Deriving Relativistic Momentum: A Guide

In summary, the formula for special relativistic momentum is p=ɣmv, where ɣ is the Lorentz factor, v is the velocity, and m is the object's mass. This differs from the classical formula, p=mv, because dx/dt is measured differently in different reference frames. This can be derived by considering time dilation and applying the Lorentz transformation. However, there are alternative ways to derive this formula, such as considering the results of collisions in all frames of reference.
  • #1
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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  • #2
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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1. What is relativistic momentum?

Relativistic momentum is a concept in physics that takes into account the effects of special relativity on the traditional concept of momentum. It describes the motion of an object with a significant fraction of the speed of light, which causes changes in its mass and energy.

2. How does relativistic momentum differ from classical momentum?

Classical momentum is calculated by multiplying an object's mass by its velocity, while relativistic momentum takes into account the object's mass, velocity, and the speed of light. This means that as an object's velocity approaches the speed of light, its relativistic momentum increases significantly compared to its classical momentum.

3. What is the equation for calculating relativistic momentum?

The equation for calculating relativistic momentum is p = m*v/√(1-(v^2/c^2)), where p is the relativistic momentum, m is the object's rest mass, v is its velocity, and c is the speed of light.

4. Why is it important to understand relativistic momentum?

Understanding relativistic momentum is crucial in accurately describing the motion of objects traveling at high speeds, such as particles in particle accelerators or spacecraft. It also helps to explain phenomena such as time dilation and mass-energy equivalence.

5. How can I apply the concept of relativistic momentum in my research or experiments?

If your research or experiments involve objects moving at high speeds, it is important to use the equation for relativistic momentum to accurately describe their motion. This will help you make more precise calculations and predictions about the behavior of these objects.

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