Relativity: Uniformly accelerated observers
1. Velocities in the two inertial frames
The Lorentz transformation of the space-time coordinates of
a moving particle between an inertial frame "S" and an
another inertial frame "S'" is:
t' = γ(t - vx/c2)
x' = γ (x - vt)
y' = y
z' = z
Where v is the velocity of the inertial frame S' with respect
to the inertial frame S, and γ = [1 - v2/c2]c-1/2.
If the particle doesn't move in S', then x' = 0 and
x = vt.
Therfore t' = γ t(1 - v2/c2) =
t/γ or t = γ t' and the time dilates in the inertial
frame S.
The velocity of the particle is u' in S',
and u in S. Therfeore:
u'x = dx'/dt' = (dx - vdt)/(dt - vdx/c2) =
(ux - v)/(1 - vux/c2)
u'y = dy'/dt' = dy/γ(dt - vdx/c2) =
uy/γ(1 - vux/c2)
u'z = dz'/dt' = dz/γ(dt - vdx/c2) =
(uz - v)/γ(1 - vux/c2)
u'x = (ux - v)/(1 - vux/c2)
u'y = (1/γ) uy/(1 - vux/c2)
u'z = (1/γ) uz/(1 - vux/c2)
2. Accelerations in the two inertial frames
The acceleration of the particle is a' in S',
and a in S. Therfeore:
a'x = du'x/dt' =
d[(ux - v)/(1 - vux/c2)]/
[γ(dt - vdx/c2)]
We have:
d[(ux - v)/(1 - vux/c2)] =
{dux(1 - vux/c2) - (ux - v)
d((1 - vux/c2))}/(1 - vux/c2)2
=
{dux - (vux/c2)dux +
ux (vdux/c2) - v2(dux/c2)}/(1 - vux/c2)2
=
{dux - dux(v2/c2)}/(1 - vux/c2)2 =
dux{ 1 - (v2/c2)}/(1 - vux/c2)2
=
(1/γ2) dux/(1 - vux/c2)2
Therefore
a'x = (1/γ2) dux/(1 - vux/c2)2/
[γ(dt - vdx/c2)] =
(1/γ3) dux/(1 - vux/c2)2(dt - vdx/c2)
With the x-component of the velocity of the particle
ax = dux/dt, we get:
a'x =
(1/γ3) ax/(1 - vux/c2)/
[(1 - vux/c2)2
=
(1/γ3) ax/(1 - vux/c2)3
a'x = (1/γ3) ax/(1 - vux/c2)3
a'y = du'y/dt' =
d[uy/γ(1 - vux/c2)]/
[γ(dt - vdx/c2)]
=
(1/γ2) d[uy/(1 - vux/c2)]/
(dt - vdx/c2)
=
(1/γ2) {duy(1 - vux/c2) - uy
d((1 - vux/c2))}/(1 - vux/c2)2(dt - vdx/c2)
=
(1/γ2) {duy - (vux/c2) duy +
uy(vdux/c2) }/(1 - vux/c2)2(dt - vdx/c2)
=
(1/γ2) {duy[1 - (vux/c2)] +
uy(vdux/c2) }/(1 - vux/c2)2(dt - vdx/c2)
With the y-component of the velocity of the particle
ay = duy/dt
a'y =
(1/γ2) {ay[1 - (vux/c2)] +
uy(vax/c2) }/(1 - vux/c3)2
a'y =
(1/γ2) {ay[1 - (vux/c2)] +
uy(vax/c2) }/(1 - vux/c2)3
Similarly
a'z =
(1/γ2) {az[1 - (vux/c2)] +
uz(vax/c2) }/(1 - vux/c2)3
3. Uniformly accelerated motion
acceleration transformation
For a particle with constant acceleration a'x fixed
in the inertial frame S' (instantaneous rest frame for the particle), we have
ux = v.
Hence
For the velocities:
u'x = 0
u'y = (1/γ) uy/(1 - v2/c2) =
γuy
u'z = γuz
u'x = 0
u'y = γuy
u'z = γuz
And for the accelerations:
a'x = (1/γ3) ax/(1 - v2/c2)3
=
γ3 ax
a'y =
(1/γ2) {ay[1 - (v2/c2)] +
uy(v ax/c2) }/(1 - v2/c2)3
=
γ4 {(ay/γ2) +
uy(v/c2)ax}
a'z =
γ4{(az/γ2) +
uz(v/c2)ax}
a'x = γ3 ax
a'y = γ4[(ay/γ2) +
uy(v/c2)ax]
a'z =
γ4[(az/γ2) +
uz(v/c2)ax]
4. Velocity of a uniformly accelerated particle
Now let's consider
a'x = γ3 ax
Remark first that
d(γ v)/dt = γ dv/dt + v dγ/dt
We have
dv/dt = ax, and
dγ/dt = d(1 - v2/c2)-1/2/dt =
-(1/2)(1 - v2/c2)-3/2 (-2vdv/dt)(1/c2) =
(1 - v2/c2)-3/2 (v ax)(1/c2) =
γ3 (v ax)(1/c2)
Then
d(γ v)/dt = γax + γ3 ax(v/c)2 =
= γax[1 + γ2(v/c)2] =
γ3 ax[1 - (v/c)2 + (v/c)2] =
= γ3 ax
d(γ v)/dt = γ3 ax = a'x
Integrating with respect t gives:
γ v = a'x t + constant
Assuming v = 0 at t= 0 leads to constant = 0. Therefore
γ v = a'x t
or
v(1 - v2/c2)-1/2 = a'x t
Squaring gives:
v2(1 - v2/c2) -1 = a'2x t2
or
v2 = a'2x t2 (1 - v2/c2)
v2[1 + a'2x t2/c2 ] = a'2x t2
Therefore
v = a'x t/[1 + a'2x t2/c2]1/2
v = dx/dt = a'x t/[1 + (a'x t/c)2]1/2
4. Position of a uniformly accelerated particle
Integrating of the above equation gives:
x = ∫ dt a'x t/[1 + (a'x t/c)2]1/2 =
a'x ∫ dt t/[1 + (a'x t/c)2]1/2 =
With
1 + (a'x t/c)2 = w
dw = 2 (a'x/c)2 t dt
so
x = [a'x/2(a'x/c)2] ∫ dw w- 1/2 =
[a'x/2(a'x/c)2] 2 w1/2 =
(c2/a'x) [1 + (a'x t/c)2]1/2
x = (c2/a'x) [1 + (a'x t/c)2]1/2
That leads to:
x2 = (c2/a'x)2 [1 + (a'x t/c)2]
= (c2/a'x)2 +
[(a'x t/c)2](c2/a'x)2 =
(c4/a'x2) +
t2 c2
or
x2 - c2t2 = c4/a'x2
x2 - c2t2 = c4/a'x2
This equation represents the hyperbolic path of the uniformly
moving particle (or moving observer) in a Minkowski space-time coordinate
system of the stationary frame S.
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