Superstring
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p^2=p_0^2+2m \Delta k
I derived it using the following:
k=\frac{1}{2}mv^2
p=mv
\frac{dk}{dv}=mv=p
\frac{dp}{dv}=m
\frac{dk}{p}=dv
\frac{dp}{m}=dv
\frac{dk}{p}=\frac{dp}{m}
mdk=pdp
\int_{k_0}^{k}mdk=\int_{p_0}^{p}pdp
m\Delta k=\frac{1}{2}(p^2-p_0^2)
p^2=p_0^2+2m \Delta k
As far as I can tell it's correct, but I though I'd get a second opinion.
I derived it using the following:
k=\frac{1}{2}mv^2
p=mv
\frac{dk}{dv}=mv=p
\frac{dp}{dv}=m
\frac{dk}{p}=dv
\frac{dp}{m}=dv
\frac{dk}{p}=\frac{dp}{m}
mdk=pdp
\int_{k_0}^{k}mdk=\int_{p_0}^{p}pdp
m\Delta k=\frac{1}{2}(p^2-p_0^2)
p^2=p_0^2+2m \Delta k
As far as I can tell it's correct, but I though I'd get a second opinion.
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