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-   -   (special relativity)Trajectory under constant ordinary force (http://www.physicsforums.com/showthread.php?t=402402)

hoyung711 May11-10 12:27 AM

(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 vo 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.

gabbagabbahey May11-10 02:11 AM

Re: (special relativity)Trajectory under constant ordinary force
 
Quote:

Quote by hoyung711 (Post 2712125)
I tried using
[tex] p_{x} = \frac{mu_{x}}{\sqrt{1-\frac{u^{2}_{x}}{c^{2}}}} = Ft[/tex]

[itex]\gamma[/itex] involves [itex]u^2[/itex], not just the x-component of the velocity:wink:

hoyung711 May11-10 02:29 AM

Re: (special relativity)Trajectory under constant ordinary force
 
Quote:

Quote by gabbagabbahey (Post 2712183)
[itex]\gamma[/itex] involves [itex]u^2[/itex], not just the x-component of the velocity:wink:

Do you mean I have to use
[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?

gabbagabbahey May11-10 03:34 AM

Re: (special relativity)Trajectory under constant ordinary force
 
Yup.

hoyung711 May11-10 08:34 AM

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?

hoyung711 May11-10 09:03 AM

Re: (special relativity)Trajectory under constant ordinary force
 
Using L.T. s.t. [tex]\bar{S}[/tex] is moving in y-direction 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?

hoyung711 May11-10 09:45 PM

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?

TingFung May12-10 06:40 AM

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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