KJW

To check whether a parametric curve xμ(λ)x^\mu(\lambda)xμ(λ) is a geodesic, I’d compute the tangent vector dxμdλdxμdλdλdxμand check whether it satisfies the geodesic equation: d2xμdλ2+Γαβμdxαdλdxβdλ=0d2xμdλ2+Γμαβdxαdλdxβdλ=0dλ2d2xμ+Γαβμdλdxαdλdxβ=0 for all components μ\muμ. I’d also compute Undefined control sequence \gto determine if the geodesic is null. If the equation holds and the norm is zero, then it’s a null geodesic. If the equation doesn’t hold, the curve isn’t a geodesic. In short- plug the curve into the geodesic equation, verify it holds, and then check the norm to classify it as null, timelike, or spacelike. Nope. Although your LaTex has crapped out, you have said enough for me to see that you are wrong. Throughout our conversation, I have been alluding to a particular notion that you have failed to realise. Firstly, you have invoked the derivative: [math]\dfrac{dx^\mu}{d\lambda}[/math] saying that if: [math]\dfrac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = 0[/math] the curve is a geodesic, otherwise it is not. But the description of the curve I gave you did not mention [math]\lambda[/math]. It did not mention any parameter. Instead, it expressed three of the coordinates in terms of the fourth coordinate: [math]x^1 = x^1(x^0)\\x^2 = x^2(x^0)\\x^3 = x^3(x^0)[/math] It's a very common way to express a trajectory, although it is usually in the form: [math]x = x(t)\\y = y(t)\\z = z(t)[/math] where the spatial coordinates and time are not on equal footing. In general relativity, the formulation is such that the coordinates [math]x^0, x^1, x^2, x^3[/math] are on equal footing (with [math]x^0[/math] or sometimes [math]x^4[/math] representing a timelike coordinate), but: [math]x^1 = x^1(x^0)\\x^2 = x^2(x^0)\\x^3 = x^3(x^0)[/math] is still a valid description of a trajectory in spacetime, even though the four coordinates are not all on equal footing. However, in this case, it is straightforward to introduce a parameter [math]t[/math] (not necessarily representing time): [math]x^0 = t\\x^1 = x^1(t)\\x^2 = x^2(t)\\x^3 = x^3(t)[/math] This places the four coordinates on equal footing (if we regard [math]x^1(t), x^2(t), x^3(t)[/math] as specified functions of [math]t[/math], as is [math]t[/math] itself). Secondly, and this is the main point of our conversation, the parameter [math]t[/math] is not in general an affine parameter. Because the geodesic equation, as expressed above, requires an affine parameter, if the parameter is not affine, then: [math]\dfrac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} \neq 0[/math] does not imply that the curve is not a geodesic. Starting from the geodesic equation with affine parameter [math]\lambda[/math]: [math]\dfrac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = 0[/math] Under new parameter [math]t[/math]: [math]\ \ \ \ \lambda = \lambda(t)\ \ \ \ ; \ \ \ \ t = t(\lambda)[/math] One obtains after a number of steps (not shown unless requested): [math]\dfrac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} = f(t) \dfrac{dx^\mu}{dt}\ \ \ \ \ \text{where:}\ \ \ \ \ \dfrac{d^2 \lambda}{dt^2} - f(t) \dfrac{d\lambda}{dt} = 0[/math] Note that the change of parameter does not alter the curve itself, only the description of the curve. Thus, if the curve, expressed in terms of parameter [math]t[/math], satisfies the above equation for any function [math]f(t)[/math], it is a geodesic. On the other hand, if the vector: [math]\dfrac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt}[/math] is not zero and is not parallel to [math]\dfrac{dx^\mu}{dt}[/math], then the curve is not a geodesic. Note that the above applies to both null and non-null curves. [If the above LaTex doesn't render, please refresh browser.]

Nope. Although your LaTex has crapped out, you have said enough for me to see that you are wrong. Throughout our converation, I have been alluding to a particular notion that you have failed to realise. Firstly, you have invoked the derivative: [math]\dfrac{dx^\mu}{d\lambda}[/math] saying that if: [math]\dfrac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = 0[/math] the curve is a geodesic, otherwise it is not. But the description of the curve I gave you did not mention [math]\lambda[/math]. It did not mention any parameter. Instead, it expressed three of the coordinates in terms of the fourth coordinate: [math]x^1 = x^1(x^0)\\x^2 = x^2(x^0)\\x^3 = x^3(x^0)[/math] It's a very common way to express a trajectory, although it is usually in the form: [math]x = x(t)\\y = y(t)\\z = z(t)[/math] where the spatial coordinates and time are not on equal footing. In general relativity, the formulation is such that the coordinates [math]x^0, x^1, x^2, x^3[/math] are on equal footing (with [math]x^0[/math] or sometimes [math]x^4[/math] representing a timelike coordinate), but: [math]x^1 = x^1(x^0)\\x^2 = x^2(x^0)\\x^3 = x^3(x^0)[/math] is still a valid description of a trajectory in spacetime, even though the four coordinates are not all on equal footing. However, in this case, it is straightforward to introduce a parameter [math]t[/math] (not necessarily representing time): [math]x^0 = t\\x^1 = x^1(t)\\x^2 = x^2(t)\\x^3 = x^3(t)[/math] This places the four coordinates on equal footing (if we regard [math]x^1(t), x^2(t), x^3(t)[/math] as specified functions of [math]t[/math]). Secondly, and this is the main point of our converstion, the parameter [math]t[/math] is not in general an affine parameter. Because the geodesic equation as expressed above requires an affine parameter, if the parameter is not affine, then: [math]\dfrac{d^2 x^\mu}{dt} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} \neq 0[/math] does not imply that the curve is not a geodesic. Starting from the geodesic equation with affine parameter [math]\lambda[/math]: [math]\dfrac{d^2 x^\mu}{d\lambda^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{d\lambda} \dfrac{dx^\sigma}{d\lambda} = 0[/math] Under new parameter [math]t[/math]: [math]\ \ \ \ \lambda = \lambda(t)\ \ \ \ ; \ \ \ \ t = t(\lambda)[/math] One obtains after a number of steps (not shown unless necessary): [math]\dfrac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt} = f(t) \dfrac{dx^\mu}{dt}\ \ \ \ \ \text{where:}\ \ \ \ \ \dfrac{d^2 \lambda}{dt^2} - f(t) \dfrac{d\lambda}{dt} = 0[/math] Note that the change of parameter does not alter the curve itself, only the description of the curve. Thus, if the curve, expressed in terms of parameter [math]t[/math], satisfies the above equation for any function [math]f(t)[/math], it is a geodesic. On the other hand, if the vector: [math]\dfrac{d^2 x^\mu}{dt^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dt} \dfrac{dx^\sigma}{dt}[/math] is not parallel to [math]\dfrac{dx^\mu}{dt}[/math], then the curve is not a geodesic. Note that the above applies to both null and non-null curves. [If the above LaTex doesn't render, please refresh browser.]

Hmmm. It looks like I'm not going to be able to coax you into providing the answer I'm looking for with my current approach, so I'll try a different approach: Suppose you are given the following description of a curve in spacetime, along with the metric of that spacetime: [math]x^1 = x^1(x^0)\\x^2 = x^2(x^0)\\x^3 = x^3(x^0)[/math] Note that [math]x^1, x^2, x^3[/math] on the left side are coordinates of the points of the curve, whereas [math]x^1(x^0), x^2(x^0), x^3(x^0)[/math] on the right side are functions of the [math]x^0[/math] coordinate. Given that you have already obtained the Christoffel symbols from the metric, how would you determine whether or not the above description of the curve in spacetime is describing a geodesic, whether null or not? [If the above LaTex doesn't render, please refresh browser.]

I've discovered a benefit in the forum software not automatically rendering LaTex code under some circumstances: If one is quoting posts that contain LaTex, it is better to quote such posts while the LaTex is not rendered because quoting rendered LaTex can be problematic. The same is true for editing one's own post that contains LaTex - it is better to edit the post while the LaTex is not rendered because it is easier to edit the LaTex code of such posts, as well as editing posts with rendered LaTex can be problematic.

To obtain that value ultimately connects to a circle.

Here are the masses (MeV/c²) of the three charged leptons: electron: 0.51100 muon: 105.658 tauon: 1776.84 These are three closely related fundamental particles. If there is any pattern among the masses of particles, it should here among the charged leptons. [Source: https://en.wikipedia.org/wiki/Lepton#Table_of_leptons]

This assumes that you know, at least approximately, the value of π.

Perhaps you can elaborate on what an affine parameter is, particularly with regards to null geodesics, which cannot be parametrised with proper time or arc length. An affine parameter is a parameter along a geodesic that preserves the geodesic equation's form. For null geodesics, proper time τ is zero, so we can't use it. Instead, we use an affine parameter λ, which labels points along the path in a way that keeps the motion equation: [math]\dfrac{d^2 x^\mu}{dλ^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{dλ} \dfrac{dx^\sigma}{dλ} = 0[/math] This ensures the particle’s path remains a true geodesic, even without proper time. In what way is this not circular? That is, you provide the null geodesic equation in which the parameter is an affine parameter, but when asked what an affine parameter is, you refer back to the null geodesic equation.

I think it is worth saying that every situation in which π appears is in some way connected to a circle. To see that this is true, for any given expression in which π appears, consider why it is π that appears in the expression. This means tracing the number back to its definition, which was originally based on the circle. For example, consider the gamma function of a half-integer: [math]\displaystyle \Gamma(\dfrac{3}{2}) = \int_{0}^{\infty} \sqrt{t}\ \exp(-t)\ dt[/math] [math]\displaystyle \text{Let } t = x^2\ \ \ \ ;\ \ \ \ dt = 2x\ dx[/math] [math]\displaystyle \Gamma(\dfrac{3}{2}) = \int_{0}^{\infty} 2x^2\ \exp(-x^2)\ dr[/math] [math]\displaystyle = \int_{0}^{\infty} \exp(-x^2)\ dx\ \ \ \ \ \text{(integration by parts:}\ u=x,dv=2x\exp(-x^2)\ \text{)}[/math] [math]\displaystyle \int_{0}^{\infty}\int_{0}^{\pi/2} r\ \exp(-r^2)\ d\theta\ dr = \dfrac{\pi}{4} \int_{0}^{\infty} 2r\ \exp(-r^2)\ dr[/math] [math]\displaystyle = \dfrac{\pi}{4} \int_{0}^{\infty} \exp(-t)\ dt = \dfrac{\pi}{4}[/math] [math]\displaystyle \int_{0}^{\infty}\int_{0}^{\pi/2} r\ \exp(-r^2)\ d\theta\ dr = \int_{0}^{\infty}\int_{0}^{\infty} \exp(-x^2-y^2)\ dx\ dy[/math] [math]\displaystyle = \int_{0}^{\infty} \exp(-x^2)\ dx\ \cdot \int_{0}^{\infty} \exp(-y^2)\ dy = \dfrac{\pi}{4}[/math] [math]\displaystyle \text{Therefore }\ \ \Gamma(\dfrac{3}{2}) = \int_{0}^{\infty} \exp(-x^2)\ dx = \dfrac{\sqrt{\pi}}{2}[/math] [If the above LaTex doesn't render, please refresh browser] In this case, the connection to a circle is the use of polar coordinates to evaluate the definite integral of the Gaussian function.

Before you become too engrossed by the more speculative aspects of your document, perhaps you could reply to the following which I posted earlier: Perhaps you can elaborate on what an affine parameter is, particularly with regards to null geodesics, which cannot be parametrised with proper time or arc length.

Have you seen A Clockwork Orange?

Reading through the opening post, I thought it was going to be about the notion that the laser point, if far enough away, would be travelling faster than the speed of light.

I think it is interesting that there are two distinct, even somewhat contrary, notions of degrees of irrationality. For example, one can consider the following sequence as a sequence of increasing irrationality: rational numbers solutions of quadratic equations with integer coefficients solutions of cubic equations with integer coefficients solutions of quartic equations with integer coefficients ... solutions of polynomial equations of finite degree n with integer coefficients ... transcendental numbers Or alternatively, one can consider how well a given number can be approximated by a rational number relative to the size of the denominator of the rational number, as indicated by the irrationality exponent [math]\mu(x)[/math] in the expression: [math]\left|x - \dfrac{p}{q}\right| < \dfrac{1}{q^{\mu}}[/math] Quite remarkably, the Liouville numbers, which are transcendental numbers, have infinite irrationality exponent and are most closely approximated by rational numbers, whereas the golden ratio, a solution of a quadratic equation, with irrationality exponent 2, is poorly approximated by rational numbers, being an extreme case of the Hurwitz inequality: [math]\left|x - \dfrac{p}{q}\right| < \dfrac{1}{\sqrt{5}q^2}[/math] Even worse than the golden ratio are other rational numbers (approximated by rational numbers that are not equal to the number), which have irrationality exponent 1.

The golden ratio is not transcendental: [math]\varphi =\dfrac{1+\sqrt{5}}{2}[/math] Or putting it another way, it is a solution of the quadratic equation: [math]\varphi^{2} - \varphi - 1 = 0[/math] [If the above LaTex doesn't render, please refresh browser] The golden ratio has an interesting property: it is representative of a family of numbers that are the least rational of all the numbers. By contrast, the Liouville numbers are numbers that are more closely approximated by rational numbers than all other numbers, and they are transcendental.

All of the mathematical expression, including the opening and closing tags, needs to be on the one line. It can wrap to multiple lines, but no new line characters.

I assumed that this was simply to eliminate boundary effects that result from trying to pack circles into a finite area. This seems unnecessary to me as one can tesselate an infinite plane with hexagons and inscribe a circle into each hexagon. Note also that the hexagon is the polygon with the largest number of sides that can tesselate a plane. And given that the ratio of the area of an inscribed circle to the area of the inscribing polygon increases with the number of sides of the polygon, the hexagon is the polygon producing the maximum packing density of circles.

The packing of spheres is different to the packing of circles in that one can't arrange spheres around a central sphere in a gap-free way, whereas one can arrange six circles around a central circle and this arrangement is gap-free. And because a flat plane can be tessellated with hexagons, it is clear that the maximum packing density for circles in a plane is the ratio of the area of an inscribed circle to the area of the inscribing hexagon: Area of inscribed circle = πr2 ≈ 3.1416 r2 Area of inscribing hexagon = 6r2 tan π/6 ≈ 3.4641 r2 Ratio of inscribed circle to inscribing hexagon = [math]\dfrac{\pi}{6}\cot\dfrac{\pi}{6} = \dfrac{\sqrt{3}\pi}{6} \approx 0.9069[/math] The maximum packing density for spheres is [math]\dfrac{\pi}{3\sqrt{2}} \approx 0.74048[/math] for a number of different arrangements, all of which has each sphere touching 12 neighbouring spheres. According to Wikipedia: "In 1611, Johannes Kepler conjectured that this is the maximum possible density amongst both regular and irregular arrangements—this became known as the Kepler conjecture. Carl Friedrich Gauss proved in 1831 that these packings have the highest density amongst all possible lattice packings. In 1998, Thomas Callister Hales, following the approach suggested by László Fejes Tóth in 1953, announced a proof of the Kepler conjecture. Hales' proof is a proof by exhaustion involving checking of many individual cases using complex computer calculations. Referees said that they were "99% certain" of the correctness of Hales' proof. On 10 August 2014, Hales announced the completion of a formal proof using automated proof checking, removing any doubt." [If the above LaTex doesn't render, please refresh browser.]

Perhaps you can elaborate on what an affine parameter is, particularly with regards to null geodesics, which cannot be parametrised with proper time or arc length.

From https://en.wikipedia.org/wiki/Greisen%E2%80%93Zatsepin%E2%80%93Kuzmin_limit "The Greisen–Zatsepin–Kuzmin limit (GZK limit or GZK cutoff) is a theoretical upper limit on the energy of cosmic ray protons traveling from other galaxies through the intergalactic medium to our galaxy. The limit is 5×1019 eV (50 EeV), or about 8 joules (the energy of a proton travelling at ≈ 99.99999999999999999998% the speed of light). The limit is set by the slowing effect of interactions of the protons with the microwave background radiation over long distances (≈ 160 million light-years)."

Don't forget that you still haven't answered my question yet.

On page 37, you said that a test particle of mass m follows: [math]\dfrac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\nu\sigma} \dfrac{dx^\nu}{d\tau} \dfrac{dx^\sigma}{d\tau} = 0[/math] However, this equation does not work for a test particle of mass zero (eg a photon). How would you adjust this equation so that it does work for a test particle of mass zero? [If the above LaTex doesn't render, please refresh browser.]

No, I'm stating that while the loopholes of local hidden variables were closed by the experimental work by John Clauser, Alain Aspect, and Anton Zeilinger, who won the Nobel prize for their work, the assumption for the closure of those loopholes is a 3+1D space-time. I demonstrate that with Feynman integrals, a 3+1+1D geometry could be hidden because that path is usually canceled out. Even the no-communication theorem assumes a 3+1D space-time. Extending quantum dynamics to an extra dimension through a quantum behavior isn't reaching, but examining, or exploring, hypothetical perspectives not approached before. I need some clarification: Are you claiming that in the extra dimension, entangled particles are local where they are measured even though they may be very distant in 3+1D spacetime? If so, then you are assuming that there is a need for the communication of measured results between the particles. Where in the proof of the no-communication theorem does it indicate the assumption of 3+1D spacetime?

One thing worth mentioning: A single photon travelling in free space cannot gravitationally collapse, regardless of how much energy it has. Any photon exists in frames of reference where is has any particular value of energy. Since this energy can be arbitrarily high, there can be no gravitational collapse. Also, a single photon travelling in free space cannot split into other particles because this would violate the conservation of energy-momentum. However, a single photon can collide with some object, or two photons travelling in opposite directions can collide with each other, to produce... whatever.

For two particles to be quantum entangled, two conditions need to be satisfied: (1): The states of the two particles are correlated. That is, the state of one particle depends on the state of the other particle. (2): The combined state of the two particles is a quantum superposition. If only (1) is satisfied, the two correlated particles may be classical particles (eg billiard balls). Or it may be the result of measuring an entangled pair of particles, causing the quantum superposition to "collapse" (Copenhagen interpretation). If only (2) is satisfied, the particles are individually a quantum superposition, and therefore the combined state of the two particles is also a quantum superposition, but the two particles are completely independent of each other, perhaps because they are in different galaxies. It is worth noting that an arbitrary two-particle state is most likely to be an entangled state. However, the entangled states usually encountered in entanglement experiments are special entangled states, not the run-of-the-mill arbitrarily chosen entangled two-particle states. (In mathematical terms, the Hilbert space of the two-particle states has a higher dimensionality than the Hilbert space of the corresponding non-entangled two-particle states.)