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I Why use units in which c=1?

  1. Jun 15, 2016 #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.
     
  2. jcsd
  3. Jun 15, 2016 #2

    Dale

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    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
     
  4. Jun 15, 2016 #3
    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.
     
  5. Jun 15, 2016 #4

    Drakkith

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    What's the K represent in that 2nd equation?
     
  6. Jun 15, 2016 #5
    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.
     
  7. Jun 15, 2016 #6

    jbriggs444

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    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.
     
  8. Jun 15, 2016 #7
    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?
     
  9. Jun 15, 2016 #8

    Dale

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    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##

    It would become ##F=kma##, as described by jbriggs444 above.
     
    Last edited: Jun 15, 2016
  10. Jun 16, 2016 #9

    vanhees71

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    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).
     
  11. Jun 16, 2016 #10

    robphy

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  12. Jun 16, 2016 #11

    Dale

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    Those are also my preferred "standard" units, although I think natural units are even better.
     
  13. Jun 16, 2016 #12
    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:

    ##\beta_v = \frac{\beta_u + \beta_w}{1 + \beta_u \beta_w}##
     
  14. Jun 16, 2016 #13

    Dale

    Staff: Mentor

    Yes. Normalized speeds are speeds in units where c=1.
     
  15. Jun 16, 2016 #14
    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.
     
  16. Jun 16, 2016 #15

    Dale

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    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.
     
  17. Jun 16, 2016 #16

    Mister T

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    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.
     
    Last edited: Jun 16, 2016
  18. Jun 16, 2016 #17
    Yes, of course ##c/c = 1##. Or am I missing what you're getting at?

    (P.S. I'm not OP.)
     
  19. Jun 17, 2016 #18
    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.
     
  20. Jun 17, 2016 #19
    By that do you mean the interval in flat space would have a c2dt2 term instead of just dt2?
     
  21. Jun 17, 2016 #20

    Nugatory

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    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.
     
    Last edited: Jun 17, 2016
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