How to relate relativistic kinetic energy and momentum

In summary, the conversation discusses solving a special relativity problem involving the graph representation of relativistic momentum as a function of relativistic kinetic energy. The participants suggest using equations for momentum and kinetic energy, but an additional equation relating gamma and velocity is needed. This leads to the conclusion that the correct graph is either C or D, with the final equation being p(K) = (1/c)√[K(K+2E0)].
  • #1
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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  • #2
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##?
 
  • #3
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##?
 
  • #4
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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  • #5
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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