Physics Problem

[PrimeCruncher]

Originally Posted by jinydu

Strangely enough, Mathworld:

http://mathworld.wolfram.com/SphericalGeometry.html

says that in a sphere, there are no parallel lines. This means that longitude lines intersect?!

In my geometry class we spent a day on this type of non-Euclidean geometry. Basically, there are no parallel lines because spherical geometry defines a line as, essentially, the intersection of a sphere and a plane (in other words, a "line" in spherical geometry is what we commonly call a circle). Their definition of a line, however, stipulates that the intersecting plane creating the "line" must pass through the center of the sphere. Hence all "lines" must intersect.

[jinydu]

Oh, that's why.

Well, it seems that the R*cos(L) formula does give the right answer. Thanks

[cheesehead]

Originally Posted by jinydu

Well, it seems that the R*cos(L) formula does give the right answer.

Of "cos" it duz!

[mfgoode]

Originally Posted by jinydu

Oh, that's why.

Well, it seems that the R*cos(L) formula does give the right answer. Thanks

I agree with most of the answers but one or two replies are totally wrong for Euclidean Geometry.
Lines of Longitude intersect at the poles.
The statement ‘Hence all lines must intersect (on a circle or sphere) is definitely wrong, regardless what Mathworld says. The lines of latitude are parallel to one another and don’t intersect one with the other. Yes they intersect with the lines of Longitude of course.

Jinydu, you must get a stronger foundation in your studies all round. At your stage (if Im not mistaken) you should be reasonably familiar with most of the 465 propositions of Euclid the foundation stone of all math studies. You should at least master 8-9 books of the 13 books of Euclid.
All his theorems and postulates/axioms have stood the test of intense scrutiny except the 5th known as the ‘parallel axiom’ which has branched off into modern geometry.

For such problems always draw a figure and work in the lowest dimension as possible. Here the circle( 2 dimensions) is the projection of the 3 dimensional sphere so convert the problem to 2 dimensions.

So draw a circle and mark off the horizontal diameter as AB the end points and also lying on the circle of centre O. Through O draw the vertical radius OY. Through O draw the line OP making an angle L degrees with OA intersecting the circle at P. This is the latitude of P. Through P draw line PQ parallel to AB and cutting vertical radius At C.
Now triangle OPC is a right angle and AOP is L degrees. Therefore angle POC is = (90-L) degrees. PC is the required radius r.
In rt.Angled triangle OPC , PC/OP(hyp) = sin (90-L)…….(1)
Now sin(90-L) =cosL
Therfore PC/OP =cosL ……from (1)
Cross multiply and we get PC ( r ) =OP cosL
But OP is the radius of the circle ( R ) assuming the earth to be a perfect sphere
Thefore PC =r=RcosL
Therefore the reqd. circumference is 2*Pi*R*cosL

If you have followed me so far take two steps further.
Archimedes proved two very important theorems. :surprised
1) The area of a sphere is the area of 4 times its great circle (The equator)
This is equivalent to the area of the circumscribed cylinder which is open at both ends.

2)The volume of a sphere is 2/3 the volume of the circumscribed cylinder.
From these two theorems we can conclude the same applies to a spherical cap of the sphere for both area and volume.
Try deriving these 2 theorems assuming formulae for area and volumes are known

Best of luck,
Mally

[ewmayer]

Originally Posted by mfgoode

The statement ‘Hence all lines must intersect (on a circle or sphere) is definitely wrong, regardless what Mathworld says. The lines of latitude are parallel to one another and don’t intersect one with the other. Yes they intersect with the lines of Longitude of course.

Depends on what you mean by "line." On the surface of a sphere there is no such thing as a Eulidean straight line (i.e. a path between any 2 distinct points which has length equal to the Euclidean distance between the points.) So to speak of length in this kind of noneuclidean setting we need to properly generalize the concept of "straight line." The most intuitive way of doing so (and also the one that leads to results that are "correct" in that they lead to no obvious self-contradictions or ambiguity) is to generalize "line" to "path of minimal length". "Minimal length" of course requires a definition of length in a noneuclidean setting, but the math here quickly gets more technical than I find desirable for the present discussion - suffice it to say there is a well-defined mathematical concept known as a metric which both defines the geometry in question and allows one to define "length" in the sense of a unique measure of distance between any 2 points, which satisfies the requisite properties of a mathematical norm. See my more-detailed footnote^* below about that, or any decent introductory text on differential geometry. (Mally, the latter should be of interest to you also in the sense that you are a devotee' of Herr Einstein's work - differential geometry is in fact the language of the theory of General Relativity.)

The formal term for such a minimal-length path in a general (i.e. not necessarily Euclidean, though including such as a special case) geometry is a geodesic. In a spherical geometry (specifically, that defined on the surface of a radius-R sphere) the geodesics are arcs of great circles (i.e. circles sharing the same radius as the sphere.) "Lines" of longitude are by definition great circles, but the only "line" of latitude which is such is the one at the equator. Non-equatorial lines of latitude are indeed circles, but having smaller radius than great circles, and as such are not geodesics, i.e. not proper analogues of "straight lines". It is in that sense that the phrase "all (nonidentical) lines on a sphere intersect" is meant.

As to the original question, if we assume that "linear velocity" really refers to the velocity change an observer standing at a given spot on the earth's surface would experience if the earth suddenly stopped rotating with respect to an inertial reference frame (a.k.a. "the fixed stars," though that's somewhat of a misnomer), if the angular rotation rate is denoted omega, the desired speed is omega*(perpendicular distance to earth's axis.) In terms of spherical coordinates with a polar angle phi (see below) that varies from 0 at the N pole to pi/2 at the equator to pi at the S pole the velocity is R*omega*sin(phi). In terms of the usual latitude L (taken in radians, 0 at the equator, +-pi/2 at the N/S poles, respectively) it's R*omega*cos(L). Since the earth has circumferece ~25000 miles at the equator and rotates at roughly one revolution per 24 hours, omega ~= 2*pi/day, which translates to a velocity of just over 1000 miles per hour at the equator. At latitude 45 degrees north or south this reduces to 1/sqrt(2) of that amount.


================

^*As an example, in Euclidean 3-space two points separated by relative coordinate difference of (dx, dy, dz) lie at a distance sqrt(dx² + dy² + dy²) from each other - the usual Euclidean straight-line distance. (The sqrt() expression here is simply the Euclidean norm.) If one writes the argument of the square root, in form of (row vector)*(3x3 matrix)*(column vector), i.e. as v * A * v^T (where v is the row vector with components dx, dy, dz and ()^T is the usual definition of transpose), the Euclidean distance is given by a matrix A with ones on the diagonal and zeros elsewhere, i.e. A is just the 3x3 identity matrix,

    / 1  0  0 \
A = | 0  1  0 |
    \ 0  0  1 / .

That is, in Euclidean coordinates we can dispense with the A-matrix and express the distance between 2 points (note in Euclidean space it doesn't matter whether dx,dy,dz are differentially small or not - in general, however, it does) as v * v^T.

In spherical polar coordinates, using the usual (r, theta, phi) radial and angular coordinates (theta is the azimuthal angle, phi, the polar), we have the following relations between spherical and Euclidean coordinates:

r = sqrt(x² + y² + y²)

x = r*sin(phi)*cos(theta),
y = r*sin(phi)*sin(theta),
z = r*cos(phi) .

Considering two nearby points that are differentially separated by v = (dx,dy,dz), we can express the separation in terms of spherical coordinates as v^T = J*w^T, where w := (dr,dtheta,dphi)^T, and J is the Jacobian matrix of the coordinate transformation:

    / dx/dr  dx/dtheta dx/dphi \   / sin(phi)*cos(theta) -r*sin(phi)*sin(theta) -r*cos(phi)*cos(theta) \
J = | dy/dr  dy/dtheta dy/dphi | = | sin(phi)*sin(theta)  r*sin(phi)*cos(theta) -r*cos(phi)*sin(theta) |
    \ dz/dr  dz/dtheta dz/dphi /   \ cos(phi)                       0            r*sin(phi)            / ,

where these derivatives are of the partial variety.

Now this is still a Euclidean calculation - the underlying metric is still the Euclidean one, we've just rewritten the quadratic form yielding the (distance)² in terms of an alternate coordinatization, in which distance is expressed as v * v^T = (w*J^T)*(J*w^T) = w * A * w^T, and the A-matrix is given by A = J^T * J, or

    /    sin(phi)*cos(theta)    sin(phi)*sin(theta)    cos(phi) \   / sin(phi)*cos(theta) -r*sin(phi)*sin(theta) -r*cos(phi)*cos(theta) \   / 1      0               \
A = | -r*sin(phi)*sin(theta)  r*sin(phi)*cos(theta)      0      | * | sin(phi)*sin(theta)  r*sin(phi)*cos(theta) -r*cos(phi)*sin(theta) | = | 0  r2*sin2(phi)  0   |
    \ -r*cos(phi)*cos(theta) -r*cos(phi)*sin(theta)  r*sin(phi) /   \ cos(phi)                       0            r*sin(phi)            /   \ 0      0           r2 / ,

i.e. the Euclidean (differential distance)² is dx² + dy² + dy² = dr² + r²*sin²(phi)*dtheta² + r²*dphi² .

On the surface of a sphere, the crucial change is that dr = 0, so we can obtain the desired metrix by zeroing the first diagonal term and replacing the variable r with the constant spherical radius R, which yields the metric consisting of a matrix with diagonal terms (0, R²*sin²(phi), R²).

[mfgoode]

To quote any part of your post Prof ewmayer will spoil the fullness and richness of the entire answer so I refrain from doing so

My apologies to you Prime Cruncher. I stand to be corrected by the very lucid and excellent explanation of Prof.ewmayer.

After I sent in my post I had a nagging doubt that I was wrong somewhere on ‘lines of latitude’ in the broader concept of modern geometry. Now Prof ewmayer has more than confirmed my ‘hunch’. I have been able to understand the difference between ‘line of minimal length’ and a geodesic (a word I could not recall in my mind). I knew we fly in great circles after 36 yrs of circling the globe and often wondered why we don’t go ‘straight’ as on the map!

Yes I am rusty on differential geometry but you Prof. have made it so interesting that it’s an impetus to go deeper and further into it. This is something I will definitely do

Thank you for your keen observation of my interest in Doktor Herr Albert Einstein (Biedermann).
Whereas I have always admired his math (and that of Minkowski) I study him for his original thoughts and the startling and daring conclusions he reached in his ‘thought experiments’ His whole epistemology to say the least is mind boggling.

In conclusion I will say Prof ewmayer IMHO that you have the natural aptitude to see the difficulty of others and the gift of making difficult concepts easy to understand.
If you do write (if not already) text books in Math or physics I can assure you I will be the first to buy.

I have treasured (literally) your marvellous explanation on ‘singularity’ and ‘renormalisation’. Please keep up the good work and be a guiding light for all of us.
Thank you,
Mally.

[ET_]

Mally, there has been a wonderful thread in Mersenneforum, where ewmayer and philmoore discussed about about ECM and P-1. It's really worth copying and studying it I did.

Luigi

[mfgoode]

Originally Posted by ET_

Mally, there has been a wonderful thread in Mersenneforum, where ewmayer and philmoore discussed about about ECM and P-1. It's really worth copying and studying it I did.

Luigi

Mille Gratia ET!.
I will do as you say as you have put me onto the giants of
mersenne forum. If I can stand on their shoulders it will enable me to see further.If they drop me I still can collect sea shells!
Buona Sera Senor! (hope thats not an abuse!)
Mally

[davar55]

Just came across this interesting thread, especially the wonderfully
lucid explanation provided by ewmayer. Want to echo the late Mally
in complimenting that post's style and clarity.

Perhaps its author could briefly make the connection to tensors,
a math topic I'd love some additional background on.

[davieddy]

Originally Posted by davar55

Just came across this interesting thread, especially the wonderfully
lucid explanation provided by ewmayer. Want to echo the late Mally
in complimenting that post's style and clarity.

Perhaps its author could briefly make the connection to tensors,
a math topic I'd love some additional background on.

Hmmm.

I'll read the whole thread in due course, but the required
answer to the OP was obviously the promptly supplied
2piR*cos(latitude)/day.

I was truly surprized to hear from Philmoore that the formula
"surface area of a sphere between two // planes h apart = 2pi*R*h"
was new to him.

Equally delightful, and simple to prove, is that the angles of
a triangle area A formed by three great circles on a sphere add up to
pi + A/R^2.

David

[davar55]

Yes, that might be a delightful result, but ewmayer at least thought
before he contributed his useful post.