Calculate Bending of Light: Find the Middle Point

[vaibhavtewari]

Hello,

while solving bending of light problem, where I shoot a light ray from one tower to other. How much light has fallen upto the middle point ? I was able to reduce the eq to

[tex]\phi=\int[1-a(\cos\theta+\frac{1}{1+\cos \theta})]^{-1/2}d\theta[/tex]

a is very small, also if there is no gravity "a" will be zero.

Is there a way I can express this integral as a series ? the first term being [tex]\phi=\theta+...[/tex]

I tried expanding the square root but the higher order integrals become ugly. I was hoping to find a neat looking series..please help

 

[Dale]

If you substitute [itex]\theta = \delta\theta + \theta[/itex] where [itex]\delta\theta[/itex] is a small quantity then you can do a Taylor series expansion about [itex]\delta\theta=0[/itex]

[tex]\frac{1}{\sqrt{a (-\cos (\theta ))-\frac{a}{\cos (\theta )+1}+1}}+\delta \theta
\left(\frac{a \sin (\theta )}{2 (\cos (\theta )+1)^2 \left(a (-\cos (\theta
))-\frac{a}{\cos (\theta )+1}+1\right)^{3/2}}-\frac{a \sin (\theta )}{2 \left(a
(-\cos (\theta ))-\frac{a}{\cos (\theta )+1}+1\right)^{3/2}}\right)+O^2[/tex]

Since both [itex]a[/itex] and [itex]\delta\theta[/itex] are small quantities then their product is second order, so this reduces to

[tex]\frac{1}{\sqrt{a (-\cos (\theta ))-\frac{a}{\cos (\theta )+1}+1}}+O^2[/tex]

 

[vaibhavtewari]

Thankyou for your effort, [tex]\theta[/tex] itself is very small as [tex]\phi[/tex], the angle towers make wrt to center of Earth is small. My question is more mathematical..is there is a way I can get a series solution to the integral equation. I don't intend to ignore any terms, no matter what order they are.

Thank You

 

[Dale]

is there is a way I can get a series solution to the integral equation. I don't intend to ignore any terms, no matter what order they are.

What do you mean by this? A series where you don't ignore any terms? That doesn't make sense to me.

 

[vaibhavtewari]

I mean something like this,

[tex]\int \frac{\sin (x)}{x}dx=\int \frac{x+x^3/3+2x^5/15+...}{x}dx=x+x^3/9+2x^5/75+...[/tex]

We can't have any analytical integration but as we see we can ha ve a series solution and if x is small we can just consider first few terms, but I have the compete series solution...series solution are useful in many ways and the problem I am trying to solve might become easier with series solution...

So I am looking for a series solution of the integral...

Thanks for putting effort and helping me out...

 

[vaibhavtewari]

I didnt mean only solution the way I have given in my example...I mean a sum over double series or any weird series looking solution...

 

[Dale]

According to Mathematica the first several terms are:

[tex]\frac{\theta }{\sqrt{1-\frac{3 a}{2}}}-\frac{a \theta ^3}{16 \left(1-\frac{3
a}{2}\right)^{3/2}}+\frac{a (3 a+16) \theta ^5}{320 \sqrt{4-6 a} (2-3 a)^2}+\frac{a
\left(-477 a^2+816 a-32\right) \theta ^7}{17920 \sqrt{4-6 a} (3
a-2)^3}+O\left(\theta ^9\right)[/tex]

but you will have to figure out the expression for arbitrary terms on your own.

 

[vaibhavtewari]

Thankyou for giving the expansion, it was helpful.