Welcome
If it is true , the hands of clock in system x'y't' ,can moov (shift) on its
scale reves , then hands of clock in the system xyt ?
E.W.

"Eugeniusz Warenda" wrote in message
news:qsbKe.163129$%K2.347@pd7tw1no...
| Welcome
| If it is true , the hands of clock in system x'y't' ,can moov (shift)
on its
| scale reves , then hands of clock in the system xyt ?
| E.W.

No.
Androcles

"Eugeniusz W" wrote in message
news:qsbKe.163129$%K2.347@pd7tw1no...
Welcome
If it is true then hands of clock in system x'y't' can move (shift) on
its
scale reverse (counter clock-wise)?
E.W.

"Eugeniusz Warenda" wrote in message
news:qsbKe.163129$%K2.347@pd7tw1no...
Welcome
If it is true , the hands of clock in system x'y't' ,can move (shift) on
its
scale reverse , then hands of clock in the system xyt ?
E.W.

"Eugeniusz W" wrote in message
news:qsbKe.163129$%K2.347@pd7tw1no...
Welcome;
If it is true the hands of clock in system x'y't' can move (shift) on
its scale reverse (or clock counter-wise)?
E.W.

In sci.physics.relativity, Eugeniusz Warenda

wrote
on Wed, 10 Aug 2005 00:00:22 GMT
qsbKe.163129$%K2.347@pd7tw1no:
Welcome
If it is true , the hands of clock in system x'y't' ,can moov (shift) on its
scale reves , then hands of clock in the system xyt ?
E.W.


Please rephrase your question. Bear in mind you only have half
of the Lorentz, which can be expressed as

x' = (x-vt)/sqrt(1-vv/cc)
t' = (t-vx/cc)/sqrt(1-vv/cc)

In more conventional notation, one might see

x' = (x-vt)/sqrt(1-v^2/c^2)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)

or, as a personal favorite,

x_A = (x_O - v*t_O) / sqrt(1-v^2/c^2)
t_A = (t_O - v*x_O/c^2) / sqrt(1-v^2/c^2)

where the coordinate matchup is a little more visible, in my opinion.
But that's just me.

Note also that the observer is anchored at x=0 or x'=0 (depending
on who is doing the observing). This means that, for example,
that the frequency change would not be 1/sqrt(1-v^2/c^2), but
would have to take into account that the light has to move
an extra distance, and one actually gets sqrt(1-v/c)/sqrt(1+v/c)
as the ratio.

Mathematically, one can show it this way.

Assume O emits two light pulses as A pulls away.
The first event, in O's space, is (0,0); the second is
(0,1). One can now use the Lorentz. The first event in
*A's* space is still (0,0), but the second event is now
at (-vg, g) in his space, where g = 1/sqrt(1-v^2/c^2).
(In more sophisticated environs one sees the greek letter
"gamma" instead of g, but this is ASCII.)

Now, that light pulse O emitted has not reached A yet.
Since it's traveling at lightspeed in A's frame[*], A will see the
second pulse at (0, g+vg/c)=(0,g(1+v/c))=(0,sqrt(1+v/c)/sqrt(1-v/c)),
since g = 1/sqrt(1-v^2/c^2) = 1/(sqrt(1+v/c)sqrt(1-v/c)).

Since frequency is speed/time, the ratio is sqrt(1-v/c)/sqrt(1+v/c),
or the frequency is lower, as the source moves away from the observer.

If A is approaching O, the second event is at (vg, g); the lightpulse
has traveled too far. One can then suptract, getting a time
of g(1-v/c) = sqrt(1-v/c)/sqrt(1+v/c) in the second case.

As for time/hand-shifting -- most of this is an artifact
of the movement between observer and event source. As far
as O, the originator of the pulses, is concerned, nothing
much is happening; he's emitted a pulse at time 0 and at
time 1, and that's pretty much it. It doesn't matter what
A does, unless O observes something from A.

Note that it is possible to correctly determine v, if O has
a rod of known length, a small mirror, and compensates for
the time it takes for light to reach him from the rod's endpoint.
[*] Yes, this should be mindblowing. Light is traveling at lightspeed
*in both frames*. There is no analogue in Newtonian physics,
really.

--
#191,
It's still legal to go .sigless.

Dear The Ghost In The Machine:

"The Ghost In The Machine" wrote
in message ...
....
Please rephrase your question. Bear in mind you only
have half of the Lorentz, which can be expressed as

x' = (x-vt)/sqrt(1-vv/cc)
t' = (t-vx/cc)/sqrt(1-vv/cc)

In more conventional notation, one might see

x' = (x-vt)/sqrt(1-v^2/c^2)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)

or, as a personal favorite,

x_A = (x_O - v*t_O) / sqrt(1-v^2/c^2)
t_A = (t_O - v*x_O/c^2) / sqrt(1-v^2/c^2)

where the coordinate matchup is a little more visible,
in my opinion. But that's just me.

Just curious, why you didn't assign "matchup notation" to v?
Because _O and _A don't get the same value... (unless v is zero).

David A. Smith

In sci.physics.relativity, N:dlzc D:aol T:com (dlzc)
N
wrote
on Wed, 10 Aug 2005 06:14:48 -0700
95nKe.310650$Qo.160748@fed1read01:
Dear The Ghost In The Machine:

"The Ghost In The Machine" wrote
in message ...
...
Please rephrase your question. Bear in mind you only
have half of the Lorentz, which can be expressed as

x' = (x-vt)/sqrt(1-vv/cc)
t' = (t-vx/cc)/sqrt(1-vv/cc)

In more conventional notation, one might see

x' = (x-vt)/sqrt(1-v^2/c^2)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)

or, as a personal favorite,

x_A = (x_O - v*t_O) / sqrt(1-v^2/c^2)
t_A = (t_O - v*x_O/c^2) / sqrt(1-v^2/c^2)

where the coordinate matchup is a little more visible,
in my opinion. But that's just me.

Just curious, why you didn't assign "matchup notation" to v?
Because _O and _A don't get the same value... (unless v is zero).

Oh? We'll see about that.

Assume that A has no idea what velocity he's moving (we'll
call it v), but he has a rod of known length L, three mirrors,
a good stopwatch, and his good buddy O is standing at
his local origin with an omnidirectional light source.

A sits on the leading edge of the rod, with the three
mirrors making a "reverse periscope"; one on the trailing
rod end, and two reflecting the beam coming from the
first mirror into A's eye.

At time 0 the leading edge of the rod is exactly at O.
A sees the first flash, no problem.

Some time later the trailing edge of the rod passes over O;
the light pulse from O reflects into the first mirror and
travels into A's eye. For convenience we assume that A's
eye is omnidirectional as well and therefore can dispense
with two of the mirrors, and the light beam hits A's eye
L/c seconds later.

If A measures a time difference between the two pulses
of t_A + L/c, can he compute the velocity using L/t_A?

The first event is of course (0,0), for both O and A.
The second event in A's coordinate space is (-L, t_A);
the third is (0, t_A + L/c). It turns out the last two
are equivalent (they're related to the same light flash)
so we'll focus on (-L, t_A).

We know that O doesn't move, therefore x_O is identically zero,
and from the Lorentz:

x_A = (x_O - v * t_O) / sqrt(1-v^2/c^2)
= (- v * t_O) / sqrt(1-v^2/c^2)

This isn't all that interesting. However, the Lorentz is
easily inverted algebraically (the manipulations are left
to the interested reader, or one can peruse one of my
older posts -- I'd have to find it myself), resulting in

x_O = (x_A + v * t_A) / sqrt(1-v^2/c^2)

This is far more useful:

0 = (x_A + v * t_A) / sqrt(1-v^2/c^2)

We know x_A = -L, and one can multiply both sides
by sqrt(1-v^2/c^2), yielding

L = v * t_A

or

t_A = L/v or v = L/t_A.

Therefore, A can correctly measure the velocity of himself
relative to O using this method. Bear in mind this is
using A's *own* rod; he cannot use anything by O except
O's unmoving (in O's coordinate space) beacon.

If O moves the beacon, that's cheating. :-)


David A. Smith




--
#191,
It's still legal to go .sigless.

Dear The Ghost In The Machine:

"The Ghost In The Machine" wrote
in message news
In sci.physics.relativity, N:dlzc D:aol T:com (dlzc)
N
wrote
on Wed, 10 Aug 2005 06:14:48 -0700
95nKe.310650$Qo.160748@fed1read01:
Dear The Ghost In The Machine:

"The Ghost In The Machine"
wrote
in message ...
...
Please rephrase your question. Bear in mind you only
have half of the Lorentz, which can be expressed as

x' = (x-vt)/sqrt(1-vv/cc)
t' = (t-vx/cc)/sqrt(1-vv/cc)

In more conventional notation, one might see

x' = (x-vt)/sqrt(1-v^2/c^2)
t' = (t-vx/c^2)/sqrt(1-v^2/c^2)

or, as a personal favorite,

x_A = (x_O - v*t_O) / sqrt(1-v^2/c^2)
t_A = (t_O - v*x_O/c^2) / sqrt(1-v^2/c^2)

where the coordinate matchup is a little more visible,
in my opinion. But that's just me.

Just curious, why you didn't assign "matchup notation" to v?
Because _O and _A don't get the same value... (unless v is
zero).

Oh? We'll see about that.

v as presented above is v_O. v_A is a different value, isn't it?

Assume that A has no idea what velocity he's moving (we'll
call it v), but he has a rod of known length L, three mirrors,
a good stopwatch, and his good buddy O is standing at
his local origin with an omnidirectional light source.

A sits on the leading edge of the rod, with the three
mirrors making a "reverse periscope"; one on the trailing
rod end, and two reflecting the beam coming from the
first mirror into A's eye.

At time 0 the leading edge of the rod is exactly at O.
A sees the first flash, no problem.

Some time later the trailing edge of the rod passes over O;
the light pulse from O reflects into the first mirror and
travels into A's eye. For convenience we assume that A's
eye is omnidirectional as well and therefore can dispense
with two of the mirrors, and the light beam hits A's eye
L/c seconds later.

If A measures a time difference between the two pulses
of t_A + L/c, can he compute the velocity using L/t_A?

The first event is of course (0,0), for both O and A.
The second event in A's coordinate space is (-L, t_A);
the third is (0, t_A + L/c). It turns out the last two
are equivalent (they're related to the same light flash)
so we'll focus on (-L, t_A).

We know that O doesn't move, therefore x_O is identically zero,
and from the Lorentz:

x_A = (x_O - v * t_O) / sqrt(1-v^2/c^2)
= (- v * t_O) / sqrt(1-v^2/c^2)

*Which* v? O does not agree with A, as to how fast A is
moving...

This isn't all that interesting. However, the Lorentz is
easily inverted algebraically (the manipulations are left
to the interested reader, or one can peruse one of my
older posts -- I'd have to find it myself), resulting in

x_O = (x_A + v * t_A) / sqrt(1-v^2/c^2)

This is far more useful:

0 = (x_A + v * t_A) / sqrt(1-v^2/c^2)

We know x_A = -L, and one can multiply both sides
by sqrt(1-v^2/c^2), yielding

L = v * t_A

or

t_A = L/v or v = L/t_A.

Therefore, A can correctly measure the velocity of himself
relative to O using this method. Bear in mind this is
using A's *own* rod; he cannot use anything by O except
O's unmoving (in O's coordinate space) beacon.

If O moves the beacon, that's cheating. :-)

I think it should be v_O. To get from O's coordinates to A's you
need the relative velocity, in O's frame of reference, of A's
velocity...

David A. Smith

"Eugeniusz W" wrote in message
news:QJeKe.164662$5V4.108312@pd7tw3no...

"Eugeniusz W" wrote in message
news:qsbKe.163129$%K2.347@pd7tw1no...
Welcome;
If it is true the hands of clock in system x'y't' can move (shift) on
its scale reverse (or clock counter-wise)?
E.W.

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