Can c be set equal to 1 in certain systems of units?

[pixel]

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.

 

[Dale]

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

 

[Battlemage!]

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 received 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]

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

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

 

[pixel]

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]

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.

 

[pixel]

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]

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.

 

[vanhees71]

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).

 

[robphy]

Famous parable: https://www.google.com/search?q="parable+of+the+surveyors"+spacetime+physics

 

[Dale]

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.

 

[SiennaTheGr8]

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:

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

 

[Dale]

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

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

 

[SiennaTheGr8]

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]

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]

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.

 

[SiennaTheGr8]

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.)

 

[Yukterez]

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.

 

[pixel]

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]

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.

 

[arydberg]

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%,

 

[robphy]

You can also express time in nanoseconds and distance in feet. The error from this is about 2%,

Grace Hopper - Nanoseconds

 

[Nugatory]

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]

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.

 

[robphy]

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?

 

[pixel]

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?

Couldn't one just as well define x⁰ = ct and continue to use meters as the unit of distance? Then the proper time interval would look like your 1) above, using the minus sign for the x⁰ term.

 

[Nugatory]

Couldn't one just as well define x⁰ = ct and continue to use meters as the unit of distance? Then the proper time interval would look like your 1) above, using the minus sign for the x⁰ term.

Yes, one could. It will work, for same reason that the analogous method of preserving the distinction between feet and fathoms ("Couldn't one just as well define z=6*depth and continue to use feet as the unit of distance? Then the interval would still look like $ds^2=dx^2+dy^2+dz^2$") will work. But why? What physical insight do we gain from either treatment? Why do we want to preserve that distinction?

The number 299,792,458 is no more fundamental than the number 6, which is the ratio of the mean distance between outstretched arms to mean length of foot averaged across the seagoing population of northern Europe about one thousand years ago. They're both useful when you need to specify a value (I've seen navigational charts that use fathoms and and charts that use meters) but neither brings any unique value to the statement of the laws of nature.

 

[ljagerman]

Think of how the speed of sound is abbreviated: Mach 1, etc.

 

[Buzz Bloom]

Are there specific recommended standard units when c = 1? I would guess these units are seconds, kilo-grams, and light-seconds. Is that correct?

 

[Buzz Bloom]

Are there specific recommended standard units for c=1?
I would guess they are

time: seconds
mass: kilograms
distance: light-seconds​

Is that correct, making speed units light-seconds per second?

 

[PeterDonis]

Are there specific recommended standard units for c=1?

No. Any units that match up correctly will work. You can use years and light-years, seconds and light-seconds, meters and "light-meters" (the time it takes light to travel 1 meter), feet and nanoseconds, etc.

Also, the unit of mass has nothing to do with this; choosing units in which $c =$ does not force you to choose any units for mass.

 

[Buzz Bloom]

Hi Peter:

Thank you for the clarification. Presumably, for any particular choice of units consistent with c=1 there is are corresponding values for various physical constants, like G for example. Are such physical constant values generally calculated and published somewhere so that someone trying to do a physical calculation, for example the amount of space distortion at a given distance from a black hole, would not have to recalculate G as part of this calculation?

Regards,
Buzz

 

[PeterDonis]

Presumably, for any particular choice of units consistent with c=1 there is are corresponding values for various physical constants, like G for example.

Not necessarily. Choosing units for which $c = 1$ doesn't fully specify a value for $G$, for example, because it doesn't set units for mass.

 

[Buzz Bloom]

Not necessarily. Choosing units for which c=1c = 1 doesn't fully specify a value for GG, for example, because it doesn't set units for mass.

Hi Peter:

Suppose I choose the following units:

time: 1 second
distance: 1 light-second:
mass: 1 kilogram​

Is there some reference source where I could find the published corresponding value for G?

Regards,
Buzz

 

[Nugatory]

Are such physical constant values generally calculated and published somewhere so that someone trying to do a physical calculation, for example the amount of space distortion at a given distance from a black hole, would not have to recalculate G as part of this calculation?

I'm sure there's a published list somewhere, but usually it's easiest to just put the units back in when you're done calculating.

We started this thread with the velocity addition formula (actually split off from a thread about velocity addition) so let's use that as an example... Say we've used the $c=1$ version of the formula to see how fast a bullet fired at one-half lightspeed from a spaceship also moving at one-half lightspeed is moving... I can do that in my head, and the answer is 4/5. Then if I want an answer in meters/sec I multiply 299792458 by 4/5; if I want an answer in miles/hr I multiply 670800000 by 4/5.