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Vadim Matveev
2006-Feb-12, 08:06 PM
1) The common case of the relative motion

Let’s consider a long rod with a letter A on one of its ends and a letter B on its body.
The rod appears as follows:

A__________B__________

Let us mentally saw off the piece of the rod right so that we obtain the remainder of the rod limited by the letters A and B – such one:

А___________В

We shall name this remainder “remainder AB”.
Let us visualize that the remainder AB, moving in a certain inertial reference system K’ with a longitudinal velocity v, possesses the length L’.
It is clear that the proper length of the remainder AB (the length in the reference system K, in which the remainder AB rests) is equal to GL’, where G is the Lorentz coefficient (G>1). We can say that the longitudinal velocity and the length of the remainder AB are relative.

2) Another case of the relative motion

Let us suppose that the rod is manufactured from combustible material and the remainder of the rod is made not by the sawing, but by the fallowing way.
Let us assume that the end, which does not have the letter designation, is set on fire at a certain moment of time, and the rod begins rapidly to burn up to its complete combustion. During the combustion of the rod the process of combustion continuously moves toward the end A of the rod. At a certain moment of time the process of combustion reaches the letter B and for a moment the remainder with the letters A and B on its ends appears – such one:

A___________B*

The mark * at the end of the objects symbolically means the continuously burning of the end of the rod.
We shall name this remainder “remainder AB”.
It is clear in this case too that the proper length L of the remainder AB is equal to GL’ and the longitudinal velocity and the length of the remainder AB are relative.

3) A case against the relative motion and relative length of the remainder AB

Let us consider a long combustible rod with letters A and B on its body moving in an inertial reference system K’ with a longitudinal velocity v, moreover the rod is burning rapidly from two ends - thus:

* ___________А__________В__________*

Suppose that at a certain moment (simultaneously in the reference system K’) the processes of combustion reach the letters A and B and for a moment appears the remainder

*A___________B*

with letters A and B on its ends. The marks * at the ends of the objects symbolically mean the continuously burning of the ends of the rod.
We shall name this remainder “remainder AB”
The length of the remainder AB of the burning rod in the inertial reference system K’ is equal to L’.
Now answer a question.
How large is the own length of the remainder AB of the rod burning from two sides (case 3)?
In my opinion, the remainder AB has no own length, if we understand by it the length of the resting remainder AB. The proper length L of the remainder AB is not equal to GL’ The remainder AB cannot rest, because in the reference systems, where the rod burning from two ends rests (or moves with a speed, which is not equal to the above-mentioned speed v) the remainder AB can not be discovered. In my opinion the speed and the length of the remainder are relative in the first case, where the remainder AB does not burn and in the second case, where the remainder AB burns from one side, but in the third case, where the remainder AB burns from both sides they (the speed and the length) are absolute.
I think that everybody who understands the cause of the difference between the first two "correct" cases on the one hand and the "abnormal" third case on the other hand, will understand much more.

hhEb09'1
2006-Feb-12, 10:07 PM
In my opinion the speed and the length of the remainder are relative in the first case, where the remainder AB does not burn and in the second case, where the remainder AB burns from one side, but in the third case, where the remainder AB burns from both sides they (the speed and the length) are absolute.but why is that your opinion?

clj4
2006-Feb-13, 06:47 AM
1) The common case of the relative motion


Let us visualize that the remainder AB, moving in a certain inertial reference system K’ with a longitudinal velocity v, possesses the length L’.
It is clear that the proper length of the remainder AB (the length in the reference system K, in which the remainder AB rests) is equal to GL’, where G is the Lorentz coefficient (G>1). We can say that the longitudinal velocity and the length of the remainder AB are relative.

1) Another case of the relative motion




You seem to have a lot of problems with:

1. Understanding special relativity in general (see your posting on one-way light speed measurements)
1. Understanding that length contraction is not length dilation. (see your first case 1)
1. Proper case numbering. (after case 1 follows case 2, not case 1 again)

Vadim Matveev
2006-Feb-13, 10:43 PM
You seem to have a lot of problems with:

1. Understanding special relativity in general


Who does not have those problems?



(see your posting on one-way light speed measurements)

You understand I have seen it. It is all correctly there.



1. Understanding that length contraction is not length dilation. (see your first case 1)


Read attentively. If the proper length L of an object is bigger then the length L’ of the same moving object (L>L’), then the length L’ of the moving object it smaller then the proper length L (L’<L).
It is length contraction - not length dilation.



1. Proper case numbering. (after case 1 follows case 2, not case 1 again)

You are right. Thank you. I have corrected the numbering.
By the way your numbering has only one number - number 1 - in three items.

clj4
2006-Feb-13, 11:10 PM
By the way your numbering has only one number - number 1 - in three items.

Yes, it was intentional.

Now, I am going to use a suggestion by Nereid and ask you a few simple questions:

1.What does burning and sawing have to do with length contraction?
2.Why do you use length dilation instead of its usual expression ?
3. Not clear where the rod is, in K or in K' ? can you repost by breaking things into shorter sentences?
4.Why do you need two frames K, K'?
5. What are you trying to prove with your question (your newly relabelled point 3 in your edited post)?

tusenfem
2006-Feb-14, 03:15 PM
1.What does burning and sawing have to do with length contraction?

Well, clj4, you cannot denie that burning off or sawing off part of the rod will to a shorter length, and who knows, in some sort of physics you may consider this also a contraction.

clj4
2006-Feb-14, 05:00 PM
...yeah, right
:dance:

Vadim Matveev
2006-Feb-15, 10:46 AM
1.What does burning and sawing have to do with length contraction?


With length contraction of the rod nothing.



2.Why do you use length dilation instead of its usual expression ?


I do not. I consider the usual expressions: L=GL’ or L’= (1/G)L, where L is the proper length and L’ – the length of moving rod. G>1, but (1/G)<1.
Where did you find the length dilation?



3. Not clear where the rod is, in K or in K' ? can you repost by breaking things into shorter sentences?


The Rod? What does mean “where - in K or in K'?
The rod is in K, in K', in K” and in the others.
Don’t mix up the rod and the remainder AB of the rod. The considered of me remainder AB of the rod is only in K’.

4.Why do you need two frames K, K'?

???



5. What are you trying to prove with your question (your newly relabelled point 3 in your edited post)?


I answered the question.
The remainder AB in case 3 has no own length, if we understand by it the length of the resting remainder AB. The remainder AB cannot rest – it can only move with speed v’ and can only have the length L’.
Lorentz’s formula does not work in this case. The proper length L of the remainder AB is not equal to GL’.
You don’t like formula L=GL’, where G>1. For you I can write formula L’=(1/G)L, where L is the proper length of the remainder AB, L’ – the length of the remainder AB of the rod moving in K’, (1/G)<1. This formula does not work too. The remainder AB (the remainder with simultaneously burning marks A and B) has no own length, because it can not be observed as a moving object. It is impossible to observe simultaneously the burning marks A and B in K.

clj4
2006-Feb-15, 02:30 PM
This formula does not work too. The remainder AB (the remainder with simultaneously burning marks A and B) has no own length, because it can not be observed as a moving object. It is impossible to observe simultaneously the burning marks A and B in K.

So what's your point? Do you think that you generated an experiment that , due to the well known lack of absolute simultaneity refutes the Lorentz transforms? this is your point?

Vadim Matveev
2006-Feb-16, 04:04 PM
So what's your point? Do you think that you generated an experiment that , due to the well known lack of absolute simultaneity refutes the Lorentz transforms? this is your point?

No. I show only a case, where Lorentz’s formula L’=(1/G)L does no work and the length of an object has not another values in another reference systems. That’s all.
From my point of view, my thought experiment does not refute Lorentz transforms.
Let us go step by step. Let us first examine two cases - the second and the third. Did you understand the essence of the experiment?
Are you agreeable to the fact that formula L’=(1/G)L is applicable to the remainder AB of the rod (2-d case) burning from one side, but not applicable to the remainder AB of the rod burning from two sides (3-d case)?
Are you agreeable to the fact that the length of the remainder AB in the third case does have only one value?

clj4
2006-Feb-16, 07:24 PM
No. I show only a case, where Lorentz’s formula L’=(1/G)L does no work and the length of an object has not another values in another reference systems. That’s all.

You haven't shown anything, there is no valid math associated with your text.




Are you agreeable to the fact that formula L’=(1/G)L is applicable to the remainder AB of the rod (2-d case) burning from one side, but not applicable to the remainder AB of the rod burning from two sides (3-d case)?

No, I don't agree with anything that you are saying. Why would that be true? You haven't explained anything mathematically. No math, no agreement on anything you are claiming. Please write down the equations



Are you agreeable to the fact that the length of the remainder AB in the third case does have only one value?
No equations=no possibility of agreement (see above) . Write down the math.

lek
2006-Feb-16, 09:03 PM
No. I show only a case, where Lorentz’s formula L’=(1/G)L does no work and the length of an object has not another values in another reference systems. That’s all.
From my point of view, my thought experiment does not refute Lorentz transforms.
Let us go step by step. Let us first examine two cases - the second and the third. Did you understand the essence of the experiment?
Are you agreeable to the fact that formula L’=(1/G)L is applicable to the remainder AB of the rod (2-d case) burning from one side, but not applicable to the remainder AB of the rod burning from two sides (3-d case)?
Are you agreeable to the fact that the length of the remainder AB in the third case does have only one value?

The burning rod idea is difficult in any, and every reference frame. To simplify things a bit, a burning rod (one end or two) is just a measured length which varies as function of time.

Whether it burns on one end or two is irrelevant. Your definition of its length in any reference frame depends on observing its length by monitorin both ends simultaniously (the problem with simultaenity is well known, as previously mentioned ...).
Since light has a finite speed, your distance "remainder AB" doesnt give same results even within same reference frame, it depends on where you are, not only which reference frame you are in, so speaking about difference with different reference frames is quite irrelevant.

Vadim Matveev
2006-Feb-19, 09:54 PM
You haven't shown anything, there is no valid math associated with your text.


Read my text. I asked: ” How large is the own length of the remainder AB of the rod burning from two sides (case 3)?” Nothing more.
I don’t prove or demonstrate anything. I ask.
If you can answer, try to answer my question with “valid math” or without it, if not, then not. It is an exercise with only one formula.
If you don’t now formula L’=(1/G)L, where L is proper length, L’ – the length of moving rod, G>1, then my question is not for you.
If you think that formula L=GL’, where G>1, is a formula of length dilation, as you wrote, then my question is also not for you.
My opinion, which I mentioned, is only my opinion – not my demonstration.

clj4
2006-Feb-20, 01:45 AM
Read my text. I asked: ” How large is the own length of the remainder AB of the rod burning from two sides (case 3)?” Nothing more.
I don’t prove or demonstrate anything. I ask.

You are NOT asking. Have another look at the title of your thread: "A case against relativity and Lorentz's Formula".
You are trying to prove a case, albeit through a misunderstood paradox.


If you can answer, try to answer my question with “valid math” or without it, if not, then not. It is an exercise with only one formula.

You have been asked to answer. But I will give you an answer: your question makes no sense. Because you do not understand the notion of relative simulataneity.


If you don’t now formula L’=(1/G)L, where L is proper length, L’ – the length of moving rod, G>1, then my question is not for you.

Your question makes no sense


My opinion, which I mentioned, is only my opinion – not my demonstration.

Look at the title of your post: "A case against relativity and Lorentz's Formula" you were trying to disprove the Lorentz transforms and relativity. It was explained to you several times that your "experiment" disregards that the process of simultaneity is relative, in other words your "experiment" is incorrectly set. You managed to prove one thing only: that you do not understand that , the act of the flames arriving in A and B simultaneously in one frame (the frame in which the rod is at rest) is NOT simultaneous in ANY other frame. This makes your so-called "experiment" (or paradox) totally irrelevant.

Ken G
2006-Feb-20, 03:44 AM
I think what these people are saying, Vadim, is that you are correct in thinking that having the rod burn at both ends is a complication that cuts to the heart of relativity concepts. But it does not say that any of those concepts are wrong, it just shows how tricky they can be. I hope you digested what lek said, because this is crucial-- the desynchronization of the clocks plays with what you mean by the length of the rod at a given time. A simple way to see this is to light both ends of the moving rod at the moment its center passes a given stationary observer. In the moving frame, the ends are lit at the same time, but in the stationary frame, the back end of the rod is lit first. So not only is the length of time the rod has been burning not agreed on by the two observers due to time dilation, but the stationary observer even has to correct for the fact that for a little while the rod was only burning from one end, when he/she calculates its length as a function of time. Furthermore, even though the back of the rod was lit first for the stationary observer, that observer will see the two flames reach the center at the same time. To make the velocity addition formula work out, and to reach the center at the same time, the back would have to be lit first! Amazingly, these things always work out when you do the full calculation, and any conclusion one draws before doing the full calculation is suspect. I suspect this is related to the problem you are having. I commend you for coming up with a new and interesting test of relativity, but your argument that there is any kind of problem is not convincing-- do the full calculation.

johntsang
2006-Feb-20, 11:46 AM
Dear Vadim:

You did offer an interesting view of how the rod is reducing in length in a moving reference frame, that's quite exciting .... with a G factor to reduce it ....

It needs more explanation from your side to connect it with the Lorentz’s formulas , if you say they are wrong, I just have difficulty in visualize it from your words .
Can you explain it by something I can understand, please.
Ask yourself this (my question ) ... it's a little mental experiment:
Before they are moving, a round ball, and ,a plane with a round hole of same radius.

Now, move against each other ... sidewise.
Right at the "time" the ball is closing to the hole........

1) you are on the ball, you have propulson, can you get into the hole ?
2) You are on the plane with that hole, move the plane, can you get the ball into the hole ?

Where is the contraction / dilation stated in that Lorentz’s formula ?

Vadim Matveev
2006-Feb-20, 10:35 PM
The burning rod idea is difficult in any, and every reference frame. To simplify things a bit, a burning rod (one end or two) is just a measured length which varies as function of time.


Why is it difficult? I think it is rather simply.



Whether it burns on one end or two is irrelevant. Your definition of its length in any reference frame depends on observing its length by monitorin both ends simultaniously (the problem with simultaenity is well known, as previously mentioned ...).


I recognize the problem of simultaneity (problem of synchronization of clocks), but in this thread I proceed from Einstein’s version of equality of the speed of lights in opposite directions.



Since light has a finite speed, your distance "remainder AB" doesnt give same results even within same reference frame,


Why not? It does.


it depends on where you are, not only which reference frame you are in, so speaking about difference with different reference frames is quite irrelevant.

Yes, my visual and subjective perception of simultaneity depends on where I am located, but the observer in SRT is not a single observer, but plenty of observers with clocks. If two events in points A and B are simultaneous in your reference frame, then they are simultaneous also when somebody of observers is standing nearer to the point A and receives the signal from the point A earlier, then from the point B. The relativity of simultaneity does not mean indeterminacy of the simultaneity in the same frame.

clj4
2006-Feb-20, 10:51 PM
This website requires that you prove your point (with math and logic) or that you get off.
You have received multiple explanations as to why your so-called experiment is wrong.
So, prove your point mathematicall or cease and desist. It is not for you to question us, it is for us to question you. And, so far, you failed to produce a valid answer.

Nereid
2006-Feb-20, 11:06 PM
This website requires that you prove your point (with math and logic) or that you get off.
You have received multiple explanations as to why your so-called experiment is wrong.
So, prove your point mathematicall or cease and desist. It is not for you to question us, it is for us to question you. And, so far, you failed to produce a valid answer.While this is a reasonably accurate summary of what this ATM section of BAUT is 'about', IMHO it is better to challenge those promoting ATM ideas to defend their cases.

If said defence is (internally) inconsistent, or inconsistent with mainstream theories where the domains of applicability overlap, or (above all) inconsistent with good observational and experimental results, then the best attack is (again, IMHO, as Nereid the BAUT member, not the mod) to highlight those inconsistencies, using the ATM proponent's own words.

In this case, I feel that Vadim Matveev has demonstrated misunderstandings of SR; perhaps it is appropriate to ask him to 'show his working' (i.e. lay out his case in sufficient detail that the 'disconnects' between the theory - SR - and his interpretation become crystal clear)?

clj4
2006-Feb-20, 11:15 PM
In this case, I feel that Vadim Matveev has demonstrated misunderstandings of SR; perhaps it is appropriate to ask him to 'show his working' (i.e. lay out his case in sufficient detail that the 'disconnects' between the theory - SR - and his interpretation become crystal clear)?

I fully agree. Unfortunately, every time we ask this he hides behind " I am only asking a question" We have asked him directly multiple times (yourself included) to prove his case. Every time he has come back with the same answer. " I am only asking a question" which is clearly not the truth. What else can we do?

clj4
2006-Feb-20, 11:57 PM
From my point of view, my thought experiment does not refute Lorentz transforms.

This statement clearly contradicts the title of your thread "A case against relativity and Lorentz's Formula".
So, one last time: what are you trying to prove? And please use math (if you can), your English is too ambigous to understand.

lek
2006-Feb-21, 06:22 PM
Why is it difficult? I think it is rather simply.

I recognize the problem of simultaneity (problem of synchronization of clocks), but in this thread I proceed from Einstein’s version of equality of the speed of lights in opposite directions.


Since light has a finite speed, your distance "remainder AB" doesnt give same results even within same reference frame,

Why not? It does.

Simultaneity doesn't exist even within one reference frame because observers at different distance to events see them at different time, so it isn't just about clocks. Although if you know the distances, you can calculate your way out of the "problem"(*)
*this same correction applied to the calculation of observed result in moving frame pretty much solves the problem



If two events in points A and B are simultaneous in your reference frame, then they are simultaneous also when somebody of observers is standing nearer to the point A and receives the signal from the point A earlier, then from the point B. The relativity of simultaneity does not mean indeterminacy of the simultaneity in the same frame.

How exactly are these events simultanious if they aren't observed at the same time?

Nereid
2006-Feb-21, 10:27 PM
Perhaps there is, as clj4 and others have already pointed out, some confusion over what this thread is about.

So let me ask Vadim Matveev, the OP, for clarification:
To the extent that it is possible to characterise a primary intention, is your primary intention to ask questions about how to analyse a particular - hypothetical - physical situation, in terms of SR? challenge SR, by providing a - hypothetical - physical situation which highlights an internal inconsistency in SR? challenge SR, by providing a - real, in principle capable of being tested in the lab - physical situation which highlights an inconsistency between SR and the real world? something else (please specify)?If it's the first, then you have posted to the wrong section (and I'll move this thread to the Q&A section); if the second or third, then you must adhere to the BAUT rules (http://www.bautforum.com/showthread.php?t=32864) (and those concerning this ATM section in particular) - which you agreed to when you became a BAUT member - and answer all direct questions relevant to your ATM claim, in a timely manner.

Vadim Matveev
2006-Feb-22, 05:30 PM
I think what these people are saying, Vadim, is that you are correct in thinking that having the rod burn at both ends is a complication that cuts to the heart of relativity concepts. But it does not say that any of those concepts are wrong, it just shows how tricky they can be.


Yes, it is so.

Vadim Matveev
2006-Feb-22, 05:39 PM
Perhaps there is, as clj4 and others have already pointed out, some confusion over what this thread is about.
So let me ask Vadim Matveev, the OP, for clarification:
To the extent that it is possible to characterise a primary intention, is your primary intention to ask questions about how to analyse a particular - hypothetical - physical situation, in terms of SR? challenge SR, by providing a - hypothetical - physical situation which highlights an internal inconsistency in SR? challenge SR, by providing a - real, in principle capable of being tested in the lab - physical situation which highlights an inconsistency between SR and the real world? something else (please specify)?If it's the first, then you have posted to the wrong section (and I'll move this thread to the Q&A section); if the second or third, then you must adhere to the BAUT rules (http://www.bautforum.com/showthread.php?t=32864) (and those concerning this ATM section in particular) - which you agreed to when you became a BAUT member - and answer all direct questions relevant to your ATM claim, in a timely manner.

Ken G. wrote: «…you are correct in thinking that having the rod burn at both ends is a complication that cuts to the heart of relativity concepts. But it does not say that any of those concepts are wrong, it just shows how tricky they can be.”
In my opinion, it cuts relativity, but doesn’t disprove the Lorentz transforms and STR.
You can move my thread to Q&A section or cancel it. It is your right.
Now personally to you. You wrote that I misunderstand SR.
What is false in my article?
I wrote.
Consider two inertial reference systems K and K’, moving with velocity v relative one to another.
Consider a rod resting in the system K (the rod is laying in the axis X). The proper length of the rod (the length of the resting rod) is equal to L. In this case the length L' of the same rod in the system K’, where the rod moves with velocity v, is L(1-v^2/c^2)^1/2. Is it not right?

Now consider a rod with marks A and B. The rod lying on the axis X in the reference systems K is burning from two sides.

* ___________А__________В__________*

At a certain moment of time the processes of combustion reach the letters A and B and for a moment the remainder with the letters A and B on its ends appears remainder AB.

*A___________B*

The length of the remainder AB resting in the system K (the proper length) is equal L.
Can we say, that the length L' of the same remainder AB moving in the system K’ is equal to L(1-v^2/c^2)^1/2?
No, we can’t! Why?
Because of relativity of simultaneity the remainder AB does not exist in the system K’. In this system the marks A and B cannot burn simultaneously.
Is it not right? What must I explain else? I don’t know. Do you?
What can I answer to the comment of clj4: “You managed to prove one thing only: that you do not understand that , the act of the flames arriving in A and B simultaneously in one frame (the frame in which the rod is at rest) is NOT simultaneous in ANY other frame”?
It is a mock, but no comment. It for little children, not for me.

clj4
2006-Feb-22, 11:37 PM
Vadim,

You keep contradicting yourself, sometimes within the same sentence:


In my opinion, it cuts relativity, but doesn’t disprove the Lorentz transforms and STR.


Could you please:
- stop evading the questions
- stop answering our questions with questions
- answer Nereid's questions asked here:
http://www.bautforum.com/showpost.php?p=687004&postcount=23

Thank you