Go Back   Physics Help Forum > College/University Physics Help > Special and General Relativity

Special and General Relativity Special and General Relativity Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old 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?
beginner is offline   Reply With Quote
Old Sep 16th 2015, 01:15 PM   #2
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,659
Originally Posted by beginner View Post
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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Special and General Relativity

Tags
energy, kinetic, variation



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Kinetic energy Fizzle Periodic and Circular Motion 2 Jul 1st 2014 12:52 AM
ratio of translational kinetic energy to rotating kinetic energy man_in_motion Periodic and Circular Motion 2 Mar 19th 2010 12:43 AM
Is the increase in kinetic energy equal to the decrease in potential energy? s3a Energy and Work 1 May 28th 2009 06:02 PM
Kinetic energy car Joene Energy and Work 5 Mar 26th 2009 04:04 AM
Kinetic Energy Morgan82 Energy and Work 1 Nov 10th 2008 05:43 PM


Facebook Twitter Google+ RSS Feed