KJW

No, it is not possible to coordinate-transform a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) to a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) In general, a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) has non-zero Ricci curvature, whereas a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) describes flat spacetime. Note that this metric can be transformed to the Minkowskian metric by the coordinate transformation: t' = t'(t) ; x' = x ; y' = y ; z' = z where t'(t) is a solution to the differential equation: dt'(t)/dt = α(t)

Note that Markus Hanke said that "uniform acceleration is locally equivalent to a uniform gravitational field" [my bold]. The distant stars are not part of a local measurement.

All redshifts and blueshifts ultimately have the same cause: Consider some arbitrary spacetime in which there is an arbitrary trajectory of an emitter and an arbitrary trajectory of a receiver. From the emitter, consider two infinitesimally separated future-directed light-like geodesics that eventually intersect the trajectory of the receiver. Then the ratio of the proper time along the trajectory of the emitter between the two light-like geodesics, and the proper time along the trajectory of the receiver between the two light-like geodesics, is the Doppler ratio (redshift or blueshift) that the receiver observes at the instant of receiving the two light-like geodesics from the emitter. From the above, one may consider more special cases where the cause of the redshift and blueshifts can be differentiated between relative velocity, acceleration, gravitation, or cosmological, as well as combinations of these. For example, constant acceleration in the Minkowskian metric can be considered in terms of relative velocity, whereas the same constant acceleration in its own frame of reference can be considered in terms of pure acceleration. In other words, the change in metric, even though the physics has not changed, leads to a change in interpretation. But in all cases, the above applies. And note that the above has been expressed entirely in terms of invariants.

In the case of the twin paradox, the redshift and blueshift are due to the relative velocity between the source and receiver. In the case of gravitational redshift and blueshift, equivalent to accelerational redshift and blueshift, there is no relative velocity between the source and receiver. The redshift and blueshift is due to the acceleration itself, and depends on the displacement between source and receiver. If one examines the metric of an accelerated frame of reference, sources that are at rest and above the receiver (note that the acceleration is directed upward) will be blueshifted, while sources that are at rest and below the receiver will be redshifted. It is worth noting that the source that is at rest below the receiver has a greater acceleration than the receiver, and that the source that is at rest above the receiver has a lesser acceleration than the receiver. However, this does not determine the redshift or blueshift, and it might not apply to gravitation.

You also have to worry about whether particular codes are recognised on this site... some aren't. So even if you write something using a LaTeX editor and everything looks fine, some of it may not be recognised when transferred to this site. Thus, it is always a good idea to test something you've not used before in The Sandbox before posting it in a thread for real.

I find it quite bizarre that some people regard it as significant that the image on a retina is upside down. It's like wondering how the images on a DVD are turned upright on a TV screen.

One other way to view rapidity it to consider three observers, A, B, and C in relative collinear motion. The Doppler ratio between observers A and C is the product of the Doppler ratios between A and B, and B and C. Taking logarithms of the Doppler ratios, the resulting quantity combines additively.

The usual presentation of the twin paradox is that the travelling twin's journey is two inertial journeys, and I have not considered anything different. But although the journey is locally inertial at almost all times during the journey, the change in direction prevents the journey from being globally inertial. So, you have to take into account whether or not what you are considering is a local or a global notion. What does this actually mean? I'll repeat: There is no discontinuity in the time of the earth in the frame of reference of the travelling twin. You have made an error in your assessment of the frame of reference of the travelling twin. Minkowski spacetime doesn't exist, except locally or perhaps as part of a multiverse. Otherwise, it is a mathematical idealisation. That's why I dropped the "Minkowski" when I said, "I think spacetime is physical". But even in general relativity with curved spacetime, local physics is the physics of Minkowskian spacetime, so Minkowskian spacetime is more than just a mathematical idealisation or approximation. I should also state that one can calculate the result of the twin paradox without reference to the notion of simultaneity of the travelling twin by considering the Doppler effect that is observed by both twins of each other during the entire journey of the travelling twin. Each twin is literally observing the clock of the other twin.

"Physical or mathematical" is a false dichotomy. In the case of the notion of simultaneity, at a given instant of a given trajectory in spacetime, there is a set of physical points in spacetime that form a physical three-dimensional space that is simultaneous to the given instant of the given trajectory. But if you examine the points themselves, you'll find nothing that physically indicates that they are simultaneous to the given instant of the given trajectory. Different instants give rise to different three-dimensional spaces, and different spacetime trajectories have their own sets of three-dimensional spaces. But you are incorrect in what you consider to be the space of simultaneity for the travelling twin of the twin paradox. You assume that the simultaneity of the travelling twin is the same as that of an outbound inertial trajectory during the outbound part of the journey and the same as that of an inbound inertial trajectory during the inbound part of the journey. But the space of simultaneity for the travelling twin is distinct from that of any inertial trajectory, taking into account the change that occurred during the journey. It should be noted that the space of simultaneity of individual points of any trajectory depends on the entire trajectory, not just a part of it. I think spacetime is physical. If I look at myself in the mirror, I am seeing what I looked like in the past. And the further away the mirror is, the further back into my past I am looking. The principle of general relativity is that all coordinate systems are equally valid. But each observer has their own natural frame of reference. This is physical in the sense that it is based on the relationship of the physical observer to the spacetime. It is not however a physical property of the spacetime itself. However, the metric of a given coordinate system of a given physical spacetime is in principle the result of physical measurements of length and time using physical rulers and physical clocks.

I'll repeat: \not \! \partial [math]\not \! \partial[/math] Perfect!!! According to a reference PDF file I have, "\!" is a "negative thin space".

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.