The Length of a Moving Rod
DAKALAMIDAS
rod which is of length L as measured by an observer O in its rest frame. This
observer measures, according to SR, the proper length of the rod at can do so
at his leisure and by any method he chooses. Suppose there is another observer
O' for which the rod moves with speed V along, say, the x-axis. O' also wants
to measure the length of the moving rod. Let's say that these two observers
have never heard of Einstein or SR etc. O' wants to answer a weird question
which he came up with: Does the length of a moving object change with respect
to its length when measured at rest? O' decides to measure the length of the
moving rod in the following way: He has ONE stopwatch at ONE location. When the
front end of the rod passes by him he starts the watch; when the back end of
the rod passes by him he stops the watch. Then he says "Ah, the length of the
rod is
L' = V * dt, where dt is the time interval recorded on my stopwatch." Now,
simultaneity plays no role here since there is only ONE clock at ONE location
being used in the whole scenario; O could have measured the proper length by
laying down unit rods way in advance. Why then should there be a length
contraction if, according to SR, length contraction follows from the relativity
of simultaneity? If one uses Einstein's contrived method of noting the
simultaneous (in the frame of O') positions of the end-points of the moving
rod, thereby being FORCED to use TWO synchronized (in the frame of O') clocks
at TWO locations on the x-axis, then of course it follows from the constancy of
c that L > L'. But in the scenario stated here the relativity of simultaneity
does not enter the measuring procedure AT ALL. SR provides no answer as to
whether L = L' or not in this case, I believe. Only by performing the actual
experiment can the question be answered. If L > L' with the ONE-CLOCK method,
then it does not follow from SR but must be attributed to some other model
(Lorentzian perhaps). Can someone explain to me how length contraction follows
from SR when the length of a moving rod is measured in the exact (no
modifications) manner stated here?
Thanks again...Demetrios Kalamidas
Barry
DAKALAMIDAS wrote:
>
> Hi...and thanks to all who might respond. Consider the following: There is a
> rod which is of length L as measured by an observer O in its rest frame. This
> observer measures, according to SR, the proper length of the rod at can do so
> at his leisure and by any method he chooses. Suppose there is another observer
> O' for which the rod moves with speed V along, say, the x-axis. O' also wants
> to measure the length of the moving rod. Let's say that these two observers
> have never heard of Einstein or SR etc. O' wants to answer a weird question
> which he came up with: Does the length of a moving object change with respect
> to its length when measured at rest? O' decides to measure the length of the
> moving rod in the following way: He has ONE stopwatch at ONE location. When the
> front end of the rod passes by him he starts the watch; when the back end of
> the rod passes by him he stops the watch. Then he says "Ah, the length of the
> rod is
> L' = V * dt, where dt is the time interval recorded on my stopwatch." Now,
> simultaneity plays no role here since there is only ONE clock at ONE location
> being used in the whole scenario; O could have measured the proper length by
> laying down unit rods way in advance. Why then should there be a length
> contraction if, according to SR, length contraction follows from the relativity
> of simultaneity?
How did you measure V?
Barry
DAKALAMIDAS
numerous textbook descrptions of SR,does one find a description of how to
determine the relative velocity V between two inertial frames? V is introduced
as ' V ' without the operational definitions used in the subsequent length and
time measurements. The accounts always begin with something like "...there is
a frame S and another frame S' that moves with velocity V with respect to
S....".
How can SR derive its conclusions regarding length and time scales for
different inertial frames if the determination of V itself requires the use of
light signals and clocks? Is the Lorentz transformation to be assumed even
before it is 'derived' from the two postulates?
Barry
It is a kind of circle.
You can start with the postulates and derive the transformations, or you
can go the other way.
It's like writing either:
x + 1 = 2 => x = 1
0r
x = 1 => x + 1 = 2
Then you ask if it's consistent and if fits the observations.
Barry
John Rennie
explains certain apparent paradoxes that arise when comparing
different frames, but it is not essential.
My favourite way of approaching SR is to use the invariance of the
metric. This states that the proper time between any two events:
dtau^2 = (c.dt)^2 - dx^2 - dy^2 - dz^2
is an invarient and will be the same for all observers. The nice think
about the metric is that it places SR on a geometric footing in the
same way that GR works.
OK, look at your problem using the metric. Start in the frame of the
rod, O. An observer in this frame sees the O' frame moving towards him
at speed V. Let the frames co-incide at (0, 0) when the O' frame
passes one end of the rod (the end of the rod being at the origin in
both frames). The O observer sees the origin of the O' frame pass the
other end of the rod at a time L/V, so the proper time, tau, as
measured in the O frame is:
tau^2 = (cL/V)^2 - L^2
Now look at things in the O' frame. Here I'm sitting at the origin and
the end of the rod passes me at time zero, i.e. (0, 0). The rod is
approaching at speed V (both frames agree on the relative speed) but
it's length will be contracted to L' in my frame. Hence the other end
of the rod passes me at a time L'/V. I measure the interval to be:
tau'^2 = (cL'/V)^2
The invarience of the metric states that tau = tau' so:
(cL'/V)^2 = (cL/V)^2 - L^2
Multiply both sides by (V/c)^2 and the expression becomes:
L' = L sqrt(1 - (v/c)^2)
which is the expression we're all familiar with, and no failures of
simultaneity have been required :-)
As you mention in your post, simultaneity problems arise when you're
making measurements at different spatial locations. In this case
because the O' observer is measuring everything at one place there is
no problem.
JR
Paul B. Andersen
"DAKALAMIDAS" <dakal...@aol.com> wrote in message news:20020429022641...@mb-fn.aol.com...
Why would you expect Einstein to explain how you measure a speed?
Speed is what it always has been, dx/dt, and it is measured
like it always has been measured: by measuring the distance
an object moves during a certain time (or the other way around).
To do that, you must measure the time at two different positions.
Paul
Paul B. Andersen
"DAKALAMIDAS" <dakal...@aol.com> wrote in message news:20020428234945...@mb-mu.aol.com...
You are wrong. Of course SR give an answer.
> Only by performing the actual
> experiment can the question be answered. If L > L' with the ONE-CLOCK method,
> then it does not follow from SR but must be attributed to some other model
> (Lorentzian perhaps). Can someone explain to me how length contraction follows
> from SR when the length of a moving rod is measured in the exact (no
> modifications) manner stated here?
> Thanks again...Demetrios Kalamidas
First of all:
If you should measure the length of a rod with unknown speed and length,
you would first have to measure the speed to use your method.
And then you must use two synchronized clocks.
But let's suppose the speed is known to be v, and the rest length
of the rod is L. What does SR predict that the length will be
measured to be, using your method?
To answer this question, we simply ask a different question,
namely: what does SR predict the clock will show when the ends
of the rod passes it?
Let's suppose the clock reads zero as the front end of
the rod passes it.
When the aft end of the rod passes it, the co-ordinates
of this event in the rod frame will be:
t' = L/v, x' = - L
Transformed to the "stationary frame":
t = gamma*(t' + x'*v/c^2)
t = (L/v - L*v/c^2)/sqrt(1 - v^2/c^2) = (L/v)*sqrt(1 - v^2/c^2)
L' = v*t = L*sqrt(1-v^2/c^2)
Paul
DAKALAMIDAS
can someone answer this question?
Bilge
Re: The Length of a Moving Rod to usenet:
>
>can someone answer this question?
Probably. Try condensing to a sentence or two and wrapping your
lines so the question is legible.
--
John Rennie
Maybe you missed my post, here it is again.
----8<--------------------
JR
Paul B. Andersen
"DAKALAMIDAS" <dakal...@aol.com> wrote in message news:20020430193837...@mb-fn.aol.com...
It has been answered.
If you missed it, here is my answer again:
Alexander Belov
>simultaneity plays no role here since there is only ONE clock at ONE
>location being used in the whole scenario
involved.
>SR provides no answer as to
>whether L = L' or not in this case, I believe.
That's wrong.
According to SR you may use any method to measure something
and you will get the same result.
>Why then should there be a length
>contraction if, according to SR, length contraction follows from the
>relativity of simultaneity?
There is no connection between derivation of length contraction
and method of it's measurement.
> Can someone explain to me how length contraction follows
>from SR when the length of a moving rod is measured in the exact (no
>modifications) manner stated here?
Length contraction follows not from ALL THE SR but from
some premises.
1)We derived length contraction using two clocks.
2)We measure length contraction using one clock.
There is no connection between 1) and 2).
Have you any questions yet?
-------------------------
Alexander Belov
To understand SR read: http://www.geocities.com/belrel
rryker1
Alexander Belov wrote:
Rod: I enjoy a scenario where two observers
in a relatively moving system are spaced 1 light second apart,
their proper length.
They are not separated via a rod.
Only space.
SR would still say there is a length contraction,
according to a rest system, even though there is
no matter separating the two moving observers.
After all, space (nothing), warps in order to produce
this effect.
Nothing (space), is sooooooooo smart!
What is interesting is that this is _NOT_ in
agreement with Lorentz Ether Theory.
And who cares?!
They're both bullshit.
Rod Ryker
DAKALAMIDAS
> You must use two clocks to measure the speed V. Simultaneity is
>involved.
Why? Everything in SR is derived from the two postulates we are told many
times: (1) the laws of nature are the same in all inertial frames (2) the
constancy of c. All consequences must follow from (1) and (2). So suppose that
the observer on the ground has knowledge of these two alleged truths but has
never heard of SR. He could then send out a light pulse to bounce off the front
of the moving rod and then say 1 second after the pulse arrives back to him he
sends another light pulse. Knowing c, the two roundtrip times, and the time
interval between the arrival of the first and the departure of the second light
pulse he can determine V with ONE clock at ONE location. Simultaneity does not
enter. So with one clock he can determine V and then later on, when the rod
arrives, he could determine L'.
Nicolaas Vroom
"DAKALAMIDAS" <dakal...@aol.com> schreef in bericht
news:20020428234945...@mb-mu.aol.com...
Length measuring of a Moving Rod goes in four steps:
1. Measuring of the length with a rod at rest.
2. Measuring of the length of a moving rod from
the rest frame with observer at rest and with a clock at rest.
3. Measuring of the length of a moving rod
with a moving observer and with clocks at rest.
4 Measuring of the length of a moving rod
with a moving observer and with a moving clock
1. In order to measure the length of a rod L0
B<-----L----->A
O ----->
you have an observer at point B
and a mirror at point A.
O sents a light signal from B to A to B.
O measures the time t
O calculates the length L0 by using the following
formula : total distance = 2*L0 = c*t
or t= 2*L0 / c
2. In order to measure the length of a moving
rod with a speed v you perform the following:
Your rod starts left from observer O and moves
to the right.
When point B (Back of the rod) meets O,
O sends a lightsignal to the front A and back to O
In order to calculate the length L
you need the formula: L+v*t = c*t
Total reflection time t1 measured
t1 = 2*t = 2 * L / (c-v)
When you calculate L based on t1 you will see that
L <> L0
To be more precise L = L0 * sqr(1-v*v/c*c)
3. In order to measure the lenght of a moving
rod with moving observer at point B and
with CLOCKS at rest you must place clocks at rest
in the rest frame along the path of the moving rod.
Those clocks have to be synchronised.
The easiest way is when all clocks are at equal distance.
To synchronise two clocks you need a point
at equal distance between those points.
From that point you send out a light signal.
The moment when that signal reaches each clock
reflects the same time in the rest frame,
ie you can than use the time on one clock to initialise
the other (slowly all the others)
In order to calculate the length of the rod
by a moving observer at B
Again when point B (Back of the rod) meets O,
O sends a lightsignal to the front A and back to
a second moving observer at B.
The time t1 that the moving observer sees
in the rest frame opposite him.
In order to calculate the length L
you need the formulas:
L+v*t = c*t (forward) and L = v*t+c*t (backward)
Total reflection time t1 measured
t1 = L / (c-v) + L/(c+v)
Again, When you calculate L based on t1 you will
see that L <> L0
To be more precise L = L0 * sqr(1-v*v/c*c)
4. In order to measure of the length of a moving rod
with a moving observer at B and with a moving clock
you do exactly the same as previous
except that the moving observer at B has his own
and that he looks at this clock at t2 when he sees
the reflected signal from the mirror (at A).
However total reflection time t2 measured
is not equal to t1 previous but smaller.
To be more precise t2=t1*sqr(1-v*v/c*c)
with t1=L *(1/(c-v) + 1/(c-v))
and L=L0 *sqr(1-v*v/c*c)
You will get t2 = 2*L0/c
Or t2 is the same as t in experiment 1.
That means the length of a rod measured at rest in rest frame
= As the length of a moving rod measured with a moving clock
= As the length of a rod at rest in a moving frame.
For more details go to:
http://users.pandora.be/nicvroom/
and select "Changing length" introduction
Specific study: "Example"
A small comment.
What experiment 2 tells you that the moving rod is contracted
"in theory".
The question is when you perform this experiment in reality
is this than also true.
The problem is there are two ways to move a rod:
You can pull the rod from the front with a speed v
You can push the rod from the back with a speed v
Both should give the same results when you perform this experiment.
I doubt this because it assumes that force (because that is what
you use in order to change the speed of the rod) propagates
instantaneous through the whole rod, which in turn results
that the whole rod has the same speed v instantaneous
and over the whole length instaneous contracts.
Nick
Eamon
< regurgitated musings snipped >
> can someone answer this question?
======================================
Can anyone question this answer?
Eamon
齯滌`偕爻,虜,齯滌`偕爻,虜,齯滌`偕爻,虜,齯滌`偕爻,虜,齯滌`偕爻,虜,齯滌
The Meek Shall Inherit the Earth.............................
................er, if that's all right with the rest of you.
chaverondier
DAKAL...@aol.com wrote in message news:<6a.1ef2de0...@aol.com>...
>
> Hi. I'd like to ask for your thoughts on the following situation: Suppose
> there is a rod of length L as measured by an observer in the rest frame of
> the rod (he measures the proper length and does so at his leisure and by any
> method he prefers). Suppose there is another observer for which th rod moves
> with velocity V along, say, the x-axis. He also wants to measure the length
> of rod. However, instead of using the Einsteinian method of noting the
> simultaneous (in his frame) positions of the end-points by way of two
> synchronized (in his frame) clocks at two locations, he does the following:
> He has ONE stopwatch at ONE location. When the front of the rod passes by he
> starts the clock, when the back end of the rod passes by he stops the clock.
> Now he says "Ah, the length of the rod is L' = V * dt, where dt is the
> interval recorded by my stopwatch". Why should there be a Lorentz contraction
> (for this observer) in this case?
> The 'relativity of simultaneity' doesn't even enter the procedure at any
> point. Why won't L = L' if the moving rod is measured by the method I
> propose? Since the proper length of the rod could have been measured by
> laying down unit rods, there is only one clock used throughout the scenario.
> If length contraction follows from the relativity of simultaneity, as
> incorporated by Eintein's admittedly contrived method of measuring moving
> rods, then why should it be observed in this case since here simultaneity
> plays no role at all? Thanks Much to all those who might reply....
> Demetrios Kalamidas email: hor...@hotmail.com
Even
* if you choose the time flowing in some preferred frame as
absolute time(for instance the time flowing in the inertial frame
where the CMBR is seen isotropic),
* if you perform an absolute synchronisation process of any clock
of any inertial frame with regard to this preferred time instead of
the classical relativist synchronisation process,
you would nevertheless get a Lorentz distance contraction.
Some people call this absolute synchronisation version of Special
Theory of Relativity the LORENTZ Ether Theory because this Special
Theory of Relativity version is compatible with velocity addition and
consequently with the existence of an aether.
Within LET framework, moving observers and observers at absolute rest
agree : the LORENTZ length contraction is no more relative. Moving
bodies contract and bodies at absolute rest don't.
Bernard CHAVERONDIER
http://membres.lycos.fr/bchaverondier
DAKALAMIDAS
>There is a rod which is of length L as measured by an observer O in its rest
frame. This observer measures, according to SR, the proper length of the rod at
can do so at his leisure and by any method he chooses. Suppose there is another
observer O' for which the rod moves with speed V along, say, the x-axis. O'
also wants to measure the length of the moving rod. Let's say that these two
observers
have never heard of Einstein or SR etc. O' wants to answer a weird question
which he came up with: Does the length of a moving object change with respect
to its length when measured at rest? O' decides to measure the length of the
moving rod in the following way: He has ONE stopwatch at ONE location. When the
front end of the rod passes by him he starts the watch; when the back end of
the rod passes by him he stops the watch. Then he says "Ah, the length of the
rod is L' = V * dt, where dt is the time interval recorded on my stopwatch."
Now,simultaneity plays no role here since there is only ONE clock at ONE
location being used in the whole scenario; O could have measured the proper
length by laying down unit rods way in advance. Why then should there be a
length contraction if, according to SR, length contraction follows from the
relativity
of simultaneity? If one uses Einstein's contrived method of noting the
simultaneous (in the frame of O') positions of the end-points of the moving
rod, thereby being FORCED to use TWO synchronized (in the frame of O') clocks
at TWO locations on the x-axis, then of course it follows from the constancy
of c that L > L'. But in the scenario stated here the relativity of
simultaneity does not enter the measuring procedure AT ALL. SR provides no
answer as to whether L = L' or not in this case, I believe. Only by performing
the actual experiment can the question be answered. If L > L' with the
ONE-CLOCK method,then it does not follow from SR but must be attributed to some
other model (Lorentzian perhaps). Can someone explain to me how length
contraction follows from SR when the length of a moving rod is measured in the
exact (no modifications) manner stated here?
The point I want to make (and the thing that confuses me) is the following:
Let's assume that both observers, O and O', know that (1) the laws of nature
are the same in all inertial frames and (2) that c is constant in all inertial
frames. In other words, they have come to know the two postulates of relativity
but have never heard of Einstein or of SR. The observer O, on the rod, measures
the length way in advance by laying down unit rods and gets value L. Observer
O' has only ONE clock.
O' first determines V: when the rod is still far off he sends a light pulse to
bounce off the front of the rod. When the pulse returns
he notes the elapsed time (roundtrip) and waits, say, 1 second before he sends
out another pulse. Knowing c, the two roundtrip times, and the interval between
the arrival of the first pulse and the departure of the second pulse O' can
determine V. O' then determines L' by the method in the original post. So there
is onle ONE clock at ONE location in ONE frame that is used in the whole
scanario.
How does length contraction follow from this EXACT METHOD of doing things?
Where is the required 'relativity of simultaneity' hidden in this procedure?
Remember, all that we have are: facts (postulates) (1) and (2) above and the
stated method for getting V and L' ; according to SR we should be able to
DERIVE length contraction and time dilation --- after all that's what we are
told all the time: that all relativistic effects follow from the two
postulates. So, there shouldn't be phrases "...a clock in the rod's frame
would read..." or "...the transformed length is..." or "...the invariant
space-time interval requires..." etc. etc.
in the given explanations... these are the things that have to follow from the
method outlined above, if SR is logically consistent, I believe.
I appreciate the responses to the original post...thanks to all of you who
responded
Demetrios
Oriel36
Demetrios,
How fast does pi as a proportion move ?
How fast does a second move, seeing that a 'clock' is 86 400 seconds
or 1 second is a proportion of 1 day of 86 400 seconds ?.
Kees Roos
> I'd like to make some clarifications on the original post (below) that I
sent:
>
> Demetrios
>
Actually you replace the relativity of simultaneity whch you want
to avoid in your experiment with another phenomenon which also results
from the Lorentz transformation.
In the traditional thought experiment the stationary observer measures
the spatial separation between two events which have zero temporal
separation, both separations relative to her/his frame of reference.
Relatively moving observers will disagree that the two events were
simultaneous: Relativity of simultaneity.
In your modified thought experiment your stationary observer measures
the temporal separation between two events which have zero spatial
separation, both separations relative to her/his frame of reference.
Relatively moving observers will disagree that the two events had zero
spatial separation, let's call it 'Relativity of standstill'
SR is not inconsistent in this respect:
Both phenomena will yield the same result - Length contraction.
--
Regards, Kees Roos
Paul Cardinale
>
> >There is a rod which is of length L as measured by an observer O in its rest
> frame. This observer measures, according to SR, the proper length of the rod at
> can do so at his leisure and by any method he chooses. Suppose there is another
> observer O' for which the rod moves with speed V along, say, the x-axis. O'
> also wants to measure the length of the moving rod. Let's say that these two
> observers
> have never heard of Einstein or SR etc. O' wants to answer a weird question
> which he came up with: Does the length of a moving object change with respect
> to its length when measured at rest? O' decides to measure the length of the
> moving rod in the following way: He has ONE stopwatch at ONE location. When the
> front end of the rod passes by him he starts the watch; when the back end of
> the rod passes by him he stops the watch. Then he says "Ah, the length of the
> rod is L' = V * dt, where dt is the time interval recorded on my stopwatch."
> Now,simultaneity plays no role here since there is only ONE clock at ONE
> location being used in the whole scenario; O could have measured the proper
> length by laying down unit rods way in advance. Why then should there be a
> length contraction if, according to SR, length contraction follows from the
> relativity
> of simultaneity?
>
derived from the postulates of relativity. Length contraction does
not come from RS, rather, both length contraction and RS come from the
postulates of SR.
Paul Cardinale
Paul B. Andersen
"DAKALAMIDAS" <dakal...@aol.com> wrote in message news:20020506194858...@mb-ml.aol.com...
> I'd like to make some clarifications on the original post (below) that I sent:
>
> >There is a rod which is of length L as measured by an observer O in its rest
> frame. This observer measures, according to SR, the proper length of the rod at
> can do so at his leisure and by any method he chooses. Suppose there is another
> observer O' for which the rod moves with speed V along, say, the x-axis. O'
> also wants to measure the length of the moving rod. Let's say that these two
> observers
> have never heard of Einstein or SR etc. O' wants to answer a weird question
> which he came up with: Does the length of a moving object change with respect
> to its length when measured at rest? O' decides to measure the length of the
> moving rod in the following way: He has ONE stopwatch at ONE location. When the
> front end of the rod passes by him he starts the watch; when the back end of
> the rod passes by him he stops the watch. Then he says "Ah, the length of the
> rod is L' = V * dt, where dt is the time interval recorded on my stopwatch."
> Now,simultaneity plays no role here since there is only ONE clock at ONE
> location being used in the whole scenario; O could have measured the proper
> length by laying down unit rods way in advance. Why then should there be a
> length contraction if, according to SR, length contraction follows from the
> relativity of simultaneity?
I have shown you the derivation of SR's prediction before.
It follows from the Lorentz transform which follows from the postulates.
Length contraction does not follow from simultaneity of relativity,
and the relativity of simultaneity does not follow from length contraction.
They both follows from the postulates; they are two sides of the same coin.
> If one uses Einstein's contrived method of noting the
> simultaneous (in the frame of O') positions of the end-points of the moving
> rod, thereby being FORCED to use TWO synchronized (in the frame of O') clocks
> at TWO locations on the x-axis, then of course it follows from the constancy
> of c that L > L'. But in the scenario stated here the relativity of
> simultaneity does not enter the measuring procedure AT ALL. SR provides no
> answer as to whether L = L' or not in this case, I believe.
You are wrong.
I have shown you the answer of SR, and I suspect others have as well.
Are you ignoring the answers?
> Only by performing
> the actual experiment can the question be answered. If L > L' with the
> ONE-CLOCK method,then it does not follow from SR but must be attributed to some
> other model (Lorentzian perhaps). Can someone explain to me how length
> contraction follows from SR when the length of a moving rod is measured in the
> exact (no modifications) manner stated here?
>
> The point I want to make (and the thing that confuses me) is the following:
>
> Let's assume that both observers, O and O', know that (1) the laws of nature
> are the same in all inertial frames and (2) that c is constant in all inertial
> frames. In other words, they have come to know the two postulates of relativity
> but have never heard of Einstein or of SR. The observer O, on the rod, measures
> the length way in advance by laying down unit rods and gets value L. Observer
> O' has only ONE clock.
> O' first determines V: when the rod is still far off he sends a light pulse to
> bounce off the front of the rod. When the pulse returns
> he notes the elapsed time (roundtrip) and waits, say, 1 second before he sends
> out another pulse. Knowing c, the two roundtrip times, and the interval between
> the arrival of the first pulse and the departure of the second pulse O' can
> determine V. O' then determines L' by the method in the original post. So there
> is onle ONE clock at ONE location in ONE frame that is used in the whole
> scanario.
You are describing a real experiment.
Possible in principle if not in practise.
> How does length contraction follow from this EXACT METHOD of doing things?
As this is a real experiment, the question is what the two
observers would measure.
Do you think they would measure the same length?
> Where is the required 'relativity of simultaneity' hidden in this procedure?
> Remember, all that we have are: facts (postulates) (1) and (2) above and the
> stated method for getting V and L' ; according to SR we should be able to
> DERIVE length contraction and time dilation --- after all that's what we are
> told all the time: that all relativistic effects follow from the two
> postulates. So, there shouldn't be phrases "...a clock in the rod's frame
> would read..." or "...the transformed length is..." or "...the invariant
> space-time interval requires..." etc. etc.
> in the given explanations... these are the things that have to follow from the
> method outlined above, if SR is logically consistent, I believe.
Of course you can derive what the observers would measure.
I have shown you the derivation before.
So what is your problem?
Do you want the derivation of the LT from the postulates?
Paul
Henry Wilson
>I'd like to make some clarifications on the original post (below) that I sent:
>
>>There is a rod which is of length L as measured by an observer O in its rest
>frame. This observer measures, according to SR, the proper length of the rod at
>can do so at his leisure and by any method he chooses. Suppose there is another
>observer O' for which the rod moves with speed V along, say, the x-axis. O'
>also wants to measure the length of the moving rod. Let's say that these two
>observers
>I appreciate the responses to the original post...thanks to all of you who
>responded
> Demetrios
Why do you need the observer O'?
What's wrong with this?
Observer O first measures the rod length by comparing it with his standard.
He then relocates the rod some distance away prior to letting it move back
towards him at a velocity that he measures using the method you stated.
Knowing v, he then calculates the length of the rod from the difference in
clock readings when its two ends pass his one clock.
The truth is he will not observe any change in rod length because none
occurs. Lorentz contractions are non-existent figments of SRian imagination
following directly from the unproven postulate that OWLS is constant for
all moving observers.
Henri Wilson's free thought Laboratory,
At the frontier of scientific invention.
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www.users.bigpond.com/rmrabb/HW.htm