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Relativistic Kinetic Energy

  • #1
In the derivation of the relativistic kinetic energy,

[tex]K=\int_{x_1}^{x_2}F\,dx = \int_{0}^{v}\frac{d}{dt}(mv)\,dx = \int_{0}^{v}(mv\,dv+v^2\,dm)[/tex]

here, my lecturer told us without showing that

[tex]mv\,dv+v^2\,dm = c^2\,dm[/tex]

Can someone please give me hints on how to combine these two integrals? I have no idea how to start.
 

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  • #2
vela
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In the derivation of the relativistic kinetic energy,

[tex]K=\int_{x_1}^{x_2}F\,dx = \int_{0}^{v}\frac{d}{dt}(mv)\,dx = \int_{0}^{v}(mv\,dv+v^2\,dm)[/tex]
You need to be a bit more careful with the limits. The integral with respect to dm doesn't have as limits 0 and v.
here, my lecturer told us without showing that

[tex]mv\,dv+v^2\,dm = c^2\,dm[/tex]

Can someone please give me hints on how to combine these two integrals? I have no idea how to start.
You can show by differentiating the expression for the relativistic mass

[tex]m = \frac{m_0}{\sqrt{1-(v/c)^2}}[/tex]

with respect to v. The LHS of the result the lecturer gave you is the integrand, so just substitute it in to get

[tex]K = \int c^2\,dm[/tex]

I'll leave it to you to figure out the proper limits.
 

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