Half of Energy: Where Does It Go? (Solving e=mv^2)

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The discussion centers on the relationship between energy and kinetic energy, specifically questioning the missing half of energy when equating e=mv^2 with the classical kinetic energy formula 1/2mv^2. It highlights that the classical kinetic energy equation is only applicable at low velocities, while the relativistic kinetic energy formula T = mc^2(γ-1) provides a more accurate representation at high speeds. Participants emphasize the importance of using correct formulas to avoid confusion between relativistic and non-relativistic physics. Clarification on the source of the formulas is requested to enhance understanding. The conversation underscores the need for precision in discussing energy equations in physics.
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combining wavelength(&)=h/mv and e=hv/& , we get e=mv^2
but kinetic energy is 1/2mv^2. where is other half?
 
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ankitpandey said:
combining wavelength(&)=h/mv and e=hv/& , we get e=mv^2
but kinetic energy is 1/2mv^2. where is other half?
Kinetic energy only takes the form 1/2mv2 in classical mechanics, which is only valid for low velocities. The full expression for relativistic kinetic energy is

T = mc^2\left(\gamma-1\right)

You should also note that your first two expression are only valid for v = c.
 
You might want to start with formulas that are actually correct, and not confuse relativistic stuff with non-relativistic stuff. Can you explain more clearly what formulas you're using and where you got them from? If you need greek symbols and don't know LaTeX, copy and paste from Redbelly98's https://www.physicsforums.com/blog.php?b=347 .
 
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THANKS, Hootenanny
 
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