(special relativity)Trajectory under constant ordinary force
1. The problem statement, all variables and given/known data
A particle is subject to a constant force F on +x direction. At t = 0, it is located at origin with velocity v_{o} in +y direction. 2. Relevant equations Determine the trajectory of the particle. x(t),y(t),z(t) 3. The attempt at a solution [tex]\vec{p}= \int \vec{F} dt[/tex] [tex]\vec{p} = \vec{F}t + constant[/tex] At t=0 [tex]\vec{p} = \gamma mv_{o}[/tex] [tex]\vec{p} = Ft\hat{x} + \gamma mv_{o}\hat{y}[/tex] what should I do next? should I integral over [tex]p_{x}[/tex] and [tex]p_{y}[/tex] separately? Is that so, what are the exact steps? I tried using [tex] p_{x} = \frac{mu_{x}}{\sqrt{1\frac{u^{2}_{x}}{c^{2}}}} = Ft[/tex] [tex]x(t) = \frac{mc^{2}}{F} (\sqrt{1+\frac{Ft}{mc}^{2}} 1) [/tex] what about y(t)? it seems it is a linear with t. I don't know where I get wrong. 
Re: (special relativity)Trajectory under constant ordinary force
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Re: (special relativity)Trajectory under constant ordinary force
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[tex]\frac{mu_{x}}{\sqrt{1\frac{u^{2}_{x}+u^{2}_{y}}{c^{2}}}} = Ft[/tex] [tex]\frac{mu_{y}}{\sqrt{1\frac{u^{2}_{x}+u^{2}_{y}}{c^{2}}}} = \gamma mu_{o}[/tex] and solve these coupling equations? 
Re: (special relativity)Trajectory under constant ordinary force
Yup.

Re: (special relativity)Trajectory under constant ordinary force
Do I have another way, because there will be messy if I solve the coupling and then take the integral.
Would Lorentz's transform help me to simplify the process? 
Re: (special relativity)Trajectory under constant ordinary force
Using L.T. s.t. [tex]\bar{S}[/tex] is moving in ydirection with [tex]u_{o}[/tex]
[tex]\bar{F_{x}}=\gamma F[/tex] [tex]\bar{F_{y}}=0[/tex] [tex]\bar{F_{z}}=0[/tex] [tex]\bar{u_{x}}=0[/tex] [tex]\bar{u_{y}}=0[/tex] [tex]\bar{u_{z}}=0[/tex] Is this correct? 
Re: (special relativity)Trajectory under constant ordinary force
I think my L.T. transform got some problems since the answer is too simple..
Can anyone give a hand? 
Re: (special relativity)Trajectory under constant ordinary force
I just sketch the method I used.
First use [tex]\frac{d\gamma mu_{x}}{dt}=F_{x}=F[/tex] [tex]\frac{d\gamma mu_{y}}{dt}=0[/tex] [tex]\frac{d\gamma mu_{z}}{dt}=0[/tex] Then shows z is always zero Hence the [tex]u^{2}[/tex] in the [tex]\gamma[/tex] becomes [tex]u_{x}^{2}+u_{y}^{2}[/tex] Afterwards, integrated out [tex]u_{x}[/tex] and [tex]u_{y}[/tex] decouple the equation s.t. one involves only [tex]u_{x}[/tex] and other [tex]u_{y}[/tex] integrate it again, you will get the answer 
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