Geometry of the Cosmos

weeman · Nov 6, 2008

I was reading an interesting page on the geometry of the cosmos (open, closed, flat). I just have a quick question from the article, which states:If you could draw a triangle by reaching far enough into space to draw lines connecting three far-flung galaxies, you could determine the curvature of the universe. The angles of a triangle in a negatively curved universe would add to greater than 180 degrees; those of a triangle in a positively curved universe, to less than 180 degrees. In a flat universe, familiar Euclidean geometry applies, and the angles of the triangle add up to exactly 180 degrees. Since it is mathematically impossible for a triangle's angles to equal more or less than 180 degrees, does this statement help prove that we live in a flat universe? Here is the link: http://archive.ncsa.uiuc.edu/Cyberia/Cosmos/CosmosShape.html <div class="Discussion_UserSignature"> Techies: We do it in the dark. "Put your hand on a stove for a minute and it seems like an hour. Sit with that special girl for an hour and it seems like a minute. That's relativity." -Albert Einstein </div>

Mee_n_Mac · Nov 6, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> In a flat universe, familiar Euclidean geometry applies, and the angles of the triangle add up to exactly 180 degrees. Since it is mathematically impossible for a triangle's angles to equal more or less than 180 degrees, does this statement help prove that we live in a flat universe? Posted by weeman</DIV> I don't think so.  The mathematical proof you speak of has implicit in it that Euclidian geometry applies.  Draw a "triangle" on the surface of a sphere and the sum of the angles totals > 180 degrees. <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

Saiph · Nov 6, 2008

yep, that's entirely the point weeman.  In a flat, euclidean universe, the triangle adds up to only 180 degrees. no more, no less.  But if the geometry isn't flat, euclids proofs don't apply anymore. an easy demonstration of a triangle with >180 degrees is, as mee_n_mac says, on a sphere.Look at a globe, and find the equator.  Look at two longitudinal lines right next to eachother.  They intersect the equator at 90 degrees.  By standard euclidean geometry this means they are parrallel, and should never intersect.  However, if you follow those longitudinal lines away from the equator, these perfectly straight lines, they will intersect at the poles.  Forming a three sided triangle.  As the angle between them and the third side is 90 degrees each, that makes a minimum of 180 degrees before you even consider the last corner they make at the pole...which can also be 90 degrees if you pick longitudinal lines that "quarter" the globe...for a maximum sum of angles of 270 degrees. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

weeman · Nov 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>yep, that's entirely the point weeman.  In a flat, euclidean universe, the triangle adds up to only 180 degrees. no more, no less.  But if the geometry isn't flat, euclids proofs don't apply anymore. an easy demonstration of a triangle with >180 degrees is, as mee_n_mac says, on a sphere.Look at a globe, and find the equator.  Look at two longitudinal lines right next to eachother.  They intersect the equator at 90 degrees.  By standard euclidean geometry this means they are parrallel, and should never intersect.  However, if you follow those longitudinal lines away from the equator, these perfectly straight lines, they will intersect at the poles.  Forming a three sided triangle.  As the angle between them and the third side is 90 degrees each, that makes a minimum of 180 degrees before you even consider the last corner they make at the pole...which can also be 90 degrees if you pick longitudinal lines that "quarter" the globe...for a maximum sum of angles of 270 degrees. Posted by Saiph</DIV> Ah. That makes sense now! Thank you for clearing that up, I was a little confused! <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-money-mouth.gif" border="0" alt="Money mouth" title="Money mouth" /> <div class="Discussion_UserSignature"> Techies: We do it in the dark. "Put your hand on a stove for a minute and it seems like an hour. Sit with that special girl for an hour and it seems like a minute. That's relativity." -Albert Einstein </div>

Saiph · Nov 7, 2008

well, small correction actually.  The maximum number of degrees on a sphere isn't 270... it's 360.  I don't know what I was thinking.  Just pick longitudinal lines opposite eachother (a "great circle" set)  They'll be 180 degrees apart at the pole, and 90 each at the equator for a net 360.  You might argue that this isn't a triangle anymore, as it's only 2 lines...but you could just cut it barely shy of 360...  Wait...can we actually make an obtuse angle at the pole...and still call that odd shape a triangle...  That would give more than 360 degrees....Oh bother, I hate spherical geometry...it makes my head hurt. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

KickLaBuka · Nov 7, 2008

If I may, please allow me to share some spherical geometry.  A charged ball has an electric field that varies in intensity in radial shells parallel to the ball.  When a current is observed approaching, the ball spins according to the right hand rule.  When that ball is spinning, it creates a magnetic field like the inside of an orange.  The magnetic field direction is determined by the direction of the rotation, and its strength is determined by the electric field at that radial shell.  The electric field and magnetic field at every point forms a 90° angle.  Those magnetic field lines reconnect at the opposite pole.  Interestingly, a magnetic field strength and direction creates a secondary magnetic field using the left hand rule.  This induces spin on other charged objects.   For this specific question, charge has a buried importance.  But let's say we know the distance to a galaxy representing our known edge of the universe; and a galaxy next to it; and a galaxy next to that.  With distance, we can certainly draw a rough representation of the universes radial shells; however, it would come only in form of a summation of arcs, what could look similar to a blastosphere.  X-ray data may indicate the shape inbetween, but we have yet to understand it.Please sift through the gemometry and the electromass consept, accept what you believe or understand, and skip over the rest. <div class="Discussion_UserSignature"> -KickLaBuka </div>

Mee_n_Mac · Nov 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>If I may, please allow me to share some spherical geometry.  A charged ball has an electric field that varies in intensity in radial shells parallel to the ball.  When a current is observed approaching, the ball spins according to the right hand rule.  When that ball is spinning, it creates a magnetic field like the inside of an orange.  The magnetic field direction is determined by the direction of the rotation, and its strength is determined by the electric field at that radial shell.  The electric field and magnetic field at every point forms a 90° angle.  Those magnetic field lines reconnect at the opposite pole.  Interestingly, a magnetic field strength and direction creates a secondary magnetic field using the left hand rule.  This induces spin on other charged objects.   For this specific question, charge has a buried importance.  But let's say we know the distance to a galaxy representing our known edge of the universe; and a galaxy next to it; and a galaxy next to that.  With distance, we can certainly draw a rough representation of the universes radial shell; however, it would come only in form of a summation of arcs, what could look similar to a blastosphere. Posted by KickLaBuka</DIV> Oh great .... now my head feels like Saiphs.  <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

DrRocket · Nov 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I was reading an interesting page on the geometry of the cosmos (open, closed, flat). I just have a quick question from the article, which states:If you could draw a triangle by reaching far enough into space to draw lines connecting three far-flung galaxies, you could determine the curvature of the universe. The angles of a triangle in a negatively curved universe would add to greater than 180 degrees; those of a triangle in a positively curved universe, to less than 180 degrees. In a flat universe, familiar Euclidean geometry applies, and the angles of the triangle add up to exactly 180 degrees. Since it is mathematically impossible for a triangle's angles to equal more or less than 180 degrees, does this statement help prove that we live in a flat universe? Here is the link: http://archive.ncsa.uiuc.edu/Cyberia/Cosmos/CosmosShape.html Posted by weeman</DIV>The angles of a triangle sum to 180 degrees only in a Euclidean geometry.  In a curved geometry, for instance the surface of a sphere (like the Earth) the angles can sum to more than 180 degrees (positively curved geometries) or less than 180 degrees (negatively curved geometries).  Note that this applies to spaces of constant curvature, and in spaces of variable curvature the sum the angles may vary depending on the specific triangle.It is believed that the universe is of constant curvature on a very large scale, but it is of variable curvature on smaller scales.  In fact it is the variable curvature that accounts for gravity in general relativity.  <div class="Discussion_UserSignature"> </div>

DrRocket · Nov 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>If I may, please allow me to share some spherical geometry.  A charged ball has an electric field that varies in intensity in radial shells parallel to the ball.  When a current is observed approaching, the ball spins according to the right hand rule.  When that ball is spinning, it creates a magnetic field like the inside of an orange.  The magnetic field direction is determined by the direction of the rotation, and its strength is determined by the electric field at that radial shell.  The electric field and magnetic field at every point forms a 90° angle.  Those magnetic field lines reconnect at the opposite pole.  Interestingly, a magnetic field strength and direction creates a secondary magnetic field using the left hand rule.  This induces spin on other charged objects.   For this specific question, charge has a buried importance.  But let's say we know the distance to a galaxy representing our known edge of the universe; and a galaxy next to it; and a galaxy next to that.  With distance, we can certainly draw a rough representation of the universes radial shells; however, it would come only in form of a summation of arcs, what could look similar to a blastosphere.  X-ray data may indicate the shape inbetween, but we have yet to understand it.Please sift through the gemometry and the electromass consept, accept what you believe or understand, and skip over the rest. Posted by KickLaBuka</DIV>Your assertions have nothing whatever to do with spherical geometry.  Spherical geometry and spherical coordinates are different things entirely.And while one might construct some sort of model representing radial mass distribution of the universe, using Earth as the center for the coordinate system perhaps, you would not get a charge distribution corresponding to the mass distribution as the universe is, on macroscopic scale, basically charge neutral.  <div class="Discussion_UserSignature"> </div>

KickLaBuka · Nov 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Your assertions have nothing whatever to do with spherical geometry.  Spherical geometry and spherical coordinates are different things entirely.  the universe is, on macroscopic scale, basically charge neutral.  Posted by DrRocket</DIV> Doctor,Nice to hear from you again.  I'm glad to help you. Your first question is a bit confusing.  How can one separate spherical geometry and spherical coordinates?  Cause when I was in school, spherical cooridnates were used exactly to describe spherical geometry.To address your second question, Electromass is showing forces that are present during "separation of charge."  That the entire universe is charge neutral doesn't stop the forces created by the separation of charge at each, dare I, sphere of reference.  <div class="Discussion_UserSignature"> -KickLaBuka </div>

MeteorWayne · Nov 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Doctor,Nice to hear from you again.  I'm glad to help you. Your first question is a bit confusing.  How can one separate spherical geometry and spherical coordinates?  Cause when I was in school, spherical cooridnates were used exactly to describe spherical geometry.To address your second question, Electromass is showing forces that are present during "separation of charge."  That the entire universe is charge neutral doesn't stop the forces created by the separation of charge at each, dare I, sphere of reference.  Posted by KickLaBuka</DIV>Mod Hat ON:----- Unfortunately, as the good Dr said, this has nothing to do with the geometry of the Universe, rather it has to do with alleged charge distribution at microscopic distances, which is irrelevant at universal scales. Please start another thread if your wish to discuss molecular level phenomena.---------Mod Hat Off <div class="Discussion_UserSignature"> But the Krell forgot one thing John. Monsters. Monsters from the Id. I really, really, really, really miss the "first unread post" function </div>

DrRocket · Nov 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Doctor,Nice to hear from you again.  I'm glad to help you. Your first question is a bit confusing.  How can one separate spherical geometry and spherical coordinates?  Cause when I was in school, spherical cooridnates were used exactly to describe spherical geometry.To address your second question, Electromass is showing forces that are present during "separation of charge."  That the entire universe is charge neutral doesn't stop the forces created by the separation of charge at each, dare I, sphere of reference.  Posted by KickLaBuka</DIV>You can use spherical coordinates to describe position in Euclidean 3-space, and that is precisely how they are normally used.  The typical treatment and use of spherical coordinates in calculus courses and undergraduate physics courses has nothing whatever to do with spherical geometry.Spherical geometry has to do with lines, and angles that are constrained to lie on the surface of a sphere.  In that context, what one calls "lines" are great circles and segments thereof -- the geodesics on a sphere.  In that context, a triangle, formed by three lines, has more than 180 degrees and there is no such thing as parallel lines -- great circles always intersect.  The geometry is non-Euclidean.  You can use spherical coordinates to describe positions in this geometry, but the "r" coordinate is fixed, usually taken as 1.  There is some anbiguity in those coordinates as well, at the poles only the latitude is important, and the longitude is not defined.  But that is only one use of spherical coordinates, and not the most usual one.Not only is the universe charge neutral as a whole, in most instances it is locally charge neutral on a macroscopic scale.  Because the electromagnetic force is so strong, it takes rather special circumstances to maintain a charge separation over any significant distance or for any significant period of time.  It can happen, but it is rather rare. <div class="Discussion_UserSignature"> </div>

SHU · Nov 16, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>It is believed that the universe is of constant curvature on a very large scale, but it is of variable curvature on smaller scales.  In fact it is the variable curvature that accounts for gravity in general relativity.  Posted by DrRocket</DIV> Not related but taking this a bit further, Can the metric, and the geometry of space change locally?  I realize there are several different proposed geometries, some related to specifics like black hole horizons but I'm having a bit of trouble understanding this.  For example:http://www.hawking.org.uk/text/physics/nut.htmlThe math is way beyond this engineering dropout but I'm trying to understand conceptually.  Wouldn't one geometry exclude another?  Or are we talking about subgeometry? (notice how I avoided Star Trek terminology)

DrRocket · Nov 17, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Not related but taking this a bit further, Can the metric, and the geometry of space change locally?  I realize there are several different proposed geometries, some related to specifics like black hole horizons but I'm having a bit of trouble understanding this.  For example:http://www.hawking.org.uk/text/physics/nut.htmlThe math is way beyond this engineering dropout but I'm trying to understand conceptually.  Wouldn't one geometry exclude another?  Or are we talking about subgeometry? (notice how I avoided Star Trek terminology) Posted by SHU</DIV>Yes the metric can change locally, in fact it is defined locally.There are two versions of the word "metric" that you might find in mathematics.One is in the definition of a "metric space" where a metric, d is a function that satisfies the intuitive notions of distance, d assigning to any two points x and y a non-negative real number, " the distance from x to y" that satisfies d(x,x)=0, d(x,y)=d(y,x)>0 if x and y are distinct and d(x,y) >/= d(x,z) + d(z,y).  A metric space has a topology generated by the metric, and you can take limits in metric spaces in the same way that you do in calculus classes. However, that definition of a metric is NOT what is meant by a metric in relativity.General relativity takes place on what is called a 4-manifold, which is a topological space that locally looks like ordinary Euclidean 4-space.  It comes equipped with a notion of differentiability and with a notion of tangent vectors (it takes some work to really define and understand these notions but think about them intuitively for the purpose at hand and you will be OK).  If you have a curve on this manifold you can measure distance along that curve by approximating it with tangent vectors and adding up the lengths of the tangent vectors -- if you have some way of determining the length of tangent vectors.So, how do you determine the length of a vector.  In ordinary vector analysis, you take the dot product of the vector with itself and then take the square root.  So what you need is a notion of a dot product, or what is also called an inner product.  What is an inner product?  It is just a quadratic form, usually required to be positive definite -- a matrix A that is positive definite.  Then if X and Y are column vectors the inner product of X and  Y is just XT A Y where XT is the transpose of X.  If you have such a matrix defined at each point of a manifold, it is called a metric, and cal measure the length of vectors based at the point corresponding to the metric.  So you see that the metric is defined locally, and since it determines the local geometry, that geometry is also local in nature and can vary from region to region over the manifold.  There are some other technicalities that basically require the quantities in question to vary smoothly, so that you can make sense of derivative of themGeneral relativity has one more wrinkle.  The metric is not positive definite, but is Lorentzian.  That complicates things a bit, but the notion that the geometry is locally defined remains valid.To  do this rigorously takes quite a bit more machinery and work, but I hope the paragraphs above communicate the basic ideas.  <div class="Discussion_UserSignature"> </div>

SHU · Nov 17, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Yes the metric can change locally, in fact it is defined locally.There are two versions of the word "metric" that you might find in mathematics.One is in the definition of a "metric space" where a metric, d is a function that satisfies the intuitive notions of distance, d assigning to any two points x and y a non-negative real number, " the distance from x to y" that satisfies d(x,x)=0, d(x,y)=d(y,x)>0 if x and y are distinct and d(x,y) >/= d(x,z) + d(z,y).  A metric space has a topology generated by the metric, and you can take limits in metric spaces in the same way that you do in calculus classes. However, that definition of a metric is NOT what is meant by a metric in relativity.General relativity takes place on what is called a 4-manifold, which is a topological space that locally looks like ordinary Euclidean 4-space.  It comes equipped with a notion of differentiability and with a notion of tangent vectors (it takes some work to really define and understand these notions but think about them intuitively for the purpose at hand and you will be OK).  If you have a curve on this manifold you can measure distance along that curve by approximating it with tangent vectors and adding up the lengths of the tangent vectors -- if you have some way of determining the length of tangent vectors.So, how do you determine the length of a vector.  In ordinary vector analysis, you take the dot product of the vector with itself and then take the square root.  So what you need is a notion of a dot product, or what is also called an inner product.  What is an inner product?  It is just a quadratic form, usually required to be positive definite -- a matrix A that is positive definite.  Then if X and Y are column vectors the inner product of X and  Y is just XT A Y where XT is the transpose of X.  If you have such a matrix defined at each point of a manifold, it is called a metric, and cal measure the length of vectors based at the point corresponding to the metric.  So you see that the metric is defined locally, and since it determines the local geometry, that geometry is also local in nature and can vary from region to region over the manifold.  There are some other technicalities that basically require the quantities in question to vary smoothly, so that you can make sense of derivative of themGeneral relativity has one more wrinkle.  The metric is not positive definite, but is Lorentzian.  That complicates things a bit, but the notion that the geometry is locally defined remains valid.To  do this rigorously takes quite a bit more machinery and work, but I hope the paragraphs above communicate the basic ideas.  Posted by DrRocket</DIV> Thanks.  Took a few reads but I think I got it.  It's the "potato" analogy explained.  Just got into this recently but that was a hurdle.  The whole was plenty without the sum of the parts. The "smoothing" from one metric to another is explained by the differential nature.

DrRocket · Nov 17, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'> Thanks.  Took a few reads but I think I got it.  It's the "potato" analogy explained.  Just got into this recently but that was a hurdle.  The whole was plenty without the sum of the parts. The "smoothing" from one metric to another is explained by the differential nature. Posted by SHU</DIV>I think you  got it.  The subject matter is what is called differential geometry.  Within differential geometry the subject of metrics come up in discussions of Riemannian geometry and to do general relativity you need to get into what are called pseudo-Riemannian spaces, and in particular Lorentzian spaces. There are text books on the subject, but none that are accessible without a fair amount of mathematical background.  Probably the best introduction is Gravitation by Charles Misner, Kip Thorne and John Archibald Wheeler.  But it is not light reading.  And it is really a bit beyond just and introduction.  It  has the twin blessing and curse of being written by physicists.  Blessing in that they try to get across the intuition as well as the formalism.  Curse in that they are sometimes rather imprecise and ambiguous to a mathematician.If there is a good, simple, intuitive but not misleading introduction to the subject I have not seen it. <div class="Discussion_UserSignature"> </div>

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