How Can Special Relativity Connect Momentum and Velocity Changes?

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SUMMARY

The discussion centers on deriving the equation \(\frac{ \partial {m \mathbf{v}}}{\partial {s}} = \frac{ \partial { \mathbf{p}}}{c \partial {t}}\), which connects momentum and velocity changes in the context of special relativity. The participants explore the relationship between infinitesimal Lorentz invariant length (ds), proper time (tau), and non-proper time (dt). A key insight involves rewriting ds in terms of dt, leading to the expression \(ds = \sqrt{(c\,dt)^2-x^2} = \sqrt{c^2-v^2}\,dt\), which simplifies the differential equation for integration.

PREREQUISITES
  • Understanding of special relativity concepts, including Lorentz invariance.
  • Familiarity with momentum (\(\mathbf{p}\)) and velocity (\(\mathbf{v}\)) in physics.
  • Basic knowledge of differential equations and calculus.
  • Experience with proper time (\(\tau\)) and its significance in relativistic physics.
NEXT STEPS
  • Study the derivation of Lorentz transformations in special relativity.
  • Learn about the relationship between momentum and energy in relativistic contexts.
  • Explore differential equations in physics, particularly in the context of motion.
  • Investigate the implications of proper time versus coordinate time in relativity.
USEFUL FOR

Students of physics, particularly those studying special relativity, as well as educators and researchers looking to deepen their understanding of the connections between momentum and velocity in relativistic frameworks.

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Homework Statement


I am trying to solve produce an equation and the last step requires that

[tex]\frac{ \partial {m \mathbf{v}}}{\partial {s} } = \frac{ \partial { \mathbf{p}}}{c \partial {t} }[/tex]

where ds is the infinitesimal Lorentz invariant length, v is velocity, and p is momentum

The problem is that I do not know how to justify this last step.


Homework Equations





The Attempt at a Solution



I tried rewriting ds in terms of tau, but then it seemed that this proper time to be equal to non-proper time which is not true.
 
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Well, why haven't you tried to write ds in terms of dt? This gives

[tex]ds = \sqrt{(c\,dt)^2-x^2} = \sqrt{c^2-v^2}\,dt[/tex]

Plug this in and you have a very easy differential equation to integrate,
 

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