Proving Relativistic Kinetic Energy: (1/2)*(gamma)mv^2

[asdf1]

How do you prove that (1/2)*(gamma)mv^2 doen't equal the kinetic energy of a particle moving at relativistic speeds?

 

[dextercioby]

Simply because

E_{kin}=m_{0}c^{2}\left(\gamma-1\right)

,which is different from your \frac{1}{2} \gamma m_{0}v^{2}...?

Daniel.

 

[Trilairian]

How do you prove that (1/2)*(gamma)mv^2 doen't equal the kinetic energy of a particle moving at relativistic speeds?

... yet another example why relativistic mass was a bad idea.

Mass doesn't change with speed.
E_{K} = (\gamma -1)mc^{2}

 

[golith]

How so mass does not change with speed? I thought that as a particle approaches the speed of light then it must lose mass. What is it i got wrong here?

 

[pervect]

There are two sorts of mass. One of them, called invariant mass, stays constant regardless of velocity and is a property of the particle itself (it doesn't depend on the particles state of motion).

This is preferred by many, probably even most, people, but there are a few vocal people who prefer the other sort of mass, relativistic mass.

Relativistic mass _increases_ with velocity according to the formula

m_r = \gamma m_0

where m_r is the relativistic mass, m_0 is the invariant mass, and \gamma = \frac{1}{\sqrt{1-(v/c)^2}} depends on the velocity of the particle.

For some more information, see for instance the sci.physics.faq "Does mass change with velocity".

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

 

[pmb_phy]

... yet another example why relativistic mass was a bad idea.

Mass doesn't change with speed.
E_{K} = (\gamma -1)mc^{2}

This question has absolutely nothing to do with the great idea of relativistic mass.

asdf1 - Its a matter of calculation. Simply calculate the kinetic energy and you'll obtain

K = (\gamma - 1)m_0 c^2

I worked out the calculation based on the work-energy theorem and placed them online at - http://www.geocities.com/physics_world/sr/work_energy.htm

Pete

 

[asdf1]

wow! thank you very much! :)