Time coordinate of the Schwarzschild metric

[Sorcerer]

I was just wondering what would change in the results of the schwarzschild metric if the time coordinate was instead measured at a finite distance from the massive body, rather than the infinite distance it is currently placed at.

 

How would the results vary as the finite distance increased or decreased?

Edited August 29, 2015 by Sorcerer

[ajb]

Your question is a little vague, but you can use different coordinate systems to describe the Schwarzschild black hole. Common ones are listed here

https://en.wikipedia.org/wiki/Schwarzschild_metric#Alternative_coordinates

 

Typically different coordinates my be better suited for specific questions.

[Sorcerer]

Thanks, which coordinate system is the best represtentation of a black hole under our current understanding of the space time. IE do any represent a black hole where space time is accelerating in expansion?

[ajb]

Thanks, which coordinate system is the best represtentation of a black hole under our current understanding of the space time. IE do any represent a black hole where space time is accelerating in expansion?

These solutions have nothing to do with the expanding Universe.

[Sorcerer]

The black hole event horizon bordering exterior region I would coincide with a Schwarzschild t-coordinate of +∞ while the white hole event horizon bordering this region would coincide with a Schwarzschild t-coordinate of −∞, reflecting the fact that in Schwarzschild coordinates an infalling particle takes an infinite coordinate time to reach the horizon (i.e. the particle's distance from the horizon approaches zero as the Schwarzschild t-coordinate approaches infinity), and a particle traveling up away from the horizon must have crossed it an infinite coordinate time in the past. This is just an artifact of how Schwarzschild coordinates are defined; a free-falling particle will only take a finite proper time (time as measured by its own clock) to pass between an outside observer and an event horizon, and if the particle's world line is drawn in the Kruskal-Szekeres diagram this will also only take a finite coordinate time in KruskalSzekeres coordinates.

 

Could you also elaborate a little more on this, or rather simplfy/explain it easier, it's similar to what I was asking in the original post I think, although I don't have much time to disect the word salad at the moment.

 

The tortoise coordinate approaches −∞ as r approaches the Schwarzschild radius r = 2GM.

When some probe (such as a light ray or an observer) approaches a black hole event horizon, its Schwarzschild time coordinate grows infinite. The outgoing null rays in this coordinate system have an infinite change in t on travelling out from the horizon. The tortoise coordinate is intended to grow infinite at the appropriate rate such as to cancel out this singular behaviour in coordinate systems constructed from it.

The increase in the time coordinate to infinity as one approaches the event horizon is why information could never be received back from any probe that is sent through such an event horizon. This is despite the fact that the probe itself can nonetheless travel past the horizon. It is also why the space-time metric of the black hole, when expressed in Schwarzschild coordinates, becomes singular at the horizon - and thereby fails to be able to fully chart the trajectory of an infalling probe.

 

And this...... long shifts at the moment, perhaps I'll disect it later.

 

One more question; The Schwarzchild metric defines the time coordinate as being measured from a clock which is an infinite distance from the massive body, can this clock actually exist under our current understanding of the universe and if not, does that mean the Schwarzchild metric only models an idealised or fantasy version of a black hole?

Edited August 31, 2015 by Sorcerer

[ajb]

Kruskal–Szekeres coordinates might be the coordinate system you are looking for. They cover the maximally extended Schwarzschild black hole and do not have 'coordinate singularities'. Also, you can draw diagrams similar to Minkowski diagrams to help with intuition.

 

Anyway, what you have to remember is that it takes an infinite amount of 'coordinate time' in Schwarzschild coordinates for a particle to reach the event horizon; basically the coordinate time here is the proper time of a distant observer. This is due to how the coordinates are set up. But as far as the particle itself is concerned, it takes only a finite amount of proper time to reach the horizon.

 

Kruskal–Szekeres coordinates are nicer in the sense that the 'coordinate time' taken to reach the event horizon is finite.

Edited August 31, 2015 by ajb

[Sorcerer]

Are there similarities between these time coordinates and those we would place from an observer to the edge of its hubble volume?

[ajb]

Are there similarities between these time coordinates and those we would place from an observer to the edge of its hubble volume?

I do not know; it may also depend on what you mean by similarities. You should compare coordinates with the coordinates used in FWR cosmologies.

[Mordred]

Are there similarities between these time coordinates and those we would place from an observer to the edge of its hubble volume?

Not really, The Hubbles sphere or Hubble horizon describes a homogeneous and isotropic fluid. It is defined as the time it takes light to reach an observer multiplied by the age of the universe without expansion.

 

The Kruskal coordinates describe and inhomogeneous and anisotropic fluid, (Bh, has a center). Whose gravititational influence causes time dilation at a rate per coordinate change that is completely different than the universe itself.

The Schwartzchild metric starts from a vacuum, then describe the density gradient toward the singularity. This has a preferred direction. The FLRW metric has no preferred location or direction.At each point in proper time the average energy/mass density throughout the universe is homogeneous and isotropic so you have no time dilation due to higher density in the past.

 

This isn't the case of a BH. At a moment in proper time you have an energy/mass gradient due to localized space time curvature. This causes a localized time dilation.

 

Universe geometry doesn't have time dilation as globally any moment in proper time the energy/mass density is uniform

Edited August 31, 2015 by Mordred

[Sorcerer]

So an observer isn't the center of its hubble volume. The hubble volume doesn't account for expansion. And the universe wasn't more dense in the past?????

Anyway a bit of digression no real need to answer that.

[Strange]

So an observer isn't the center of its hubble volume. The hubble volume doesn't account for expansion. And the universe wasn't more dense in the past?????

 

Did anyone say that? Yes, no, and yes.

[Sorcerer]

Not really, The Hubbles sphere or Hubble horizon describes a homogeneous and isotropic fluid. It is defined as the time it takes light to reach an observer multiplied by the age of the universe without expansion.

 

The Kruskal coordinates describe and inhomogeneous and anisotropic fluid, (Bh, has a center). Whose gravititational influence causes time dilation at a rate per coordinate change that is completely different than the universe itself.

The Schwartzchild metric starts from a vacuum, then describe the density gradient toward the singularity. This has a preferred direction. The FLRW metric has no preferred location or direction.At each point in proper time the average energy/mass density throughout the universe is homogeneous and isotropic so you have no time dilation due to higher density in the past.

 

This isn't the case of a BH. At a moment in proper time you have an energy/mass gradient due to localized space time curvature. This causes a localized time dilation.

 

Universe geometry doesn't have time dilation as globally any moment in proper time the energy/mass density is uniform

 

So the first implies a hubble volume doesn't account for the expansion of the universe, which is odd because I thought the hubble volume was created by the universe having expanded faster than the speed of light, not allowing us to observe parts which haven't had time for the light from them to reach us yet.

 

The second implies that a hubble volume doesn't have a center, which is also odd, because every observer has its own independent hubble volume, shifted by a distance of c/t (light years, seconds, etc). So each observer is the center of its own independent hubble volume.

 

The third(x2) implies that observations of the past, (once allowed as the hubble volume increases to such a scale of moments soon after the big bang and recombination, where the universe was smaller and denser), won't show any effects of the density of matter on observations of the passage of time. I find this odd, since if we obey the laws of conservation of mass and energy, yet the size of the universe is increasing, then the energy+mass density of the universe must decrease over time, why then would there be no relativistic effect?

 

This also leads me to another question, but I'll create a seperate thread for that.

Edited September 1, 2015 by Sorcerer

[Strange]

So the first implies a hubble volume doesn't account for the expansion of the universe, which is odd because I thought the hubble volume was created by the universe having expanded faster than the speed of light, not allowing us to observe parts which haven't had time for the light from them to reach us yet.

 

The Hubble volume is defined as "is a spherical region of the Universe surrounding an observer beyond which objects recede from that observer at a rate greater than the speed of light due to the expansion of the Universe"

https://en.wikipedia.org/wiki/Hubble_volume

 

This is caused by expansion but (oddly) doesn't take expansion into account, which is why it is smaller than the observable universe.

 

The second implies that a hubble volume doesn't have a center

 

No, it says that the FLRW metric doesn't have a centre; i.e. the universe doesn't have a centre. Both the observable universe and the Hubble volume are centred on the observer, but the universe isn't.

 

The third(x2) implies that observations of the past, (once allowed as the hubble volume increases to such a scale of moments soon after the big bang and recombination, where the universe was smaller and denser), won't show any effects of the density of matter on observations of the passage of time.

 

Correct. You can only define/measure time dilation by comparing clocks. So you can compare gravitational time dilation in one part of space with that at another location at the same time. But how can you compare a clock yesterday with a clock today?

[Mordred]

Strange pretty much covered your questions on my post.

 

Time dilation requires to a non accelating observer an energy/mass gradient. This occurs with a BH.

 

However no matter where the observer is located in the universe, the energy/mass distribution is the same everywhere at any moment in proper time. Yes the universe becomes denser in the past. However at any moment in proper time The average mass density is uniform, which means there is no time dilation to that observer.