Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Speed of Light, Squared

  1. Jun 29, 2005 #1
    Special relativity says that the speed of light is a constant regardless of the speed of the observer.

    I've always been troubled with the idea of squaring that speed (C ^ 2) unless speed of light to the zero power (C ^ 0) also has meaning at the same time and in the same space; the net effect to an observer fixed in time being 1 = 1 where 0 = 0.

    To an observer fixed in space, 0 would appear equal to 1 at the same time that 1 appeared equal to 0.

    To a real observer, part of one appears to be in zero and part of zero appears to be in one.

    Please destroy this view of special relativity.
    Last edited: Jun 29, 2005
  2. jcsd
  3. Jun 29, 2005 #2
    What's wrong with squaring a number? If this is an assault on SR you might want to post this in the theory development section.
  4. Jun 29, 2005 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    The speed of light squared is not a velocity. It appears in some expressions where velocity^2 appears, like E=mc^2 (compare to E=(1/2)mv^2).

    The speed of light is constant for all observers regardless of their state of motion, this is however totally unrelated to the properties of c^2.
  5. Jun 29, 2005 #4
    Nothing wrong with squaring a number.

    What I have a problem understanding is squaring a constant that cannot change for an observer who is fixed in both time and space.
  6. Jun 29, 2005 #5
    My exact problem. Speed of light squared is a concept that depends on your time/space perspective.
  7. Jun 29, 2005 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Why would you have a problem squaring a constant? (I have no idea what you mean by an observer "fixed in both time and space". The speed of light is an invariant: it has the same value as measured by anyone regardless of their motion.)

    Do you have a problem with the formula for (non-relativistic) kinetic energy: [itex]{KE} = 1/2 m v^2[/itex] ? In that formula speed is not a constant, but is frame dependent. I trust that you are fine with that. So, if anything, squaring c should be even less of an issue.

    Actually, the speed of light, being frame independent, is less dependent on "time/space perspective" than ordinary, sub-light speeds. (But, even if something is frame-dependent, what prevents you from squaring it?)
  8. Jun 29, 2005 #7
    Maybe a calculus class may help. I dont fully understand the logic in squaring c either :pondering:
  9. Jun 29, 2005 #8
    Whenever you find a [itex]c^2[/itex] there's probably a way to write it differently, making the source of the square more obvious.
    Look e.g. at [itex]E=mc^2[/itex] which is usually written more correctly as [itex]E/c=\sqrt{(m_0c)^2+p^2}[/itex]. The [itex]c[/itex] on the left part is a scaling factor while the [itex]c[/itex] in the right part is a velocity.
    In [itex] t'=\gamma(t-vx/c^2)=\gamma(t-(v/c)(x/c))[/itex], [itex]v/c[/itex] is a ratio of velocities while in [itex]x/c[/itex] the [itex]c[/itex] is a correction in the scale of [itex]x[/itex].
    Try it. You'll find for most cases either this separation in a velocity and a scaling factor, while in other cases you may find the square to have its roots in a Minkowski equivalent of a "Pythagorean" operation ([itex]A^2=C^2-B^2[/itex] for Minkowski space, equivalent to [itex]C^2=A^2+B^2[/itex] for Euclidean space).
  10. Jun 29, 2005 #9
    Do you have a problem with square rooting pi? [tex]r=\frac{\sqrt{A}}{\sqrt{\pi}}[/tex] This is simply a rearranged version of [tex]A={\pi}r^2[/tex], which can be used to find the radius of a circle if you know the area.

    What about [tex]\sqrt{E}=\sqrt{m}c[/tex]? You don't have to square c; [tex]E=mc^2[/tex] is just a little more elegant than the former.
    Last edited: Jun 29, 2005
  11. Jun 29, 2005 #10


    User Avatar

    Staff: Mentor

    Its exctly the same as squaring velocity in the kinetic energy equation - thats what you have to do to turn a velocity into an energy (well, that and multiply by mass and a proportionality constant).

    Bernoulli's equation also uses velocity squared -- to calculate pressure. So it isn't an unusual thing to square a velocity..
    The speed of light is a constant, but it is also a real, physical speed.

    Lets look at it from the other angle: why would it be a problem to square a constant (others have provided examples...)?
    Last edited: Jun 29, 2005
  12. Jun 29, 2005 #11


    User Avatar
    Staff Emeritus
    Science Advisor

    c is not a "number" in the standard MKS unit system, it is a velocity. Squaring a velocity makes sense in certain cirumstances, as when one squares the velocity and multiplies by the mass to calculate the energy. The units work out correctly

    energy - force * distance = mass * acceleration * distance = kg * (m/s^2) * m =
    kg * (m/s)^2

    Thus we see that unitwise, energy is naturally the product of a mass multipled by the square of a velocity.

    Squaring a constant velocity makes sense under the same circumstances that squaring a non-constant velocity makes sense. The fact that the velocity is constant is really irrelevant to the units.
  13. Jun 29, 2005 #12


    User Avatar
    Science Advisor

    Let's see. How about h**0 = c **0= 1 = e**0, etc, where h is Planck's constant, and e is the electron's charge. I'm 6'2 tall. As long as observers are moving perpendicular to me, I will be 6'2 for all those observers. If my height is H, then all those observers will see the same value for H**1/7.

    The usual rules of algebra proclaim that the values of c**N, with N any real number from - infinity to + infinity are the same for all inertial observers. Hint: the rules of algebra are Lorentz invariant.

    Reilly Atkinson
  14. Jul 2, 2005 #13
    No Problem Squaring pi

    Speed of light involves time; pi involves space.

    We comprehend time as a non-zero because nothing can move at that speed.

    Yet something appears to change between objects separated by space.

    To both the similarity and change between objects, time is zero.
    Last edited: Jul 2, 2005
  15. Jul 2, 2005 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This makes no sense what-so-ever.

    You have a problem with squaring properties that involve time? What about acceleration which is distance/time²?
  16. Jul 2, 2005 #15


    User Avatar
    Staff Emeritus
    Science Advisor

    I suppose we'd better not square Planck's constant, either.

    The Bohr radius


    however does involve the square of Planck's constant. Since h-bar has units of

    kg m^2 / sec

    I suppose this must be a bad thing since it's even more complicated in its units than a velocity, and it's a universal constant, to boot.

    (Personally, I still don't see the problem).

    There are numerous other places in physics where fundamental constants are squared. I would just try and tell Swampbeast to "deal with it"
    Last edited by a moderator: Apr 21, 2017
  17. Jul 2, 2005 #16
    I believe that the OP has an issue with the invariant C being expressed, mathematically, beyond the constants' immutable value.
    In a sense the OP is correct is such concern, as a constant value can not be un-reasonably altered in any equation.
  18. Jul 2, 2005 #17

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Nowhere in any equation of special relativity is 'c' anything other than 'c'. Squaring the speed of light doesn't give you a new value of 'c', it gives you a value of 'c2'

    There is nothing correct about the OP.

    (edit: except the first line, that is)
    Last edited: Jul 2, 2005
  19. Jul 2, 2005 #18
  20. Jul 2, 2005 #19

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No I'm not, and no it doesn't. Squaring the speed of light is not in any way contrary to the speed of light postulate. I have no idea of why anyone would think that it is.

    This makes absolutely no sense.
  21. Jul 2, 2005 #20
    Are you then suggesting that c-squared is a valid condition? If so, it's inclusion in mathematics is valid.
    What evidence, then, does anyone have for a c-squared phenomenon?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook