Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?
Thanks,
David Seppala

wrote:
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?
Thanks,
David Seppala

This is a homework problem. Either show us your work or be damned to
ignorance.

wrote in message
...
| Would someone please be kind enough to post the formula for the
| following problem? A space craft is on the x-axis. At time t0=0 as
| measured by the craft's clock, the craft fires a laser in the
| x-direction, and simultaneously starts accelerating along the x-axis.
| The accelerometer on the craft reads g. The acceleration stops at
| time t1= N seconds as measured by an ideal clock on board the craft.
| How far away is the pulse of light from this craft (as measured by
| observers on the craft) when the acceleration stops?
| Thanks,
| David Seppala

D(ship) = 1/2gt^2
D(light) = ct.
t = N.
D(light) - D(Ship) = cN - 1/2gN^2.
(I'm a Galilean relativist, so I find it easy.)
The ship catches up to its own pulse when
1/2gt^2 - ct + 0 = 0.
Solving for t, we see we have a quadratic in the form ax^2 +bx+ c = 0.
I'll leave you to that one. Don't get confused with c the constant
and c the speed of light.

Androcles

wrote in message ...
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

Equations at bottom of
http://users.pandora.be/vdmoortel/di...eleration.html

T is proper time (what you call t1 = N).
Proper acceleration is a (what you call g).

At proper time T the ship is inertial and has velocity
w.r.t. the original frame:
v(T) = c tanh(aT/c)
with corresponding gamma-value
gamma(T) = cosh(aT/c)

The time in the original frame is
t(T) = c/a sinh(aT/c)
so the light pulse is at distance
ct(T) = c^2/a sinh(aT/c)

W.r.t. the original frame the craft is at distance
x(T) = c^2/a [ cosh(aT/c) - 1 ]

The distance (as measured in the original frame)
between the craft and the pulse is
ct(T) - x(T)

Divide by gamma(T) to find the distance in the
craft's final inertial frame:
( c^2/a sinh(aT/c) - c^2/a [ cosh(aT/c) - 1 ] ) / cosh(aT/c)
giving something like
c^2/a [ sinh(aT/c) - cosh(aT/c) + 1 ] / cosh(aT/c)

Dirk Vdm

"bluechip" wrote in message ...
wrote:
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?
Thanks,
David Seppala

This is a homework problem. Either show us your work or be damned to
ignorance.

Homework?
Seppala?
Nah...

Dirk Vdm

wrote in message t...
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?
Thanks,
David Seppala

People would provide an answer if they thought you were really
interested in learning something. But it's well known that you're a
willfully ignorant jackass.

Paul Cardinale

"Dirk Van de moortel" wrote
in message ...

wrote in message
...
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

Equations at bottom of
http://users.pandora.be/vdmoortel/di...eleration.html

T is proper time (what you call t1 = N).
Proper acceleration is a (what you call g).

At proper time T the ship is inertial and has velocity
w.r.t. the original frame:
v(T) = c tanh(aT/c)
with corresponding gamma-value
gamma(T) = cosh(aT/c)

The time in the original frame is
t(T) = c/a sinh(aT/c)
so the light pulse is at distance
ct(T) = c^2/a sinh(aT/c)

W.r.t. the original frame the craft is at distance
x(T) = c^2/a [ cosh(aT/c) - 1 ]

The distance (as measured in the original frame)
between the craft and the pulse is
ct(T) - x(T)

Divide by gamma(T) to find the distance in the
craft's final inertial frame:
( c^2/a sinh(aT/c) - c^2/a [ cosh(aT/c) - 1 ] ) / cosh(aT/c)
giving something like
c^2/a [ sinh(aT/c) - cosh(aT/c) + 1 ] / cosh(aT/c)

Dirk Vdm

I'm wondering about the last part of your derivation. The final distance
between the craft and the pulse (as measured in the craft's frame) will be
determined by simultaneity according to the craft's frame rather than
simultaneity in the original frame. When I take this into account I get

Distance = (c^2/g) [sinh(gT/c)+cosh(gT/c) - 1]

where g is the proper acceleration.

I noticed something kind of interesting (if I didn't make a mistake). Let K
be the original frame and K' the final inertial frame of the craft. As
usual, assume that the origins of the frames coincide at zero time in both
frames and the craft starts from the origin of the original frame at t = 0.
Let E be the event where the craft first "arrives" at rest in frame K'.
Then the time of this event has the same value in both K and K'.

Simplicio

"J.J. Simplicio" wrote in message
news:eLNZc.226986$8_6.112745@attbi_s04...

I noticed something kind of interesting (if I didn't make a mistake). Let
K
be the original frame and K' the final inertial frame of the craft. As
usual, assume that the origins of the frames coincide at zero time in both
frames and the craft starts from the origin of the original frame at t =
0.

"J.J. Simplicio" wrote in message news:eLNZc.226986$8_6.112745@attbi_s04...

"Dirk Van de moortel" wrote
in message ...

wrote in message
...
Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

Equations at bottom of
http://users.pandora.be/vdmoortel/di...eleration.html

T is proper time (what you call t1 = N).
Proper acceleration is a (what you call g).

At proper time T the ship is inertial and has velocity
w.r.t. the original frame:
v(T) = c tanh(aT/c)
with corresponding gamma-value
gamma(T) = cosh(aT/c)

The time in the original frame is
t(T) = c/a sinh(aT/c)
so the light pulse is at distance
ct(T) = c^2/a sinh(aT/c)

W.r.t. the original frame the craft is at distance
x(T) = c^2/a [ cosh(aT/c) - 1 ]

The distance (as measured in the original frame)
between the craft and the pulse is
ct(T) - x(T)

Divide by gamma(T) to find the distance in the
craft's final inertial frame:
( c^2/a sinh(aT/c) - c^2/a [ cosh(aT/c) - 1 ] ) / cosh(aT/c)
giving something like
c^2/a [ sinh(aT/c) - cosh(aT/c) + 1 ] / cosh(aT/c)

Dirk Vdm

I'm wondering about the last part of your derivation. The final distance
between the craft and the pulse (as measured in the craft's frame) will be
determined by simultaneity according to the craft's frame rather than
simultaneity in the original frame. When I take this into account I get

Distance = (c^2/g) [sinh(gT/c)+cosh(gT/c) - 1]

where g is the proper acceleration.

Yes, I think you are right.

The sanity checks I did, were first to verify that the
limit of my distance divided by T for T to zero is c,
which is okay. Your distance satisfies that as well.
Then I had verified that the limit of the distance for
T to infinity is zero, which I intuitively figured was
okay, and the limit for a to infinity was zero as
well. With your expression both these limits are
infinity. Intuition is clearly a bad advisor here :-)

See below...


I noticed something kind of interesting (if I didn't make a mistake). Let K
be the original frame and K' the final inertial frame of the craft. As
usual, assume that the origins of the frames coincide at zero time in both
frames and the craft starts from the origin of the original frame at t = 0.
Let E be the event where the craft first "arrives" at rest in frame K'.
Then the time of this event has the same value in both K and K'.

Indeed, with the K-coordinates for event E(T)
x(T) = c^2/a ( cosh(aT/c) - 1 )
t(T) = c/a sinh(aT/c)
the Lorentz Transformation and the expressions
v(T) = c tanh(aT/c)
gamma(T) = cosh(aT/c)
give with a bit of algebra
t'(T) = gamma(T) ( t(T) - v(T) x(T) )
= ...
= c/a sinh(aT/c)
= t(T)
x'(T) = gamma(T) ( x(T) - v(T) t(T) )
= ...
= c^2/a ( 1 - cosh(aT/c) )
= - x(T)
Surprising and very neat :-)

The event L where the line of simultaneity of event E(T)
with equation
t' = c/a sinh(aT/c)
crosses the light line with equation
x' = c t'
has x'-coordinate
x'_L = c^2/a sinh(aT/c)
So the distance between event L and event E is indeed
distance = x'_L - x'(T)
= c^2/a [ sinh(aT/c) + cosh(aT/c) - 1 ]
as given by you.

Tricky beasts, these accelerated frames :-)

Dirk Vdm

Dirk Van de moortel wrote:
"J.J. Simplicio" wrote in message news:eLNZc.226986$8_6.112745@attbi_s04...

"Dirk Van de moortel" wrote
in message ...

wrote in message

et...

Would someone please be kind enough to post the formula for the
following problem? A space craft is on the x-axis. At time t0=0 as
measured by the craft's clock, the craft fires a laser in the
x-direction, and simultaneously starts accelerating along the x-axis.
The accelerometer on the craft reads g. The acceleration stops at
time t1= N seconds as measured by an ideal clock on board the craft.
How far away is the pulse of light from this craft (as measured by
observers on the craft) when the acceleration stops?

Equations at bottom of
http://users.pandora.be/vdmoortel/di...eleration.html

T is proper time (what you call t1 = N).
Proper acceleration is a (what you call g).

At proper time T the ship is inertial and has velocity
w.r.t. the original frame:
v(T) = c tanh(aT/c)
with corresponding gamma-value
gamma(T) = cosh(aT/c)

The time in the original frame is
t(T) = c/a sinh(aT/c)
so the light pulse is at distance
ct(T) = c^2/a sinh(aT/c)

W.r.t. the original frame the craft is at distance
x(T) = c^2/a [ cosh(aT/c) - 1 ]

The distance (as measured in the original frame)
between the craft and the pulse is
ct(T) - x(T)

Divide by gamma(T) to find the distance in the
craft's final inertial frame:
( c^2/a sinh(aT/c) - c^2/a [ cosh(aT/c) - 1 ] ) / cosh(aT/c)
giving something like
c^2/a [ sinh(aT/c) - cosh(aT/c) + 1 ] / cosh(aT/c)

Dirk Vdm

I'm wondering about the last part of your derivation. The final distance
between the craft and the pulse (as measured in the craft's frame) will be
determined by simultaneity according to the craft's frame rather than
simultaneity in the original frame. When I take this into account I get

Distance = (c^2/g) [sinh(gT/c)+cosh(gT/c) - 1]

where g is the proper acceleration.


Yes, I think you are right.

The sanity checks I did, were first to verify that the
limit of my distance divided by T for T to zero is c,
which is okay. Your distance satisfies that as well.
Then I had verified that the limit of the distance for
T to infinity is zero, which I intuitively figured was
okay, and the limit for a to infinity was zero as
well. With your expression both these limits are
infinity. Intuition is clearly a bad advisor here :-)

Another way to get the result is to note that the speed of the light
pulse (measured in the accelerated coordinate system) is e^t (c=1, g~1),
so after N secs the distance from the origin is approx. e^N - 1 light-secs.

Interestingly, e^N - 1 N for N 0 so the distance to the light pulse
after accelerating in its direction is greater than it would be with no
acceleration. It takes a hard acceleration in the opposite direction of
the light pulse in order to remain "close" to it.

Re. inuition - suppose the initial acceleration were instantaneous to v
= tanh N. After N secs. the light pulse would then be at a distance of
N light secs, but since the specified acceleration rate is less than
that, the distance to the pulse after N secs must be greater than N
light secs (i.e. approaches infinity for increasing N).