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Taylor, Einstein and Newton

  1. Oct 5, 2007 #1
    I have E(v) = (mc^2)/(sqrt(1-(v^2/c^2)).

    I have found the second-order Taylor-polynomial for v=0, and I get:

    T_2_E(v) = mc^2 + ½mv^2.

    My teacher asks me, why this equation must be true - what is so special about the second order Taylor-polynomial for v = 0 for E(v)?
    Last edited: Oct 5, 2007
  2. jcsd
  3. Oct 5, 2007 #2
    Sorry, I didn't post in the homework-section. My bad
  4. Oct 5, 2007 #3


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    I've moved it for you.

    Let me ask you this with regards to your original question. How can one go about reconciling the relativistic expression for energy with the Newtonian expression for energy for "slow moving" bodies, where Newtonian theory works well?
  5. Oct 5, 2007 #4
    Well, the expression in my first post is the total kinetic energy a particle has - it unites both Newtons kinetic energy - for "slow moving" bodies - and Einsteins expression for kinetic energy in rest?
  6. Oct 5, 2007 #5


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    It is not that relationship which is so imprtant, but the following:
    where p is the momentum.

    If an object has zero momentum, say zero velocity, we get the relation sought.

    However, an equally fascinating relationship is seen, namely the possible existence of objects whose rest mass m=0, but still having both energy and momentum!
    In that case, we have E=cp.

    Essentially, one revolutionary shift in the conception of physics with Einstein was that no longer were the basic stuff of which objects are characterized by "mass" and "velocity", rather, "energy" and "momentum" becomes the proper stuff of which objects are made.

    Mass becomes a secondary concept; prior to Einstein, it was a primary concept.

    The E=mc^2 relation then contains the possibility that mass might be "reconverted" into pure energy, something frankly unthinkable prior to Einstein.
  7. Oct 5, 2007 #6
    - but I am a little confused, because the expression in #1 equals E(v) = (mc^2)/(sqrt(1-(v^2/c^2)) for v=0, which is the point the Taylorpol. is evolved abot?
  8. Oct 5, 2007 #7


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    Consider the relativistic total energy E of an object moving at some velocity v. Subtract from this energy E the relativistic total energy E_0 of an object moving at a velocity of zero - the result is 1/2 m v^2. This is the Newtonian kinetic energy. Do you see any physical significance in this result, or might you say "that's just a coincidence"?
  9. Oct 5, 2007 #8


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    Indeed, the relativistic "kinetic" energy will in the slow limit equal the classical kinetic energy+a constant.
  10. Oct 5, 2007 #9
    So you mean the total energy E = T_2_E(v) = mc^2 + ½mv^2 - which is then subtracted total energy E_0 for v=0 = mc^2, so we get ½mv^2?

    If yes, then it proofs Newton right, but only for small velocities, but not when they are equal to zero?
  11. Oct 5, 2007 #10
    - but wait, for v=0, we then get that E=0, which it isn't?
  12. Oct 5, 2007 #11
    The effect of the m c^2 term is to add a constant term to the energy. Consider e.g. a collision of two particles and write down the equation for conservation of energy. The m c^2 terms will cancel. This is similar to the freedom to add a constant term to the potential energy. The potential energy of a particle in a constant gravitational field of g is m g h, but it doesn't matter relative to which point you measure the height h.

    So, in the nonrelativistic limit, special relativity reduces to theory that says that the energy of a free particle is m c^2 + 1/2 m v^2, but such a theory is indistinguishable from a theory that says that the energy is 1/2 m v^2.

    The only way one could notice the effect of the m c^2 is if m were to change. This can happen even in the nonrelativistic realm, e.g. in nuclear reactions. Consider e.g. nuclear fission of U-235 in which several MeVs of energy is leberated. The fission products will still be nonrelativistic to a good approximation, but the masses will not sum up equally on both sides of the reaction. The mc^2 terms then don't cancel and the nonrelativistic theory doesn't apply. If you bring all the m c^2 terms to one side, then you see that you can interpret it classically as the energy that was stored in the U-235 that is converted to kinetic energy of the fission products.

    Another example. Consider a particle of mass m1 moving with velocity v1, colliding with a particle of mass m2 at rest and then sticking to that particle. Compute the velocity of this particle usoing classical mechanics. You know that in classical mechanics energy is not conserved here but momentum is conserved. You can calculate the enrgy loss.

    Next, solve this problem using special relativity. Note that now total energy is conserved (collision is inelastic leading to a loss of kinetic energy, but the heat from the collision is assumed to stay in the system and is thus accounted for by the m c^2 term). You can compute the mass of the particle after the collision and compute the difference of that with m1 + m2 and verify that this indeed accounts for the loss in kinetic energy.
  13. Oct 6, 2007 #12
    Ahh, I see. I'll just do the calculations ..

    Thanks everybody
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