# Simultaneous equation from relativity

• Feb 9th 2010, 05:22 PM
Kiwi_Dave
Simultaneous equation from relativity
I need to solve the equations:

$\displaystyle ct_1=l+\frac 12 \alpha t_1^2$

$\displaystyle ct_2=l+\frac 12 \alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

First I try adding and subtracting the two equations to get:
$\displaystyle c(t_1+t_2)=2l+\alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

$\displaystyle c(t_1-t_2)=\frac 12 \alpha (t_1+t_2)^2$

But I can make no real progress from there.

Now the context of the question means that:

$\displaystyle \alpha << 1$, $\displaystyle l << 1$ and I am more than happy to make a few aproximations. I expect the binomial aproximation may be useful.

The required solution is:

$\displaystyle t_1+t_2 = \frac{2l}c (1+\frac {\alpha l}2)$

I have posted the whole question on Physics help forum at:
http://www.physicshelpforum.com/phys...lleration.html
• Feb 9th 2010, 08:56 PM
Kiwi_Dave
Finally solved it myself.

eqn1 $\displaystyle ct_1=l+\frac 12 \alpha t_1^2$

eqn2 $\displaystyle ct_2=l+\frac 12 \alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

eqn3 $\displaystyle c(t_1+t_2)=2l+\alpha t_1^2-\frac 12 \alpha (t_1+t_2)^2$

Let
$\displaystyle T=t_1+t_2$

From eqn1:
$\displaystyle t_1=\frac{c-\sqrt{c^2-2 \alpha l}}{\alpha}=\frac{c-c \sqrt{1-\frac {2 \alpha l}{c^2}}}{\alpha}$

$\displaystyle t_1 \approx \frac {c-c+\frac {\alpha l}{c}+\frac{\alpha^2l^2}{2c^2}}{\alpha}=\frac lc+\frac{\alpha l^2}{2c^2}$
So
$\displaystyle t_1^2 \approx \frac {l^2}{c^2}+\frac{\alpha l^3}{c^3}$

now substituting into eqn3

$\displaystyle cT \approx 2l-\frac 12 \alpha T^2+ \alpha t_1^2=2l-\frac 12 \alpha T^2+\frac {\alpha l^2}{c^2}+\frac{\alpha^2 l^3}{c^3}$
Rearanging
$\displaystyle cT+\frac 12 \alpha T^2 \approx 2l+\frac {\alpha l^2}{c^2}+\frac{\alpha^2 l^3}{c^3}$

Dumping the smallest terms:

$\displaystyle cT \approx 2l+\frac {\alpha l^2}{c^2}$

And finally

$\displaystyle T \approx \frac {2l}{c}(1+\frac {\alpha l}{2c^2})$