Black Hole Question

[Mark_Heli]

I was just thinking about black holes and had two questions to try and help visualise the distance & speed of an incoming particle on a human understandable scale:

1. What is the "vertical distance" that an incoming particle would have to travel to get from a diameter of 25km down to 1mm, and then the plank length?

2. How long it would take for the incoming particle to travel these two distances?

Hopefully the attached picture help to explain these questions clearly...

Thanks,
Mark

[Eratosthenes]

....where is the event horizon in your diagram? Time slows down space is stretched, what happens to the shape of your particle? Are you assuming it doesnt change? The Planck length is many orders of magnitude smaller than the smallest known particle.

where did you get the black hole from - I want one.

:D

[sjastro]

Since you seem to be boasting a knowledge of BHs that somehow qualifies you to ridicule the OP's perfectly valid questions, let me redirect the questions to you in a slightly more mathematical format.

(1) Explain why the maximum coordinate velocity for a particle accelerating towards a non rotating BH is 0.39c.
(2) Show why the maximum coordinate velocity is reached at a distance of 3 times the event horizon radii.
(3) Explain how the coordinate velocity of the particle is zero at the event horizon.
(4) Why does the proper velocity of the particle reach c at the event horizon?
(5) Explain the differences of the properties of a measuring stick inside and outside the horizon.

You should have no difficulties in educating us on the subject.

[Eratosthenes]

Allow me to ask Mark if he felt ridiculed by the posting of the above questions.

Mark, did you feel ridiculed by the posting of the above questions?

:D

[Max Vondel]

Children, Children......be nice!

:P

[Eratosthenes]

he started it teacher, he always starts it....

send him to the principal's office

:D

[Stonius]

In an attempt to help the original poster find some clarity, I believe the answer is different depending on where you observe the particle from. Perhaps a little elaboration might help focus the brains trust? :-)

[Mark_Heli]

Ok good, I have learnt that the question is not quite right - that is a start. Thank you Peter, Steven & Markus. I also find the banter quite amusing :lol: If anything, based on my questions I should be ridiculed. Ha Ha....

While I don't have a physics background, I enjoy reading popular physics books and watching Leonard Susskind lectures on YouTube.

Apart from other possible issues, I understand that the size of the black hole and position of the observer would need to be specified and there will be impacts caused by length contraction and time dilation.

Let me try again with maybe a simpler question where I don't put my foot in it (although I almost certainly will).

How long would it take the midpoint of particle approaching a stellar mass black hole (3km event horizon) to travel from 25km to 5km away from the singularity when observed from a distance of (1) 100km away and (2) 10 000 km away.

Updated picture again.

Thanks,
Mark

[sjastro]

Hello Mark,

I'll give you a very general outline of the calculation and the theory without giving a specific answer as the resulting integral is very difficult to solve. Also the derivation of the integral requires a working knowledge of the Schwarzschild metric which is a static Black Hole solution for General Relativity.

I'll leave it as an exercise to solve the integral although I suspect there is no direct solution and would need to be solved by numerical methods.

In my response to Peter I used embolden terms coordinate and proper velocity.
To solve your problem we need to look at coordinate and proper time by incorporating a third observer who is located outside the gravitational well of the black hole.
To simplify matters we will assume the object is released from rest and falls into a non rotating black hole.
This third observer will measure the coordinate time it takes for the object to be released from rest 25 km from the singularity and free falls to 5 km from the singularity.
The coordinate time is the elapsed time measurement based from the third observer's frame of reference. The object is sending a signal to this observer.
If on the other hand there is a clock attached to the moving object, this clock is measuring proper time as the clock is in the object's frame of reference.

The mathematics for a non rotating black hole is based on the Schwarzschild metric and the time taken for an object initially at rest to fall from 25 km to 5 km in the gravitational field of a one solar mass black hole is given by the integral.

http://members.iinet.net.au/~sjastro/astrophysics/freefall.gif
(attachment 1)

c is the speed of light. G is the gravitational constant, r is the radial distance.

As stated this is one very difficult integral to solve and the answer is based on the coordinate time of the distant third observer.
We will simply call it t.

The next step is to calculate the proper time T for the observers located at 100 and 10000 km from the singularity respectively.
Since these observers are located in different locations in the gravitational well their clocks will run slower compared to the coordinate time measured by the distant third observer.
One is comparing the proper times of these observers based on how slow their local clocks run against the distant observers coordinate time.

The equation in this case is much more simple.

http://members.iinet.net.au/~sjastro/astrophysics/grav_time_dilation.gif
(attachment 2)

rs is the event horizon and has a value of 2.95 km.
r is the radial distance of the observers from the singularity.
t is the coordinate time for the third observer.

For the observer at 10000 km, the proper time T1 = t(1-(2.95/10000)^0.5)
For the observer at 100 km the proper time T2 = t(1-2.95/100)^0.5)

Hope this helps.

Regards

Steven

[Mark_Heli]

Hi Steven,

Thank you very much for the detailed response. I have read through the answer and I am going to read up further on the Schwarzschild metric.

From my original slightly ill-formed question, you have given me a lot of information.

There are some truly bright people on the forum...

Thanks,
Mark

[sjastro]

Using the on line integral calculator (http://www.integral-calculator.com/), the solution to the equation http://members.iinet.net.au/~sjastro/astrophysics/freefall.gif reveals the coordinate time for an object dropping from 25 to 5 km for the third observer is 1.97 X 10^9 years.
Comparing this to my own calculation which required me to butcher the equation using linear approximations in order to integrate it, I obtained a value of 2.02 X 10^10 years.

Since computers do a far better job than humans when it come to brute force calculations, I'll accept the computer result as more accurate.:P

Using the computer result the proper time for the observer 10000km away from the singularity is 1.94 X 10^9 years and the observer 100 km away is
1.63 X 10^9 years.

This indicates the gravitational time dilation occurring.
If the calculation was for freefall from 25 km to the event horizon at 2.95 km, the coordinate time would be infinite.

[billdan]

Hi Steven,

I'm not questioning your formula that you used, but I find it hard to believe it takes nearly 2 Billion years for an observed object to travel 20Km falling into a black hole.
If you had said it was 20 Sec's vs 16 Sec's, I would have said that sounds reasonable.

I understand that it is from a close observers point of view where time slows down, but its still mind boggling.

Regards
Bill

[sjastro]

Bill,

The result looks quite incredulous but at or near the event horizon, time literally stands still to an observer outside the gravitational well.
To this observer photons emitted from the object are infinitely red shifted, and if it was possible to observe the object it would be forever stuck on the event horizon without passing through it.

The falling object on the other hand would freely pass through the horizon.
If the object is a clock and assuming it is not destroyed by tidal forces, the maximum proper time elapsed during the 20 km fall is given by the equation.
http://members.iinet.net.au/~sjastro/astrophysics/freefall_propertime.gif
This works out to being about 6.67 X 10^-5 seconds.
Over this distance in the vicinity of the event horizon the object is travelling at around 0.995c.

The differences are quite staggering when comparing the clock's proper time to the elapsed coordinate time of 1.94 X 10^9 years when measured by an outside observer

On a much more subtle scale we observe these effects in our own backyard.
The Shapiro time delay (https://en.wikipedia.org/wiki/Shapiro_delay) is one such example.

Steven

[xelasnave]

Steven if objects are frozen in time, if I have it correct, would this not mean that we should "see" a collection of various stars or gas that have fallen into a black hole over the life time of that black hole?

Alex

[sjastro]

Alex,

The closer the object is to the event horizon, the greater the gravitational redshift.
While it may take an infinite time for the object to free fall to the event horizon to a distant observer, in reality the observer will never see the object frozen at the event horizon as the emitted photons are infinitely redshifted.

Steven

[Atmos]

If we were able to speed up time so as to watch someone fall into the black hole over the course of a minute or two as opposed to the billions of years required, we wouldn't see said volunteer fall in, merely fade out of existence.

[Eratosthenes]

....to reach the actual centre (singularity) it's even greater - the time taken is approximately in the vicinity of infinity give or take a few trillion years.

:D

(as far as questioning the formula - well you should question absolutely everything in the religion of Physics before you arrive at your own conclusions concerning reality. There are holes all over the place in the Holy Book of Physics - especially The Book of Quantum, verse 6:11)

[Eratosthenes]

.....are you saying that black holes don't exist?

[xelasnave]

Thanks Steven.
Alex

[sjastro]

In my previous calculations I inadvertently expressed the value of M in the general equation for freefall http://members.iinet.net.au/~sjastro/astrophysics/general_equation.gif as being one (one solar mass) instead of the value 1.989 X 10^30 kg.:P

Needless to say the results are somewhat different now and Bill's suspicions of the results were quite justified.:)

The coordinate freefall time interval t for the third observer is given by equation (1).
http://members.iinet.net.au/~sjastro/astrophysics/freefall.gif
This is now 126 sec.

The clocks located at 100 and 10000 km would record this time interval by equation (2) http://members.iinet.net.au/~sjastro/astrophysics/grav_time_dilation.gif as 124.13 and 125.98 seconds respectively.

To illustrate the coordinate time to fall from 25 km to 2.95km the event horizon is infinite, I graphed equation (1).
http://members.iinet.net.au/~sjastro/astrophysics/Black_Hole_Time_Dilation.png

The red shaded area is the calculated time of 126 sec when the particle falls from 25 to 5 km as seen from a distant observer.
As graph shows the event horizon is a mathematical limit where the change in coordinate time becomes infinitely large.

Steven

[sil]

While Mark is sleeping off the great answers can I sidetrack a little please.



Spaghettification. I can accept this what what would appear to an observer as an object approaches/crosses an event horizon of a black hole. What I've been wondering recently is does this ACTUALLY occur to an object itself. From its own reference frame would it even be noticeable approaching/crossing the event horizon since space-time is stretched etc wouldnt observations (visual/felt/heard/etc) stil be observed/recorded by the falling object the same as if it was floating free in space or falling under gravity on earth? Just because an observer standing off to the side sees the falling object stretch I don't see why we can say with certainty it gets "ripped apart" it will just bend to the shape of space time its falling through. Does this make sense?

[Eratosthenes]

Twice as big as Mars: Hubble spots molten ‘cannonballs’ blasting through space

https://www.rt.com/viral/361940-fireballs-space-hubble-vhydrae/

...I wonder if black holes are directly or indirectly involved with these cannonballs?