imatfaal

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

logx(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] 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.

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!

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

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.

I've said it before and I will say it again - I like the strict no nonsense way this forum is run (and your moderation is a significant part of that ethos). There are a good couple of dozen posters who are clearly highly knowledgeable and willing to teach/discuss/chat, and that is an invaluable resource; I am convinced that a good number of those posters would not be so forthcoming if they were constantly needing to fend off attacks from crackpots, moody teens, and single-issue fanatics.

Slinkey - quite apart from Migl's arguments that I need to think over - this section of your argument I think has a lot of problems You are making the incorrect assumption that for an observation to be valid it must be able to be made from all frames of reference - the whole concept of SR, lightcones, simultaneity etc showed this to be false. This is a hangover from newtonian physics; the reality of Bob's observations from an accelerated reference frame do not compromise the reality of Alice's observations from the free-falling reference frame. I don't believe you are questioning the nature of blackholes - it is just a lack of engagement with altered frames of reference

This question is definitely doing the rounds of the physics teachers of world - I have seen it crop up several times recently. It seems a difficult one - because whilst none of the posts I have seen have had trouble working out the speed of contact with the ground many posters haven't made the link to carry on. And there is an implicit assumption that must be made that I think makes the question a little dodgy; unless it is that assumption that the questioner is looking to teach.

I will read back the thread if I can find time for 6 pages - and from your summary its tough stuff I must admit I find myself at odds with your very opening; nothing can enter =/= nothing can be seen to enter The fact that Bob in an accelerated exterior reference frame sees nothing cross the event horizon does not mean that nothing can enter a black; all it means is that nothing can be seen from that frame to enter a black hole. Alice in freefall enters the blackhole from her reference frame; Bob sees this in slower and slower motion, dimmer and redder due to the gravity of the blackhole. In the classical two diamond space time diagram of a simple blackhole whilst Bob follows a hyperbolic fixed radius line Alice follows a straight line; Bob never sees Alice cross not because she doesn't but because he is in a different reference frame and no 45deg light-like signal from Alice can ever intersect with Bobs space time line that is asymptotic to the 45deg light-like line. But I will read the thread to see how you have dealt/argued with this. I will also dig out a diagram to better explain the above paragraph.

And if they fail, they almost certainly will be brought to trial (although possible not a very fair one - perhaps only a military tribunal in a deniable off-shore location).

And what are the chances that one of us is actually a computer AI participating in a long-term Turing Test? I bet it's...um...Me. Oh Bugger! What a giveaway; failed again!

Chris There is a great calculator here A black hole of 4*10^6 solar masses (similar to that at centre of milky way) has a Hawking Radiation temperature of around 10^-14 Kelvin. Alan I havent read through the thread but to make it clear - if the background temp is higher than the temp of the black hole the blackhole not only doesnt evaporate, it increases in mass; this increase in mass causes its temperature to DROP, which in turn causes it to gain even more mass (ie slight positive feedback). The same applies when the blackhole temperature goes above background, it loses mass which causes it to gain temperature, which in turn increases the rate of evaporation. ie it will not reach an equilibrium, it will either get bigger or smaller. I haven't read what went before - but whilst a distant observer in an accelerating frame will never see the poor guy fall through an event horizon - from the frame of the victim there is no change and no noticeable affect of the event horizon. Poor old victim will sail straight through and get ripped up by the tidal gravitational forces

I would define the floor as what I walk upon - so perhaps not. But you can potentially walk normally on the inside surface of the outer rim - and you can call that what you will.

Thats why the sci-fi films have the space stations of the future as enormous doughnut shapes. If the doughnut is spinning, in the rim you would feel as if there was gravity towards the outer edge - at the hub one would be weightless.

I am convinced that any decent amount of postgraduate research changes ones attitude to the material and breaks a vital bond with the undergrad learner; which is why senior year undergrads can connect and teach in a different and complementary manner to postgrads (from full tenured profs to phd candidates). There is something in the depth, the different weighting given to information and the new ability to be sceptical that creates a marked change; it's liberating but also quite scary when you realise that there is no real authority in the area you are studying. I cannot help but think that the process of teaching lower year students is also a great experience for the undergrad teachers - there is no better way of organising your thoughts than having to prepare to teach them.

When the size of the animal starts to approximate the size of the habitat then Antoine might have had a point but even with the most terrible of lizards that is not even close. The change in mass between the dinosaurs and creatures today is miniscule when compared to the mass of the earth; the difference between a 11000kg of a Bull Elephant and 60000kg of a Brachiosaurus is unimportant in the context of a 5 million million million million kg earth. And even though it is in the sea the largest animal ever known is still around today.