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Inquiry about kinetic force in STR

  1. Aug 9, 2011 #1
    I have recently bought Einstein's book on the General and Special theory of Relativity and I have finished reading the first half (STR). I understood the ideas and concepts being discussed, as well as the Lorentz transformation, but I have only on question, how did he obtain the relativistic formula for kinetic force ([itex]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]) ?
     
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  3. Aug 9, 2011 #2
    Not sure what you mean by 'kinetic force'. From the energy equation
    [itex]E^2=p^2c^2+m^2c^4[/itex]​
    the kinetic energy is
    [itex]mc^2\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)[/itex]​
    whereas the total energy, E, is
    [itex]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]​
     
  4. Aug 9, 2011 #3

    pervect

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    Assuming you mean the ordinary force, sometimes called the 3-force ( to distinguish it form the relativistic 4-force), it's just

    F = dp/dt

    where p is the relativistic momentum. I'm assuming you're familiar with the concept of a derivative, if not - well, I'm not sure if the explanation can be made any simper, but it'd be good to know.
     
  5. Aug 9, 2011 #4
    This is the chapter of the book that I got that from http://www.bartleby.com/173/15.html.

    I am familiar with the concept of derivatives, but I'm just learning calculus and can only do some basic ones.

    What I was wondering was what is the expression of kinetic energy (of a mass m) from STR, how did they get from the original formular for kinetic energy to this one, and can the Lorentz transformation tell us anything about mass, velocity, acceleration?
     
  6. Aug 9, 2011 #5

    WannabeNewton

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    You can actually go from that formula to the original one. You know that [itex]\mathbf{P} = m\mathbf{U} = m\frac{\mathrm{d} \mathbf{x}}{\mathrm{d} \tau }[/itex] and if you lorentz boost to some moving frame, then [itex]P^{0} = E = m(1 - v^{2})^{-1/2} \approx m + \frac{1}{2}mv^{2}[/itex] to first order in [itex]v[/itex].
     
    Last edited: Aug 9, 2011
  7. Aug 9, 2011 #6

    jtbell

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    You can derive the relativistic kinetic energy from the work-energy theorem. It starts out just like in classical mechanics, but you have to use relativistic momentum instead of classical momentum. See this post:

    https://www.physicsforums.com/showpost.php?p=458653&postcount=2

    (m is the rest mass and [itex]\gamma = 1 / \sqrt{1 - v^2/c^2}[/itex])
     
  8. Aug 9, 2011 #7

    pervect

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    You can find Einstein's derivation in his original paper:

    http://www.fourmilab.ch/etexts/einstein/specrel/www/
    (an online source). But you may need more than a passing familiarity with calculus to follow it - it's not super hard, but you'd need to be comfortable with transforming the differential equations of motion from one frame to another using the Lorentz
    transforms.

    Basically Einstein takes the POV that we know Newton's laws work at zero velocity, and uses that to find the relativistic expression that works at all velocities.

    Also, if your goal is to understand special relativity better, you'd be better off with a more modern textbook (such as Taylor & Wheeler's space-time physics), dealing with the more modern idea of the energy- momentum 4-vector, rather than reading Einstein's original approach involving the so-called transverse and longitudinal masses, which arise from transforming the differential equations as above.

    As others have noted, once you've gotten the expression for relativistic momentum (by whatever means), you can find the relativistic energy by work = force * distance.

    force = dp/dt, the rate of change of momentum with time

    Equivalently, power = dE/dt = force * dx/dt = force * velocity
     
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