# Stuck on derivation of gravitational deflection of light, need help with calculus

I'm trying to show that my theory can explain the gravitational deflection of light by a different method to Einstein, I just can't complete the derivation. This is what I have so far... (it may help to write this out properly yourself)

$$\int_{0}^{a} a \, da=-\int_{0}^{\pi} \frac{GM}{r^2 c^2} \frac{\sin b}{\frac{2GM}{r^2 c^2}+1} \, ds$$

We are calculating the deflection of a beam of light bending around a large object of mass, such as the sun. The final result can be tested against Einstein's value of $1.7$ seconds of arc for a beam of light passing our sun at grazing distance, i.e. the distance equal to the radius of the sun.

In this equation,

$a$ = angle of deflection

$G$ = the Gravitational constant

$M$ = Mass of object

$c$ = speed of light

$r$ = distance from centre of the object being passed to the light beam

$b$ = angle between $r$ and the beam of light

$s$ = distance along beam of light

$d$ (unseen here) = unmodified closest approach of the beam of light, in the case of the test case of the sun, this would be the radius of the sun

I can solve this equation when $r$ does not vary as a function of $a$. i.e. when the beam comes no closer to the object of mass being passed as a result of its deflection, i.e. keeping $r$ as a constant when it is in fact a variable. This yields a result the same as Einstein achieved when he made the same mistake, $0.85$ seconds of arc, however Einstein realised his mistake and corrected it before it could be tested, increasing the result by a factor of $2$.

If you want a better explanation, my article can be downloaded from here (free of course)... http://www.smashwords.com/books/view/540505

Please send help. I'm very stuck.

• What I mean when I say not treating r as a variable, what I meant to say was pretending that r does not vary as a function of a. – M.Spears Aug 30 '16 at 5:50
• To solve by that (incorrect) method... s=(r)(cos(b)) therefore ds=(-)(r)(sin(b))db and r=d/sin(b) and substitute into equation..... – M.Spears Aug 30 '16 at 5:51
• The left-hand integral makes no sense, because the upper limit is the same as the variable of integration. Simply changing the variable to another letter (e.g. $\theta$) would make sense: then the integral is $\frac12a^2$. If that is what you mean, just write it. The second integral makes no sense, because you do not explain what is a constant and what is a variable, or how the variables are defined in terms of the variable of integration. – John Bentin Aug 30 '16 at 9:20
• Sorry, I wrote that as an answer, it was supposed to be a comment. I've only signed up here because I was hoping someone knew how to do calculus, it's been a long time between drinks for me. – M.Spears Aug 31 '16 at 11:04

Yeah ok. I only added the integral on the left hand side to make it easier. Change it to a e or something if you like, it makes no real difference. I just had the "a" before, which was supposed to be an "alpha". Can you solve it? G, M, c, are constants. So in the case where I ignore that the beam of light comes closer to the object of mass it is passing (for example the sun) this equation can be solved thus...

s=(r)(cos(b)) ds=-(r)(sin(b))(db) r=d/(sin(b))

Therefore in the case where 1/((2)(G)(M)/(((r)(c^2))+1))=1 Or close enough to it... a= integral (-)((G)(M)(sin(b))/((r^2)(c^2)))(ds) from b=0 to b=(pi)

Becomes a= integral ((G)(M)(sin(b))/((d)(c^2)))db from b=0 to b=(pi)

Which equals a=(2)(G)(M)/((d)(c^2)) Which is Einstein's exact equation from 1911. However the beam of light does come closer to the object of mass being passed as a result of its deflection and the final equation Einstein arrived at in 1915 before ir could be tested and confirmed was a=(4)(G)(M)/((d)(c^2))

The complete equation I've derived for the first part without the adjustment is actually a=(1/2)((pi)-(((c^2)(d)(pi))/((c^4)(d^2)-(4)(G^2)(M^2))^(1/2))+((2)(c^2)(d)(arctan((2)(G)(M)/((c^4)(d^2)-(4)(G^2)(M^2))^(1/2)))/((c^4)(d^2)-(4)(G^2)(M^2))^(1/2))

Hopefully I've written that correctly. But like I said, the problem is that the beam of light comes closer to the object of mass being passed as a result of its deflection. I don't know how to do this. I'm more of a philosopher than a mathematician or physicist. So please, no more pedantic crap about the way I've written something, can anyone solve the problem?

Thank you. Much appreciated.