How to Derive the Lorentz Transformation for the x Component of Momentum?

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SUMMARY

The discussion focuses on deriving the Lorentz transformation for the x component of momentum, specifically the equation Px' = γ(Px - vE/c²). The user initially applies the Lorentz transformation for velocity, ux' = (ux - v)/(1 + vux/c²), and attempts to relate momentum using Px = γm₀ux. However, they encounter difficulties in simplifying the resulting expression, particularly with an additional factor that complicates the derivation. The conversation highlights the importance of correctly applying the dependence of γ on velocity in the transformation process.

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  • Understanding of Lorentz transformations in special relativity
  • Familiarity with momentum and its relation to velocity
  • Knowledge of the gamma factor (γ) and its dependence on speed
  • Basic proficiency in algebraic manipulation of equations
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  • Learn about the implications of relativistic mass and rest mass in momentum calculations
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Students of physics, particularly those studying special relativity, and anyone interested in understanding the mathematical foundations of momentum transformations in relativistic contexts.

insynC
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Homework Statement



Derive the Lorentz transformation for the x component of momentum, i.e.

Px' = [tex]\gamma[/tex] (Px - vE/(c[tex]^{}2[/tex]))

I've used Px = x component of momentum (not very good with latex, sorry!)

Homework Equations



I thought the best place to start was the Lorentz transformation for velocity (which was given):

ux' = [ux - v] / [1 + v ux/(c[tex]^{}2[/tex])]

The Attempt at a Solution



Applying this, I used the fact Px = [tex]\gamma[/tex] m0 ux - where m0 is rest mass - and then fiddled around with it.

I was able to almost get the answer, except on the RHS I got what is required multiplied by a factor of:

1 / [ [tex]\gamma[/tex] - [tex]\gamma[/tex] ux v /(c[tex]^{}2[/tex]) ]

Unfortunately I couldn't show this was equal to 1 and am not even convinced it is. Was the approach I took the easiest way to the answer? I've tried it again and got the same problem, so maybe there is a better way to tackle it.
 
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insynC said:
Applying this, I used the fact Px = [tex]\gamma[/tex] m0 ux - where m0 is rest mass - and then fiddled around with it.

Hi insynC! :smile:

(have a gamma: γ)

(and use the X2 tag just above the reply field: Px :wink:)

I'm not sure that you used γ's dependence on speed:

Px = γ(u)ux , Px' = γ(v)vx
 
Good point, I hadn't taken that into consideration.

I have to now introduce γ(ux') and γ(ux), but ultimately the answer has a γ(v) in it. I'm struggling to remove the excess γ's introduced to convert the velocity transform into the momentum transform.
 

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