Why relativistic momentum equals the following?

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The discussion centers on the derivation of relativistic momentum, specifically how to express momentum in terms of energy and mass. It highlights the equation P = √(γ² - 1) * mc, derived from the relationship between momentum and the Lorentz factor γ. Participants note that while using p = γmv is valid, eliminating velocity from the equations can lead to the same result. However, it is suggested that using the energy-momentum relation E² - p²c² = m²c⁴ is a more straightforward method to solve for momentum. The conversation emphasizes the importance of understanding these fundamental equations in relativistic physics.
Foruer
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In a solution to a problem we were given, it is written that a positron momentum with energy of 2mc2
(where γ=2) is √(γ2-1)*mc = √(4-1)*mc = √3*mc

How did they get that P=√(γ2-1)*mc?
 
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You are familiar with p = ϒmv right?
And also with ϒ = 1/√(1-(v/c)2) right?

Eliminate v from the two equations and what do you get?
 
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What I get is the answer ?:) thank you :wink:
 
While it gives you the right answer, it is much simpler to use the energy-momentum relation ##E^2 - p^2 c^2 = m^2 c^4## and just solve for ##p##.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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