Deriving Relativistic Momentum

soothsayer · Dec 17, 2010

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?

bcrowell · Dec 17, 2010

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

bobc2 · Dec 17, 2010

Multi-panel space-time diagram with axes X4 and X1 showing blue and red rockets moving apart, a Pythagorean-style triangle with the Einstein hypotenuse equation, and vector diagrams e) with momentum p = mv/sqrt(1-v^2/c^2) and energy p4 = mc.

Scanned page of typed text with two equations: c squared times Delta t double-prime squared equals c squared t'^2 minus v1' squared t'^2 (equation 1), and Delta t'' = Delta t' times the square root of (1 minus v1'^2 over c^2), equal to Delta t' over gamma, with gamma = 1 over the square root of (1 minus v1'^2 over c^2) (equation 2).

Scanned page of text with equations 3, 4 and 5 deriving 4-velocity relations from the spacetime interval, using proper time and the speed of light c.

Typed page of relativity derivation showing equation 6, c^2 = (γc)^2 - (γv'_1)^2; equation 7, P_4'^2 = P_4'^2 - P_1'^2; equation 8, (mc)^2 = (γmc)^2 - (γmv'_1)^2; and equation 9, γmc = γmc^2/c = E'/c.

Two equations on a white background: equation 10 gives E' = γmc² = mc² / sqrt(1 − v'₁²/c²), and equation 11 expands it as mc²[1 + ½ v'₁²/c² + ⅜ v'₁⁴/c⁴ + ...] ≈ mc² + ½mv'₁², with the surrounding text in red and black.

Deriving Relativistic Momentum

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