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    Hi all!

    I'm doin a 3rd year physics parctice paper and I'm stuffin up a question about great circles:

    (Original post by Exam Q)
    Show that the mean great circle distance between two points on a unit sphere is \pi but that the means shortest distance is \pi/2
    Now, I'm gonna confess, I'm very rusty on any kinda maths that isn't calculus and I'm pretty shoddy at that too but this seem like it should be trivial.

    From what I can remember, the great circle distance (d) between 2 points is d=Ra where R is the radius and a is the angle between the chords to the centre (please forgive the inaccurate vocab) and that the total circumference of a great circle is 2 \pi r

    So, I figures, all I need to do is integrate d over the possible values for a and divide by the total circumference. I can't seem to get it to work

    and ideas? (solutions welcome, it's only personal revision)

    Thanks x
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    We're on the unit sphere, so R = 1.

    So yeah, that's what you're doing. Just do the integral and the answer drops out. (Although I get 3pi/2 and pi/2 (which makes sense as the expectation is linear))
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    Well, I figured R being a constant, it wouldn't really matter what it was til the end... though now you come to mention it...

    as for the working I've got so far: integrating between 0 and is giving me then I'm dividing by and then I get 1.

    Am I using the wrong limits or something? I can't help but feel that to get the answer of I'll need an integral solution of something of the form
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    You need to integrate max(a, 2pi-a) and min(a, 2pi-a)
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    Sorry, I don't understand, I take it those are the limits for the integral, but if I'm integrating a, doesn't thisjust put a back into the problem?

    Thanks for replying btw ^_^
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    No, for the maximum distance, you need to integrate max(a, 2pi-a) between 0 and 2pi. This is because max(a,2pi-a) is the maximum distance!
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    Again, I'm confused.

    I'm just gonna mention here (because it the only time I've seen that max(), min() terminology), that I'm doing this by hand and not programming a caluclator/ to do it for me.

    How would I go about integrating a max(a,2pi-a) function?

    ((I get the impression that I'm being dense here))
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    Split the integral up into the bit where a > 2pi-a and the other bit
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    I've just re-read the previous messages. you got 3\pi/2 and \pi/2? these aren't the solutions provided...
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    I still don't understand. All I've managed to work out is that the reason the measn great distance is twice that of the mean shortest distance is the the MGD takes all the paths and for every path greater than \pi there is a shorter distance. so the MGD=2*MSD.
    Am I right?

    Still can't get the integral to work.
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    (Original post by loopylouise)
    I still don't understand. All I've managed to work out is that the reason the measn great distance is twice that of the mean shortest distance is the the MGD takes all the paths and for every path greater than \pi there is a shorter distance. so the MGD=2*MSD.
    Am I right?

    Still can't get the integral to work.
    Nope. The MGD isn't even twice the MSD.

    Suppose we have two points separated by angle \theta. We know that there are two routes between the points (along the great circle). One tracing the angle, and one going the opposite way around. The one tracing the angle has length \theta, the one going the other way 2\pi - \theta

    Therefore the shortest distance for a given \theta will be the smaller of these two distances, \min(\theta, 2\pi-\theta).

    Now, \theta ranges from to 2\pi.

    When \theta < 2\pi - \theta, ie \theta < \pi, \min(\theta,2\pi-\theta) = \theta. When \theta > \pi, the opposite is true \min(\theta, 2\pi-\theta) = 2\pi-\theta

    Therefore the integral becomes:

    \int_0^{2\pi} \min (\theta, 2\pi-\theta ) \, d\theta  = \int_0^{\pi} \theta \, d\theta + \int_{\pi}^{2\pi} (2\pi-\theta) \, d\theta
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    Ah, now I see. So for the an great circle distance:

    \int_{\pi}^{2\pi}\theta\,d\theta + \int_0^{\pi}(2\pi- \theta)\,d\theta ?

    if this the case I get 3/2 \pi which isn't the answer the question wants
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    (Original post by SimonM)
    ..
    (Original post by loopylouise)
    ..
    I'm a bit confused here.

    What do you understand "the great circle distance between two points on a unit sphere" to be? (I would have thought you'd draw a great circle between the two points, and then take the length of the shorter of the arcs connecting them).

    What do you understand "the shortest distance between 2 points on a unit sphere" to be? (I would have thought it was the length of the straight line connecting them (even though it goes inside the sphere)).

    Although I confess, I'm really not seeing where the given answers come from in any event.
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    (Original post by DFranklin)
    I'm a bit confused here.

    What do you understand "the great circle distance between two points on a unit sphere" to be? (I would have thought you'd draw a great circle between the two points, and then take the length of the shorter of the arcs connecting them).

    What do you understand "the shortest distance between 2 points on a unit sphere" to be? (I would have thought it was the length of the straight line connecting them (even though it goes inside the sphere)).

    Although I confess, I'm really not seeing where the given answers come from in any event.
    To be honest with you, I'm not entirely sure. I didn't even know what a great circle was when I came across this past question so after a bit of web-surfing I've assumed that the great circle distance is the distance between two points on the great circle and that there are two values of the great circle distance for each two points and that the shortest distance is the smallest of the two.

    Having said that, I've not found anywhere that says it specifically.
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    The only way I can see the answers making sense is if you have some inherent orientation for the direction you go along for the great circle (which you could do, for example by taking the cross product to get a signed vector), and "the great circle distance" is the distance you need to go in that direction.

    Whether that's what is actually meant I have no idea.
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    (Original post by DFranklin)
    The only way I can see the answers making sense is if you have some inherent orientation for the direction you go along for the great circle (which you could do, for example by taking the cross product to get a signed vector), and "the great circle distance" is the distance you need to go in that direction.

    Whether that's what is actually meant I have no idea.
    It is a wonderfully vague question, isn't it? It's not like we do much spherical anything that isn't applied to a electromagnetic or quantum mechanical system and then the most we need to do is integral over a volume element. Sheer bloodyminded geometry never comes up

    Direction doesn't seem to feature as it's talking about any 2 points on the sphere.:confused:
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    (Original post by loopylouise)
    It is a wonderfully vague question, isn't it? It's not like we do much spherical anything that isn't applied to a electromagnetic or quantum mechanical system and then the most we need to do is integral over a volume element. Sheer bloodyminded geometry never comes up

    Direction doesn't seem to feature as it's talking about any 2 points on the sphere.:confused:
    Well, it can do. Given a great circle between the two points, there are two possible directions you can go along the circle, giving two possible distances. But something like the cross-product rule I suggested would be a way of choosing the direction, and hence the distance, in a way that would make the expected distance pi.

    Whether that's what was intended, I have no idea.
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    (Original post by DFranklin)
    Well, it can do. Given a great circle between the two points, there are two possible directions you can go along the circle, giving two possible distances. But something like the cross-product rule I suggested would be a way of choosing the direction, and hence the distance, in a way that would make the expected distance pi.

    Whether that's what was intended, I have no idea.
    I'm not sure what I would be cross-product-ing
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    The vectors of the two points.

    (Edit: although I'm having a sudden misgiving that this works - I think it might always give you the shorter distance. You can probably fudge something even more arbitrary than the cross product though).
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    Woo! Fudge! the best flavoured maths

    At least I know now that it isn't a completely trival question that I'm too thick to work out with a click of the fingers.
 
 
 
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