6 ms·

The Privileged Character of 3+1 Spacetime (2011)

32 points by Bootvis 12y ago

transfire 12y ago
The problem with this analysis is that it is basing its anthropic conclusion on anthropic physical theory. In other words, it compares what would happen if the number of dimensions changed using theories devised for a 3+1 universe. Obviously if the dimensions changed, the theories have to change. And in an alternate universe with a different number of dimensions the theories would not be the same.
- aetherson 12y ago
  Well, that's always the problem with anthropic arguments. But "electrons can not maintain orbitals" sounds pretty damning. Could we imagine that there is some other kind of stable state that allows progression up from fundamental particles to complex macroscopic objects big enough to be some kind of observer? Yeeessss.... but that would be more convincing if we had some idea of what fundamental-particle relationship that would be.
- nhaehnle 12y ago
  I agree. I like this topic, but I have to say that there are better, more fundamental arguments in favour of 3 space dimensions. For example, two space dimensions only allow planar graphs, which seems pretty limiting for the development of complex structures. On the other hand, as you increase the number of spatial dimensions, the curse of dimensionality becomes more important and space in general becomes "emptier". Also, knots are only interesting in 3 dimensions. Arguing about time dimensions is harder, because time is tied more strongly to our subjective experience of consciousness, which is something that we really don't understand yet.
drostie 12y ago
Another fun fact is that the +−−− metric allows relativity to be described in a spinor calculus (which are in turn isomorphic to quaternions). The basic idea looks like this: annex the identity matrix as σ_0 = onto the 3 Pauli spin matrices σ_1, σ_2, σ_3 and then form the product with any 4-vector v^μ to find that every 4-vector is a 2x2 Hermitian matrix, V = v^μ σ_μ. Of course there's nothing special about the +−−− metric here; but when you calculate the determinant of V you find: det V = v^μ v_μ so that the determinant is the Lorentz metric. This means that Lorentz transforms preserve determinants in this system. There is the 4-reversal C = \v → -v as well as the parity transform P = \v → (det v) v^(-1), with time reflection T = CP as usual. Lorentz transforms which do not have those reflections in them instead have some complex matrix det L = 1 such that they take the form \v → L v L†. Now think about the vectors in this system. A good way to poke inside the vectors is to look at matrices which are projections, det V = 0 so that V = [α] [α* β*] = u u† [β] If you actually work it out, things like α/β work out to be stereographic projections of that null vector on the celestial sphere. The Lorentz transform takes the vector u above to the vector L u. So when you look at, say, incoming light rays to a point (say, yourself), those points are shifted by this L matrix, performing a bilinear/Mobius transform of the incoming light rays. By ignoring rotations (which are unitary matrices!) and going for boosts, you can quickly convince yourself that as you boost in a direction, the stars all "tilt forwards", and you can dream that maybe if you were a photon, in addition to living a timeless existence, maybe only the event which emitted you is behind you, while the entire universe lies in front of you.
- HCIdivision17 12y ago
  This is a classic example of something I feel I should be able to follow and understand, but in fact cannot interpret as anything other than technobabble. I'm going to point to this as an example of how there are different levels of familiarity and expertise. And to be sure: this seems to be a fantastic comment. I'm going to spend some time slowly working through it until I grok the terms.
  - axus 12y ago
    σ_0, indeed
  - v64 12y ago
    That's the power of mathematical notation: lots of very complex ideas compressed down into a minimal number of characters. I have a math degree and can follow along, but I would need to pull out a few of my old algebra textbooks to really grok what's happening in an intuitive sense.
  - semi-extrinsic 12y ago
    If it makes you feel any better, this would take up at least one two-hour lecture in a decent graduate-level university course on GR. If you are going to try and comprehend it on you own, I am personally fond of Simmonds' "A brief on tensor analysis" as an introduction to the way vectors are understood here.
  - drostie 12y ago
    Okay, so, four-vectors have four components: v0, v1, v2, v3. The last three components are "spatial" and the first is "temporal". An example of a four-vector is a 4-position of a particle moving from position (0, 0, 0) with speed u in the x-direction: r0 = c * t r1 = u * t r2 = 0 r3 = 0 This is actually a family of points, one for each number t. The 4-vector is a thing which everyone can agree on. (Mind you, you might disagree on the components of the 4-vector, but everyone agrees on the "arrow" in 4d space which the vector represents. That's a little abstract and philosophical but it means that there is one objective reality which people are seeing in different ways. In particular if one person says "these two particles collided at this place and time," everybody else has some place and time where they also saw those two particles colliding.) In particular, everyone agrees on a number based on the 4-vector, which is the number v0² − v1² − v2² − v3². This is called the "magnitude" of V under the "Lorentz metric". In our case above, the magnitude is (c² − u²) t². In the particle's own "rest frame" (v1 = v2 = v3 = 0) it must agree with these coordinates so we must find that its coordinates in that frame are: [√(c² − u²) t, 0, 0, 0]. But of course this must be equal to [c τ, 0, 0, 0] for the particle's own time coordinate τ (Greek letter tau), because everybody agrees on the speed of light c and in this formalism it's a conversion factor between space-dimensions and time dimensions. The conclusion is that t = τ / √(1 − u²/c²). This conversion factor occurs so much that it's usually just written with a γ (Greek letter gamma) as t = γ τ. The general transform which preserves the Lorentz metric in this direction is the "x-boost": v0' = γ * (v0 − (u/c) * v1) v1' = γ * (v1 − (u/c) * v0) v2' = v2 v3' = v3 We can perform this "boost" twice, once with speed u = +w, then again with speed u = -w, to find that v0'' = v0, v1'' = v1, which proves that the mathematics is totally consistent. We can also compute (v0')² − (v1')² to confirm that it is the same as v0² − v1². There are other possibilities too for preserving the Lorentz metric: there are 3d space rotations with no change to v0, and boosts in other directions, and combinations of the two (what are called 4-screws). There are also reflections (flipping vi to -vi). But that's about it. When we want to ignore the reflections we talk about "proper" Lorentz transforms, with the idea that reflections are somehow improper (mostly because they're not a continuous family of transforms). All of the stuff about the Pauli matrices is basically about forming the 2x2 complex matrix V = [v0 + v3 v1 − i v2] [v1 + i v2 v0 − v3 ] The determinant of a matrix [a b; c d] is the product of its eigenvalues and can be calculated as a * d − b * c. This has a nice property: the determinant of a product of matrices is the product of the determinants; det(A * B) = det(A) * det(B). This matrix is also nice because it is Hermitian, which means that it is its own conjugate transpose. Hermitian matrices are complex matrices which have real eigenvalues. Every proper Lorentz transform can be thought of as a 2x2 complex (non-Hermitian in general) matrix L. We take its conjugate-transpose L† and convert the matrix V above to be V' = L V L†. This arrangement keeps V' Hermitian, and we can work out v0, v1, v2, and v3 from its above components. The only problem is that the Lorentz metric may have changed, unless det(V') = det(L) det(L†) det(V) = det(V). This amounts to demanding that det(L) = ±1. However, we can throw away the det(L) = -1 solutions as redundant, because if det(L) = -1 then det(i * L) = +1, but when we complex-conjugate that we get: L V L† = 1 * L V L† = (-i * i) L V L† = (i L) V (i L)† so they have the exact same effect on these four-vector matrices. The Lorentz boost in the v3 direction has a very nice Lorentz matrix which looks like: [e^(k/2) 0 ] [ 0 e^(-k/2)] Running through the motions gives v0' + v3' = e^k (v0 + v3), v0' − v3' = e^-k (v0 − v3), which works out to v0' = cosh(k) v0 + sinh(k) v3 while v3' = sinh(k) v0 + cosh(k) v3, where sinh and cosh are the hyperbolic sines/cosines. Conclusion: the v3 boost by a speed u is defined by that matrix with k = atanh(-u/c). Again, lightlike vectors (det V = 0) are projections of some 1x2 complex vector called a spinor, V = s s†. Expanding V out with s = [α; β] we find that β/α = (v1 + i v2) / (v0 + v3). This is essentially a stereographic projection. To fill in the gaps, let us assume that we've got future-pointing vectors (v0 > 0) which are "coming in towards us" and need to be projected to a point on a sphere at radius 1 from which they came. After scaling the null vector by dividing by R it has some components (1, -x/R, -y/R, -z/R) where (x, y, z) is the point on the sphere of radius R which the null vector seems to be "coming out of". Then -β/α is an actual stereographic projection of (x, y, z) onto the complex plane; if we take the last formula with R = 1 we get back just (x + i y) / (1 - z). Lorentz matrices then transform these components directly. In the case of the v3 boost, we're transforming -β/α to (-β/α) * e^k, so we're just scaling the complex plane by a constant, k = -atanh(u/c). If u > 0 then we are boosting towards the North pole and k is negative, so all of the stars that we see (lightlike vectors) get compressed in the stereoscopic projection towards z=0 on the complex plane. In other words, all of the stars tilt in the direction we're going. Final freebie: if something has a round outline when it is stationary, then it has a round outline when it's moving past you, because bilinear transforms map circles to circles. The image that you see within that circle is distorted (you can imagine a bullseye target, the interior circles "shift" towards one of the walls of the outer circle uniformly) but the outline is still circular. In this respect, Lorentz length contraction is "invisible".
abecedarius 12y ago
Greg Egan has a whole lot to say about 4+0 spacetime: http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGONAL.html http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGON...

32 points by Bootvis 12y ago

The problem with this analysis is that it is basing its anthropic conclusion on anthropic physical theory. In other words, it compares what would happen if the number of dimensions changed using theories devised for a 3+1 universe. Obviously if the dimensions changed, the theories have to change. And in an alternate universe with a different number of dimensions the theories would not be the same.

aetherson 12y ago
Well, that's always the problem with anthropic arguments. But "electrons can not maintain orbitals" sounds pretty damning. Could we imagine that there is some other kind of stable state that allows progression up from fundamental particles to complex macroscopic objects big enough to be some kind of observer? Yeeessss.... but that would be more convincing if we had some idea of what fundamental-particle relationship that would be.
nhaehnle 12y ago
I agree. I like this topic, but I have to say that there are better, more fundamental arguments in favour of 3 space dimensions. For example, two space dimensions only allow planar graphs, which seems pretty limiting for the development of complex structures. On the other hand, as you increase the number of spatial dimensions, the curse of dimensionality becomes more important and space in general becomes "emptier". Also, knots are only interesting in 3 dimensions. Arguing about time dimensions is harder, because time is tied more strongly to our subjective experience of consciousness, which is something that we really don't understand yet.

drostie 12y ago

Another fun fact is that the +−−− metric allows relativity to be described in a spinor calculus (which are in turn isomorphic to quaternions). The basic idea looks like this: annex the identity matrix as σ_0 = onto the 3 Pauli spin matrices σ_1, σ_2, σ_3 and then form the product with any 4-vector v^μ to find that every 4-vector is a 2x2 Hermitian matrix, V = v^μ σ_μ. Of course there's nothing special about the +−−− metric here; but when you calculate the determinant of V you find: det V = v^μ v_μ so that the determinant is the Lorentz metric. This means that Lorentz transforms preserve determinants in this system. There is the 4-reversal C = \v → -v as well as the parity transform P = \v → (det v) v^(-1), with time reflection T = CP as usual. Lorentz transforms which do not have those reflections in them instead have some complex matrix det L = 1 such that they take the form \v → L v L†. Now think about the vectors in this system. A good way to poke inside the vectors is to look at matrices which are projections, det V = 0 so that V = [α] [α* β*] = u u† [β] If you actually work it out, things like α/β work out to be stereographic projections of that null vector on the celestial sphere. The Lorentz transform takes the vector u above to the vector L u. So when you look at, say, incoming light rays to a point (say, yourself), those points are shifted by this L matrix, performing a bilinear/Mobius transform of the incoming light rays. By ignoring rotations (which are unitary matrices!) and going for boosts, you can quickly convince yourself that as you boost in a direction, the stars all "tilt forwards", and you can dream that maybe if you were a photon, in addition to living a timeless existence, maybe only the event which emitted you is behind you, while the entire universe lies in front of you.

HCIdivision17 12y ago
This is a classic example of something I feel I should be able to follow and understand, but in fact cannot interpret as anything other than technobabble. I'm going to point to this as an example of how there are different levels of familiarity and expertise. And to be sure: this seems to be a fantastic comment. I'm going to spend some time slowly working through it until I grok the terms.
- axus 12y ago
  σ_0, indeed
- v64 12y ago
  That's the power of mathematical notation: lots of very complex ideas compressed down into a minimal number of characters. I have a math degree and can follow along, but I would need to pull out a few of my old algebra textbooks to really grok what's happening in an intuitive sense.
- semi-extrinsic 12y ago
  If it makes you feel any better, this would take up at least one two-hour lecture in a decent graduate-level university course on GR. If you are going to try and comprehend it on you own, I am personally fond of Simmonds' "A brief on tensor analysis" as an introduction to the way vectors are understood here.
- drostie 12y ago
  Okay, so, four-vectors have four components: v0, v1, v2, v3. The last three components are "spatial" and the first is "temporal". An example of a four-vector is a 4-position of a particle moving from position (0, 0, 0) with speed u in the x-direction: r0 = c * t r1 = u * t r2 = 0 r3 = 0 This is actually a family of points, one for each number t. The 4-vector is a thing which everyone can agree on. (Mind you, you might disagree on the components of the 4-vector, but everyone agrees on the "arrow" in 4d space which the vector represents. That's a little abstract and philosophical but it means that there is one objective reality which people are seeing in different ways. In particular if one person says "these two particles collided at this place and time," everybody else has some place and time where they also saw those two particles colliding.) In particular, everyone agrees on a number based on the 4-vector, which is the number v0² − v1² − v2² − v3². This is called the "magnitude" of V under the "Lorentz metric". In our case above, the magnitude is (c² − u²) t². In the particle's own "rest frame" (v1 = v2 = v3 = 0) it must agree with these coordinates so we must find that its coordinates in that frame are: [√(c² − u²) t, 0, 0, 0]. But of course this must be equal to [c τ, 0, 0, 0] for the particle's own time coordinate τ (Greek letter tau), because everybody agrees on the speed of light c and in this formalism it's a conversion factor between space-dimensions and time dimensions. The conclusion is that t = τ / √(1 − u²/c²). This conversion factor occurs so much that it's usually just written with a γ (Greek letter gamma) as t = γ τ. The general transform which preserves the Lorentz metric in this direction is the "x-boost": v0' = γ * (v0 − (u/c) * v1) v1' = γ * (v1 − (u/c) * v0) v2' = v2 v3' = v3 We can perform this "boost" twice, once with speed u = +w, then again with speed u = -w, to find that v0'' = v0, v1'' = v1, which proves that the mathematics is totally consistent. We can also compute (v0')² − (v1')² to confirm that it is the same as v0² − v1². There are other possibilities too for preserving the Lorentz metric: there are 3d space rotations with no change to v0, and boosts in other directions, and combinations of the two (what are called 4-screws). There are also reflections (flipping vi to -vi). But that's about it. When we want to ignore the reflections we talk about "proper" Lorentz transforms, with the idea that reflections are somehow improper (mostly because they're not a continuous family of transforms). All of the stuff about the Pauli matrices is basically about forming the 2x2 complex matrix V = [v0 + v3 v1 − i v2] [v1 + i v2 v0 − v3 ] The determinant of a matrix [a b; c d] is the product of its eigenvalues and can be calculated as a * d − b * c. This has a nice property: the determinant of a product of matrices is the product of the determinants; det(A * B) = det(A) * det(B). This matrix is also nice because it is Hermitian, which means that it is its own conjugate transpose. Hermitian matrices are complex matrices which have real eigenvalues. Every proper Lorentz transform can be thought of as a 2x2 complex (non-Hermitian in general) matrix L. We take its conjugate-transpose L† and convert the matrix V above to be V' = L V L†. This arrangement keeps V' Hermitian, and we can work out v0, v1, v2, and v3 from its above components. The only problem is that the Lorentz metric may have changed, unless det(V') = det(L) det(L†) det(V) = det(V). This amounts to demanding that det(L) = ±1. However, we can throw away the det(L) = -1 solutions as redundant, because if det(L) = -1 then det(i * L) = +1, but when we complex-conjugate that we get: L V L† = 1 * L V L† = (-i * i) L V L† = (i L) V (i L)† so they have the exact same effect on these four-vector matrices. The Lorentz boost in the v3 direction has a very nice Lorentz matrix which looks like: [e^(k/2) 0 ] [ 0 e^(-k/2)] Running through the motions gives v0' + v3' = e^k (v0 + v3), v0' − v3' = e^-k (v0 − v3), which works out to v0' = cosh(k) v0 + sinh(k) v3 while v3' = sinh(k) v0 + cosh(k) v3, where sinh and cosh are the hyperbolic sines/cosines. Conclusion: the v3 boost by a speed u is defined by that matrix with k = atanh(-u/c). Again, lightlike vectors (det V = 0) are projections of some 1x2 complex vector called a spinor, V = s s†. Expanding V out with s = [α; β] we find that β/α = (v1 + i v2) / (v0 + v3). This is essentially a stereographic projection. To fill in the gaps, let us assume that we've got future-pointing vectors (v0 > 0) which are "coming in towards us" and need to be projected to a point on a sphere at radius 1 from which they came. After scaling the null vector by dividing by R it has some components (1, -x/R, -y/R, -z/R) where (x, y, z) is the point on the sphere of radius R which the null vector seems to be "coming out of". Then -β/α is an actual stereographic projection of (x, y, z) onto the complex plane; if we take the last formula with R = 1 we get back just (x + i y) / (1 - z). Lorentz matrices then transform these components directly. In the case of the v3 boost, we're transforming -β/α to (-β/α) * e^k, so we're just scaling the complex plane by a constant, k = -atanh(u/c). If u > 0 then we are boosting towards the North pole and k is negative, so all of the stars that we see (lightlike vectors) get compressed in the stereoscopic projection towards z=0 on the complex plane. In other words, all of the stars tilt in the direction we're going. Final freebie: if something has a round outline when it is stationary, then it has a round outline when it's moving past you, because bilinear transforms map circles to circles. The image that you see within that circle is distorted (you can imagine a bullseye target, the interior circles "shift" towards one of the walls of the outer circle uniformly) but the outline is still circular. In this respect, Lorentz length contraction is "invisible".

abecedarius 12y ago

Greg Egan has a whole lot to say about 4+0 spacetime: http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGONAL.html http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGON...