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Kinetic energy 1/2 mv^2 looks similar to E = mc^2

  1. Jul 10, 2007 #1
    Last year in physics, i noticed that the formula for kinetic energy is ( E=1/2mv^2 ) looks an awful lot like the famous E=mc^2 , since c is a speed...

    How then, are these two equations related?

    I'm guessing that E=mc^2 only has to do with things travelling at the speed of light, but that still doesn't explain the 1/2 in the equation for kinetic energy.

    It isn't terribly important or anything I am just curious and don't understand it :P thanks
     
  2. jcsd
  3. Jul 10, 2007 #2

    berkeman

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    I'm no expert in relativity (there's an understatement!), but E=mc^2 is the "rest energy" of a mass. It is not related to the kinetic energy of motion. Hopefully you'll get some better answers than that from others here on the PF. Here is a wikipedia.org page to give you some background.

    http://en.wikipedia.org/wiki/E=mc^2

    I'm moving this thread to another general forum where you should get some good answers.
     
  4. Jul 10, 2007 #3

    Dick

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    Your guess there is a close relation is accurate. The full form of Einsteins relation is E=m*c^2*sqrt(1+v^2/c^2). For v<<c this is approximately, E=m*c^2*(1+(1/2)*v^2/c^2)). Or E=m*c^2+(1/2)*m*v^2. Pretty close relation, huh?
     
  5. Jul 10, 2007 #4

    cristo

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    The relativistic kinetic energy of a particle can be written as [tex]K=mc^2(\gamma-1)[/tex] where here gamma is the Lorentz factor, which is equal to [tex]\left(1-\frac{v^2}{c^2}\right)^{-1/2}[/tex]. For small velocities, i.e. v<<c, [tex]\gamma=1+\frac{1}{2}\frac{v^2}{c^2}+\cdots[/tex] and hence [tex]K\rightarrow\frac{1}{2}mv^2[/tex].
     
  6. Jul 10, 2007 #5
    Just to add one more note to the previous responses (which were good), the big surprise about E = mc^2 was that previously we'd thought that energy went to zero as velocity went to zero, based on KE = 1/2 mv^2. Then we learned that there's this stubborn non-zero term that doesn't go away. That's why it's called the "rest energy" - it's the energy you have left when velocity has gone to zero.
     
  7. Jul 11, 2007 #6
    You can think of the mc^2 term as a sort of potential energy associated with the mass of an object.
     
  8. Jul 11, 2007 #7
    If it isn't terribly important or anything and you are just curious, it means you have a more genuine interest on physics.
     
  9. Jul 11, 2007 #8
    To add another more note, I would say it in another way: if you heat a stationary body, it weighs more!
     
  10. Jul 11, 2007 #9
    A good introductory question to ponder is why physics was able to progress so far without ever taking this rest energy term into account. After all, energy calculations are a vital component of most problems in physics, yet we have been leaving (what is usually) the biggest chunk of energy out. How is this possible?

    Hint: We have always been free to vary the magnitude of a system's potential energy by changing our choice of origin for the system. The same goes for kinetic energy, except it is not the location of the origin that we change but the entire speed of the reference frame. Since Newton's Laws are valid in any inertial reference frame this impies the dynamics of the system should be the same as well, even though the energy calculated is different for each frame. Come to understand why this is true and you will know the answer to the original question.
     
    Last edited: Jul 11, 2007
  11. Jul 11, 2007 #10
    This is wrong. Only massless objects can travel at the speed of light. Since m=0 this means they have no rest energy (E = mc^2 = 0). E=mc^2 actually only applies to things not travelling at the speed of light. For massless particles all of the energy is in its momentum: E=pc.
     
  12. Jul 12, 2007 #11
    thanks for the answers :)

    Thanks for all the answers ^_^

    wow that's *weird* :) and by weird i mean i wouldn't have thought of that.

    So if they are massless p must not =mv ... i'm assuming that p=mv is for bigger slower things.
     
  13. Jul 13, 2007 #12
    As with kinetic energy, momentum has a more general form given by:

    [itex]\vec{p}=\gamma m\vec{v}[/itex]

    where [itex]\gamma[/itex] is again the Lorentz factor. If you use the same power series expansion as cristo described earlier, then this reduces to the classical result for momentum (just as kinetic energy did):

    [itex]\vec{p}=m\vec{v}[/itex]

    Notice that the classical result implies a massless particle has no momentum. However, the more general and accurate equation states that the momentum is indeterminate (0\0) for massless particles iff their velocity is the speed of light. Which, as it turns out, is always the case!
    Consequently, we cannot use this equation to determine the momentum of massless particles either. So where does the result [itex]p=\frac{E}{c}[/itex] come from? As it turns out, there is another important equation from relativity that relates the energy of a particle to its momentum:

    [itex]E^{2}=(pc)^{2}+(mc^{2})^{2}[/itex]

    and it is from this that we are finally able to obtain our result for massless particles. I should warn you though, that this equation actually sufferes from the same 0/0 fiasco when applied to massless particles. You just can't tell when it is written in this polished form. The actual (and rarely seen) form is:

    [itex]E^{2}=(pc)^{2}+\left(\frac{mc^{2}\gamma}{\gamma}\right)^{2}[/itex]

    By ignoring the [itex]\frac{\gamma^{2}}{\gamma^{2}}[/itex] factor, technically, we are implying that the invaraince of our 4-momentum applies to all reference frames, including even those frames moving at the speed of light. This claim is definitely overstepping our bounds. Hence this equation can only suggest that [itex]p=\frac{E}{c}[/itex] is true. It is experiments that have actually born this out (for photons and neutrinos anyways).
     
    Last edited: Jul 13, 2007
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