Re: Two derivations (was Re: I'd like to understand this better.)
From: Eleaticus (no_reply_at_yahoo.co.uk)
Date: 07/14/04
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- In reply to: David McAnally: "Two derivations (was Re: I'd like to understand this better.)"
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Date: Wed, 14 Jul 2004 19:20:35 GMT
Nice one, McAnally. But equations ultimately can't conceal the falsity of
the set of concepts taken as a whole. The ideas behind relativity started
off with the presumption that the M-M result was null, but it wasn't. The
original literary sources tell us so.
Are you still abusing them, David?
"David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
news:cd2707$2pt$1@bunyip.cc.uq.edu.au...
> Below, I take the case of an inertially moving body and an accelerating
> body in Special Relativity, where the two bodies meet at two distinct
> events on their trajectories. I worked in a coordinate system in which
> the spatial coordinates of the accelerating body remain constant and the
> spatial coordinates of the inertially moving body are not linear functions
> of time, so that I was working relative to the accelerating body. The
> conclusion is that the proper time between the two events for the
> accelerating body is smaller than the proper time for the inertially
> moving body, in agreement with the conclusion which is derived from
> Special Relativity by more traditional means.
>
> I have also included an analysis of the case where a person shines a light
> straight up from a point on the equator, with the light being reflected
> back to earth by a mirror six light hours away. My analysis was performed
> in the frame which rotates with the earth. The conclusion of my analysis
> is that the light returns to earth at a point on its surface diametrically
> opposite the point on the surface from whence it was emitted.
>
> My analysis is presented in the following:
>
> If a body is accelerating, then its "hyperplanes of simultaneity" can no
> longer be used to partition spacetime. By "hyperplane of simultaneity",
> I mean the convex 3-manifold (a submanifold of spacetime) orthogonal to
> its worldline, or in simpler language, the three-dimensional hyperplane
> in spacetime which passes through the instantaneous event occupied by the
> body, and which is orthogonal to the instantaneous 4-velocity of the body.
> To put it another way, if we take an event on the worldline of the body,
> and if we take the rest frame of the body at the event, i.e. the inertial
> frame in which the body is stationary at the event, then the "hyperplane
> of simultaneity" for the body at that event is the hyperplane of
> simultaneity in that inertial frame (i.e. the hyperplane of events
> simultaneous with the given event relative to the inertial frame).
>
> If a body is moving inertially, then the "hyperplanes of simultaneity"
> do not intersect, and their union is all of spacetime (i.e. they form
> a partition of spacetime), and so they can be used to coordinatize
> spacetime. But if a body is accelerating, then the "hyperplanes of
> simultaneity" at different events on its worldline do intersect (with
> the additional phenomenon that far enough away in the direction opposite
> to the acceleration, the "hyperplane of simultaneity" is locally
> travelling backwards in time). Further, there are cases such that the
> union of the "hyperplanes of simultaneity" is not all of spacetime (the
> case of uniform acceleration is such an example). Specifically, if a
> body is instantaneously accelerating with acceleration g relative to its
> rest frame, then the "hyperplane of simultaneity" does not move forward
> in time at a distance of c^2/g behind the body in the direction opposite
> to the direction of the acceleration, within its instantaneous rest frame.
> More generally, for events in the "hyperplane of simultaneity",
>
> (1) For events such that the direction from the body to the
> event, relative to the rest frame, makes an acute angle
> with the acceleration, the "hyperplane of simultaneity"
> moves very quickly forward in time.
>
> (2) For events such that the direction from the body to the
> event, relative to the rest frame, makes an obtuse angle
> with the acceleration, and the scalar product of the
> acceleration and displacement vector is greater than -c^2,
> the "hyperplane of simultaneity" moves slowly forward in time.
>
> (3) For events such that the direction from the body to the
> event, relative to the rest frame, makes an obtuse angle
> with the acceleration, and the scalar product of the
> acceleration and displacement vector is equal to -c^2, the
> "hyperplane of simultaneity" does not move in time at all
> (i.e. infinitesimally distant "hyperplanes of simultaneity"
> intersect at those events).
>
> (4) For events such that the direction from the body to the
> event, relative to the rest frame, makes an obtuse angle
> with the acceleration, and the scalar product of the
> acceleration and displacement vector is less than -c^2,
> the "hyperplane of simultaneity" moves *backwards* in time.
>
> Clarifications of the above can be made available on request. The
> important point here is that "simultaneous" loses its significance for
> an accelerating body. The "hyperplanes of simultaneity" cannot be used
> to coordinatize spacetime. Consequently, other means must be employed
> to coordinatize spacetime.
>
> Let an inertial frame have coordinates (t',x',y',z'), and suppose that
> relative to the frame, the body, A, follows a trajectory x' = X(t'),
> y' = Y(t'), z' = Z(t'), where X, Y and Z are C^2 functions of t'.
> Another set of coordinates for spacetime is given by (t,x,y,z), where
> t = t', x = x'-X, y = y'-Y, z = z'-Z. In the new coordinate system, the
> coordinates of A satisfy x(t) = 0, y(t) = 0, z(t) = 0. Calculations now
> show that the invariant line element is determined by
>
> ds^2 = [c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt^2 - 2 (dX/dt) dx dt
>
> - 2 (dY/dt) dy dt - 2 (dZ/dt) dz dt - dx^2 - dy^2 - dz^2,
>
> and so the metric is given by
>
> g_tt = c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2, g_tx = g_xt = - dX/dt,
>
> g_ty = g_yt = - dY/dt, g_tz = g_zt = - dZ/dt, g_xx = g_yy = g_zz = -1,
>
> with all other components equal to zero. The inverse matrix to (g_ij)
> is given by
>
> g^tt = 1/c^2, g^tx = g^xt = - (dX/dt)/c^2, g^ty = g^yt = - (dY/dt)/c^2,
>
> g^tz = g^zt = - (dZ/dt)/c^2, g^xx = - 1 + (dX/dt)^2/c^2,
>
> g^xy = g^yx = (dX/dt) (dY/dt)/c^2, g^xz = g^zx = (dX/dt) (dZ/dt)/c^2,
>
> g^yy = - 1 + (dY/dt)^2/c^2, g^yz = g^zy = (dY/dt) (dZ/dt)/c^2,
>
> g^zz = - 1 + (dZ/dt)^2/c^2.
>
> This can be verified directly by matrix multiplication.
>
> The components of the gradient of the metric are given by
>
> g_tt,t = - 2 (dX/dt) (d^2X/dt^2) - 2 (dY/dt) (d^2Y/dt^2)
>
> - 2 (dZ/dt) (d^2Z/dt^2),
>
> g_xt,t = g_tx,t = - d^2X/dt^2, g_yt,t = g_ty,t = - d^2Y/dt^2,
>
> g_zt,t = g_tz,t = - d^2Z/dt^2,
>
> with all other derivatives equal to zero.
>
> The Christoffel symbols are given by
>
> Gamma_ttt = - (dX/dt) (d^2X/dt^2) - (dY/dt) (d^2Y/dt^2)
>
> - (dZ/dt) (d^2Z/dt^2),
>
> Gamma_xtt = - d^2X/dt^2, Gamma_ytt = - d^2Y/dt^2,
>
> Gamma_ztt = - d^2Z/dt^2,
>
> with all other symbols equal to zero, and so
>
> Gamma^x_tt = d^2X/dt^2, Gamma^y_tt = d^2Y/dt^2, Gamma^z_tt = d^2Z/dt^2,
>
> with all other quantities of the form Gamma^i_jk being equal to zero.
>
> The equations for a geodesic are therefore given by
>
> d^2t/du^2 = 0, d^2x/du^2 = - (d^2X/dt^2) (dt/du)^2,
>
> d^2y/du^2 = - (d^2Y/dt^2) (dt/du)^2,
>
> d^2z/du^2 = - (d^2Z/dt^2) (dt/du)^2,
>
> where u parametrizes the events on the geodesic.
>
> Because d^2t/du^2 = 0, then t = au+b for constants a and b, and so the
> equations are
>
> d^2x/dt^2 = - d^2X/dt^2, d^2y/dt^2 = - d^2Y/dt^2,
>
> d^2z/dt^2 = - d^2Z/dt^2.
>
> The solution of these equations is given by
>
> x = - X + a_x t + b_x, y = - Y + a_y t + b_y, z = - Z + a_z t + b_z,
>
> for constants a_x, b_x, a_y, b_y, a_z and b_z. It follows that the
> trajectory of an inertially moving body satisfies
>
> x = - X + a_x t + b_x, y = - Y + a_y t + b_y, z = - Z + a_z t + b_z,
>
> for some constants a_x, b_x, a_y, b_y, a_z and b_z.
>
> Substituting these formulae into the expression for the line element, it
> follows that (ds/dt)^2 = c^2 - a_x^2 - a_y^2 - a_z^2 for the inertially
> moving body.
>
> Suppose a second body, B, moves inertially, and that it meets A at times
> t_1 and t_2 (where t_1 < t_2). It follows that at t = t_1 and t = t_2,
> the coordinates of B are x = 0, y = 0 and z = 0. Solving for the
> arbitrary constants, the trajectory of B is given by
>
> x = - X + t [X(t_2)-X(t_1)]/[t_2-t_1] + [t_2X(t_1)-t_1X(t_2)]/[t_2-t_1],
>
> y = - Y + t [Y(t_2)-Y(t_1)]/[t_2-t_1] + [t_2Y(t_1)-t_1Y(t_2)]/[t_2-t_1],
>
> z = - Z + t [Z(t_2)-Z(t_1)]/[t_2-t_1] + [t_2Z(t_1)-t_1Z(t_2)]/[t_2-t_1].
>
> Substitutions will verify that x = 0, y = 0 and z = 0 at t = t_1, and
> also at t = t_2. During the intervening interval, the proper time for
> A is
>
> T_A = 1/c int_{t_1}^{t_2} sqrt[c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2]
>
> dt,
>
> and the proper time for the inertially moving body B is
>
> T_B = 1/c sqrt[c^2 (t_2-t_1)^2 - (X(t_2)-X(t_1))^2 - (Y(t_2)-Y(t_1))^2
>
> - (Z(t_2)-Z(t_1))^2].
>
> By the integral form of Schwartz's inequality,
>
> (X(t_2) - X(t_1))^2
>
> = [int_{t_1}^{t_2} (dX/dt) dt]^2
>
> = [int_{t_1}^{t_2} 1 (dX/dt) dt]^2
>
> <= [int_{t_1}^{t_2} 1^2 dt] [int_{t_1}^{t_2} (dX/dt)^2 dt]
>
> = (t_2 - t_1) int_{t_1}^{t_2} (dX/dt)^2 dt,
>
> where 'f <= g' denotes that f is less than or equal to g. Alternatively,
> use the fact that for all real h, int_{t_1}^{t_2} [(dX/dt) - h]^2 dt >= 0
> (where 'f >= g' denotes that f is greater than or equal to g). It follows
> that
>
> int_{t_1}^{t_2} (dX/dt)^2 dt >= (X(t_2)-X(t_1))^2/(t_2-t_1).
>
> Similarly,
>
> int_{t_1}^{t_2} (dY/dt)^2 dt >= (Y(t_2)-Y(t_1))^2/(t_2-t_1),
>
> int_{t_1}^{t_2} (dZ/dt)^2 dt >= (Z(t_2)-Z(t_1))^2/(t_2-t_1).
>
> So
>
> (t_2 - t_1) int_{t_1}^{t^2} [c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt
>
> <= [c^2 (t_2-t_1)^2 - (X(t_2)-X(t_1))^2 - (Y(t_2)-Y(t_1))^2
>
> - (Z(t_2)-Z(t_1))^2].
>
> By the integral form of Schwartz's inequality,
>
> {int_{t_1}^{t_2} sqrt[c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt}^2
>
> = {int_{t_1}^{t_2} 1 sqrt[c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt}^2
>
> <= {int_{t_1}^{t_2} dt} {int_{t_1}^{t_2} [c^2 - (dX/dt)^2 - (dY/dt)^2
>
> - (dZ/dt)^2] dt}
>
> = (t_2-t_1) int_{t_1}^{t_2} [c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt,
>
> so that
>
> {int_{t_1}^{t_2} sqrt[c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt}^2
>
> <= [c^2 (t_2-t_1)^2 - (X(t_2)-X(t_1))^2 - (Y(t_2)-Y(t_1))^2
>
> - (Z(t_2)-Z(t_1))^2].
>
> It follows that
>
> int_{t_1}^{t_2} sqrt[c^2 - (dX/dt)^2 - (dY/dt)^2 - (dZ/dt)^2] dt
>
> <= sqrt[c^2 (t_2-t_1)^2 - (X(t_2)-X(t_1))^2 - (Y(t_2)-Y(t_1))^2
>
> - (Z(t_2)-Z(t_1))^2],
>
> and so T_A <= T_B, i.e. no more time elapses for A than elapses for the
> inertially moving body B. Furthermore, since the proof was by two
> applications of Schwartz's inequality, then equality of T_A and T_B
> occurs iff dX/dt, dY/dt and dZ/dt are constant for t_1 <= t <= t_2, and
> so T_A = T_B iff A is inertially moving for t_1 <= t <= t_2, which holds
> iff A and B have coincident paths from t = t_1 to t = t_2.
>
> So, without invoking the Principle of Equivalence, and using only Special
> Relativity, we have a demonstration that if B is an inertially moving
> body, and another body A coincides with B at two events in its trajectory,
> then the proper time for A between the two events is strictly less than
> the proper time for B between the events, unless A is inertially moving
> throughout the period between the two events (and therefore coincident
> with B, throughout). Furthermore, this result was demonstrated in a
> coordinate system which treated things from the point of view of A (by
> having x = 0, y = 0, z=0, throughout the entire trajectory of A).
>
> =============================================
>
> Now, we come to the case of a rotating frame of reference. Imagine a set
> of x, y and z axes, which is rotating at a constant angular velocity
> about the z axis. Suppose that at time t = 0, the axes coincide with a
> non-rotating set with coordinates, x', y' and z. So
>
> x' = x cos(wt) - y sin(wt),
>
> y' = x sin(wt) + y cos(wt),
>
> where w is the angular velocity. Calculation demonstrates that the line
> element is determined by
>
> ds^2 = (c^2 - w^2 x^2 - w^2 y^2) dt^2 + 2wy dx dt - 2wx dy dt - dx^2
>
> - dy^2 - dz^2,
>
> so that the metric is given by
>
> g_tt = c^2 - w^2 x^2 - w^2 y^2, g_tx = g_xt = wy, g_ty = g_yt = -wx,
>
> g_xx = g_yy = g_zz = -1,
>
> and all other components are zero.
>
> The inverse of the matrix (g_ij) has elements
>
> g^tt = 1/c^2, g^tx = g^xt = wy/c^2, g^ty = g^yt = -wx/c^2,
>
> g^xx = - 1 + w^2 y^2/c^2, g^xy = g^yx = - w^2 x y/c^2,
>
> g^yy = - 1 + w^2 x^2/c^2, g^zz = -1,
>
> and all other components are zero. This can be verified directly by
> matrix multiplication.
>
> The derivatives of the components of the metric are given by
>
> g_tt,x = - 2 w^2 x, g_tt,y = - 2 w^2 y, g_tx,y = g_xt,y = w,
>
> g_ty,x = g_yt,x = - w,
>
> and all other derivatives are zero.
>
> It follows that
>
> Gamma_ttx = Gamma_txt = - w^2 x, Gamma_xtt = w^2 x,
>
> Gamma_tty = Gamma_tyt = - w^2 y, Gamma_ytt = w^2 y,
>
> Gamma_xty = Gamma_xyt = w, Gamma_ytx = Gamma_yxt = - w,
>
> and all other symbols are zero, and so
>
> Gamma^y_tx = Gamma^y_xt = w, Gamma^x_ty = Gamma^x_yt = - w,
>
> Gamma^x_tt = - w^2 x, Gamma^y_tt = - w^2 y,
>
> and all other symbols are zero.
>
> Therefore the equation for a geodesic is given by
>
> d^2t/du^2 = 0, d^2x/du^2 = 2 w (dy/du) (dt/du) + w^2 x (dt/du)^2,
>
> d^2y/du^2 = - 2 w (dx/du) (dt/du) + w^2 y (dt/du)^2, d^2z/du^2 = 0,
>
> where u parametrizes the events on the geodesic.
>
> Because d^2t/du^2 = 0, then t = au+b for constants a and b, and so the
> equations are
>
> d^2x/dt^2 = 2 w (dy/dt) + w^2 x, d^2y/dt^2 = - 2 w (dx/dt) + w^2 y,
>
> d^2z/dt^2 = 0.
>
> The solution for the equation for z is easy: z = a_z t + b_z, for some
> constants a_z and b_z. The equations for x and y are simultaneous
> differential equations, which will be rewritten as
>
> d^2x/dt^2 - 2 w (dy/dt) - w^2 x = 0, d^2y/dt^2 + 2 w (dx/dt) - w^2 y = 0.
>
> It follows that
>
> d^4x/dt^4 - 2 w (d^3y/dt^3) - w^2 (d^2x/dt^2) = 0,
>
> d^3y/dt^3 + 2 w (d^2x/dt^2) - w^2 (dy/dt) = 0,
>
> and so
>
> d^4x/dt^4 + 2 w^2 (d^2x/dt^2) + w^4 x = 0.
>
> This is a fourth-order linear differential equation with constant
> coefficients. The characteristic equation is v^4 + 2 w^2 v^2 + w^4 = 0,
> which has solutions v = iw twice and v = -iw twice. The solution to the
> differential equation is therefore given by
>
> x = (At + B) cos(wt) + (Dt + E) sin(wt),
>
> for some constants A, B, D and E. The simultaneous solution for y is
>
> y = - (At + B) sin(wt) + (Dt + E) cos(wt).
>
> So inertial motion is determined by
>
> x = (At + B) cos(wt) + (Dt + E) sin(wt),
>
> y = - (At + B) sin(wt) + (Dt + E) cos(wt),
>
> z = a_z t + b_z.
>
> Substituting these formulae into the expression for the line element, it
> follows that (ds/dt)^2 = c^2 - A^2 - D^2 - a_z^2 (after much calculation
> if you try to attack it directly, or a little calculation if you know the
> trick). This means that an inertially moving material body satisfies
> A^2+D^2+a_z^2 < c^2, and a light ray satisfies A^2+D^2+a_z^2 = c^2.
>
> Now, place the origin at the centre of the earth, the x axis towards zero
> degrees longitude on the equator, the y axis towards 90 degrees East on
> the equator, and the z axis towards the North Pole, and w = pi/12 radians
> per hour (or, more accurately, 2 pi radians every 23 hours and 56
> minutes). Suppose that at t = 0, a person at zero degrees longitude on
> the equator (in reality, this is a spot in the Gulf of Guinea, off the
> coast of Africa, but let us take an ideal case where there is land, or a
> ship, available) shines a light in a direction directly from the centre of
> the earth. Let the radius of the earth be R, then the trajectory of the
> light is given by
>
> x = A (t + R/c) cos(wt) + D (t + R/c) sin(wt),
>
> y = - A (t + R/c) sin(wt) + D (t + R/c) cos(wt),
>
> z = 0,
>
> as the light's path goes through the centre of the earth. At t = 0,
> y = 0, so that D = 0. Since c^2 = A^2 + D^2 (as the 4-velocity is
> necessarily null), then A = c for the outward bound journey. So the
> trajectory is given by
>
> x = c (t + R/c) cos(wt),
>
> y = - c (t + R/c) sin(wt).
>
> The light travels for one quarter of a sidereal day, at which point, it
> is reflected by a stationary mirror, and returns to earth. The mirror
> starts at x = R + pi c/(2w), y = 0, and z = 0 (i.e. directly overhead).
> Since it is presumably stationary in the frame of the earth, its motion
> in the rotating frame is given by
>
> x = (R + pi c/(2w)) cos(wt),
>
> y = - (R + pi c/(2w)) sin(wt).
>
> At t = pi/(2w), the light ray is at
>
> x = 0,
>
> y = - (R + pi c/(2w)),
>
> and the mirror is at
>
> x = 0,
>
> y = - (R + pi c/(2w)).
>
> It follows that the light ray meets the mirror and is reflected. At the
> time of the meeting, dy/dt = 0 for the mirror, dx/dt = - wR - pi c/2
> for the light ray, and dy/dt = - c. After reflection,
> dx/dt = - wR - pi c/2, and dy/dt = c, for the light ray. The equation
> for the reflected light ray is
>
> x = (A t + B) cos(wt) + (D t + E) sin(wt),
>
> y = - (A t + B) sin(wt) + (D t + E) cos(wt).
>
> At t = pi/(2w), the formulae above yield
>
> pi D/(2w) + E = 0,
>
> - pi A/(2w) - B = - R - pi c/(2w),
>
> - pi A/2 - wB + D = - wR - pi c/2,
>
> - A - pi D/2 - wE = c,
>
> so that D = 0 from the second and third equations, and A = -c from the
> first and fourth equations. From the first equation, it follows that
> E = 0, and from the second equation, it follows that B = R + pi c/w.
>
> So the trajectory of the light ray is
>
> x = (- ct + R + pi c/w) cos(wt),
>
> y = (ct - R - pi c/w) sin(wt).
>
> When t = pi/w, then x = -R and y = 0, i.e. at 180 degrees longitude on
> the equator, exactly opposite from the position on the earth's surface
> from which the light was sent.
>
> Since the physical system is cylindrically symmetric, then this is true
> for any point on the equator, i.e. if a light is shone vertically from
> the surface of the earth, and overhead there is a stationary mirror
> oriented to reflect the light back to earth, at a distance equal to that
> travelled by light in a quarter of a sidereal day, then the light will
> return to earth at the spot on the earth's surface exactly opposite that
> from which it was sent.
>
> David
>
> And all dared to brave unknown terrors, to do mighty deeds,
> to boldly split infinitives that no man had split before -
> and thus was the Empire forged.
>
> -----
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