, for example, then the observed velocity of the light beam would be: $\frac{c_0+c_0}{1+\frac{c_0^2}{c_0^2}}=c_0\neq0$... But I won't -1, just in case you didn't mean that... – centralcharge Jul 6 '13 at 9:59
Observed velocity of a light beam is always c, no matter the reference frame you're in. However, observed velocity of one's own reference frame is always 0 for an observer in that frame. An observer could choose a reference frame that's moving with respect to themselves, but it's usually easier (for the math) to consider the origin of the observer's reference frame fixed on the observer. Good instincts keeping an eye out, but internal reference frame velocity is typically zero. Semantics, heh. – B. Elliott Jul 6 '13 at 21:45
with respect to oneself, now I see what you're talking about. P.S. In future, use "@" when replying in comments, for example "@Dimension10". I wasn't notified when you replied, since you didn't put it. – centralcharge Jul 17 '13 at 13:33

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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Despite the indoctrination by high school physics teachers, physics is dimensionless. So, all units and dimensionfull constants can indeed be deleted. Now, this looks to be impossible as it then looks like you are throwing away information from equations that you can then no longer get back from using only those equations. This is not true, but recovering the original equations does require one to study the appropriate scaling limits of the theory in question, in this case special relativity in a way that is not usually done in textbooks.

Properly restoring c in the equations of special relativity should proceed as follows. We need to study what happens in the limit of zero velocities when one imposes conservation of energy and momentum. Of course, one can say that then everything stops moving. But the more interesting question is what happens if we make a movie of processes happening at slower and slower velocities and we simultaneously speed up the playback speed of the movie so that we keep on seeing the motion of the involved objects. We can then take the limit of scaling all the velocities to zero while the movie will still appear to show the objects moving at more or less the same speeds on the screen. The way the object appear to move and interact with each other on the screen will then be described by classical laws of physics.

The equations for energy and momentum of a free particle are:

$$E = \gamma(v) E_{0}$$

$$\vec{P} = \gamma(v) E_{0} \vec{v}$$

where

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

We put $\vec{v} = \vec{v'}/c$ and assume that v' is kept finite while c is sent to infinity. Note that here c is just a dimensionless scaling parameter. This amounts to sending the velocity to zero while zooming in into the low velocity world so that it remains visible. To make the velocity dependence of the energy visible we need to expand it to second order in $\vec{v'}/c$, while the velocity dependence of momentum appears at zeroth order. Replacing the expressions for the energy and momentum by these expansions, gives:

$$E = E_{0}\left(1 + \frac{v'^2}{2c^2}\right)$$

$$\vec{P} = E_{0} \frac{\vec{v'}}{c}$$

Now, consider an elastic collision for which the above expressions for the energy and momentum are applicable, i.e. v' is finite. Then, demanding that we have conservation of momentum in any arbitrary frame, yields that both the sum of the rest rest energies and the sum of the momenta are separately conserved. Energy conservation then implies that that the the sum of the kinetic energies

$$T = E_{0}\frac{v'^2}{2c^2}$$

is conserved. The two terms in the energy equation scale in different ways, so we have to introduce a new variable to make sure that both terms remain visible in the c to infinity limit. We can e.g. make the kinetic energy finite by putting

$$E_{0} = m c^2$$

where m is assumed to stay finite in the c to infinity limit. We then have to rescale the momentum:

$$\vec{P'}= \frac{\vec{P}}{c}= m \vec{v'}$$

The total rescaled momentum, the kinetic energy and the "mass" m are then finite quantities that are conserved in the c to infinity limit. Note that we end up with this result irrespective of how we make the quantities finite. We could just as well have kept $E_{0}$ finite. If we then call this the mass m:

$$E_{0} = m$$

then we need to rescale the momentum according to:

$$\vec{P'}= \vec{P} c = m \vec{v'}$$

and the kinetic energy needs to be rescaled according to:

$$T' = T c^2 = \frac{1}{2} m v'^2$$

What then happens is that the rest energy $E_{0}$ does not scale in the same way as the kinetic energy. We can define a rescaled rest energy $E_{0}'$ that does scale in the same way as the kinetic energy:

$$E_{0}' = E_{0} c^{2} = m c^{2}$$

So, we see that in the scaling limit we end up with three finite independent conserved quantities: mass, kinetic energy and momentum and that if we formally keep the rescaling parameter c, the relation between the rest energy expressed in the finite kinetic energy units and the mass is $E_{0} = m c^2$.

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## protected by Qmechanic♦Jul 6 '13 at 12:48

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