Sunday, 23 January 2011

One side of hand-written notes only...

It's finally over! My week of dreaded exams... Not that they went to plan. Now with most exams, it is normally customary for the paper to be proof read, i.e before the actual exam is sat. Well that certainly does NOT seem to be the case with the folks here at the University of Portsmouth! Thats right, we were due to sit our exam at 11:15 on a Thursday... At 11:16 - I raised my hand with a puzzled look on my face.
'I'm afraid theres a mistake with the paper'
Looking through, every symbol had been replaced with a black dot. Every bracket, $\pi$, $\rho$, $\Omega$,.. well you get the idea. What with this being an exam on astrophysics, those symbols are pretty much used through out, and so the whole exam was a write off. after half an hour, it was finally decided that the exam would be cancelled, and so we all had to leave one at a time (as there were still exams happening) which took another 10 minutes. It was finally decided that another exam would be held tomorrow (the Friday).

After we vented our rage upon the Head of Department, and various other people, we eventually managed to sit our exam. As I went to the trouble of learning it so much, I'm actually going to include one of the questions I had to answer in this blog, and I'll also show the sheet of notes I took in... (the blog title is a clue).

1) Show that a sphere of uniform density will collapse to zero radius in a "free-fall" time of
\[t_{FF} = \left[\frac{3\pi}{32G\rho}\right ] ^{\frac{1}{2}}\]





We can start solving this by stating the following formula:
\[\rho\frac{d^{2}r}{dt^{2}}=-\frac{GM}{r^{2}}\rho-\frac{dP}{dr}\]

We are told in the question that we are working in uniform density. As of that, $\rho$ can be 'ignored' as it is a constant. WE also know that we are in free-fall, so $P=0$. That leaves us with the following:
\[\frac{d^{2}r}{dt^{2}}=-\frac{GM}{r^{2}}\]
We can start by multiplying through by $\frac{dr}{dt}$, to give:

\[\frac{dr}{dt}\left(\frac{d^{2}r}{dt^{2}\right)}=-\frac{GM}{r^{2}}\frac{dr}{dt}\]

We can integrate both sides of this equation with respect to $t$, however before we do that, it is possible to make some substitutions on the Left hand side.
\[x=\frac{dr}{dt}\]
\[\frac{dx}{dt}=\frac{d^{2}r}{dt^{2}}\]

We can now re-write as:

\[\int x\left(\frac{dx}{dt}\right)dt=- \int \frac{GM}{r^{2}}\left(\frac{dr}{dt}\right)dt\]

We can cancel the $dt's$ on both sides of the equation, which leaves us with:

\[\int x dx=- \int \frac{GM}{r^{2}}dr\]

\[\Rightarrow\frac{x^{2}}{2}=\frac{GM}{r}+c\]

We can state that when $t=0, r=r_{0}$, where $r_{0}$ is the starting radius. We stated above that $x=\frac{dr}{dt}$, and so now $t=0$, we can rewrite the above formula to show that:

\[c=-\frac{GM}{r_{0}}\]

\[\therefore \frac{x^{2}}{2}=\frac{GM}{r}-\frac{GM}{r_{0}}\]

\[\Rightarrow x^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\]

\[\Rightarrow \left(\frac{dr}{dt}\right)^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\]

Now we need to integrate, with $t=t_{FF}$, and $r=0$

\[t_{FF}\Rightarrow \int_0^{t_{FF}} dt= \int_{r_{0}}^0 \left(\frac{dt}{dr}\right)dr\]

By inversing the right hand side, we can show that
\[\int_{r_{0}}^0 \left(\frac{dr}{dt}\right)^{-1}dr\]

 Now we are basically going to perform some shuffling about with the equations, which involves square-rooting $\frac{dr}{dt}$, and then inversing. This can be shown:

\[\left(\frac{dr}{dt}\right)^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\Rightarrow \frac{dr}{\sqrt{\frac{2GM}{r}-\frac{2GM}{r_{0}}}}=dt\]

We can no integrate this, though we can take out some constants to make life a bit easier.
Take out:
\[-\frac{1}{\sqrt{2GM}}\]
To give
\[-\frac{1}{\sqrt{2GM}}\int_{r_{0}}^0 \left(\frac{r}{1-\frac{r}{r_{0}}}\right)^{\frac{1}{2}}dr\]

We can now introduce some more variables in which we can sub in:
\[r=r_{0}\sin^{2}\theta\]
\[dr=2r_{0}\cos\theta\sin\theta d\theta\]

If we substite $r$ into $\left(\frac{r}{1-\frac{r}{r_{0}}}\right)$, we get the following:
\[\left(\frac{r}{1-\frac{r}{r_{0}}}\right) \Rightarrow\left(\frac{r_{0}\sin^{2}\theta}{1-\sin^{2}\theta}\right)=\left(\frac{r_{0}\sin^{2}\theta}{\cos^{2}\theta}\right)=\sqrt{r_{0}}\tan\theta\]

Now that has been simplified we can substitute our new values into the intergral, which gives us the following:

\[t_{FF}=-\frac{1}{\sqrt{2GM}}\int_{r_{0}}^0\left(2r_{0}\cos\theta\sin\theta\sqrt{r_{0}}\tan\theta d\theta\]

this can be simplified further by saying:
\[2r_{0}\times r_{0}^{\frac{1}{2}}=2r_{0}^{\frac{3}{2}}\]
and
\[\tan\theta\times\cos\theta\times\sin\theta=\sin^{2}\theta\]
\[\sin^{2}\theta=\frac{1-\cos 2\theta}{2}\]

We can take out from the intergral that $2r_{0}^{\frac{3}{2}}$ is a constant, leaving us with:

\[-\frac{2r_{0}^{3}}{\sqrt{2GM}}\int_{r_{0}}^0\frac{1}{2}\left(1-\cos 2\theta\right)d\theta\]

so when $r=r_{0}$, $\sin^{2}\theta=1\Rightarrow\theta=\frac{\pi}{2}$

\[\therefore =-\frac{2r_{0}^{3}}{\sqrt{2GM}}\left[\frac{1}{2}\theta-\frac{1}{4}\sin2\theta\right]_{2\pi}^{0}\]

As we know both of the limits would cancel out $\sin$, we can cancel through on the second half of the intergral, which we then get left with:

\[\frac{\pi}{4}\left(\frac{-2r_{0}^{3}}{GM}\right)^\frac{1}{2}\]

Once again stating formula, we can say that:

\[M=\frac{4}{3}\pi r_{0}^{3}\rho\]

We can substitute that into above to give:

\[\frac{\pi}{4}\left( \frac{-2r_{0}^{3}}{{G\left(\frac{4}{3}\pi r_{0}^{3}\rho}}\right)}\right)^{\frac{1}{2}}\]

This is where the magic happens - you can see that the $r_{0}^{3}$'s can cancel out, we can take the $\frac{\pi}{4}$ into the square root section by squaring (and then we can cancel the bottom $\pi$, giving:

\[\left(\frac{-2\pi}{16G{\frac{4}{3}}\rho}\right)\]

taking the $3$ upto the top, and then dividing through by $2$ can give:

\[\left(\frac{3\pi}{32G\rho}\right)^{\frac{1}{2}}\]

Which is what we were after!

16 comments:

Craig said...

Nice, very thorough working! Maybe this will help someone studying this subject in the future...

Rondariel said...

Looks pretty complicated but wow. Astrophysics looks interesting...

TheDankMan said...

hey sean,
nice blog thanks for following me,
im having trouble i cant seem to find a follow button on your page?
please fix this dont think im not following you back.
TheDankMan

Fruffles said...

Are you a wizard?

Festooner said...

I don't understand any of this, but I love to look at it.

KB said...

AAAAHHHH! MATH! I thought I was done after DiffyQ!!

Frosty said...

Fucking hell man, interesting but you just broke me

FOllowed
http://frostytnm.blogspot.com/

Killuvogel said...

Wow respect man, that was heavy.

Viewed and followed!


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Hoop Jones said...

I have seen these equations far too many times, and they upset me always :p

follow back!
http://randomawesomeness34.blogspot.com

Cindir13 said...

I need to brush up on my math.

Icky Ray said...

That's pretty impressive...

Cpt Pownzor said...

exams are the worst Ive got one this week im not looking forward to :(

BMRMike said...

That looks very interesting. I'm a econ/math major and that stuff is just pure entertainment.

stan said...

thats some very good maths

AbstractReality said...

Oh god.. hope I don't have to see this.. although I probably will..

pleyame said...

Just wanted to say:
tl;dr