A dim glimmer

As I write this, I am sitting in Sheffield’s fine Victorian railway station, on my way home from visiting the astronomers here. Yesterday I sang for my supper, giving a double bill seminar on the radio background and on the big blue bump. Supper duly followed, and was exceeding pleasant. The Sheffield group is a small but lively one. As normal, the postdocs were worried about job prospects and the academics were moaning about writing exam questions. To quote P.C. : “I love my job but I hate writing exam questions.” With you there, Professor C.

The Crowther mentioned my recent blog post about how the Universe is almost empty. He said he likes to set his classes ballpark estimate exercises. He tried one out on me. You can have a go too. Its quite good to start by using your instinct to make a guess, before gathering a few facts to do the quick mental calculation. That way you can get your frisson of surprise. Suppose, said Paul, you take the material of the Earth and stretch it out from here to the Sun – how wide would that rope be ? His students guess a wide range of answers, but usually around a mm. Give it a go.

Here is another one, for which I will give you the answer – how powerful is a one kg accreting black hole ? As we all know, quasars are immensely luminous because they have supermassive black holes at their hearts – those accreting black holes are the most efficient energy sources we know. We are talking big numbers. At the Eddington limit, that billion solar mass black hole can be radiating up to 10^47 erg/s. When I first came to giving an undergraduate course with some of this stuff in, I felt duty bound to do things in SI units. So I got a formula for the Eddington limit : L_Edd=6.37M. Wuh ? A one kg black hole gives me 6 Watts ? My electric fire can do better than that ! My electric fire is only a few kg,  but it gives me 2 kW.

Then comes the epiphany. You realise that accretion is not the slightest bit efficient per unit of mass of accretor ; what is impressive is the energy per unit mass of the accretee. To get 2kW out and so heat your bedsit, you need a black hole of at least 314 kg – about as big as two motorbikes. However, once you have it, the rate you have to burn fuel as it were, i.e. the rate at which you need to drop matter into the black hole, is about 2 x 10^-13 kg/s; one kg of fuel would last you roughly 140,000 years… I now leave it as an exercise for the reader how long a kg of coal would give that same 2kW. Scaling up to the most luminous quasars, you need to eat about 20 solar masses a year. Peanuts.

Then it made me think again about that electric bar fire. Its not really producing 2kW all by itself. It is plugged in and sucking that energy out of a giant power station miles away. Its really a similar story. First you need a vast power factory, and all the surrounding infrastructure. Only once you have it can you burn some fuel and give the illusion of that tiny fire producing energy. Likewise, before you can get that accretion energy goodness, you somehow have to assemble those billion solar masses. But thats another story…

28 Responses to A dim glimmer

  1. Here’s my favourite: Think of the Hubble Deep Field and Tony Tyson’s images of galaxies essentially covering the sky. All those galaxies, all those stars. On the other hand, we understand nuclear matter quite well. So, ignoring the fact that it would collapse into a black hole, calculate the radius of a ball with the density of nuclear matter which contains the entire visible universe.

    It’s about the same as that of the orbit of Mars.

    And then there is the one about the total energy received from extraterrestrial sources by all the radio telescopes since the dawn of radio astronomy, and how it compares to the energy of a falling snowflake.

  2. Paul says:

    My favourite vaguely related thing (with advanced follow questions for the enthused):

    1) Guess / calculate the kinetic energy in TeV of:
    a) a rifle bullet
    b) golf ball
    c) draft of a paper you handed to your supervisor, screwed up into a ball and thrown back at you.

    2) what’s the incoming cosmic ray flux at those energies (from Pierre Auger et al) in events / square km / year incident on the earths atmosphere

    3) probability of particle making it through the atmosphere to sea level

    4) and of hitting someone

    5) and getting absorbed by their head

    Finally – how many years do you have to wait for someone to (or alternatively, how many people per year) get a) their brains blown out b) knocked unconscious, and c) mildly surprised, by incoming high energy cosmic ray particles?

  3. Phil Uttley says:

    Hope you had time at the station to enjoy the Sheffield Tap? Pub on a railway platform – genius!

  4. Paul says:

    PS. “Gathering a few facts to do the quick mental calculation”?

    Seriously, this isn’t the 1990s you know.

    • andyxl says:

      ok ok, but whether or not you had help from Mr Wolfram, did the answer surprise you ?

      • Aaron Robotham says:

        It’s well known that we mere humans struggle to scale dimensions rapidly Even when using the same dimensions we can be easily baffled. I’ve won many pints with the “diameter of pint glass is larger than the height” bet. Also, some pretty “big names” have been out by factors of a 1,000+ when asked: I have a rope that is tight around the Earth’s equator and I want to lengthen it so it is 1m above the surface the whole way round. How much longer does the rope need to be? You have 1 second! Note that I would never accept a written response to this question.

  5. Mrs Trellis says:

    Dear Reg,

    That Crowther chap has certainly come a long way since the time he used to appear on Crackerjack.

    Yours sincerely

    Mrs Trellis

  6. John Peacock says:

    Andy, The 314-kg BH would do a lot better than 2kw. For one that small, the power output is dominated by Hawking radiation, and would be about 10^27.5 W (but this would only last a nanosecond). To get 2kw from Hawking alone, you need 6 x 10^14 kg; Hawking power scales as 1/M^2, so Eddington wins only above about 10^10.5 kg. Interestingly, the power from such a black hole is approximately equal to what is needed to keep the UK supplied (60 GW); strangely, this alternative is not mentioned in Dave Mackay’s otherwise wonderful “sustainable energy without the hot air”…

  7. Martyn Wells says:

    When I was working at Imperial with Bob Joseph he had a number of ballpark type questions that needed a bit of lateral thinking. The ones I remember are –
    There is a pole star. How many stars are visible with the naked eye?
    Your car windscreen has ice on it in the morning. What is the minimum age of the universe?

  8. andyxl says:

    I am waiting for others to try answering Martyn’s questions, and Paul’s…

  9. Michael Merrifield says:

    There is a pole star. How many stars are visible with the naked eye?

    If the pole star is visible to the naked eye, then at least one.

    • I think the idea is that given about 6000 stars are visible to the naked eye, how likely is it that there is a pole star (i.e. a star quite near the pole)? Of course, due to the precession of the equinoxes, there is not always a pole star, which is how we know that Francis Bacon did not write Shakespeare’s plays. How’s that for lateral thinking?

      • The sky has about 40 thousand square degrees, or on average almost 7 stars per square degree. Thus, the pole star is closer to the pole than is typically the case, meaning that we are lucky to have a pole star now.

        Easier is to just gauge the typical distance between stars and note that it is larger than the separation between the pole star and the pole.

      • Nick Cross says:

        I think you have it the wrong way around Phillip. If you see a star close to the pole (i.e. within a few degrees), how many stars are there likely to be. To have a good chance of a star within a few degrees then there must be at least 1 star per 10 square degrees — about 4000 stars in the sky.

      • Martyn Wells says:

        The question is posed the other way round so the way I work it is
        How close does a star need to be to the pole to be navigationally useful and therefore labelled as a pole star? Lets say 1 degree giving a density of ~1 star/3 square degrees. (The average you quote below should be 1 star/7 square degrees). This would give the total stars visible = 40000/3 ~13000, not too bad an estimate! But really its luck that the result is that good because it assumes a uniform distribution of stars. If the pole was nearer the galactic plane then the problem would be identifying which star is the pole star rather than the chance of there being a star near the pole.

  10. Monica Grady says:

    Depends whether the Sun is out or not….

  11. Albert Zijlstra says:

    1 degree seems a bit demanding. Even Polaris will only be that close to the Pole for a few hundred years. For a seafaring journey of say 1000 km, 1 degree gives an error of only 17 km. If you relax the limit a bit and allow for a few degrees distance from the pole, your calculation would be even closer. But any fainter than 4-5 mag would leave it naked-eye but not too useful in practice.

    Since when is Polaris called Polaris?

    I remember a note in a book claiming that 450,000 yr ago, Capella and Aldebaran put in a performance as a double pole star, and Aldebaran had magnitude -1.5 at that time. That would have been an interesting time to ask this question.

    • Michael Merrifield says:

      You are off by a factor of two, you bunch of hemisphereists!

      • Albert Zijlstra says:

        Actually, there is a southern pole star as well at the moment so the numbers should be ok. It is a bit on the faint side (naked eye, but barely), and as Polaris, it is a variable. Now calculate the number of variable naked-eye stars..

  12. andyxl says:

    I guess we are all Bayesians these days. Is nobody going to attempt the argument that the existence of a pole star does not constrain any models because it is where it is, and we have not measured a frequency of any kind ?

  13. andyxl says:

    Aron – took me at least ten seconds. I won’t write the answer down of course.

    • Aaron Robotham says:

      I’ve often wondered why people’s guts get thrown so far off the correct answer with these sorts of questions. We do really well at seemingly much more complicated problems all the time. But I guess “all the time” is the key thing- I don’t think many lives on have ever depended on rapid scaling of the circumference of a circle. Or, indeed, squishing spheres a lot. I’d be interested to hear other theories…

  14. belesar says:

    A point in the sky that corresponds with Earth’s own North Pole. Because this point always lies directly above the Earth’s pole, the bright star that lies close to it, Polaris, always lies due north from an observer’s point of view. From the surface of the Earth, it appear that all celestial objects, from the Sun to the stars, rotate around this point once each day.

  15. […] Here is another one of those occasional posts with mazin facks bout stronomy. (See also Almost Nothing and A Dim Glimmer). […]

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