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.

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.

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.

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

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).

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.

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

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.

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.

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.

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.