# Why use units in which c=1?

• I
You can simplify the formula a bit by choosing to measure time in seconds and distances in light-seconds so c=1

Going off on a tangent here, and I know this is done a lot, but I have never understood the advantage. Just to save some typing? Showing c explicitly in the equation seems to be more instructive, especially to someone new to the subject.

• MeJennifer

## Answers and Replies

Dale
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Just to save some typing?
It also focuses on the physics instead of the unit conversions. It makes the formulas more concise and with less physically meaningless clutter.

It is the same reason why we write F=ma instead of F=kma

• vanhees71
Going off on a tangent here, and I know this is done a lot, but I have never understood the advantage. Just to save some typing? Showing c explicitly in the equation seems to be more instructive, especially to someone new to the subject.
I recieved some enlightment from Dale and a couple others regarding the permissivity/permitivity constants in a conversation that eventuslly went along the same line. Ultimately the constants we choose in a physics equations are basically not much more than unit conversion factors chosen for convenience. If you think about it you can always make up any unit you want, so getting hung up on raw numbers is counterproductove to learning physics, I gather.

Drakkith
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It is the same reason why we write F=ma instead of F=kma

What's the K represent in that 2nd equation?

Yes, but based on the B label for the OP's post, it just seems an unnecessary complication to start redefining the unit of distance.

jbriggs444
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What's the K represent in that 2nd equation?
A unit conversion factor. It is needed if force is defined as an independent unit rather than a derived unit.

For instance, if one measures mass in pounds-mass, force in pounds-force and acceleration in feet per second squared then ##f=\frac{1}{32.7}ma##

If one measures mass in stones, force in dynes and acceleration in furlongs per microfortnight per second then ##f=kma## for some value of k that I do not wish to compute.

• Stephanus, Vedward, Battlemage! and 4 others
It also focuses on the physics instead of the unit conversions...

How does that focus on the physics more than using v/c, which shows that what's important is the velocity relative to the speed of light? Especially for someone at the B level.

It is the same reason why we write F=ma instead of F=kma

And what would happen to F=ma if we don't use meters as the unit of distance?

Dale
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How does that focus on the physics more than using v/c, which shows that what's important is the velocity relative to the speed of light? Especially for someone at the B level.
For example, it is easier to see that ##(u+w)/(1+uw)=1## for ##w=1## than it is to see that ##(u+w)/(1+uw/c^2)=c## for ##w=c##

And what would happen to F=ma if we don't use meters as the unit of distance?
It would become ##F=kma##, as described by jbriggs444 above.

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vanhees71
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Gold Member
It also focuses on the physics instead of the unit conversions. It makes the formulas more concise and with less physically meaningless clutter.

It is the same reason why we write F=ma instead of F=kma
Yes, and this implies also that one shouldn't use the SI in electrodynamics. It's just "physically meaningless clutter". That's OT in this thread, but it's a pest that many modern textbook authors of introductory theory texts insist on using the SI units, including Jackson with the newest edition of his textbook. Of course, when doing the relativistic formulation of electrodynamics (the only adequate one in the 21st century, if you ask me), he's switching back to the good old Gaussian units (although I'd prefer the rationalized version of it, the Heaviside-Lorentz units).

• Dale
robphy
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Dale
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I'd prefer the rationalized version of it, the Heaviside-Lorentz units
Those are also my preferred "standard" units, although I think natural units are even better.

• vanhees71
How does that focus on the physics more than using v/c, which shows that what's important is the velocity relative to the speed of light? Especially for someone at the B level.

I actually think that it's better not to use c=1 at first, unless you're already perfectly comfortable with the arbitrariness of units and constant-values.

For example, it is easier to see that ##(u+w)/(1+uw)=1## for ##w=1## than it is to see that ##(u+w)/(1+uw/c^2)=c## for ##w=c##

But the velocity-composition formula is quite elegant when you use normalized speeds:

##\beta_v = \frac{\beta_u + \beta_w}{1 + \beta_u \beta_w}##

• weirdoguy
Dale
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But the velocity-composition formula is quite elegant when you use normalized speeds:
Yes. Normalized speeds are speeds in units where c=1.

Yes. Normalized speeds are speeds in units where c=1.

By normalized speed, I just mean ##\beta = v/c##, which works whether or not you set c=1 (if you do, then ##\beta = v##). Sorry if I wasn't clear before.

Dale
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By normalized speed, I just mean ##\beta = v/c##, which works whether or not you set c=1 (if you do, then ##\beta = v##). Sorry if I wasn't clear before.
I understood. I was just pointing out that ##\beta## is itself a speed in units where c=1. Even where v and c have other units ##\beta## does not. In fact, c is the conversion factor between the units of v and units where c=1.

Writing it with the symbol ##\beta## does not mean that it is not a speed.

Mister T
Science Advisor
Gold Member
By normalized speed, I just mean ##\beta = v/c##, which works whether or not you set c=1 (if you do, then ##\beta = v##). Sorry if I wasn't clear before.

When you divide ##v## by ##c## you are normalizing. The resulting ratio has a value identically equal to ##1## for a beam of light in a vacuum.

To answer the original question, setting ##c=1## makes the physics clearer. If you don't do that you measure distance and time in different units. Could you imagine teaching the Pythagorean theorem to students where you insisted on measuring each leg of a right triangle in different units? For example, a triangle has legs measuring 3 ft and 4 m. Find the length of the triangle's hypotenuse. You would certainly insist that it's easier to measure each leg's length in the same units, because otherwise you must first convert one or the other distance so that each leg's length is measured in the same units. It is precisely the same with spacetime geometry. Measuring distance and time in the same units spares one the labor of having to first convert one or the other measurement so that the two are expressed in the same units.

In the Pythagorean theorem you find the sum of the squares of the two legs. In the spacetime geometry of special relativity you find the difference between the squares of the length and time measurements.

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• Battlemage! and FactChecker
When you divide ##v## by ##c## you are normalizing. The resulting ratio has a value identically equal to ##1## for a beam of light in a vacuum.

Yes, of course ##c/c = 1##. Or am I missing what you're getting at?

(P.S. I'm not OP.)

It's not only c, you can even set G=M=c=1, then all the times you get are in units of GM/c³, all distances in GM/c² and all velocities are in terms of c. Then the results are valid for any given initial values, you just have to multiply the numbers with the constants.

To answer the original question, setting ##c=1## makes the physics clearer. If you don't do that you measure distance and time in different units.

By that do you mean the interval in flat space would have a c2dt2 term instead of just dt2?

Nugatory
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By that do you mean the interval in flat space would have a c2dt2 term instead of just dt2?

Yes. And to see how that obscures the physics, compare two forms of the metric for ordinary three-dimension Euclidean space:
1) ##ds^2=dx^2+dy^2+dz^2##
2) ##ds^2=dx^2+dy^2+36dz^2##

In the first I am using the same units for all three axes. In the second I am measuring distances along the x and y axes in feet and distances along the z axis in fathoms. That factor of 36 appears for the same reason that ##c^2## terms appear when you don't choose units that make ##c## equal to one.

Which one makes it more clear that we are in Euclidean space and just using the Pythagorean theorem? Which form makes it more clear that there is no preferred direction in this space? Is any physical insight contributed by that factor of 36?

Now it is true that the time axis is different from the spatial axes in Minkowski space, and we do want that real physical difference to show up in the metric... And indeed it does, because there is a minus sign on the ##dt^2## term (if we use the -1,1,1,1 signature). That minus sign is telling us something useful about the geometry of spacetime, completely independent of our choice of units.

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• Orodruin, m4r35n357, vanhees71 and 2 others
arydberg
Gold Member
Going off on a tangent here, and I know this is done a lot, but I have never understood the advantage. Just to save some typing? Showing c explicitly in the equation seems to be more instructive, especially to someone new to the subject.
You can also express time in nanoseconds and distance in feet. The error from this is about 2%,

• m4r35n357
robphy
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You can also express time in nanoseconds and distance in feet. The error from this is about 2%,
Grace Hopper - Nanoseconds

• Battlemage! and m4r35n357
Nugatory
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You can also express time in nanoseconds and distance in feet. The error from this is about 2%,
Yep - that's one of many happy approximations in life. The scale is very convenient for anything that's going on inside a lab, but there is one pedagogical disadvantage - it looks as arbitrary as the speed of light in meters/sec. When you say "time in seconds, distances in light-seconds" it's obvious what you're doing to make c equal to one.

Dale
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When you say "time in seconds, distances in light-seconds" it's obvious what you're doing to make c equal to one.
I wonder why the light-year and light-second are recognized units of distance, but the light-meter is not a recognized unit of time. By recognized I mean unofficial but still widely used.

• Battlemage! and Nugatory
robphy
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I wonder why the light-year and light-second are recognized units of distance, but the light-meter is not a recognized unit of time. By recognized I mean unofficial but still widely used.
... continued resistance to the metric system? 