From c150075066fa904ebb719ca7ecf097124f94e6d6 Mon Sep 17 00:00:00 2001 From: Francis Russell Date: Thu, 1 Dec 2011 03:00:50 +0000 Subject: [PATCH] Replace $$s with amsmath's align environment. --- convert_tex4ht | 2 +- relat10.tex | 225 ++++++++++++++++++++++++++++++++++++------------- 2 files changed, 168 insertions(+), 59 deletions(-) diff --git a/convert_tex4ht b/convert_tex4ht index fdd6415..35e6194 100755 --- a/convert_tex4ht +++ b/convert_tex4ht @@ -2,7 +2,7 @@ set -e # Output is XHTML, document character encoding will be UTF-8. -OPTIONS1="xhtml,charset=utf-8" +OPTIONS1="xhtml,charset=utf-8,pic-align" # Yes, that first whitespace character is meant to be there. # unicode support, generate unicode characters instead of escape codes, environment file. diff --git a/relat10.tex b/relat10.tex index 57c5bc7..ceb5b32 100644 --- a/relat10.tex +++ b/relat10.tex @@ -78,6 +78,7 @@ \usepackage{ucs} \usepackage[utf8x]{inputenc} +\usepackage{amsmath} % Declare "longrightarrow" as "rightarrow" as the fomer never renders well. \DeclareUnicodeCharacter{10230}{→} @@ -653,7 +654,9 @@ carriage. The velocity $w$ of the man relative to the embankment is here replaced by the velocity of light relative to the embankment. $w$ is the required velocity of light with respect to the carriage, and we have - $$w = c-v.$$ +\begin{align*} +w = c-v. +\end{align*} The velocity of propagation ot a ray of light relative to the carriage thus comes cut smaller than $c$. @@ -1123,7 +1126,9 @@ reference-body $K$ and for the reference-body $K'$. A light-signal is sent along the positive $x$-axis, and this light-stimulus advances in accordance with the equation - $$x = ct,$$ +\begin{align*} +x = ct, +\end{align*} \noindent {\it i.e.} with the velocity $c$. According to the equations of the Lorentz transformation, this simple relation between $x$ and $t$ involves a @@ -1138,7 +1143,9 @@ t' &=& \frac{(I-\frac{v}{c})t}{\sqrt{I-\frac{v^2}{c^2}}} \noindent from which, by division, the expression - $$x' = ct'$$ +\begin{align*} +x' = ct' +\end{align*} \noindent immediately follows. If referred to the system $K'$, the propagation of light takes place according to this equation. We thus see that the @@ -1211,18 +1218,24 @@ the origin ($x'=0$) of $K'$. $t'=0$ and $t'=I$ are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks: -$$t = 0$$ +\begin{align*} +t = 0 +\end{align*} \noindent and -$$t' = \frac{I}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +t' = \frac{I}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ As judged from $K$, the clock is moving with the velocity $v$; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but -$$\frac{I}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +\frac{I}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ \noindent seconds, {\it i.e.} a somewhat larger time. As a consequence of its motion @@ -1251,14 +1264,18 @@ Galilei transformation (Section 11). In place of the man walking inside the carriage, we introduce a point moving relatively to the co-ordinate system $K'$ in accordance with the equation -$$x' = wt'.$$ +\begin{align*} +x' = wt'. +\end{align*} ~ By means of the first and fourth equations of the Galilei transformation we can express $x'$ and $t'$ in terms of $x$ and $t$, and we then obtain -$$x = (v + w)t.$$ +\begin{align*} +x = (v + w)t. +\end{align*} ~ This equation expresses nothing else than the law of motion of the @@ -1281,7 +1298,9 @@ x'=wt' first and fourth equations of the Lorentz transformation. Instead of the equation \ref{eqnA} we then obtain the equation -$$W = \frac{v+w}{I+\frac{vw}{c^2}}$$ +\begin{align*} +W = \frac{v+w}{I+\frac{vw}{c^2}} +\end{align*} ~ \noindent which corresponds to the theorem of addition for velocities in one @@ -1442,11 +1461,15 @@ accordance with the theory of relativity the kinetic energy of a material point of mass m is no longer given by the well-known expression -$$m\frac{v^2}{2}$$ +\begin{align*} +m\frac{v^2}{2} +\end{align*} \noindent but by the expression -$$\frac{mc^2}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +\frac{mc^2}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ This expression approaches infinity as the velocity $v$ approaches the @@ -1455,7 +1478,9 @@ than $c$, however great may be the energies used to produce the acceleration. If we develop the expression for the kinetic energy in the form of a series, we obtain -$$mc^2 + m\frac{v^2}{2} + \frac{3}{8} m \frac{v^4}{c^2} + \cdots$$ +\begin{align*} +mc^2 + m\frac{v^2}{2} + \frac{3}{8} m \frac{v^4}{c^2} + \cdots +\end{align*} ~ When $v^2/c^2$ is small compared with unity, the third of these terms is @@ -1491,24 +1516,32 @@ with the velocity $v$, which absorbs\footnotemark\ an amount of energy $E_0$ in the form of radiation without suffering an alteration in velocity in the process, has, as a consequence, its energy increased by an amount -$$\frac{E_0}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +\frac{E_0}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ In consideration of the expression given above for the kinetic energy of the body, the required energy of the body comes out to be -$$\frac{\left(m+\frac{E_0}{c^2}\right)c^2}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +\frac{\left(m+\frac{E_0}{c^2}\right)c^2}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ \noindent Thus the body has the same energy as a body of mass -$$\left(m+\frac{E_0}{c^2}\right)$$ +\begin{align*} +\left(m+\frac{E_0}{c^2}\right) +\end{align*} ~ \noindent moving with the velocity $v$. Hence we can say: If a body takes up an amount of energy $E_0$, then its inertial mass increases by an amount -$$\frac{E_0}{c^2}$$ +\begin{align*} +\frac{E_0}{c^2} +\end{align*} ~ \noindent the inertial mass of a body is not a constant but varies according to @@ -1519,7 +1552,9 @@ the conservation of energy, and is only valid provided that the system neither takes up nor sends out energy. Writing the expression for the energy in the form -$$\frac{mc^2+E_0}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +\frac{mc^2+E_0}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ \noindent we see that the term $mc^2$, which has hitherto attracted our attention, @@ -1532,7 +1567,9 @@ the changes in energy $E_0$ to which we can Subject a system are not large enough to make themselves perceptible as a change in the inertial mass of the system. -$$\frac{E_0}{c^2}$$ +\begin{align*} +\frac{E_0}{c^2} +\end{align*} ~ \noindent is too small in comparison with the mass $m$, which was present before @@ -1625,7 +1662,9 @@ experiences a contraction in the direction of motion in consequence of that motion. the contracted length being proportional to the expression -$$\sqrt{I-\frac{v^2}{c^2}}.$$ +\begin{align*} +\sqrt{I-\frac{v^2}{c^2}}. +\end{align*} This, hypothesis, which is not justifiable by any electrodynamical facts, supplies us then with that particular law of motion which has @@ -1755,7 +1794,9 @@ on the theory of relativity, since according to this theory time is robbed of its independence. This is shown by the fourth equation of the Lorentz transformation: -$$t' = \frac{t-\frac{v}{c^2}x}{\sqrt{I-\frac{v^2}{c^2}}}$$ +\begin{align*} +t' = \frac{t-\frac{v}{c^2}x}{\sqrt{I-\frac{v^2}{c^2}}} +\end{align*} ~ Moreover, according to this equation the time difference $\Delta t'$ of two @@ -1980,8 +2021,10 @@ field), \noindent where the ``gravitational mass" is likewise a characteristic constant for the body. From these two relations follows: -$$\mbox{(acceleration)} = \frac{\mbox{gravitational mass}}{\mbox{inertial mass}} - \times \mbox{intensity of the gravitational field}$$ +\begin{align*} +\mbox{(acceleration)} = \frac{\mbox{gravitational mass}}{\mbox{inertial mass}} + \times \mbox{intensity of the gravitational field} +\end{align*} ~ If now, as we find from experience, the acceleration is to be @@ -2615,7 +2658,9 @@ indicate the distance (line-interval) between $P$ and $P_1$, as measured with a little rod, by means of the very small number $ds$. Then according to Gauss we have - $${ds}^2 = g_{11}{du}^2 + 2g_{12}dudv = g_{22}{dv}^2$$ +\begin{align*} +{ds}^2 = g_{11}{du}^2 + 2g_{12}dudv = g_{22}{dv}^2 +\end{align*} \noindent where $g_{11}, g_{12}, g_{22}$, are magnitudes which depend in a perfectly definite way on $u$ and $v$. The magnitudes $g_{11}$, $g_{12}$ and $g_{22}$, @@ -2626,7 +2671,9 @@ continuum with reference to the measuring-rods, but only in this case, it is possible to draw the $u$-curves and $v$-curves and to attach numbers to them, in such a manner, that we simply have: - $${ds}^2 = {du}^2 + {dv}^2$$ +\begin{align*} +{ds}^2 = {du}^2 + {dv}^2 +\end{align*} Under these conditions, the $u$-curves and $v$-curves are straight lines @@ -2648,14 +2695,18 @@ distance $ds$ is associated with the adjacent points $P$ and $P_1$, this distance being measurable and well defined from a physical point of view, then the following formula holds: -$${ds}^2 = g_{11}{dx_1}^2 + 2g_{12}dx_1dx_2 . . . . g_{44}{dx_4}^2$$ +\begin{align*} +{ds}^2 = g_{11}{dx_1}^2 + 2g_{12}dx_1dx_2 . . . . g_{44}{dx_4}^2 +\end{align*} \noindent where the magnitudes $g_{11}$, etc., have values which vary with the position in the continuum. Only when the continuum is a Euclidean one is it possible to associate the co-ordinates $x_1 \ldots x_4$. with the points of the continuum so that we have simply -$$ds^2 = {dx_1}^2 + {dx_2}^2 + {dx_3}^2 + {dx_4}^2$$ +\begin{align*} +ds^2 = {dx_1}^2 + {dx_2}^2 + {dx_3}^2 + {dx_4}^2 +\end{align*} In this case relations hold in the four-dimensional continuum which are analogous to those holding in our three-dimensional measurements. @@ -2719,19 +2770,25 @@ second Galileian system we shall suppose that the corresponding differences for these two events are $dx'$, $dy'$, $dz'$, $dt'$. Then these magnitudes always fulfil the condition\footnotemark. - $${dx}^2 + {dy}^2 + {dz}^2 - c^2{dt}^2 = {dx'}^2 + {dy'}^2 + {dz'}^2 - c^2{dt'}^2$$ +\begin{align*} +{dx}^2 + {dy}^2 + {dz}^2 - c^2{dt}^2 = {dx'}^2 + {dy'}^2 + {dz'}^2 - c^2{dt'}^2 +\end{align*} The validity of the Lorentz transformation follows from this condition. We can express this as follows: The magnitude - $${ds}^2 = {dx}^2 + {dy}^2 + {dz}^2 - c^2{dt}^2$$ +\begin{align*} +{ds}^2 = {dx}^2 + {dy}^2 + {dz}^2 - c^2{dt}^2 +\end{align*} \noindent which belongs to two adjacent points of the four-dimensional space-time continuum, has the same value for all selected (Galileian) reference-bodies. If we replace $x$, $y$, $z$, $\sqrt{-I} \cdot ct$ , by $x_1$, $x_2$, $x_3$, $x_4$, we also obtaill the result that - $${ds}^2 = {dx_1}^2 + {dx_2}^2 + {dx_3}^2 + {dx_4}^2$$ +\begin{align*} +{ds}^2 = {dx_1}^2 + {dx_2}^2 + {dx_3}^2 + {dx_4}^2 +\end{align*} \noindent is independent of the choice of the body of reference. We call the magnitude ds the ``distance'' apart of the two events or @@ -3187,7 +3244,9 @@ to Euclidean geometry of the plane, equal to a constant value $\pi$, which is independent of the diameter of the circle. On their spherical surface our flat beings would find for this ratio the value - $$\pi \frac{\sin \frac{r}{R}}{\frac{r}{R}}$$ +\begin{align*} +\pi \frac{\sin \frac{r}{R}}{\frac{r}{R}} +\end{align*} {\it i.e.} a smaller value than $\pi$, the difference being the more considerable, the greater is the radius of the circle in comparison with the radius $R$ of the ``world-sphere." By means of this relation @@ -3319,7 +3378,9 @@ it. \footnotetext{For the radius R of the universe we obtain the equation - $$R^2=\frac{2}{\kappa p}$$ +\begin{align*} +R^2=\frac{2}{\kappa p} +\end{align*} The use of the C.G.S. system in this equation gives $2/k = 1^.08 \cdot 10^{27}$; $p$ is the average density of the matter and $k$ is a constant connected @@ -3346,7 +3407,9 @@ and the time $t'$. We require to find $x'$ and $t'$ when $x$ and $t$ are given. A light-signal, which is proceeding along the positive axis of $x$, is transmitted according to the equation - $$x = ct$$ +\begin{align*} +x = ct +\end{align*} or \begin{equation} \label{eqn:a1} @@ -3386,11 +3449,15 @@ By adding (or subtracting) equations (\ref{eqn:a3}) and (\ref{eqn:a4}), and intr convenience the constants $a$ and $b$ in place of the constants $\lambda$ and $\mu$, where - $$a = \frac{\lambda+\mu}{2}$$ +\begin{align*} +a = \frac{\lambda+\mu}{2} +\end{align*} \noindent and - $$a = \frac{\lambda-\mu}{2}$$ % ?? +\begin{align*} +a = \frac{\lambda-\mu}{2} +\end{align*} % ?? \noindent we obtain the equations @@ -3405,7 +3472,9 @@ and $b$ were known. These result from the following discussion. For the origin of $K^1$ we have permanently $x' = 0$, and hence according to the first of the equations (\ref{eqn:a5}) - $$x = \frac{bc}{a}t$$ +\begin{align*} +x = \frac{bc}{a}t +\end{align*} If we call $v$ the velocity with which the origin of $K^1$ is moving relative to $K$, we then have @@ -3430,7 +3499,9 @@ require to take a ``snapshot'' of $K^1$ from $K$; this means that we have to insert a particular value of $t$ (time of $K$), {\it e.g.} $t = 0$. For this value of $t$ we then obtain from the first of the equations (5) - $$x' = ax$$ +\begin{align*} +x' = ax +\end{align*} Two points of the $x'$-axis which are separated by the distance $\Delta x' = I$ when measured in the $K^1$ system are thus separated in our instantaneous @@ -3445,18 +3516,24 @@ photograph by the distance from the equations (\ref{eqn:a5}), taking into account the expression (\ref{eqn:a6}), we obtain - $$x' = a \left( I - \frac{v^2}{c^2} \right) x$$ +\begin{align*} +x' = a \left( I - \frac{v^2}{c^2} \right) x +\end{align*} \noindent From this we conclude that two points on the $x$-axis separated by the distance $I$ (relative to $K$) will be represented on our snapshot by the distance - $$\Delta x' = a \left( I - \frac{v^2}{c^2} \right) \quad . \quad . \quad . \quad \mbox{(7a)}$$ +\begin{align*} +\Delta x' = a \left( I - \frac{v^2}{c^2} \right) \quad . \quad . \quad . \quad \mbox{(7a)} +\end{align*} But from what has been said, the two snapshots must be identical; hence $\Delta x$ in (7) must be equal to $\Delta x'$ in (7a), so that we obtain - $$a = \frac{I}{I-\frac{v^2}{c^2}} \quad . \quad . \quad . \quad \mbox{(7b)} $$ +\begin{align*} +a = \frac{I}{I-\frac{v^2}{c^2}} \quad . \quad . \quad . \quad \mbox{(7b)} +\end{align*} The equations (\ref{eqn:a6}) and (7b) determine the constants $a$ and $b$. By inserting the values of these constants in (\ref{eqn:a5}), we obtain the first @@ -3473,7 +3550,9 @@ and the fourth of the equations given in Section 11. Thus we have obtained the Lorentz transformation for events on the $x$-axis. It satisfies the condition - $$x'^2 - c^2t'^2 = x^2 - c^2t^2 \quad . \quad . \quad . \quad \mbox{(8a)} $$ +\begin{align*} +{x'}^2 - c^2{t'}^2 = x^2 - c^2t^2 \quad . \quad . \quad . \quad \mbox{(8a)} +\end{align*} The extension of this result, to include events which take place outside the $x$-axis, is obtained by retaining equations (\ref{eqn:a8}) and @@ -3492,7 +3571,9 @@ manner. We suppose a light-signal sent out from the origin of $K$ at the time $t = 0$. It will be propagated according to the equation - $$r = \sqrt{x^2+y^2+z^2} = ct$$ +\begin{align*} +r = \sqrt{x^2+y^2+z^2} = ct +\end{align*} \noindent or, if we square this equation, according to the equation @@ -3506,18 +3587,22 @@ the postulate of relativity, that the transmission of the signal in question should take place---as judged from $K^1$---in accordance with the corresponding formula - $$r' = ct'$$ +\begin{align*} +r' = ct' +\end{align*} \noindent or, - $$x'^2 + y'^2 + z'^2 - c^2t'^2 = 0 \quad . \quad . \quad . \quad \mbox{(10a)} $$ +\begin{align*} +{x'}^2 + {y'}^2 + {z'}^2 - c^2{t'}^2 = 0 \quad . \quad . \quad . \quad \mbox{(10a)} +\end{align*} In order that equation (10a) may be a consequence of equation (\ref{eqn:a10}), we must have \begin{equation} \label{eqn:a11} - x'^2 + y'^2 + z'^2 - c^2t'^2 = \sigma (x^2 + y^2 + z^2 - c^2t^2) + {x'}^2 + {y'}^2 + {z'}^2 - c^2{t'}^2 = \sigma (x^2 + y^2 + z^2 - c^2t^2) \end{equation} @@ -3545,7 +3630,9 @@ transformation thus: It expresses $x', y', x', t'$, in terms of linear homogeneous functions of $x, y, x, t$, of such a kind that the relation - $$x'^2 + y'^2 + z'^2 - c^2t'^2 = x^2 + y^2 + z^2 - c^2t^2 \quad . \quad . \quad . \quad \mbox{(11a)} $$ +\begin{align*} +{x'}^2 + {y'}^2 + {z'}^2 - c^2{t'}^2 = x^2 + y^2 + z^2 - c^2t^2 \quad . \quad . \quad . \quad \mbox{(11a)} +\end{align*} \noindent is satisfied identically. That is to say: If we substitute their expressions in $x, y, x, t$, in place of $x', y', x', t'$, on the @@ -3572,7 +3659,9 @@ accordance with this, we insert and similarly for the accented system $K^1$, then the condition which is identically satisfied by the transformation can be expressed thus: -$${x'_1}^2 + {x'}_2^2 + {x'}_3^2 + {x'}_4^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 \quad . \quad . \quad . \quad \mbox{(12)}.$$ +\begin{align*} +{x'_1}^2 + {x'}_2^2 + {x'}_3^2 + {x'}_4^2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 \quad . \quad . \quad . \quad \mbox{(12)}. +\end{align*} \noindent That is, by the afore-mentioned choice of ``coordinates," (11a) [see the end of Appendix II] is transformed into this equation. @@ -3596,7 +3685,9 @@ x'_2, x'_3$) with the same origin, then $x'_1, x'_2, x'_3$, are linear homogeneous functions of $x_1, x_2, x_3$ which identically satisfy the equation - $${x'}_1^2 + {x'}_2^2 + {x'}_3^2 = x_1^2 + x_2^2 + x_3^2$$ +\begin{align*} +{x'}_1^2 + {x'}_2^2 + {x'}_3^2 = x_1^2 + x_2^2 + x_3^2 +\end{align*} The analogy with (12) is a complete one. We can regard Minkowski's ``world'' in a formal manner as a four-dimensional Euclidean space (with @@ -3684,7 +3775,9 @@ in such away, that the angle described by the radius sun-planet between one perhelion and the next should exceed that corresponding to one complete revolution by an amount given by - $$+ \frac{24\pi^3a^2}{T^2e^2(I-e^2)}$$ +\begin{align*} ++ \frac{24\pi^3a^2}{T^2e^2(I-e^2)} +\end{align*} \noindent (N.B. -- One complete revolution corresponds to the angle $2\pi$ in the absolute angular measure customary in physics, and the above @@ -3723,7 +3816,9 @@ a heavenly body would be deviated towards the latter. For a ray of light which passes the sun at a distance of $\Delta$ sun-radii from its centre, the angle of deflection (a) should amount to - $$a = \frac{1.7 \mbox{seconds of arc}}{\Delta}$$ +\begin{align*} +a = \frac{1.7 \mbox{seconds of arc}}{\Delta} +\end{align*} It may be added that, according to the theory, half of Figure 05 this deflection is produced by the Newtonian field of attraction of the @@ -3819,7 +3914,7 @@ the calculated deviations of the stars (in seconds of arc) are set forth in the following table of results: % Table 01: -$$ +\begin{align*} \begin{array}{r|rr|rr} \mbox{Number of the Star} & \mbox{First} & \mbox{Co-ordinate~~} & \mbox{Second} & \mbox{Co-ordinate~~} \\ \hline @@ -3832,7 +3927,7 @@ $$ 10 & -0'08 & +0'09 & +0'35 & +0'32 \\ 2 & +'095 & +0'85 & -0'27 & -0'09 \end{array} -$$ +\end{align*} \section{Displacement of Spectral Lines Towards the Red} @@ -3844,7 +3939,9 @@ positions of the clocks. We shall now examine this dependence quantitatively. A clock, which is situated at a distance $r$ from the centre of the disc, has a velocity relative to $K$ which is given by - $$V = \omega r$$ +\begin{align*} +V = \omega r +\end{align*} \noindent where $\omega$ represents the angular velocity of rotation of the disc $K^1$ with respect to $K$. If $v_0$, represents the number of ticks of the @@ -3853,15 +3950,21 @@ clock is at rest, then the ``rate'' of the clock ($v$) when it is moving relative to $K$ with a velocity $V$, but at rest with respect to the disc, will, in accordance with Section 12, be given by - $$v = v_2\sqrt{I-\frac{v^2}{c^2}}$$ +\begin{align*} +v = v_2\sqrt{I-\frac{v^2}{c^2}} +\end{align*} \noindent or with sufficient accuracy by - $$v = v_0 \left( I-\frac{1}{2} \frac{v^2}{c^2} \right)$$ +\begin{align*} +v = v_0 \left( I-\frac{1}{2} \frac{v^2}{c^2} \right) +\end{align*} \noindent This expression may also be stated in the following form: - $$v = v_0 \left( I-\frac{1}{c^2} \frac{\omega^2r^2}{2} \right)$$ +\begin{align*} +v = v_0 \left( I-\frac{1}{c^2} \frac{\omega^2r^2}{2} \right) +\end{align*} If we represent the difference of potential of the centrifugal force between the position of the clock and the centre of the disc by $\phi$, @@ -3870,11 +3973,15 @@ unit of mass against the centrifugal force in order to transport it from the position of the clock on the rotating disc to the centre of the disc, then we have - $$\phi = \frac{\omega^2r^2}{2}$$ +\begin{align*} +\phi = \frac{\omega^2r^2}{2} +\end{align*} \noindent From this it follows that - $$v = v_0 \left( I + \frac{\phi}{c^2} \right)$$ +\begin{align*} +v = v_0 \left( I + \frac{\phi}{c^2} \right) +\end{align*} In the first place, we see from this expression that two clocks of identical construction will go at different rates when situated at @@ -3902,7 +4009,9 @@ ought to take place for spectral lines produced at the surface of stars as compared with the spectral lines of the same element produced at the surface of the earth, the amount of this displacement being - $$\frac{v_0-v}{v_0} = \frac{K}{c^2} \frac{M}{r}$$ +\begin{align*} +\frac{v_0-v}{v_0} = \frac{K}{c^2} \frac{M}{r} +\end{align*} For the sun, the displacement towards the red predicted by theory amounts to about two millionths of the wave-length. A trustworthy -- 2.47.3