# Kinetic energy

1. Oct 4, 2005

### asdf1

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

2. Oct 4, 2005

### 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.

3. Oct 4, 2005

### Trilairian

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

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

4. Oct 4, 2005

### 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?

5. Oct 4, 2005

### pervect

Staff Emeritus
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

Last edited: Oct 4, 2005
6. Oct 4, 2005

### pmb_phy

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

7. Oct 6, 2005

### asdf1

wow! thank you very much! :)