B How to relate relativistic kinetic energy and momentum

greg_rack
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Hi guys,

a special relativity problem requested to choose the right graph representing relativistic momentum ##p## as a function of rel. kinetic energy ##K##, from these four:
IMG_C89C1901D709-1.jpeg
At first, I tried writing ##p## as a function of ##K##, in order to then analyze the function's graph and see if it matches one of the four above, being ##p=\gamma mv## and ##K=mc^2(\gamma - 1)##, but I couldn't rearrange those two in such a way.
By deduction, I believe the graph should be C or D, since momentum would reasonably tend to infinity in a non-linear way(A) due to the presence of factor ##\gamma##, nor as indicated by B...
 
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You have an equation for ##p(\gamma, v)## and an equation for ##K(\gamma)##, so of course, you cannot use them to find an equation ##p(K)##, you need an extra equation, do you know of any equation that relates ##\gamma## and ##v##?
 
Gaussian97 said:
You have an equation for ##p(\gamma, v)## and an equation for ##K(\gamma)##, so of course, you cannot use them to find an equation ##p(K)##, you need an extra equation, do you know of any equation that relates ##\gamma## and ##v##?
Aren't ##\gamma## and ##v## simply related by ##\gamma##'s definition ##\gamma=\frac{1}{\sqrt{1-\beta ^2}}##, being ##\beta## the velocity in terms of ##c##?
 
greg_rack said:
Aren't ##\gamma## and ##v## simply related by ##\gamma##'s definition ##\gamma=\frac{1}{\sqrt{1-\beta ^2}}##, being ##\beta## the velocity in terms of ##c##?
Yes. So you should be able to write ##p(v)## and ##K(v)## and hence ##K(p)##.

Alternatively, do you know anything about ##E^2## and ##p^2c^2##?
 
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Ibix said:
Alternatively, do you know anything about ##E^2## and ##p^2c^2##?
Yes, ##E^2=p^2c^2+m^2c^{4}##.
Rearranging, it indeed gets ##p(E)=\frac{1}{c}\sqrt{E^2-E_0^2}##, hence:
$$p(K)=\frac{1}{c}\sqrt{K(K+2E_0)}$$
which corresponds to graph C :)
 
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