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A simple diferential equation!

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Just out of pure curiosity, can anyone here give me any advice on the problem of solving the following differential equation

[math]\left\{\sum_i \vec{\alpha}_i \cdot \nabla_i + \sum_{i neq j}\beta_{ij}\delta(x_i -x_j)\delta(\tau_i - \tau_j) \right\}\vec{\psi} = K\frac{\partial}{\partial t}\vec{\psi} = iKm\vec{\psi}.[/math]

 

where,

 

[math][\alpha_{ix} , \alpha_{jx}] \equiv \alpha_{ix} \alpha_{jx} + \alpha_{jx}\alpha_{ix} = \delta_{ij}[/math]

 

[math][\alpha_{i\tau} , \alpha_{j\tau}] = \delta_{ij}[/math]

 

[math][\beta_{ij} , \beta_{kl}] = \delta_{ik}\delta_{jl}[/math]

 

[math][\alpha_{ix}, \beta_{kl}]=[\alpha_{i\tau}, \beta_{kl}] = 0 \text{ where } \delta_{ij} =

\left\{\begin{array}{ c c }

0, & \text{ if } i \neq j \\

1, & \text{ if } i=j

\end{array} \right.

[/math]

 

Any advice on approaching this problem would be greatly appreciated.

 

Thank you very much!

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This is the Dirac equation. I would first change notation, but that doesn't matter.

 

The usual thing to do is pick a representation for the Clifford algebra. Doesn't matter which one as we are doing relativistic quantum mechanics.

 

Or you could look at any book on QFT which will treat this equation (almost) fully.

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This is the Dirac equation. I would first change notation, but that doesn't matter.

Been a while since I posted here but thought I might look around again. I actually have a Ph.D. in Theoretical Nuclear Physics from Vanderbilt University. I posted this thread because I wanted to see if there was anyone here who understood things at that level. I am afraid that is not Dirac's equation; Oh it looks a lot like Dirac's equation (and it is relatively easy to show Dirac's equation is an approximation to it) but there are important differences with profound consequences. Apparently there is no one here capable of comprehending the consequences of those subtle differences.

 

One exact solution I discovered leads to a rather significant geometric proof.

 

The proof concerns a careful examination of the projection of a trivial geometric structure on a one dimensional line element.

Also shown is an interesting corollary consisting of the fact that three (and only three) such projections can be made on three orthogonal axes thus a three dimensional collection of absolutely arbitrary points can always be seen as a projection of that same trivial n dimensional structure. It follows that any collective motion of such a collection of points can be seen as a projection of that trivial structure under complex rotation in that n dimensional space.

 

Another post which seems to have gone totally over everyone's head.

 

Have fun -- Dick

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I am afraid that is not Dirac's equation; Oh it looks a lot like Dirac's equation (and it is relatively easy to show Dirac's equation is an approximation to it) but there are important differences with profound consequences. Apparently there is no one here capable of comprehending the consequences of those subtle differences.

 

Nope, just those of us that are have lots of other things to worry about. Feel free to elaborate on this equation for us. Simply writing down a differential equation is not enough to get us "hooked".

 

One exact solution I discovered leads to a rather significant geometric proof.

 

Geometric proof of what?

 

 

Also shown is an interesting corollary consisting of the fact that three (and only three) such projections can be made on three orthogonal axes thus a three dimensional collection of absolutely arbitrary points can always be seen as a projection of that same trivial n dimensional structure. It follows that any collective motion of such a collection of points can be seen as a projection of that trivial structure under complex rotation in that n dimensional space.

 

We are not privy to the theorem.

 

Another post which seems to have gone totally over everyone's head.

 

Well I suggest that it is your fault. You may be an expert in this field, "Dirac-like equations", but the rest of us are not. It will be up to you to motivate the discussion and point out the non-trivial results of your work. I can't see how you expect us to get at this from just quoting a differential equation.

 

Anyway, feel free to point to you preprints or published papers for us. See you again in a few years...

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Geometric proof of what?

(The following has been corrected for typo errors in the original)

 

With regard to "weak emergence" (that is with regard to the definition of "weak emergence") I feel it can also be dispensed with via the following proof. That is, emergence is emergence and there is nothing either weak or strong about it! People begin to think about weak "emergence" when what they are looking at is more complex than what they can deduce from "known laws". What they seem to forget is that physics is applicable only to problems which can be reduced to one body problems by some procedure.

 

 

I am of the opinion that the following proof is of great significance when one goes to consider "emergent" phenomena and the complexity achievable from simple constructs. The proof concerns a careful examination of the projection of a trivial geometric structure on a one dimensional line element.

 

 

The underlying structure will be an n dimensional rigid entity defined by a collection of n+1 points connected by lines (edges) of unit length embedded in an n dimensional Euclidean space (i.e., a minimal n dimensional equilateral polyhedron; the generalized concept of a higher dimensional equilateral triangle with unit edges). The universe (the collection of information to be analyzed) of interest will be the projection of the vertices of a that polyhedron on a one dimensional line element. The logic of the analysis will follow the standard inductive approach: i.e., prove a result for the cases n=0, 1, 2 and 3. Thereafter prove that if the description of the consequence is true for n-1 dimensions, it is also true for n dimensions. The result bears very strongly on the range of complexity of "emergent" phenomena given an extremely simple source.

 

 

First of all, the projection will consist of a collection of points (one for each vertex of that polyhedron) on the line segment of interest. Since motion of that polyhedron parallel to the given line segment is no more than uniform movement of every projected point, we can define the projection of the center of the polyhedron to be the center of the line segment: i.e., linear motion of the polyhedron has no real consequences. Furthermore, as the projection will be orthogonal to that line segment and the n dimensional space is Euclidean, any motion orthogonal to that line segment introduces no change in the projection whatsoever. It follows that the only motion of the polyhedron which provides any interesting changes in the distribution of points on the line segment will be rotations of the polyhedron in the n dimensional space.

 

 

The assertion which will be proved is that every conceivable distribution of points on the line segment is achievable by a specifying a particular rotational orientation of the polyhedron relative to the line segment of interest. Before we proceed to the proof, one issue of significance must be brought up. That issue concerns the scalability of the distribution. I referred to the collection of points on the line segment as the "universe of interest" as I want the student to think of that distribution of points as a universe: i.e., any definition of length must be arrived at via some defined characteristic of the the distribution itself or some subset of the distribution. Thus any two distributions which differ only by a scale factor will be considered to be identical distributions.

 

 

Case n=0 is trivial as the polyhedron consists of one point (with no edges) and resides in a zero dimensional space. It's projection on the line segment is but one point (which, from the above constraints, is at the center of the line segment by definition) and no variations in the distribution of any kind are possible. Neither is it possible to define length within that "universe". It follows trivially that every conceivable distribution of a lone point on a line segment where the center of the distribution is defined to be the center of the line segment is achievable by a particular rotational orientation of the polyhedron (of which there are none). Thus the theorem is valid for n=0 (or at least can be interpreted in a way which makes it valid). I said it was trivial; it is only here for continuity in that it lets me begin with one point.

 

 

Case n=1 is also trivial as the polyhedron consists of two points and one edge residing in a one dimensional space. Since the edge is to have unit length, one point must be a half unit from the center of the polyhedron and the other must be a half unit from the center in the opposite direction. Since rotation is defined as the trigonometric conversion of one axis of reference into another, rotation can not exist in a one dimensional space. It follows that our projection will consist of two points on our line segment. We can now define both a center (defined as the midpoint between the two points) and a length (define it to be the distance between the two points) in this universe but there is utterly no use for our length definition because there are no other lengths to measure. It follows trivially that every conceivable distribution of two points on a line segment (which is one) is achievable by a particular rotational orientation of the polyhedron (of which there are none). Thus the theorem is valid for n=1.

 

 

Case n=2 is the first case which is not utterly trivial. Fabrication of an equilateral n dimensional polyhedron is not (in general) a trivial endeavor. In order to keep our life simple, let us construct our equilateral polyhedron in such a manner so as to make the initial orientation of the lower order polyhedron orthogonal to the added dimension. Thus we can move the lower order entity up from the center of our new coordinate axis and add a new point on the new axis below the center. In this case, the coordinates of previous polyhedron (as displayed in the n Euclidean space) remain exactly what they were for the n-1 established coordinates and are all shifted by the same distance from zero along the new axis. The new point has a position zero in all the old coordinates (it is on the new axis) and an easily calculated position on in the negative direction on that new axis (that distance must be equal to the new radius of the vertices of the old polyhedron as measured in the new n dimensional space).

 

 

The proper movement is quite easy to calculate. Consider a plane through the new axis and a line through any vertex on the lower order polyhedron. If we call the new axis the x axis and the line through the chosen vertex the y axis, the y position of that vertex will be the old radius of the vertex in the old polyhedron. The new radius will be given by the square root of the sum of the old radius squared and the distance the old polyhedron was moved up in the new dimension squared. That is exactly the same distance the new point must be from the new center. Assuring the new edge length will be unity imposes a second Pythagorean constraint consisting of the fact that the old radius squared plus (the new radius plus the distance the old polyhedron was moved up) squared must be unity.

 

 

[math]r_n = \sqrt{x_{up}^2 + r_{n-1}^2}[/math]

 

 

and

 

 

[math] 1 = \sqrt{r_{n-1}^2 + (x_{up} + r_n)^2 }[/math]

 

 

The solution of this pair of equations is given by

 

 

[math]r_n = \sqrt{\frac{n}{2(n+1)}} [/math]

 

 

and

 

 

[math]r_{up} = \frac{1}{\sqrt{2n(n+1)}}[/math]

 

 

 

The case n=0 was a single point in a zero dimensional space. The case n=1 can be seen as an addition of one dimension x_1 (orthogonal to nothing) where point #1 was moved up one half unit in the new dimension and a point #2 was added at minus one half in the new dimension (both the new radius and "distance to be moved up" are one half). The case n=2 changes the radius to one over the square root of three and the line segment (the result of case n=1) must be moved up exactly one half that amount. A little geometry should convince you that the result is exactly an equilateral triangle with a unit edge length. Projection of this entity upon a line segment yields three points and the relative positions of the three points are changed by rotation of that triangle.

 

 

In this case, we have two points to use as a length reference and a third point who's distance from the center can be specified in terms of that defined length reference. Using those definitions, two of the points can be defined to be one unit apart and the third point's position can vary from any specific position from plus infinity to minus infinity. The infinities are approached when the edge defined by the two vertices being used as our length reference approaches orthogonality to the line segment upon which the triangle is being projected (in which case the defining unit of measure falls towards zero). Plus infinity would be when the third point is on the right (by convention) and minus infinity when the third point is on the left (by common convention, right is usually taken to be positive and left to be negative). It thus follows that every conceivable distribution of three points on a line segment is achievable by a particular rotational orientation of the polyhedron (our triangle). Thus the theorem is valid for n=2.

 

 

Case n=3 consists of a three dimensional equilateral polyhedron consisting of four points, six unit edges and four triangle faces: i.e., what is commonly called a tetrahedron. If you wish you may show that the radius of vertices is given by one half the square root of three halves and the altitude by the radius plus one over two times the square root of six (as per the equations given above).

 

 

In examining the consequences of rotation, to make life easy, begin by considering a configuration where a line between the center of our tetrahedron and one vertex is parallel to the axis of projection on our reference line segment. Any and all rotations around that axis will leave that vertex at the center of our line segment and actually consist of rotation in the plane of the face opposite to that point. Essentially, except for that particular point, we obtain exactly the same results which were obtained in case n=2 (that would be projection of the triangle face opposite the chosen vertex). Using two of the points on that face to specify length, we can find an orientation which will yield the third point in any position from minus infinity to plus infinity while the forth point remains at the center of the reference line segment.

 

 

Having performed that rotation, we can rotate the tetrahedron around an axis orthogonal to the first rotational axis and orthogonal to the line on which the projection is being made. This rotation will end up doing nothing to the projection of the first three points except to uniformly scale their distance from the center. Since we have defined length in terms of two of those points, the referenced configuration obtained from the first rotation does not change at all. On the other hand, the forth point (which was projected to the center point) will move from the center towards plus or minus infinity depending on the rotation direction (the infinite positions will correspond to the orientation where the line of projection lies in that face opposite the fourth point: i.e., the scaled reference distance approaches zero). It follows that all possible configurations of the four points in our projection can be reached via rotations of the tetrahedron and the theorem is valid for n=3.

 

 

The final part of the proof (if it is true for an n-1 dimensional figure, it is true for an n dimensional figure) requires a little thought:

 

 

Since the space in which the n dimensional polyhedron is embedded is Euclidean, we can specify a particular orientation of that polyhedron by listing the n coordinates of each vertex. That coordinate system may have any orientation with respect to the orientation of the polyhedron. That being the case, we are free to set our coordinate system to have one axis (we can call it the x axis) parallel to the line on which the projection is to be made. In that case, except for a scale factor (which must be obtained from the distribution), a list of the x coordinates of each point correspond exactly to the apparent positions of the projected points on our reference line. Thus I will henceforth use the x axis in the n dimensional space as a surrogate for my reference line segment.

 

 

If the theorem is true for an n-1 dimensional polyhedron, there exists an orientation of that polyhedron which will correspond to any specific distribution of n points on a line (where scale is established via some procedure internal to that distribution of points). If that is the case, we can add another axis orthogonal to all n-1 axes already established, move that polyhedron up along that new axis a distance equal to

 

 

[math]x_n = \frac{1}{\sqrt{2n(n+1)}}[/math]

 

 

 

and add a new point at zero for every coordinate axis except the nth axis where the coordinate will be set to

 

 

[math]-r_n = x_n = - \sqrt{\frac{n}{2(n+1)}}[/math].

 

 

 

The result will be an n dimensional equilateral polyhedron with unit edge which will project to exactly the same distribution of points obtained from the previous n-1 dimensional polyhedron with one additional point at the center of our reference line segment.

 

 

If our n dimensional polyhedron is rotated on an axis perpendicular to both the reference line segment and the nth axis just added, the only effect on the original distribution will be to adjust the scale of every point via the scale factor [math]cos\theta[/math]. That is, the new x_i is obtained by

 

 

[math]x_{1i New} = x_{1i Old}cos\theta + \sqrt{\frac{n}{2(n+1)}}sin\theta [/math],

 

 

 

where theta is the angle of rotation (notice that the sin term yields a simple shift exactly the same for all points which is quite meaningless as far as the pattern of those points is concerned). Meanwhile, the x_1 position of the added point will be given exactly by [math]-r_n sin \theta[/math] (the cos term vanishes as it started on the origin of x_1). Once again, the added point may be moved to any position between plus and minus infinity which occur at plus and minus ninety degrees of rotation. Once again, the length scale is to be established via some procedure internal to the distribution of points. It follows that the theorem is valid for all possible n.

 

 

QED

 

 

There is an interesting corollary to the above proof. Notice that the rotation specified in the final paragraph changes only the components of the collection of vertices along the x axis and the nth axis. All other components of that collection of vertices remain exactly as they were. Since the order used to establish the coordinates of our polyhedron is immaterial to the resultant construct, the nth axis can be a line through the center of the polyhedron and any point except the first and second (which essentially establish the x axis under our current perspective). It follows that for any such n dimensional polyhedron for n greater than three (any x projection universe containing more than four points) there always exists n-2 axes orthogonal to both the x and y axes. These n-2 axes may be established in any orientation of interest so long as they are orthogonal to each other and the x,y plane. Thus, by construction, for any point (excepting the first and the second which establish the x axis) there exists an orientation of these n-2 axes such that one will be parallel to the line between that point and the center of the polyhedron. Any rotation in the plane of that axis and the y axis will do nothing but scale the y components of all the points and move that point through the collection, making no change whatsoever in the projection on the x axis.

 

 

We can go one step further. Within those n-2 axes orthogonal to the x and y axes, one can choose one to be the z axis and still have n-3 definable planes orthogonal to both the x and the y axes. That provides one with n-3 possible rotations which will leave the projections on the x and y axes unchanged. Since, in the construction of our polyhedron no consequences of rotation had any effect until we got to rotations after addition of the third point, these n-3 possible rotations are sufficient to obtain any distribution of projected points on the z axis without altering the established projections on the x and y axes.

 

 

Thus it is seen that absolutely any three dimensional universe consisting of n+1 points for n greater than four can be seen as an n dimensional equilateral polyhedron with unit edges projected on a three dimensional space. That any means absolutely any configuration of points conceivable. Talk about "emergent" phenomena, this picture is totally open ended. Any collection of points can be so represented! Consider the republican convention at noon of the second day (together with every object and every person in the rest of the world; and all the planets; and all the galaxies ...) where the collection of the positions of all the fundamental particles in the universe is no more than a projection of some n dimensional equilateral polyhedron of unit size on a three dimensional space. Talk about emergent phenomena!

 

 

On top of that, if nothing in the universe can move instantaneously from one position to another, it follows that the future (another distribution of that collection of positions of all the fundamental particles in the universe) is no more than another orientation of that n dimensional polyhedron and the evolution of the universe in every detail must correspond to continuous rotation of that figure. Think about that view of a rather simple geometric construct and the complex phenomena which is directly emergent from the fundamental perspective.

 

 

Have fun -- Dick

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(The following has been corrected for typo errors in the original)

 

 

With regard to "weak emergence" (that is with regard to the definition of "weak emergence") I feel it can also be dispensed with via the following proof. That is, emergence is emergence and there is nothing either weak or strong about it! People begin to think about weak "emergence" when what they are looking at is more complex than what they can deduce from "known laws". What they seem to forget is that physics is applicable only to problems which can be reduced to one body problems by some procedure.

 

 

I am of the opinion that the following proof is of great significance when one goes to consider "emergent" phenomena and the complexity achievable from simple constructs. The proof concerns a careful examination of the projection of a trivial geometric structure on a one dimensional line element.

 

 

The underlying structure will be an n dimensional rigid entity defined by a collection of n+1 points connected by lines (edges) of unit length embedded in an n dimensional Euclidean space (i.e., a minimal n dimensional equilateral polyhedron; the generalized concept of a higher dimensional equilateral triangle with unit edges). The universe (the collection of information to be analyzed) of interest will be the projection of the vertices of a that polyhedron on a one dimensional line element. The logic of the analysis will follow the standard inductive approach: i.e., prove a result for the cases n=0, 1, 2 and 3. Thereafter prove that if the description of the consequence is true for n-1 dimensions, it is also true for n dimensions. The result bears very strongly on the range of complexity of "emergent" phenomena given an extremely simple source.

 

 

First of all, the projection will consist of a collection of points (one for each vertex of that polyhedron) on the line segment of interest. Since motion of that polyhedron parallel to the given line segment is no more than uniform movement of every projected point, we can define the projection of the center of the polyhedron to be the center of the line segment: i.e., linear motion of the polyhedron has no real consequences. Furthermore, as the projection will be orthogonal to that line segment and the n dimensional space is Euclidean, any motion orthogonal to that line segment introduces no change in the projection whatsoever. It follows that the only motion of the polyhedron which provides any interesting changes in the distribution of points on the line segment will be rotations of the polyhedron in the n dimensional space.

 

 

The assertion which will be proved is that every conceivable distribution of points on the line segment is achievable by a specifying a particular rotational orientation of the polyhedron relative to the line segment of interest. Before we proceed to the proof, one issue of significance must be brought up. That issue concerns the scalability of the distribution. I referred to the collection of points on the line segment as the "universe of interest" as I want the student to think of that distribution of points as a universe: i.e., any definition of length must be arrived at via some defined characteristic of the the distribution itself or some subset of the distribution. Thus any two distributions which differ only by a scale factor will be considered to be identical distributions.

 

 

Case n=0 is trivial as the polyhedron consists of one point (with no edges) and resides in a zero dimensional space. It's projection on the line segment is but one point (which, from the above constraints, is at the center of the line segment by definition) and no variations in the distribution of any kind are possible. Neither is it possible to define length within that "universe". It follows trivially that every conceivable distribution of a lone point on a line segment where the center of the distribution is defined to be the center of the line segment is achievable by a particular rotational orientation of the polyhedron (of which there are none). Thus the theorem is valid for n=0 (or at least can be interpreted in a way which makes it valid). I said it was trivial; it is only here for continuity in that it lets me begin with one point.

 

 

Case n=1 is also trivial as the polyhedron consists of two points and one edge residing in a one dimensional space. Since the edge is to have unit length, one point must be a half unit from the center of the polyhedron and the other must be a half unit from the center in the opposite direction. Since rotation is defined as the trigonometric conversion of one axis of reference into another, rotation can not exist in a one dimensional space. It follows that our projection will consist of two points on our line segment. We can now define both a center (defined as the midpoint between the two points) and a length (define it to be the distance between the two points) in this universe but there is utterly no use for our length definition because there are no other lengths to measure. It follows trivially that every conceivable distribution of two points on a line segment (which is one) is achievable by a particular rotational orientation of the polyhedron (of which there are none). Thus the theorem is valid for n=1.

 

 

Case n=2 is the first case which is not utterly trivial. Fabrication of an equilateral n dimensional polyhedron is not (in general) a trivial endeavor. In order to keep our life simple, let us construct our equilateral polyhedron in such a manner so as to make the initial orientation of the lower order polyhedron orthogonal to the added dimension. Thus we can move the lower order entity up from the center of our new coordinate axis and add a new point on the new axis below the center. In this case, the coordinates of previous polyhedron (as displayed in the n Euclidean space) remain exactly what they were for the n-1 established coordinates and are all shifted by the same distance from zero along the new axis. The new point has a position zero in all the old coordinates (it is on the new axis) and an easily calculated position on in the negative direction on that new axis (that distance must be equal to the new radius of the vertices of the old polyhedron as measured in the new n dimensional space).

 

 

The proper movement is quite easy to calculate. Consider a plane through the new axis and a line through any vertex on the lower order polyhedron. If we call the new axis the x axis and the line through the chosen vertex the y axis, the y position of that vertex will be the old radius of the vertex in the old polyhedron. The new radius will be given by the square root of the sum of the old radius squared and the distance the old polyhedron was moved up in the new dimension squared. That is exactly the same distance the new point must be from the new center. Assuring the new edge length will be unity imposes a second Pythagorean constraint consisting of the fact that the old radius squared plus (the new radius plus the distance the old polyhedron was moved up) squared must be unity.

 

 

 

7b614f91c0577967fd6fd04358c0c778-1.png

 

 

and

 

 

1cb79abd082753edc914a9d0d9801ce7-1.png

 

 

The solution of this pair of equations is given by

 

 

5584b695a05cc7d0d29a30df254febe3-1.png

 

 

and

 

 

b4a48263d5ea72ba63fbaba4cf9c97c4-1.png

 

 

 

 

The case n=0 was a single point in a zero dimensional space. The case n=1 can be seen as an addition of one dimension x_1 (orthogonal to nothing) where point #1 was moved up one half unit in the new dimension and a point #2 was added at minus one half in the new dimension (both the new radius and "distance to be moved up" are one half). The case n=2 changes the radius to one over the square root of three and the line segment (the result of case n=1) must be moved up exactly one half that amount. A little geometry should convince you that the result is exactly an equilateral triangle with a unit edge length. Projection of this entity upon a line segment yields three points and the relative positions of the three points are changed by rotation of that triangle.

 

 

In this case, we have two points to use as a length reference and a third point who's distance from the center can be specified in terms of that defined length reference. Using those definitions, two of the points can be defined to be one unit apart and the third point's position can vary from any specific position from plus infinity to minus infinity. The infinities are approached when the edge defined by the two vertices being used as our length reference approaches orthogonality to the line segment upon which the triangle is being projected (in which case the defining unit of measure falls towards zero). Plus infinity would be when the third point is on the right (by convention) and minus infinity when the third point is on the left (by common convention, right is usually taken to be positive and left to be negative). It thus follows that every conceivable distribution of three points on a line segment is achievable by a particular rotational orientation of the polyhedron (our triangle). Thus the theorem is valid for n=2.

 

 

Case n=3 consists of a three dimensional equilateral polyhedron consisting of four points, six unit edges and four triangle faces: i.e., what is commonly called a tetrahedron. If you wish you may show that the radius of vertices is given by one half the square root of three halves and the altitude by the radius plus one over two times the square root of six (as per the equations given above).

 

 

In examining the consequences of rotation, to make life easy, begin by considering a configuration where a line between the center of our tetrahedron and one vertex is parallel to the axis of projection on our reference line segment. Any and all rotations around that axis will leave that vertex at the center of our line segment and actually consist of rotation in the plane of the face opposite to that point. Essentially, except for that particular point, we obtain exactly the same results which were obtained in case n=2 (that would be projection of the triangle face opposite the chosen vertex). Using two of the points on that face to specify length, we can find an orientation which will yield the third point in any position from minus infinity to plus infinity while the forth point remains at the center of the reference line segment.

 

 

Having performed that rotation, we can rotate the tetrahedron around an axis orthogonal to the first rotational axis and orthogonal to the line on which the projection is being made. This rotation will end up doing nothing to the projection of the first three points except to uniformly scale their distance from the center. Since we have defined length in terms of two of those points, the referenced configuration obtained from the first rotation does not change at all. On the other hand, the forth point (which was projected to the center point) will move from the center towards plus or minus infinity depending on the rotation direction (the infinite positions will correspond to the orientation where the line of projection lies in that face opposite the fourth point: i.e., the scaled reference distance approaches zero). It follows that all possible configurations of the four points in our projection can be reached via rotations of the tetrahedron and the theorem is valid for n=3.

 

 

The final part of the proof (if it is true for an n-1 dimensional figure, it is true for an n dimensional figure) requires a little thought:

 

 

Since the space in which the n dimensional polyhedron is embedded is Euclidean, we can specify a particular orientation of that polyhedron by listing the n coordinates of each vertex. That coordinate system may have any orientation with respect to the orientation of the polyhedron. That being the case, we are free to set our coordinate system to have one axis (we can call it the x axis) parallel to the line on which the projection is to be made. In that case, except for a scale factor (which must be obtained from the distribution), a list of the x coordinates of each point correspond exactly to the apparent positions of the projected points on our reference line. Thus I will henceforth use the x axis in the n dimensional space as a surrogate for my reference line segment.

 

 

If the theorem is true for an n-1 dimensional polyhedron, there exists an orientation of that polyhedron which will correspond to any specific distribution of n points on a line (where scale is established via some procedure internal to that distribution of points). If that is the case, we can add another axis orthogonal to all n-1 axes already established, move that polyhedron up along that new axis a distance equal to

 

 

 

1c594e53791c6c61ef5003c36e166258-1.png

 

 

 

 

and add a new point at zero for every coordinate axis except the nth axis where the coordinate will be set to

 

 

 

a3452c69fd935527e8d5dd465d4dd3b6-1.png.

 

 

 

 

The result will be an n dimensional equilateral polyhedron with unit edge which will project to exactly the same distribution of points obtained from the previous n-1 dimensional polyhedron with one additional point at the center of our reference line segment.

 

 

If our n dimensional polyhedron is rotated on an axis perpendicular to both the reference line segment and the nth axis just added, the only effect on the original distribution will be to adjust the scale of every point via the scale factor 2c98dfeb901d979b5f090b18e0cd00ff-1.png. That is, the new x_i is obtained by

 

 

 

2aad1352ba9cecf51e0189e41c836867-1.png,

 

 

 

 

where theta is the angle of rotation (notice that the sin term yields a simple shift exactly the same for all points which is quite meaningless as far as the pattern of those points is concerned). Meanwhile, the x_1 position of the added point will be given exactly by c6615716d81ba13e5aa824180ee5b4a3-1.png (the cos term vanishes as it started on the origin of x_1). Once again, the added point may be moved to any position between plus and minus infinity which occur at plus and minus ninety degrees of rotation. Once again, the length scale is to be established via some procedure internal to the distribution of points. It follows that the theorem is valid for all possible n.

 

 

QED

 

 

There is an interesting corollary to the above proof. Notice that the rotation specified in the final paragraph changes only the components of the collection of vertices along the x axis and the nth axis. All other components of that collection of vertices remain exactly as they were. Since the order used to establish the coordinates of our polyhedron is immaterial to the resultant construct, the nth axis can be a line through the center of the polyhedron and any point except the first and second (which essentially establish the x axis under our current perspective). It follows that for any such n dimensional polyhedron for n greater than three (any x projection universe containing more than four points) there always exists n-2 axes orthogonal to both the x and y axes. These n-2 axes may be established in any orientation of interest so long as they are orthogonal to each other and the x,y plane. Thus, by construction, for any point (excepting the first and the second which establish the x axis) there exists an orientation of these n-2 axes such that one will be parallel to the line between that point and the center of the polyhedron. Any rotation in the plane of that axis and the y axis will do nothing but scale the y components of all the points and move that point through the collection, making no change whatsoever in the projection on the x axis.

 

 

We can go one step further. Within those n-2 axes orthogonal to the x and y axes, one can choose one to be the z axis and still have n-3 definable planes orthogonal to both the x and the y axes. That provides one with n-3 possible rotations which will leave the projections on the x and y axes unchanged. Since, in the construction of our polyhedron no consequences of rotation had any effect until we got to rotations after addition of the third point, these n-3 possible rotations are sufficient to obtain any distribution of projected points on the z axis without altering the established projections on the x and y axes.

 

 

Thus it is seen that absolutely any three dimensional universe consisting of n+1 points for n greater than four can be seen as an n dimensional equilateral polyhedron with unit edges projected on a three dimensional space. That any means absolutely any configuration of points conceivable. Talk about "emergent" phenomena, this picture is totally open ended. Any collection of points can be so represented! Consider the republican convention at noon of the second day (together with every object and every person in the rest of the world; and all the planets; and all the galaxies ...) where the collection of the positions of all the fundamental particles in the universe is no more than a projection of some n dimensional equilateral polyhedron of unit size on a three dimensional space. Talk about emergent phenomena!

 

 

On top of that, if nothing in the universe can move instantaneously from one position to another, it follows that the future (another distribution of that collection of positions of all the fundamental particles in the universe) is no more than another orientation of that n dimensional polyhedron and the evolution of the universe in every detail must correspond to continuous rotation of that figure. Think about that view of a rather simple geometric construct and the complex phenomena which is directly emergent from the fundamental perspective.

 

 

Have fun -- Dick

 

 

 

1) I have reproduced this load of crap to avoid letting you edit it after the fact. But it is still just a load. There is absolutely nothing here of any mathematical or physical interest. In fact it is not clear that there is even a lucid thought.

 

2) To anyone interested in simplicial conplexes in n-space, and the application to topology, I direct them to CRF Maunder's book Algebraic Topology. One might also profitably read Convex Polytopes by Grunbaum or Regular Polytopes by Coxeter.

 

3) You are well named, Dick.

 

4) Have you met owl ?

Edited by DrRocket

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... I direct them to CRF Maunder's book Algebraic Topology.

 

That one is already on my shelf. I think it is one of the easier books on algebraic topology to read. Maybe one to add to my blog...

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ajb - talking about your reviews, could you (a/o DocRock) recommend an introductory text(s) for the maths needed for physics? If I had to give an idea of complexity I would say around the level of preparing a student with maths and science a' levels (18 yo final school exams for non-Brits) for their first undergraduate year - or possibly at the easier end of the range of undergraduate texts.

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ajb - talking about your reviews, could you (a/o DocRock) recommend an introductory text(s) for the maths needed for physics? If I had to give an idea of complexity I would say around the level of preparing a student with maths and science a' levels (18 yo final school exams for non-Brits) for their first undergraduate year - or possibly at the easier end of the range of undergraduate texts.

 

For basic mathematics anything with a title like mathematics for engineers and scientists will be a great place to start. From there, depending on your tastes and interests you can find more focused books.

 

The two books by Glyn James are good: Modern Engineering Mathematics and Advanced Modern Engineering Mathematics. I also like Engineering Mathematics: A Programmed Approach by C.W. Evans. Jenny Olive's book, Maths: A Student's Survival Guide: A Self-Help Workbook for Science and Engineering Students comes highly recommended, though I personally have not used this one.

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Jenny Olive's book is great - been there, seen it, and have the t-shirt - it was the next step I was after; I will check out Glyn James.

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ajb - talking about your reviews, could you (a/o DocRock) recommend an introductory text(s) for the maths needed for physics? If I had to give an idea of complexity I would say around the level of preparing a student with maths and science a' levels (18 yo final school exams for non-Brits) for their first undergraduate year - or possibly at the easier end of the range of undergraduate texts.

 

 

Paul Bamberg and Schlolmo Sternberg -- A Course in Mathematics for Students of Physics

 

This is an excellent undergraduate text aimed at future physicists. It does require introductory calculus. It is excellent from a pedagogical perspective, does not require much background beyond calculus and gets to the heart of things needed by physicists. Sternberg in particular is quite famous for mathematics related to physics. His text on differential geometry is a classic.

 

For students below that level, what is needed is simply to come to understand calculus at the introductory level and any calculus text is applicable. However, calculus texts are notoriously poor and the only calculus text that I can recommend without reservation is Calculus by Mike Spivak -- but there are as many opinions on calculus texts and there are people to hold those opinions.

 

For basic mathematics anything with a title like mathematics for engineers and scientists will be a great place to start. From there, depending on your tastes and interests you can find more focused books.

 

The two books by Glyn James are good: Modern Engineering Mathematics and Advanced Modern Engineering Mathematics. I also like Engineering Mathematics: A Programmed Approach by C.W. Evans. Jenny Olive's book, Maths: A Student's Survival Guide: A Self-Help Workbook for Science and Engineering Students comes highly recommended, though I personally have not used this one.

 

Not all such books are a good place to start. Kreyszig's Advanced Engineering Mathematics is just awful. I know one guy who is acknowleldged in the forward. He was consulted by the publisher, and his advice to the publlisher was "don't". He also wrote a book on functional analysis -- I was given a copy by a friend, but I think that in this case "free" is far too expensive.

 

I don't know a lot about Wyle's book, also widely used in the U.S., but I know a bit about Wyle, and am not impressed.

 

Arfken's Mathematical Methods for Physicists, on the other hand, is pretty good.

 

I don't know the books by Glyn James. If you like them I imagine that they are superb.

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Paul Bamberg and Schlolmo Sternberg -- A Course in Mathematics for Students of Physics

 

This is an excellent undergraduate text aimed at future physicists. It does require introductory calculus. It is excellent from a pedagogical perspective, does not require much background beyond calculus and gets to the heart of things needed by physicists. Sternberg in particular is quite famous for mathematics related to physics. His text on differential geometry is a classic.

 

For students below that level, what is needed is simply to come to understand calculus at the introductory level and any calculus text is applicable. However, calculus texts are notoriously poor and the only calculus text that I can recommend without reservation is Calculus by Mike Spivak -- but there are as many opinions on calculus texts and there are people to hold those opinions.

 

 

 

Not all such books are a good place to start. Kreyszig's Advanced Engineering Mathematics is just awful. I know one guy who is acknowleldged in the forward. He was consulted by the publisher, and his advice to the publlisher was "don't". He also wrote a book on functional analysis -- I was given a copy by a friend, but I think that in this case "free" is far too expensive.

 

I don't know a lot about Wyle's book, also widely used in the U.S., but I know a bit about Wyle, and am not impressed.

 

Arfken's Mathematical Methods for Physicists, on the other hand, is pretty good.

 

I don't know the books by Glyn James. If you like them I imagine that they are superb.

 

 

Many thanks

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That one is already on my shelf. I think it is one of the easier books on algebraic topology to read. Maybe one to add to my blog...

 

You might also consider Singer and Thorpe's Lecture Notes on Elementary Topology and Geometry

Massey's Algebraic Topology: An Introduction. The former requires no real background and is a true undergraduate text, while the latter requires just a bit of point-set toplogy. Both are very accessible.

 

Many thanks

 

General tip (of course there are exceptions) -- quite often the very best mathematics books are written by the guys with the biggest baddest reputations. They tend to have those reputations because they have deep insight and can get to the heart of the matter quickly and clearly. Sternberg and

Singer and two such guys and their books show it.

 

One of the very best and most clear talks that I ever sat through was given by John Milnor -- and mathematicians don't come any better or with a more impressive reputation for deep research. On the other hand, God save me from talks by brand-new PhDs.

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