KJW

From what I can tell, "\displaystyle" does nothing. It was the "\not" placed in braces "{}" that you need, as well as the small space "\,". I actually tried to do something similar not long ago, so I have prior experience (which I didn't explicitly use). Let's try this: \not \! \partial [math]\not \! \partial[/math] Perfect!!! According to a reference PDF file I have, "\!" is a "negative thin space".

This is what I come up with: {\not}\tiny\,\normalsize\partial [math]{\not}\tiny\,\normalsize\partial[/math] This didn't render exactly as it did in Online LaTeX Equation Editor. Try this: {\not}\small\,\normalsize\partial [math]{\not}\small\,\normalsize\partial[/math] Try this: {\not}\,\partial [math]{\not}\,\partial[/math] Testing: X \tiny X \normalsize X [math]X \tiny X \normalsize X[/math] How about this: \displaystyle{\not}\tiny\,\normalsize\partial [math]\displaystyle{\not}\tiny\,\normalsize\partial[/math] Or this: \not \partial [math]\not \partial[/math] This: \not{\partial} [math]\not{\partial}[/math]

As a result of this thread, I did some exploration of factoring semiprimes. I even came up with a solution of sorts, but this didn't seem to be any more efficient than brute force.

Accelerated objects can be described perfectly well in special relativity. But accelerated frames of reference are outside the scope of standard special relativity. That's because accelerated frames of reference involve some of the mathematics of general relativity (though not the mathematics of spacetime curvature). Standard special relativity limits itself to the Minkowskian metric. The Minkowskian metric is invariant to Lorentz transformations, and inertial trajectories in spacetime transform to inertial trajectories under Lorentz transformations. Thus, all inertial trajectories in Minkowskian spacetime are on equal footing in that they all observe the same spacetime metric. The invariance of the Minkowskian metric to Lorentz transformations implies that it is not possible to measure one's velocity relative to Minkowskian spacetime, and that only velocities relative to other objects can be measured, which is made possible because symmetry to Lorentz transformations is broken. In the case of an accelerated frame of reference, the transformation from an inertial frame of reference to the accelerated frame of reference is not a Lorentz transformation, it is a transformation under which the Minkowskian metric is not invariant. That is, the metric of an accelerated frame of reference is not a Minkowskian metric. Thus, an observer in an accelerated frame of reference can distinguish between being in an accelerated frame of reference and being in an inertial frame of reference. Even though velocity is only relative, acceleration is absolute because one can measure one's current velocity relative to one's past velocity. Thus, absolute acceleration does not imply absolute velocity.

This is correct. I have used this to explain the asymmetry in the twin paradox. This asymmetry is often explained by the fact that only the travelling twin accelerates, whereas the above explains precisely how the fact that only the travelling twin accelerates manifests the asymmetry. Indeed, one can calculate the twin paradox result from the Doppler effect observed by each of the twins, which naturally agrees with the result obtained by calculating the proper time for each of the twins. Unlike velocity, which is only relative, acceleration is absolute. In the twin paradox, only the travelling twin accelerates. Not only that, the travelling twin knows that they accelerated, while the stay-at-home twin knows that they did not accelerate, and both the travelling twin and the stay-at-home twin knows that only the travelling twin accelerated. Thus, both the travelling twin and the stay-at-home twin observe the Doppler effect change from redshift to blueshift when they observe the travelling twin accelerate. .

According to Wikipedia, the peculiar velocity of the Sun relative to the comoving cosmic rest frame is 369.82 ± 0.11 km/s towards the constellation Leo. That doesn't seem "almost immobile" to me.

I suppose that depends on what is meant by "brighter". In terms of radiation intensity, I'd have thought one could increase that beyond the intensity of the source, if it is an extended source. But clearly one can't change the frequency of the photons merely by focusing a beam, so the effective temperature (if is black body radiation) of the radiation can't be altered. in that way. Is that what you meant? By "brighter", I mean "higher intensity". It should be noted that black body radiation is not just a frequency distribution but also an intensity, and that any radiation that deviates from black body radiation, either by frequency distribution or by intensity, is radiation that can do work. In the case of the perpetual motion machine, the ability to focus black body radiation from an object into a smaller space increases the intensity of the image so that it is no longer black body radiation and can do work. If one wishes to capture the entire radiation output of a small spherical object as an image, one can place the object at the focus of an internally reflective prolate spheroid. The image forms at the other focus. But note that symmetry demands that the image is the same size as the object, and hence the same intensity. On the other hand, if one uses a magnifying glass to focus sunlight onto a piece of paper, the image is much smaller than the sun, but then one is capturing only a tiny proportion of the total output of the sun.

Have you taken into account the hysteresis exhibited by ferromagnetic materials?

The design to which I am referring was also a perpetual motion machine of the 2nd kind. It consisted of a large diameter pipe radiating thermal radiation at room temperature, this radiation being focused onto a smaller diameter pipe, the increased intensity resulting in an elevated temperature of the smaller diameter pipe. I could see nothing wrong with this design and was forced to conclude that it is impossible to focus radiation to an image that is brighter than the source. I raised this on a forum I regularly visited at the time, and my hypothesis was confirmed by another member.

I'm strongly opposed to forum moderators locking threads about perpetual motion machines. I think perpetual motion machines can be interesting as well as instructive. I have even discovered an interesting principle as a result of considering a design that could be used to construct a perpetual motion machine: Neither lenses nor mirrors can produce an image that is brighter than the source.

Proper time, the time that clocks measure, is absolute, the same for all observers. What an observer does has no effect on the age of the earth. This indicates a misunderstanding of relativity. You appear not to understand what "time is relative" actually means. I suggest you read the Wikipedia article "Proper time".

No. Only c is invariant to Lorentz transformations. Bear in mind that c is derived from the electromagnetic constants in Maxwell's equations.

One thing I like to say is that [math]ds[/math] is not an exact differential. If [math]ds[/math] were an exact differential, there would exist a function of the coordinates, [math]s(x, y, z, t)[/math], and the spacetime distance between any two points would simply be the difference of the two values of [math]s[/math] at the endpoints and independent of the path between them. But [math]ds[/math] is not an exact differential and the spacetime distance between two points does depend on the path between them. For distances between pairs of locations in three-dimensional space, it seems rather obvious that this depends on the path between them, so it's rather natural that the same applies in four-dimensional spacetime and that the notion of universal time does not exist. Mathematically, if [math]ds[/math] were an exact differential: [math]ds = \dfrac{\partial s}{\partial x} dx + \dfrac{\partial s}{\partial y} dy + \dfrac{\partial s}{\partial z} dz + \dfrac{\partial s}{\partial t} dt = \dfrac{\partial s}{\partial x^{\mu}} dx^{\mu}[/math] and: [math](ds)^2 = g_{\mu\nu} dx^{\mu} dx^{\nu} = \dfrac{\partial s}{\partial x^{\mu}} \dfrac{\partial s}{\partial x^{\nu}} dx^{\mu} dx^{\nu}[/math] [math]g_{\mu\nu} = \dfrac{\partial s}{\partial x^{\mu}} \dfrac{\partial s}{\partial x^{\nu}}[/math] which, as the tensor product of two vectors, is singular, in violation of the requirement that the metric tensor be invertible.

No, the correct method is to choose a spacetime coordinate system, establish the metric for that coordinate system, establish the spacetime trajectory of the sword in that coordinate system, then integrate the expression for ds along that trajectory. Bear in mind that the sword has been around the sun a couple of thousand times and around the earth hundreds of thousands of times if one chooses a heliocentric coordinate system. One could choose a geocentric coordinate system but then the metric will be more complicated. One could even create a "sword-centric" coordinate system in which the spacetime trajectory and integration are trivial at the expense of an even more complicated metric.

The Minkowskian metric can be coordinate-transformed to an FLRW metric of a constantly expanding hyperboloidal space. That I have proven above. I do not claim that the Minkowskian metric can be coordinate-transformed to an FLRW metric of an expanding flat space. But I do claim that an FLRW metric of an expanding flat space can be coordinate-transformed to a spacetime in which the big bang singularity is a single point, but this spacetime is not flat.

That inference is incorrect because I explicitly said that the Schwarzschild metric cannot be either the Minkowskian metric or the FLRW metric of an expanding flat space.

Where did I claim that?

I did. But you did not explain how it relates to what I posted. I demonstrated the equivalence between an FLRW metric and an explosion metric by demonstrating the existence of a coordinate transformation between them. I believe that is sufficient. Please explain to me why it is not. Because the Riemann curvature tensor field is non-zero. Actually, I can go further and say that the Schwarzschild metric cannot be an FLRW metric of an expanding flat space (I've never examined this question about expanding curved spaces). That's because an FLRW metric of an expanding flat space is conformally flat, i.e. the Weyl conformal tensor field is zero, in contrast to the Schwarzschild metric for which the Weyl conformal tensor field is the only non-zero curvature tensor field.

It's not the square root of a vector. It's the square root of the t-component as a function of the new coordinates. The three-dimensional space of constant proper time that is created by the explosion is homogeneous and isotropic, as is the equivalent three-dimensional space described by the FLRW metric. I didn't say "homogeneous and isotropic explosion", I said "homogeneous and isotropic explosion metric", for the sake of brevity.

Neither the FLRW metric nor the corresponding explosion metric have a preferred direction. But I assume you are referring to a perturbation of the FLRW metric. Perhaps the perturbation can be decomposed into irreducible representations of the symmetry group of the FLRW metric. In any case, a coordinate transformation that transforms a homogeneous and isotropic FLRW metric to a homogeneous and isotropic explosion metric will also transform a perturbation of the FLRW metric to a perturbation of the explosion metric. Indeed, any vector or tensor field on the FLRW metric will transform covariantly to the corresponding vector or tensor field on the explosion metric. The point is that the explosion metric describes the exact same physical spacetime as does the expanding space. One further point is worth noting: When I considered the coordinate transformation from the FLRW metric: [math](ds)^2 = c^2 (dt)^2 - a^2(t) ((dr)^2 + r^2 ((d\theta)^2 + \sin^2(\theta) (d\phi)^2))[/math] the coordinate transformation: [math]r = \dfrac{r'}{a(t(t',r'))}[/math] included the scale function [math]a(t)[/math]. But the coordinate transformation doesn't actually need to be custom created for the particular FLRW metric. One could use a much simpler generic coordinate transformation to contract the big bang singular space to a single point. This is because one is still applying a coordinate transformation along with the equivalence that implies.

I can do better than that. Using the description of a continuum of trajectories given earlier, I can provide an explicit FLRW metric corresponding to an explosion in Minkowskian spacetime. The description of a continuum of trajectories: [math]x(v_x, v_y, v_z, \tau) = v_x t(v_x, v_y, v_z, \tau)[/math] [math]y(v_x, v_y, v_z, \tau) = v_y t(v_x, v_y, v_z, \tau)[/math] [math]z(v_x, v_y, v_z, \tau) = v_z t(v_x, v_y, v_z, \tau)[/math] [math]t(v_x, v_y, v_z, \tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_x^2 - v_y^2 - v_z^2}}[/math] Note that the description of a continuum of trajectories is actually a coordinate transformation in disguise. Let's change to spherical coordinates: [math](ds)^2 = c^2 (dt)^2 - (dr)^2 - r^2 ((d\theta)^2 + \sin^2(\theta) (d\phi)^2)[/math] Then the description of a continuum of trajectories becomes (noting that [math]\beta = \dfrac{1}{c} \sqrt{v_x^2 + v_y^2 + v_z^2}[/math] ) : [math]t(\tau, \beta, \theta, \phi) = \tau\ (1 - \beta^2)^{-\frac{1}{2}}[/math] [math]r(\tau, \beta, \theta, \phi) = \beta\ c\ t(\tau, \beta, \theta, \phi) = \beta\ c\ \tau\ (1 - \beta^2)^{-\frac{1}{2}}[/math] [math]\theta(\tau, \beta, \theta, \phi) = \theta[/math] [math]\phi(\tau, \beta, \theta, \phi) = \phi[/math] [math]dt(\tau, \beta, \theta, \phi) = (1 - \beta^2)^{-\frac{1}{2}}\ d\tau + \tau\ \beta\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta[/math] [math]dr(\tau, \beta, \theta, \phi) = \beta\ c\ dt(\tau, \beta, \theta, \phi) + c\ t(\tau, \beta, \theta, \phi)\ d\beta[/math] [math]= \beta\ (1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau + c\ \tau\ \beta^2\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta + c\ \tau\ (1 - \beta^2)^{-\frac{1}{2}}\ d\beta[/math] [math]= \beta\ (1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau + c\ \tau\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta[/math] [math]d\theta(\tau, \beta, \theta, \phi) = d\theta[/math] [math]d\phi(\tau, \beta, \theta, \phi) = d\phi[/math] [math](c\ dt(\tau, \beta, \theta, \phi))^2 - (dr(\tau, \beta, \theta, \phi))^2[/math] [math]= (c\ dt(\tau, \beta, \theta, \phi) + dr(\tau, \beta, \theta, \phi))\ (c\ dt(\tau, \beta, \theta, \phi) - dr(\tau, \beta, \theta, \phi))[/math] [math]= (1 + \beta)((1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau + c\ \tau\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta)\ (1 -\beta)((1 - \beta^2)^{-\frac{1}{2}}\ c\ d\tau - c\ \tau\ (1 - \beta^2)^{-\frac{3}{2}}\ d\beta)[/math] [math]= (1 - \beta^2)((1 - \beta^2)^{-1}\ c^2 (d\tau)^2 - c^2 \tau^2\ (1 - \beta^2)^{-3}\ (d\beta)^2)[/math] [math]= c^2 (d\tau)^2 - c^2 \tau^2\ (1 - \beta^2)^{-2}\ (d\beta)^2[/math] Substituting this expression and the expression for [math]r[/math] into the spherical coordinates Minkowskian metric gives the following intermediate metric: [math](ds)^2 = c^2 (d\tau)^2 - c^2 \tau^2\ ((1 - \beta^2)^{-2}\ (d\beta)^2 - \beta^2\ (1 - \beta^2)^{-1}\ ((d\theta)^2 + \sin^2(\theta) (d\phi)^2))[/math] A coordinate transformation to a more appropriate radial coordinate (note that [math]\beta[/math] is dimensionless, and that [math]t_{\text{ref}}[/math] is a constant reference time corresponding to the time at which the expansion scale is unity): [math]\dfrac{dr'}{c\ t_{\text{ref}}} = (1 - \beta^2)^{-1}\ d\beta[/math] [math]c\ t_{\text{ref}}\ \dfrac{d\beta}{dr'} = 1 - \beta^2 \ \ \ \ \ ;\ \ \ \ \ \beta = 0 \ \ \text{at} \ \ r' = 0[/math] [math]\beta = \tanh\left(\dfrac{r'}{c\ t_{\text{ref}}}\right) \ \ \ \ \ ;\ \ \ \ \ \dfrac{\beta^2}{1 - \beta^2} = \sinh^2\left(\dfrac{r'}{c\ t_{\text{ref}}}\right)[/math] [math]\tau = t' \ \ \ \ \ ;\ \ \ \ \ \theta = \theta' \ \ \ \ \ ;\ \ \ \ \ \phi = \phi'[/math] [math](ds)^2 = c^2 (dt')^2 - \dfrac{t'^2}{t_{\text{ref}}^2} \left((dr')^2 + c^2\ t_{\text{ref}}^2\ \sinh^2\left(\dfrac{r'}{c\ t_{\text{ref}}}\right)\ \Big((d\theta')^2 + \sin^2(\theta') (d\phi')^2\Big)\right)[/math] Thus, a description of an explosion from a single point in Minkowskian spacetime has been coordinate-transformed to an FLRW metric of a constantly expanding hyperboloidal three-dimensional space. Q.E.D.

I do not see the point of your previous post. It doesn't matter if you can provide mathematics that shows that an explosion is not homogeneous and isotropic. I have already mathematically proven that under the particular conditions which I described, an explosion that I have already defined will produce a three-dimensional space that is homogeneous and isotropic. Any mathematics that you provide that shows otherwise will not satisfy the conditions I have described and can be dismissed on that basis. But note that I am choosing observers who are equivalent to observers at rest in an FLRW metric.

I think it needs to be said that in the case of the explosion in Minkowskian spacetime, although it doesn't appear to be so in a spacetime diagram, all the spacetime trajectories are normal to the three-dimensional hyperboloidal space of constant proper time. In other words, all the particles are at rest in the three-dimensional space. This three-dimensional space really does appear to be expanding. One could even apply to the explosion in Minkowskian spacetime the inverse of the coordinate transformation I wrote above for the FLRW metric to obtain an FLRW metric for an expanding hyperboloidal space.

I defined it in this post as: An explosion is simply a set of straight trajectories in spacetime that originate from a single point in spacetime. It's relative speed, not just speed. This entire discussion is from the perspective of one of the fragments.

Can Mathematica do it? Actually, I'm far more interested in the general theoretical aspects than dealing with particular metrics. But if I really needed to invert a non-diagonal metric, I could do it. And I do know how to express the inverse of the metric tensor in terms of the metric tensor of arbitrary dimensions in tensor notation.