Deriving Relativistic Momentum

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SUMMARY

The derivation of relativistic momentum is defined by the formula p=ɣmv, where ɣ (gamma) is the Lorentz factor, calculated as ɣ=1/(1-(v/c)^2)^1/2. This formula arises from the transformation of time and space in different inertial reference frames, specifically through time dilation and length contraction. The discussion highlights confusion regarding the relationship between classical momentum and relativistic momentum, emphasizing that while classical momentum is p=mv, relativistic momentum incorporates the Lorentz factor to account for the effects of high velocities approaching the speed of light.

PREREQUISITES
  • Understanding of Lorentz transformations
  • Familiarity with the concept of time dilation
  • Basic knowledge of classical mechanics, specifically momentum
  • Knowledge of special relativity principles
NEXT STEPS
  • Study the derivation of the Lorentz factor in detail
  • Explore alternative derivations of relativistic momentum
  • Learn about relativistic energy and its relationship to momentum
  • Investigate the implications of relativistic effects in high-energy physics
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Students of physics, educators teaching special relativity, and anyone interested in understanding the principles of relativistic momentum and energy.

soothsayer
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I'm having some problems trying to figure out how to derive relativistic momentum. The way it was explained to me, classically, p=mv=m(dx/dt),but dx/dt is measured differently in different reference frames. So, if you look at time dilation, t=ɣt' where ɣ= 1/(1-(v/c)^2)^1/2 and t' is time in the moving inertial reference frame. So, dt/dt' = ɣ. m(dx/dt') = m(dx/dt)(dt/dt') = ɣmv. So, p=ɣmv, which is the given formula for special relativistic momentum (though usually u is used instead of v to distinguish velocities of frames and objects). I get all of this.

My question is, don't we have to find p=m(dx'/dt')? if dx/dx' = ɣ and dt/dt' = ɣ, then
m(dx'/dt') = m(dx/dt)(dt/dt')(dx'/dx) = mv(ɣ/ɣ) = mv.
Or, basically, x'=x/ɣ and t'=t/ɣ, so dx/dt = dx'/dt'. Length contracts at the same rate time dilates (dictated by value of ɣ). So where does p=ɣmv come from?
 
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soothsayer said:
I'm having some problems trying to figure out how to derive relativistic momentum. The way it was explained to me, classically, p=mv=m(dx/dt),but dx/dt is measured differently in different reference frames. So, if you look at time dilation, t=ɣt' where ɣ= 1/(1-(v/c)^2)^1/2 and t' is time in the moving inertial reference frame. So, dt/dt' = ɣ. m(dx/dt') = m(dx/dt)(dt/dt') = ɣmv. So, p=ɣmv, which is the given formula for special relativistic momentum (though usually u is used instead of v to distinguish velocities of frames and objects). I get all of this.
Hmm...this doesn't make a lot of sense to me. I think either someone gave you a bad explanation or your notes on it are garbled.

soothsayer said:
Or, basically, x'=x/ɣ and t'=t/ɣ
But this isn't how the Lorentz transformation actually works.

There are lots of different ways of approaching relativistic energy and momentum. If the approach your instructor used isn't working for you (or for me), why not just look at a different derivation? For example, one can get there by requiring that the results of collisions make sense in all frames of reference: http://www.lightandmatter.com/html_books/6mr/ch01/ch01.html#Section1.3
 
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