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Why is c even in the equation?

  1. Mar 21, 2004 #1
    why is "c" even in the equation?

    i was hoping someone could explain why light speed even enters the e=mc2 equation. energy equals mass times the speed of light squared. i get that, and what the formula predicts. but why the speed of light? i was never one to go into the mathematics of all of this.

    its just one of those things that you read and never really fully answer for yourself, i guess. i have read many books on relativity, and understand some things about it as a whole. but why is it that light-speed squared just happens to work out as the number that gives the right figure to multiply mass against? does it have something to do with the units that are used to calculate energy?
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  3. Mar 21, 2004 #2


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    Re: why is "c" even in the equation?

    It all goes back to the fact that c (the speed of light) is a constant for all observers. becuase of this, measurements of length and time between moving systems have to be adjusted by formulas that contain c (the Lorentz transformations).

    Now if you take these formulas to calculate the energy of a moving object (which depends on these measurements) you get a formula that looks like this:

    [tex]E = \frac{mc^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

    Note that if v =0, the formula reduces to

    [tex] E=mc^{2}[/tex]

    which is the energy equivalence of the object's rest mass.
  4. Mar 24, 2004 #3
    There are many books that give a derivation of the E=mc^2 forumla all based upon c as the velocity of light. There is another approach - a global derivation based upon expansion, where c is the expansion velocity of the Hubble sphere (which has the same value as the velocity of light)... accelerations of mass leads to forces, and force times distance is energy - when you calculate the energy required to remove a mass from the universe against this force - vola - you also get E = mc^2
  5. Mar 24, 2004 #4


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    Re: why is "c" even in the equation?

    It seems to me that it does have something to do with the units.
    Maybe the units do not explain everything but they might help give some understanding.

    In any consistent system of units the unit of energy is
    always equal to the unit of mass multiplied by the square of
    the unit speed

    take for example the normal metric system where the energy unit is the joule and the unit mass is the kilogram

    the joule is equal to one kilogram multiplied by the square of the unit speed

    that is one kilogram x (meter/second)2

    So if you have a mass and you want to get an energy quantity you are algebraically required (by the units system) to multiply by the square of some speed.

    So suppose Einstein had figured out that the energy content of an object is proportional to the object's inertia but he had not yet discovered that the ratio was c2

    he would think: if E is proportional to m, then there must be SOME speed that I can square and multiply m by, to get E. What could that speed be? What speed could give the constant of proportionality?

    If this proportion between E and m is a universal proportion then the speed I need must be some universal fundamental speed constant. Well the only universal constant speed I know is c, could it maybe be that?

    I guess the basic help that units give here is to show that if E and m are going to be proportional to each other there must be SOME speed whose square is the constant proportion between them.

    This could have been deduced 100 years before Einstein, people have known for as long as they have had energy units that energy units are always equal to the mass unit multiplied by the square of a speed.
    Newton could have told you that. Energy (work) = F x d = force x dist
    force = mass x accel
    force = mass x dist/time2
    force x dist = mass x dist2/time2
    force x dist = mass x speed2
    Last edited: Mar 24, 2004
  6. Mar 24, 2004 #5
    Einstein used thought experiments involving photons, so from studying these, one can understand where c, the speed of light, enters the story.

    There is one experimental fact you will have to acknowledge; photons carry besides energy (E) also an amount of momentum p=E/c.

    Imagine a box with mass M and length L on which no forces work, so the center of gravity will remain fixed. One the left side a photon is emitted and will cause the box to move to the other side. Because of the conservation of momentum the box will move with a velocity v wich obeys: Mv=p=E/c -> v=E/Mc. After a time t=L/c the photon will reach the oher side and the box will be moved a distance d=vt.

    The center of gravity will be shifted by an amount mL-Md. But wait, because the are no external forces at work the center of gravity must be fixed. So: mL=Md=Mvt=EL/c^2 -> m=E/c^2 or E=mc^2.

    So this shows that a photon can be assigned a 'mass' of m=E/c^2...

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    Last edited: Mar 24, 2004
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