Could a microscopic black hole suck the whole Earth?

[Sorcerer]

We spoke earlier of asymtotes. An asymptote starting from any 4d coordinate approaching the origin/ intersection of x,y,z,t with the origin, that is the point where space time is singular, will only ever be at the origin after an infinite amount if spacetime

Edited August 18, 2015 by Sorcerer

[Strange]

We spoke earlier of asymtotes. An asymptote starting from any 4d coordinate approaching the origin/ intersection of x,y,z,t with the origin, that is the point where space time is singular, will only ever be at the origin after an infinite amount if spacetime

 

Please show, in mathematical detail, how this conclusion is derived from the Schwarzschild metric.

[Sorcerer]

 

Indeed. That time is the proper time (i.e. in its own frame of reference) for an object in free fall.

 

 

Photons do not have a (valid) frame of reference. I do wish science popularizers would stop this "photons don't experience time" nonsense...

Perhaps but it also makes little sense to speak of events within an event horizon. Yet you're giving the times in seconds, from what observer?

[Strange]

Perhaps but it also makes little sense to speak of events within an event horizon.

 

It makes sense for observers inside the event horizon. It is only from outside the event horizon that it is not possible to observe what happens.

 

Yet you're giving the times in seconds, from what observer?

It is the proper time (i.e. in its own frame of reference) for an object in free fall.

 

More detail here: http://casa.colorado.edu/~ajsh/schwp.html

Edited August 18, 2015 by Strange

[Sorcerer]

 

An observer outside the black hole cannot see us beyond this point - we would appear to take an infinite amount of time to pass through, becoming slower and more redshifted as time goes by. But from our own point of view space and time continue normally.

 

http://casa.colorado.edu/~ajsh/singularity.html

 

But from our own point of view

 

 

Photons do not have a (valid) frame of reference.

 

Relative to an observer freely falling radially from rest at infinity,

lol sorry I have to look up what that last part actually means now

Sheesh u must be butthurt to -1. What did I do?

[Strange]

All of that confirms what I have said. So I'm not sure what your point is. (Tempting though it was, the -1 wasn't from me.)

[DrP]

It might not have been him that gave it to you - it may have been someone else - others read the conversation... It wasn't me this time either. Try reading what the guy has said to you and look up the links he gave you. It is apparent from the conversation that he know what he is talking about and you not so much - so read around what he says and trust him - if you don't understand it then he will explain it to you - if you ignore him and speculate randomly without trying to understand what he wrote then I can predict more -1s from others.

Hope that helps.

[Sorcerer]

Not really I'm not good with analogies. If space is curved so there is no path then there is no path.... Surely there must be a path it's just impossible for a photon to reach the end of it from one direction.

Ie a path that's stretched to infinity.

You could've said my original statement here had no valid frame of reference...... I mean how do you observe a photon falling away from you towards a singularity, it will never be observable to you once it has passed you. I thought it would be obvious therefore I was talking about it from a photons point of view.

It might not have been him that gave it to you - it may have been someone else - others read the conversation... It wasn't me this time either. Try reading what the guy has said to you and look up the links he gave you. It is apparent from the conversation that he know what he is talking about and you not so much - so read around what he says and trust him - if you don't understand it then he will explain it to you - if you ignore him and speculate randomly without trying to understand what he wrote then I can predict more -1s from others.

Hope that helps.

That's what I was doing and I found statements that directly contradict him.

Edited August 18, 2015 by Sorcerer

[Strange]

That's what I was doing and I found statements that directly contradict him.

 

You will have to explain that, because I see nothing that contradicted what I said. You seem to be picking words and phrases out of context and investing them with some sort of hidden meaning.

Edited August 18, 2015 by Strange

[DrP]

reference them and ask him about them then - don't just say he's wrong - show it with references to what you have read - probably it will just be the case you have misinterpreted something. Anyway - I'll stay out of the conversation now as I do not know enough about BHs. I still imagine them to physically be really dense matter (rather than a singularity) - much like a neutron star, but denser - made from squished fundamental particals of matter - like a giant macro nucleus... but I am probably wrong about that. ;-).

[Sorcerer]

 

A photon (or anything else) will reach the singularity in finite time.

What observer can observe this reaching of the singluarity, the photon will never return to be observed.

[Strange]

What observer can observe this reaching of the singluarity, the photon will never return to be observed.

 

How is that relevant? All we know about what occurs inside (or around) a black hole is calculated from the equations of general relativity.

[Sorcerer]

 

What observer can observe this reaching of the singluarity, the photon will never return to be observed.

Looking inward, we see the same Schwarzschild surface (red grid lines) apparently still ahead. Persons who fell through before us would continue to appear here. Already redshifted and slowed, such persons would continue to grow ever more redshifted and slow.

 

http://casa.colorado.edu/~ajsh/singularity.html

 

Through the screen formed by the outward Schwarzschild surface, we see more distant parts of the Schwarzschild surface from above, curved into our view by the gravity of the black hole.

We never get to see the central singularity. Nor do we see anyone else ever hit the central singularity.

 

Edited August 18, 2015 by Sorcerer

[Strange]

Yes. And ... ?

You asked how long it would take to reach the singularity (or rather, you said you never would). That page shows the results of the calculation (for a Schwarzschild black hole of a given mass). And explains in some detail (with pictures and animations) how this might appear - slightly artificially, of course, because the event horizon, for example, is not actually visible.

 

For some reason you don't want to accept his calculation that you would reach the singularity in a fraction of a second. But you are happy to post random quotes from that web page. Would you like to explain their significance to you?

[Sorcerer]

 

How is that relevant? All we know about what occurs inside (or around) a black hole is calculated from the equations of general relativity.

Observers see an object ahead of them towards the singularity frozen in time and they never see it reach it. So it takes an infinite amount of time or has an infinite path inwards. Similarly, at the singularity (or just outside of it if you want), any photon would have an infinite path to travel to exit the blackhole.

Edited August 18, 2015 by Sorcerer

[Strange]

Observers see an object ahead of them towards the singularity frozen in time and they never see it reach it.

 

You do not see it "frozen in time". It is heading towards the singularity faster than you are. You will never see them reach it, nor will you see the singularity.

 

But you (and it) will still reach the singularity in finite time.

 

Please stop making stuff up and read the linked pages (or others if you think I just happened to stumble across an erroneous source).

[Sorcerer]

And don't forget we were talking about photons, because we were talking about how the gravitational field of a black hole doesn't let photons escape the event horizon. This http://casa.colorado.edu/~ajsh/singularity.html is (a story, because it's impossible, it makes no mistake saying so) about a sentient observer.

 

 

(but we are already dead anyway from the tides).

I'm guessing here but I don't think the Schwarzschild equations deal with infinity, or if it existed in them it was re-normalised.

I will look into that later.

[Strange]

And don't forget we were talking about photons, because we were talking about how the gravitational field of a black hole doesn't let photons escape the event horizon.

 

And as photons will be racing ahead of you at the speed of light, they will get to the singularity even sooner.

[Sorcerer]

• is the proper time (time measured by a clock moving along the same world line with the test particle),
• c is the speed of light,
• t is the time coordinate (measured by a stationary clock located infinitely far from the massive body),
• is the radial coordinate (measured as the circumference, divided by 2π, of a sphere centered around the massive body),
• θ is the colatitude (angle from North, in units of radians),
• φ is the longitude (also in radians), and
• is the Schwarzschild radius of the massive body, a scale factor which is related to its mass M by r_s = 2GM/c², where G is the gravitational constant.^[1]

 

 

So can the test particle be a photon?

 

 

In metric theories of gravitation, particularly general relativity, a test particle is an idealized model of a small object whose mass is so small that it does not appreciably disturb the ambient gravitational field.

 

It seems to fit the criteria. While a sentient observer doesn't.

 

-From Wiki

 

 

And as photons will be racing ahead of you at the speed of light, they will get to the singularity even sooner.

Ahead of you? What if they crossed the event horizon at the same time as you.

Relative to an observer stationary in the Schwarzschild metric, our velocity has now reached the speed of light. Relative to an observer freely falling radially from rest at infinity, our velocity is (8/9)^1/2c = 0.94 c.

http://casa.colorado.edu/~ajsh/singularity.html

 

freely falling radially from rest at infinity

 

I can't find an exlanation for what this means after alot of searching, the term comes up alot but the definition is scarce. It also sounds like a fictional observer (or one at the singularity).

 

t is the time coordinate (measured by a stationary clock located infinitely far from the massive body),

 

Which would be outside the observable universe, beyond another event horizon. I guess there's nothing wrong with that, but it just seems like an odd concept. I guess it's to control for the relativistic effects of gravity. Maybe like the concept of there being a clock measuring the time outside the event horizon of the Black Hole, to the singularity, which is infinitely far away? (actually thinking about that is its inverse, the first clock is outside the observable universe, the second is inside.)

Edited August 18, 2015 by Sorcerer

[Strange]

So can the test particle be a photon?

 

I don't believe so. Photons always follow null geodesics, the purpose of that is to calculate the path of a massive particle. (meaning "particle with mass" rather than a really big one )

 

Ahead of you? What if they crossed the event horizon at the same time as you.

 

Light will still travel at the speed of light (in your frame of reference).

 

I can't find an exlanation for what this means after alot of searching, the term comes up alot but the definition is scarce.

 

Free fall: falling purely under the effects of gravity; i.e. not experiencing any forces or acceleration.

Radially: directly towards the centre of the object (black hole, in this case)

From rest: starting with a velocity of zero (relative to the object it is falling towards)

At infinity: it starts at rest an infinite distance away (or sufficiently far away that it doesn't make a significant difference)

 

It also sounds like a fictional observer (or one at the singularity).

 

All observers (in this context) are fictional!

[Sorcerer]

 

The solution is a useful approximation

 

 

I don't believe so. Photons always follow null geodesics, the purpose of that is to calculate the path of a massive particle. (meaning "particle with mass" rather than a really big one )

 

Photons have mass right? You mean a particle with rest mass?

 

 

 

Light will still travel at the speed of light (in your frame of reference).

Yes but you're travelling at the speed of light too.

 

 

Free fall: falling purely under the effects of gravity; i.e. not experiencing any forces or acceleration.

Radially: directly towards the centre of the object (black hole, in this case)

From rest: starting with a velocity of zero (relative to the object it is falling towards)

At infinity: it starts at rest an infinite distance away (or sufficiently far away that it doesn't make a significant difference)

 

How can it "START" at rest an infinite distance away, that would mean it was never at rest.

To travel an infinite distance would take an infinite amount of time, there would always be a preceeding point on the timeline when it was in motion. What's more important, that the particle was originally at rest or that it was orginally an infinite distance away?

 

You could also look at that, if the observer DID start at rest an infinite distance away, it would not have yet reached a place to be able to observe, because the age of the universe is finite.

 

 

 

 

All observers (in this context) are fictional!

 

The metric only deals with test particles.

If this metric requires "t is the time coordinate [to be] (measured by a stationary clock located infinitely far from the massive body)", (which is impossible unless there is an infinite distance to the singularity) if instead the solution included those two values as being from a finite distance, which is a far more reasonable assumption, how would it change the results?

 

Would finite values for time to reach a singularity become infinite?

Edited August 18, 2015 by Sorcerer

[Strange]

Photons have mass right? You mean a particle with rest mass?

 

No, photons do not have mass.

 

Yes but you're travelling at the speed of light too.

 

To echo your earlier question, in what frame of reference. But it doesn't matter. in your frame of reference photons will be rushing ahead of you at the speed of light.

 

How can it "START" at rest an infinite distance away, that would mean it was never at rest.

 

Huh? This is mathematics we are talking about. There is no problem with it being infinitely far away.

 

Imagine you are sufficiently far away holding the test particle stationary above the black hole. You let it go and it starts to move (very, very slowly at first) under the effect of the black hole's gravity. As you are not infinitely far away, then the test particle will not quite reach the speed of light.

 

However, you can move further away and the final velocity at the event horizon will be closer to the speed of light. You can make it get as close to the speed of light as you like by moving sufficiently far away. That is what "infinitely far away" means.

 

Have you done a basic calculus course?

 

The metric only deals with test particles.

 

As a simplifying assumption. As the text you quoted earlier said, this is to minimize the effect on the gravitational field you are calculating. If you need to take the mass of your particle into account, then things get much more complicated. In fact, there may not even be an analytical solution. For example, calculating the orbits or merger of two black holes has to use numerical methods (simulation).

 

But, for a large black hole as in this example, you could probably consider the Earth to be a "test particle".

[Sorcerer]

lol never mind on that last one about t, it's self defeating. If it needs to be a finite distance to show it's an infinite distance, then it's not a finite distance.

 

wow, my bad

 

No, photons do not have mass.

 

 

 

Photons have relativistic mass / inertial mass. https://en.wikipedia.org/wiki/Mass#Special_relativity https://van.physics.illinois.edu/qa/listing.php?id=1424

 

 

 

To echo your earlier question, in what frame of reference. But it doesn't matter. in your frame of reference photons will be rushing ahead of you at the speed of light.

 

 

And in their frame of reference you are? It doesn't seem correct to me, objects travelling at the speed of light shouldn't experience this relativistic effect, only objects travellng below. I'm probably wrong though, can you back this up?

 

 

Huh? This is mathematics we are talking about. There is no problem with it being infinitely far away.

 

Imagine you are sufficiently far away holding the test particle stationary above the black hole. You let it go and it starts to move (very, very slowly at first) under the effect of the black hole's gravity. As you are not infinitely far away, then the test particle will not quite reach the speed of light.

 

However, you can move further away and the final velocity at the event horizon will be closer to the speed of light. You can make it get as close to the speed of light as you like by moving sufficiently far away. That is what "infinitely far away" means.

 

Exactly, too much math and not enough reality, there is a huge problem with that assumption modeling the reality.

 

The universe is a finite age, this will never happen, the universe may come to an end at some point in time, if it doesn't there will always be another unit of time to follow, so it will never be infinitely old.

Also, something infinitely far away began outside hubble volume of the BH, the universe is expanding it will never reach the BH ever.

 

This model makes no physical sense. Remove the infinities, see what happens.

 

 

Have you done a basic calculus course?

 

 

I got an A+ in the last year of highschool for Calculus, never really touched it much since. I understand limits as they approach 0. The key word is APPROACH. They never actually ever reach that limit. EVER.

 

As a simplifying assumption. As the text you quoted earlier said, this is to minimize the effect on the gravitational field you are calculating. If you need to take the mass of your particle into account, then things get much more complicated. In fact, there may not even be an analytical solution. For example, calculating the orbits or merger of two black holes has to use numerical methods (simulation).

 

But, for a large black hole as in this example, you could probably consider the Earth to be a "test particle".

 

 

 

Well you contradict yourself there, if you want to "minimise the effect on the gravitational field you are calculating" you would choose a photon. The relative mass would have an effect and would need to be taken into consideration, because by definition, you then wouldn't be using proper time.

From wiki:

 

is the proper time (time measured by a clock moving along the same world line with the test particle),

 

 

 

The concept of a test particle often simplifies problems, and can provide a good approximation for physical phenomena.

I guess it is a more simplistic way of thinking of a black hole and infinity, unfortunately not a logically correct one, when you use it to state factual information about real black holes.

Edited August 18, 2015 by Sorcerer

[Strange]

Photons have relativistic mass / inertial mass. https://en.wikipedia.org/wiki/Mass#Special_relativity https://van.physics.illinois.edu/qa/listing.php?id=1424

 

We are only concerned with rest mass. "Relativistic mass" is a misleading phrase and usually discouraged (because it is just a source of confusion).

 

And in their frame of reference you are? It doesn't seem correct to me, objects travelling at the speed of light shouldn't experience this relativistic effect, only objects travellng below. I'm probably wrong though, can you back this up?

 

In your own frame of reference you are not moving at the speed of light.

 

In someone else's frame of reference, you might be moving slightly slower than the speed of light, so photons are just inching away from you. But in your frame of reference, photons travel at the speed of light.

 

I understand limits as they approach 0. The key word is APPROACH. They never actually ever reach that limit. EVER.

 

That is not the point: you can get arbitrarily close. You can make the answer as accurate as you wish by approaching infinity. In some cases 1 metre might be enough. In this example, maybe 1 million kilometres is good enough.

 

This is basically the inverse of escape velocity: that is defined as the speed you would need to escape gravity to infinity (without further propulsion). No one says escape velocity is meaningless because you can't get to infinity. (Actually, I think there was someone here a while ago saying that, but no sane person would say it!)

 

It is just a way of calculating the velocity that a free falling body will have as it reaches the event horizon.

 

Exactly, too much math and not enough reality, there is a huge problem with that assumption modeling the reality.

 

This is purely mathematical. You can't go out and measure these things. It is full of approximations and assumptions. The Schwarzschild metric is completely unrealistic: it is for a non-rotating, static black hole that has existed for eternity in an otherwise empty universe.

[Sorcerer]

 

We are only concerned with rest mass. "Relativistic mass" is a misleading phrase and usually discouraged (because it is just a source of confusion).

 

 

I said:

 

 

Photons have mass right? You mean a particle with rest mass

 

In your own frame of reference you are not moving at the speed of light.

Earlier you said photons have no (valid) frame of reference, why would any other thing travelling at the speed of light have one?

 

I guess it doesn't matter since no observer to have that frame of reference could be stationary in the metric anyway (dont use impossible spaceship stories).

 

 

That is not the point: you can get arbitrarily close. You can make the answer as accurate as you wish by approaching infinity. In some cases 1 metre might be enough. In this example, maybe 1 million kilometres is good enough.

Exactly and when you use arbitary values for infinites to model objects which actually include infinities, you can get finite answers to infinite realities.

 

 

This is basically the inverse of escape velocity: that is defined as the speed you would need to escape gravity to infinity (without further propulsion). No one says escape velocity is meaningless because you can't get to infinity. (Actually, I think there was someone here a while ago saying that, but no sane person would say it!)

Oh it's not meaningless, especially when dealing with objects whos parameters are otherwise finite, it's a great way to model things, it's a excellent approximation, in regards to modelling finite things. Just not Black Holes, it becomes a rather poor approximation then.

 

 

This is purely mathematical. You can't go out and measure these things. It is full of approximations and assumptions. The Schwarzschild metric is completely unrealistic: it is for a non-rotating, static black hole that has existed for eternity in an otherwise empty universe.

Exactly.

Edited August 18, 2015 by Sorcerer