(special relativity)Trajectory under constant ordinary force

  • Thread starter Thread starter hoyung711
  • Start date Start date
  • Tags Tags
    Constant Force
hoyung711
Messages
5
Reaction score
0

Homework Statement



A particle is subject to a constant force F on +x direction. At t = 0, it is located at origin with velocity vo in +y direction.

Homework Equations



Determine the trajectory of the particle. x(t),y(t),z(t)

The Attempt at a Solution



\vec{p}= \int \vec{F} dt
\vec{p} = \vec{F}t + constant

At t=0
\vec{p} = \gamma mv_{o}
\vec{p} = Ft\hat{x} + \gamma mv_{o}\hat{y}

what should I do next? should I integral over p_{x} and p_{y} separately? Is that so, what are the exact steps?

I tried using
p_{x} = \frac{mu_{x}}{\sqrt{1-\frac{u^{2}_{x}}{c^{2}}}} = Ft
x(t) = \frac{mc^{2}}{F} (\sqrt{1+\frac{Ft}{mc}^{2}} -1)

what about y(t)? it seems it is a linear with t. I don't know where I get wrong.
 
Last edited:
Physics news on Phys.org
hoyung711 said:
I tried using
p_{x} = \frac{mu_{x}}{\sqrt{1-\frac{u^{2}_{x}}{c^{2}}}} = Ft

\gamma involves u^2, not just the x-component of the velocity:wink:
 
gabbagabbahey said:
\gamma involves u^2, not just the x-component of the velocity:wink:

Do you mean I have to use
\frac{mu_{x}}{\sqrt{1-\frac{u^{2}_{x}+u^{2}_{y}}{c^{2}}}} = Ft
\frac{mu_{y}}{\sqrt{1-\frac{u^{2}_{x}+u^{2}_{y}}{c^{2}}}} = \gamma mu_{o}

and solve these coupling equations?
 
Last edited:
Yup.
 
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?
 
Using L.T. s.t. \bar{S} is moving in y-direction with u_{o}

\bar{F_{x}}=\gamma F
\bar{F_{y}}=0
\bar{F_{z}}=0

\bar{u_{x}}=0
\bar{u_{y}}=0
\bar{u_{z}}=0

Is this correct?
 
Last edited:
I think my L.T. transform got some problems since the answer is too simple..
Can anyone give a hand?
 
I just sketch the method I used.
First use
\frac{d\gamma mu_{x}}{dt}=F_{x}=F
\frac{d\gamma mu_{y}}{dt}=0
\frac{d\gamma mu_{z}}{dt}=0

Then shows z is always zero
Hence the u^{2} in the \gamma becomes u_{x}^{2}+u_{y}^{2}

Afterwards, integrated out u_{x} and u_{y}
decouple the equation s.t. one involves only u_{x} and other u_{y}
integrate it again, you will get the answer
 
Last edited:
Back
Top