High School How to relate relativistic kinetic energy and momentum

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The discussion focuses on determining the correct graph for relativistic momentum as a function of relativistic kinetic energy. Initial attempts to express momentum in terms of kinetic energy using the equations for momentum and kinetic energy were unsuccessful without an additional relationship. Participants confirmed that the Lorentz factor, gamma (γ), relates to velocity (v) and can be used to derive expressions for momentum and kinetic energy in terms of velocity. The equation relating energy and momentum was also introduced, leading to the conclusion that the correct graph is C, as it aligns with the derived relationship for momentum in terms of kinetic energy. Overall, the discussion emphasizes the importance of understanding the relationships between these relativistic quantities.
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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Einstein said, when describing someone falling off a building, that the Earth accelerating up to meet him/her. Without the Earth getting larger in all directions as the paradox goes, it curvature of space-time which is why you can have the acceleration up without the surface moving up as you follow a geodesic path. Any deviation from that geodesic will requires a force which is what causes you to have weight on a scale on earth. However, what if we consider an orbiting satellite which is...

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