Re: Time Dilation disappears

On Nov 30, 7:00 am, "Paul B. Andersen"
<paul.b.ander...@xxxxxxxxxxxxxxx> wrote:
zarm...@xxxxxxxxx wrote:
On Nov 28, 1:31 pm, "Paul B. Andersen"
<paul.b.ander...@xxxxxxxxxxxxxxx> wrote:
zarm...@xxxxxxxxx wrote:
On Nov 25, 11:23 pm, Jürgen Clade <cl...@xxxxxxxxxx> wrote:
zarm...@xxxxxxxxx schrieb:
Suppose there is a source which has head light and back light (or has
two torches, ahead and behind) while traveling at
1- Speed of light (theoretically)
2- higher percentage of light speed
Would the speed of light that emit in opposite direction be c, if not
why? Would one be able to see light that shine in opposite direction
from torches in case 1.
The speed of the light will always be c. The Doppler effect would change
the *frequency* of the light such that the head light becomes blue
shifted and the back light becomes red shifted. In the first case (which
is impossilbe) the frequency of the head light would become infinite and
the frequency of the back light would drop to zero.
Where did light go that travel at c for thousand million years after
touching the extreme boundary of closed universe? Is there any study?
I don't know what you mean with "the extreme boundary of closed
universe". The universe is not closed, neither it has a boundary.
As Einstein said, you will end up back where you started. Therefore I
assumed that light travelled for 100 millions of yrs would end up
back, where it started.
In a closed, stationary universe this would be the case. But the
universe is neither closed nor stationary.
MfG,
Juergen
Here is my question which could not succinct earlier.
http://www.youtube.com/watch?v=KHjpBjgIMVk
Three  spaceships 1, 2 and 3 are moving close the speed of light in a
galaxy but in a straight line instead of as shown in the link.
Distance b/t 1 and 2 = b/t 2 and 3. Like
1                          2                            3
So the ships are stationary in their common rest frame.
Let's call the distance between them d.

Mounted on middle spaceship2 a lazar (back and front) that can fire a
pulses of light and there are two mirror one on the back of ship1 and
the other on forefront of ship3 . The pilot of the middle spaceship2
fires a pulses of light at both mirrors of 1 and 3 and watches as
reflected back. A clock on board measured  how long the round trips
take.
And the round trips would take T = 2d/c

As it is said speed of light is constant irrespective of its
frequencies. Therefore round trip should take the same time for both
cases.
Now how would you describe time dilation wrt to the observer on
Astrid, pilot of the space ship 1, 2 and 3 while watching the
pulses.
?
Do you possibly mean that the three spaceships are moving at a speed v
close to c relative to an asteroid, and what would an observer on
that asteroid measure?

He would measure the distance between the spaceships to be
Lorentz contracted d' = d sqrt(1-v^2/c^2)
He would see the speed of light to be c in his rest frame.
So he would see that the light pulse would take the time
d'/(c-v) in one direction, and d'/(c+v) in the other direction,

Read again:

where (c-v) and (c+v) are the differences between the speed of
light and the speed of the ships in the two directions respectively.

The pulse is travelling with the speed c, the ships with the speed v.
Nothing is moving with the speed (c+v) or (c-v).





So the round trip time, as measured on synchronized clocks in
his rest frame, would be:

T' = d'/(c-v) + d'/(c+v)= 2d'c/(c^2-v^2) = (2d/c)/sqrt(1-v^2/c^2)
T' = T/sqrt(1-v^2/c^2)

That's time dilation.

http://video.google.com/videoplay?docid=6322511432077219124&hl=en
Similarly, would the pilot of the middle ship2 observe the
simultaneous lighting (reflection of both pulses) strike to forefront
and back of his ship,
Yes, obviously.

if not then speed of light is not constant for a
round trip which is assumed to be the same in round trip in experiment
for measuring c.
But it is.

if yes, then length contraction by a Lorentz factor
is not true and so the time dilation.
Now look:
Time dilation and Lorentz contraction only occur when
an observer measure durations and lengths associated with
a _moving_ object.

Of course the existence of an arbitrary asteroid cannot
affect the spaceships in any way. The ships don't contract,
and clocks on the ship don't slow down just because
an arbitrary observer is passing by.

But the speed of the ships relative to the observer
will affect what the observer measures!
By comparing the readings on the clock passing by
to two synchronized clocks in his own frame of reference,
he will _measure_ the rate of the passing clock as slowed down.
Likewise, by noting the simultaneous (in his frame!) positions
of the ends of the moving ship, and measure the length between them,
he will _measure_ the moving ships to be contracted.

Understand this:
An arbitrary observer cannot affect the observed object in any way.
But the speed of the object relative to the observer can affect
the observer's _observations_ of the object.

Example:
The speed of the Earth relative to a cosmic muon doesn't affect
the muon in any way.
But the speed of the muon relative to the Earth affects our
_measurement_ of the muon's lifetime.

Time dilation is a real, measurable phenomenon. But it is
not a property of the observed clock, it is a property of
the relationship between the observer and the clock.

This is what the cranks in this NG invariably fails to understand.

--
Paul

http://home.c2i.net/pb_andersen/-Hide quoted text -

- Show quoted text -

As light always travel with c and nothing can travel faster than c
(except pull of BH) therefore c+v and c-v is impossible for a pulse to
travel.

Indeed.
That's why no pulse is travelling with either of those speeds.
Read my posting again please.
Carefully this time.

An observer on the asteroid would see that pulse fired by pilot of
ship2 at mirror of ship1, covered an extra distance of magnitude viz
covered by ship with c while moving in a uniform motion frame of
reference. This is because both pulse and ship1 moving in the same
forward direction and c > v. So this is a longer trip. However
reflection of this pulse would be a shorter trip as explained below.

Pulse viz fired at mirror of ship3 would also travel with c but it
covered a shorter distance of magnitude viz traveled by ship3 with v
in forword direction (opposite to the direction of pulse). Because
both pulse and ship3 are  moving opposite to each other. Thus this is
a shorter trip. However reflection of this would be a longer trip as
explained earlier.

Quite right.
I calculated the round trip time measured in the Asteroid frame to be:
T' = (2d/c)/sqrt(1-v^2/c^2)

This round trip time is equal for both pulses, so the observer
on the Asteroid would therefore see the pulses hit ship 2 simultaneously.

However, time taken by both pulses for their total round trip will be
same   and hence pilot of ship 2 would observe simultaneous lighting
(reflection of both pulses) strike to forefront and back of his ship
and thus there is no contraction of length and dilation of time . If
not, then there is something wrong with c.

Quite.
Important point:
If we consider the lengths of the ship to be zero, or an least
negligible compared to the distance between the ships, then
the pulses hit each other simultaneously at the same point.
All observers _must_ agree that the pulses hit each other simultaneously!

Did you think that SR says otherwise?

But your (somewhat imprecise) question was:
  "Now how would you describe time dilation wrt to the observer on
  Astrid, pilot of the space ship 1, 2 and 3 while watching the
  pulses."

And there is a time dilation.
If the observer on ship 2 measure the round trip time to be T,
the observer on the asteroid would see the round trip time
dilated to T' = T/sqrt(1-v^2/c^2).

Now come to link
http://www.youtube.com/watch?v=KHjpBjgIMVk

Pulse viz fired by the pilot of upper ship at the mirror of the lower
ship has two components one is forward (v) and the other is downward
(c) . Observer on asteroid would observe both motion with the same
clock (time) and should not be confused with extra time viz required
to cover a length with v or time required to cover a length with c.
Thus time required by v to covered a length by pulse in its forward
direction is actually included in the time required for c in downward
direction. An observer would find a pulse at any interval of time in b/
w two ship on the same vertical line. Thus actual time is
misunderstood in the theory by adding a time to cover a length with v
in its forward direction.

The question is not what the observer on the asteroid would
understand or not understand or what might confuse him.
The question is what he would measure with real clocks.
-------------------------------------------------------
And he _must_ use two clocks to do the measurement!
To measure what a clock shows at an event, the clock
must be present at the event. And since his clock must
be stationary to him, he must use two clocks.

   ----------------------------
    |          /\
    d      ct'/  \ct'
    |        /    \
    |       / 2vt' \
   --------|--------|----------> x'
           C1      C2

He measure the time of reflection from the lower mirror with
the clock C1, and the time of the next reflection from the lower
mirror with the clock C2. If the clocks are synchronized,
and if the the speed of light is c, then have:
   d^2 + (vt')^2 = (ct')^2
where t' is the time the pulse uses from the lower to the upper mirror
The "round trip ...

read more »- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

I am not a physicist but still want to make clear the following
point.

A pulse has two components of its motion. Horizontal v and vertical c.
Now analyze both componenets of a motion separately as shown in the
link

After 1st reflection a pulse move with c in vertical direction and
also drag with v in horizontal direction (along the ship)
simultaneously.

..As v < c therefore for a pulse to cover a total treading (from C1 to
C2) with v in its slant would be smaller as compared to vertical round
trip covered by a pulse. Although time required to cover a distance
between C1 and C2 (the total treading in slant) = time required for
total vertical round trip (2d)

Now come to triangle

Distance between C1 and C2 is 2vt’and half is vt’
vertical side of the triangle is d.
The slant side of the triangle shouldn't not be ct’ but the actual
slant side should be vt’ + ct’ because a pulse has two components as
mentioned above. If you resolved slant motion of pulse into its two
component you will get both c and v separately. Therefore vt’ should
be added alongwith ct’ in order to represent the real slant side.


What about the clock C3 which is synchronized with C1 and C2 and
fixed with the lower mirror and recorded both reflection. Time
recorded by C3 would be different from those of C1 and C2 in your case
because clock C3 would always find a pulse above the mirror at any
interval of time and would measured time for c only not for v.
In nutshell, in a triangle posted by you
The slant side represent ct’ only and therefore missed to include
vt’ (horizontal component v of a pulse)
Also, as speed of light is always c and should cover specified
distance per second. If longer then c is not valid as shown in slant
side
.

Relevant Pages

  • Re: Time Dilation disappears
    ... Now how would you describe time dilation wrt to the observer on ... Astrid, pilot of the space ship 1, 2 and 3 while watching the ... So he would see that the light pulse would take the time ... That's why no pulse is travelling with either of those speeds. ...
    (sci.physics.relativity)
  • Re: Time Dilation disappears
    ... Now how would you describe time dilation wrt to the observer on ... Astrid, pilot of the space ship 1, 2 and 3 while watching the ... So he would see that the light pulse would take the time ... That's why no pulse is travelling with either of those speeds. ...
    (sci.physics.relativity)
  • Re: Time Dilation disappears
    ... Now how would you describe time dilation wrt to the observer on ... Astrid, pilot of the space ship 1, 2 and 3 while watching the ... So he would see that the light pulse would take the time ... That's why no pulse is travelling with either of those speeds. ...
    (sci.physics.relativity)
  • Re: Time Dilation disappears
    ... Now how would you describe time dilation wrt to the observer on ... The pulse is travelling with the speed c, the ships with the speed v. ... By comparing the readings on the clock passing by ... That's why no pulse is travelling with either of those speeds. ...
    (sci.physics.relativity)
  • Re: Is the speed of light really constant ?
    ... > seen in A' and B's frame of reference.Since this would specify the ... > and B as well,it therefore it must also be the time when the light pulse ... > observer B ... >> M' bounces a signal off the clock at A' and times its return. ...
    (sci.physics.relativity)