Relativistic momentum and gamma factor - differential equation

AI Thread Summary
The discussion focuses on deriving the force equations for a relativistic particle under different conditions, specifically when the force is perpendicular and parallel to the particle's velocity. There is confusion regarding the time derivative of the gamma factor, particularly why it is zero in the perpendicular case and the use of v_x instead of v. Participants suggest revisiting the time derivative of the full momentum vector to clarify the relationship. Additionally, there is a challenge in integrating the kinetic energy equation, with hints provided to explore hyperbolic functions for a solution. The conversation emphasizes breaking down complex problems into manageable parts for better understanding.
kapitan90
Messages
32
Reaction score
0

Homework Statement


I am supposed to show that the force on a relativistic particle when
a) it's perpendicular to particle's velocity is F=γm_0\frac{dv}{dt}
b) it's parallel to particle's velocity is F_x=m_0γ^3\frac{dv_x}{dt}

I may make use of the fact that \frac{dγ}{dt}=\frac{v_xγ^3}{c^2}*\frac{dv_x}{dt}
2. Relevant equation
I don't understand why \frac{dγ}{dt}is 0 in case a) (this follows from the last formula, but I don't understand it either), why do we use v_x instead of v which γ depends on ?
 
Last edited:
Physics news on Phys.org
I am also supposed to show that the kinetic energy of a particle accelerated from rest to v_xisW =∫F_xdx=m_0c^2(γ-1) but I am stuck with the integral∫ (1-v^2/c^2)^{-0.5} dx I tried to integrate it by parts and to use Wolfram, but it couldn't solve it either. Any ideas?
 
One problem at a time. First problem
Your formula for d(gamma)/dt is not quite right. Well, it's right for your scenario, but it's also confusing you. Try going over the time derivative again. Maybe first try to take the time derivative of the full momentum vector, and then analyze the two cases when it comes time. Then to predict a future complication, don't be worried if you have a v^2/c^2, you can rewrite as some common terms.

Okay, because I don't know when I'll check back, I'll give you a clue to the second problem, it starts with hyper and ends with trig.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top