Physics Help Forum Variation of kinetic energy in SR

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 Sep 16th 2015, 10:17 AM #1 Junior Member   Join Date: Sep 2015 Posts: 2 Variation of kinetic energy in SR I was playing around on Mathematica with the kinetic part of the La Grangean for special relativity (with the speed of light equal to one): m/Sqrt[1-v^2]. I took the derivative with respect to velocity and then took the derivative of the resulting term with respect to time and then simplified. I got the following. (m*(1+2v^2)*a)/((1-v^2)^(5/2)) I then assumed that there was no potential energy and hence set the resulting equation equal to zero. There are four solutions: m=0 a=0 v=I/Sqrt[2] v=-I/Sqrt[2] The first two solutions make sense to me. But what do the last two mean? What does it mean for the velocity to equal an imaginary number. I do know that the numbers in question are (or are close to?) the eigenvectors for the spin matrix of the photon. But what does it mean for the velocity to equal such numbers? Did I do the variations improperly? Or do the solutions somehow make sense?
Sep 16th 2015, 01:15 PM   #2

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 Originally Posted by beginner I was playing around on Mathematica with the kinetic part of the La Grangean for special relativity (with the speed of light equal to one): m/Sqrt[1-v^2]. I took the derivative with respect to velocity and then took the derivative of the resulting term with respect to time and then simplified. I got the following. (m*(1+2v^2)*a)/((1-v^2)^(5/2)) I then assumed that there was no potential energy and hence set the resulting equation equal to zero. There are four solutions: m=0 a=0 v=I/Sqrt[2] v=-I/Sqrt[2] The first two solutions make sense to me. But what do the last two mean? What does it mean for the velocity to equal an imaginary number. I do know that the numbers in question are (or are close to?) the eigenvectors for the spin matrix of the photon. But what does it mean for the velocity to equal such numbers? Did I do the variations improperly? Or do the solutions somehow make sense?
They don't mean anything, just like when you square an equation in the process of solving it...you sometimes get extra solutions.

When you solved this you set the potential energy equal to 0 and did not include a generalized force. Thus you essentially assumed that a = 0, which is why you got the result you got. If you had put a generalized force term into the mix you would have found you derived F = ma.

-Dan
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