# (special relativity)Trajectory under constant ordinary force

1. May 11, 2010

### hoyung711

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

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

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: May 11, 2010
2. May 11, 2010

### gabbagabbahey

$\gamma$ involves $u^2$, not just the x-component of the velocity

3. May 11, 2010

### hoyung711

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: May 11, 2010
4. May 11, 2010

### gabbagabbahey

Yup.

5. May 11, 2010

### hoyung711

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?

6. May 11, 2010

### hoyung711

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: May 11, 2010
7. May 11, 2010

### hoyung711

I think my L.T. transform got some problems since the answer is too simple..
Can anyone give a hand?

8. May 12, 2010

### TingFung

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: May 12, 2010