# Black Hole Essay

Discussion in 'Physics & Math' started by Reiku, Nov 23, 2011.

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1. ### ReikuBannedBanned

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In my class we have been asked to write an essay on Black Holes. I have chosen the Flat Approximation of the Local Neighbourhood of a Black Hole as my topic. Would the members here be so kind to read over it and express their criticisms?

Close Encounters of the Black Hole Kind

The idea as you get closer to a Black Hole, the hyperbolic acceleration increases. The Hyperbolic angle is $\omega$. You can model this as being increased by $\omega = 1, \omega = 2, \omega = 3...$ and so on as you check reference the grid. The Hyperbolic coordinates are then $\omega$ and $r$ where $r$ is the distance from the center of the Black Hole.

If $x$ and $t$ are your spatial and time coordinates, then you can state

$x = \rho cos h \omega$

and

$t = \rho sin h \omega$

They are similar to polar coordinates. Our metric is then:

$d\tau^2 = \rho^2 d\omega^2 - d\rho^2$

Here our $\omega$ term is playing the role of time and $\rho$ is playing the role of space, so you may want to add the extra coordinates of space to this

$d\tau^2 = \rho^2 d\omega^2 - d\rho^2 - dy^2 - dz^2$

$d\tau$ is called the proper time.

We are going to take an imaginary journey to near the neighbourhood of a Black Hole. You must first imagine that we can get so close to the black hole, that you are tangent to the Horizon. The Horizon is given as $2MG = r$.

To do this, we need to use the Schwarzchild Metric (or Schwarzchild Vacuum) which is used in physics to describe a perfectly spherical distrubuted mass of a non-rotating body. The surface of such an object is called the Event Horizon. In our case, we will bw considering the exotic body of a Black Hole, a situation in physics where matter has been compressed to such a high density that the escape velocity exceeds that of what could be acheived by currently the fastest known particle; a photon. This does however include any kind of massless radiation. The Schwartzchild Metric is given as:

$d \tau^2 = (1 - \frac{2GM}{r}) dt^2 - \frac{dr^2}{(1-\frac{2GM}{r})} - r^2 d \Omega^2$

Here $d\Omega^2 = d\phi^2 + (sin \phi)^2 d \theta$ and this suffices a 2-dimensional coordinate sphere. This is all there really is to a metric of a black hole. It can give you as much information as you are likely to require to describe them. This is the general solution ''outside'' of where the mass is associated - outside the boundary of a black hole, or indeed, any object you might consider, like Earth or a star.

There are two singular regions which can be considered in the metric for a Black Hole. One is a coordinate artefact where our coordinate do something rather odd and the other one is located inside the Black Hole which will be encountered if someone was to pass through the Event Horizon.

When $r=0$, $dt^2$ blows up to infinity and even worse, blows up with a negative sign. $dr$ and $dt^2$ interchange signs so the first $(1 - \frac{2GM}{r})$ is negative and the second $(1 - \frac{2GM}{r})$ becomes positive.

This shows that spacelike events become timelike and vice versa; This doesn't mean that space literally becomes time, or that time literally becomes space, it is just a strange coordinate artefact... the behaviour of moving in space will become restricted and moving in time will become more free. In the second bit $(\frac{dr^2}{(1 - \frac{2MG}{r})})$ when $r=2MG$ the $(\frac{2MG}{r})=1$ and if you follow through the rest of the expression, you simply find that $1-1 = 0$ (no surprise), but the cooefficient of $dr^2$ becomes infinite.

Similar things can be found in our original example of the hyperbolic metric

$d\tau^2 = \rho^2 d\omega^2 - d\rho^2$

This is actually the same as

$= Rd\omega^2 - \frac{dR^2}{4\rho^2}$

Here $\rho^2 = R$. When you apply the same conditions as before, you will see that $= Rd\omega^2$ and $\frac{dR^2}{4\rho^2}$ swap signs for when $R=0$ when $d\omega \rightarrow 0$.

It might be strange to think that $\rho^2 < 0$. This is just changing the way you are looking at the sign of $R$. Positive $R$ is where we began assuming the equations of the metric. Negative $-R$ is now in the region above this. (See Diagram).

Now we can ask whether there is an approximation of the Schwarzchild Metric which will suit the metric we began with:

$d\tau^2 = \rho^2 d\omega^2 - d\rho^2$

We should begin to rearrange terms.

$d\tau^2 = (\frac{r - 2MG}{r}) dt^2 - \frac{dr^2r}{r-2MG}$

So, what have I done? Well keep in mind this equation is incomplete. We are taking this step-by-step. The term $(\frac{r - 2MG}{r})$ is simply been rearranged from $(1 - \frac{2GM}{r})$. That's nothing special, just a bit of algebra. Here we must remember not to set the numerator $r=2MG$. That assume we are touching the Horizon. What we are assuming is that we are just by the horizon and no more.

However, the $r$ in the denominator does not change as drastically as the $r$ in the numerator, so we can actually set this equal to $2MG$. Continue for the rest of the equation will give you:

$d\tau^2 = (\frac{r - 2MG}{2MG}) dt^2 - \frac{dr^22MG}{r-2MG} - (2MG)^2\Omega$

Is now a complete equation again (but realize this is not a metric of a spherical Black Hole any more), modified slightly nonetheless. You can actually modify this further by replacing the last term with $(dy^2 + dz^2)$.

$d\tau^2 = (\frac{r - 2MG}{2MG}) dt^2 - \frac{dr^22MG}{r-2MG} - dz^2 - dy^2$

This is very good for a Flat Space Approximation. It does work well for the Neighbourhood of the local surface of a Black Hole which may appear to calculations as pretty much as a flat metric.

Now, reminding ourselves that $d\tau^2 = \rho^2d\omega^2$, the proper distance $\rho^2$ from the Horizon $r=2MG$ then this without the sign is

$ds^2 = dr^2 \frac{2MG}{r - 2MG}$

Take the square root of this

$ds = dr \sqrt{\frac{2MG}{r - 2MG}}$

Now we need to take the integral in respect to $r=2MG$ to $r$ (the point in which we are sitting just by the Horizon) then

$\rho = \int_{r=2MG}^{r} dr' \frac{\sqrt{2MG}}{\sqrt{r'-2MG}}$

The $\sqrt{2MG}$ is a constant so we can bring that out

$\sqrt{2MG} \int_{r=2MG}^{r} dr'{\frac{1}{\sqrt{r - 2MG}}$

Which yields

$\rho = 2\sqrt{2MG}\sqrt{r - 2MG}$

Now we need to rewrite the metric in harmony of this new equation. This is of course, the proper distance from the Horizon.

Rearranging gives

$\frac{\rho}{2\sqrt{2MG}} = \sqrt{r - 2MG}$

Simplifying makes

$\frac{\rho}{\sqrt{8MG}} = \sqrt{r - 2MG}$

and now we are going to square it

$\frac{\rho^2}{8MG} = r - 2MG$

Knowing there is another $2MG$ when we compare this to the metric gives

$\frac{\rho^2}{16(MG)^2}dt^2 - d\rho^2 - dz^2 - dy^2$

Now this is not quite the same as

$\rho^2d\omega^2 - d\rho^2 - dz^2 - dy^2$

To sort this problem, we need to know that

$\omega = \frac{t}{4MG}$

$d\omega = \frac{1}{4MG}dt$

In light of this we easily realize that $d\omega$ is really just $\frac{dt^2}{16(MG)^2}$ so we finally have

$\rho^2(d\frac{t}{4MG})^2 - d\rho^2 - dz^2 - dy^2$

which now makes a perfect approximation of the Schwarztchild Metric. So we have just taken a journey very close to a black hole in a Flat Approximated Spacetime.

Last edited: Nov 23, 2011

3. ### TachBannedBanned

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Err, no. This is not even wrong.

This is not even wrong either, it borders on hilarious. I stopped reading but there is enough nonsense to fill several pages pointing out your goofs. I am sure that many others will have a field day with your essay. It is so reminescent of Reiku, it is surreal.

ETA: Sorry, I couldn't resist this "pearl"

...and this one:

F-

Last edited: Nov 23, 2011

5. ### ReikuBannedBanned

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11,238
Oh I am reporting you. If you are going to call my by any name, you can call me by my real name here. Learn it.

It's Mister.

(as a side note, how do you think the Schwartzchild Metric is written. This is how we have learned it in class??)

7. ### ReikuBannedBanned

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11,238
Read the ******* work. I explained that that equation was incomplete you twat. :bugeye:

8. ### TachBannedBanned

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Based on what you are writing , I don't think that you go to any class.

9. ### TachBannedBanned

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It is not that it is "incomplete", it is downright hilarious.

10. ### ReikuBannedBanned

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11,238
And why is it hilarious? These are serious equations, nothing about them is wrong or incomplete. I hope they ban your ass soon.

Hopefully sooner than later.

11. ### TachBannedBanned

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For someone with basic knowledge of the subject, your errors are glaring. And I did not spend more than three minutes looking over your work. I am sure that many others will have a field day with it.

12. ### ReikuBannedBanned

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11,238
I asked for criticism. Shut up or put up. You are simply trolling tach. I know nothing is wrong with them!!!!

13. ### ReikuBannedBanned

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I see I missed out a squared term in that object. I've written a lot of equations, that's a lot of tex tach. Just shut up and move along if you aren't going to contribute properly. I see you do it in so many threads here.

14. ### ReikuBannedBanned

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11,238
Big deal, missed another squared sign. That's because I copied and pasted. Anything more.. harder to offer?

15. ### ReikuBannedBanned

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11,238
There fixed. Breath tach, breath!!!!

16. ### TachBannedBanned

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Have fun with another couple of hilarious ones, they are harder to spot (and to fix) :

Bye.

Last edited: Nov 23, 2011
17. ### ReikuBannedBanned

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11,238
How is it hilarious, when I copied and pasted from basically the same equation, and I missed out on one squared sign??????

You must be pleased very easily.

Right good, **** off back to your doppler dissitation, and arsing up your integrals.

18. ### ReikuBannedBanned

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11,238
oh I see what tach is getting his knickers in a twist about. The denominator in the second term of the Schwarzchild metric has a ''c'' instead of r, which, was of course carried out to other parts of the work because I copied and pasted.

See, tach is just a troll. He couldn't have came in here, been a gentleman and pointed the fact out. Instead he posted abusive posts one after the other as if it was all one big joke. The difference between me and tach, is if I notice a mistake, I won't try and go back and edit the work without making a notification about it. NOT only that, but I doubt tach even recognizes his mistakes even after he has need pointed them out.

Thanks though tach, as useless as your ability is to have a conversation about a subject, I needed to get this in by tomorrow. Unfortunately for you, the part where I hadn't sqaured two of my terms and the ''c'' doesn't even appear in my work. These are typo's. So that is good for me. I am quite happy now that you couldn't even spot a proper mistake.

19. ### TachBannedBanned

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5,265
Nah, you have a lot more errors. Many of them you can't pass as typos.

Don't worry about it, you have plenty of "proper" mistakes. I just pointed a couple, I don't want to spoil the fun for the others.

20. ### ReikuBannedBanned

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11,238
Don't hold back tach. Typo's are one thing.

And what mistakes are you so clearly seeing? Common... I want to know. Afterall, I need to present this.

21. ### ReikuBannedBanned

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This is confusing me I will add though. Maybe tach has a problem with the terminology? Maybe, just maybe he is confused with the part of me saying

''Now, reminding ourselves that [eq] , the proper distance [eq] from the Horizon [eq] then this without the sign is''

I am guessing but he might have thought the $d\tau$ part is strictly proper time. Is he aware of the following?

You can calculate from the Schwarzschild metric

$ds^2=(1-\frac{2GM}{r}) dt^2 - r^2d\phi^2$,

where $ds$ means the proper time and length intervals as a function of the coordinate intervals.

But then who knows with Tach. He comes into peoples threads and acts as conspicuous and as mysterious as he can without ever giving a direct answer.

edit: removed a c^2 here. Didn't see why I should have it in after neglecting it using natural units throughout my work.

22. ### TachBannedBanned

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What you put together ain't fixable. Besides, I doubt that there is any class where you are going to present any of this "stuff".

23. ### ReikuBannedBanned

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LOL... that comment made my night. It really did tach. So It's not wrong? Why all the hoolah?

Mod note : Language!

Last edited by a moderator: Nov 23, 2011