Wolfram Alpha:

```Hyperbolic Motion and Rindler Coordinates
-----------------------------------------

Accelerated Coordinate Systems
------------------------------

In flat space Newtonian mechanics, the motion of a uniformly
accelerated object is described by a parabola and follows the
familiar SUVAT equation, s = (1/2)at2.

Things are more complicated in Special Relativity.  Since
v = at, as t increases, v would eventually be greater than
the velocity of light, which is not allowed.  Instead, we
must think of the motion of an object in terms of hyperbolas
through flat space.  The hyperbola is asymptotic to the light
cone and, therefore, the velocity of light cannot be exceeded.
(A light cone is the path that a flash of light, emanating
from a single event (localized to a single point in space and
a single moment in time) and traveling in all directions,
would take through spacetime).

The equation of the hyperbola is:

X2 - c2T2 = R2

Where X, T are inertial coordinates in Minkowski space and R
is the distance along the horizontal axis from the event
horizon when T = 0.  The PROPER ACCELERATION along a particular
curve, R, is:

α = c2/R ∴ R = c2/α

Dimensional check:

[α] = [(L2/T2).(1/L)] = [L/T2] ✓

[R] = [(L2/T2)/(L/T2)] = [L] ✓

The proper acceleration along each curve is constant.

It is clear from this equation that as R gets smaller, α
increases.

Thus,

X2 - c2T2 = c4/α2

When T = 0, X = R = c2/α.

The hyperbola is drawn in Minkowski spacetime and represents
a worldline.  It is Lorentz invariant because X2 - c2T2 is
an invariant quantity.  Therefore, such a transformation will
just move the observer to a different point on the worldline
where they will experience exactly the same acceleration as
before.  By analogy with the polar coordinate conversions for
the circle, x = rcosθ and y = rsinθ, it is possible to write
the following transformations in hyperbolic space.

X = Rcosh((α/c)ω) and cT = Rsinh((α/c)ω)

Where (α/c)ω is the rapidity, ξ.

Dimension check: [(α/c)] = ((L/T2)/(L/T)) = 1/T

[ω] = T

[(α/c)ω] = ((L/T2)/(L/T).T)

= dimensionless. ✓

The corresponding inverse transformations are:

R = √(X2 - c2T2)

ω = (c/α)arctanh(cT/X)

R and ω are the RINDLER COORDINATES of the accelerating
object.  They are the coordinates for an observer moving with
proper acceleration, α.  It is important to note that
ω are lines of constant time and are NOT the proper time
measured along the curve of constant R.

Unlike the Newtonian case, a uniformly accelerated motion in
a Minkowski spacetime cannot cover the entire spacetime.  It
is restricted to the wedge of spacetime shown, bounded by the
light cone.  This is referred to as the RINDLER WEDGE or
RINDLER SPACE.

The Rindler Metric
------------------

Since sinh2(αω/c) - cosh2(αω/c) = -1  where -∞ < ω < +∞

R2cosh2(αω/c) - R2sinh2(αω/c) = R2    (X2 - cT2 = R2)

Taking the differentials we get:

dT = dRsinh(αω/c) + (Rαdω/c)cosh(αω/c)

∴ dT2 = dR2sinh2(αω/c) + (Rαdω/c)2cosh2(αω/c)

+ 2(Rαdω/c)sinh(αω/c)cosh(αω/c)

and,

dX = dRcosh(αω/c) + (Rαdω/c)sinh(αω/c)

∴ dX2 = dR2cosh2(αω/c) + (Rαdω/c)2sinh2(αω/c)

+ 2(Rαdω/c)sinh(αω/c)cosh(αω/c)

The metric is:

dτ2 = c2dT2 - dX2

= (Rα/c)2dω2 - dR2

-          -  -   -
dτ2 = | (Rα/c)2  0 || dω2 |
|    0    -1 || dR2 |
-          -  -   -

Dimension check:  [(Rα/c).dω] = L.(L/T2)/(L/T).T

= L ✓

This the basic version of the RINDLER METRIC in flat
spacetime that is analogous to the metric for the circle
in polar coordinates, ds2 = dr2 + r2dθ2.

The invariant spacetime interval is:

dτ2 = d(ct)2 - dx2

= R2sinh2(αω/c) - R2cosh2(αω/c)

= -R2

Time Dilation
-------------

The proper time along a worldline in inertial coordinates
can be found as follows:

The first step is to parameterize the hyperbola using λ.

cT(λ) = Rsinhλ   X(λ) = Rcoshλ

The tangent vector to the worldline (hyperbola) is:

ds/dλ = (d(cT)/dλ)(∂s/∂(cT)) + (dX/dλ)(∂s/∂X)

= (Rcoshλ)eT + (Rsinhλ)eX

∴ (ds/dλ)2 = [(Rcoshλ)eT + (Rsinhλ)eX][(Rcoshλ)eT + (Rsinhλ)eX]

From the Minkowski metric:

-    -
| 1  0 |
| 0 -1 |
-    -

eT.eT = 1; eX.eT = 0; eX.eX = -1

(ds/dλ)2 = R2

cτ = ∫|ds/dλ|dλ

= ∫√[(ds/dλ)(ds/dλ)]dλ

= ∫Rdλ

= R[λfinal - λinitial]

The proper time along a worldline in Rindler coordinates:

R = √(X2 - c2T2)

ω = (c/α)arctanh(cT/X)

= (c/α)arctanh(Rsinhλ/Rcoshλ)

= (c/α)arctanh(tanhλ)

= (c/α)λ

ds/dλ = (dω/dλ)(∂s/∂ω) + (dR/dλ)(∂s/∂R)

= (c/α)eω + 0eR

(ds/dλ)2 = (c/α)2eω.eω

From the Minkowski metric:

-          -
| (Rα/c)2  0 |
|    0    -1 |
-          -

eω.eω = (Rα/c)2; eR.eω = 0; eR.eR = -1

(ds/dλ)2 = (c/α)2(Rα/c)2

= R2

cτ = ∫√[(ds/dλ)(ds/dλ)]dλ

= R∫dλ

= R[λfinal - λinitial]

Therefore, the conclusion is that the proper time, τ, for
a given Rindler observer is the same in both coordinate
systems.

Let λ = ω

cτ = ∫|ds/dω|dω

= ∫√[(ds/dω)(ds/dω)]dω

= ∫√[eω.eω]dω

= ∫√[(Rα/c)2]dω

= (Rα/c)∫dω

= (Rα/c)[ωF - ωI)

If ωI = 0 then:

cτ = (α/c)Rω

Or,

τ = (α/c2)Rω

If all Rindler observers set their clocks to 0 at T = 0
(i.e. T = ω = τ = 0) then there is a choice of which Rindler
observer's proper time, τ, will be equal to the Rindler time,
ω.  It is conventional to define the Rindler coordinate
system such that the Rindler observer whose proper time, τ,
matches the Rindler time, ω.  This corresponds to the curve
where (α/c2)R = 1.  For other Rindler observers at different
distances from the Rindler horizon, the proper time will
equal some constant multiple of the proper time at the
Rindler coordinate (1,1).

Ex:  Let c = 1.  R = 1/α ∴ α = 1 => R = 1

α    ω  (R,ω)     τ = αRω
---   -  -----     -------
1.0   0  (1,0)       0.0
1  (1,1)       1.0
2  (1,2)       2.0

Therefore, from the perspective of an observer at B, A's
clock would appear to be running at 1/2 the rate of his
own.  Likewise, from A's perspective, B's clock would
appear to be running at twice the rate of his own.  Now
consider,

The distance from the origin to where the 1st hyperbola
intersects the X axis (T = 0) is D.  The proper acceleration
along this hyperbola is by convention set to c2/α = D = 1.

D2 = c2T2 - X2

= [(c2/α)2sinh2((α/c)τ)]2 - [(c2/α/c)cosh2((α/c)τ)]2

= c4/α2

∴  D = c2/α

1st hyperbola:  D = c2/α

2nd hyperbola:  2D = 2c2/α  (≡ α/2)

3rd hyperbola:  3D = 3c2/α  (≡ α/3)

We can replace the initial distance, D, with general distance,
R, which is a constant multiple of D and τ with the general
coordinate, ω.  Therefore,

cT = (Nc2/α)sinh((α/c)τ) => Rsinh((α/c)ω)

X = (Nc2/α)cosh((α/)τ) => Rcosh((α/)ω)

From the point of a stationary observer in the inertial
reference frame at X = 0, we get:

cT = (c2/α)sinh(ατ/c)

and,

cτ = (c2/α)arcsinh(αT/c)

∴ τ= (c/α)arcsinh(αT/c)

Note:  arcsinh(x) = ln(x + √(x2 + 1))

For hyperbola 1, let α = 0.1c => c/α = 10 and α/c = 0.1.

τ = (10)arcsinh(0.1T)

For hyperbola 2, α = 0.05c => c/α = 20 and α/c = 0.05.

τ = (20)arcsinh(0.05T)

This gives:

α = 0.1c    α = 0.05c
T       τ1           τ2
-  -----------  -----------
1  0.998340789  0.999583801
2  1.986901103  1.996681578

Notes:

- τ2 at 2s. is exactly twice τ1 at 1s.

- The stationary observer at the event horizon would see
a clock in the accelerated frame moving progressively
faster and faster as the object accelerated towards
the horizon and progressively slower and slower as the
object moves away from the event horizon.  Likewise,
a stationary observer well away from event horizon
will see the opposite effect.

- It should be emphasized that all observers see time
running normally in their own frames. There is nothing
wrong with their own clocks.  Time itself is slowing
down or speeding up because of the because of the
effects of acceleration.

Length Contraction
------------------

Consider 2 stationary spaceships one at A and the other at B.
The ships are attached to each other by a tight string of
length, D.  If both ships start accelerating at the same
time with the same constant rate they will have identical
instantaneous velocities.  According to SR, a stationary
observer would see the length of the string connecting the
ships themselves contracting in length causing the string
to break.  Now consider the motion of the ships in the
uniformly accelerated reference frame.  If the 2 ships
follow their respective hyperbolae then the distance as
measured from point of view of each spaceship will always
remain the same.  This can be seen as follows:

A = (X1,cT1)

B = (X2,cT2)

Distance AB = √[(X2 - X1)2 + (cT2 - cT1)2]  (Pythagoras)

T = 0 => Distance AB = √[(X2 - X1)2] = Distance 23

In Rindler coordinates:

X1 = R1coshξ1 cT = R1sinhξ1

X2 = R2coshξ2 cT = R2sinhξ2

ξ1 = ξ2 = 0

X1 = R1 cT = 0

X2 = R2 cT = 0

(X2 - X1)2 = (R2 - R1)2

Or,

(X2 - X1) = (R2 - R1)

Therefore, the distances 23 = 34 = AB = BC = EF = FG remain
the same no matter how long the acceleration proceeds.
Effectively, this means that the trailing ship needs to
accelerate just a little bit harder to maintain the correct
distance between the ships.

Now from the point of view of a stationary observer traveling
along the T axis (X = 0) it appears that the the back ship is
catching up with the front ship and the distance between the
two is diminishing.  Again we can see this from the following:

cT1 = cT2 = cT.

X12 - cT2 = R12

X22 - cT2 = R22

∴ X22 - R22 = X12 - R12

∴ X22 - X12 = R22 - R12

∴ (X2 - X1)(X2 + X1) = R22 - R12

But R22 - R12 is fixed = K.

∴ (X2 - X1) = K/(X2 + X1)

At very large T, (X2 + X1) will be very large and (X2 - X1)
will be very small.

So the conclusion is that the string will break if both
spaceships accelerate with the same proper acceleration
but if the ship in the back accelerates just a little
harder than the one in front so that distance, D, remains
constant, the string will not break.  This is BELL'S

The Equivalence Principle
-------------------------

Take a particular curve, R, and displace it by a tiny
amount ΔR.  We can write:

dτ2 = (α/c)2(R + ΔR2)2dω2 - d(R + ΔR)2

= (α/c)2(R2 + 2RΔR + ΔR2)dω2 - d(R + ΔR)2

= (α/c)2(1 + 2ΔR/R + ΔR2/R2)R2dω2 - dΔR2  (dR = 0)

= (Rα/c)2(1 + 2ΔR/R)dω2 - dΔR2 since R >> ΔR

= (1 + 2ΔR/R)c2dT2 - dΔX2

Now, α = c2/R.  Therefore,

dτ2 = (1 + 2ΔRg)dT2 - dΔX2

= (1 + 2φ)dT2 - dΔX2

Where φ = ΔRg is defined as the GRAVITATIONAL POTENTIAL
ENERGY per unit mass (c.f. mgh with m = 1 and h ≡ ΔR).

Equations of motion:

d2xμ/dτ2 = -Γρσμ(dxρ/dτ)(dxσ/dτ)

For a small spatial velocity this becomes:

d2x1/dτ2 = -Γ001(dx0/dτ)(dx0/dτ)

= -Γ001(1)(1)

= (1/2)∂g00/∂x1

= -g   [-g00 = (1 + 2ΔRg)]

Rindler coordinates show the correspondence between
uniformly accelerated reference frames and uniform
gravitational fields.  In other words they show the
equivalence of an accelerated observer with no gravity,
to an observer in a local gravitational field.  This
is the original statement of the Equivalence Principle.
They also demonstrate that an event horizon is not
unique to General Relativity, and gravity is not a
requirement.  An observer that is accelerating at a
constant rate will see an apparent horizon.

The Schwarzchild Metric
-----------------------

The analysis to date is for flat spacetime in the absence
of matter.  Uniformally accelerated reference frames
correspond to homogeneous gravitational fields and do not
produce tidal forces.  Tidal forces are a consequence of
curved spacetime produced by the presence of gravitating
masses.  Therefore, to see the effects of curved spacetime
it is necessary to use a metric that contains a gravitating
mass term.  Consider φ = -GM/r where M is the mass and
r is the distance from the center of M.  The above metric
becomes:

dτ2 = (1 - 2MG/rc2)dt2 - (1/c2){dx2 + .... }

In spherical coordinates this becomes:

dτ2 = (1 - 2MG/rc2)dt2 - (1/c2){dr2 + r2Ω2}

[i.e. dx2 + dy2 + dz2 = dr2 + r2(dθ2 + sin2θdφ2)]

The spatial term is split into a radial component, dr2,
and an angular term, r2(dθ2 + sin2θdφ2).

The problem with this metric is that at some point the
(1 - 2MG/rc2) term changes its sign and the time coordinate
becomes spacelike.  Under these circumstances the metric
will have 4 negative terms which violates spacetime.  It
turns out from Einstein's equations that the correct form
of the metric is:

dτ2 = (1 - 2MG/rc2)dt2 - (1/(1 - 2MG/rc2))dr2 - (1/c2)r2Ω22

Sometimes this is written as:

dτ2 = (1 - rS/r)dt2 - (1/(1 - rS/r))dr2 - (1/c2)r2Ω22

Where rS = 2MG/c2 is called the SCHWARZCHILD RADIUS.

Dimension check:

[2GM/rc2] = [kg/(m2/s2).(m3/kg.s2).(1/m)] = [1/1]

Now when a sign change occurs the time and spatial terms
flip to maintain the signature of the metric.

This is the famous Schwarzchild metric that describes
the spacetime around a spherically symmetric gravitating
object such as a black hole (or earth for that matter).

Consider a light ray traveling radially towards the mass.
Light is a massless particle that travels along the edges
of the light cone referred to as the NULL GEODESIC.  It
corresponds to the light like case, dτ2 = 0.  The above
equation becomes:

(1 - 2MG/rc2)dt2 = (1/(1 - 2MG/rc2))dr2

Or,

dr/dt = ±(1 - 2MG/rc2) =  ±(1 - rs/r)

Thus, as the light ray approaches the event horizon it
appears to a distant observer to become progressively
slower and slower compared to their coordinate time, t,
eventually becoming 'frozen' at the Scwartzchild radius.

However, a person running along with the light wave would
not see this slowing and would measure the velocity of
the ray to be that of the speed of light using their
proper time, τ.

The apparent slowing of the light ray is a consequence of
gravitational time dilation.  Mathematically.

dτ2 = (1 - 2MG/rc2)dt2 - (1/(1 - 2MG/rc2))dr2 - (1/c2)r2Ω22

Focus on 1st term on RHS.

dτ2 = (1 - 2MG/rc2)dt2

∴ dτ = √(1 - 2MG/rc2)dt

∴ dt = dτ/√(1 - 2MG/rc2)

This has a similar form to the Lorentz Transform from SR:

dt = dτ/√(1 - v2/c2)

By comparison v = √(2MG/r).

Time dilation causes light to oscillate at a lower frequency
so the light becomes more and more shifted towards a longer
wavelength as it approaches the event horizon.

λ↑ = c/f↓ = ct(↑).

At the horizon, the frequency becomes so low, its wavelength
so large, that it appears to get stuck and never pass through
the surface.

Interestingly enough, an observer inside the event horizon
can easily see outside.  The event horizon prevents light
from traveling to someone outside, but it does not prevent
light from traveling from outside to the observer. It also
does not prevent two infalling observers from exchanging
light signals on their way to doom.

Gravitational Redshift
----------------------

Consider a photon in an elevator undergoing a uniform
constant acceleration:

-------
|       |
| .x=Δx |  K' = hf', U' = mgΔx = gΔxhf/c2
|       |
|       |
|       |
| .x=0  |  K = hf, U = 0
|       |
-------
|
v
α

Conservation of energy requires:

K + U = K' + U'

∴ hf + 0 = hf' + gΔxhf/c2

∴ f + 0 = f' + gΔxf/c2

∴ f' = f(1 - gΔx/c2)

Let f' = fO  (O = observer) and f = fS (S = source).
The Doppler equation is:

fO = [(c - vO)/(c + vS)]fS

= [(c - vO)/c]fS   (vS = 0)

= [(1 - vO/c)]fS

By comparison:

fO = (1 - gΔx/c2)fS   (m = 1)

Dimension check:  [gΔx/c2)] = (L/T2).L/(L2/T2)

[vO/c] = (L/T)/(L/T)

fO = (1 - mgΔx/c2)fS

= (1 - φ/c2)fS

Using the Schwarzchild metric gives:

dt = dτ/√(1 - v2/c2)

dt/dτ = tO/tS = λO/λS = fS/fO = 1/√(1 - 2MG/rc2)

dτ/dt = tS/tO = λS/λO = fO/fS = √(1 - 2MG/rc2)

~ (1 - MG/rc2) (Binomial expansion)

= (1 - φ/c2)

By convention:

λO/λS - 1 = z

Therefore, the gravitation redshift is written as:

1 + z = λO/λS = 1/√(1 - 2MG/rc2)

In the weak field case, this becomes

z ~ -αΔx/c2

Therefore,

1 + z = (1 - αΔx/c2)

= (1 - φ/c2)

Length Contraction in GR?
-------------------------

A meter stick falling towards a black hole would seem to
be length contracted from the point of view of a distant
observer because the front end of the stick would appear
to be moving slower and slower such that the back end
effectively 'catches up'.

What is really happening is that light from the front end
of the stick is becoming increasingly redshifted with
respect to the back end.  Ultimately, the stick would
appear to have no length and would be 'smeared' on the
surface of the event horizon in a sedimentary way.

However, because there is now a gravitating mass present,
a meter stick will experience tidal forces that will
stretch it in the radial direction.  These tidal forces
exist because the gravitational field is not homogeneous
(uniform).  The stretching is an effect that can be
measured by an observer thatis comoving with the object.
If it is stretched by a factor, s, where s > 1 then its
length will measure to be sL.  However, determining the
length contraction from afar depends on the ability of
the observer to measure length in curved spacetime.  If
the observer is close to the stick and stationary, it is
possible to measure length if the velocity measurement
is made in a time that is very short compared to the
reciprocal of the acceleration.  Under these conditions
curvature can be ignored and the length can be calculated
as sL/γ.  The two effects are, therefore, in a competition
but are independent.  This means that in some cases the
tidal forces will prevail, while in other cases the length
contraction will prevail.

Because of the ambiguity associated with measuring length
at distances in curved spacetime, referring to length
contraction in GR without specifying how it is measured
is not that meaningful.  Consequently, length contraction
is not a central concept in GR.  In other words, the
the length contraction of a distant object only really
makes sense in the context of flat spacetime and
Special Relativity.

Spaghettification
-----------------

The point at which tidal forces destroy an object or kill
a person will depend on the black hole's size.  For a
supermassive black hole, the spaghettification point lies
inside the event horizon, so a person may cross the event
horizon without noticing any real squashing and pulling.
However, it will be only a matter of time before the person
is 'spaghettified'.  In contrast, for small black holes,
the tidal forces would kill even before the person reaches
the event horizon.

The Rindler Metric versus the Schwarzchild Metric
-------------------------------------------------

ξ, defined as follows.

R2 = 4ξ ∴ 2RdR = 4dξ ∴ dR = 4dξ/2R ∴ dR2 = dξ2/ξ

ω = t/2 ∴ dω = dt/2 ∴ dω2 = dt2/4

dτ2 = (Rα/c)2dω2 - dR2

= (α/c)2R2dω2 - dξ2/ξ

= (α/c)24ξdω2 - - dξ2/ξ

= (α/c)24ξdt2/4 - dξ2/ξ

= (α/c)2ξdt2 - dξ2/ξ

Let (α/c)2 = 1.  This corresponds to α = c.  Therefore,

R = c2/α = c2/c = c = 3 x 108m = 186,000 miles

The metric becomes:

dτ2 = ξdt2 - dξ2/ξ

Dimension check:  [dξ2/ξ] = L4/L2 = L2

[(α/c)2ξdt2] = (L2/T4)/(L2/T2).L2.T2

= L2

dτ2 = (1 - 2MG/rc2)dt2 - (1/(1 - 2MG/rc2))dr2 - (1/c2)r2Ω22

Rewrite (1 - 2MG/c2r) as (1 - rS/r) and set rS = 1.  This
changes r in the metric so it now measures the distance
from the event horizon rather that the distance from the
center of the gravitating mass.

|<-- R -->
|
.<--- | -- r -->.
COM    |
|
rS = 2MG/c2

COM = center of gravitating mass.

The Schwarzchild metric becomes:

dτ2 = (r - 1)dt2 - dr2/(r - 1)

Let ξ = (r - 1) ∴ dξ = dr

dτ2 = ξdt2 - dr2/ξ

This metric has exactly the same form as the Rindler metric
in flat spacetime.  However, to determine the true flatness
of the spacetime requires computation of the curvature
tensor, which involves taking second derivatives of the
metric.

Appendix
--------

4-velocity and 4-acceleration Along the Hyperbola
-------------------------------------------------

The general form of the 4-velocity and 4-acceleration are:

U = dX/dτ = (γc,γv)

A = dU/dτ = (cdγ/dτ,d(γv)/dτ)

= γ(cdγ/dt,d(γv)/dt)

= (γ4(v.a)/c,γ2a + γ4v(v.a)/c2) (not derived)

Where v, a are the 3-velocity and 3-acceleration respectively.

For the hyperbola these become:

X = (c2/α)cosh((α/c)τ) and cT = (c2/α)sinh((α/c)τ)

UT = dcT/dτ = ccosh((α/c)τ)

UX = dX/dτ = csinh((α/c)τ)

U2 = (UT)2 - (UX)2 = c2

AT = d2cT/dτ2 = αsinh((α/c)τ)

AX = d2X/dτ2 = αcosh((α/c)τ)

A2 = (AT)2 - (AX)2 = -α2

d(U.U)/dτ = (dU/dτ)U + UdU/dτ = 2U.A

But U.U = c2 ∴ d(U.U)/dτ = 0 so U.A = 0.

Rapidity
--------

Mathematically, rapidity can be defined as the hyperbolic
angle that differentiates two frames of reference in relative
motion, each frame being associated with distance and time
coordinates.  The rapidity of an object passing constantly
from one inertial frame to another in such a way that its
change of speed in a fixed time interval is always the same,
must be increasing at a constant rate with respect to the
object's proper time.  This rate of change of rapidity, ξ,
with respect to proper time, τ, is the proper acceleration,
α.  Therefore,

α = cdξ/dτ = cd(ατ/c)/dτ = α

The rapidity is related to the coordinate velocity, v, by
the equation:

v = ctanhξ

= ctanh(ατ/c)

Proof:

X = √(R2 + c2T2)

v = dX/dT = c2T/√(R2 + c2T2)

= cRsinh(ατ/c)/√(R2 + R2sinh2(ατ/c))

= csinh(ατ/c)/√(cosh2(ατ/c))

= ctanh(ατ/c)

= ctanhξ

The 'proper' velocity, w, is given by:

w = dX/dτ = (dX/dT)(dt/dτ)

= γctanhξ

= γv

Therefore, w is made up of the spacelike components of the
object's 4-vector velocity (i.e. it is a 3-vector).

Now γ = dt/dτ = coshξ (see note on Lorentz Transform).
Therefore,

w = coshξctanhξ

= csinhξ

The name proper velocity is a misnomer since it is not
Lorentz invariant.  Its name derives from being a derivative
with respect to proper time, τ.  To avoid this confusion,
it is often referred to as the CELERITY.

Note:  U.U = c2 is invariant where U is the 4-velocity.

The coordinate acceleration, a, is given by:

a = dv/dT = c2T/√(R2 + c2T2)

= c2R2/(R2 + c2T2)3/2

= c2R2/(R2 + R2sinh2(ατ/c))3/2

= c2R2/(R2)3/2(1 + sinh2(ατ/c))3/2

= c2/Rcosh3(ατ/c)

= α/cosh3(ατ/c)

= α/cosh3ξ

= α/γ3

or,

α = γ3a

The proper acceleration, α, is given by:

U = (γc,γv)

= (γc,w)

A = dU/dτ = (cdγ/dτ,dw/dτ)

= (γ4(v.a)/c,γ2a + γ4v(v.a)/c2)

Therefore, dw/dτ is made up of the spacelike components
of the object's 4-vector acceleration (i.e. a 3-vector).
This can be better understood by introducing the idea of
a MOMENTARY COMOVING INERTIAL REFERENCE FRAME (MCRF).

Lorentz transformations only work with inertial frames
and cartesian coordinate systems where the worldlines
are straight.  Therefore, they don't work in accelerated
reference frames because the world lines are not straight
and the coordinate systems are curvilinear.  MCRFs are
inertial frames traveling at the same instantaneous
velocity as the object at any moment.  In other words,
for a specific point on a non-inertial worldline, they
have the same 4-velocity and 4-position for an infinitesimal
period of time.  A curved worldline can then be thought of
as a series of these instantaneous inertial reference
frames.  This means that a Lorentz transformation from a
stationary inertial frame to an MCRF at each point on the

Consider an accelerating rocket, R.

S           S'
|           |
|          -|-  -> u' (wrt S')
|         | R | -> u (wrt S)
|          -|-  -> α
--|-----------|-------------- x
|           |
|           | -> v (wrt S)
| -> a

S' is the instantaneous inertial MCRF that always moves
along at the same speed and direction as the accelerated
frame.  Therefore, u' = 0 and u = v.  Now,

-  -          -  -        -                    -
A = d| γc |/dτ = γd| γc |/dt = |       cγ(dγ/dt)      |
| γv |        | γv |      | vγ(dγ/dt) + γ(dv/dt) |
-  -          -  -        -                    -

In the MCRF, v = 0 ∴ γ = 1 ∴ dγ/dt = 0.  Therefore,

-     -
A = |   0   | = α
| dv/dt |
-     -

So,

A.A = α2

Also

ds2 = c2dτ2 - dx2  in the accelerating frame.

= c2dt'2 - dx'2  in the MCRF.

Then at the moment the particle is at rest dx2 = dx'2 = 0.
Then,

c2dτ2 = c2dt'2

Or,

dτ = dt'

In other words, the accelerated clock's rate is identical
to the clock rate in a MCRF.

Vector Formulation
------------------

The tangent to the worldline at any point, p, is a straight
line.  These vectors can be identified in terms of basis
vectors in Minkowski space as:

(x0,x1,x2,x3) <=> x0e0|p + x1e1|p + x2e2|p + x3e3|p

Where eμ = ∂/∂xμ = ∂μ

- -
| 1 |
e0 = | 0 | etc.
| 0 |
| 0 |
- -

A distance between 2 events in Minkowski spacetime as
defined by the invariant interval (ct)2 - x2, is the same
in any orthonormal spacetime basis.  Proof:

t
|      (t=5,x=3) ≡ (t'=4,x'=0)
|     /
|    /S
|   /
et ^  ^et'
|/  ex
--->-------------- x
\ ex'
v

(t,x) basis:

S2 = (ct)2 - x2

= 52 - 32

= 16

(t',x') basis:

S = 4et' + 0ex'

S2 = (ct)2 - x2

= 42 - 02

= 16

In the MCRF:

U = ceω + 0eR (since U.U = c2)

U is always along the worldline, parallel to eω and
eR = 0.  Therefore, the observer in each of the MCRFs
thinks they are stationary in space (i.e. their spatial
velocity is 0).

Now, A.U = 0 implies that:

A = 0eω + aReR

The observer in each of the series of instantaneous f
rames has a purely spatial acceleration of aR.  If this
is the constant proper acceleration, α, then, A = αeR.

For the observer in the stationary (not MCRF) reference
frame the time and space component will both change.
Therefore,

A = aTeT + aXeX

Because the dot product of 4-vectors is invariant regardless
of the reference frame, the observer in the stationary
reference frame will see the same acceleration as the
observer in the instantaneous frame.  Therefore,

A = aTeT + αeX

and,

A.A = (aT)2et2 + α2ex2 = -α2

Acceleration = Rate of Change of Rapidity
-----------------------------------------

Proof that the proper acceleration is the rate of change
of the rapidity with respect to the proper time can be
obtained by using the chain rule.

dw/dt = (dw/dξ)(dξ/dτ)(dτ/dt)

∴ α = (dw/dξ)(dξ/dτ)(1/γ)

Now, w = csinhξ ∴ dw/dξ = ccoshξ = cγ

∴ α = cγ(dξ/dτ)(1/γ)

= cdξ/dτ

Now, ξ = ατ/c

∴ dξ/dτ = α/c

So,

α = cdξ/dτ = α Q.E.D
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