imatfaal

OK Here's a quote from the book you said supports your case

"Bob never sees her fall through the horizon; to him it takes an infinite time for Alice to reach the point of no return. But in Alice's frame of reference, she falls right past the horizon and begins to feel funny only when she approaches the singularity.

 

The horizon of the Schwarzschild Black hole is at the Schwarzschild Radius. Alice may be doomed when she crosses the horizon, but just like the pollywogs, she still has some time before being destroyed at the singularity"

Leonard Susskind -"The Black Hole War: My Battle with Stephen Hawking to Make the World Safe for Quantum Mechanics" Back Bay Books/NYC 2008 - Paperback edition 2009. page 45

Which bit of that makes you think that Alice doesn't cross the event horizon?

Your flippant answer to my serious question of from where and how Charlie is watching the black hole evaporate makes it quite clear that you have no understanding of different reference frames and seem to think there is a privileged reference frame from which everything is uniquely objective. If you did understand this you would realise that you can say nothing about observations of blackholes without specifying the frame of reference .

If you are asking do I have an alternative theory, then no I don't, but I do have a question about the article. It states

 

"In fact, recent observations[15] indicate that the radius is no more than 6.25 light-hours, about the diameter of Uranus' orbit."

 

I worked out the Schwarzchild radius for 4.1 million solar masses and it came out 12,109,072,691.52m, or 40.39 light seconds (feel free to check that), so if the mass is not confined within that radius it isn't a BH. Agreed?

 

6.25 light hours is substantially larger than 40.39 light seconds.

So, you tell me. Are you sure there's a BH there?

That's selective quoting - a few lines down the same article explains.

The only known type of object which can contain 4.1 million solar masses in a volume that small is a black hole.

While, strictly speaking, there are other mass configurations that would explain the measured mass and size, such an arrangement would collapse into a single supermassive black hole on a timescale much shorter than the life of the Milky Way.

The article also makes it quite clear that the 6.25 light hours is an upper bound based on orbital data - not an estimate on the actual size.

I have quoted below a section from the article upon which the wikipedia figure of 6.25 light hours is taken - to make it clear that the weight of scientific opinion agrees with the existence of the supermassive BH

 

Together, the stellar motions reveal a central dark mass of 3.7(±0.2) × 10^ 6 (R_o/8kpc)³M⊙ and confine it to within a radius of a mere 45 AU or equivalently 600 Rsh, dramatically strengthening the case for a supermassive black hole, the location of which is now determined to within ± 1.3 mas (10 AU). Consequently, the dark mass at the center of the Milky Way has become the most ironclad case of a supermassive black hole at the center of any normal type galaxy

A M Ghez et al "Stellar Orbits Around the Galactic Center Black Hole" Astrophys.J.620:744-757,2005. page 22 available at http://arxiv.org/PS_...6/0306130v2.pdf

ok I am giving you too much help - but here goes, from here on though it hints only ;-)

your answer is correct but clumsy ( and that line starting with "so ..." is just plain bad!!!) - but this is how I would do it

[math] log_2(9)*log_3(4)*log_4(8) [/math]

[math] \frac{log9}{log2}*\frac{log4}{log3}*\frac{log8}{log4}[/math] change to a single base

[math] \frac{log9}{log3}*\frac{log4}{log4}*\frac{log8}{log2} [/math] rearrange

[math] 2 * 1 * \frac{log8}{log2} [/math] simplify the obvious division of logs to numbers

[math] \frac{2*log(8)}{log(2)} = \frac{2*log(2^3)}{log(2)} = \frac{2*3*log(2)}{log(2)} = \frac{6 log(2)}{log(2)} = 6 \frac{log2}{log2} = 6 [/math]

No need for a calculator. I haven't specified which base - because with these operations it doesn't matter, at no point did I need to evaluate a log

If I was helpful, let me know by clicking the [+] sign ->

During the panic about 20 years that there was a NEO that was a bit too near-earth for comfort - the leading runner to deflect it from earth was a huge array of mirrors focussing suns light on a small spot - this vapourised and expanded causing a reactive force

If you can get the time it is in the air - then you can work out the max height and the vertical speed. the horizontal speed - of course you need to measure how far away the bombs are landing. Come back with those and someone will help you out

Two things - Firstly in general, you need to learn how to convert log base x into log base y, cos dealing with separate bases is a real pain. I would convert both to natural logs. Secondly, in particular [math] log_3(9) = ? [/math]

I think any multiplication or division you come across will be looking for a re-arrangement into something more manageable rather directly tackling the problem

the question was from where are you observing it - and how are you observing it.

Hawking radiation.

the question was from where are you observing it - and how are you observing it.

He says it quite early on. I have the audio book so don't have a page number.

I have the book and have read the opening of it (been delayed) - he is quite clear that Alice goes through the EH.

But as Klaynos said the light year is the distance travelled in a vacuum - this is very closely approximated in space but not the same.

The confusion is that a lightyear is not how far light travels in outerspace in a year - but how far light travels in a vacuum in a year. To all intents it is the same thing - but as soon as space gains a measurable density and refractive index then there is a difference; but this does not change the speed of light in vacuum which is constant, nor the lightyear, but it does change how much distance light will travel in space in a certain time.

Oh Yeah - I understood what you where doing, I was saying you should make it clear that you do. At the level I learnt up to , and the level you are at, much of the credit given by teachers is for you demonstrating knowledge of methods and understanding the implementation and the logic as to why they work. If you leave this out then you are risking the marker assuming that you do not understand the methodology especially if you make an arithmetical error and your final answer is awry.

The event of falling toward the EH is outside the EH and is in principle observable by anyone outside the EH.

Yes - and from outside the EH we observe Alice falling towards the EH- what we do not observe is crossing. The 'event' in question is crossing not falling towards.

And you also accept that BHs evaporate, yes? So if you never see Alice cross the EH and the BH evaporates then on which side of the EH was/is Alice?

From where are you observing the black hole evaporate? and how?

It's not a fact. It's a conjecture that has never been evidenced except by math that is known to be incomplete.

This is grasping at straws. The same theory that states that an outside observer in an accelerated reference frame will not see Alice cross the EH says that Alice's space time sends her through the EH without her noticing any local change.

I suggest you read Leonard Susskind's "The Black Hole War" because he disagrees with you. In fact the only way he can reconcile it is by invoking "black hole complementarity" in the same vein as particle-wave complementarity.

Well in Susskind's lectures at Stanford he quite clearly states that Alice goes straight through the EH. This is repeated in the black hole war - I will dig out a a page number. The information that Alice entails and how to deal with it is highly debatable and contested - but not crucial here.

Actually this is still hotly debated.

Not in the sense that you are debating it. This is covered in undergrad level text books (look at Hartle An Intro to E's GR ) . I notice you haven't bothered to comment on my notes on proper time v coordinate time and your incorrect choice of coordinates.

The current thoughts on the great prehistoric monuments in NW Europe suggested that the building of the henges and avenues was the work of the entire community (admittedly prob not the leaders) in a form of social enterprise. Whilst the early tribes did hold slaves it is not thought that the monuments were built quickly by forced labour, but more likely done over generations in a festive spirit. As someone who has saved up cash and holiday to go and build a bridge and a school - I can empathise with this concept of a community project.

Vastor - I don't open unknown websites on my company pc so won't be able to look over your work. In general; unless space is at a premium put in every stage and keep each stage simple, do not assume that any stage is simple (ie explain why you feel you have the right to do anything that isn't very straightforward), do not hide any side calculations, use good layout like latex or others - if not check your bracketing obsessively, and always go back through to check minus signs and numbers of like terms.

I have noticed you put in a numerical check after the algebraic work - this is good, but be sure to make it obvious. Also be really careful to get this right; its frighteningly easy to assume the answer to check sum which negates the whole point (like you did in your log example when you assumed it showed you were wrong).

Ahoy Capt

You are complicating too much. To start with, try working out exactly what is being requested

[math] the\ sum\ of\ the\ reciprocals = \frac{1}{a} + \frac{1}{b} [/math]

Perhaps try rearranging/simplifying the RHS - once you do this the answer will jump out and hit you

Hey Vastor - you are not being so lucky with people telling you that you have made an error! Again I think you maths is ok - if difficult to follow. To explain to the pirate Captain

-11 = 3(x^2 - 2x)

-11 + [3(2/2)^2] = 3[x^2 - 2x + (2/2)^2]

Perhaps if I put in an intermediate stage or two

-11 = 3(x^2 - 2x)

-11 + 3(2/2)^2 = 3(x^2 - 2x) + 3(2/2)^2

-11 + 3(2/2)^2 = 3[(x^2 - 2x) + (2/2)^2]

-11 + 3 = 3[(x^2 - 2x) + 1]

-8 = 3(x^2-2x +1)

It becomes clear you have just added 3 to each side of the equation - quite legitimate. If I was marking this I would require at least the first intermediate stage and I would need to see how you came up with the formulation (2/2)^2.

Have a bash at the other problem and I am sure that someone will step up to help

1. No. That wasn't my point.

 

 

 

2. No, it's not. It's actually where you are failing because you are still failing to understand that you will never see anyone reach the EH if you are outside the EH. It will take an infinite amount of time for them to reach the EH from any observer outside the EH. ie. never.

 

 

 

3. No, that's an assumption that you have no evidence for and the point under discussion.

 

4. Consider this:

I measure the mass of a BH and find it is 5 solar masses. I take Hawking's equations and work out that the BH will exist for 8.268976e+85 seconds. I then dive towards the BH. As I fall toward the BH I look back to where I came from and see that as I get closer to the BH the Universe is speeding up relative to me. At some point before I reach the EH - where time dilation is infinite - the dilation will be so great that 8.268976e+85 seconds will pass in a tiny fraction of a second as measured by me and the BH will have had enough time to evaporate.

 

So when did I cross the EH?

 

I'm sure you agree that 8.268976e+85 seconds is a lot less than infinite, no?

 

 

 

I'm well aware that clocks in the same reference frame tick at the same rate. I think rather it is you that is not understanding what I am saying.

1. if it wasn't your point why did you say

The event is outside the EH thus it is in principle observable from any reference frame outside the EH

Events outside EH are observable - those inside are not.

2. you are erroneously linking the event and its observation by a distant third party observer. I know that an observer in an accelerated reference frame will never see Alice cross the event horizon; but I also understand that this does not preclude the fact that Alice does fall through the barrier. It is not two separate histories causing a paradox as you are insisting upon it is a single history and an observation. An observation of a signal by a distant observer does not affect the passage of Alice through space - it is the signal that is affected not the history.

3. No - it is widely accepted physics and the reason why you cannot reconcile yourself to agreed views. In order to understand what is happening to Alice in a physical sense you do not calculate her progress using coordinate time what you must do is calculate her progress using a coordinate independent proper time - and she will reach the singularity (let alone the EH) in a finite amount of time.

[math] \tau = \left( \frac{\pi}{2\sqrt{2m}} \right) r^{3/2}[/math]

4. Immaterial - the black hole could evaporate in a couple of months and my logic would remain. By the way - you are thinking in terms of Schwarzchild coordinate system and that is a bit of a problem; whilst Gravitational time dilation in a the vicinity of a gravitating spherical nonrotating mass can use schwarzchild coordinates with a black hole one must use eddington-finkelstein coordinates. In Schwarzchild coordinate the sums at R=2m are very dodgy indeed as the result is singular and this might be the root of your misunderstanding.

log_x(x) = 1

I would put a bigger bracket around it to be sure - but I think you are sound

[math] log_2 \left( \frac{2p}{(1-3p)} \right) = log_2 2[/math]

when i checked

 

[math] log_2 2\frac{1}{4} - log_2 (1-3\frac{1}{4}) = log_2 2[/math]

 

[math] log_2 \frac{1}{2} - log_2 \frac{1}{4} = log_2 2[/math]

 

[math] log_2 \frac{1}{4} = log_2 2[/math] and yes, this is fail, i miss something somewhere?

 

from my understanding, the question of 'solve the equation' would mean to be find the value of the unknown 'p', doesn't ?

The reason you got this bit wrong is that

[math] log_2 \frac{1}{2} - log_2 \frac{1}{4} \not= log_2 \frac{1}{4}[/math]

If you think about it log₂ (1/2) has to be -1 as 2^-1 = 1/2 and similarly log₂(1/4) has to be -2 as 2^-2=1/4

and -1-(-2) = 1 = log₂(2)

You sure DJB - looks fine to me.

Err, no, it's not. The event is outside the EH thus it is in principle observable from any reference frame outside the EH. Secondly, to all outside observers an entity will never be seen to cross the EH the event of crossing the EH does not happen for any observer outside the EH. To argue that in the reference frame of the entity falling towards the EH that he experiences the event of crossing the EH is to argue there are two distinct histories for the entity falling towards the EH. History 1, for all observers outside the EH, and history 2, for the observer falling towards the EH are at odds with each other.

 

Remember the premise is that an object falling toward an EH will continue to fall towards the EH at an ever decreasing rate for all outside observers and will never be seen to cross the EH. As BHs evaporate then the object will not cross the EH before the BH evaporates.

 

So, you have to reconcile two distinct and contradictory histories.

Cool.

Your points in order

The events outside the EH are observable - the crossing of the EH is not as it is not outside the EH.

An event that happens to Alice does not happen to a remote entity Bob in any circumstances; it happens to Alice and is observed by Bob - the difference between the two and the necessity of transmission of this information is where you are failing. There is no history for Bob apart from that communicated from the distant Alice. Bob and all outside observers knows what has happened because they understand the principle that stops any observation from exiting the BH and provides an ever slowing picture of the near approach.

Your premise is fine for outside observers - what they see is affected by the gravitational time dilation, but it is only the message/the observation that is affected; Alice ploughs on without even noticing the EH. Time dilation is relative to outside observers it is not internal to Alice's FoR - you do realise that Alice never sees her own clock get slower!

The question now is: Does Cap'n Refsmmat agree with you ??

Dollars to donuts he does. Your first question of the thread was ambiguous - your post had two phrases both clear. That's one of the benefits of using latex as soon as you have multiple levels of brackets or division

Hi Sofia

That's not the expansion that I got - and just as importantly the question asks for correct factorisation not expansion. It doesn't look like HKUS is coming back so I don't mind answering his homework now

(x-10)[(x+4)(x+1) - 24] - 3[(-11x - 11) + 24] + 8[-21 + 3x] I have put in extra spaces to exaggerate the sections

(x-10)[(x^2+5x+4)-24] +33x+33-72 + 24x -168 (multiply out innermost bracket (x+4)(x+1) - and simplify other terms

(x-10)[(x^2+5x-20] +33x+33-72 + 24x -168 (add in constant)

x^3+5x^2-20x-10x^2-50x+200 +33x+33-72 + 24x -168 (multiply bracket by (x-10)

x^3 + 5x^2 - 10x^2 -20x-50x+33x+24x +200-72-168+33 (gather like terms)

X^3 -5x^2 -13x -7

To factorise you know you need three constants that multiply together to give 7 (ie its gotta be 7 and two ones - but you need to check minus signs) a little work gives

(x+1)(x+1)(x-7)

Hope that helps

 

No, it's nothing to do with reference frames. Events happen regardless of reference frame. We might disagree about their duration, position, or order, but events still happen.

Slinkey - it's everything to do with reference frames. The outside observer is in an accelerated reference frame compared to the free-falling victim. The event happens - the outside observer can never see any form of EMR that comes from that event. The whole point of spacetime diagrams with lightlike lines is to show that some events are not observable by measurement from a distance position or different FoR. You mentioned that you had read the black-hole war - although how you got anything out of it if you cannot believe these foundations - Susskind regularly uses the pond with a drain analogy. If the drain is fast flowing enough there is a point at which the boat with fastest engine cannot escape the drag - if the fastest engine (EMR) is too slow what can send a message? It isn't that a message isn't sent from the striken boat nor that the event didnt take place it's that after a certain point the message itself which has a finite speed cannot get to any observer.

If the circuit is such that resistance will cause temperature rise and if the coils are close enough to stop heat dissipation as Parth suggested above then as the material is less able to lose heat, the temperature increase will be greater, which for most metals at room temperature will raise resistance and generate more heat...

I now think that was one big set up and that I was the unwitting straight-guy. nice one

How are you pronouncing it to get multiple (any) diphthongs? /sud.ɔː.ɡæz.əm/ - and if you mean digraphs then diphthongs has just as many in fewer letters

Lemur - that's not plagiarism. Plagiarism has a actual component - repeating without reference/attribution someone else's ideas, phrasing, research, etc and a mental component that this was done with a deliberate intention to deceive and pass the work off as one's own. What you are describing is commonplace and the germ of many fruitful collaborations; that two independent minds can come to the same conclusion from the available data is affirming and immensely gratifying.