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Help understanding e=mc^2

  1. Aug 15, 2009 #1
    I'm new to the forum and I'm not being purposefully thick, but has anybody wondered how this equation works? I mean if c is the fastest anything can go, then what sense does it make to use a quantity like c^2? Thanks!
     
  2. jcsd
  3. Aug 15, 2009 #2
    *Sigh* physicists are constantly asked about this equation.

    First off , c^2 is not even a speed. Speed has the dimension L/T = length/Time.
    c^2 obviously has the dimension L^2/T^2 and that is not the dimension of speed.


    The equation means that energy is proportional to the mass of an object. And a little mass has a LOT of energy. For example, you could take a book and extract the energy to power a town for a year. Provided, technology doesn't allow us to do that.

    Also, there should be an E=mc^2 sticky so people don't post these threads anymore.
     
  4. Aug 15, 2009 #3

    JesseM

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    As Pinu7 said, c^2 has units of length^2/time^2, and you can see that you'd need a constant of this type in an equation relating mass to energy because energy has units of mass*length^2/time^2.

    One way of thinking about it is in terms of reactions that convert rest mass into kinetic energy or vice versa--for example, when matter and antimatter particles collide they convert into a pair of photons, and energy is conserved in spite of the fact that the kinetic energy of the photons is much greater than the kinetic energy of the original particles, as long as you include the particles' rest mass as a type of energy which was converted to kinetic energy for the photons (which have zero rest mass).

    Another way of thinking about it is that if you use something like a scale to measure the mass of a bound system (like an atom, or a bunch of particles in a box, or a solid object), all forms of energy in the system combine to give the reading you get--thus for example a brick will weigh slightly more if you heat it up, because the heat increases the average kinetic energy of the particles that compose it, and likewise an electron and proton weigh slightly less when bound into a hydrogen atom than they do when weighed individually, because when bound into an atom their potential energy is lower.
     
  5. Aug 15, 2009 #4
    Ok, thanks. I now understand what you mean by the relationship in terms of dimensions. But is it just the "beauty of reality" expressed in math that allows the equation to be exactly c^2 and not c X some fraction of C, or is there something more intrinsic involved.
     
  6. Aug 15, 2009 #5

    russ_watters

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    The equation has the units of energy....if it was just C or if it was multiplied by something else, it wouldn't (necessarily) have units of energy.

    Note the similarity between that equation and the kinetic energy equation....e=1/2mv^2
     
  7. Aug 15, 2009 #6
    "The equation has the units of energy...." that's the part I get. It's the part of the specific quantity of (300,000 m/s)^2 which I find unique. Like why that number and not another?
    Separate question: Does e=mc^2 have anything to do with the principle that the resting mass of a particle changes continually as the particle approaches the speed of light. Is that special relativity effect due to a mass to energy conversion?
     
  8. Aug 15, 2009 #7

    JesseM

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    To understand why it's c^2 as opposed to some other velocity squared, you need to look at an actual derivation of the equation. Here is one for the special case of a photon being absorbed by the walls of a box, though you'd need a more general derivation for arbitrary bound systems.
     
  9. Aug 15, 2009 #8

    Janus

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    Special Relativity, from which e=mc² comes, is built on the postulate that c is an invariant speed. It is a fundamental constant of the universe. As a result, measurements of time and space differ depending on relative motion. Taking this into account, you find that the energy of a moving mass is not e=mv²/2 but
    [tex]e= \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
    Note that if v=0, the equation reduces to e=mc²
    So basically, it is the specific quantity because c is a fundamental property of the universe that has a profound effect on its operation.

    The resting mass never changes. What changes is what is called the relativistic mass(though is some circles even that term is not popular). Yes, there is connection. After all, they both come from the application of the same theory.
     
  10. Aug 15, 2009 #9
    Thanks
     
  11. Aug 16, 2009 #10
    Another derivtion of E=mc^2
    m=m0(1-v^2/c^2)^-1/2
    By binomial expansion,
    m=mo(1+1/2v^2/c^2+...) [neglect higher powers of v/c]
    m=m0+1/2m0v^2/c^2
    1/2m0v^2=(m-m0)c^2
     
    Last edited: Aug 16, 2009
  12. Aug 16, 2009 #11

    Fredrik

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    That's not a derivation Vin.
     
  13. Aug 16, 2009 #12
    What's wrong? I'd read it somewhere
     
  14. Aug 16, 2009 #13

    Fredrik

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    You can't derive an equality from an approximate equality. To call your calculation a derivation of E=mc2 is like saying that 3*3≈10 implies that 10/3=3.

    Note that I didn't say that your calculations are rubbish. I'm just saying that they're not a derivation of E=mc2. The terms you neglected are the ones that vanish the fastest when you take the limit c→∞, so what you're doing is to show that the relativistic expression for energy turns into the non-relativistic expression in that limit. That's an important result, but it's not the result you said you were deriving.
     
  15. Aug 16, 2009 #14
    v<<c, better
    What turns out from that is that kinetic energy is not exactly equal to 1/2m0v^2, otherwise the result is still E=mc^2
     
    Last edited: Aug 16, 2009
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