# Ngô Quốc Anh

## June 14, 2013

### MuPAD: Drawing asymptotically flat spacetime

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 23:06

Today, I will show how to draw an asymptotically flat spacetime with two ends using MuPad. To draw ends, we use hyperbolic functions.

First, we talk about the end. The function that I am going to use is $h(t)=\frac{1-4t}{t-2}+12$. It is clear that $h$ blow up at $t=2$ and approaches $-4$ at infinity. To use $h$, we use the following

h := proc(t)
begin
(1-4*t)/(t-2)+12
end_proc

I have added $12$ to the function $h$ so that $h$ approaches $8$ at infinity. Then to draw the (upper) end, I use the following function

f := proc(x, y)
begin
if
x^2 + y^2 > 3.4
then
h(x^2+y^2)
else
end_if
end_proc

I have used the number $3.4$ because I do not want my end is too tall, keep in mind that $2$ is the blow-up number. We can calculate

h(3.4)

to see that this number is nothing but $3$ which is close to $1$ as I need. We are now able to draw the first end using the following

plot(
plot::Function3d(f, x = -6 .. 6, y = -6 .. 6, Submesh = [3, 3]),
ViewingBox = [Automatic, Automatic, 3 .. 8],
Scaling = Constrained)

What we obtain is the following picture

By using the negative of $f$, i.e. $-f$, we can draw the lower end as follows

plot(plot::Function3d(f, x = -6 .. 6, y = -6 .. 6, Submesh = [3, 3]),
plot::Function3d(-f, x = -6 .. 6, y = -6 .. 6, Submesh = [3, 3]),
ViewingBox = [Automatic, Automatic, -8 .. 8],
Scaling = Constrained)

We obtain the following picture.

The only thing left is to draw the “compact set” of the spacetime. In order to make the whole smooth, I need to know the tangent line at the point $3.4$ of the function $h$. I use

D(h)(3.4)

To draw the upper half compact core, I use an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ for some $a,b>0$.

Suppose $a \in (0,2)$ and $b>0$, let us consider the function

$\displaystyle y=b\sqrt{\frac{x^2}{a^2}-1}.$

By solving $y=3$, we find that

$\displaystyle x = a \sqrt{\frac{9}{b^2}+1}.$

Since

$\displaystyle y'=\frac{bt}{a^2\sqrt{\frac{t^2}{a^2}-1}}$

we know that

$\displaystyle y'(\sqrt{\frac{17}{5}})=\sqrt{\frac{17}{5}}\frac{b}{a^2\sqrt{\frac{17}{5a^2}-1}}.$

Therefore, the condition $y'(\sqrt{\frac{17}{5}})=287.304$ gives

$\displaystyle b=155.812a^2 \sqrt{\frac{17}{5a^2}-1} .$

Using $a=\sqrt 2$, we obtain $b=260.724$.

In this case, we can also use a cylinder via

plot::Surface([sqrt(3.4)*cos(v), sqrt(3.4)*sin(v), u], u = -3 .. 3, v = 0 .. 2*PI)

Eventually, this is what we have

Note that in the above picture, I have used the point $2.5$ instead of the point $3.4$.