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E=mc^2 as a measurement tool?

  1. Sep 7, 2004 #1
    Hey all,

    Just discovered your informative forum. In these times of stupidity, its good to see people with a love of knowledge.

    A friend asked me the following question:

    "if you know the mass & speed of a plane, can you use this theory
    to calculate it's energy?"

    I assume the answer is no because c has to be a constant (the speed of light). :confused:

    Can someone please enlighten me as to whether the equation can be used in this manner?

    Thanks alot
     
  2. jcsd
  3. Sep 7, 2004 #2
    Another question: Where did e=mc^2 come from? How was it derived?
     
  4. Sep 7, 2004 #3
    Sure, use your formula and add :

    [itex]
    \[
    KE = \frac{1}{2}m\,v^2
    \]
    [/itex]

    The rest energy from E=mc^2 and the kinetic energy from above.
     
  5. Sep 7, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What energy are you talking about? The potential energy depends only on the conservative forces (mostly gravity) acting on the airplane. Its Kinetic energy is
    (1/2)mv2 where v is the speed of the airplane (relative to the observer). mc2 gives what might be called its "relativistic" energy which is seldom relevant in classical problems.

    I'm trying to remember the simplest derivation of "E= mc2. It can be derived, if I remember correctly, from considering a particle absorbing and then emitting light (calculating both energy and momentum in between absorption and emission).
     
  6. Sep 7, 2004 #5
    The complete formula is
    [tex]E^2 = \vec{p}\,^2 + m^2[/tex] when one uses units in wich c=1. In this formula, [tex]m[/tex] is a constant, the rest mass, and [tex]\vec{p}[/tex] is the linear momentum [tex]\vec{p}=m\vec{v}[/tex]. Alternatively, one can define the mass to be running with speed, such that [tex]E=m[/tex] applies. This is physically motivated by the interpretation in terms of inertia. I do not wanna go into semantics here. Try to find Pet's posts (pmb_phys is his user name)
    check also wofram

    The E=m comes from the nature of 4-vectors : the energy is a component of the energy-momentum 4-vector. 4-vectors transform under a change of referential frame as described by Lorentz transformations (see link above)

    The way it was derived by Einstein himself... I am not too sure, I forgot.
     
  7. Sep 7, 2004 #6
    Einstein himself derived it from a 'gedankenexperiment' involving the movement of a photon in a box, and considering the center of mass as a photon moves from the one side to the other. But this argument relies on another equation, namely that of the relation between energy and momentum in EM waves [itex]E=pc[/itex]. So if you're more comfortable with this equation (following from the Maxwell equations) you could try reading: http://www.geocities.com/physics_world/sr/einsteins_box.htm

    Ofcourse using p=mv=mc (in case of a photon) E=pc is basically E=mc^2, but it is a little more subtle than just using the classical formula for momentum.

    Another simple derivation is the following (this was the derivation used in my relativity textbook):

    In Special relativity (SR) an objects inertia increases as it approaches the speed of light, making it more and more difficult to increase the speed. You could assign this extra inertia to the mass of the object by saying the mass (this is actually called 'relativistic mass' , but I'll call it just mass) increases. Working this out it turns out the mass increases with a factor [itex] \gamma (v)[/itex]. This is a velocity dependent function increasing to infinity as the speed v approaches c thus making it impossible to acquire a speed large than c. I SR time slows down and length is shortened by the same factor!

    So mathematically this means the mass (m(v)) in terms of it's rest mass (m) will be:
    [tex]m(v) = \gamma m[/tex] with [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]



    For low speeds this can be approximated mathematically by:
    [tex]m(v)=m+\frac{1}{c^2}(\frac{1}{2}m \gamma v^2)[/tex]

    But this last term is a particles low speed kinetic energy divided by c^2! So the kinetic energy of a particle contributes to its mass in a way wich is consistent with:

    [tex]E=m \gamma c^2[/tex]


    Or in terms of relativitsic mass: [tex]E=m(v) c^2[/tex]. Einsteins famous equation!

    See also: https://www.physicsforums.com/showthread.php?t=41354
     
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