Derive a formula for momentum in terms of kinetic energy

Click For Summary

Homework Help Overview

The discussion revolves around deriving a formula for momentum in terms of kinetic energy within the context of relativistic physics. Participants are exploring relationships involving particle velocity, momentum, total energy, and the Lorentz factor.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate equations relating total energy and kinetic energy to express momentum as a function of kinetic energy. There are questions about the correctness of their approaches and the complexity of the equations being derived.

Discussion Status

Some participants have made progress in their derivations and have identified relationships between momentum and kinetic energy. However, there remains uncertainty about the origins of certain equations and the best methods to simplify their expressions. Guidance has been sought regarding the derivation process.

Contextual Notes

Participants are working under the constraints of using specific variables and relationships provided in class, including the need to eliminate certain mass terms from their equations. There is also a reference to a resource that may aid in understanding the derivation of momentum from total energy.

martinhiggs
Messages
22
Reaction score
0

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.


[tex]E^{2} = p^{2}c^{2} + m^{2}c^{4}[/tex]

[tex]KE = E - m_{0}c^{2}[/tex]

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

[tex]p^{2} = \frac{KE^{2}}{c^{2}} - m^{2}c^{2} - m_{0}^{2}c^{4}[/tex]

The only thing I could think of doing next is:

[tex]KE = \frac{p^{2}}{2m_{0}} , m_{0} = \frac{p^{2}}{2KE}[/tex]

[tex]p^{2} = \frac{KE}{c^{2}} - m^{2}c^{2} - \frac{p^{4}}{4KE^{2}}c^{2}[/tex]

[tex]p^{2} + \frac{p^{4}}{4KE^{2}}c^{2} = \frac{KE}{c^{2}} - m^{2}c^{2}[/tex]

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

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 :)
 
Physics news on Phys.org
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:

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

[tex]= \frac{\sqrt{E_{tot}^{2} - m_{0}^{2}c^{4}}}{c}[/tex]

From this final equation, I noticed that

[tex]KE = \sqrt{E_{tot}^{2} - m_{0}^{2}c^{4}}[/tex]

So this means that I have the relation:

[tex]p = \frac{KE}{c}[/tex]

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! :)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
3
Views
3K
Replies
46
Views
3K
Replies
8
Views
2K
Replies
13
Views
2K