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Rocky1775
2012-Mar-20, 06:37 PM
If its true that nothing falls through the event horizon because time slows to a stop there, then at what distance does an object falling towords a black hole stop accelerating and start decelerating? Lets assume a two solar mass black hole.

WayneFrancis
2012-Mar-21, 02:44 AM
Don't photons mediate the forces of magnetic fields ... or in other words it takes photons for a magnetic field to be created?

If this is true then how do the photons escape the blackhole to create a magnetic field?

The key is "virtual photons" and as I understand them they are not restricted to the same rules as normal photons. You can't use them to transmit anything but the information they do. IE the charge, mass of the black hole.

It's kind of like saying if nothing can escape the black hole then how does gravity escape the black hole. Gravity isn't under the same restrictions and charge doesn't seem to be under that restriction either.

Selfsim
2012-Mar-21, 02:59 AM
If its true that nothing falls through the event horizon because time slows to a stop there, then at what distance does an object falling towords a black hole stop accelerating and start decelerating? Lets assume a two solar mass black hole.My answer attempt is here .. (http://www.bautforum.com/showthread.php/128971-Acceleration-vs-Velocity-Black-Hole-Event-Horizon?p=1998737#post1998737)
The Summary is:

For an observer outside the Event Horizon (EH), if an object is released from rest, and falls towards a Schwarzschild BH, it will reach a maximum co-ordinate velocity of 0.39c at a distance of 3 times the Schwarzschild radius, after which it appears to slow down, (from the outside observer's perspective), due to the effects of gravitational redshift.
The mathematical derivation is about 3 pages long (so, I won't clog up the thread with it .. ).

Regards

pzkpfw
2012-Mar-21, 11:05 AM

caveman1917
2012-Mar-21, 11:11 PM
The mathematical derivation is about 3 pages long

3 pages? That seems awfully long, how did you calculate this? Even from first principles it shouldn't take more than about 5-10 lines of calculation, depending on how full you write things out.

I'm just asking because in that other post you say you derived it by "substituting the schwarzschild metric in the field equations" and i'm not sure what you mean by that, or why it would take 3 pages, though you did get the correct result.

Selfsim
2012-Mar-22, 02:19 AM
3 pages? That seems awfully long, how did you calculate this? Even from first principles it shouldn't take more than about 5-10 lines of calculation, depending on how full you write things out.

I'm just asking because in that other post you say you derived it by "substituting the schwarzschild metric in the field equations" and i'm not sure what you mean by that, or why it would take 3 pages, though you did get the correct result. Well, I didn't say I knew what I was doing ! :) … I was 'finding out' as I went along !

The steps were:

i) take the Schwarzschild metric and find the non-vanishing Christoffel terms, considering only terms for objects free falling radially, from an infinite distance from the EH;
ii) apply these (in differential eqn form), to the generalised geodesic acceleration eqn, and calculate the solutions specific for a Schwarzschild BH;
iii) one of these solutions is the eqn for the geodesic acceleration. Integrating it, results in the geodesic velocity (proper time). The relationship between proper time and coordinate time, also turns up by integrating the other one (from (ii) above). Combining that with the geodesic velocity eqn, gives the coordinate velocity.
iv) maximising the velocity at r, and solving for r, then results in 3 x rs.
v) using (iv) you can then solve for the maximum velocity, which comes out to be 0.39c

(I've tried to leave out some in-between steps, so I hope all that still makes sense (?) ).

Regards

caveman1917
2012-Mar-22, 03:06 AM
Well, I didn't say I knew what I was doing ! :) … I was 'finding out' as I went along !

The steps were:

i) take the Schwarzschild metric and find the non-vanishing Christoffel terms, considering only terms for objects free falling radially, from an infinite distance from the EH;
ii) apply these (in differential eqn form), to the generalised geodesic acceleration eqn, and calculate the solutions specific for a Schwarzschild BH;
iii) one of these solutions is the eqn for the geodesic acceleration. Integrating it, results in the geodesic velocity (proper time). The relationship between proper time and coordinate time, also turns up by integrating the other one (from (ii) above). Combining that with the geodesic velocity eqn, gives the coordinate velocity.
iv) maximising the velocity at r, and solving for r, then results in 3 x rs.
v) using (iv) you can then solve for the maximum velocity, which comes out to be 0.39c

(I've tried to leave out some in-between steps, so I hope all that still makes sense (?) ).

Regards

I see. There's no need to calculate the Christoffel terms, you can take advantage of the symmetry to get steps i to iii in a couple of calculations by noting the timelike killing vector (if none of the coefficients depends on a coordinate, that coordinate is a killing vector, they are therefor easy to spot if they are there).
If there are killing vectors it's a lot easier to get your constants of motion through them rather than calculating Christoffel symbols.

Since none of the coefficients depends on t, \vec{t} = (1,0) is a killing vector and therefor its dot product with the four-velocity is a conserved quantity.

So (1,0) \cdot (\frac{dt}{d\tau}, \frac{dr}{d\tau}) = (1-r_s/r) \frac{dt}{d\tau} is a conserved quantity, which is equal to 1 (take the limit where r goes to infinity).

ETA: actually you could even leave this step of calculating the dot product out, since \vec{t} is a killing vector, and there are no dt cross terms in the metric, you can just set the other terms in the schwarzschild metric to zero, giving you d\tau^2 = (1-r_s/r)^2dt^2 \Rightarrow 1 = (1-r_s/r)\frac{dt}{d\tau}. (end ETA)

So just plug (1-r_s/r)\frac{dt}{d\tau} = 1 in the reduced schwarzschild metric and you get \frac{dr}{d\tau} = -\sqrt{\frac{r_s}{r}} \Rightarrow \frac{dr}{dt} = \frac{dr}{d\tau}\frac{d\tau}{dt} = -(1-r_s/r)\sqrt{\frac{r_s}{r}} as the velocity as seen by the schwarzschild observer.

Then you can immediately do your steps iv and v \frac{d[-(1-r_s/r)\sqrt{r_s/r}]}{dr} = 0 \Rightarrow r = 3r_s and plugging that into the velocity gives -(1-r_s/3r_s)\sqrt{\frac{r_s}{3r_s}} = \frac{-2}{3\sqrt{3}} \sim -0.38 as you have it.

Calculating the Christoffel symbols is the most general (and mathematically daunting) way of doing it, applicable when you have no killing vector fields (which you don't have in general). However in practice most ideal "nice" spacetimes you'll work with will admit some killing vector fields so calculating the Christoffel symbols for these is overkill.

Selfsim
2012-Mar-22, 04:35 AM
Cool .. thanks for that caveman … certainly makes it a LOT simpler !
Ya learn somethin' every day, eh … :)

Cheers

caveman1917
2012-Mar-22, 11:00 PM
Cool .. thanks for that caveman … certainly makes it a LOT simpler !
Ya learn somethin' every day, eh … :)

Cheers

You're welcome, i had a feeling you were taking some long hard route if you had to go through 3 pages. Though you should really be commended for getting the right answer by finding it out as you went along and taking what is probably the hardest way to solve the problem. Quite a feat i'd say.

Cougar
2012-Mar-23, 12:35 AM
That was pretty amazing, caveman. I can't even see how one gets that last equivalence. But back to Rocky's initial assumption...

If its true that nothing falls through the event horizon because time slows to a stop there, then....

No, that's not true. :D From the viewpoint of that falling thing, it surely 'flies' right through the event horizon, at near light speed I'd expect. (Not sure what happens after that.) But it is the falling object's signal that you are receiving, as a distant observer, that appears to slow, as caveman so precisely described above.

Selfsim
2012-Mar-23, 01:01 AM
You're welcome, i had a feeling you were taking some long hard route if you had to go through 3 pages. Though you should really be commended for getting the right answer by finding it out as you went along and taking what is probably the hardest way to solve the problem. Quite a feat i'd say.Whilst the commendation is much appreciated, the original aim was to demonstrate that 'faith' is not needed when it comes to accepting the end product of mathematical derivations in Physics.

There is a disturbing trend thesedays to reject mathematics in science, as something to be regarded cynically, to be mistrusted, and to be used to propagate political viewpoints.

In this case, the answer was quite straightforward, but this unfortunately, betrays much complexity.

Most of what is dicussed in these forums, is the physical resultant of similar mathematical operations. The resultants are not merely a verbalised stories, passed from one poster to another. This seems to be very much forgotten around the web thesedays resulting in way too much cynicism of mathematics in science (IMHO).

(And this statement may be overkill on my part also, for which I apologise in advance). :)

Regards

Selfsim
2012-Mar-23, 01:08 AM
That was pretty amazing, caveman. I can't even see how one gets that last equivalence. To be perfectly honest, just understanding why caveman's solution works, calls for a very deep understanding of GR, Killing vector fields, and the role of symmetry in GR.
The other method might end up looking fairly ugly, but I found it easier to understand.

Regards

astromark
2012-Mar-23, 01:33 AM
.. Even in this science based forum where a fine understanding of mathematics is necessary.

Just to comprehend the riddle as asked. More importantly it is said that the tools of trade is mathematics.. and are.

That it can be told and explained to a interested party that does not understand that level of maths.. important.

Problem solving by expression of good science is more than the maths..of it. This is a good platform for this argument..

That the in falling object seems to slow and stop is a function of the mass distortion of the BH.. on space time..

That in reality of that object is that it now is part of that BH. It has been added to the mass.

Selfsim
2012-Mar-24, 12:56 AM
That the in falling object seems to slow and stop is a function of the mass distortion of the BH.. on space time..

That in reality of that object is that it now is part of that BH. It has been added to the mass.How do you know this ?

Cougar
2012-Mar-24, 02:18 AM
...calls for a very deep understanding...

Oh darn. I only have vague understandings of those things. But a Killing vector field is essentially just a vector field that preserves the metric, apparently. Yau's recent book, The Shape of Inner Space: String Theory and the Geometry of the Universe's Hidden Dimensions, gets one used to these sorts of ideas and operations. Just the simple concept of the metric - how you measure distance in the space you're investigating - and operations upon it, seems to be pretty fundamental to the enterprise....

Selfsim
2012-Mar-24, 02:19 AM
That the in falling object seems to slow and stop is a function of the mass distortion of the BH.. on space time..

That in reality of that object is that it now is part of that BH. It has been added to the mass.How do you know this ?
How do Supermassive Black Holes at the centres of galaxies get to be so massive, given their slow feeding rate relative to the age of our observable universe ? (Ie: how did they get so big, in such a comparatively small time ?)

Selfsim
2012-Mar-24, 02:45 AM
Oh darn. I only have vague understandings of those things. But a Killing vector field is essentially just a vector field that preserves the metric, apparently. Yau's recent book, The Shape of Inner Space: String Theory and the Geometry of the Universe's Hidden Dimensions, gets one used to these sorts of ideas and operations. Just the simple concept of the metric - how you measure distance in the space you're investigating - and operations upon it, seems to be pretty fundamental to the enterprise....I think the answer to the why Killing vector fields can be used in this particular case, calls for a very involved explanation.

If caveman, (or anyone else), is up to the challenge of explaining it in simple terms however .. for future reference purposes, I'd love to see it done in this medium .. (and, I'd still have the utmost respect for anyone who even attempts such a challenge, regardless of measures of successful knowledge transfer … ).

'Tis really a topic for another thread, however.

Cheers

caveman1917
2012-Mar-24, 07:08 PM
I think the answer to the why Killing vector fields can be used in this particular case, calls for a very involved explanation.

It of course depends on the amount of background knowledge one already has, but i'll give it a shot :). A geodesic minimizes proper time(1), meaning that the Lagrangian for our motion (the "thing" that is being extremized) is proper time. Using 2d schwarzschild as an example, the metric is

d\tau^2 = (1-r_s/r)dt^2 - (1-r_s/r)^{-1}dr^2

so minimizing our lagrangian over the path means minimizing
\int \mathcal{L} d\tau

and since our Lagrangian is proper time
\mathcal{L} = \sqrt{(1-r_s/r)\frac{dt^2}{d\tau^2} - (1-r_s/r)^{-1}\frac{dr^2}{d\tau^2}}.

because of the square root that then appears in the integrand, we can change the Lagrangian to a simpler form (minimizing the square root of a quantity is the same as minimizing the quantity itself, the factor of 1/2 comes in handy later but also doesn't affect the minimizing behaviour)

\mathcal{L} = \frac{1}{2}\left[ (1-r_s/r)\frac{dt^2}{d\tau^2} - (1-r_s/r)^{-1}\frac{dr^2}{d\tau^2} \right]

Note that this doesn't work for null paths where proper time is zero, you'll need another parametrization rather than proper time, but the same concepts apply.

What it means for a lagrangian to be extremized over a particle's path is a constraint set by the Euler-Lagrange equation.

\frac{d}{d\tau}\frac{\partial \mathcal{L}}{\partial \dot{\lambda}} = \frac{\partial \mathcal{L}}{\partial \lambda}}

In words this says that two things have to be the same. The second is how the langrangian changes when we change one of the variables on which it depends. The first is two bits, the second bit is how the lagrangian changes when we change the time derivative of that variable, the first bit says how this second bit changes as we go along the path of the particle. So the change over the path of the change of the lagrangian wrt the change of the time derivative of a variable must be the same as the "base" change of the lagrangian wrt that variable.

So if our Langrangian actually does extremize over the path, these two have to equal. But what if our Lagrangian doesn't depend on some variable, even though it does depend on the time derivative of that variable? That means that the second part of that equation is zero, if the Lagrangian doesn't depend on a variable then a change in that variable will obviously not change the Lagrangian. So that means that the change of the Lagrangian with respect to a change in the time derivative of that variable (the second bit of the first part) remains constant over the path (the first bit of the first part).

Let's do this for our Lagrangian we have for the schwarzschild metric. We see that it doesn't depend on t, though it does depend on its time derivative dt/d\tau. So we know that the change of the Lagrangian wrt dt/d\tau is constant over the path of the particle, meaning that
(1-r_s/r)\frac{dt}{d\tau} is constant over the entire geodesic, it is a so-called constant of motion.

Since we let the particle start from rest at infinity dt/d\tau will be 1, there is at that moment no time dilation between the particle and the stationary observer at infinity since both are at rest at infinity.

So \lim_{r \to \infty}(1-r_s/r)\frac{dt}{d\tau} = 1 and since this is constant, it will always be equal to 1.

So we can plug (1-r_s/r)\frac{dt}{d\tau} = 1 into the schwarzschild metric and we're good to go.

The added thing is that the Lagrangian is very much dependent on a given coordinate system. We could for example write the schwarzschild metric in different coordinates that it does depend on t. A Killing vector is in a sense a coordinate-independent generalization of this. A Killing vector remains a Killing vector, even if it doesn't seem obvious from looking at the metric. The symmetry remains a symmetry, irrespective of how we choose to put coordinates over the surface.

(1) strictly speaking it doesn't minimize proper time over the path. It only does this locally, there might be a globally shorter path. Basically the path is in a stable local minimum, so small variations will "pull it back", but that doesn't mean that there is not a lower global minimum somewhere else.

Cougar
2012-Mar-24, 07:23 PM
Thanks again, caveman. Actually, I was mainly talking about....

-(1-r_s/3r_s)\sqrt{\frac{r_s}{3r_s}} = \frac{-2}{3\sqrt{3}}

Come to look at this a little closer, I don't notice r_s defined anywhere....

caveman1917
2012-Mar-24, 08:16 PM
Come to look at this a little closer, I don't notice r_s defined anywhere....

You're right, i never defined it, r_s is the schwarzschild radius, ie r_s = \frac{2GM}{c^2}. It's just a shorthand notation for a constant, it cancels out in that equation.

Selfsim
2012-Mar-25, 04:56 AM
I cannot speak for others, but as far as I’m concerned, the thinking behind all this is so seriously cool .. I’m frankly in stunned awe of it all.

Introducing Lagrangian mechanics, continuous symmetry and Killing vectors, all at the same time, in order to simplify the mathematics, without losing track of the original problem, just blows me away.

Visualising the synergy between a geodesic, and something like a piece of bent wire in the traditional ‘brachistochrone curve1’ problem (http://en.wikipedia.org/wiki/Brachistochrone), is a kind of a good starting point.

One feeling I’m left with over all this, is sheer and utter respect for the depths of thinking and abstraction, all those who originally worked all this out, must have gone to.

That they left behind knowledge for us, of how to use the tools they created for solving similar problems in the future, is a testament to their determination, forethought and expertise.

Thank you so much for this caveman ... your description is excellent .. many ‘thank yous’ for accepting such a difficult challenge … it is very much appreciated. :) :)

Best Regards

1. This is the question of: ‘What shape can a piece of wire can be bent into, in order for a bead sliding along it, to get the end, in the shortest possible time. The answer is a cycloid shape.

caveman1917
2012-Mar-25, 05:28 AM
A couple of errors got in the above (though nothing that changes anything fundamentally).

ETA: actually you could even leave this step of calculating the dot product out, since \vec{t} is a killing vector, and there are no dt cross terms in the metric, you can just set the other terms in the schwarzschild metric to zero, giving you d\tau^2 = (1-r_s/r)^2dt^2 \Rightarrow 1 = (1-r_s/r)\frac{dt}{d\tau}. (end ETA)

This bit is incorrect, there's an incorrect squared term there that makes this untrue. Setting the other terms to zero will give you
d\tau^2 = (1-r_s/r)dt^2
(note that the coefficient isn't squared). You still need to differentiate wrt dt/d\tau, so this way doesn't work, doing the actual dot product does however.

A geodesic minimizes proper time(1)

A timelike geodesic maximizes proper time, it's a spacelike geodesic that minimizes proper distance. There is a minus sign connecting these two, which is the reason for that difference. This doesn't change anything further on though, the full action is
S = -m\int \mathcal{L} d\tau, the Lagrangian is still extremized.

So wherever i said that proper time is minimized, change it to maximized :)

caveman1917
2012-Mar-25, 07:06 AM
Introducing Lagrangian mechanics, continuous symmetry and Killing vectors, all at the same time

Well they are intimately connected. A Killing vector is something that leaves the metric unchanged in a certain "direction", meaning that we can choose a coordinate system such that a coordinate "points in that direction". This means that the metric in those coordinates doesn't depend on that coordinate. And since the Lagrangian is simply the metric, it means the Lagrangian doesn't depend on that coordinate, and we get a constant of motion out of it. The connection of these is through Noether's theorem.

in order to simplify the mathematics

I wouldn't as much say that it is to simplify the mathematics. It's certainly true that the geodesic equation is usually first derived using Christoffel symbols but, at least to me, it seems that the Lagrangian approach is more fundamental (you can just as well get the general geodesic equation out of it). It's also something that will always work, be it classical mechanics, newtonian gravity, maxwell, SR, GR, QM, ... It's a very fundamental concept underpinning all of physics. That it simplifies the math seems, again at least to me, more of a side-benefit.

without losing track of the original problem

We didn't need to apply the initial conditions as given by the original problem until the end (where we derived that the actual value of this constant of motion was 1, it is equal to the energy per unit mass of the particle as seen by the observer at infinity).

Once we know that (1-r_s/r)\frac{dt}{d\tau} is a constant of motion, we can solve all problems concerning radial (geodesic) motion with it, by setting it to an appropriate value.

Cougar
2012-Mar-25, 01:23 PM
...it cancels out in that equation.

:doh: When all else fails, consider the obvious. :doh: