Derive a formula for momentum in terms of kinetic energy

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SUMMARY

The discussion focuses on deriving a formula for momentum (p) in terms of kinetic energy (KE) using relativistic physics concepts. The key equations utilized include the total energy equation E² = p²c² + m²c⁴ and the kinetic energy equation KE = E - m₀c². The final derived relationship is p = KE/c, indicating that momentum is directly dependent on kinetic energy. The participant successfully navigates through complex equations and arrives at a clear conclusion with the assistance of provided resources.

PREREQUISITES
  • Understanding of relativistic energy equations
  • Familiarity with the Lorentz factor (gamma)
  • Knowledge of kinetic energy (KE) and its relation to total energy (E)
  • Basic concepts of particle momentum (p) and velocity (beta)
NEXT STEPS
  • Study the derivation of the total relativistic energy formula
  • Explore the implications of the Lorentz factor in momentum calculations
  • Learn about the relationship between kinetic energy and momentum in different frames of reference
  • Investigate advanced topics in relativistic mechanics and their applications
USEFUL FOR

Students of physics, particularly those studying relativistic mechanics, educators teaching advanced physics concepts, and anyone interested in the mathematical relationships between kinetic energy and momentum.

martinhiggs
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Homework Statement



Using:
particle velocity, beta
particle momentum, p
total energy, E
Lorentz factor, gamma
kinetic energy, KE

Derive an equation for momentum as a function of kinetic energy. The functions have to depend either on the variable in the bracket, p(KE), or on a constant.

The Attempt at a Solution



This is what I've done so far, and I am now stuck, and unsure if the way I am doing it is correct or if there is a different approach.


E^{2} = p^{2}c^{2} + m^{2}c^{4}

KE = E - m_{0}c^{2}

KE = \sqrt{p^{2}c^{2} + m^{2}c^{4}} - m_{0}c^{2}

p^{2} = \frac{KE^{2}}{c^{2}} - m^{2}c^{2} - m_{0}^{2}c^{4}

The only thing I could think of doing next is:

KE = \frac{p^{2}}{2m_{0}} , m_{0} = \frac{p^{2}}{2KE}

p^{2} = \frac{KE}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4KE^{2}}c^{2}

p^{2} + \frac{p^{4}}{4KE^{2}}c^{2} = \frac{KE}{c^{2}} - m^{2}c^{2}

p^{2}(1 + \frac{p^{2}}{4KE^{2}}c^{2}) = \frac{KE}{c^{2}} - m^{2}c^{2}

I'm not sure if this is the best or easiest way to do this, as it seems to be pretty messy, and I also have one more m in the equation that I need to get rid of but am not sure of the best way of doing so.

Any help will be greatly appreciated :)
 
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Ok, so I've been working on this problem for about 24 hours and I think I'm finally getting somewhere with it. In class we were given a sheet of useful formulae, and this included:

p = \gamma \beta m_{0} c = \frac{m_{0} \beta c}{\sqrt{1 - \beta^{2}}}

= \frac{\sqrt{E_{tot}^{2} - m_{0}^{2}c^{4}}}{c}

From this final equation, I noticed that

KE = \sqrt{E_{tot}^{2} - m_{0}^{2}c^{4}}

So this means that I have the relation:

p = \frac{KE}{c}

Which is momentum which is only dependent on KE or a constant!

The only problem I have now is working out where that equation from p comes from, can anybody help?
 
Great, finally I've figured it all out! Thank you for your help! :)
 

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