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

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Discussion Overview

The discussion centers on the proof regarding the relativistic kinetic energy of a particle, specifically addressing the expression (1/2)*(gamma)mv^2 and its validity at relativistic speeds. Participants explore the definitions and implications of kinetic energy in the context of special relativity.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that the correct expression for relativistic kinetic energy is E_{kin}=m_{0}c^{2}(\gamma-1), which differs from (1/2)*(gamma)mv^2.
  • There is a contention regarding the concept of mass, with some arguing that mass does not change with speed, while others reference the idea of relativistic mass, which increases with velocity according to m_r = gamma m_0.
  • One participant emphasizes that the question of relativistic mass is irrelevant to the calculation of kinetic energy, suggesting that the focus should be on the work-energy theorem to derive the correct kinetic energy expression.
  • Another participant expresses confusion about the concept of mass change with speed, indicating a misunderstanding of the definitions of invariant mass and relativistic mass.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the expression (1/2)*(gamma)mv^2 for relativistic kinetic energy. There are competing views regarding the concept of mass and its relationship to velocity, with some advocating for invariant mass and others for relativistic mass.

Contextual Notes

Participants reference different interpretations of mass and kinetic energy without resolving the underlying assumptions or definitions. The discussion highlights the complexity of the topic and the varying perspectives on relativistic concepts.

asdf1
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How do you prove that (1/2)*(gamma)mv^2 doen't equal the kinetic energy of a particle moving at relativistic speeds?
 
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Simply because

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

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

Daniel.
 
asdf1 said:
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.
[tex]E_{K} = (\gamma -1)mc^{2}[/tex]
 
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?
 
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

[itex]m_r = \gamma m_0[/itex]

where [itex]m_r[/itex] is the relativistic mass, [itex]m_0[/itex] is the invariant mass, and [itex]\gamma = \frac{1}{\sqrt{1-(v/c)^2}}[/itex] 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:
Trilairian said:
... yet another example why relativistic mass was a bad idea.

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

[tex]K = (\gamma - 1)m_0 c^2[/tex]

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
 
wow! thank you very much! :)
 

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