\usepackage{ucs}
\usepackage[utf8x]{inputenc}
+\usepackage{amsmath}
% Declare "longrightarrow" as "rightarrow" as the fomer never renders well.
\DeclareUnicodeCharacter{10230}{→}
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 $c$.
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
\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
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
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
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
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
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
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
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,
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
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
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
\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
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 $g_{22}$,
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
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.
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
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
\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
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}
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
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
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
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
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
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
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}
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
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.
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
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
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
forth in the following table of results:
% Table 01:
-$$
+\begin{align*}
\begin{array}{r|rr|rr}
\mbox{Number of the Star} & \mbox{First} & \mbox{Co-ordinate~~} & \mbox{Second} & \mbox{Co-ordinate~~} \\
\hline
10 & -0'08 & +0'09 & +0'35 & +0'32 \\
2 & +'095 & +0'85 & -0'27 & -0'09
\end{array}
-$$
+\end{align*}
\section{Displacement of Spectral Lines Towards the Red}
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
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 $\phi$,
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
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
projects / francis / relativity.git / commitdiff