View Full Version : black hole?

mike842

2008-Oct-13, 06:27 AM

Hi, I have a simple question:

if a solar mass is 2x10^30 kg

if there are 1x10^11 stars in a galaxy

if there are 1.25x10^11 galaxies in the universe

Then there about 2.5x10^52 kg of "normal" matter in the universe

since normal matter comprises 4% of total mass and 22% of the universe mass is dark matter there's about 5.5 times as much dark vs normal matter

multiplying 2.5x10^52 kg of "normal" matter by 6.5 give as estimate for total mass or 1.625x10^53 kg

using the formula for Schwarzschild radius

r = 2Gm/c^2

where G = 6.67300 × 10-11 m3 kg-1 s-2

velocity of light is 3x10^8 m/s

then r = 2.41E+26 m

since a light year is about 9.46*10^15 meters

Schwarzschild radius for the known mass of the universe is about

25.5*10^9 light years

hummmmmm

looks like we're living in a bubble universe that's a black hole as seen from outside

a bit strange but I guess you get this question frequently

what's the explanation?

thorkil2

2008-Oct-13, 08:34 AM

I don't see it as strange. For the photon or the wave front, the speed of light is infinite. Zero distance from point a to point b (because it takes zero time to get there). Not hard then to see the Universe as a singularity from that frame of reference.

timb

2008-Oct-13, 09:33 AM

Hi, I have a simple question:

if a solar mass is 2x10^30 kg

if there are 1x10^11 stars in a galaxy

if there are 1.25x10^11 galaxies in the universe

Then there about 2.5x10^52 kg of "normal" matter in the universe

since normal matter comprises 4% of total mass and 22% of the universe mass is dark matter there's about 5.5 times as much dark vs normal matter

multiplying 2.5x10^52 kg of "normal" matter by 6.5 give as estimate for total mass or 1.625x10^53 kg

using the formula for Schwarzschild radius

r = 2Gm/c^2

where G = 6.67300 × 10-11 m3 kg-1 s-2

velocity of light is 3x10^8 m/s

then r = 2.41E+26 m

since a light year is about 9.46*10^15 meters

Schwarzschild radius for the known mass of the universe is about

25.5*10^9 light years

hummmmmm

looks like we're living in a bubble universe that's a black hole as seen from outside

a bit strange but I guess you get this question frequently

what's the explanation?

Cosmology is way over my head but the observable universe is over 40 Gly in radius so your conclusion fails.

RussT

2008-Oct-13, 10:20 AM

I don't see it as strange. For the photon or the wave front, the speed of light is infinite. Zero distance from point a to point b (because it takes zero time to get there). Not hard then to see the Universe as a singularity from that frame of reference.

Ahhhhh, So, when 'science'/we/they/mainstream does the Lunar Ranging experiments, the photon/wave front is traveling 'instantly' to and from the moon??? Either? Neither?

pzkpfw

2008-Oct-13, 11:49 PM

...the photon/wave front is traveling 'instantly' to and from the moon???...

Who is measuring it? (Whose clock is used?)

The scientists? Or the photons themselves?

Cougar

2008-Oct-14, 12:40 AM

Hi, I have a simple question:

if a solar mass is 2x10^30 kg

if there are 1x10^11 stars in a galaxy

if there are 1.25x10^11 galaxies in the universe

Then there about 2.5x10^52 kg of "normal" matter in the universe

since normal matter comprises 4% of total mass and 22% of the universe mass is dark matter there's about 5.5 times as much dark vs normal matter

multiplying 2.5x10^52 kg of "normal" matter by 6.5 give as estimate for total mass or 1.625x10^53 kg

using the formula for Schwarzschild radius

r = 2Gm/c^2

where G = 6.67300 × 10-11 m3 kg-1 s-2

velocity of light is 3x10^8 m/s

then r = 2.41E+26 m

since a light year is about 9.46*10^15 meters

Schwarzschild radius for the known mass of the universe is about

25.5*10^9 light years

hummmmmm

looks like we're living in a bubble universe that's a black hole as seen from outside

a bit strange but I guess you get this question frequently

what's the explanation?

One thing I expect is not right is your hypothesis:

if there are 1.25x10^11 galaxies in the universe...

You're apparently talking about the visible universe. Observational support for an early rapid expansion would also seem to support the proposition that our universe is much, much bigger than what we can currently see of it, which would make its actual mass and extent hugely larger.

And if indeed the expansion is accelerating as measurements imply, then the curtain on our light horizon is slowly closing, and the outer edge of our visible universe will constantly be disappearing.

looks like we're living in a bubble universe that's a black hole as seen from outside

a bit strange but I guess you get this question frequently

what's the explanation?

Even if there was a "mass-radius coincidence," I think there are lots of problems with such an idea. Though I don't have citations, I expect many people have considered it and quickly ran into problems with it, not the least of which is detailing the meaning of "as seen from outside."

thorkil2

2008-Oct-14, 04:38 AM

Ahhhhh, So, when 'science'/we/they/mainstream does the Lunar Ranging experiments, the photon/wave front is traveling 'instantly' to and from the moon??? Either? Neither?

So far as the observer is concerned, no. We measure a finite time of travel. But the photon, in it's own frame of reference, takes zero time to get from any point a to any point b anywhere in the Universe. It is only seen as finite speed from a frame of reference that is not moving at c.

(edit) I just re-read my original statement, and it was not put as clearly as it should have been. Sorry. I can see RussT's question as being justified, but the answer here is correct. (end edit)

RussT

2008-Oct-14, 09:01 AM

So far as the observer is concerned, no. We measure a finite time of travel. But the photon, in it's own frame of reference, takes zero time to get from any point a to any point b anywhere in the Universe. It is only seen as finite speed from a frame of reference that is not moving at c.

(edit) I just re-read my original statement, and it was not put as clearly as it should have been. Sorry. I can see RussT's question as being justified, but the answer here is correct. (end edit)

So, when 'science' measures photons to be traveling at a Constant "c", 186,282.40 miles per second, that means that when we send a light signal to the moon, in 1 tenth of a second, that 'wave front' has traveled ~18,628 miles......and will reach the 186,282.40 mile mark in 9 more tenths of a second

SO, how do you explain those photons in your definition outracing the 'wave front' to get to the moon "Instantaneously"/in 0 time?

matt.o

2008-Oct-14, 09:40 AM

RussT - you need to familiarise yourself with time dilation (the special relativistic version) and then think about how time works in the photon's reference frame as compared to the earth-bound observers frame.

thorkil2

2008-Oct-14, 09:40 AM

The zero travel time is only from the frame of reference of the photon, not from your position as a measuring entity. You experience and measure the time it takes for a photon to get from the earth to the moon. From the photon's frame of reference, traveling at c, no time elapses and there is no distance. If you could travel at c, no time would elapse for you either, for any distance traveled. It's a consequence of time dilation. The photon you see takes a finite travel time as you measure it. If you could place a hypothetical clock at the wavefront, traveling with the wavefront, it would take your identical rest frame clock an infinite number of seconds to equal the traveling clock's one second (or any fraction thereof, for that matter). Of course, travel at c is impossible for anything but the photon/light wave, but it still is a definable frame of reference, and what happens there is not the same as what you observe from your rest frame.

To get a clearer picture, try this: You have two identical clocks. One travels at 0.5c, the other remains with you in your rest frame. It will take your clock 1.1547 seconds to equal one second on the traveling clock. At 0.9c, it will take your clock 2.2941 seconds to equal one second of the traveling clock, and the interval grows as the speed increases toward c. At c (if you could achieve c) the difference would be infinite. Length contraction does not apply only to the moving object, but (because of time dilation) to the subjective distance traveled. If you're the traveler, the Universe gets shorter for you as you approach c. Again, if you could achieve c, distance (from your moving frame) would disappear. It would take zero time to get from any point a to any point b in the Universe. That's not ATM; I think you'll find the effect described in any physics book.

thorkil2

2008-Oct-14, 02:11 PM

To put this another way, for c to be constant, time and space must be variable in the ways noted above.

timb

2008-Oct-14, 10:17 PM

The zero travel time is only from the frame of reference of the photon, not from your position as a measuring entity. ...

Because it is impossible for an observer to be in the frame of reference in question, you can at least be sure that your hypothesis will never be contradicted by experiment.

For his next act thorkil2 will demonstrate that an infinite number of angels can stand on the head of a pin, but only within the event horizon of a black hole. :)

alainprice

2008-Oct-14, 11:33 PM

The zero travel time is only from the frame of reference of the photon, not from your position as a measuring entity. You experience and measure the time it takes for a photon to get from the earth to the moon. From the photon's frame of reference, traveling at c, no time elapses and there is no distance. If you could travel at c, no time would elapse for you either, for any distance traveled. It's a consequence of time dilation. The photon you see takes a finite travel time as you measure it. If you could place a hypothetical clock at the wavefront, traveling with the wavefront, it would take your identical rest frame clock an infinite number of seconds to equal the traveling clock's one second (or any fraction thereof, for that matter). Of course, travel at c is impossible for anything but the photon/light wave, but it still is a definable frame of reference, and what happens there is not the same as what you observe from your rest frame.

To get a clearer picture, try this: You have two identical clocks. One travels at 0.5c, the other remains with you in your rest frame. It will take your clock 1.1547 seconds to equal one second on the traveling clock. At 0.9c, it will take your clock 2.2941 seconds to equal one second of the traveling clock, and the interval grows as the speed increases toward c. At c (if you could achieve c) the difference would be infinite. Length contraction does not apply only to the moving object, but (because of time dilation) to the subjective distance traveled. If you're the traveler, the Universe gets shorter for you as you approach c. Again, if you could achieve c, distance (from your moving frame) would disappear. It would take zero time to get from any point a to any point b in the Universe. That's not ATM; I think you'll find the effect described in any physics book.

How do you suggest we discuss redshift? Perhaps we call it an instant shift in the energy of the photon, which never had time nor place to gain or lose energy in the first place.

Let's teach it starting tomorrow, sounds like fun.

The only way we can say a photon is instantaneous is if it has an instant effect, which has not been found.

Back to the OP. The event horizon you found is still smaller that the observable universe(assuming the current radius is 46 billion light years).

thorkil2

2008-Oct-15, 03:54 AM

For timb and alainprice--come on, this is text book SR. If you have a complaint, take it to Einstein. The red shift is measured from some relative rest frame. Your measurements are valid from that frame and only that frame. Change frames, and you get different results. I'll say it differently: if I could use the word "experiences" loosely in this context, for lack of a better word, I would say the photon "experiences" no elapsed time from source to destination. That has nothing to do with what you measure because you are doing the measuring from a different frame of reference. It isn't instantaneous from any frame of reference you have access to, so no, there won't be instant effect. To the photon, however, the effect is instantaneous. Different frames of reference. Now, I've been dealing with this for 50 years, but I won't make you take my word for it. A quote from Paul Davies (Professor of Natural Philosophy at the University of Adelaide in Australia at the time this was written) in his book "About Time" (1995), p.190:

From the point of view of the [light] pulse, no time at all elapses as, in our frame of reference, it sweeps across the solar system.

Will that do, or do I need to dig out more references?

RussT

2008-Oct-15, 10:24 AM

To put this another way, for c to be constant, time and space must be variable in the ways noted above.

First of all, THIS is absolutely False!

To make a flat out statement like this is absolutely irresponsible!

"c" can be Constant and Time can be Constant as well as 'Space'.

ETA: I am being very careful because I am in Q&A to only 'ask questions', BUT...

RussT

2008-Oct-15, 11:17 AM

The zero travel time is only from the frame of reference of the photon, not from your position as a measuring entity. You experience and measure the time it takes for a photon to get from the earth to the moon. From the photon's frame of reference, traveling at c, no time elapses and there is no distance. If you could travel at c, no time would elapse for you either, for any distance traveled. It's a consequence of time dilation. The photon you see takes a finite travel time as you measure it. If you could place a hypothetical clock at the wavefront, traveling with the wavefront, it would take your identical rest frame clock an infinite number of seconds to equal the traveling clock's one second (or any fraction thereof, for that matter). Of course, travel at c is impossible for anything but the photon/light wave, but it still is a definable frame of reference, and what happens there is not the same as what you observe from your rest frame.

To get a clearer picture, try this: You have two identical clocks. One travels at 0.5c, the other remains with you in your rest frame. It will take your clock 1.1547 seconds to equal one second on the traveling clock. At 0.9c, it will take your clock 2.2941 seconds to equal one second of the traveling clock, and the interval grows as the speed increases toward c. At c (if you could achieve c) the difference would be infinite. Length contraction does not apply only to the moving object, but (because of time dilation) to the subjective distance traveled. If you're the traveler, the Universe gets shorter for you as you approach c. Again, if you could achieve c, distance (from your moving frame) would disappear. It would take zero time to get from any point a to any point b in the Universe. That's not ATM; I think you'll find the effect described in any physics book.

SO, when we send a light beam to the moons reflector (Let's assume for ease that the moon is exactly 186,282.40 Miles) we wind up with 2 maths for the 'same' beam IE; that beam is both traveling at "c" and takes one second to reach the moon AND that same beam traveled 0 distance (Because 'space was contracted) in 0 time, umm or is it 186,282.40 miles traveled "Instantly"?and BOTH maths are considered "Simultaneous"?

Don't you think that this causes a "Duality" problem?

IF, there was a mirror at 186,282.40 miles, and if those photons were 'Really' getting there 'instantaneously', then would light travel time there and back only be 1/2 of what it should be ie; 1 second rather than the round trip time of 2 seconds?

captain swoop

2008-Oct-15, 12:39 PM

this photon stuff is hijacking someone elses question thread. maybe one of you should start a new one.

Jeff Root

2008-Oct-15, 03:28 PM

Russ, what are you doing? You are writing as though you know nothing

at all about special relativity.

-- Jeff, in Minneapolis

thorkil2

2008-Oct-15, 05:11 PM

this photon stuff is hijacking someone elses question thread. maybe one of you should start a new one.

MY apologies, and thank you for pointing that out. It wasn't my intention to take this so far away from the OP. What started as a fairly simple comment has gone far afield. I'll think about starting another thread--so I can answer the arguments, if nothing else.

Argos

2008-Oct-15, 05:49 PM

Schwarzschild radius for the known mass of the universe is about 25.5*10^9 light years

Only if that mass was squeezed into a singularity [A key concept when talking Black Holes]. So the OP´s point seems moot.

thorkil2

2008-Oct-15, 07:22 PM

Schwarzschild radius for the known mass of the universe is about 25.5*10^9 light years

Only if that mass was squeezed into a singularity [A key concept when talking Black Holes]. So the OP´s point seems moot.

Which takes this back to my observation that there is a frame of reference from which the Universe might be interpreted as a singularity. As for the validity of the calculations, they would depend greatly on the range of accuracy of estimated values in the input data. I didn't follow through with the calculations, and I have no way of determining the validity of the original numbers, but have long considered the idea of the Universe as a singularity interesting. (I have opened a separate thread to continue the discussion of photon time. Apologies again for hijacking the OP).

alainprice

2008-Oct-15, 07:27 PM

Is the sun a singularity? It has an event horizon!

If the EH is smaller than the surface, I'm pretty sure that does not qualify the object as a singularity.

Argos

2008-Oct-15, 07:32 PM

Which takes this back to my observation that there is a frame of reference from which the Universe might be interpreted as a singularity.

Well, as far as GR is concerned, a singularity is a "point" in the universe where the density of matter and the gravitational field are infinite. It has a well-defined meaning within the scope of the [mainstream] theory.

timb

2008-Oct-15, 08:10 PM

I assumed the OP meant "pre singularity black hole" when he said "singularity".

thorkil2

2008-Oct-15, 09:08 PM

Well, as far as GR is concerned, a singularity is a "point" in the universe where the density of matter and the gravitational field are infinite. It has a well-defined meaning within the scope of the [mainstream] theory.

On the contrary, a singularity (the speculative possibility of naked singularities aside) is a point that is cut off from the Universe (you can access the singularity from the Universe, but you can't access the Universe from the singularity). But I agree in principle. The suggestion that the Universe might be a singularity is purely speculative; an interesting thought exercise. I have no intention of opening an ATM discussion here; only to note that there is a perspective from which it might be interpreted as such. No claims as to whether it might actually be.

As for solar event horizon in the previous post, I have no idea what that is supposed to mean or how it bears on this.

alainprice

2008-Oct-15, 09:21 PM

If the EH is smaller than the surface, I'm pretty sure that does not qualify the object as a singularity.

I'm pretty sure this is saying the same thing as the OP. We have an event horizon, but we find the event horizon is smaller than the total space being considered. Ergo, there is no singularity as a result.

I've quoted the statement to reiterate what many others have already said.

Cougar

2008-Oct-16, 03:01 AM

Well, as far as GR is concerned, a singularity is a "point" in the universe where the density of matter and the gravitational field are infinite. It has a well-defined meaning within the scope of the [mainstream] theory.

And as your characteristics imply, the meaning of the infinities is that the equations become inoperative and undefined at that 'point'.

Personally, I don't think the interior of a black hole can be too radically different than the interior of a neutron star, which is already pretty darn extreme. The fact that we are unaware of any quantum interaction that can halt the collapse of such concentrated mass does not necessarily mean that there isn't one. After all, the gravitational effect of a black hole is (from a safe distance) largely proportional to its mass, like any other body.

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