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enWhere Lies the Optical Properties of Matter in General Relativity?
http://bernardhlavenda.com/node/217
<span class="field field--name-title field--type-string field--label-hidden">Where Lies the Optical Properties of Matter in General Relativity?</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Sun, 09/23/2018 - 10:14</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>In the December 12 1936 issue of <em>Nature</em>, Ludwik Silberstein publishes a letter entitled "Minimal lines and geodesics within matter: A fundamental difficulty of Einstein's theory."</p>
<p>The issue Silberstein address was that Einstein equates geometry $G$ with a material tensor $T$ so that one should know what the other is talking about. Otherwise, you would be equating apples and oranges. </p>
<p>Now according to Einstein's generalization of the Minkowski indefinite metric, </p>
<p>$$ds^2=g_{\nu\mu}dx^{\nu}dx^{\mu}$$</p>
<p>its vanishing should still represent "light-lines". Its banal to show that <em>in vacuo</em> the light-lines $ds=0$ represent light trajectories that propagate at speed $c$, while in matter that is characterized by a material tensor $T_{\nu\mu}$, "the minimal lines manifestly <em>cannot represent light propagation</em>, even to a rough approximation." In such an isotropic medium, the light velocity is not $c$ but $c/n$, where $n$ is the index of refraction in the material medium. </p>
<p>He uses the case of a Schwarzschild incompressible liquid sphere where the components of the mixed matter tensor are</p>
<p>$$T_1^1=T_2^2=T_3^3=-p/c^2, \hspace{30pt} T_4^4=\varrho,$$</p>
<p>where $p$ is the hydrostatic pressure and $\varrho$ is the constant density of the liquid, Since the $T$'s determine the $g$'s via Einstein equation, if there is no trace of $n$, the index of refraction, in $T$, there can be no trace of it in the $g$'s. Wrong!</p>
<p>The Schwarzschild metric is determined by Einstein's condition of emptiness, the vanishing of all the components of the Ricci tensor, $R_{\n\mu}=0.$ Due to the fact that the $g$'s depend on $r$ since the medium is isotropic and static, there will be a nonunity index of refraction. But, if we want to consider the Schwarzschild metric immersed in an incompressible liquid sphere who tells $T$ how to react since the index of refraction lies within the $g$'s and not in $T$ which cannot account for gravitational energy at all. Therefore, you are, in effect, equating apples and oranges. </p>
<p>Moreover, the radius of curvature of the sphere is $c\surd(3/8\pi G\varrho)$, which is the product of $c$ and the Newtonian free-fall time. So says $T$, but what does $G$ say? It says that the radius of curvature is $c\surd(r^3/GM)$. So which is it? Are we working with a model of constant density, $\varrho$, or one with constant mass, $M$? </p>
<p>According to Dirac, the "emptiness" of the system is unaffected by the gravitational field whereas other fields do. But, then there can be no mass, $M$ and the gravitational field lies in the $g$'s. For if there is mass, then there is a density of matter defined as $M/V$, where $V$ is the volume of the incompressible fluid sphere. </p>
<p>It is therefore a logical inconsistency to consider the gravitational field to be buried in the metric and not reflected in the matter tensor which is equated to it via Einstein's equations. But if the gravitational field is in $G$ how can it simultaneously be in $T$? The linearized Einstein equations supposedly describe gravitational waves whose source is a pseudo-tensor density. Now the right-hand side of the equation accounts for the source of the gravitational field. So which is it: Is gravity geometry or gravity a force that can do work and should be included in the matter tensor? You can't have your cake and eat it too!</p>
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Sun, 23 Sep 2018 08:14:31 +0000adminlavenda217 at http://bernardhlavenda.comWhy Black Holes Don't Exist in the Schwarzschild Metric, and How to Create One
http://bernardhlavenda.com/node/216
<span class="field field--name-title field--type-string field--label-hidden">Why Black Holes Don't Exist in the Schwarzschild Metric, and How to Create One</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Sun, 09/16/2018 - 15:00</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>We are told to imagine that within the Schwarzschild radius, a particle will spiral in to its doom hitting the singularity in an undetermined amount of time. All this comes from extrapolating the Schwarzschild metric beyond its domain of validity, $r<2GM/c^2$. This is undoubtedly why Karl Schwarzschild did not find any such sucking in of material, and why he went beyond his outer solution to create an inner solution for $r<\surd(G\varrho)/c$, where $\varrho$ is the density and $1/\surd(G\varrho)$ is the Newtonian free-fall time.</p>
<p>In our last blog we wrote the generalized Newtonian force as</p>
<p>$$F_N=L^2u^2\left(u+\frac{d^2u}{d\vartheta^2}\right),$$</p>
<p>where $L$ is the angular momentum per unit mass and $u=1/r$. Now the terms in parentheses can be written as a power series</p>
<p>$$\frac{d^2u}{d\vartheta^2}+u=A+Bu+Cu^2+Du^3+Eu^4+Fu^5\cdots$$</p>
<p>where all coefficients are positive. </p>
<p>The Schwarzschild solution as terms up to $u^2$ with $B=0$, i.e., no in-spiraling. The first term $A=\mu/L^2$, where $\mu=GM$, the gravitational parameter, is the Keplerian ellipse</p>
<p>$$u=\frac{\mu}{L^2}\left(1+\epsilon\cos(\vartheta-\vartheta_0)\right),$$</p>
<p>where $\epsilon<1$ is the eccentricity. The famous square term is responsible for the advance of the perihelion and the deflection of light, where $C=3\mu/(Lc)^2$. Newton's law of inverse square corresponds to an ellipse where the inverse fourth power is a cardioid. The pinch in the cardioid has the effect of slightly rotating the ellipse, as seen below.</p>
<img alt="limacon" data-entity-type="file" data-entity-uuid="8653713b-6230-4be6-96a8-5ddaf625ae13" src="http://bernardhlavenda.com/sites/default/files/inline-images/Pascal.gif" class="align-center" /><p>The force at work is therefore a combination of Newton's inverse square and an inverse quartic.</p>
<p>There are then three operative forces in the outer Schwarzschild metric: Newton's inverse square force, the centrifugal repulsion and an attractive inverse power law. However, it is not as Baez claims:</p>
<p>"It is as if on top of Newtonian gravity we had another attractive force obeying an inverse fourth power law. This overwhelms the others at shorter distance, so if you get too close to a black hole, you spiral in to your doom." </p>
<p>He refers to a Wikipedia article, but the only thing that I found incorrect is the statement that "circular orbits are possible when the effective forces are zero." By "effective" forces it is meant the sum of the above. Moreover, where are the black holes? The coefficient $C=3\mu/c^2$ is so small that the rate of angular change of the line of apsides,</p>
<p>$$\Delta\omega=6\pi\left(\frac{\mu}{cL}\right)^2,$$</p>
<p>per period represents $43"$ per century in the case of Mercury! This hardly can be considered as the source of a black hole!</p>
<p>How then can we create a black hole? It arises from the squared term above where $B=\mu/L^2$. This is a Cotes spiral, $r=a/\vartheta$. For simplicity, consider all other terms zero. The equation of the trajectory becomes</p>
<p>$$\frac{d^2u}{d\vartheta^2}+\left(1-\frac{\mu}{L^2}\right)u=0.$$</p>
<p>When gravitational attraction overcomes centrifugal repulsion,</p>
<p>$$L^2>\mu$$</p>
<p>the particle will spiral into the center, s shown below.</p>
<img alt="Cotes spirals" data-entity-type="file" data-entity-uuid="4403b7e8-cb94-4e43-be9e-3d99e319d2e3" src="http://bernardhlavenda.com/sites/default/files/inline-images/Cotes_0.jpg" class="align-center" /><p>The Archimedes spiral $r=a\vartheta$ is a combination of inverse cube and inverse fifth. Thus, we can expect unstable orbital behavior from odd inverse powers of the force. Since there are no such terms in the Schwarzschild metric, there is no spiraling in of matter. QED</p>
<p>In the next blog we will discuss the centrifugal term, $L^2/r^2$, and its role in eliminating all angular dependencies in the laws of force. We can outline it briefly here. </p>
<p>Consider the equation for the ellipse,</p>
<p>$$r=(A+B\cos\vartheta)^{-1}.$$</p>
<p>The square of the velocity is</p>
<p>$$v^2=\dot{r}^2+\dot{z}^2,$$</p>
<p>where $\dot{r}=B L\sin\vartheta$ and $\dot{z}=L/r$, having used $L=r^2\dot{\vartheta}$. The force is the negative of one half the derivative with respect to $r$, viz.,</p>
<p>$$F=-\frac{L^2}{r^2}\left(B\cos\vartheta-\frac{1}{r}\right)=\frac{\mu}{r^2},$$</p>
<p>where $\mu=AL^2$. Without the centrifugal force term we would get an angular dependency in Newton's inverse square law. All other forces are derivable in the same way. So instead of adding angle dependencies, the second term in the square of the velocities eliminates them.</p>
<p>This seemingly contradicts what Baez has to say about Newton's propositions 43-45 in Book I. According to Baez, Newton showed that if you add $L^2/r^3$ to the force, "you can find a motion with the same radial dependence but with a different angle dependency." The above example clearly shows that it eliminates the angle dependence!</p>
<p> </p>
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Sun, 16 Sep 2018 13:00:43 +0000adminlavenda216 at http://bernardhlavenda.comA Weberian Force For the Generalized Newtonian Force
http://bernardhlavenda.com/node/215
<span class="field field--name-title field--type-string field--label-hidden">A Weberian Force For the Generalized Newtonian Force</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Thu, 09/13/2018 - 11:27</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>There are more parallels between Newton's force and the Weberian force of electrodynamics than meets the eye. Newton derived his inverse square law from the centrifugal force</p>
<p>$$F_c=F_s\sin\alpha=\frac{v^2}{\varrho},$$</p>
<p>where $F_s$ is the force directed at the source, $v$ the velocity and $\varrho$, the radius of curvature. Newton then introduced Kepler's II,</p>
<p>$$rv\sin\alpha=L,$$</p>
<p>to obtain</p>
<p>$$F_s=\frac{L^2}{r^2\sin^3\alpha}.$$</p>
<p>Newton developed, early in his carrier an expression for the radius of curvature, which reads</p>
<p>$$F_s=\frac{L^2}{r^3}\left\{1+\frac{2}{r^2}\left(\frac{dr}{d\theta}\right)^2-\frac{1}{r}\frac{d^2r}{d\theta^2}\right\},$$</p>
<p>although he did not write it in this form. From the way we have written the centrally directed force,</p>
<p>it is very strange that Newton didn't consider a centripetal force varying as the inverse cube of the distance. Introducing the transform $u=1/r$, the force becomes</p>
<p>$$F_s=L^2u^2\left[u+\frac{d^2u}{d\theta^2}\right].$$</p>
<p>In this form, it is appear that the equation of the trajectory,</p>
<p>$$\frac{d^2u}{d\theta^2}+u=\mbox{const.},$$</p>
<p>equivalent to $\varrho\sin^3\alpha=\mbox{const}$, gives his inverse square law for the <em>particular</em> conic section of an ellipse. But, this is not the only conic section that can be obtained, and will explain the dual force laws for the magic pairs</p>
<p>$$[1,-2]\hspace{20pt}[-4,-7]\hspace{20pt}[-5,-5]$$</p>
<p>that we have discussed in previous blogs. The first pair is the Hooke-Newton pair, while the other two are unknown. This is where out generalized force law comes in, that is not limited to these pairs, but, moreover, to the elusive inverse cubic law as well.</p>
<p>First, we establish that the force expression is conservative, by being able to derive it from a potential</p>
<p>$$W=\frac{1-(d\ln r/d\theta)^2}{2r^2},$$</p>
<p>much like Weber derived his force. The difference here is that the azimuthal angle $\theta$ replaces the current elements, $ids$ and $i'ds'$ in Weber's force:</p>
<p>$$F_w=\frac{ids\cdot i'ds'}{r^2}\left(1+2r\frac{\partial^2 r}{\partial s\partial s'}-\frac{\partial r}{\partial s}\frac{\partial r}{\partial s'}\right).$$</p>
<p>The angular dependencies in the generalized Newtonian force are therefore analogous to Ampere's law of which Weber's force synthesizes with Coulomb's force.</p>
<p>It is readily apparent that the generalized Newtonian force can derived from the Euler-Lagrange equation</p>
<p>$$\frac{d}{d\theta}\frac{\partial W}{\partial(dr/d\theta)}-\frac{\partial W}{\partial r}=F_s/L^2.$$</p>
<p>Second, we see that Einstein's equation for the advance of the perihelion of Mercury,</p>
<p>$$\frac{d^2u}{d\theta^2}+u=3u^2.$$</p>
<p>This corresponds to an inverse quartic force, and corresponds to a cardoid,</p>
<p>$$r=1+\cos\theta.$$</p>
<p>Third, the inverse seventh corresponds to a Cassini oval, and, in particular the lemniscate,</p>
<p>$$r=\surd\sin 2\theta.$$</p>
<p>This turns out give the force,</p>
<p>$$F_s=\frac{3L^2}{r^7},$$</p>
<p>one of the magic trio, since</p>
<p>$$\frac{d^2u}{d\theta^2}+u=3u^5.$$</p>
<p>Finally, the unstable spirals, $r=a/\theta$,and $r=a\theta$, the logarithmic and Archimedes spirals, respectively, correspond to the inverse cubic</p>
<p>$$F_s=\frac{L^2}{r^3},$$</p>
<p>and the combination of inverse cubic and fifth,</p>
<p>$$F_s=\frac{L^2}{r^3}\left(1+\frac{2a}{r^2}\right),$$</p>
<p>respectively. The latter belongs to one of the magic trio, but seems to have no problems to appear with the solo inverse cubic.</p>
<p>According to de Sitter's appraisal of GR, "gravity is not a force." This, in essence, relegates gravity to the time dimension which it has no right of being there. That is why Einstein's determination of the "gap" in the numerical value of the advance of the perihelion, was just that---gap fitting. Here we can appreciate it as the result of the combination of the inverse square and inverse quartic forces. We will have more to say about this in coming blogs.</p>
</div></div>
Thu, 13 Sep 2018 09:27:54 +0000adminlavenda215 at http://bernardhlavenda.comWhat are Gravitational Waves Anyway?
http://bernardhlavenda.com/node/214
<span class="field field--name-title field--type-string field--label-hidden">What are Gravitational Waves Anyway?</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Sun, 09/09/2018 - 13:53</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>We all know what sound waves are. A mechanical wave, which a sound wave is, vibrates in the medium by causing compression and rarefactions, as in the diagram below.</p>
<p></p>
<img alt="propagation of sound" data-entity-type="file" data-entity-uuid="2a0cf2c4-3fd5-4173-8e51-b3e080af8d6d" src="http://bernardhlavenda.com/sites/default/files/inline-images/sound%20waves_0.gif" class="align-center" /><p><img alt="" height="15" src="data:image/gif;base64,R0lGODlhAQABAPABAP///wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==" title="Click and drag to move" width="15" /></p>
<p>Compression causes the particles of the medium to bunch up, while rarefaction causes them to separate. If gravitational waves are "the ripples in the fabric of space time" then they must be considered mechanical waves.Since rarefactions and compressions occur in the direction of propagation the ripples in space time must be longitudinal waves. </p>
<p>Longitudinal waves, unlike transverse waves, cannot be polarized. Light is a good example of a transverse wave, which is a dipole wave, where the polarizations occur in directions normal to the direction of propagation. Polarization can go from being linear to circular.</p>
<p>The important difference between longitudinal and transverse waves is that the latter does not need a medium to propagate in. It is as if the electric and magnetic fields, directed normally to the direction of propagation, ride piggy-back on one another. This means that not only can light propagate in a medium, like air or a solid material, but also can propagate in the vacuum of space. </p>
<p>Are gravitational waves the intergalactic sound waves of the universe, which in ancient times we referred to as the "music of the heavens"? If they are then how do you explain the tensorial polarization of the waves? Then it wouldn't be possible to derive the spectra that LIGO has in the chirping and merging of two black holes by a non-general relativistic approach which uses the gravitational analogs of the electric and magnetic fields. Gravitation acceleration has been derived by Hilbron as the time derivative of the gravitational analog of the vector potential. That it reproduces known (supposedly, observed results) the gravitational waves must be a kind of electromagnetic wave. And it is supported by the fact (?) that gravitational waves also travel at the speed of light.</p>
<p>For if gravitational waves were longitudinal waves, their propagation would use up energy in making ripples in space-time, and it would be hard to rationalize the signal obtained by LIGO of the "merger of two black holes some 1.3 billion light years away." They would have dissipated away their energy a long time ago. The fact that the merger of a pair of neutron stars in August 2017, observed by LIGO, was so exactly confirmed by electromagnetic astronomers, rather than a confirmation makes it suspicious that gravitational waves are a form of electromagnetic waves, that is electromagnetic waves affected by gravity. But then, why do they need the fabric of space-time to propagate on, whereas electromagnetic waves do not?</p>
<p>Gravitational wave luminosity is of the same nature as electrodynamic, black body radiation, except for a difference in orders of magnitude. What cancels out the lower third-order term that would match thermal black body radiation to a tee? It is often said that conservation of mass and momentum prevent lower that quadrupole radiation from appearing. But where does that enter the expression for gravitational luminosity except to raise aberration to fifth order in the relative velocity?</p>
<p>We have no way of determining the velocity of a moving car from the inside without reference to observations made on the outside. Not true with acceleration, where we can distinguish between acceleration and deceleration by forces acting on our bodies. Michelson and Morley showed---using a Michelson interferometer--that motion through the ether was imperceptible.</p>
<p>What if uniform acceleration were imperceptible and gravity belonged to such a frame? We are supposedly living in a steadily expanding universe according to Hubble. However, consider the Milky Way itself which is a large spiral galaxy that is part of a cluster of galaxies known as the Local Group. If the Milky Way could be thought of as a uniformly rotating disc at an angular speed, $\sqrt{G\varrho}$, where $\varrho$ is some average density. This is the inverse of Newton's free fall time. The Milky Way rotates as a pinwheel at $270$ km/sec, and takes $200$ million light years to complete one rotation. This would give an angular speed of $\omega\approx 10^{-15}$ sec$^{-1}$. </p>
<img alt="milky way" data-entity-type="file" data-entity-uuid="1c5486ce-6909-4f03-9ee2-3030aed60a20" src="http://bernardhlavenda.com/sites/default/files/inline-images/milky-way-differential-rotation.jpg" class="align-center" /><p>The uniformly rotating disc could thus be thought of as a hyperbolic plane, and planet earth would be one of the Poincarites that inhabit such a two-dimensional world. It is a well-known property of the hyperbolic plane that we would be unable to determine our position in the plane because if we proceeded in any direction our rulers and clocks would increase or decrease along with us so that it would be impossible to distinguish our position with say an observer on the rim of the disc. Recall that Einstein believed that the watch of an observer on the rim would go slower than the inertial observer located at the center of the disc. This, however, is impossible in a hyperbolic world. There are no privileged positions on the disc. </p>
<p>Perhaps the simplest way to understand the effect of gravity is to transform the Poincare disc model to the upper half-plane. An isometry will take the disc to the upper half-plane which can be likened to an unevenly heated region where the $x$-axis represents "infinite" cold [cf. <em>A New Perspective in Relativity</em>, Ch. 2]. The strength of gravitation attraction could then be measured in terms of temperature of the unevenly heated half-plane---an entirely scalar operation.</p>
<p>Gravity in such a world would appear to us static just like the ether in the Michelson-Morley experiment. The only way it could be brought alive is by the way is to shine light from the outside which would see as a refraction due a medium with an index of refraction different than one. So light can measure the effects that gravitational fields have on it.</p>
<p>Albeit all this is speculative, but no less speculative than gravitational waves which need a medium to propagate in like mechanical waves, yet are polarized, albeit a putative tensorial polarization. Just by the fact that they can be derived from a gravitational analog of electrodynamics makes that tensorial character precarious as well as the fact that they arise from a linearization of Einstein's equation for weak fields.</p>
<p>And pinpointing the source of gravitational waves is no less precarious. If light cannot escape the event horizon of a black hole neither should gravity. How can there exist a binary black holes if there separation is greater than their event horizons? The last phase before merger consists of ringing, implying a rapid change in frequency, and radiation. But is this frequency that of the beam in the interferometer or that appearing in Kepler's III to describe the evolution of the merger. And how is this compatible with Kepler's laws for don't forget that Kepler's laws are compatible with an inverse square law, and not an inverse cubic that would result, say, in Cotes' spirals.</p>
<p>Newton foresaw such a possible by adding on the cubic to his inverse square law. That would modify the angular dependency while keeping the radial part constant. In fact, it preceded the former: Proposition 9 dealt with motion along an equal angular spiral under the action of a central force which must be cubic. Proposition 11 dealt with motion to a focus in an ellipse where the force law was inverse square.</p>
<p> </p>
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Sun, 09 Sep 2018 11:53:14 +0000adminlavenda214 at http://bernardhlavenda.comHow to Use A Michelson Interferometer to Measure the Newtonian Force
http://bernardhlavenda.com/node/213
<span class="field field--name-title field--type-string field--label-hidden">How to Use A Michelson Interferometer to Measure the Newtonian Force</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Sat, 09/08/2018 - 18:00</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>LIGO chose to measure the strength of gravitational waves by using a Michelson type interferometer while recognizing that the "ripples in the fabric of spacetime cause the frequency of the laser light to fluctuate "ever so slightly" as well as admitting to the fact that "a gravitational wave does stretch and squeeze the wavelengths of light in the arms." However, LIGO adds that "it turns out that it doesn't matter. What matters is how long the laser beam spends traveling in each arm. When a gravitational wave passes, it changes the length of the arms,which changes how far each laser beam needs to travel before being reunited with its partner beam."</p>
<p>How then can you get interference, either constructive or destructive if the wavelength of the laser beam changes, and the lengths of the arms also change? Are we back to Weber's dumbbells which does not use the principle of the interferometer? LIGO also insists that the speed of light will always be $c$---which is impossible if the wavelength changes. The frequency of the beam must always be the same in any Michelson interferometer. So what is going on? Why haven't we seen any interference patterns and fringes? Obviously, you can't if everything is affected by the passage of a gravitational wave.</p>
<p>I believe it is possible to measure the strength of the gravitational (or electrical) force using a Michelson interferometer. Gravity will affect the passage of a ray of light like a medium with an index of refraction, $n>1$. A Michelson interferometer is know to measure the index of refraction, say of air, by comparing one arm with nothing in it and the other arm filled with air. Instead of air, two masses can be placed in one arm, and this will change the index of refraction. If $n$ changes by an amount $\Delta n$, the path length, which was originally $2nL$ changes to $2\Delta n L$, $L$ is the length of the arm and the $2$ means that light passes twice because it is reflected from the far mirror.</p>
<p>Moving the masses together, the pattern will shift by an amount $\Delta n=\lambda/2L$, where $\lambda$ is the wavelength of the laser light that is used: it must be constant for fringes to occur, contrary to what LIGO claims. A shift of $m$ fringes will occur when $\Delta n=m\lambda/2L$.</p>
<p>Consider a refractive index of the form</p>
<p>$$n=1+\frac{GM}{r},$$</p>
<p>compatible with Kepler's laws. When the space between the two masses changes, the index of refraction will be changed to </p>
<p>$$|\Delta n|=F_N|\Delta r|,$$</p>
<p>where $F_N=GM/r^2$, Newton's inverse square law. This can be related to the number of fringes shifted $m$ by</p>
<p>$$F_N|\Delta r|=m\lambda/2L.$$</p>
<p>The phase shift will be $\Delta\phi=c\Delta t/\lambda$ so that the strength of the force can be measured by the phase shift according to</p>
<p>$$F_N=\frac{m}{2L\Delta\phi}.$$</p>
<p>This is completely static as it must be, and shows that the gravitational force changes the index of refraction of the medium. Light cannot propagate at speed $c$ when $n\neq 1$.</p>
<p>A similar expression can be found in Saulson's article "If light waves are stretched by gravitational waves, how can be use light as a ruler to detect gravitational waves" [AJP 65 (1997) 501]. After a lengthy discussion about how gravitational waves do stretch and squeeze both the laser beam and interferometer arm, he merely claims that "the phase shift is related to the strength of the gravitational wave by the relation, $\Delta\phi=2\pi L h_0/\lambda$", without specifying what $\lambda$ is, and where $h_0$ is the amplitude of the wave.</p>
<p>This is in contradiction with his earlier statement that "there is no direct sense in which we observe the wavelength of light in the arms. Our observations are instead of the phase of light, that is, the arrival time of the wave crests." But,"[t]he distance between successive wave crests (i.e., the wavelength) is $\lambda=c\nu^{-1}$ throughout the arm." So on the one hand the distance between successive wave crests is the wavelength, but because gravitational waves squeeze and stretch the wavelength, it is not!</p>
<p>It is clear from Saulson's discussion that the speed of light cannot be constant, traveling at $1\pm h(t)$ times $c$ in each of the orthogonal arms. It is therefore difficult to see how "we can use the travel time of light as a reliable ruler under most conditions, in spite of the stretching of light that goes on when space expands." </p>
<p>Other unsubstantiated statements are: "the gravitational wave causes no phase shift at the beam splitter immediately after its arrival." As we have seen above, the Michelson interferometer is completely stationary. It cannot distinguish at time $\tau$ when "[a]ll the other wave crests suddenly arrive at $t=\tau$ become farther from the beam splitter than they were before. Gravitational wave or no, light travels through the arm at the speed of light, $c$. The physically observable meaning of the stretching of space is that the light in it has to cover extra distance, and so will arrive late. And, since each successive wave crest has to cover a larger extra distance to make it back to the beam splitter, the total time delay (or phase shift) builds up steadily until all of the light that was in the interferometer are at $t=\tau$ finally makes it back to the beam splitter. This phase shift is observable, and it builds up over the storage time $\tau_s=2L/c$ of the interferometer arm."</p>
<p>Whatever type of interferometer accomplishes this, it is certainly not of the Michelson type! </p>
</div></div>
Sat, 08 Sep 2018 16:00:25 +0000adminlavenda213 at http://bernardhlavenda.com Schwarzschild's Outer Metric Is Incompatible with Kepler's III Law
http://bernardhlavenda.com/node/212
<span class="field field--name-title field--type-string field--label-hidden"> Schwarzschild's Outer Metric Is Incompatible with Kepler's III Law</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Tue, 08/28/2018 - 11:20</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>Brown, in his <em>Reflections on Relativity</em> attempts "to give a very plausible (if not entirely rigorous) derivation of Schwarzschild's metric purely from [a] knowledge of the inverse square characteristic of gravity, Kepler's third law for circular orbit, and the null intervals of light paths."</p>
<p>Brown bases his derivation on the space-time metric</p>
<p>$$d\tau^2=g_{tt}dt^2-r^2d\phi^2,$$</p>
<p>for purely circular motion in the equatorial plane ($\theta=\pi/2$), and at constant radius ($dr=0$). However, for null intervals of light paths $d\tau=0$ so his demonstration reduces to $g_{tt}=r^2\omega^2$, where we introduced the angular speed as $\omega=d\phi/dt$, with $t$ lab time. There is no constant of integration, and, even if we were to introduce Kepler's law</p>
<p>$$r^3\omega^2=GM,$$</p>
<p>we would get the desired result, $g_{tt}=1-2GM/c^2r$, which is the time component of the Schwarzschild metric. </p>
<p>Poor Kepler didn't know the difference between local time, $\tau$, and coordinate time $t$, and, even if he had known, it wouldn't have changed any of his law, except maybe for small relativistic corrections. But, we want to use Kepler's III as it stands. Consequently, we work with the space part of the metric,</p>
<p>$$dt^2=g_{rr}dr^2+r^2d\phi^2.$$</p>
<p>One could object at this point and say we put $g_{tt}=1$ to derive the above equations. However, it will turn out that</p>
<p>$$g_{tt}g_{rr}=1,$$</p>
<p>so that multiplying each of the metric coefficient by the same factor does not change the curvature. </p>
<p>For the metric</p>
<p>$$dt^2=E(r)dr^2+G(r)d\phi^2,$$</p>
<p>where the metric coefficient depend only on the radial coordinate $r$, the Gaussian curvature is</p>
<p>$$K_G=-\frac{1}{2\sqrt{EG}}\frac{d}{dr}\left[\frac{dG/dr}{\sqrt{EG}}\right].$$</p>
<p>Putting $E=(1-2GM/r)^{-1}$ and $G=r^2$ into the expression for $K_G$, we find</p>
<p>$$K_G=-\frac{GM}{r^3c^2}=-\frac{\omega^2}{c^2}<0,$$</p>
<p>where we used Kepler's III to get the second equality.</p>
<p>We want to compare this with Kepler's laws. Newton found the radius of curvature is given by</p>
<p>$$\rho=r/\sin\alpha=c/\omega>0,$$</p>
<p>where $\sin\alpha=r\omega/c$. The radius of curvature appear's in Newton's expression for the centripetal acceleration, $v^2/\rho$. (Please refer to a previous blog for the derivation.) </p>
<p>The expression for Gaussian curvature says that Schwarzschild's outer metric is dealing with a hyperbolic point, whereas Newton's radius of curvature says that it is an elliptic point. This is apparent from Kepler's I. The two are incompatible with one another, and we had no right to use Kepler's III to reduce the Gaussian curvature to a (negative) constant. Schwarzschild's outer solution is a hyperbolic metric of non-constant curvature.</p>
<p>Instead, if we set $E=(1-r^2\omega^2/c^2)^{-1}$ and $G=r^2$, we come out with a <em>positive </em>Gaussian curvature,</p>
<p>$$K_G=\frac{\omega^2}{c^2}>0.$$</p>
<p>Thus, we can set </p>
<p>$$K_G=1/\rho^2,$$</p>
<p>showing that the metric of the inner solution is compatible with Kepler's laws. In fact, we were wrong to apply Kepler's III to the outer solution, which has a non-constant curvature---and there is absolutely nothing that can make it constant. </p>
<p>We have remarked that if we divide the metric coefficients by the same factor, it will not change the curvature. Conformal transformations of the metric keep angles constant but change the magnification so that should not change the curvature. So if we take $E=(1-r^2\omega^2)^{-2}$ and $G=r^2/(1-r^2\omega/c^2)$, we come out with the well-known Beltrami metric, which I have shown to describe a uniformly accelerating disc. There is no need to apply Kepler's III since the metric is one of constant (positive) curvature.</p>
<p> </p>
<p> </p>
<p> </p>
</div></div>
Tue, 28 Aug 2018 09:20:48 +0000adminlavenda212 at http://bernardhlavenda.comHow Kepler's Laws Destroy the LIGO Analysis of the Interferometer Strain Signal
http://bernardhlavenda.com/node/211
<span class="field field--name-title field--type-string field--label-hidden">How Kepler's Laws Destroy the LIGO Analysis of the Interferometer Strain Signal</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Mon, 08/27/2018 - 10:38</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>We will show that Kepler's III destroys the independence of the frequency $\omega$ and its rate of change in time, $\dot{\omega}$ so that they cannot be varied independently to determine the 'chirp' mass $\eta^{3/5}M$ from</p>
<p>$$ G(\eta N)^{3/5}M=c^3\left(\frac{\dot{\omega}}{3\omega^{11/3}}\right)^{3/5}.$$</p>
<p>where $N$ is a numerical factor, $32/5$ for linear GR, and $N=2/5$ for the EM-like model discussed by Hilborn. For equal masses $\eta^{3/5}M=M/\sqrt[5]{2}$ so that Kepler's third law</p>
<p>$$GM=a^3\omega^2,$$</p>
<p>will always apply, where $a$ is the distance of separation between the two masses. However, Kepler's III is used when convenient, and, at other times, forgotten.</p>
<p>Kepler's III implies that the right-hand side of the equation be constant, </p>
<p>$$\frac{\dot{\omega}}{\omega^{11/3}}=\kappa=\mbox{const.}$$</p>
<p>It can be immediately integrated to yield the period, $P=2\pi/\omega$</p>
<p>$$P(t)=\left(P_0^{8/3}-\frac{8}{3}\kappa(t-t_0)\right)^{3/8},$$</p>
<p>where $P_0$ is the period of the orbit at time $t=t_0$. The constant $\kappa$, is a constant times $(a\omega/c)^5$ that we discussed in previous blogs.</p>
<p>From the LIGO-VIRGO data for GW950914, and using the linear GR value of $N$ they found a chirp mass of approximately $30 M_{\circledcirc}$. According to the collaboration, different values of $M$ will lead to different results for the mass separation, $a(t)$, <em>for a given orbital frequency</em>. This is incorrect because of the Keplerian constraint,</p>
<p>$$3\frac{\dot{a}}{a}=-2\frac{\dot{\omega}}{\omega}; $$</p>
<p>that is a change in $a(t)$ imposes a change in $\omega(t)$. But let's not worry about this for the moment.</p>
<p>The waveform plots "strain" versus time and looks like the one in the diagram that LIGO obtained for the event GW150914 (magnified $10^{21}$).</p>
<img alt="strain vs time" data-entity-type="file" data-entity-uuid="fc5a216f-25ed-448b-bd7b-f4fd5db2e792" src="http://bernardhlavenda.com/sites/default/files/inline-images/waveform%201.jpg" class="align-center" /><p>The question, among many others, is how they got an explicit time dependency for the wave strain? Not being able to follow the logic behind numerical relativity, I will limit myself to the EM-like model discussed by Hilborn, which captures the salient features except for the angular dependencies.</p>
<p>The EM-like model uses retarded times and positions to extract explicit time dependencies. Let $R$ be the distance between the center of mass and the point of observation. The gravitational analogue of the vector potential is claimed to be</p>
<p>$$\vec{A}=\frac{G\eta M\omega^2}{c^3 R}\vec{r}(\hat{R}\cdot\vec{r}).$$</p>
<p>Retarded and present times are related by</p>
<p>$$c(t-t_R)=R.$$</p>
<p>And the field generated at $\vec{x}_0$ is related to it at time $t$ by</p>
<p>$$\vec{r}(t)=\vec{x}_0-\vec{v}t.$$</p>
<p>The position and velocity of one of the masses is</p>
<p>$$\vec{r}=r\left[\hat{x}\cos(\omega t_R)+\hat{y}\cos(\omega t_R)\right],$$</p>
<p>and</p>
<p>$$\vec{v}=\omega r\left[-\hat{x}\sin(\omega t_R)+\hat{y}\cos(\omega t_R)\right],$$</p>
<p>where $\hat{x}$ and $\hat{y}$ are unit normals in the $x$- and $y$- directions, respectively. It is apparent that $r$ and $\omega$ have nothing whatsoever to do with $a$ and $\omega$ of the previous discussion.</p>
<p>Is the assumption that $\vec{r}$ and $\vec{v}$ vary periodically? True the latter is the time derivative of the former, but they must also satisfy</p>
<p>$$r\left[\hat{x}\cos(\omega t_R)+\hat{y}\sin(\omega t_R)\right]=\vec{x}_0-\omega r\left[-\hat{x}\sin(\omega t_R)+\hat{y}\cos(\omega t_R)\right]t.$$</p>
<p>For simplicity, we place the source at the origin, and compare components. Along the $x$-axis, we have</p>
<p>$$\cos(\omega t_R)=\omega t\sin(\omega t_R)t.$$</p>
<p>From this, and the relation between past and present times we get</p>
<p>$$\cot(\omega t_R)-\omega t_R=\omega R/c,$$</p>
<p>which cannot be valid for all times, and values of $\omega$ since $R>0$, by definition. Yet, this is the origin of the explicit time dependencies in the EM-like model which is claimed to be consistent with linear GR.</p>
<p>Hilborn then claims that on the strength of the periodicities in $\vec{r}$ and $\vec{v}$ that</p>
<p>$$\frac{d\vec{v}}{dt}=-\omega^2\vec{r},$$</p>
<p>apparently not realizing that this is the condition for the vanishing of the radial force,</p>
<p>$$\vec{F}_{rad}=\ddot{\vec{r}}-\omega^2\vec{r}=\vec{0}.$$</p>
<p>But, this is, indeed, a consequence of the assumption that position and velocity are periodic in time.</p>
<p>It is, therefore, no wonder that the gravitational vector potential is written as</p>
<p>$$\vec{A}=\frac{G\eta M\omega^2r^2}{2c^3R}\sin\theta\left[\cos\theta(1+\cos(2\omega t_R)\hat{\theta}+\sin(2\omega t_R)\hat{\phi}\right],$$</p>
<p>where $\hat{\theta}$ and $\hat{\phi}$ are the polarization unit vectors. The fundamental expression for the gravitational acceleration, $\vec{g}$, is the negative time derivative. Since we intend to project along the $x$-axis, where $\hat{x}\cdot\hat{\phi}=0$, the expression for the gravitational acceleration vector is</p>
<p>$$\vec{g}=\frac{\partial\vec{A}}{\partial t_R}=-\frac{G\eta M\omega^3r^2}{3c^3R}\sin(2\theta)\sin(2\omega t_R)\hat{\theta}.$$</p>
<p>This means that $r$ and $\omega$ have absolutely nothing to do with $a$ and $\omega$ that we introduced previously, and claimed satisfies Kepler's laws.</p>
<p>Assuming the mass is free to move in the $x$-direction, projecting $\vec{g}$ along $x$, $\hat{x}\cdot\hat{\theta}=\sin\beta$, Hilborn get the harmonic oscillator equation</p>
<p>$$\ddot{x}=\vec{g}\cdot\hat{x}=\frac{G\eta M\omega^3r^2}{3c^2R}\sin\beta\sin(2\theta)\sin(2\omega t),$$</p>
<p>conveniently forgetting to indicate retarded time of which all the variables are function of. Now, of all things, we are told to integrate this equation to get</p>
<p>$$x(t)=-\frac{\vec{g}\cdot\hat{x}}{(2\omega)^2},$$</p>
<p>where the $2$ comes from the fact that the wave frequency is twice the orbital frequency. We are told that "the minus sign simply tells us that the displacement and forcing term are π radians out of phase for the sinusoidal driving of an object at a frequency far above its natural oscillation frequency." But the radial force is zero!</p>
<p><em>The above integration in time is valid provided</em></p>
<p>$$ \omega^3a^2=\mbox{const}.$$</p>
<p><em>This is not Kepler's III, $a^3\omega^2=\mbox{const}.$ </em>We can stop here, but let us continue.</p>
<p>By a series of unscrupulous procedures, the strain equation finally emerges,</p>
<p>$$\Delta L=\Delta x_2(t)-\Delta x_1(t)=\frac{G\eta M\omega^2r^2}{4c^3R}\Delta t\sin\beta\sin(2\theta)\cos(2\omega t).$$</p>
<p>Anyone who has persevered this far will be rewarded: Introducing $\Delta t=\Delta R/c=L\cos\beta/c$, where $L$ is the length of the arm of the interferometer, and <em>using "Kepler's law to replace $r$ in favor of $\omega$" </em> result in</p>
<p>$$\Delta L=\frac{(GM)^{5/3}\eta\omega^{2/3}}{8c^3R}L\sin(2\beta)\sin(2\theta)\cos(2\omega t).$$</p>
<p>Regarding the first substitution, $\Delta t=t-t_R=R/c$, and no one has varied the past position, while regard the second only confirms our suspicions that $r$ is $a(t)$ and $\omega$ is $\omega(t)$. Holding them constant in above integration is in violation of Kepler's III.</p>
<p>Now the procedure used to calculate the gravitational waveform is</p>
<p>1. Select an initial value of the separation of the masses that is a few times the combined Schwarzschild radii of the binary objects. (The strong in-spiral of the binary objects begins when the separation of the masses is a few times the combined Schwarzschild radii.) </p>
<p>2. Use</p>
<p>$$\dot{r}=r_s\left[\left(\frac{r_0}{r_s}\right)^4-\frac{N\eta c}{r_s}(t-t_0)\right]^{1/4},$$</p>
<p>which we discussed in the previous blog, to find $r(t)$ for a sequence of time values.</p>
<p>3. From $r(t)$, use Kepler’s III to find $\omega(t)$ .</p>
<p>4. Use that value of $\omega(t)$ in the above expression for $\Delta L$, and repeat steps 3 and 4 for the next time value.</p>
<p>5. Plot the results.</p>
<p>It's step 4 that ruins the procedure for it has been derived using $\omega^3a^2=\mbox{const}$, and not Kepler's III, $a\omega^2=GM/a^2$.</p>
<p>The conclusion that "the relative displacement oscillates at twice the orbital frequency of the source and the amplitude depends on the combination $\eta M^{5/3}$ multiplied by $\omega^{2/3}$, exactly the dependence predicted by linear GR" is inaccurate. The explicit periodic time dependencies comes from a misuse of retarded time and position, i.e. $\vec{r}$ and $\vec{v}$ cannot be periodic functions of time because past will become present and present will become past on a regular basis. Moreover, Kepler's law cannot be used when it proves convenient: either it applies, or doesn't apply. The expression for the chirp mass explicitly depends on Kepler's III, but the analysis leading up to the final result that relates the strain $\Delta L$ does not--nor does the analysis of the LIGO-VIRGO data.</p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
</div></div>
Mon, 27 Aug 2018 08:38:24 +0000adminlavenda211 at http://bernardhlavenda.comGravitation Waves Aberrate!
http://bernardhlavenda.com/node/210
<span class="field field--name-title field--type-string field--label-hidden">Gravitation Waves Aberrate!</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Sun, 08/26/2018 - 13:34</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>$\ldots$ so say LIGO [Abbott <em>et al</em>, "Observation of gravitational waves from a binary black hole merger", (Hulse) Taylor and Weisberg, "The relativistic binary pulsar B1913+16: thirty years of observation and analysis", and Peters and Mathews, "Gravitational radiation from point masses in a Keplerian orbit".</p>
<p>In a tutorial Hilborn, in his tutorial on the derivation of gravity waves without general relativity, asks if the luminosity of gravitational waves,</p>
<p>$$L=\frac{2}{5}\frac{\mu}{c^5}a^4\omega^6$$</p>
<p>where $\mu=GM$, can be understood in simple terms. He asks " Come up with a better explanation of the physics contained in [the above] and share it with the author of this tutorial. He is always looking for better ways of understanding important results." How about writing it as</p>
<p>$$L=\frac{2}{5}\mathcal{T}\cdot v\left(\frac{v}{c}\right)^5?$$</p>
<p>The first term is the rate of energy loss: $GM^2/a^2\cdot v$, where $a$ is the separation between the binaries, and $v=a\omega$ is the angular speed, $\omega=2\pi/P$, with $P$ the period of the orbit. The last term is the aberrational factor---which is of fifth order! This says that gravitational waves aberrate just like electromagnetic waves---except weaker than because it is of fifth order.</p>
<p>Why they insist on the fifth order is because of strained analogy with the source of the radiation: a quadrupole. The quadrupole, $Q$, is proportional to the moment of inertia, $Mr^2$ and its third order derivative is $\ddot{Q}\sim Mr^2\omega^3$. Hence, the rate of energy loss is proportional to $|\ddot{Q}|^2$, and this is the connection with Einstein's calculation of a rotating dumbbell. Conservation of mass prohibits monopole radiation, conservation of momentum that of dipole radiation so the next best thing is quadrupole radiation. However, viewed as an aberrational effect there is nothing that cancels out the third order term, which is</p>
<p>$$L\sim\mathcal{T}\cdot v\left(\frac{v}{c}\right)^3=\frac{\mu}{a^2}a\omega\left(\frac{a\omega}{c}\right)^3.$$</p>
<p>By the same token, this term would be proportional to the square of the second time derivative of the dipole moment, $\mu a\omega^2$.</p>
<p>Setting this term equal to the surface luminosity would result in a temperature,</p>
<p>$$kT=\sqrt[4]{\frac{\mu}{a}(\hbar\omega)^3\left(\frac{a\omega}{c}\right)},$$</p>
<p>which would show a weak aberrational dependence. However, the analogy is flawed, and this has nothing to do with dipole radiation.</p>
<p>The luminosity is given by the thermal black-body formula</p>
<p>$$L=\sigma T^4 4\pi a^2,$$</p>
<p>where $\sigma=k^4/\hbar^3c^2$, leaving out inessential numerical factors. Equating the two expressions we derive an expression for the absolute temperature,</p>
<p>$$kT=\sqrt[4]{\frac{GM}{a}(\hbar\omega)^3\left(\frac{a\omega}{c}\right)^2}.$$</p>
<p>This is, indeed, surprising because it says that the temperature will undergo aberration, appearing as a squared term as if for a potential. First order aberration would also not work. But, for third-order we get</p>
<p>$$kT=\sqrt[4]{\frac{GM}{a}(\hbar\omega)^3},$$</p>
<p>which shows that temperature does not aberrate. It also tells you the composition of factor that make up the absolute temperature: one part gravitational and three parts electromagnetic. The latter can only be turned into the former by considering specific examples as we do below.</p>
<p>For black body cavity travelling at the speed $v$, the temperature increases by the factor $1/\sqrt{1-v^2/c^2}$, where $v$ is the translational speed of the cavity. If light is traveling at speed $c$ in a given direction, and the cavity at speed $v$ in another, then the decrease in velocity is proportional to the distance $c\sqrt{1-v^2/c^2}$, since $c$ is the hypotenuse of a right triangle with base $u$. This is not aberration.</p>
<p>Thermodynamically, the temperature can be increased by the motion because the heat absorbed increases by the same factor, but their ratio, the entropy, must remain invariant. Said plainly, disorder cannot be used to tell how fast the cavity is moving. But, as $v\rightarrow 0$, so the temperature returns to its stationary value. Not so if the temperature is affected by aberration.</p>
<p>Consider the Hertzprung--Russell diagram, where there is a known relation between luminosity and mass for binary stars. For the lower part of the curve, the mass $M$ increase from $0.1 M_{\circledcirc}$ to $M_{\circledcirc}$ as $L$ increases by a factor of $!0^3$. This means $L\sim M^3$ in this part of the curve. In contrast in the upper part of the curve $M$ changes by a factor of $2$ results in a change in $L$ by a factor of $20$. This means that $L\sim M^4$ for binary stars on this part of the curve.</p>
<p>Rather than considering the surface luminosity, $L$, we consider the total energy density (again omitting inessential numerical factors),</p>
<p>$$E=\frac{(kT)^4}{(\hbar c)^3}a^3,$$</p>
<p>and equate it with the energy </p>
<p>$$\frac{\mu}{a}\left(\frac{a\omega}{c}\right)^3,$$</p>
<p>which is the effect that aberration has due to a finite speed of propagation of electromagnetic waves</p>
<p>$$\frac{GM}{a}\psi,$$</p>
<p>where the angle</p>
<p>$$\psi=\theta-\tan^{-1}\theta=\theta-\theta+\frac{1}{3}\theta^3-\frac{1}{5}\theta^5+\cdots$$</p>
<p>and $\theta=a\omega/c$, since $a/c$ is the time it takes light to propagate a distance $a$. Recall that $\psi$ is the difference between where the planet is presently at orbiting a much larger mass, $M$, and where it would have been had it been linearly extrapolated from its past position. The first non-vanishing term is the cubic, not the fifth order as LIGO has assumed. But LIGO took their expression from Hulse and Taylor, and they took it from Peters and Mathew.</p>
<p>Equating the radiation energy to the gravitational one leads to a temperature of</p>
<p>$$kT=\sqrt[4]{\frac{\mu}{a}(\hbar\omega)^3},$$</p>
<p> without aberration, as it should be. The temperature is one quarter gravitational and three-quarters electromagnetic. For the binaries in the upper part of the HR curve, $\hbar\omega=GM/a$, while in the lower part of the curve, $\hbar\omega=(GM/a)^{2/3}$. This means that for all binaries on the main sequence the ratio of the mass to radius is constant and the core temperature is of the order $10^7 K$.</p>
<p>All the analyses use Keplerian dynamics, so why shouldn't we? Newton wrote the centrifugal force as</p>
<p>$$\mathcal{T}=\mathcal{F}\cos\vartheta=\mathcal{F}\sin\alpha=\frac{v^2}{\varrho},$$</p>
<p>where $\vartheta$ is the angle between the centripetal, $\mathcal{T}$, and centrally pointed force, $\mathcal{F}$, and $\alpha$ is the complementary angle. Since it is is given by</p>
<p>$$\sin\alpha=\frac{a\omega}{v_G},$$</p>
<p>Newton considered a first-order aberrational effect. $\varrho$ is the radius of curvature, and Newton found it to be $\varrho=a/\sin\alpha$. Hidden away is the fact that the radius of curvature depends upon the speed at which gravity propagates, $v_G$. For an infinite speed the curvature tends to zero.</p>
<p>If we use the above expression in conjunction with Kepler's II,</p>
<p>$$av\sin\alpha=K,$$</p>
<p>the conservation of angular momentum, which is impervious to the speed at which gravity propagates. Introducing the expression for $\sin\alpha$, converts Kepler's second into Kepler's third</p>
<p>$$a^3\omega^2=\mu.$$</p>
<p>In fact, Newton's law of inverse square,</p>
<p>$$\mathcal{F}=a\omega^2=\frac{\mu}{a^2},$$</p>
<p>implies $av^2=\mbox{const}$, which is just Kepler's III. Of course we can always multiply </p>
<p>$$\mathcal{F}\sin\alpha=\frac{\mu}{\varrho}$$</p>
<p>by any power of $\sin\alpha$ without disturbing the Keplerian nature of the problem. What it does is to change the expression for the centripetal force.</p>
<p>Consider the secular expression for the square of the semi-major axis,</p>
<p>$$a^2(t)=a_0^2+4\frac{\mu}{v_G}(t-t_0).$$</p>
<p>we can always arrange it that $a_0^2=4\mu t_0/v_G$, at the initial time. Taking the equation to the power $3/2$ and multiplying it by $\omega^2$, we get</p>
<p>$$\frac{v}{v_G}=\frac{1}{M},$$</p>
<p>where</p>
<p>$$M=\omega t=E-\varepsilon\sin E$$</p>
<p>is the mean anomaly, $E$ (not to be confused with total energy used above), the eccentric anomaly, and $\varepsilon$ the eccentricity. To verify the validity of the above expression for the mean anomaly, we differentiate the equation with respect to time we find</p>
<p>$$\frac{\dot{v}}{v_G}=-\frac{v}{a} v_G\left(\frac{v}{v_G}\right)^2=-\frac{\mu}{a^2}\frac{v}{v_G}=-\mathcal{F}\sin\alpha,$$</p>
<p>which is the first equality in Newton's relation, except for the minus sign, where we have used</p>
<p>$$\dot{E}=\sqrt{\frac{\mu}{a}}(1-e\cos E)^{-1}.$$</p>
<p>It is not hard to verify that Kepler's III is satisfied for <strong>all</strong> powers of the aberration, but the first non-vanishing term in the expansion of $\psi$. </p>
<p>That term has very simple and important properties with regard to luminosity. The secular equation is</p>
<p>$$a^3(t)=a_0^3+4\frac{\mu^2}{v_G^3}(t-t_0),$$</p>
<p>and again we cancel initial conditions. Taking the time derivative leads to</p>
<p>$$\dot{a}=\frac{4}{3v_G^3}\left(\frac{\mu}{a}\right)^2.$$</p>
<p>Using Kepler's III in the form</p>
<p>$$\frac{3}{2}\frac{\dot{a}}{a}=-\frac{\omega}{\omega}=\frac{\dot{P}}{P},$$</p>
<p>where $P$ is the period of the orbit, we come out with</p>
<p>$$\dot{P}=12\pi\psi,$$</p>
<p>where the angle $\psi$ is given by the cubic term in the above expansion. This says that the rate of change of the period is entirely due to the retarded action of the force which causes the motion and travels at a finite speed, $v_G$.</p>
<p>I hasten to add that, except for the connection with the luminosity expression and absolute temperature, the basic formulas and general conclusion that gravity does not aberrate and gravitational waves, if they exist, do not travel at speed $c$, can be found in van Flandern's paper, "The speed of gravity: What the experiments say". This paper was unduly criticized by Carlip and Kevin Brown. It will not due to replace the permittivity $\epsilon_0$ and permeability $\mu_0$ of free space by</p>
<p>$$\frac{1}{4\pi\epsilon_0}\rightarrow-G\hspace{20pt}\mbox{and}\hspace{20pt}\frac{\mu_0}{4\pi}\rightarrow-\frac{G}{c^2},$$</p>
<p>and use electromagnetism to derive the expression for the rate of decrease of energy for "gravitational waves". There is no conclusive experiment that warrants such a substitution. When comparing the observed and predicted periods and their rates of change of the binary pulsars PSR1913+16 and PSR1534+12, van Flandern wrote that "[a]t a glance, we see there is no possible match. The predicted period changes that would result if gravity propagated at the speed of light in a manner analogous to electromagnetic forces, are orders of magnitude larger than the observed period changes. For PSR1913+16, they have opposite signs as well. From PSR1534+12, we can set a lower limit to the speed of gravity as an electromagnetic-type propagating force $2800c$." </p>
<p>These words are as true as when they were uttered two decades ago. I wouldn't be concerned about the sign of the rate of change of the period for we obtain a sign change when considering Newton's relation from a different viewpoint. However, what we should be concerned about is that if we use the above expression for $\dot{P}$, the order of magnitude is twice that what has been observed, while if the fifth power</p>
<p>$$\dot{P}\sim\left(\frac{a\omega}{c}\right)^5,$$</p>
<p>is used by all the above authors, turns out to be seven powers smaller than what is observed,<em> if gravitational waves propagate at the speed of light.</em> There is nothing in between to fill the gap: aberration is always odd powers in the relative speeds. Even if the latter expression were correct, which it isn't, they would still have to explain why the cubic term is missing. That too satisfies Kepler's law, and even more important it is congruous with the black-body expression for luminosity, which the other expressions are not.</p>
<p>The connection between the third order aberration and the total radiation energy of black bodies is too good to be a mere coincidence. When dealing with rotational motion we note a peculiarity insofar as the index of refraction is proportional to the speed rather than inversely proportional as it it for the usual form of Snell's law. Here too, there is a peculiarity in that the mean anomaly, $M$, which would correspond to the angle $\theta$ in the above expansion of $\psi$ should be proportional to one another, and not inversely proportional to each other as it turns out. Recall that $\theta$ is the angle of the sector traced out by the planetary motion, from past to present positions.</p>
<p>As a final point, claims have been made that there is a miraculous cancellation of the first order aberration terms that was made by Carlip, "Aberration and the speed of gravity" and Ibison <em>et al, </em>"The speed of gravity revisited". They concluded that time-retardation only affects the magnitude of the force, but not the direction, of the field. However, there is one small point, they considered the time-retarded <em>radially directed </em>force, the electric field, analogous to $\mathcal{F}$, and, of course, this does not manifest aberration, as Newton well-knew.</p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
<p><br />
</p>
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Sun, 26 Aug 2018 11:34:26 +0000adminlavenda210 at http://bernardhlavenda.comGravitational waves from orbiting binaries without general relativity: a tutorial on not how to do physics
http://bernardhlavenda.com/node/209
<span class="field field--name-title field--type-string field--label-hidden">Gravitational waves from orbiting binaries without general relativity: a tutorial on not how to do physics</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Thu, 08/23/2018 - 15:20</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>The best way to show how ridiculous is the theory of gravitational waves is to present the arguments used by Hilbron in a tutorial on gravitational waves that does not use general relativity. Although his expressions for the polarization, angular distributions, and overall power results differ from those of GR, waveforms that are very similar to the pre-binary-merger portions of the signals observed by the Laser Interferometer Gravitational-Wave Observatory (LIGO-VIRGO) collaboration.</p>
<p>The power radiated by gravitational waves without keeping track of numerical factors is</p>
<p>$$\dot{E}=\mathcal{T}v$$</p>
<p>where the transverse force per unit mass is</p>
<p>$$\mathcal{T}=\frac{\mu}{a^2}\sin^5\alpha,$$</p>
<p>with</p>
<p>$$\sin\alpha=\frac{v}{c},$$</p>
<p>is the aberration factor and $v=a\omega$ is the transverse velocity. </p>
<p>Everyone claims that this is compatible with Keplerian motion. This is inaccurate. Newton wrote the transverse force per unit mass as</p>
<p>$$\mathcal{T}=\mathcal{F}\sin\alpha=\frac{v^2}{\rho},$$</p>
<p>where $\mathcal{F}$, is the force per unit mass, directed to the center of the source, $\rho$ is the radius of curvature. Introducing Kepler's II,</p>
<p>$$av\sin\alpha=K,$$</p>
<p>into the above results in</p>
<p>$$\rho\sin^3\alpha=\frac{K^2}{\mathcal{F}a^2}.$$</p>
<p>Newton calculated the radius of curvature using his theory of fluxons and found that the left-hand side is constant. Hence, he established his inverse square law,</p>
<p>$$\mathcal{F}=\frac{\mu}{a^2}.$$</p>
<p>QED</p>
<p>The transverse force is therefore not given by the above expression but rather by</p>
<p>$$\mathcal{T}=\mathcal{F}\sin\alpha,$$</p>
<p>as Newton reasoned. </p>
<p>If we plug in $\sin\alpha=a\omega/c$ in the expression for the conservation of angular momentum, we get</p>
<p>$$av^2=Kc,$$</p>
<p>which amalgamates $Kepler's II with his third, </p>
<p>$$a^3\omega^2=GM.$$</p>
<p>The secular variation of the semi-major axis of the ellipse is</p>
<p>$$a^2=a^2_0+4\frac{\mu}{c}(t-t_0).$$</p>
<p>If we agree that $a_0^2=4\mu t_0/c$ at the initial instant, then on the strength of Kepler's II or III, which are now equivalent,</p>
<p>$$\omega^{-4/3}=\omega_0^{-4/3}+\left(\frac{\mu}{cK}\right)^{1/3}(t-t_0).$$</p>
<p>Hibron finds the differential equation for the semi-major axis</p>
<p>$$\dot{a}=-2N\eta\frac{(GM)^3}{c^5r^3},$$</p>
<p>where $N\eta$ are numerical factors. He "repackages" it to read</p>
<p>$$\dot{a}=-\frac{\eta Nc}{4}\left(\frac{r_S}{a}\right)^3,$$</p>
<p>by introducing the Schwarzschild radius, $r_S=2\mu/c^2$. This he integrates to</p>
<p>$$a^4=a_0^4-N\eta r_S^3c(t-t_0),$$</p>
<p>which definitely does not correspond to the Keplerian result. </p>
<p>We can see this more clearly by considering the equation for the angular speed, $\omega$, which he finds as</p>
<p>$$\dot{\omega}=3\eta N\mu^{5/3}\omega^{11/3}/c^5$$.</p>
<p>Rather than integrating the last equation, he expresses the so-called chirp mass, $\eta^{5/3}M$ in terms of it, viz.,</p>
<p>$$(\eta N)^3 M=\frac{c^3}{3^{3/5}G}\left(\dot{\omega}/\omega^{11/3}\right)^{3/5},$$</p>
<p>which is void of any meaning. Rather, we choose to integrate it, and come out with</p>
<p>$$-\omega^{-8/3}=-\omega_0^{-8/3}+8N\eta\mu^{8/3}(t-t_0).$$</p>
<p>Notice the incorrect negative sign. It is also apparent that when the time is eliminated by the equation for $a^4$, Kepler's III does not result.</p>
<p> </p>
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Thu, 23 Aug 2018 13:20:07 +0000adminlavenda209 at http://bernardhlavenda.comThe Luminosity Attributed to Gravitational Waves is Incompatible with a Black-Body Spectrum
http://bernardhlavenda.com/node/208
<span class="field field--name-title field--type-string field--label-hidden">The Luminosity Attributed to Gravitational Waves is Incompatible with a Black-Body Spectrum</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/user/1" typeof="schema:Person" property="schema:name" datatype="" xml:lang="">adminlavenda</span></span>
<span class="field field--name-created field--type-created field--label-hidden">Tue, 08/21/2018 - 10:43</span>
<div class="clearfix text-formatted field field--name-body field--type-text-with-summary field--label-hidden field__item"><div class="tex2jax_process"><p>In our last blog we derived the speed of gravity from the observational period and variation in the period of the binary pulsar PSR 1913+16 that was discovered by Taylor and Hulse. The luminosity spectrum they used went back to a 1963 paper of Peter and Mathews who took the rate of energy emission of Einstein's rotating dumbbell,</p>
<p>$$L=\frac{G}{5c^5}\left(\frac{d^3Q_{ij}}{dt^3}\frac{d^3Q_{ij}}{dt^3}-\frac{1}{3}\frac{d^3Q_{ii}}{dt^3}\frac{d^3Q_{jj}}{dt^3}\right),$$</p>
<p>and evaluated the quadrupole moments using a Keplerian ellipse</p>
<p>$$d=\frac{a(1-e^2)}{1+e\cos\psi},$$</p>
<p>and the angular "velocity"</p>
<p>$$\dot{\psi}=\frac{G(m_1+m_2)a(1-e^2)]^{1/2}}{d^2}.$$</p>
<p>However, $\dot{\psi}$ will only be the angular <em>speed</em>, which implies that</p>
<p>$$\dot{\psi}^2a^3=\frac{GM(1+e\cos\psi)^4}{(1-e^2)^3},$$</p>
<p>which should be independent of both the eccentricity $e$ and the true anomaly, $\psi$, for then it would reduce to Kepler's III. </p>
<p>Be that as it may, they find a luminosity given by</p>
<p>$$L=\frac{2}{5}\frac{GM^2}{c^5}a^4\omega^6,$$</p>
<p>where $\omega$ is the angular speed of a binary system of two equal masses, each with mass $M/2$ in a circular orbit of radius $a/2$ about the system's center of mass. The LIGO team took this as the "gravitational wave luminosity". How electromagnetic radiation became gravitational radiation is anyone's guess.</p>
<p>However, their expression is incompatible with the luminosity of a star considered as a perfect black body. Omitting numerical factors, the Luminosity is</p>
<p>$$L\sim\omega E\sim \omega\frac{(kT)^4a^3}{(\hbar c)^3},$$</p>
<p>where $E$ is the total radiated energy. In comparison their proposed luminosity can be written as</p>
<p>$$L\sim\omega\frac{GM}{a}\left(\frac{a\omega}{c}\right)^3.$$</p>
<p>In order that this be associated with gravitational waves, we must assume with Poincare' that gravitational waves travel at the speed of light. Over a decade later, Einstein made the same assumption but forgetting to reference the source.</p>
<p>Equating the two expressions, we come to a blatant contradiction: the absolute temperature cannot depend on the speed of light, or for that matter, any other speed. We have point out in a previous blog, "An error in the LIGO calculation of the 'chirp' mass" that the luminosity must be given by</p>
<p>$$L\sim\omega\frac{GM}{a}\left(\frac{a\omega}{c}\right)^3,$$</p>
<p>for in this case the temperature will be given by</p>
<p>$$kT=\left[\frac{GM}{a}(\hbar\omega)^3\right]^{1/4},$$</p>
<p>independent of $c$. </p>
<p>We recall from yesterday's blog that the cubic term in the expression for the luminosity is a third-order aberration effect. Einstein would raise it by a power of two. For then the temperature would be a function of $c$, and temperature measurements would allow the determination of the speed of light. </p>
<p><em>Gravitational waves are not the same as electromagnetic waves confined to a black-body cavity! And even if it were, it would not be given by the Peters-Mathew expression. </em></p>
<p>We may venture to derive the expression for the luminosity of gravitational waves that should (and must) propagate at a velocity $v_G$, which we have no reason to suppose is the same as $c$. The luminosity would be given by</p>
<p>$$L\sim\omega\frac{GM}{a}\frac{a\omega}{v_G}.$$</p>
<p>The corresponding thermal luminosity would be</p>
<p>$$L\sim\omega\frac{(kT)^2}{\hbar v_G}a.$$</p>
<p>This would correspond to a one dimensional system of dimension $a$. The temperature of such a system would be</p>
<p>$$kT=\left(\frac{GM}{a}\hbar\omega\right)^{1/2}.$$</p>
<p>We would also venture so far as the say that the dynamics of a rotating dumbbell is incompatible with Keplerian dynamics, as it is with black-body radiation.</p>
<p> </p>
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Tue, 21 Aug 2018 08:43:39 +0000adminlavenda208 at http://bernardhlavenda.com