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Tutorial on Hawking radiation

  1. May 16, 2003 #1


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    to keep things simple the units are c=G=hbar=k=1 and
    all black holes are non-rotating uncharged.

    I think everybody here probably has a qualitative notion of how the Hawking radiation arises from the event horizon of a black hole---there has been tons of pop-sci journalism about this. What may be lacking is a corresponding quantitative feel---an ability to estimate magnitudes.

    If a black hole surface area is on the order of a square angstrom, for instance, what is the temperature compared with things you know about----a hot sidewalk, the surface of the sun, the core of the sun, the big bang, etc?

    Again, if the surface area is a square angstrom or so, what is the wattage roughly speaking? One should have a rough quantitative grasp to fill out the intuitive picture.

    History: in 1974 Bekenstein said the entropy S is given by
    S = A/4, where A is the area

    And according to Hawking (1975) the temperature T is given by

    We should do an example to get a feel for that----say area is 1050 planck, which is about a square angstrom.
    then 4piA is 12.6E50 and sqrt(4piA) is 3.5E25 and the temperature (one over that) is 0.28E-25

    This is about one quarter of the temperature at the core of the sun----suncore temp is 10-25, roughly 1200 eevee.
    Being a quarter of that one can think of it as soft xray or hard UV or whatever. So it isnt big bang temperature which would be just one, or even suncore, but about a quarter of that.

    What is the luminosity or wattage of this thing? Well the Boltzmann law says take the fourth power of the temp and multiply by pi2/60. That gives the radiant power per unit area and then you have to multiply by A to get the overall power.

    The Hawking temperature, remember, is 1/sqrt(4piA). And so
    the fourth power is 1/(4piA)2

    And multiplying by the Stefan-Boltzmann const pi2/60
    gives us (1/960)(1/A2) which is power per unit area

    And then multiplying by A to get the total power gives

    (1/960)(1/A) which if you want you can write 1/(960A)

    *****THAT'S ALL, the tutorial is over*****

    But there is one more little thing having to do with SI units.
    Suppose you love metric units and want to put back in all the hbar and cees and stuff and do a purely metric calculation of the same thing. Here is how to get the metric formula back:

    Planck power is c5/G
    Planck area is Ghbar/c3

    That is all you need to know! You just take the 1/(960A) formula and instead of A you put in (A'/planckarea) where A' is the area in metric. Then the formula becomes

    1/(960 A') x (planckarea in metric) x (planckpower in metric),

    in other words (1/960A') x hbar c2.

    This hbar ceesquared is just what is needed to
    make the units agree, if you choose to work in SI metric.
  2. jcsd
  3. May 16, 2003 #2
    I appreciate the time you put into this:smile: It is an excellent tutorial and much needed, it's not often described in any further detail as how the event occurs with no or very little mathematical accompaniment.

    Thanks for enlightening me on the realistic approach to Hawking Radiation:smile:
  4. May 17, 2003 #3


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    You are most welcome Kyle and I thank you for the kind words. Actually your response helps me in what I am doing.

    I am looking for potential textbook examples of interesting things that natural units (c=G=hbar=k) are good for calculating.
    I figure there could be a dynamite kickass physics textbook either online or hardcopy that would show things that are neat to do with natural units.

    The relation between a black hole area and brightness is one
    possible example-----simply 1/960A

    Another is the Chandrasekhar limit (the threshhold mass for having a supernova explode) it is incredibly easy to calculate
    the Chandra limit just from knowing one thing: the mass of a proton. Would this interest you?

    BTW in planck the proton's mass is 1/(13E18)

    actually instead of 13 one can say 12.99 some. But 13 is
    plenty close for back of envelope calculation. A lot of neat stuff
    can be calculated just from the mass of a proton.

    Anyway, I want to find stuff that grabs people's interest and which also is extremely simple to calculate in planck and is
    tedious and involved to calculate in metric.

    The chandrasekhar limit is a good example---it is a pain in the butt to calculate in metric.

    Another example is the light-bending formula of GR. By how much is a ray of light bent that passes a million miles from the sun (from the sun's center). Yucky to calculate in metric but a snap in planck.

    I want to see if a textbook-like thing can be put together of such things. Also it has to be fun. Your reactions help guide me in this.
  5. May 17, 2003 #4
    This would interest me very much indeed! Epecially the SNe threshold. You don't come across many (none in my experience) of down to earth, real life (meaning common experience) examples in a not-so-mathematical explantion.
  6. May 17, 2003 #5


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    It's 11:30 PM and I'm getting sleepy so shd probably start on that tomorrow. You in Illinois---must be 1:30AM, nightowl.

    Thing about Ch. limit. Everybody says it is 1.4 solar masses
    but nobody says why.

    Well it happens that in natural units solar mass is 0.93E38
    so 1.4 solar is 1.3E38

    So here we have this unexplained number. Why does a no-longer-fusing ball explode if its mass is 1.3E38
    and not explode if its mass is less, say 1.2E38?

    Where does this number come from.

    Turns out it comes from the mass of the proton which is
    one over 13E18.

    the chandra limit is (pi/4)(13E18)2

    really not all that hard to calculate in the c=hbar=G=1 system
    but gets messy if you have to include a whole lot of G's and hbar's and cees in it.
    Here you just have to square 13E18---the proton compton wavelength actually, reciprocal of its mass----and then multiply by pi/4.

    Should get to bed, dream of exploding stars.
  7. May 17, 2003 #6
    Yep, approaching 2 am...where exactly, please excuse my foolishness, does pi/4 come from? And is there something that actually makes 1.3E38 special, a reason for the explosion?
  8. May 17, 2003 #7


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    foolishness! :wink: (false modesty) it is one of the more astute questions I have heard on PF

    I can only share with you where I am on this.
    Frank Shu (UC Berkeley) has a remarkable textbook called
    "The Physical Universe---An Introduction to Astronomy"
    (expensive, so use interlibrary loan and get it free) and on page 128 it gives a formula for the Chandrasekhar mass as a certain number times the mass of the proton.

    Even tho it's messy I will copy out this number.
    It has a term (Z/A)^2 which is where the 1/4 comes from.
    After the star stops fusing it has some chemical composition
    with an average atomic number Z and average atomic weight A.

    Like Carbon12 has Z=6 and A=12, so its (Z/A)=1/2 and the square is 1/4.

    What this is talking about is the "proton fraction"
    the mass is made up of protons and neutrons and if it is already neutrony it is easier to crush into neutronmatter, so the threshhold for collapse is a little lower
    but if it is very protony( high Z/A, like carbon) then the threshhold is higher. Maybe I should not be so informal but that is the intuitive content of the 1/4 in the formula. It is really (Z/A)^2 but that is approximately 1/4.

    The most extreme case of a star that has finished fusing is iron---the whole core is iron---and that has Z/A = 26/56, which is not quite 1/2 but still close enough for rough estimates.

    So rather than get into the Z/A detail I just say 1/4.

    Here is Frank Shu verbatim

    0.20 (Z/A)2 (hc/Gmprot2)3/2

    That is a pure dimensionless number and the mass threshhold is what you get by multiplying that number by the mass of the proton.

    This is hard. My guess is that unless you are as smart as Chandrasekhar it will take a few days to assimilate this.
    You might want to tinker with it algebraically.

    It conceals the planck mass = sqrt(hbar c/G), or rather, it
    conceals the square of the planck mass = hbar c/G

    It conceals, if you scramble around algebraically a bit,
    the RATIO of the proton mass, mprot to the
    planck mass.

    What would happen if you rewrote the formula to be in terms of
    the planck mass instead of (as Frank Shu has it) in terms of the proton mass?----after all that's just a changes of units.

    Anyway mess with it a bit if you have time and we will talk later.
  9. May 17, 2003 #8


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    Why 1.3E38 is special

    Oh yes, you asked why that mass is special.

    It is the strength limit of ordinary matter.

    Ordinary matter (nuclei made of protons and neutrons, plus a sea of electrons) will crush itself by its own weight and turn into
    neutron matter (roughly a thousandth the diameter) if there is too much of it.

    this mass 1.3E38 is how much it takes for this to happen

    Imagine a ball of iron with that mass. The pressure is so
    great that it causes the protons to eat the electrons and become neutrons
    then the neutrons clump together into neutron matter which occupies only a billionth as much space (very roughly speaking).

    So the whole ball of iron is suddenly in free fall towards its own center. A jillion loaded freight elevators have suddenly all at the same moment busted their cables. Gravitational potential energy is being released on a mammoth scale and being converted into
    kinetic energy.

    The nerd's term for forcing the neutrons to eat the electrons is
    "reverse beta decay" and it incidentally releases a neutrino.
    So suddenly a jillion neutrinos blow out of the center of this ball of iron.

    Various other things take place and everybody has a great time.

    They say we are made of the material blown out into space by these explosions---because without them most of the carbon nitrogen oxygen etc that is cooked by fusion in starcores would be trapped in dead cores and not get out into space where it can condense into planets and stuff. So these explosions are a necessary part of life as well as being beautiful.
  10. May 17, 2003 #9
    Let's see if I understand this...if instead of Carbon or any other close to 1/2 element, you use Radon instead and get (86/222)2 and come much closer to 2/13 or for rough estimate purposes, 1/7. This would change π/4 into π/7?

    Basically, a fourth isn't a set number, but simply a rough estimate of what (Z/A)2 comes out to?
    Last edited: May 17, 2003
  11. May 17, 2003 #10


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    By Marcus:
    The Chandrasekhar limit is not a limit above which a star will explode. Very few supernovae have anything at all to do with the "Chandra limit" of 1.44 Sm for He cores, and 1.79 sm for Fe cores.

    The rest of your math post didn't catch my eye...:smile:

    (Edited to correct to the 1.79 figure above)
    Last edited: May 17, 2003
  12. May 17, 2003 #11


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    What I thought I said was that it was a mass limit for no-longer-fusing stars. Perhaps you did not see that condition.

    At the end of fusion, when the core is (say) iron which cant fuse anymore, then the star has no source of heat to fight against gravity. Then, if it exceeds the limit, it can collapse.

    I have been over this so many times I may occasionally skip details but don't believe I said anything misleading.

    Of course by the time a star has finished fusing it will in many cases have already blown off most of its mass (in red giant etc) and be essentially a core of its former self.

    There are a lot of scenarios----like the type Ia binary star mass transfer onto a white dwarf etc----but the Chandra limit is
    the important triggering mass threshhold that always comes in somewhere in the story.

    If I made an error somewhere please quote and discuss.
  13. May 17, 2003 #12


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    Yes! You got it! (unless Labguy wants to contradict)

    Here is something interesting you can check if you just have a periodic table of the elements. The highest a star can fuse to is IRON. It cannot go to Radon!!!! So there is a limit on how low Z/A can get. For rough and ready I just take it to be 1/2 because it is around that or a little less, for iron.

    Here is how to check it. It is called "the curve of binding energy" and you calculate it say for iron-56 by dividing the atomic weight
    55.935 by the nucleon number 56. this gives the mass-energy per nucleon (per proton or neutron).

    It is at a minimum for iron. So nuclear fusion reactions can be exothermic (yielding energy) up to iron. But it actually COSTS energy to squash iron nucleuses together to get still heavier things.

    The elements higher than iron are actually synthesized in the violent explosion of a supernova when there is a lot of extra energy around to waste on side-reactions.

    the heavier-than-irons are not made in normal slow thermonsuclear burning that goes on throughout the stars life.

    My attitude is very back-of-envelope about this. I say 1/4 instead of (Z/A)^2. But I am glad your impulse is to be more careful. By being deliberate and analytical you have uncovered this thing about iron being the minimum of the curve of energy. Or maximum depending on how you draw the curve.
  14. May 17, 2003 #13


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    No critical error, it is just that the Chandar limit(s) are masses above where the mass must collapse to a different state. Stars explode or not based on many other criteria, and only then does the Chandra limit(s) come into play. In some stars, usually Type II supernovae, the iron core can cause the implosion/rebound/explosion if the core exceeds the limit. In smaller stars a white dwarf may be the end result and do nothing but cool over many years. In most binaries, an accretting white dwarf will not explode even if the white dwarf exceeds the limit. No explosion; just a collapse to a neutron star. Type Ia supernovae are extremely rare cases, regardless of most oversimplified texts and websites.
  15. May 17, 2003 #14


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    Hello Labguy it interests me very much that you edited your
    post just now to change from 1.39 solar (for a core of iron)
    to 1.79 solar (for a core of iron). Do you have a web resource
    giving chandra mass for various chemical makeups that you would like to share with us?
  16. May 17, 2003 #15


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    Ah! I am reassured that there was no critical error you were pointing to! Your point is that sometimes there can be a collapse (say down to neutron matter) without an explosion.

    I agree. the gravitational energy that is released can be carried off in various ways. I believe (correct me if I am mistaken) that some very bright gamma ray bursts are not so far thoroughly understood and there is theorizing that they may be due to
    a star's sudden collapse----the model being in some ways similar to a supernova but in other essential ways different. If you have some websites to add to the discussion please contribute them. It is an interesting topic. Some collapses apparently two-stage---first down to neutron matter and then after a few days of cooling or something proceeding to black hole---or so I'm told. Like to know more about this.
  17. May 17, 2003 #16


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    The edit was just because I typed the wrong number the first time, just a typo.

    I have not yet found a website (haven't hunted either) with enough detail to go into the specifics of the two different "Chandra Limits". My source is SUPERNOVAE by Paul and lesley Murdin, as well as several other books by Wheeler, S.E. Woosley and others.

    Chandra did calculate another, you may know, and this is the 3.2 solar masses beyond where any matter, usually neutron stars, would have to collapse further to a black hole. Lately, there is good evidence for "Strange Quark" stars also.

    I only jumped in here because of the statement that "Another is the Chandrasekhar limit (the threshhold mass for having a supernova explode)..". I know, that you know, there are many other factors involved. Just got bugged--- :smile:-- by the general oversimplification. It is a bad habit of mine..:frown:

    Added Edit:
    Damn, you can sure type long responses. I am a two-finger guy, very slow.
    Last edited: May 17, 2003
  18. May 17, 2003 #17
    Let me see how this works, as I want to graph this curve:


    A=56 Z=26





    -Sign Reversal-


    EB=8.79033263 MeV (per A)

    is all this work correct? Just want to make sure:smile: Of course I could just go rip off some binding energy curves that somebody else did, but where is the fun in this?
  19. May 18, 2003 #18


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    Hi Kyle, I just got back to the computer and saw your post. I have not checked the math but am responding impulsively impromptu.
    I believe we should look at a few sample points on the curve first.
    Like, be very empirical.
    I have my fat CRC Handbook out and Im opening it to
    the table of isotopes. Everybody should have the CRC Handbook of chemstry and physics----or is this stuff on the web?

    It fell open to sodium-23----22.9898
    Flip the page oxygen-16----15.9949
    Have to have hydrogen-1---1.00797
    And of course carbon-12----12.0000
    Already have iron-56-----55.9349
    What about calcium-40---39.9626
    Another at random silver-107---106.9051

    I chose these more or less randomly but
    they are common isotopes in each case---stable.
    Now I could be wrong. What I expect is that if
    in each case I divide the weight by the nucleon number
    I will get a minimum at iron.


    So there are some sample points on the curve. you can fill in
    some more. Personally I expect the curve to not be a curve!
    I expect it to be jagged and jittery because so much is going
    on as you proceed up the nuclear weight scale. But they CALL
    it a curve, as you would expect of people for whom everything
    is curves----unless its a Lie Group. Just kidding.
    Do you have a web source for the nuclear binding energy?
    It might be a pretty thing to see online.
    I should really be saying the binding energy for iron is maximal
    because the binding energy is the deficit---how much the nucleons are in hock for. But I rushed into it and did things
    upside down I guess. For me it is more intuitive to think of
    iron as minimal.
    Do you have a table of the weights of the isotopes?
    Or know of one online?
  20. May 18, 2003 #19
    Clicking on the elements brings up information, which includes isotopes.

    Also, I see what you calculated, what I calculated was different, it was this...


    Your method is much easier, I will do both. Tomorrow will be a boring day in school, I'll just do it during school:smile:

    EDIT: As the second link shows, I was calculating the binding energy per nucleon.
  21. May 18, 2003 #20


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    Frankly I am thrilled to be dealing with someone who has this
    much gumption. Go for it. I am confident your results will be
    right and (partly for lack of time) will not check them.
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