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Speed of light squared got to do with the production of energy

  1. Jul 8, 2004 #1
    Whas wondering what has the speed of light squared got to do with the production of energy what is this convienient relationship or is it in fact a number close to c2 and in fact has nothing to do with c2 but its used to make the equation look good ?
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  3. Jul 8, 2004 #2


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    The equation is a result of Special Relativity and can be derived. The 2 is indeed a 2. Note the similarity between this expression and the expression for kinetic energy.
  4. Jul 8, 2004 #3
    In the expression, C^2 doesn't really have anything to do with the production of energy. It is a factor for indicating how much energy is contained within a certain quantity of mass. You cannot move mass at the speed of light squared and produce energy.
  5. Jul 8, 2004 #4
    It's a result of a calculation. For a simple calculation see

    Notice also that mc2 has the same dimensions as energy. If the equation was E = mc then the dimensions would be incorrect (i.e. it wouldn't have the dimensions of energy). Notice the resemblance between E = mc2 and K = (1/2)mv2?

    But it was not put in there to make the dimensions come out right. It is a result of calculation.

  6. Jul 8, 2004 #5
    I think your missing my point, imagine one is primitve at physics ( like me ) and didnt know the equation than what are the odds that it would happen to be c2 ( oh is it exactly c2 ) . Ive actually begun to notice a few coincidentals and similiarties in the maths of physics.

    Is there something going on behind it all.
  7. Jul 8, 2004 #6
    Special relativity is the result of measuring and combining space and time in the same unit of distance. (One second in meters is the distance a light pulse covers in one second, 299792458 meters.) In special relativity energy is easiest expressed in units of mass, like kilograms. What the [tex]E=mc^2[/tex] is actually saying is that rest energy equals mass.
    By multiplying this by [tex]c^2[/tex] we simply change unit from kilograms to conventional joules of energy. It has something to do with energy being the time part of some 4-vector combining amount of movement and energy... can't remember how right now though, I'm just starting to get into this stuff again...

  8. Jul 8, 2004 #7


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    The formula for relativistic energy is [tex]e^2 = p^2c^2 + m^2c^4 [/tex]. e is the energy, p is the magnitude of the three dimensional momentum, m is the invariant mass, and c is the speed of light in a vacuum.

    In the particle's rest frame, its three-momentum is zero and the first term drops out, giving the famous relation.

    The energy momentum four-vector has a timelike component e, a spacelike component pc, and its magnitude is [tex]mc^2[/tex]; when you compute the magnitude as hypotenuse squared from the components you have to use the Minkowski version of Pythagoras, with the time component squared minus the space component squared. Thus: [tex] (mc^2)^2 = e^2 - (pc)^2 [/tex], which you can rearrage to give the energy formula.
  9. Jul 8, 2004 #8
    I think the point being made, or the question asked, from bozo, is WHY c^2 exists in an equation if c^2 is impossible in reality, regardless of mathematical convenience.
    That is, if a factor is expressed mathematically that is not possible to produce in reality, is there a problem here? I think that is what bozo is getting at.
  10. Jul 8, 2004 #9
    It is not an easy deduction from relativity; it is a somewhat tough deduction. The simple result E=mc2 just tumbles out from the math at the end. It often happens that simple-looking results require a heavier investment in math. There is a special case or two where the result comes out more easily.

    As for the dimension of energy:

    energy is equivalent to work; work has dimension force*distance; force has dimension mass*distance/time2; so work has dimension mass*distance2/time2;so work has dimension mass*(distance2/time2);(distance2/time2) is velocity2;so work is mass*velocity2;so energy is mass*velocity2.

    Therefore E = mc2 is at least viable, consisting of a mass and a velocity2. But this is NO proof that the equation is actually true. Notice from the previous paragraph that the velocity2 factor came NOT from the force or the distance separately, but from the combination of the two, somehow intricately joined. So c in this case doesn't represent the speed of the mass.
  11. Jul 9, 2004 #10
    Certainly c doesn't represent the speed of the mass itself. It represent the speed of circulation of energy inside mass.

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