No sir.. we can’t get through the singularity (part 1)

Few days ago I suddenly remembered a movie titled Sunshine that was released in 2007.. It was directed by Danny Boyle (Slumdog Millionaire) and there are many stars in the movie including Chris Evans, Cillian Murphy, Michelle Yeoh and so on. In the movie, for unexplained reason, sun, the nearest star to earth is dying. Therefore a group of scientists are sent to drop a nuclear bomb to sun to reignite it. It was a great movie by the way, but that’s not why I am writing this post.. I remembered a scene where the physicist (played by Cillian Murphy) cannot calculate the trajectory or speed (or whatever) because in his calculation sun plays the role of singularity..

From wikipedia, gravitational singularity is:

 “ A gravitational singularity or spacetime singularity is a location where the quantities that are used to measure the gravitational field become infinite in a way that does not depend on the coordinate system. ”

Note the words “does not depend on the coordinate system”. This particular scene remind me of my own project on Newtonian (and non-Newtonian Mechanics). In a part of the project we also meet similar problem, where the sun plays the role of singularity and our efforts to calculate everything goes to waste.

I will give you a simple formulations to illustrate.. Remember the 2nd Newton Law: $F=ma$. We would like to model the conservative system of a single center of gravity.. So if you’re a particle at position $X$ we will have a system of differential equations:

$X'=V$ and $V'=-F(X)$ where $V$ is the velocity vector and $F(X)$ is the force vector. From Newton’s law of gravitation we know that  $F(X)= -c \frac{X}{|X^3|}$. If we simplify further that the constant is 1 and the graphic is two dimensional (that is $X=(x_1,x_2)$) we get 4 equations in our system. Changing our coordinates into polar coordinates we get the following system:

$r'= v_r$

$\theta'= v_{\theta}/r$

$v_r'=-\frac{1}{r^{2}}+\frac{v_{\theta}^2}{r}$

$v_{\theta}'=-\frac{v_r v_{\theta}}{r}$

With this coordinate, we try to plot the trajectory of a single particle in this system.. Okay, we actually draw a more complex system (it’s gonna be clear if you plot the system yourself) but what I’m trying to show you is the same

and here is another plot

So in the model above I don’t introduce you the variable $a$ which existed in the plot since we took $a=1$, but the total energy $h$ and the angular momentum $l$ can be computed from the system. And yes, for the earth $a=1$ is about correct, but for Mercury the value is about $a=1.1$. So why we plot $a=1.4$ and $a=1.6$.. naaah, we just wanna get more dramatic plots 😛

So the thing is if we consider the origin as the sun, it means the particle goes collide with and through the sun. This is a computational mistake since I suppose we don’t see Mercury collide with the sun every now and then right? It was because when $r \rightarrow 0$ all of $\theta,v_r,$and $v_{\theta}$ gets infinitely large for there are division by $r$ (see the system of equations).

But actually, physicists don’t consider the sun as gravitational singularity.. which means we should be able to get the right plot somehow. If you read the quote from wikipedia carefully you may have guessed that the answer is.. yes we should, and the method is by another change of coordinate.

In the next posting I will show you what we did and Mathematica will strike back, with some more accurate plots of course 😀