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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 g

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".
smoyer 12y ago
It's a foreign language ... but you can't simply learn words one at a time as it won't make sense until you know most of it.

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".

smoyer 12y ago

It's a foreign language ... but you can't simply learn words one at a time as it won't make sense until you know most of it.