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enUnfortunate Similarities Between Geometrical Optics and General Relativity
http://bernardhlavenda.com/node/187
<span class="field field--name-title field--type-string field--label-hidden">Unfortunate Similarities Between Geometrical Optics and 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">Thu, 06/21/2018 - 12:44</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>General relativity generalizes the line element</p>
<p>$$-c^2d\tau^2=-Bc^2dt^2+Adr^2+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2),$$</p>
<p>to the case where the coefficients, $A$ and $B$, appearing the metric can be functions of $r$, and perhaps also $t$. However, if they are functions of $r$, the coefficients will necessarily be functions of $t$ on account of the geodesic equations. The time variables $\tau$ and $t$ distinguish between 'proper', or local, time, and coordinate time, respectively. In the former, the clock is attached to the object in motion, while in the latter, it measures 'laboratory' time. Potentials, whether they be electromagnetic or gravitational, appear in the coefficient of $dt^2. Neglecting the space part, one arrives at conclusions that the ticks of a clock become further apart in a gravitational field, or the red-shift in the frequency is due to the presence of a massive body. These are unfortunate conclusions.</p>
<p>Geometrical optics has a lagrangian of the form</p>
<p>$$\mathcal{L}=\frac{1}{2}\left(A\dot{r}^2+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2)-n^2(r)\right),$$</p>
<p>where $n(r)$ is a varying index of refraction and $A$, a coefficient. The dot denotes a differential with respect to an affine parameter $\tau$, and the index of refraction can be written in terms of the differential of one affine parameter $t$ with respect to it,</p>
<p>$$n^2(r)=B\dot{t}^2,$$</p>
<p>where $B$ is another coefficient. The similarity between coordinate $t$ and 'proper' times $\tau$ is striking---but extremely misleading. Explicit time dependencies does not enter into geometric optics, and any difference in the affine parameters will show up in wavelength changes, and not frequency changes. Time does appear however in the rate of change of $r,\vartheta,\varphi,$ and $t$.</p>
<p>Geometrical optics has been likened to a zero-energy mechanics, by requiring the difference $\dot{r}^2-n^2$, in a spherically symmetric system to vanish [Evans and Rosenquist, <em>AJP </em><strong>54</strong> (1986) 876]. This is incorrect since it does not introduce any limiting speed. Rather, we have shown in a previous blog that the index of refraction can be determined by the Legendre transform of the kinetic energy, $T=(1/2)A\dot{r}^2$, viz.,</p>
<p>$$\Psi(\dot{t})=\dot{r}\frac{\partial T}{\partial\dot{r}}-T,$$</p>
<p>where a new variable $\dot{t}$ is defined by</p>
<p>$$\dot{t}=\frac{\partial T}{\partial\dot{r}}=A\dot{r},$$</p>
<p>so that the new potential $\Psi$ is</p>
<p>$$\Psi(\dot{t})=\frac{\dot{t}^2}{2A}.$$</p>
<p>Subtracting this off the total kinetic energy gives the lagrangian</p>
<p>$$\mathcal{L}=\frac{1}{2}\left(A\dot{r}^2+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2)-\frac{\dot{t}^2}{A}\right).$$</p>
<p>Comparing this with the expression for $n$, we find $B=1/A$.</p>
<p>The Euler-Lagrange equations are given by</p>
<p>$$\ddot{r}+\frac{1}{2}A^{\prime}\dot{r}^2-r\dot{\varphi}^2-\frac{A^{\prime}}{2A^2}\dot{t}^2=0,$$</p>
<p>for the radial coordinate in the plane $\vartheta=\pi/2$, and the first integrals, $r^2\dot{\varphi}=L$ and $\dot{t}/A=\kappa$. </p>
<p>Note that in the definition of the Legendre transform we have used the definition of $\dot{t}=A\dot{r}$, whereas the variational equations identified $\dot{t}/A=\mbox{const.}$ so it is natural to set that constant equal to the limiting speed $\kappa$. Introducing $c$ into the radial equation of motion results in</p>
<p>$$\ddot{r}-r\dot{\varphi}^2=-\frac{A^{\prime}}{2}\left(\dot{r}^2-c^2\right).$$</p>
<p>Since the radial acceleration must be negative, $A^{\prime}<0$. For the Weber force, $A=1/r$ and the vanishing of the force field is</p>
<p>$$\ddot{r}-\frac{1}{2}\frac{\dot{r}^2}{r^2}+\frac{c^2}{r}=0,$$</p>
<p>where $\kappa=\surd 2 c$, is Weber's constant. The index of refraction is $n=c\surd(2/r)$, as those for light orbits.</p>
<p>The vanishing of the Weber force is compatible with the geodesic equations resulting from the Euler-Lagrange equation</p>
<p>$$\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot{r}}-\frac{\partial\mathcal{L}}{\partial r}=0.$$</p>
<p>It is to be emphasized that $\tau$ is <strong>not</strong> proper time, but just some affine parameter that measures the advance along the trajectory $r$. Similarities can often times be very deceiving!</p>
<p>Now, the index of refraction has its foot in two worlds: the world of affine variations and the world of real time changes because of its definition, </p>
<p>$$n=c/v,$$</p>
<p>where $v$ is the velocity of light in a medium with an index of refraction $n$. The angular momentum has been defined with respect to the affine parameter, $\tau$. Let us use $\lambda$ and $\lambda^{\prime}$ for the two affine parameters, and $t$ for time--this time. We can write for any affine parameter $\lambda$ as</p>
<p>$$\frac{c}{n^2}=\frac{d\lambda}{dt}=\frac{d\lambda}{d\lambda^{\prime}}\frac{d\lambda^{\prime}}{dt}.$$</p>
<p>One definition of the index of refraction is $n=d\lambda^{\prime}/d\lambda$, while the other is $n=c/v=c/d\lambda^{\prime}/dt$. If we consider the angular momentum</p>
<p>$$n^2r^2\frac{d\varphi}{dt}=L,$$</p>
<p>we find for the Weber field that the angular velocity $rd\varphi/dt$ is a constant of the motion. This agrees with the interpretation of Evans and Rosenquist.</p>
<p>Regarding the first definition of the index of refraction as the variation of one affine parameter with respect to the other, we have</p>
<p>$$d\lambda^{\prime}=n d\lambda,$$</p>
<p>indicating that a shift in the affine parameter occurs for $n\neq 1$. </p>
<p>For the Schwarzschild metric, the coefficient $A=(1-\mathcal{R}/r)^{-1}$ so that the index of refraction varies as $n=1/(1-\mathcal{R}/r)^{1/2}$. This will cause a shift in the affine parameters by an amount</p>
<p>$$d\lambda^{\prime}=\left(1-\mathcal{R}/r\right)^{-1/2}d\lambda.$$</p>
<p>In a medium of varying index of refraction, the affine parameter, which can be measured in a number of wavelengths of light that is used, will vary. However, nothing can be said about the frequency since in a medium of varying index of refraction the wavelength varies but the frequency does not. If the gravitational shift of spectral lines is determined in terms of wavelength, everything is fine. But, no reddening can occur because the frequency of light is invariant. </p>
<p>In other words, there is no mechanism that will allow the frequency to vary, as in the case of a moving mobile, while light propagates through a completely static field whose index of refraction changes according to the strength of the gravitational potential.</p>
<p> </p>
<p> </p>
<p> </p>
</div></div>
Thu, 21 Jun 2018 10:44:55 +0000adminlavenda187 at http://bernardhlavenda.comHow to Derive the Correct Equations of Motion for the Electro-and Gravito-Fields
http://bernardhlavenda.com/node/186
<span class="field field--name-title field--type-string field--label-hidden">How to Derive the Correct Equations of Motion for the Electro-and Gravito-Fields</span>
<span class="field field--name-uid field--type-entity-reference field--label-hidden"><span lang="" about="http://bernardhlavenda.com/index.php/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, 06/19/2018 - 08:36</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 noticed (at least) two fallacies in the derivation of the equations of motion of the gravitational field equations: </p>
<p>I. The constraint $A\dot{t}$, where $A$ is any metric coefficient depending on $r$, and $\dot{t}$ is the derivative of time (only one time!) with respect to an affine parameter, $lambda$, cannot be introduced into the variational equations more than once, and</p>
<p>II. Since $t$ is a cyclic (ignorable, kinosthenic, or whatever you want to call it) variable the Lagrange equation reads</p>
<p>$$\frac{d}{d\lambda}\left(A\frac{dt}{d\lambda}\right)=0,$$</p>
<p>which leads to the first integral</p>
<p>$$A\frac{dt}{d\lambda}=\mbox{const.}$$</p>
<p>All the literature uses the above condition, but, with $\dot{t}=\mbox{const.}$, meaning that $A\neq A(\lambda)$. However, since $A=A(r)$, this implies that $r\neq r(\lambda)$, which clearly contradicts the equations of motion. Moreover, since $r=r(\lambda)$, it is also a function of $t$ because of the constraint, $A\dot{t}=\mbox{const}$. Finally, this violates the so-called condition of a <em>static</em> field in that the components of the Ricci tensor no longer vanish; in particular, $R_{01}=-\frac{\dot{A}}{Ar}$, and, consequently, all the rest.</p>
<p>We also used the condition that the metric coefficients, $A$ and $B$, satisfy $AB=1$, in the Lagrangian</p>
<p>$$\mathcal{L}=\frac{1}{2}\left(A\dot{r}^2+r^2\dot{\varphi}^2-B\dot{t}^2\right),$$</p>
<p>in the simple case of motion in the plane, $\vartheta=\pi/2$. </p>
<p>Now, the last term in the above expression for the Lagrangian can be looked upon as a constraint on the motion. That term is the Legendre transform of the first term, considered as a kinetic energy, This is the same way as we fransform from velocity to momentum in mechanics. Consider the function,</p>
<p>$$\Psi(\dot{t})=\dot{r}\frac{\partial\mathcal{L}}{\partial\dot{r}}-\mathcal{L}(\dot{r}).$$</p>
<p>Now, define $\dot{t}=A\dot{r}$, and introducing this gives</p>
<p>$$\Psi(\dot{t})=\frac{\dot{t}^2}{2A}.$$</p>
<p>When this is subtracted from the Lagrangian, $\mathcal{L}=(1/2)(A\dot{r}^2+r^2\dot{\varphi}^2)$, we get</p>
<p>$$\mathcal{L}=\frac{1}{2}\left(A\dot{r}^2+r^2\dot{\varphi}^2-\frac{\dot{t}^2}{A}\right),$$</p>
<p>proving that </p>
<p>$$A=1/B.$$</p>
<p>Finally, when the constraint $\dot{t}/A=\dot{r}$ is introduced into the variational equation, we have to choose that speed $\dot{r}$ which is the maximum speed, namely $c$, or in Weber's case $\surd 2 c$.</p>
<p>For $A=1/r$ we get the Weber field; for $A=1/(1+\mathcal{R}/r)$, we get the Schwarzschild field, where $\mathcal{R}$ is the Schwarzschild radius, and for $A=1+\mathcal{R}/r$ we get the Gerber field, which he used to determine the advance of the perihelion of Mercury. In the case of the Schwarzschild field, the equation of motion is</p>
<p>$$\ddot{r}-r\dot{\varphi}^2=g\left(\frac{\dot{r}^2-c^2}{1-\alpha/r}-2v_S^2\right),$$</p>
<p>where in the planar case $v_S=r^2\dot{\varphi}^2.$ <em> This shows definitively that there is no gravitational repulsion if the second postulate of the special theory holds.</em></p>
<p>We can also treat $\varphi$ as a kinosthenic, (or ignorable) variable which can be eliminated in favor of the angular momentum, $L$, which is defined by $L=r^2\dot{\varphi}=\partial\Phi/\partial\dot{\varphi}$. Subtracting off $\Phi=L^2/2r^2$ from the Lagrangian gives</p>
<p>$$\mathcal{L}=\frac{1}{2}\left(A\dot{r}^2-\frac{L^2}{r^2}-\frac{\dot{t}^2}{A}\right).$$</p>
<p>Choosing $A=\mathcal{R}/r$, the radial equation is</p>
<p>$$\frac{2Gm}{c^2}\left(\frac{\ddot{r}}{r}-\frac{\dot{r}^2}{r^2}\right)=\frac{L^2}{r^3}-\frac{Gm}{r^2},$$</p>
<p>after introducing $\dot{t}=Ac=2\frac{Gm}{c^2r}c$. For transverse motion ($\dot{r}=0$), we get the condition,</p>
<p>$$\frac{L^2}{r^3}=\frac{Gm}{r^2},$$</p>
<p>or, equivalently, $r\dot{\varphi}^2=Gm/r^2$ for orbital motion. This clearly shows that the gravitational potential which was introduced through the term</p>
<p>$$-A^{\prime}\dot{t}^2/2A^2=-A^{\prime}c^2/2=Gm/r^2$$</p>
<p>has nothing at all to do with time. The factor of $2$ in the definition of the Schwarzschild radius $\mathcal{R}$, is required for the variational principle, and not the asymptotic time limit since there is no time involved. In this sense, the Newtonian potential is a completely static field which does not enter into the coefficient of the $dt^2$ in the metric. For if clocks slow down in a gravitational field, they will also do so for the electric field for, in fact, the gravitational scalar potential was introduced by the analogy with the electric field [Moller, <em>The Theory of Relativity,</em> p. 246.]</p>
</div></div>
Tue, 19 Jun 2018 06:36:22 +0000adminlavenda186 at http://bernardhlavenda.comWhy Gravitational Repulsion Doesn't Exist
http://bernardhlavenda.com/node/185
<span class="field field--name-title field--type-string field--label-hidden">Why Gravitational Repulsion Doesn't Exist</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, 06/16/2018 - 12:21</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 has been much discussion in the literature of anti-gravity machines based on gravitational repulsion that would be the source for propulsion of payloads in the foreseeable future. The possibility of gravitation repulsion comes from an analysis of the Schwarzschild metric by Droste and Hilbert which predict that "a large mass moving faster that $c/\surd 3$ could serve as a driver to accelerate a much smaller mass from rest to a good fraction of the speed of light" (Felber: http://arxiv.org/abs/0910/1084).</p>
<p>One thing that gravitational repulsion would do for sure is to make Newton turn over in his grave. The possibility of gravitational repulsion is the result of misconceived use of eliminating a cycle, or ignorable, variable, and then reintroducing it into the Euler-Lagrange equations of motion. There is nothing special about the time variable we use to clock time, or the angle variable we use to measure angles. Both are "kinosthenic", in the terminology of J J Thomson, who thought of the possibility of developing a mechanics without forces by assuming the presence of such coordinates. They are also referred to as speed coordinates which can be eliminated.</p>
<p>The morale of the story is you can't use a constraint <em>twice </em>in the same variational principle. A good example can be found in <em>Thermodynamics of Irreversible Processes</em>, p. 100. What was a minimum to start with then becomes a maximum, and vice-versa. A double dose of a constraint led Onsager to conclude that heat flow can only lower, or cause no change at all, the entropy production. We would expect just the opposite to happen, and so, too, with the variational principles of GR.</p>
<p>Consider the metric,</p>
<p>$$-d\tau^2=-Bdt^2+Adr^2+r^2d\varphi^2,$$</p>
<p>in the plane $\vartheta=\pi/2$. $\tau$ is referred to as 'local' or 'proper' time, and $t$, coordinate time, the time that would be measured in the laboratory. If we divide through by $d\tau^2$ and consider the Lagrangian,</p>
<p>$$L=\frac{1}{2}\left\{A\dot{r}^2+r^2\dot{\varphi}^2-B\dot{t}^2\right\},$$</p>
<p>where the dot denotes differentiation with respect to $t$, and $A$ and $B$ are functions of $r$. It makes not difference if we exchange $t$ for $\tau$ because $t$, or $\tau$ are ignorable coordinates like $\varphi$; their differentials appear but not the variables themselves. The Euler-Lagrange equations are</p>
<p>$$\frac{d}{d\tau}\left(A\dot{r}\right)-\frac{1}{2}A^{\prime}\dot{r}^2-r\dot{\varphi}^2+B^{\prime}\dot{t}^2=0,$$</p>
<p>$$\frac{d}{d\tau}=r^2\dot{\varphi}=0,$$</p>
<p>$$\frac{d}{d\tau}B\dot{t}=0.$$</p>
<p>The last two equations are examples of ignorable, or cyclic, coordinates. They can be eliminated by the first integrals</p>
<p>$$r^2\dot{\varphi}=L \hspace{20pt}\mbox{and} \hspace{20pt}B\dot{t}=c.$$</p>
<p>This is a way of introducing the constants, $L$ and $c$, which are angular momentum and the speed of light, respectively. </p>
<p>Introducing them into the radial equation of motion leads to</p>
<p>$$\ddot{r}+\frac{A^{\prime}}{2A}\dot{r}-\frac{L^2}{Ar^3}+\frac{B^{\prime}}{2AB^2}c^2=0.$$</p>
<p>So far, so good. Now comes the error. There must be something very special about $t$---which we just eliminated---to want to reinsert it again! In the words of McGruder [Phys, Rev. D 25 (1982 3191] "[t[o obtain the Schwarzschild radial acceleration as a function of the Schwarzschild radial velocity $\dot{r}=dr/dt$, and the transverse velocity $v_S=(\dot{r}^2\dot{\vartheta}^2+r^2\sin^2\vartheta\dot{\varphi}^2)^{1/2}$ we first insert [$\dot{t}=c/B$] into the identities</p>
<p>$$\frac{dx^i}{d\tau}=\frac{dx^i}{dt}\frac{dt}{d\tau}, \hspace{10pt} i=1,2,3,"$$</p>
<p>where McGruder uses the parameter $p$ for our $\tau$. He then obtains</p>
<p>$$a_s\equiv\ddot{r}-\frac{r\dot{\vartheta}^2}{A}-\frac{r\sin^2\vartheta\dot{\varphi}^2}{A}=\left[\frac{B^{\prime}}{B}-\frac{A^{\prime}}{2A}\right]\dot{r}^2-\frac{B^{\prime}c^2}{2A}.$$</p>
<p>There is absolutely no reason to favor $p$, $\tau$, $t$, etc., but what is important is the introduction of $c^2$, as Weber fully recognized.</p>
<p>For it is the first term on the right-hand side that shouldn't be there, for what should have been $\dot{r}^2-c^2<0$ has now become</p>
<p>$$a_S=\frac{g}{c^2}\left\{\frac{3\dot{r}^2}{1-\alpha/r}-2v^2_S-c^2(1-\alpha/r)\right\},$$</p>
<p>where $g=GM/r^2$, and $\alpha=2GM/c^2$. This leads McGruder to the conclusion that: "In Newton's theory the acceleration of gravity is always negative, indicating gravitational attraction. In Einstein's theory, however, [the above equation] shows that for</p>
<p>$$\dot{r}^2>\frac{1}{3}c^2(1-\alpha/r)+\frac{2}{3}c^2(1-\alpha/r),$$</p>
<p>we have $a_S>0, implying gravitational acceleration." This violates the second postulate of special relativity, allowing the radial velocity to become even infinite.</p>
<p>There is nothing scared about the time $t$, and introducing the first integral, $B\dot{t}=c$ <em>twice </em>has led to the error resulting in the above inequality. </p>
<p>We can show this explicitly by deriving Einstein's equation for the deflection of light which involves setting the Lagrangian </p>
<p>$$A\dot{r}^2+r^2\dot{\varphi}^2-A^{-1}\dot{t}^2=0,$$</p>
<p>where we set $A=B^{-1}$. Eliminating $\dot{t}$ through the constraint, $A\dot{t}=c$, results in</p>
<p>$$A\dot{r}^2+r^2\dot{\varphi}^2-Ac^2=0.$$</p>
<p>It doesn't make <em>one iota</em> of a difference what the dot stands for because on dividing through by $\dot{\varphi}^2$ there results</p>
<p>$$\left(\frac{dr}{d\varphi}\right)^2=\frac{c^2}{L^2}r^4-\frac{r^2}{A},$$</p>
<p>where we eliminated $\dot{\varphi}$ by the expression for the <em>conserved</em> angular momentum. Notice that if we had used</p>
<p>$$\frac{d}{dp}r^2\frac{d\varphi}{dp}=0,$$</p>
<p>we would have come out with</p>
<p>$$\frac{r^2\dot{\varphi}}{1-\alpha/r}=L,$$</p>
<p>which is no longer the conservation of angular momentum! True, the second term in the denominator is small, but to neglect it like Moller did would be to avoid using the transform from $p$ to $t$ which introduces the extraneous term, $B^{\prime}\dot{r}^2/B$, in the equation of motion for the radial coordinate given above.</p>
<p>Having set $dr/dp/d\varphi/dp=dr/d\varphi$, and all traces of the affine parameter $p$ have completely disappeared leaving the equation for the trajectory. Had we used the above transform from $p$ to $t$, given by Gruder, who referenced it from Weinberg's <em>Gravitation and Cosmology, </em>we would be led to an erroneous result. Differentiating the equation for $dr/d\varphi$ by $\varphi$ and transform to the variable, $u=1/r$, give</p>
<p>$$\frac{d^2u}{d\varphi^2}+u=3gu^2,$$</p>
<p>which is exactly the same equation that GR found for the deflection of light by a massive body. </p>
<p>Parenthetically, we might add that to get the shift in the perihelion of Mercury, all we have to do is to set the Lagrangian to a constant, say $-1$, instead of zero. This has falsely led credence to the distinction between proper and coordinates times. Although this introduces the constant term $\alpha/L^2$, we can't believe that proper time has led to the introduction of a static potential. It has the effect of adding a constant to the Lagrangian which does not affect the Euler-Lagrange equations.</p>
<p>Weber never had any trouble in defining the time variable, he transformed the currents into charges, where the velocity time charge $ve=c\times ids$, $c$ times the current element, $ids$. The $c$ so introduced was defined as the number of statical units of electricity which are transmitted by a unit electric current in a unit of time. The transformation from $s$ to $t$ was</p>
<p>$$\frac{dr}{dt}=v\frac{dr}{ds},$$</p>
<p>but<em> no one ever stopped to ask which $t$</em>? It was merely a <em>deux ex machina </em>for introducing $\surd 2 c$, which was Weber's constant.</p>
<p>Hence, there is no such thing as gravitation repulsion and the second postulate of the special theory is upheld.</p>
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Sat, 16 Jun 2018 10:21:29 +0000adminlavenda185 at http://bernardhlavenda.comThe Need for a Road Map in General Relativity
http://bernardhlavenda.com/node/184
<span class="field field--name-title field--type-string field--label-hidden">The Need for a Road Map 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">Thu, 06/14/2018 - 10:48</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 have seen that the same formulas may have entirely different meanings and the same same physical phenomena may be explained in many ways. Two such examples are the general relativistic calculation of the deflection of light and the advance of the perihelion of Mercury. Both these phenomena were known before the advent of general relativity, and even the numerical gap in the advance of the perihelion was known, but not to the degree it is known today.</p>
<p>Weber taught us that when static and motional forces are combined, it is necessary to introduce a limiting velocity. In fact, in 1856 Weber and Kohlrausch determined its value. According to Weber, $c$ is "that relative velocity which electrical masses $e$ and $e'$ have and must retain, if they are not to act on each other at all." His fundamental law was</p>
<p>$$\frac{ee'}{r^2}\left(1-\frac{1}{c'}\dot{r}^2+\frac{2r}{c'}\ddot{r}\right),$$</p>
<p>where it turned out that $c'$ exceeded Maxwell's constant, which is the inverse of the square root of the product of the dielectric and magnetic permeabilities of the vacuum, by $\surd 2$, We may see how this arises by considering the Lagrangian,</p>
<p>$$\mathcal{L}=-r(c\dot{t})^2+\frac{\dot{r}^2+r^2\dot{\varphi}^2}{r},$$</p>
<p>in the plane $\vartheta=\pi/2$. The EL equations,</p>
<p>$$\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot{x}}-\frac{\partial\mathcal{L}}{\partial x}=0,$$</p>
<p>are explicitly given by</p>
<p>$$2\frac{d}{d\tau}\left(\frac{\dot{r}}{r}\right)+\frac{\dot{r}^2}{r^2}-\dot{\varphi}^2+(c\dot{t})^2=0,$$</p>
<p>$$\frac{d}{d\tau}r\dot{\varphi}=0,$$</p>
<p>and</p>
<p>$$\frac{d}{d\tau}c^2r\dot{t}=0.$$</p>
<p>The second and third equations lead to the first integrals, $r\dot{\varphi}=v=\mbox{const.}$, and $c^2r\dot{t}=H=\mbox{const.}$. Whereas $v$ has the physical significance of a velocity, $H$ doesn't have one so we can set it equal to unity. That is to say, $\dot{t}$ does not have any physical significance regarding a 'coordinate' time, nor does $\tau$ refer to 'proper' time. It is put there only to provide a limiting velocity, as Weber very well knew. </p>
<p>For upon introducing the first integrals into the radial equation there results</p>
<p>$$\frac{1}{c^2-v^2}\left(2\frac{\ddot{r}}{r}-\frac{\dot{r}^2}{r^2}\right)+\frac{1}{r^2}=0.$$</p>
<p>This is the condition that the Weber force vanish with the only difference that Weber's constant, $c'$, is now given by $\surd(c^2-v^2)$, which explicitly states that $c$ is a limiting velocity. It also points to the fact that motion causes a contraction in the radial direction, which predates relativity by almost a half a century.</p>
<p>In the previous blog we noticed a difference between the way general relativity treats the advance of the perihelion as opposed to the deflection of light by a massive body. The latter case could be derived from the same procedure as above adopting the Lagrangian</p>
<p>$$\mathcal{L}=\frac{1}{2}\left\{\frac{\dot{r}^2}{(1-\mathcal{R}/r)^2}+\frac{r^2\dot{\varphi}}{1-\mathcal{R}/r}-c^2\dot{t}^2\right\},$$</p>
<p>which is obtained by dividing the metric</p>
<p>$$-c^2d\tau^2=-(1-\mathcal{R}/r)c^2dt^2+\frac{dr^2}{1-\mathcal{R}/r}+r^2d\varphi^2,$$</p>
<p>through by $(1-\mathcal{R}/r)^{-1}$, $d\tau^2$, and considering the Lagrangian to be half the right-hand side. Even if subtracted a constant from the Lagrangian it would have no effect on the EL equations. Yet, if we were to solve</p>
<p>$$-c^2=-(1-\mathcal{R}/r)c^2\dot{t}^2+\frac{\dot{r}^2}{1-\mathcal{R}/r}+r^2\dot{\varphi}^2,$$</p>
<p>obtained by dividing through by $d\tau^2$, and solving the equation with respect to the first integrals,</p>
<p>$$(1-\mathcal{R}/r)\dot{t}=c/v=\mbox{const.},\hspace{20pt}\mbox{and}\hspace{20pt}r^2\dot{\varphi}=L=\mbox{const.},$$</p>
<p>the result is entirely different. In fact, the left-hand side is necessary for introducing the Newtonian potential, and the fact that there is a limiting velocity. This apparently gives some credence in treating two times: one coordinate and the other a proper time. However, appearance can be deceiving.</p>
<p>Solving the metric with the aid of the first integrals, and introducing $u=1/r$ gives the equation,</p>
<p>$$\varphi=\varphi_0+\int^u_{u_0}\frac{du}{\surd\left(W+\mathcal{R}u/L^2-u^2+\mathcal{R}u^2\right)}.$$</p>
<p>After multiplying through by $1-\mathcal{R}/r$ in the metric, the constant term, which represented the square of the increment in proper time, introduces a $-1$ in the first term in the term</p>
<p>$$W=\frac{(c/v)^2-1}{L^2},$$</p>
<p>and the linear term $\mathcal{R}u=2GM/rc^2$. Closed orbits having negative total energy require $W<0$, which seemingly violates the fact that $c$ is a limiting velocity. The linear term is essentially twice the Newtonian potential which wouldn't be there if we did not consider the "proper" time. Why should "proper" time be necessary to introduce a static potential?</p>
<p>Be this as it may, we continue to follow conventional "reasoning". Quoting from the classic text, Adler, Bazin, and Schiffer, <em> Introduction to General Relativity,</em> "Unfortunately even though this [the above expression for $\varphi$] is an <em>exact</em> solution to the problem, it expresses the angle $\varphi$ as an integral of $u=1/r\ldots$ To make the problem more transparent differentiate with respect to $\varphi$ so that it looks like a classical Kepler problem."</p>
<p>The question is why bother when we could have begun directly with the Lagrangian,</p>
<p>$$\mathcal{L}=\frac{1}{2}\left\{\frac{\dot{r}^2}{1-\mathcal{R}/r}+r^2\dot{\varphi}^2-(1-\mathcal{R}/r)(c\dot{t})^2+c^2\right\},$$ </p>
<p>where the last term is $c^2d\tau^2$, after having divided through by $d\tau^2$? It is a constant, and as such, it will not influence the EL equations of motion. The EL equation for the radial coordinate is easily found to be</p>
<p>$$(1-\mathcal{R}/r)\ddot{r}-g\frac{\dot{r}^2}{c^2}=\frac{L^2}{r^3}(1-\mathcal{R}/r)^2-g(c/v)^2,$$</p>
<p>where $g=GM/r^2$. It doesn't at all look like the equation obtained by setting the difference in the metric equal to zero. Moreover, the fact that $\dot{r}/\dot{\varphi}=dr/d\varphi$ has been used in the derivation. This contradicts the assumption that the "field of a static body cannot depend on time which is ensured by $R_{01}=-\dot{A}/rA=0$, which implies $A=A(r)$", where $R_{01}$ is a component of the Ricci metric and $A$ a coefficient in the metric. </p>
<p>The metric coefficients have to be a function of time if $r$ is. Moreover, the need to consider proper time, or for that matter time at all, for introducing a static potential is absurd. The need for introducing time at all is analogous to the way Weber combined Ampere's force with Coulomb's. Time itself does not come into play except for determining the rate of change of the radial coordinate connecting the two current elements. Hence, in order to get the sought after results in general relativity, you have to follow a road map!</p>
</div></div>
Thu, 14 Jun 2018 08:48:37 +0000adminlavenda184 at http://bernardhlavenda.comWeber versus Schwarzschild: GR Much Ado About Nothing
http://bernardhlavenda.com/node/183
<span class="field field--name-title field--type-string field--label-hidden">Weber versus Schwarzschild: GR Much Ado About Nothing</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, 06/12/2018 - 08:47</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 proposed a metric,</p>
<p>$$-c^2d\tau^2=-rc^2dt^2+\frac{1}{2r}\left(dr^2+r^2d\varphi^2\right),$$</p>
<p>in the plane, $\vartheta=\pi/2$, for the Weber force,</p>
<p>$$F_W=\mbox{const.}\times\left(\ddot{r}-\frac{\dot{r}^2}{2r^2}+\frac{c^2H^2}{r^2}\right),$$</p>
<p>whose vanishing coincides with the Euler-Lagrange equations, or, equivalently, the geodesics. </p>
<p>The Lagrangian describing this geometry is</p>
<p>$$\mathcal{L}=\frac{1}{2}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{nu}=\frac{1}{2}\left\{-rc^2\dot{t}^2+\frac{1}{2r}(\dot{r}^2+r^2\dot{\varphi}^2)\right\},$$</p>
<p>where the dot denotes differentiation with respect to $\tau$. There is no distinction between "coordinate", $t$, and "proper", $\tau$ times here: The introduction of $dt^2$ is just a way of introducing $c$ which appears in Ampere's theory as the ratio of the electromagnetic to electrostatic unit of electricity. The surprise was it could be represented as a velocity. That's Ampere's theory. Now in Weber's theory, which is dynamical, the same constant $c$ is the number of units of statical electricity which are transmitted by a unit electric current in a unity of time. This is Maxwell's definition. We will not get into the discussion of whether the two are equal, that's another story.</p>
<p>Now, it does appear odd that we have claimed that the metric coefficient $g_{00}=-r c^2$ accounts for the Coulomb force, and the only way to unite it with the kinetic energy terms in the above equation is through $c^2$. The potential</p>
<p>$$\Phi=\frac{1}{4\pi\epsilon_0 r}\cdot 4\pi r^2=\frac{r}{\epsilon_0},$$</p>
<p>represents the surface charge density $\sigma$ on a ball of radius $r$, where $\sigma\propto 1/r$ is the surface charge density. The potential satisfies Poisson's equation,</p>
<p>$$\nabla^2\Phi=\frac{1}{r^2}\frac{d}{r^2}\left(r^2\frac{d\Phi}{dr}\right)=\frac{2\sigma}{\epsilon_0},$$</p>
<p>in a spherically symmetric environment. The inverse square dependency of the Coulomb force will be brought in by the constraint that $\dot{t}$ is a cyclic coordinate which leads to an immediate first integral. The Coulomb potential, like its gravitational counterpart, has nothing whatsoever to do with the slowing down of clocks in an electrostatic field! The formal similarity of equations can, at times, be most misleading!</p>
<p>The Euler-Lagrange (EL) equations are</p>
<p>$$\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot{x}^{\mu}}-\frac{\partial\mathcal{L}}{\partial x^{\mu}}=0.$$</p>
<p>The reason for using $\lambda$, and not $\tau$, is that the differential of the latter vanishes on light cones, a condition that determines the speed of light.</p>
<p>Now, $\varphi$ and $t$ are cyclic variables because only their derivatives appear in the Lagrangian. This immediately leads to first integrals of the motion,</p>
<p>$$r\dot{t}=H, \hspace{20pt}\mbox{and}\hspace{20pt} r\dot{\varphi}=V,$$</p>
<p>where both $H$ and $V$ are arbitrary constants of the motion.</p>
<p>It is easy to check that the geodesic equations</p>
<p>$$\frac{d}{d\tau}\left(g_{\alpha\beta}\frac{dx^{\beta}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{\mu\nu}}{\partial\alpha}\dot{x}^{\mu}\dot{x}^{\nu}=0,$$</p>
<p>coincide with the EL equations. For instance the equation for radial motion is</p>
<p>$$\ddot{r}-\frac{\dot{r}^2}{2r^2}+2c^2 r\dot{t}^2=0.$$</p>
<p>Introducing the first integral of the motion for $\dot{t}$ gives</p>
<p>$$\frac{1}{r^2}\left(r\ddot{r}-\frac{1}{2}\dot{r}^2+(cH)^2\right)=0$$</p>
<p>which is the vanishing of the Weber force, apart from a multiplicative coefficient which cannot be determined. The last term is the Coulomb potential which needs $c^2$ to appear along side the other two terms. Let us now compare this with general relativity's handling of the Schwarzschild field.</p>
<p>The Schwarzschild metric has the form</p>
<p>$$ds^2=Bdt^2-Adr^2-r^2d\Omega^2,$$</p>
<p>where the last term is the element of a two-sphere which will not concern us because we are considering radial motion. The metric coefficients are initially considered functions of both $r$ and $t$, where the "radial coordinate $r$ defines the area of a $2-$sphere and hence it has proper physical meaning." The quote like the others we will use comes from a paper by Dadhich, "On the Schwarzschild field." But, isn't it a little premature to give $r$ such a "proper" physical meaning when we will find that it can't be extended to the origin of the coordinate system?</p>
<p>Einstein's condition for "emptiness" means that the components of the Ricci tensor vanish. For instance, the time component is</p>
<p>$$R^0_0=-\frac{1}{2AB}\left\{\nabla^2B-\frac{B^{\prime}}{2}\left(\frac{A^{\prime}}{A}+\frac{B^{\prime}}{B}\right)-\ddot{A}+\frac{\dot{A}^2}{2A}+\frac{\dot{A}\dot{B}}{2B}\right\},$$</p>
<p>where the prime and dot denote differentiation w.r.t. $r$ and $t$ respectively. </p>
<p>We are now told that from a "physical point of view it is obvious that [the] field of a static body cannot depend upon $t$ which is ensured by $R_{01}=-\dot{A}/rA=0\ldots$ which implies $A=A(r)$" But how is this ever to be congruent with the geodesic equations which are an immediate consequence of the variational principle?</p>
<p>The requirement that both $R_0^0$ and</p>
<p>$$R_1^1=R_0^0+\frac{1}{rA}\left(\frac{A^{\prime}}{A}+\frac{B^{\prime}}{B}\right)$$</p>
<p>vanish requires $AB=f(t)=\mbox{const.}$. The same hold true for the Weber metric, but not for the reason that the effect of $A$ in "curving space" exactly cancels out the contribution to the field energy density contained in $B=1+\phi$, where $\phi$ is the Newtonian potential. "In the usual derivation it is here the asymptotic flatness is invoked to set $f(t)=1$." However, wasn't it agreed upon that $A$ and $B$ are independent of $t$ because we are treating a "static" gravitational problem? "Even if $f(t)$ is retained the field will be asymptotically flat and the solution with $f\neq 1$ is physically indistinguishable from the one with $f=1$. So we apply no boundary condition to set $f=1$." What differential equation is the author referring to? to the Laplace equation in the expression for $R_0^0$? And who needs it? when we have the equation for the geodesics readily available?</p>
<p>The claim that $A$ compensates $B$, for otherwise "one does not see how non-linear aspect of gravity is accounted for in the Schwarzschild solution" doesn't hold water since the same condition is found in the Weber metric and there is no talk of "warping" space by the presence of $A$ and the field energy density which appears as a "gravitational charge density." There $A$ and $B$ are definitely functions of time, and $\Phi$, the surface charge density, does not satisfy Laplace's equation. Why should we ever expect it to do so where there is an electrostatic field?</p>
<p>Schrodinger, in his article "The possibility of fulfillment of the relativity requirement in classical mechanics", didn't pick up the fact that "the total interaction energy of two mass point with masses $\mu$ and $\mu'$ with separation $r$,</p>
<p>$$W=\gamma\frac{\mu\mu'\dot{r}^2}{r}-\frac{\mu\mu'}{r},$$</p>
<p>where $\gamma$ is a fitting parameter, does not curve space, $A=1/2r$, but he, moreover, didn't recognize it as Weber's potential with $\gamma=1/2$. (He needed $\gamma$ $6$ times that in order to account for the advance of the perihelion of Mercury.)</p>
<p>Consulting the expression for $R_0^0$, the time derivatives don't even give the time-dependent terms in the Weber force. The condition for the vanishing of this term is</p>
<p>$$\frac{\ddot{r}}{2r}-\frac{\dot{r}^2}{r^2}+2c^2=0,$$</p>
<p>which is not even close to the vanishing of the Weber force given above. If $A$ and $B$ are allowed to depend upon time, the vanishing of $R_0^0$ is</p>
<p>$$\ddot{r}-2\frac{\dot{r}^2}{r}=0,$$</p>
<p>where all effects of the potential $B=1+2\phi$ vanishes, since it solves Laplace's equation. But, then there is the $R_{01}=-\dot{A}/rA$ term which does not vanish. </p>
<p>Instead of the Schwarzschild metric, it would be much easier (and much more instructive) to begin with</p>
<p>$$\mathcal{L}=\frac{\dot{r}^2-(c\dot{t})^2}{2r}.$$</p>
<p>Then, with the first integral, $r\dot{t}=\surd 2$, the EL equation is simply</p>
<p>$$\frac{\ddot{r}}{r}-\frac{1}{2}\left(\frac{\dot{r}}{r}\right)^2+\frac{c^2}{r^2}=0,$$</p>
<p>which is the vanishing of the Weber force, whether it be for Coulomb, or Newton.</p>
<p>Dadhich also claims that even when $B=1$, $A=1/(1-\mathcal{R}/r)$, where $\mathcal{R}=2GM/c^2$ will create a space curvature that will "manifest in tidal acceleration for non-radial motion." This is clearly inaccurate as any tangential acceleration must come from the term $r^2\dot{\varphi}^2$ in the metric. As such there is no linear term, but, we can invent one by dividing the Lagrangian through by $(1-\mathcal{R}/r)^{-1}$ to get</p>
<p>$$ \mathcal{L}=\frac{1}{2}\left(-c^2\dot{t}^2+\frac{\dot{r}^2}{(1-\mathcal{R}/r)^2}+\frac{r^2\dot{\varphi}^2}{(1-\mathcal{R}/r}\right).$$</p>
<p>This was the trick Moller used to derive the geodesic equations---not from the $4$-vector of the velocity---but from the relativistic expression for the orbiting particle, which doesn't exist. Setting the Lagrangian equal to zero, and changing the differentiation variable to $\varphi$ through the first integral,</p>
<p>$$\frac{r^2\dot{\varphi}}{1-\mathcal{R}/r}=L,$$</p>
<p>give</p>
<p>$$\frac{dr}{d\varphi}=\pm\frac{r^2}{L}\sqrt{E-\frac{L^2}{r^2}\left(1-\frac{\mathcal{R}}{r}\right)},$$</p>
<p>where we have used $\dot{t}=H$ [cf. Eqn (7.5.10) in <em>A New Perspective on Relativity</em>]. This identifies the enigmatic $\dot{t}^2$ with the total energy, $E=H^2$.The last term, supposedly, gives the exact expression for the deflection of light by a massive body. The funny thing is that the Newtonian potential is missing; for otherwise it would also explain the advance of the perihelion of Mercury. Why the Coulomb potential was accounted for by the $\dot{t}^2$ term in the Lagrangian above, and why the proper time interval $d\tau^2$ must be included in what was the condition for the vanishing of the Lagrangian appears mysterious. This will be treated in the next blog.</p>
<p>However, there now there is a tangential acceleration,</p>
<p>$$a_{tan}=r\ddot{\varphi}+2\dot{r}\dot{\varphi}=\frac{\mathcal{R}}{r}\frac{\dot{r}\dot{\varphi}}{1-\mathcal{R}/r},$$</p>
<p>which cannot be wished away. In Moller's words, $L$ cannot in general be interpreted as angular momentum, since the notion of a 'radius vector' occurring in the definition of the angular momentum has an unambiguous meaning only in a Euclidean space." This clashes with Dadhich's assertion that "the radial coordinate $r$ defines the area of a $2$-sphere and hence has proper physical meaning", which we quoted above.</p>
<p>The truth of the matter is that by dividing the Lagrangian through by $(1-\mathcal{R}/r)^{-1}$ converts the space part into the Beltrami metric which is a hyperbolic metric of constant curvature when $\mathcal{R}/r$ is replaced by the mass density $\varrho$ according to $GM/c^2r\rightarrow (4\pi/3)G\varrho/c^2r^2$. This, in fact, applies to the inner solution of Schwarzschild solution [cf. Sec. 9.19.4 in <em>A New Perspective on Relativity</em>].</p>
</div></div>
Tue, 12 Jun 2018 06:47:50 +0000adminlavenda183 at http://bernardhlavenda.comThe Metric for the Weber Force
http://bernardhlavenda.com/node/182
<span class="field field--name-title field--type-string field--label-hidden">The Metric for the Weber 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">Sun, 06/10/2018 - 11:21</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 have seen, in the last blog, that there is a rather strict analogy between the Schwarzschild metric and Gerber's potential, both of which give the same expression to the perihelion shift of Mercury to order $1/c^2$. It has long been known that to this order the force derived from Gerber's potential is the same as the Weber force. There is a slight difference, however. Whereas the Weber force acts radially between the two masses so that there is no tangential force acting on the orbiting particle, the Schwarzschild metric has a finite tangential force that destroys the conservation of angular momentum.</p>
<p>All accounts of this rely on the misconception that the term that does not allow for the conservation of energy is small so that conservation can be assumed in the first approximation. This is false. The tangential acceleration for the Schwarzschild metric is</p>
<p>$$r\dot{\omega}+2\dot{r}\omega=2gr\frac{\dot{r}\omega}{1-\alpha/r},$$</p>
<p>where $\alpha\equiv 2GM$ and $g=\alpha/2r^2$. This destroys the general relativistic equation for the orbit as</p>
<p>$$\frac{d^2u}{d\varphi^2}+u=\frac{\mathcal{R}}{L^2}+3\mathcal{R}u^2,$$</p>
<p>where $u=1/r$, $\mathcal{R}=2GM/c^2$ is the Schwarzschild radius for the gravitating mass $M$, and the angular momentum is</p>
<p>$$L=\frac{r^2\omega}{1-\alpha/r},$$</p>
<p>which is hardly a constant. The Weber force has no such problems, although the derivation of the advance of the perihelion rests on equating the Weber force with radial acceleration, $\ddot{r}=r\omega^2$, which, we will see, is highly unscrupulous. This is because you are equating a radial acceleration with a radial acceleration.</p>
<p>The Weber force can be derived from the metric</p>
<p>$$-d\tau^2=-rdt^2+\frac{d{r}^2+r^2d\varphi^2}{2rc^2}.$$</p>
<p>Rather than considering the equations of the geodesics, we can consider the Lagrangian</p>
<p>$$\mathcal{L}=\frac{\dot{r}^2+r^2\dot{\varphi}^2}{2r}-c^2r\dot{t}^2,$$</p>
<p>where the dot stands for some parameter which measures the distance along the path, not necessarily $\tau$, which is constant for light rays. </p>
<p>The Euler-Lagrange equations are</p>
<p>$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}}-\frac{\partial\mathcal{L}}{\partial x}=F,$$</p>
<p>which vanishes along geodesics. $x$, and its derivative $\dot{x}$ stand for any of the three variables in $\mathcal{L}$. It will be appreciated that both $\varphi$ and $t$ are cyclic variables since</p>
<p>$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}}=0.$$</p>
<p>Hence, the velocity (and not the angular momentum) $r\dot{\varphi}=V=\mbox{const.}$$, and $r\dot{t}=H=\mbox{const.}$$ Introducing these into the Euler-Lagrange equation for the radial coordinate,</p>
<p>$$\frac{\ddot{r}}{r}-\frac{\dot{r}^2}{r^2}+c^2\dot{t}^2=F,$$</p>
<p>gives</p>
<p>$$F_W=\frac{H^2}{r^2}-\frac{1}{c^2}\left(\frac{\dot{r}^2}{r^2}-\frac{\ddot{r}}{r}\right).$$ The terms in $\dot{\varphi}$ dropped out on their own accord.</p>
<p>The first term in the Weber force is the Coulomb potential; it enters in exactly the same way as the Newtonian potential does in general relativity, i.e., through the metric coefficient $dt^2$. This has nothing to do with a hyperbolic metric or the propagation of waves! It is, however, somewhat prophetic that $c^2$ enters in the same way as it does in Ampere's law, as the factor that allows a static force to be added onto Ampere's law. </p>
<p>What allows us to equated $F_W$ with the radial acceleration, $\ddot{r}-r\dot{\varphi}^2$? Nothing! The angular velocity appeared as a conserved quantity in the derivation, so that if we were to equate the Weber force with $\ddot{r}+V\omega$, it would not lead to</p>
<p>$$(1+6\mathcal{R}u)\frac{d^2u}{d\varphi^2}+u=\frac{\mathcal{R}}{L^2}-3\mathcal{R}\left(\frac{du}{d\varphi}\right)^2,$$</p>
<p>since it is not $L$ which is constant. The second term in the coefficient of the second derivative on the left-hand side is what gives the same advance as $3\mathcal{R}u^2$ in the general relativistic equation. </p>
<p>These are remarkable coincidences! Equating acceleration to acceleration seems a little redundant to put it mildly, but to come out with the correct numerical result is prophetic! Then again what has Coulomb's potential to do in the metric coefficient of the $dt^2$ is also unexplainable. Yet, it introduces $c^2$ which is necessary to relate it to the kinetic energy term in the Lagrangian. Moreover, the hyperbolic nature of the metric allows for the propagation of waves at speed $c$. Yet, as we know from the Weber force, these waves will be purely longitudinal. </p>
<p>Finally, it is interesting to remark that the non-conservation of the angular momentum in the Schwarzschild metric comes from a totally extraneous origin in the relativistic mass of the orbiting mass. The three momentum is [cf. Moller, <em>The Theory of Relativity</em> Eq. (12) p. 348]</p>
<p>$$m_0\Gamma\left(\frac{\dot{r}}{1-\alpha/r},r^2\dot{\vartheta},r^2\sin^2\vartheta\dot{\varphi}\right),$$</p>
<p>where $m_0$ is the rest mass of the peripheral body and $\Gamma=1/\surd(1-\alpha/r-v^2/c^2)$ is the relativistic factor that depends on the three-velocity, </p>
<p>$$v^2=\frac{\dot{r}^2}{1-\alpha/r}+r^2\dot{\vartheta}^2+r^2\sin^2\vartheta\dot{\varphi}^2.$$</p>
<p>We have set $\vartheta=\pi/2$, but, to quote Rabi. "who ordered this?" Not only is the central mass non-existent in the Schwarzschild metric, insofar as it is there but not there (emptiness!), we now have to do with an orbiting mass that is speed dependent! It is the $\Gamma=1/(1-\alpha/r)$ factor that destroys the conservation of angular momentum since $\Gamma r^2\omega=\mbox{const.}$, and not $r^2\omega$ by itself. </p>
<p>The same result for the expression of angular momentum can be obtained from Fermat's principle of least time [cf. <em>A New Perspective on Relativity, </em>Sec. 7.3] so we know it is correct. However, it is amazing to see the degree to which the general relativists go to get what they want. </p>
<p>The nagging questions remain as to why Coulomb's potential should appear in the metric coefficient of the squared time interval when propagation is not even being contemplated, and why the metric needs to be hyperbolic. But these are no stranger than the remarkable coincidences that appear in general relativity.</p>
</div></div>
Sun, 10 Jun 2018 09:21:04 +0000adminlavenda182 at http://bernardhlavenda.comIs Gravitation a Longitudinal or Transverse Force--Or Both?
http://bernardhlavenda.com/node/181
<span class="field field--name-title field--type-string field--label-hidden">Is Gravitation a Longitudinal or Transverse Force--Or Both?</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">Fri, 06/08/2018 - 12:05</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 idea behind Ampere's investigations was that charges in motion exert forces on each other and the force is directed along the line connecting the charges or the current elements which he dealt with. Historically, Ampere's work was anteceded by the experimentation of Biot-Savart who studied the interaction of currents and magnetics. It was Ampere's idea that a magnetic could be modeled as a small current loop. It is commonly knowledge that the two laws are equivalent when one of the current loops is closed. </p>
<p>As common as knowledge is, this is wrong. A force which acts in the direction of the line connecting two current elements cannot be likened to the force normal to the current element and the line connecting the elements. Ampere's force is a longitudinal law, while Biot and Savart defined the magnetic field which is normal to the direction of flow of electric charges. In fact, the discovery of cathode rays led to the demise of Ampere's force since longitudinal currents were deemed disruptive to the flow of conduction currents. As a matter of fact, Ampere's law is related to the displacement current, but that's another story. </p>
<p>About two decades later, Grassmann, unhappy with the angle dependencies in Ampere's law, developed his own law According to Grassmann's view, a pair of current elements do not attract or repel each there; each experience a force that is perpendicular to itself which is caused by the other circuit element. The force on a current element, due to the other element, lies in a plane containing the direction of the current element and the line connecting the two elements. And because of Ampere's rule, the interaction of current elements can always be reduced to a two dimensional analysis in which all components normal to the plane produce no forces on the element in question. Therefore the interacting components of the Grassmann force all lie in the same plane.</p>
<p>So Grassmann's force, which was later to become the magnetic component of the Lorentz force abandons the description of the mutual attraction or repulsion of current elements, and violates Newton's third which states that for every action there is an equal and opposite reaction. This sounded the death knell of classical physics. And if special theory is an outgrowth of Lorentz's force, and general relativity an outgrowth of special relativity then gravity does not describe the attraction of masses at all.</p>
<p>In fact, general relativity is as close to Newton's law of gravitation as Grassmann's force is to Neumann's force. Grassmann's force is the triple vector product</p>
<p>$$\vec{F}_G=\frac{1}{r^3}\left(i'd\vec{s}'\wedge(id\vec{s}\wedge\vec{r}\right),$$</p>
<p>acting on $ids$ for current elements, $id\vec{s}$ and $i'd\vec{s}'$, and $\vec{r}$ is the radial vector that connects them. There are two components to Grassmann's force,</p>
<p>$$\vec{F}_G=\frac{ii'}{r^2}\left(d\vec{s}\cos\theta'-\hat{r}\cos\epsilon\right),$$</p>
<p>where $\theta'$ is the angle that element $id\vec{s'}$ makes with the unit normal, $\hat{r}$, along the line connecting them. The angle $\epsilon$ is the angle formed from the continuation of the lines directed along the two current elements. It is the second term that is related to the Neumann force</p>
<p>$$F_N=\frac{ids\cdot i'ds'}{r^2}\cos\epsilon.$$</p>
<p>Whittaker in his <em>A History of the Theories of Aether and Electricity</em> shows that the other term in Grassmann's force vanishes when summed over a close circuit. If this term is related to relativistic effects then what Whittaker showed was that all relativistic effects vanish upon closing the circuit. This is independent of the magnitude of the relative velocities.</p>
<p>What we shall now show is that Ampere's law gives the exact perihelion shift for Mercury if the ratio of electromagnetic to electrostatic units is reduced by a factor of 3. Now Ampere's law is a longitudinal, ponderomotive force, and not an electromotive force, The adjective 'longitudinal' is superfluous since all ponderomotive forces lie along the line connecting the masses. </p>
<p>When Einstein determined the advance of the perihelion of Mercury, it was well known the non-Newtonian advance of the orbital perihelion per revolution was $6\pi M/ac^2$, where $M$ is the mass of the sun, $a$ the semi-latus rectum of the orbit, and $c$ a propagation constant that relates the Newtonian potential to one that is a function of the velocity of the motion of the planet. So the challenge was to find a potential which gives this value.</p>
<p>According to A. A. Vankov (which we wholeheartedly agree) in his manuscript entitled "General relativity problem of Mercury’s perihelion advance revisited",<br />
</p>
<p>"... Mercury’s relativistic eﬀect has never been directly observed and even not evaluated from circumstantial astronomical evidence. The matter is that the GR theory, at least as it given in literature, does not provide a clue about distinguishing between the classical drag along with the equinoxes precession, on the one hand, and relativistic eﬀect, on the other hand. There is no other way but look for an admissible anomaly gap to be ﬁlled with the predetermined perihelion advance of 43′′ per century as tight as possible, no matter of what kind the eﬀect is. Such “a gap ﬁtting” cannot be termed “the conﬁrmation of the GR prediction”. Another issue is a statistical meaning of the gap ﬁtting. There is no single publication devoted to the treatment of observations of Mercury perihelion advance; the claimed numbers are stated in diﬀerent works on empirical data not treated in rigorous terms of statistical theory. A bad ﬁtting practice and the precision concept abuse should be noticed. At the same time, the usage of standard “precise” initial conditions in ephemerides calculations makes the results stable what creates an illusion of their high-precision, while their real precision remains unknown."</p>
<p>"Gap-fitting" is an academic exercise so that if one knew the potential that is responsible for the effect one might try to rationalize it physically. A high school teacher named Paul Gerber did just this, and anteceded Einstein by about two decades. The potential he proposed was</p>
<p>$$V(r,\dot{r})=-\frac{GM}{r}\left(1-\frac{\dot{r}}{c}\right)^{-2}.$$</p>
<p>The second factor, appended on to Newton's law has led to some heated debates so as not to detract from Einstein's result. Many an author has given short shrift to it, as the statement "what Gerber brings forth as physical considerations appears unintelligible" made by the staunch Einstein supporter, von Laue. If the second factor were reduced to $-1$, one might argue that it was due to taking into account retarded time. But the additional factor escapes all rationale. </p>
<p>It will become apparent that Gerber\s potential is<em> identical </em>to Weber's,</p>
<p>$$V_W=-\frac{GM}{r}\left(1-\frac{1}{2}\frac{\dot{r}^2}{c^2}\right),$$</p>
<p>at least to order $1/c^2$.</p>
<p>The force that is to be equated with the radial acceleration $\ddot{r}-r\omega^2$, is given by the Lagrangian term</p>
<p>$$F=\frac{d}{dt}\frac{\partial V}{\partial\dot{r}}-\frac{\partial V}{\partial r}$$</p>
<p>$$=-\frac{GM}{r^2}\left(1-\frac{\dot{r}}{c}\right)^{-4}\left\{\frac{6r\ddot{r}}{c^2}-\frac{2\dot{r}}{c}\left(1-\frac{\dot{r}}{c}\right)+\left(1-\frac{\dot{r}}{c}\right)^2\right\}.$$</p>
<p>When this is expanded in powers of $\dot{r}/c$, there results</p>
<p>$$F=-\frac{GM}{r^2}\left(1-3\frac{\dot{r}^2}{c^2}+6\frac{r\ddot{r}}{c^2}-8\frac{\dot{r}^3}{c^3}+24\frac{r\dot{r}\ddot{r}}{c^3}-\cdots\right).$$</p>
<p>This represents the force per unit mass that acts in he direction of the central gravitating mass $M$ and is equal to the radial acceleration, as Ampere-Weber would assume. In fact, if we define $\tilde{c}^2=c^2/6$, and retain terms up to $1/c^2$, we get</p>
<p>$$F_A=-\frac{GM}{r^2}\left(1-\frac{1}{2}\left(\frac{\dot{r}^2}{\tilde{c}^2}\right)^2+\frac{r\ddot{r}}{\tilde{c}^2}\right),$$</p>
<p>which is exactly Weber's law with the Newtonian force replacing the Coulomb one where $\tilde{c}$ is the number of units of statical mass (electricity) which are transmitted by a unit material (electric) current in a unit of time, to borrow Maxwell's definition of $c$. It presupposes some unit of mass just light in Le Sage's (1784) explanation of gravity in terms of a tiny, rapidly moving particles in space. These particles would travel at the speed $\tilde{c}$=c/\surd 6$, or around 17/% the speed of light.</p>
<p>There is no tangential acceleration for the vanishing of $r\dot{\omega}+2\dot{r}\omega$ leads to the conservation of angular momentum, $L$. Calling $u=1/r$, Einstein got the trajectory of the motion as</p>
<p>$$\frac{d^2}{d\varphi^2}+u=\frac{GM}{L^2}+3\mathcal{R}u^2,$$</p>
<p>whereas Gerber got</p>
<p>$$\frac{d^2u}{d\varphi^2}+u=\frac{GM}{L^2}-\frac{1}{2}\frac{GM}{\tilde{c}^2}\left(\frac{du}{d\varphi}\right)^2-\frac{GM}{\tilde{c}^2}\frac{d^2u}{d\varphi^2}.$$</p>
<p>Any correction to the equation for the orbit,</p>
<p>$$\frac{d^2u}{d\varphi^2}+u=\frac{GM}{L^2},$$</p>
<p>will lead to an 'opening' of the orbit and some type of precession. The second term on the right of Gerber's equation does not result in a first-order correction, and it is by providence that the last terms in the general relativity equation give the same result for the precession where the speed is $c/\surd 6$. But, speed of what? The nonlinear term in Gerber's equation does not enter the first-order correction to the orbit so it is impossible to distinguish between $k=-1/2$ and $k=-1$, where $k$ is the ratio of the longitudinal to the parallel force, taking the parallel force as unity. These are two extremes orientations of the current elements that always interact along the line connecting them. There are no tangential components to the acceleration.</p>
<p>What Gerber says, in essence, is that gravity is far-action (according to Newton) and push (according to Le Sage). Only be a quirk in the numerical equivalence of the first-order correction term can we say that gravity is due to warped space-time (according to Einstein). Even if the latter would be true, it gives us no understanding of the mechanism that causes the perturbation in the orbit, which should also hold for the other planets, and not Mercury alone.</p>
<p> </p>
</div></div>
Fri, 08 Jun 2018 10:05:01 +0000adminlavenda181 at http://bernardhlavenda.comIs Gravity Incompatible with Geodesic Motion?
http://bernardhlavenda.com/node/180
<span class="field field--name-title field--type-string field--label-hidden">Is Gravity Incompatible with Geodesic Motion?</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, 06/04/2018 - 07:30</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>Accelerating masses are repudiated to be the sources of gravitational waves just like accelerating charges in an antenna produce electromagnetic waves with the exception that the former needs a quadrupole, because masses have no charge, while the dipoles will suffice for the latter. Now, it is a matter of fact that all particles, whether they be 'test' particles or real ones, follow geodesics. Geodesics are paths of constant velocity so how can general relativity predict the existence of gravitational waves if masses don't accelerate?</p>
<p>Let us return to our discussion of the Weber force. The first two decades of the nineteenth century saw a profound interest in trying to understand the interaction of magnets with current loops. Notable among the experimentalists were Biot and Savart. But, it was Ampere who had the idea of replacing a magnet by a current, and began to experiment on the force between one current and another. That is, he hypothesized that magnetism may be the result of electrical currents; a magnetic 'molecule' in a current loop, which Gauss would latter refer to as a "galvano-electric orbit" in his 1832 paper on magnetism.</p>
<p>Ampere's force between any two current elements, $ids$ and $i'ds'$, had to be a function of the current strengths, $i$ and $i'$, the distance of their separation, $r$, and the area enclosed by the loop, $dsds'$. However, Ampere's force was still be be a force of "action-at-a-distance", like Coulomb's and Newton's force which preceded it. However, it would be able to account for all phenomena between current elements in motion save that of Faraday's law of induction which is related to accelerating charges. </p>
<p>It is commonly believed that Weber's force rectified this situation by including the acceleration of moving charges. It consisted of three terms: the static Coulomb force, a term involving the radial velocities of the moving charges, and finally the relative acceleration between the charges. The latter was able to account for induction. Implicit in this is the transfer from current elements $ids$ and $i'ds'$ to charges $e$ and $e'$. The time increment $dt$ was introduced by setting $ids=edt/c^2$, which required the intervention of a new constant, $c$, in order to amalgam a completely static force with a moving one. Weber understood his constant as the "relative velocity which electrical masses $e$ and $e'$ have, and must retain, if they are not to act on each other at all." In 1855, Weber and Kohlrausch set out to determine the numerical value of this constant.</p>
<p>So action at a distance was replaced by the continuous propagation of a force and the only distinction between $ids$ and $i'ds'$ was in the difference in charges $e$ and $e'$. This, according to Maxwell's derivation of Weber's law was the result of Fencher's hypothesis that positive and negative charges travel at the same speeds but in opposite directions. This was not the first time that a flawed hypothesis was to lead to a correct result, but, it avoided the possibility of taking retarded motion into consideration which Lienard and Wiechert would do in the last decade of the nineteenth century. This seemingly minor point has been completely overlooked in the literature.</p><p></p>
<p>So, it really makes no difference in writing down the force law, whether it be Ampere's or Weber's. So let us consider the former. Ampere's experiments clearly led him to believe that there will be an attraction or repulsion between the current elements that depended on the direction of flow. From his first experiment he surmised that the current elements will attract or repel each other depending on their relative orientations. In other words, how does the force differ, say, when the current elements are placed longitudinally, or parallel to the line connecting them, than when they are parallel to each other, normal to the line connecting them? </p><p></p>
<p>The answer to this question was found in Ampere's second experiment. From this experiment Ampere was able to conclude that "an infinitely small portion of electrical current exerts no action on another infinitely small portion of current situated in a plane which passes through its midpoint, and which is perpendicular to its direction." So the force depends on the relative orientation of the current elements. This led Ampere to write down a law of force in the form</p><p></p>
<p>$$\frac{ii'ds\cdot ds'}{r^n}\Phi(\theta,\theta'),$$</p><p></p>
<p>where $\Phi$ represents an unknown function of the angles, $\theta$ and $\theta'$, which the current elements make with the line $r$ joining them. Undoubtedly, Ampere was led to the determination of the exponent $n$ by drawing an analogy between Coulomb's potential and that of Newton. For, in matter of fact, his law must reduced to that of Coulomb when the relative velocities tended to zero.</p><p></p>
<p>Ampere represented the parallel components of the current elements by $ids\sin\theta$ and $i'ds'\sin\theta'$, while $ids\cos\theta$ and $i'ds'\cos\theta'$ were the longitudinal components. According to the results of his second equilibrium experiment, there would be no product of longitudinal and parallel components, like $ids\sin\theta\cdot i'ds'\cos\theta'$, since they do not interact with each other. However, "longitudinal" and "parallel" are relative because the current elements were not constrained to lie in the same plane. Although his second equilibrium experiment assumed planarity, any non-coplanar element can always be decomposed into two components, one lying in the plane and the other normal to the plane. The fact that the normal component would not contribute to the force allows the assumption of planarity to hold sway.</p><p></p>
<p>Ampere's law would thus consist of a combination of parallel,</p><p></p>
<p>$$\frac{ids\cdot i'ds'}{r^2}\sin\theta\sin\theta',$$</p><p></p>
<p>and longitudinal,</p><p></p>
<p>$$\frac{ids\cdot i'ds'}{r^2}\cos\theta\cos\theta'$$</p><p></p>
<p>interactions. However, there is no reason why these force components should enter the force with equal weights. This was a consequence of Ampere's first equilibrium experiment. So Ampere introduced the constant $k$, which represented the ratio of the force between the current elements in the longitudinal position to those which are parallel. If the parallel force is taken as unity, then Ampere wrote</p><p></p>
<p>$$\frac{ids\cdot i'ds'}{r^2}\left(\sin\theta\sin\theta'+k\cos\theta\cos\theta'\right).$$</p><p></p>
<p>An additional equilibrium experiment, recorded in 1823, allowed Ampere to fix $k=-1/2$; that is, the longitudinal force is only half as great as the parallel component. The negative sign meant that the difference in angles, $\varepsilon=\theta'-\theta$ would play a role, and this was later associated with the phenomenon of induction, since it would result from the extension of the directions of the two current elements.</p><p></p>
<p>When Weber tried to convert Ampere's law into a dynamical one, he was unsure about the longitudinal force, $k\cos\theta\cos\theta'$, which he was inclined to drop. It was Gauss who dissuaded him from doing so, because, if Ampere's force was to stand or drop, it must do so in its entirety. And that is what Weber did---accept Ampere's law in its entirety.</p><p></p>
<p>We have seen how the angle $\eta$ between the two planes in which the current element lied in, in the force law,</p><p></p>
<p>$$\frac{ids\cdot i'ds'}{r^2}\left(\sin\theta\sin\theta'\cos\eta+k\cos\theta\cos\theta'\right),$$</p><p></p>
<p>could safely be put to zero without jeopardizing the generality of the law. Maxwell also saw fit to exclude the angular dependencies in order to come out with a propagation equation for waves that travelled at a speed $c/\surd 2$. It is incomprehensible that such a phenomenon did not pass through the mind of Ampere, seeing that his colleague Fresnel was sharing his Paris apartment at the same time that he was performing his experiments. And then there was the nagging question of where should gravity fit in seeing that both force laws would start out with an inverse square law. </p><p></p>
<p>Even before Weber began his investigations, Mossotti thought of a way of including gravitation under the banner of electrodynamics. If the universe consisted of equal amounts of positive and negative charges then unlike charges would be attracted by the same force that like charges would be repelled. Mossotti questioned this hypothesis and considered the possibility that the attractive force between unlike charges should slightly exceed the repulsive force between like charges. Weber later gave credence to such a hypothesis, and in a posthumously published note queried whether it would be physically possible to measure such a small difference.</p><p></p>
<p>Others were thinking of applying Weber's force law to phenomena that seemed to defy Newton's law of action-at-a-distance. Seegers in 1864 proposed to analyze the recently discovered anomaly of the advance of the perihelion of Mercury in terms of Weber's law. Both Scheibner and Tisserand later evaluated the anomaly using Weber's law. </p><p></p>
<p>Although this was the first time that a critical speed was introduced, and corrections introduced of the order $1/c^2$, there was no rationale for taking over Ampere's, or Weber's, law lock stock and barrel. In particular, why should the parallel components be weighed twice as important as the longitudinal components? We are not treating charged particles that can show both <i>repulsion</i> as well as attraction. Without any reason for considering such a distinction we can set Ampere's constant, $k=-1$, based on the premise of "insufficient reason" to do so otherwise. This choice implies equal magnitudes to current elements that are longitudinal and parallel, but with a force of the opposite sign!</p><p></p>
<p>Weber's theorem that the electrostatic force must be reduced when the particles are in relative motion, then when they are at rest, led him to append a term involving the radial velocity, $\dot{r}$, onto Coulomb's law. Since his theorem applies both to the cases where the particles are approaching each other and receding from each other, this term must be at least quadratic in the radial velocities since the force would be impervious to the direction of flow, i.e.,</p><p></p>
<p>$$\frac{ee'}{r^2}\left(1+\kappa\dot{r}^2\right),$$</p><p></p>
<p>where $\kappa$ is a constant, which should turn out negative since motion decreases the strength of the Coulomb force.</p><p></p>
<p>Attraction between the particles would make them speed up when approaching each other, making $\dot{r}$ positive, and when receding from each other this velocity will become negative. At the point where the current elements are opposite one another, the relative velocity vanishes, but not their relative acceleration. This led to the introduction of a term of relative acceleration in Weber's law. However, Weber also notice that since particles in parallel current elements do not travel along the same line, the acceleration should be decreased by the distance of their separation and stand in the ratio to the square of the velocity as, viz.,</p><p></p>
<p>$$\ddot{r}=-\frac{\kappa}{r}\dot{r}^2.$$</p><p></p>
<p>But $\kappa$ must be proportional to the negative the negative of Ampere’s constant. If we call the constant proportionality, $1/c^2$, then Weber’s force can be written as</p><p></p>
<p>$$\frac{ee’}{r^2}\left\{1+\frac{1}{c^2}\left(r\ddot{r}-\dot{r}^2\right)\right\},$$</p><p></p>
<p>the terms in the round parentheses being Ampere’s law for $k=-1$.</p><p></p>
<p>The smallness of the factor $1/c^2$ led Weber to believe that it is possible “to grasp why the electrodynamic effect of electrical masses$\dots$ compared with electrostatic$\ldots$ always seems infinitesimally small, so that in general the former only remains significant, when as in galvanic currents, the electrostatic forces completely cancel each other in virtue of neutralization of the positive and negative electricity.”</p><p></p>
<p>The same will not be true when the “electrical masses” are replaced by uncharged masses, so that the most we can expect is that there will be small corrections to Newton’s law like that of the perihelion shift of Mercury. </p><p></p>
<p>Thus, the force law for gravity between masses $m$ and $m’$ becomes</p><p></p>
<p>$$\frac{Gmm’}{r^2}\left\{1+\frac{1}{c^2}\left(r\ddot{r}-\dot{r}^2\right)\right\}.$$</p><p></p>
<p>The Ampere term can now be written as a complete differential</p><p></p>
<p>$$\frac{d}{ds’}\left(\frac{1}{r}\frac{dr}{ds}\right),$$</p>
<p></p><p>with $k=-1$ instead of his original expression,</p>
<p>$$\frac{1}{\surd 2}\frac{d^2\surd r}{ds'ds}.$$</p>
<p>Our new expression the same equation as that of an optical ray with index of refraction, $n=1/r$. The claim is that these rays follow geodesics. According to Stokes’ theorem,</p><p></p>
<p>$$\oint \vec{G}\cdot d\vec{s}’=\int\int\nabla\wedge \vec{G} dsds’=0,$$</p><p></p>
<p>if $G$ is a total derivative like the expression above. Such behavior we should also expect of gravity. Simply said: Gravity has no curl.</p><p></p>
<p>In contrast, the most that could be expected from Ampere’s law with $k=-1/2$ would be the component of the force perpendicular to the plane containing the line connecting the two elements and the direction of the other current element to vanish. </p><p></p>
<p>In time, Weber’s correspondence to Ampere’s force is</p><p></p>
<p>$$\frac{d}{dt}\left(\frac{\dot{r}}{r}\right).$$</p><p></p>
<p>Its vanishing,</p><p></p>
<p>$$\ddot{r}-\frac{\dot{r}^2}{r}=0,$$</p><p></p>
<p><span style="mso-spacerun:yes"> </span>corresponds to geodesic motion as we shall now show. Weber’s analog of Ampere’s force will vanish when $\dot{r}=\mbox{const.}\times r$. This implies that the index of refraction will be proportional to $1/r$. It will prove more convenient to consider motion in the plane, and restrict the index of refraction to a function of a single parameter, say $y$, the distance from the x-axis, which if we consider the unevenly heated plane as in the Poincare half-plane model of hyperbolic geometry, represents the axis of infinite cold. The index of refraction will be a decreasing function of height indicating ascending temperatures.</p><p></p>
<p>The null cones will be given by</p><p></p>
<p>$$c^2dt^2=n^2\left(dx^2+dy^2\right),$$</p><p></p>
<p>where the index of refraction is $n=1/y$. The metric will be diagonal with components, $1/y^2$. The equations of the geodesics are</p><p></p>
<p>$$\ddot{x}-\frac{2}{y}\dot{x}\dot{y}=0,$$</p><p></p>
<p>and</p><p></p>
<p>$$\ddot{y}-\frac{1}{y}(\dot{y}\dot{y}-\dot{x}\dot{x})=0.$$</p><p></p>
<p>These equations are precisely the geodesic equations</p>
<p>$$\ddot{z}^m+\Gamma^{m}_{ij}\dot{z}^i\dot{z}^j=0,$$</p>
<p>where</p>
<p>$$\Gamma_{ij}^m=\frac{1}{2}g^{mn}\left(g_{in,j}+g_{jn},i-g_{ij,n}\right),$$</p>
<p>and the commas denote differentiation with respect to the argument. In the present case all the Christoffel symbols are equal to $-1/y$. Consequently, the presence of an Amperian force indicates a deviation from geodesic motion, or constant velocity. General relativity, by its very nature, is incapable of treating such motion.</p>
<p> </p>
<p> </p>
</div></div>
Mon, 04 Jun 2018 05:30:37 +0000adminlavenda180 at http://bernardhlavenda.comWhen Ampere Meets Doppler
http://bernardhlavenda.com/node/179
<span class="field field--name-title field--type-string field--label-hidden">When Ampere Meets Doppler</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, 06/02/2018 - 08:04</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>It has often been claimed that Ampere's force is equivalent to Grassmann's force, which is a precursor of the Lorentz force, when the force is integrated round a circuit, in which the current element forms a part, vanishes. This is taken to mean that a force exerted by a complex circuit element is a right angles to the element. As such it was taken to be equivalent to the earlier discovered Biot-Savart law. The Biot-Savart law,</p>
<p>$$\vec{B}=\frac{\mu}{4\pi r^2}(\vec{v}\wedge\hat{r}),$$</p>
<p>is just the definition of the magnetic field. Obviously, it does not act along the unit vector $\hat{r}$ from one charge to the other. Neither does the Grassmann force,</p>
<p>$$\vec{F}^G=\frac{\mu}{4\pi r^2}[\vec{v}'\wedge(\vec{v}\wedge\hat{r})],$$</p>
<p>which acts also in the direction $\vec{v}$, from the charge $e$ to the charge $e'$.</p>
<p>J J Thomson, and other illuminaries of late nineteenth century could not digest Lorentz's law because it violated Newton's law of action-and-reaction, i.e, something is being picked-up by its own bootstraps. Now Ampere's law acts along the line connecting the two circuit elements, which Grassmann's law has a component in the direction of one of the two circuit elements which is acting on the other. Hence, they do not in general point in the same direction. How then can they be considered equivalent in any respect?</p>
<p>Moon and Spencer [<em>J Franklin Inst. <strong>257 </strong></em>(1954) 305] go even farther and claim that Grassmann's force "is not derivable from the force between charges. Thus <em>the equation is untenable, and the whole process of defining a magnetic field from this equation is unsound.</em> We wouldn't go so far as to agree with them, because it is backed-up experimentally, but it is interesting to note that in the same paper, these authors claim "Einstein's famous paper of 1905,$\ldots$, which enunciated the theory of special relativity, was an attempt to remedy this non-relativistic aspect of electromagnetic theory. Looking back from the vantage point of present knowledge [1954], we may wonder if Einstein's work was not directed up a blind alley." This statement we find more sympathy for, but wonder how it could have come from a tenured professor at MIT?</p>
<p>Ampere's law, which was an addendum to the static Coulomb law, depends on the acceleration as well as the relative velocities of the two current elements. Shouldn't it, therefore, be amenable to the longitudinal Doppler law? </p>
<p>The Weber force acting from charge $e$ to $e'$ along $r$ is</p>
<p>$$\vec{F}^W=\hat{r}\frac{ee'}{4\pi\epsilon r^2}\left\{1+\frac{1}{c^2}\left(\vec{v}\cdot\vec{v}-\frac{3}{2}(\hat{r}\cdot\vec{v})^2+\vec{r}\cdot\vec{a}\right)\right\},$$</p>
<p>where $c=1/\surd(\epsilon\mu)$, which is Maxwell's constant, not necessarily coinciding with the speed of light.</p>
<p>If $\theta$ is the angle between $\hat{r}$ and $\vec{v}$ as shown in the figure below, then if $e$ is moving at constant velocity $\vec{v}$ with respect to $e'$, the Moon and Spencer claim that the Ampere force resulting from this motion is</p>
<p>$$\vec{F}'=\hat{r}\frac{ee'}{4\pi\epsilon r^2}\beta^2\left(1-\frac{3}{2}\cos^2\theta\right),$$</p>
<figure role="group" class="caption caption-img align-center"><img alt="The force acting on charge $e'$ due to $e$ at a distance $r$ moving with uniform velocity." data-entity-type="file" data-entity-uuid="39058785-7862-465e-859e-bdcdbae77f09" src="http://bernardhlavenda.com/sites/default/files/inline-images/Page1.jpg" /><figcaption>The force acting on charge $e'$ by a charge $e$ moving with velocity $\vec{v}$ and acceleration $d\vec{v}/dt$.</figcaption></figure><p>where $\beta=v/c$, for $d\vec{v}/dt=\vec{a}=0$.If the motion is in the direction toward $e'$, the combination of the force with Coulomb's force is</p>
<p>$$F'=\hat{r}\frac{ee'}{4\pi r^2}\left(1-\frac{1}{2}\beta^2\right),$$</p>
<p>which looks like an approximation to $\surd(1-\beta^2)$. But, that makes no sense since the force is acting longitudinally, and $\surd(1-\beta^2)$ would be a second-order Doppler shift in the direction normal to the motion. So it would appear that Ampere is at odds with Doppler!</p>
<p>From an historical perspective, Biot and Savart experimented with forces between currents and magnets. Ampere was the first to experiment on forces between two current elements $ids$ and $i'ds'$. He first came up with the force</p>
<p>$$F^A=\frac{ids\cdot i'ds'}{r^2}\left\{\sin\theta\sin\theta'\cos\eta+k\cos\theta\cos\theta'\right\},$$</p>
<p>somewhere around 1820, but he was not able to determine the unknown constant $k$ until 1823 when he set it equal to $-1/2$. $\theta$ and $\theta'$ are the angles that the velocities of the two current elements make with $\hat{r}$, and $\eta$ is the angle between the two planes that the current elements find themselves in. </p>
<p>Now, in terms of the radial velocity $\dot{\vec{r}}$ and radial acceleration $\ddot{\vec{r}}$, the combined Coulomb and Ampere laws is</p>
<p>$$F^W=\frac{ee'}{r^2}\left\{1+\frac{1}{c^2}\left(k\dot{r}^2+r\ddot{r}\right)\right\},$$</p>
<p>in the radial direction uniting the two charges. In terms of the relative velocity, $\vec{v}$ and relative accelerations $\vec{a}$, the force acting on $e'$ has magnitude</p>
<p>$$F^W=\frac{ee'}{r^2}\left\{1+\frac{1}{c^2}\left(\vec{v}\cdot\vec{v}+(k-1)(\hat{r}\cdot\vec{v})^2+\vec{r}\cdot\vec{a}\right)\right\}.$$</p>
<p>Now, having traced the origin of $k$ this far, insteading of choosing $k=-1/2$, we set $k=-1$, we get the numerical coefficient in the numerator of the Moon and Spencer expression as $(1-2\cos^2\theta)$, so that it becomes</p>
<p>$$F=\frac{ee'}{r^2}\left(1-\beta^2\right),$$</p>
<p>when combined with Coulomb's law and the motion is along $\hat{r}$ so that $\theta=0$. Now things are looking a bit brighter for if we replace $r$ by its retarded value $r'$, according to $r=(1-\hat{r}\cdot\vec{\beta})r'$, we get</p>
<p>$$\vec{F}=\hat{r}\frac{ee'}{r'^2}\frac{1-\beta^2}{(1-\beta)^2}=\frac{ee'}{r'^2}\left(\frac{1+\beta}{1-\beta}\right).$$</p>
<p>This is none other than (the square) of the longitudinal Doppler shift! This says that<em> the Weber force is the result of a Doppler shift of the Coulomb force when Ampere's constant is chosen $k=-1$.</em></p>
<p>This is corroborated by the Lienard-Wiechert expression for the electric field, found in all texts on electrodynamics,</p>
<p>$$\vec{E}=e\frac{(\vec{r}'-r'\vec{\beta})(1-\beta^2)}{(r'-\vec{r}'\cdot\vec{\beta})^3},$$</p>
<p>when $\vec{\beta}\parallel\vec{r}'$. When they are not parallel to one another we have aberration; it's as simple as that.</p>
<p>This brings to mind the conclusion arrived at by Ibison, Puthoff, and Little in the paper "The speed of gravity revisited" in their attempt to refute Van Flandern that gravity propagates much faster than the speed of light, by explaining "the origin of apparently instantaneous connections, first within EM, and then within strong-field GR."</p>
<p>For EM, they express the electric field as</p>
<p>$$\vec{E}=e\frac{\hat{r}}{r^2}\times\left\{\frac{1-\beta^2}{K^3}\right\},$$</p>
<p>where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$ and $K=\surd(1-\beta^2\sin^2\theta)$. Into this expression we can use their relation</p>
<p>$$r'/r=K/(1-\beta\cos\theta'),$$</p>
<p>to obtain</p>
<p>$$\vec{E}=e\frac{\hat{r}}{(r'-\vec{r}'\cdot\vec{\beta})^2}\frac{1-\beta^2}{K}=\vec{E}'\frac{c_0}{c},$$</p>
<p>where $c_0$ is one-way speed of light. This would have been the ratio of frequencies shifted by a Doppler effect had considered only one way propagation. In terms of the Kennedy Thorndike interferometer, the time taken to run along an optical path of length $\ell$ is</p>
<p>$$t_1=\frac{\ell}{c}\left\{\frac{K+\beta\cos\theta}{1-\beta^2}\right\}.$$</p>
<p>On the return leg, the only thing that changes is the sign of $\beta$ so that the time to complete both legs is</p>
<p>$$\frac{2\ell}{c}\frac{K}{1-\beta^2}.$$</p>
<p>But, this time should be equal to the ratio of the distance covered, $2\ell$, to the speed, $c_0$. This speed will equal $c$ only when $\beta=0$. The factor in the braces of the time for the outward journey is Doppler shift, and since the electric field cannot depend on the sign of $\beta$, it is suppressed in the Ibison <em>et al.</em> expression, just like considering the round-trip in the Kennedy-Thorndike interferometer. In other words, the factor $\gamma^2K$, where $\gamma=1/\surd(1-\beta^2)$, can be considered as the two-way Doppler shift, obtained by reflecting the light beam by a mirror placed at a distance $\ell$ from the source.</p>
<p> </p>
<p>The fact that electric field is pointing in the present direction was loaded from the start,</p>
<p>$$\vec{r}=\vec{r}'-r'\vec{\beta},$$</p>
<p>so that should raise no eyebrows. But listen to their conclusion: "This result proves the claim that the electric field from a uniformly moving source is not aberrated. The force on a test charge is directed towards the instantaneous--not the retarded--position of the source. The factor in braces affects the magnitude, but not the direction, of the field." How can it when it was put in from the very start?! Moreover, the factor </p>
<p>$$\frac{c_0}{c}=\frac{1-\beta^2}{K(\theta)},$$</p>
<p>is the Doppler shift for "there and back". Moreover, it can be generalized to aberration from the relation</p>
<p>$$\cos\theta=\frac{K(\cos\theta'-\beta)}{1-\beta\cos\theta'},$$</p>
<p>when $\cos\theta'\neq\beta$, as in the Einstein convention. </p>
<p>We can even generalize further. Any relation between past and present space coordinates, $r'$ and $r$, past and present times, $t'$ and $t$, or frequencies, or time and space increments, are related by the same Doppler shifts. This means that the Einstein metric,</p>
<p>$$ds^2=g_{00}dt^2+g_{ij}dx^idx^j,$$</p>
<p>has only meaning for $ds^2=0$, along a null, or light, cone. As such it cannot accommodate accelerations since it would no longer have a quadratic form. As we have always said, accelerations have nothing to do with Doppler shifts and aberration. When included they would raise the metric to a cubic!</p>
<p>This brings to mind Carlip's claims, in his manuscript "Aberration and the speed of gravity", where he tries to extend the preceding electromagnetic argument to general relativity. According to him, "we cannot simply require by fiat that a massive source accelerate. The Einstein field equations are consistent only when all gravitational sources move along trajectories determined by their equations of motion, and in particular, we can consistently represent an accelerated source only if we include the energy responsible for its acceleration." Masses, according to Einstein's equations move along geodesics--trajectories of constant velocity. Energy can't be localized in general relativity, so it is an enigma how energy is to added to make a mass accelerated. Newton had the answer to that! Admitting to the fact that a "test particle in spacetime will travel along a geodesic, $\ldots$, the 'acceleration' of such a particle, in Newtonian language, is determined by the connection $-\Gamma^i_{00}$", if the particle was initially at rest. How can this be if the "test" particle follows a geodesic?</p>
<p>The answer, Carlip contends, is to incorporate 'acceleration", $Gm/r^2$, in the $g_{00}$ term. However, this is inconsistent with the structure of the metric, as we now show. Neglecting terms quadratic in the acceleration term, $\alpha dt$, where $\alpha=a/c$, the KT metric can be written as</p>
<p>$$c^2dt^2(1-\beta^2)=dr^2+(2\beta+\alpha dt)\cos\theta drdt.$$</p>
<p>Completing the square, we find that to lowest order</p>
<p>$$\frac{ds}{dt}=\frac{1-\beta^2}{\sqrt{K^2+\alpha\beta\cos\theta dt}+\beta\cos\theta}.$$</p>
<p>But, this makes no sense since the rhs depends on the time interval $dt$ on account of the fact that the metric contained a cubic term, $dt^2dr$. Geodesics, metrics, Doppler, and aberration abhor terms that depend on the acceleration. The key to incorporating accelerations is found in Ampere's law, and it will be the subject of our next blog.</p>
<p>However, the point to emphasize is that space $ds$ and $dt$ behave in the same way as past and present times $t'$ and $t$, past and present positions, $r'$ and $r$, and Coulomb and Ampere forces for $k=-1$. They are all related through Doppler shifts caused by the inclusion of velocities.</p>
<p> </p>
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Sat, 02 Jun 2018 06:04:46 +0000adminlavenda179 at http://bernardhlavenda.comCosmic Coincidences
http://bernardhlavenda.com/node/178
<span class="field field--name-title field--type-string field--label-hidden">Cosmic Coincidences</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">Wed, 05/30/2018 - 11:33</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 critical density, as we have seen, is determined by introducing the Hubble parameter, $H$, defined from the velocity distance relation, $v=HR$. Rather than being a 'receding speed', $v$, and $R$ the distance the galaxy and the observer, the former appears as a revolving speed, so that $H$ is an angular speed of rotation, and the latter the radius measuring the distance between the galaxy and the cosmic center. Although the latter interpretation of Hubble's constant has not been open to direct observations, it lends itself to a series of numerical coincidences.</p>
<p>First, and foremost, is the determination of the critical density of the universe from---of all places---the Schwarzschild radius,</p>
<p>$$\mathcal{R}=\frac{2GM}{c^2}=\frac{c}{\omega}.$$</p>
<p>If the angular speed is identified with Hubble's constant, and the density is introduced as $\rho=3M/4\pi\mathcal{R}^3$, we immediately obtain the critical density,</p>
<p>$$\rho_c=\frac{3H^2}{8\pi G},$$</p>
<p>which is enough to raise eyebrows.</p>
<p>Second, if the angular velocity is $H\mathcal{R}$, the centripetal acceleration is</p>
<p>$$a=H^2\mathcal{R}.$$</p>
<p>Again using the critical density, the acceleration is found to be</p>
<p>$$a=Hc,$$</p>
<p>which is numerical $6.48\times 10^{-10}$ m/sec. This is the same order of magnitude as the MOND acceleration, $a_0$, in </p>
<p>$$v\simeq\sqrt[4]{2GMa_0},$$</p>
<p>where $a_0\simeq (1.2\pm.03)\times 10^{-10}$ m/sec. Milgrom determined this value by fitting flat rotation curves of galaxies obeying the Tully-Fisher relation,</p>
<p>$$v\propto L^{1/4},$$</p>
<p>where the luminosity $L$ is proportional to the mass $M$ of the spiral galaxy.</p>
<p>Finally, if we use Milgrom's relation, </p>
<p>$$v=\sqrt[4]{2GMa_0},$$</p>
<p>introduce $v=H\mathcal{R}$, and the critical density, we find</p>
<p>$$a_0=H^2\mathcal{R}.$$</p>
<p>Alternatively, if we know $a_0$, from fitting, and $H$ from the results of the 2015 Planck survey, we find $\mathcal{R}\sim 10^{26}$ m/sec, which is the order of the radius of the universe!</p>
<p>This is truly remarkable in view of the fact that we are dealing with very high speed relativistic phenomena, and, yet, $a_0$, is about a 100 billion times smaller than the acceleration due to gravity at the earth's surface!</p>
<p>Universal constants, like Planck's constant, Boltzmann's constant, Newton's constant, charge, and the speed of light are often used as benchmarks for determining the limits of applicability of a theory. Consider the 'classical' radius of an electron, $r_c=e^2/mc^2$. It contains the charge and the speed of light so we know that it is electrodynamic, ultra-relativistic, and 'classical'. If we divide this radius by the fine structure constant $\alpha=e^2/\hbar c$, we get the Compton wavelength, $\hbar/mc$, which is still relativistic, but non-classical since it contains Planck's constant. Another division by $\alpha$ results in the 'classical' Bohr radius, $\hbar^2/me^2$, which is quantum and non-relativistic since $c$ has disappeared. The same procedure applies to gravity by substituting $e\rightarrow \sqrt{G}M$.</p>
<p>In reference to the above 'coincidences', the critical mass is classical and non-relativistic, although the ultra-relativistic Schwarzschild radius was used to derived it. Moreover, Hubble's law is classical but the centrifugal acceleration used to derive Milgrom's constant, $a_0$ is ultra-relativistic. The big question is: How can the critical density be non-relativistic when all that depends on it is ultra-relativistic?</p>
<p>All we can conclude from this is that if there is any physics in this, which is extremely doubtful, it is very well hidden.</p>
<p> </p>
<p> </p>
<p> </p>
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Wed, 30 May 2018 09:33:23 +0000adminlavenda178 at http://bernardhlavenda.com