1 "DAKALAMIDAS" |
The Length of a Moving Rod | maandag 29 april 2002 5:49 |

2 "Barry" |
Re: The Length of a Moving Rod | maandag 29 april 2002 6:22 |

3 "DAKALAMIDAS" |
Re: The Length of a Moving Rod | maandag 29 april 2002 8:26 |

4 "Paul B. Andersen" |
Re: The Length of a Moving Rod | maandag 29 april 2002 14:06 |

5 "Paul B. Andersen" |
Re: The Length of a Moving Rod | maandag 29 april 2002 14:19 |

6 "Barry" |
Re: The Length of a Moving Rod | maandag 29 april 2002 15:51 |

7 "John Rennie" |
Re: The Length of a Moving Rod | maandag 29 april 2002 20:42 |

8 "DAKALAMIDAS" |
Re: The Length of a Moving Rod | woensdag 1 mei 2002 1:38 |

9 "Bilge" |
Re: The Length of a Moving Rod | woensdag 1 mei 2002 5:38 |

10 "John Rennie" |
Re: The Length of a Moving Rod | woensdag 1 mei 2002 19:59 |

11 "Paul B. Andersen" |
Re: The Length of a Moving Rod | donderdag 2 mei 2002 10:41 |

12 "Alexander Belov" |
Re: The Length of a Moving Rod | donderdag 2 mei 2002 20:43 |

13 "rryker1" |
Re: The Length of a Moving Rod | vrijdag 3 mei 2002 1:21 |

14 "DAKALAMIDAS" |
Re: The Length of a Moving Rod | vrijdag 3 mei 2002 6:03 |

15 "Nicolaas Vroom" |
Re: The Length of a Moving Rod | vrijdag 3 mei 2002 11:41 |

16 "Eamon" |
Re: The Length of a Moving Rod | vrijdag 3 mei 2002 13:53 |

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? 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

> |
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? |

How did you measure V?

Barry

How is V measured in SR accounts? Where in Einstein's writings, or in the
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?

"DAKALAMIDAS"

> |
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? 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. 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"

> | How is V measured in SR accounts? Where in Einstein's writings, or in the 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? |

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

> |
How is V measured in SR accounts? Where in Einstein's writings, or in the 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? |

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

Failure of simultaneity is not essential for length contraction. It 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

On 29 Apr 2002 03:49:45 GMT, dakalamidas@aol.com (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 |

> |
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? |

can someone answer this question?

DAKALAMIDAS said some stuff about 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.

--

> | can someone answer this question? |

Maybe you missed my post, here it is again.

----8<--------------------

Failure of simultaneity is not essential for length contraction. It 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

On 29 Apr 2002 03:49:45 GMT, dakalamidas@aol.com (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 |

"DAKALAMIDAS"

> > |
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? |

> |
can someone answer this question? |

It has been answered. If you missed it, here is my answer again:

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

> | 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 |

> | SR provides no answer as to whether L = L' or not in this case, I believe. |

> | Why then should there be a length contraction if, according to SR, length contraction follows from the relativity of simultaneity? |

> | 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? |

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

> > | 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 |

> | You must use two clocks to measure the speed V. Simultaneity is 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 |

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

> |
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'.

"DAKALAMIDAS"

> |
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? 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 |

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

dakalamidas@aol.com (DAKALAMIDAS) wrote in message news:<20020430193837.02356.00003537@mb-fn.aol.com>...

< regurgitated musings snipped >

> | can someone answer this question? |

======================================

Can anyone question this answer?

Eamon

ø¤º°`°º¤ø,¸¸,ø¤º°`°º¤ø,¸¸,ø¤º°`°º¤ø,¸¸,ø¤º°`°º¤ø,¸¸,ø¤º°`°º¤ø,¸¸,ø¤º°

The Meek Shall Inherit the Earth.............................

................er, if that's all right with the rest of you.

Created: 6 May 2002

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