Be careful, check it out ;-)

Dirk Vdm

"Dirk Van de moortel" schreef in bericht news: wHti8.245624$rt4.23988@afrodite.telenet-ops.be...

>	"Nicolaas Vroom" wrote in message news: QOoi8.244807$rt4.23850@afrodite.telenet-ops.be...

> >	If rigid rods do not exist what happens then if you take a rod (2l) in the middle at point m and you move the rod such that point m has a speed v, are both parts (the front part (l) and the back (l)) contracted in the same manner ?

>	As soon as the rod is not accelerating anymore, yes.

Acceleration means that the speed of point m (the point in the middle) increases. Suppose this takes place during a period t At t acceleration stops. This means that the speed of point m becomes fixed. Before that moment the situation below exists. The front part is shorter as the back part (But IMO the situation is more complicated) If you claim that immediate after that moment the length of both parts become equal than some form of FTL is involved.

See also https://www.nicvroom.be/length4.htm (Changing length part 4)

>	During the acceleration the leading part will be a bit shorter than the trailing part because the point where you pull will be moving towards the front and away from the back before the front and the back start moving. (I assume you want the rod to be moving along its own length).

> >	If rigid rods exists I expect (accept) that the length of both parts will be equal. Because rigid rods do not exist I "see" a problem.

>	So you think that the lengths cannot be equal? That is Henry Wilson logic. P ==> Q is not equivalent with not P ==> not Q but with: not Q ==> not P

Sorry.
I do not understand.

Nick.

"Nicolaas Vroom" wrote in message news:3mMi8.247864$rt4.23837@afrodite.telenet-ops.be...

>	"Dirk Van de moortel" schreef in bericht news:wHti8.245624$rt4.23988@afrodite.telenet-ops.be..

> >	"Nicolaas Vroom" wrote in message news:QOoi8.244807$rt4.23850@afrodite.telenet-ops.be...

> > >

If rigid rods do not exist what happens then if you take a rod (2l) in the middle at point m and you move the rod such that point m has a speed v, are both parts (the front part (l) and the back (l)) contracted in the same manner ?

> >	As soon as the rod is not accelerating anymore, yes.

>

Acceleration means that the speed of point m (the point in the middle) increases. Suppose this takes place during a period t At t acceleration stops. This means that the speed of point m becomes fixed. Before that moment the situation below exists. The front part is shorter as the back part (But IMO the situation is more complicated) If you claim that immediate after that moment the length of both parts become equal than some form of FTL is involved.

Not immediately after, that's right. There will be some vibration. Just think of a spring. Every rod you make will be made of atoms. An influence on some part of the rod will take time to travel to the other parts of the rod. The travelling takes time.

>	See also https://www.nicvroom.be/length4.htm (Changing length part 4)

Sorry, don't have the time to check it all out.

> >	During the acceleration the leading part will be a bit shorter than the trailing part because the point where you pull will be moving towards the front and away from the back before the front and the back start moving. (I assume you want the rod to be moving along its own length).

> > >

If rigid rods exists I expect (accept) that the length of both parts will be equal. Because rigid rods do not exist I "see" a problem.

> >	So you think that the lengths cannot be equal? That is Henry Wilson logic. P ==> Q is not equivalent with not P ==> not Q but with: not Q ==> not P

>	Sorry. I do not understand.

Elementary logic. From what you wrote, you seem to think
that "P ==> Q" is equivalent with "not P ==> not Q"

Do you understand that

Dirk Vdm

"Dirk Van de moortel" schreef in bericht news:WrRi8.248827$rt4.24079@afrodite.telenet-ops.be...

>	"Nicolaas Vroom" wrote in message news:3mMi8.247864$rt4.23837@afrodite.telenet-ops.be..

> >	"Dirk Van de moortel" schreef in bericht news:wHti8.245624$rt4.23988@afrodite.telenet-ops.be..

> > >

"Nicolaas Vroom" wrote in message news:QOoi8.244807$rt4.23850@afrodite.telenet-ops.be...

> > > >

If rigid rods do not exist what happens then if you take a rod (2l) in the middle at point m and you move the rod such that point m has a speed v, are both parts (the front part (l) and the back (l)) contracted in the same manner ?

> > >

As soon as the rod is not accelerating anymore, yes.

> >

Acceleration means that the speed of point m (the point in the middle) increases. Suppose this takes place during a period t At t acceleration stops. This means that the speed of point m becomes fixed. Before that moment the situation below exists. The front part is shorter as the back part (But IMO the situation is more complicated) If you claim that immediate after that moment the length of both parts become equal than some form of FTL is involved.

>	Not immediately after, that's right. There will be some vibration. Just think of a spring. Every rod you make will be made of atoms. An influence on some part of the rod will take time to travel to the other parts of the rod. The travelling takes time.

I was expection such a comment. That is why I wrote above:
IMO the situation is more complicated.
I like your idea (analogy) of using springs.
Consider a rod of 20 springs of equal length.
Point m is between the end of spring 10 and the beginning of spring 11.

1. move point m a distance dl/dt towards spring 11
spring 11 will contract with that amount dl
spring 10 expand with that same amount dl

2. move point m a distance 2dl/dt towards spring 11
spring 11 will contract with that amount 2dl and expand with an amount dl (total contraction 2dl)
spring 12 will contract with amount dl
spring 10 will expands with that same amount 2dl and contract with an amount dl (total expansion 2dl)
spring 9 will contract with amount dl.

3. move point m a distance 3dl/dt towards spring 11
Spring 11 total contraction 3dl, spring 12: 2dl and spring 13 dl
Spring 10 total expansion 3dl, sping 9: 2dl and spring 8: 1dl

4. move point m a distance 3dl/dt towards spring 11
Spring 11 total contraction 3dl, spring 12: 3dl
spring 13 2dl spring 14 1dl
Spring 10 total expansion 3dl, sping 9: 3dl
spring 8: 2dl and spring 7 1dl.

Finally spring 11 to 20 are all contracted each with 3dl
Sping 1 to 10 are all expanded with 3dl
To get a better idea go to:

https://www.nicvroom.be/length4.htm (Changing length part 4) and study the paragraph: Description of an Experiment https://www.nicvroom.be/length4.htm#exp1 (you have to combine the middle parts of each of the two sketches)

In fact the front part is contracted in total with 30 dl and the back part is expanded with 30 dl BUT the total length stays the same.

Thats not what we "want". Accordingly to SR both parts should contract.

Nick

Nicolaas says...

>	I like your idea (analogy) of using springs. Consider a rod of 20 springs of equal length. Point m is between the end of spring 10 and the beginning of spring 11. 1. move point m a distance dl/dt towards spring 11 spring 11 will contract with that amount dl spring 10 expand with that same amount dl

[stuff deleted]

I don't think that your description of the situation in https://www.nicvroom.be/length4.htm#exp1 is completely correct. Your web page seems to say that length contraction only occurs if you push on one end of a spring, but that if you pull on the other end, you get length expansion, instead. That's true of springs, but it isn't the same thing as SR length contraction.

Qualitatively, here is what happens when you push on the left end of a long spring and then let go.

1. The left end of the spring moves faster than the right end, so the spring compresses (slowing down the left end).
2. The compression wave moves down the spring to the right.
3. When the wave reaches the right end, that end moves forward.
4. Typically at this point, the right end starts moving faster than the left end. So the spring expands (slowing the left end).
5. The expansion wave moves down the spring to the left.
6. When the wave reaches the left end, that end starts accelerating.
7. Typically at this point, the left end starts moving faster than the right end, and so the spring compresses again.

So, if you shove the left end, the spring will compress, then expand, then compress, then expand, etc. If instead you pulled on the right end, the spring will *first* expand, then compress, then expand, etc.

These expansions and compressions are *not* to be confused with the "length contraction" of Special Relativity. In either case, eventually internal friction will stop the expansion/compression cycle, and the spring will have a new equilibrium length. It is this equilibrium length that is predicted by SR to be square-root(1 - (v/c)^2) times the size of the original spring (as measured from the inertial reference frame in which the spring is originally at rest).

Relativistic length contraction is a prediction about the *equilibrium* length, as measured in some frame, of an object that is moving at velocity v relative to that frame. It isn't about transient compressions or expansions.

-- Daryl McCullough
CoGenTex, Inc.
Ithaca, NY

Nicolaas Vroom wrote:

>	ande452@attglobal.net schreef in bericht news: 3C805A4D.755B@attglobal.net...

> >	Eugene Shubert wrote:

> > >

I recall reading ages ago why rigid rods can't exist in SR. Is there an easy to read proof that their existence is impossible? I understand why there's an acceleration limit g = (c^2)/L in being able to accelerate a rigid rod with length L. With that constraint being satisfied, I assume someone can show me how it would be possible for a future space scientist to build a simple apparatus that could send simple yes/no messages into the past. Any takers?

> >

There's no theoretical proof of the non-existence of anything. You're missing the point. If you can experimentally demonstrate the existence of a rigid rod (or even a rod in which the speed of sound is > c) then you have experimentally falsified SR. SR doesn't "prove" the non-existence of rigid rods. SR is false if you can show that a rigid rod exists.

>

If rigid rods do not exist what happens then if you take a rod (2l) in the middle at point m and you move the rod such that point m has a speed v, are both parts (the front part (l) and the back (l)) contracted in the same manner ? If rigid rods exists I expect (accept) that the length of both parts will be equal. Because rigid rods do not exist I "see" a problem.

SR doesn't describe how long something is before and after you accelerate it. It describes how the length of the SAME object appears to different observers moving at different velocities wrt the object.

If you accelerate an extended object, the length before and after acceleration depends on how you accelerate it, not just on what frame of reference you observe it in. This is all described in the FAQ entry on Bell's "paradox".

An extended object before and after acceleration is not the same object as seen in two different frames of reference.

John Anderson

"Daryl McCullough" schreef in bericht news:a6j8nr01i9p@drn.newsguy.com...

>

I don't think that your description of the situation in https://www.nicvroom.be/length4.htm#exp1 is completely correct. Your web page seems to say that length contraction only occurs if you push on one end of a spring, but that if you pull on the other end, you get length expansion, instead. That's true of springs, but it isn't the same thing as SR length contraction. Qualitatively, here is what happens when you push on the left end of a long spring and then let go.

First you push slowly, then faster and finally you push and keep the speed constant. (for the other spring, first you pull slowly, then faster and finally you pull with a constant speed)

>

1. The left end of the spring moves faster than the right end, so the spring compresses (slowing down the left end). 2. The compression wave moves down the spring to the right. 3. When the wave reaches the right end, that end moves forward. 4. Typically at this point, the right end starts moving faster than the left end. So the spring expands (slowing the left end). 5. The expansion wave moves down the spring to the left. 6. When the wave reaches the left end, that end starts accelerating. 7. Typically at this point, the left end starts moving faster than the right end, and so the spring compresses again. So, if you shove the left end, the spring will compress, then expand, then compress, then expand, etc. If instead you pulled on the right end, the spring will *first* expand, then compress, then expand, etc.

I agree that a spring is not the same as a rod to explain and discuss SR. Consider a rod of 1 light min. Suppose you push a distance dl in dt against that rod. (in the next dt etc you do the same) Q1 When will you see the front end moving ? IMO when this disturbance moves at speed c after 1 min.

In stead of 1 rod of 1 light min use now 2 rods of 1 light min. Those rods make contact at point m. The same as above at point m you push a distance dl in dt against the first rod and you pull the second rod. (in the NEXT dt etc you do the same) Q2 When will you see the back end of the second rod moving ? IMO when this disturbance moves at speed c after 1 min.

That means during the first minute the total length of the two rods stays constant.

Q3 How much does the front end of the first rod move in the first dt after the front start moving ?

If the answer is dl and if dt is 1 min than this means that the front rod contracts with a distance dl. (1)

Q4 How much does the back end of the second rod move in the first dt after the back start moving ?

If the answer is dl and if dt is 1 min than this means that the back rod expands with a distance dl. (2)

If the front rod is contracted with a distance dl and the back rod expanded with a distance dl than the total length of the rod stays constant.

However this is in conflict with SR. Both rods should contract. Specific the back rod (the pulled one) should contract. Most probably (2) is wrong it should be more than dl. But (1) can also be wrong.

>

These expansions and compressions are *not* to be confused with the "length contraction" of Special Relativity. In either case, eventually internal friction will stop the expansion/compression cycle, and the spring will have a new equilibrium length. It is this equilibrium length that is predicted by SR to be square-root(1 - (v/c)^2) times the size of the original spring (as measured from the inertial reference frame in which the spring is originally at rest).

And this should be the case if you push against the back or pull at the front with a speed v in the rest frame.

My point is that this is difficult to understand in physical sense.

Nick.

ande452@attglobal.net schreef in bericht news: 3C8DA025.DCE@attglobal.net...

>	Nicolaas Vroom wrote:

> >

>	SR doesn't describe how long something is before and after you accelerate it. It describes how the length of the SAME object appears to different observers moving at different velocities wrt the object.

My interest is to the length of a moving object as it appears to an observer at rest. Under two conditions: If you push and if you pull with the same speed v in the rest frame.

>	If you accelerate an extended object, the length before and after acceleration depends on how you accelerate it, not just on what frame of reference you observe it in. This is all described in the FAQ entry on Bell's "paradox".

Which FAQ ? Bell's Spaceship Paradox? http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html
Can Special Relativity handle accelerations? http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

I doubt if "how you accelerate" has anything to do with it. If the rest length is lo and the final speed is v than the the final length accordingly to SR is "a function of Lorentz transformation" independent of "how you accelerate" it.

The issue is physically how do both rods reach that length if you push one and and pull the other. (Assuming the rods are not rigid)

>	An extended object before and after acceleration is not the same object as seen in two different frames of reference.

Only the rest frame is studied.

Nicolaas says...

>	"Daryl McCullough" wrote:

>>	So, if you shove the left end, the spring will compress, then expand, then compress, then expand, etc. If instead you pulled on the right end, the spring will *first* expand, then compress, then expand, etc.

>	I agree that a spring is not the same as a rod to explain and discuss SR.

I didn't say that. A rod *is* the same thing as a spring. The only difference is that the "spring constant" for a rod is so large that the stretching and contracting is tiny. But it isn't nonzero. If you push one end of a rod, the rod will contract slightly (and then vibrate---expand then contract then expand). If you pull one end, it will stretch (and then vibrate).

>	Consider a rod of 1 light min. Suppose you push a distance dl in dt against that rod. (in the next dt etc you do the same) Q1 When will you see the front end moving ? IMO when this disturbance moves at speed c after 1 min.

Right. At that moment, the rod will be compressed slightly. But afterwards, it will expand again, and then compress, etc.

>

In stead of 1 rod of 1 light min use now 2 rods of 1 light min. Those rods make contact at point m. The same as above at point m you push a distance dl in dt against the first rod and you pull the second rod. (in the NEXT dt etc you do the same) Q2 When will you see the back end of the second rod moving ? IMO when this disturbance moves at speed c after 1 min. That means during the first minute the total length of the two rods stays constant.

Yes, you're right. The total length is constant for the first minute.

>	Q3 How much does the front end of the first rod move in the first dt after the front start moving ? If the answer is dl and if dt is 1 min than this means that the front rod contracts with a distance dl. (1)

>

Q4 How much does the back end of the second rod move in the first dt after the back start moving ? If the answer is dl and if dt is 1 min than this means that the back rod expands with a distance dl. (2) If the front rod is contracted with a distance dl and the back rod expanded with a distance dl than the total length of the rod stays constant.

>	However this is in conflict with SR.

No, it's not! As I said, the transient compression and stretching of a rod following being pushed or pulled has *nothing* to do with the length contraction of SR. It's only

>	Both rods should contract.

SR doesn't say any such thing. A spring doesn't contract when you pull one end---it stretches. The same is true of a rod. But if you wait, then the spring will later contract again, and then vibrate (expanding, contracting, etc.) until internal friction causes its vibrations to die down.

>	Specific the back rod (the pulled one) should contract.

SR doesn't say that. If you displace the middle of a rod to the right by 1 centimeter, then the right half of the rod will compress, while the left half will expand. Then later the whole rod will vibrate (expanding and contracting etc.). Eventually, the length of the rod will settle down to its equilibrium length.

>	Most probably (2) is wrong it should be more than dl. But (1) can also be wrong.

>>

These expansions and compressions are *not* to be confused with the "length contraction" of Special Relativity. In either case, eventually internal friction will stop the expansion/compression cycle, and the spring will have a new equilibrium length. It is this equilibrium length that is predicted by SR to be square-root(1 - (v/c)^2) times the size of the original spring (as measured from the inertial reference frame in which the spring is originally at rest).

>	And this should be the case if you push against the back or pull at the front with a speed v in the rest frame.

I believe that it is true.

>	My point is that this is difficult to understand in physical sense.

It is not difficult to understand this: If you pull on one end of a rod, it will stretch slightly. When it reaches its maximum length, it will "spring back" and start to compress again. It then will reach its minimum length, and will then start to expand again.

So, when you pull a rod from one end, the rod's length will vibrate back and forth between a minimum length and a maximum length. Eventually, it will settle down to a length that is somewhere between the two.

-- Daryl McCullough CoGenTex, Inc. Ithaca, NY

"Daryl McCullough" schreef in bericht news:a6lc3902pcp@drn.newsguy.com...

>	Nicolaas says...

> >	"Daryl McCullough" wrote:

>

> > >

So, if you shove the left end, the spring will compress, then expand, then compress, then expand, etc. If instead you pulled on the right end, the spring will *first* expand, then compress, then expand, etc.

> >	I agree that a spring is not the same as a rod to explain and discuss SR.

>	I didn't say that. A rod *is* the same thing as a spring.

I agree

>

The only difference is that the "spring constant" for a rod is so large that the stretching and contracting is tiny. But it isn't nonzero. If you push one end of a rod, the rod will contract slightly (and then vibrate---expand then contract then expand). If you pull one end, it will stretch (and then vibrate).

I agree

> >	Consider a rod of 1 light min. Suppose you push a distance dl in dt against that rod. (in the next dt etc you do the same) Q1 When will you see the front end moving ? IMO when this disturbance moves at speed c after 1 min.

>	Right. At that moment, the rod will be compressed slightly. But afterwards, it will expand again, and then compress, etc.

I hope you realize that "I" do not push (pull) dl in dt once (and then stop) but continue to do that. ie in the next dt "I" push the rod again with dl etc etc. That means in the first dt there is acceleration in the next dt "I" push with a constant speed

"You" could also push in each dt: dl, 2*dl, 3*dl, 3*dl, 3*dl etc

> >	In stead of 1 rod of 1 light min use now 2 rods of 1 light min. That means during the first minute the total length of the two rods stays constant.

>	Yes, you're right. The total length is constant for the first minute.

> >

Q3 How much does the front end of the first rod move in the first dt after the front start moving ? Q4 How much does the back end of the second rod move in the first dt after the back start moving ? If the front rod is contracted with a distance dl and the back rod expanded with a distance dl than the total length of the rod stays constant.

>

> >	However this is in conflict with SR.

>	No, it's not! As I said, the transient compression and stretching of a rod following being pushed or pulled has *nothing* to do with the length contraction of SR. It's only

Yes it is (or at least it seems) when v>0

> >	Both rods should contract.

>	SR doesn't say any such thing. A spring doesn't contract when you pull one end---it stretches. The same is true of a rod. But if you wait, then the spring will later contract again, and then vibrate (expanding, contracting, etc.) until internal friction causes its vibrations to die down.

Yes it does (or at least it seems) when v>0 The problem is that both rods do not behave symmetric The front rod has the tendancy/preference to be contracted. The back rod has the tendancy/preference to be expanded. Accordingly to SR they both should be contracted.

> >	Specific the back rod (the pulled one) should contract.

>

SR doesn't say that. If you displace the middle of a rod to the right by 1 centimeter, then the right half of the rod will compress, while the left half will expand. Then later the whole rod will vibrate (expanding and contracting etc.). Eventually, the length of the rod will settle down to its equilibrium length.

This is true if you push (pull) for a fixed interval = n*dt and then make v = 0 (ie dl=0).
The point of this exercise is that the speed of point m is constant and v>0
"You" also can not speak of one whole rod. When v=0 point m (front second, back first) does not move.
When v becomes zero only the back of second rod and the front of first rod still can move (ie vibrate)
When v>0 there is "no communication" between front and back.

> >	My point is that this is difficult to understand in physical sense.

The true issue is how long after, when v becomes >0 and constant, become the length of both rods equal and in accordance to SR.

The same issue exists when v becomes 0.

(For RIGID RODS the answer is instantaneous ?)

Nick

"Nicolaas Vroom" wrote in message news:...

>	I doubt if "how you accelerate" has anything to do with it.

How you accelerate an object affects how it is stressed, and that affects the deformations of the object (which must occur because the rod isn't rigid).

Paul Cardinale

"Paul Cardinale" schreef in bericht news:64050551.0203131103.4217e2ba@posting.google.com...

>	"Nicolaas Vroom" wrote in message news:..

> >	I doubt if "how you accelerate" has anything to do with it.

>	How you accelerate an object affects how it is stressed, and that affects the deformations of the object (which must occur because the rod isn't rigid).

I (can) expect that deformation is a function of the speed v but it is difficult that this deformation is a function of the amount of time it took to reach that speed. ie that the deformation is different if you slowly reached a certain speed v versus if you did it fast.

I can understand that the amount of deformation is different for different types of rods (iron versus wood)

However this particle discussion does not solve the issue first that initially it makes a difference if you push or pull the rod. The pushed rod is initially contracted. The pulled rod is initially expanded.

Secondly that finally that the length of both rods should be the same, assuming that the speed of the point where the pushed rod is pushed (at the end) and where the pulled rod is pulled (at the beginning) is the same and constant. (previous this is called point m)

Nick

Nicolaas Vroom wrote:

>	"Paul Cardinale" schreef in bericht news:64050551.0203131103.4217e2ba@posting.google.com..

> >	"Nicolaas Vroom" wrote in message news:..

> > >

I doubt if "how you accelerate" has anything to do with it.

> >	How you accelerate an object affects how it is stressed, and that affects the deformations of the object (which must occur because the rod isn't rigid).

>	I (can) expect that deformation is a function of the speed v but it is difficult that this deformation is a function of the amount of time it took to reach that speed.

That's not what he is saying. He's saying that the deformation depends on how you accelerate each part of the object. If you start accelerating different parts of the object at different times in some frame, then the object is deformed.

This deformation has nothing to do with relativistic length contraction which doesn't compare an object before and after acceleration. It compares an object in the same state of motion as observed in different frames of reference.

Again, this is discussed in the FAQ entry on Bell's "paradox".

John Anderson

ande452@attglobal.net schreef in bericht news:3C902E09.6C7@attglobal.net...

>	Nicolaas Vroom wrote:

> >	"Paul Cardinale" schreef in bericht news:64050551.0203131103.4217e2ba@posting.google.com..

> > >

"Nicolaas Vroom" wrote in message news:..

> > > >

I doubt if "how you accelerate" has anything to do with it.

> > >

How you accelerate an object affects how it is stressed, and that affects the deformations of the object (which must occur because the rod isn't rigid).

> >	I (can) expect that deformation is a function of the speed v but it is difficult that this deformation is a function of the amount of time it took to reach that speed.

>	That's not what he is saying. He's saying that the deformation depends on how you accelerate each part of the object. If you start accelerating different parts of the object at different times in some frame, then the object is deformed.

I agree that If you accelerate different parts (The front the back) of the same rod at different times (first front later back) in the rest frame then the object is deformed.

But that is not the issue of my posting

I want to study two rods with in the rest frame make contact at point m

      rod2         m      rod 1
<----------------><---------------->
                    -----> pull/push direction.

During that whole period (a>0) front rod is contracted and back rod is expanded.
When v > 0 and constant (a=0) both rods should contract. How is this explained ?

A whole different issue is what happens if "you" accelerate all points of one rod at the same time (simultaneous) in the rest frame

>	This deformation has nothing to do with relativistic length contraction which doesn't compare an object before and after acceleration. It compares an object in the same state of motion as observed in different frames of reference. Again, this is discussed in the FAQ entry on Bell's "paradox".

Which FAQ ? Bell's Spaceship Paradox? http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html
Can Special Relativity handle accelerations? http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

Nick

"Nicolaas Vroom" wrote in message news:htKj8.426$DE4.147@afrodite.telenet-ops.be...

>	Yes it does (or at least it seems) when v>0 The problem is that both rods do not behave symmetric The front rod has the tendancy/preference to be contracted. The back rod has the tendancy/preference to be expanded. Accordingly to SR they both should be contracted.

You are confusing two different phenomena. One thing is the elastic deformations of the rod when it is set in motion. These are physical deformations which can be measured in a frame of reference in which the - say - midpoint of the rod is stationary. Note that these deformations will lead to mechanical stresses in the rod.

A quite different matter is the Lorentz contraction due to the motion of the rod. This is NOT a physical deformation, but is only due to that the length of the rod is measured in a frame in which the rod is moving. (It's a consequence of the relativity of simultaneity.) The Lorentz contraction does not lead to mechanical stresses in the rod; it does in fact not physically affect the rod in any way.

Think of it this way: When you set the rod in motion by pulling it from its midpoint, different part of the rod will have different speeds, and different parts will be physically deformed to a different degree. When you measure the length of the rod, the different parts of the rod will be Lorentz contracted according to the speed that part of the rod has, and note that it is the deformed length that is Lorentz contracted. So the part of the rod that is physically stretched will still be Lorentz contracted, e.g. measured to be a little shorter than the stretched length measured in the frame of the rod. Thus the Lorentz contracted rod may still be longer than the rest length of the not deformed rod.

Paul

"Nicolaas Vroom" wrote in message news:...

>	ande452@attglobal.net schreef in bericht news:3C902E09.6C7@attglobal.net..

> >	Nicolaas Vroom wrote:

> > >

"Paul Cardinale" schreef in bericht news:64050551.0203131103.4217e2ba@posting.google.com..

> > > >

"Nicolaas Vroom" wrote in message news:..

> > > > >

I doubt if "how you accelerate" has anything to do with it.

> > > >

How you accelerate an object affects how it is stressed, and that affects the deformations of the object (which must occur because the rod isn't rigid).

> > >

I (can) expect that deformation is a function of the speed v but it is difficult that this deformation is a function of the amount of time it took to reach that speed.

You've got it backwards. You may be confusing "contraction" with "deformation". Speed contracts, acceleration (more properly, the force necessary to accelerate) deforms (unless the acceleration can be applied uniformly).

> >	That's not what he is saying. He's saying that the deformation depends on how you accelerate each part of the object. If you start accelerating different parts of the object at different times in some frame, then the object is deformed.

>	I agree that If you accelerate different parts (The front the back) of the same rod at different times (first front later back) in the rest frame then the object is deformed. But that is not the issue of my posting

Yes, it is. See below.

>	I want to study two rods with in the rest frame make contact at point m rod2 m rod 1 <----------------><----------------> -----> pull/push direction. At point m (back of rod1, front of rod 2) "I" push against rod 1 and "I" pull rod 2 When I push rod 1 is compressed and rod 2 expands.

Note that the ends of the rods opposite m, will accelerate later than m.
This is because it takes time for the force of acceleration to propagate down the rods.

>

(This is the case when a<>0) In the first dt point m moves a distance dl In the second dt point m moves a distance 2dl In the third dt point m moves a distance 3dl Starting from the fourth dt we keep v>0 and constant ie distance is 3dl In first, second and third dt a>0 During that whole period (a>0) front rod is contracted and back rod is expanded. When v > 0 and constant (a=0) both rods should contract. How is this explained ?

Deformation: Since no rods are perfectly elastic, you may expect that their will be some residual deformation from the acceleration.

Contraction: Length contraction as a function of velocity is a prediction of SR.
Are you asking how this is derived from the postulates?

>	A whole different issue is what happens if "you" accelerate all points of one rod at the same time (simultaneous) in the rest frame

The rod would be stretched or torn apart. If the rod could endure no stretching, each point where acceleration is applied would rip off one piece.

Paul Cardinale

"Nicolaas Vroom" wrote in message news:...

>

I want to study two rods with in the rest frame make contact at point m rod2 m rod 1 <----------------><----------------> -----> pull/push direction. At point m (back of rod1, front of rod 2) "I" push against rod 1 and "I" pull rod 2 When I push rod 1 is compressed and rod 2 expands. (This is the case when a<>0) In the first dt point m moves a distance dl In the second dt point m moves a distance 2dl In the third dt point m moves a distance 3dl Starting from the fourth dt we keep v>0 and constant ie distance is 3dl In first, second and third dt a>0 During that whole period (a>0) front rod is contracted and back rod is expanded. When v > 0 and constant (a=0) both rods should contract. How is this explained ?

When the shock waves hit the far ends of the rods, the one hitting the front end will be more severe, because of the Doppler effect. Thus, the front rod will expand more rapidly than the back rod contracts. This will tend to cancel out the initial differences in length, so that when the two rods have settled down, they will be the same length.

Nicolaas Vroom wrote:

>	ande452@attglobal.net schreef in bericht news:3C902E09.6C7@attglobal.net..

> >	Nicolaas Vroom wrote:

> > >

"Paul Cardinale" schreef in bericht news:64050551.0203131103.4217e2ba@posting.google.com..

> > > >

"Nicolaas Vroom" wrote in message news:..

> > > > >

I doubt if "how you accelerate" has anything to do with it.

> > > >

How you accelerate an object affects how it is stressed, and that affects the deformations of the object (which must occur because the rod isn't rigid).

> > >

I (can) expect that deformation is a function of the speed v but it is difficult that this deformation is a function of the amount of time it took to reach that speed.

> >	That's not what he is saying. He's saying that the deformation depends on how you accelerate each part of the object. If you start accelerating different parts of the object at different times in some frame, then the object is deformed.

>

I agree that If you accelerate different parts (The front the back) of the same rod at different times (first front later back) in the rest frame then the object is deformed. But that is not the issue of my posting I want to study two rods with in the rest frame make contact at point m rod2 m rod 1 <----------------><----------------> -----> pull/push direction. At point m (back of rod1, front of rod 2) "I" push against rod 1 and "I" pull rod 2 When I push rod 1 is compressed and rod 2 expands. (This is the case when a<>0) In the first dt point m moves a distance dl In the second dt point m moves a distance 2dl In the third dt point m moves a distance 3dl Starting from the fourth dt we keep v>0 and constant ie distance is 3dl In first, second and third dt a>0 During that whole period (a>0) front rod is contracted and back rod is expanded. When v > 0 and constant (a=0) both rods should contract. How is this explained ? A whole different issue is what happens if "you" accelerate all points of one rod at the same time (simultaneous) in the rest frame

> >	This deformation has nothing to do with relativistic length contraction which doesn't compare an object before and after acceleration. It compares an object in the same state of motion as observed in different frames of reference. Again, this is discussed in the FAQ entry on Bell's "paradox".

>	Which FAQ ?

The FAQ for this newsgroup.

>	Bell's Spaceship Paradox?

Yes, I sort of remember that this is what it's called there. The idea is to look at two spaceships attached by a cable and accelerated "identically" in some particular frame. The cable is the same length in the original rest frame, but stretched in the locally comoving frame.

That means that the acceleration process has changed the length of this system in its local rest frame. It's a different object as you acclerate it. The internal stresses in it change.

Also, as I and others have pointed out, this has nothing to do with length contraction in SR which considers the length of something in the same state of motion wrt different frames of reference. The object itself is not accelerated.

John Anderson

"Paul B. Andersen" schreef in bericht news:a6q27p$lb9$1@snipp.uninett.no...

>	"Nicolaas Vroom" wrote in message news:htKj8.426$DE4.147@afrodite.telenet-ops.be..

> >	Yes it does (or at least it seems) when v>0 The problem is that both rods do not behave symmetric The front rod has the tendancy/preference to be contracted. The back rod has the tendancy/preference to be expanded. Accordingly to SR they both should be contracted.

>

You are confusing two different phenomena. One thing is the elastic deformations of the rod when it is set in motion. These are physical deformations which can be measured in a frame of reference in which the - say - midpoint of the rod is stationary. Note that these deformations will lead to mechanical stresses in the rod.

Does that imply that the rod which is pushed with a constant speed v is contracted in the rest frame ? Does that imply that the rod which is pulled with a constant speed v expands in the rest frame ? IMO it does. This contraction is may be not constant all the time (when v = constant) but it exists. The same for expansion.

>	A quite different matter is the Lorentz contraction due to the motion of the rod. This is NOT a physical deformation, but is only due to that the length of the rod is measured in a frame in which the rod is moving. (It's a consequence of the relativity of simultaneity.)

In the frame of the rod there are 2 effects: Length contraction and Time dilation (rel to rest frame) Both effects "cancel" each other that is why you measure the length of that rod in the space ship (moving frame) the length is the same when the space ship is at rest.

In fact SR works the other way around. It starts from the concept that you should measure the same (equivalence principle) As a result length contraction and time dilation should cancel each other.
(I should say the "rules" of SR, Lorentz transformation, are such that this is the case)

IMO both length contraction and time dilation are physical effects.

>	The Lorentz contraction does not lead to mechanical stresses in the rod; it does in fact not physically affect the rod in any way.

Please have a look at https://www.nicvroom.be/calc1.htm and study the time t1 ie question 1 The time t1 is the time that a pulse transmitted from the back reaches the front of a moving rod which has a speed v. The time t1 is measured in the rest frame. Is this value equal to the first value of line three, or to the second value of line five (in the table) ?

(This url also demonstrates that in the moving frame you measure "the same" as in the rest frame)

>	Think of it this way: When you set the rod in motion by pulling it from its midpoint, different part of the rod will have different speeds, and different parts will be physically deformed to a different degree.

Okay
IMO the front rod is physical deformed (with v = constant) that its length is contracted with a distance dl1
IMO the back rod is physical deformed (with v = constant) that its length is expanded with a distance dl2
Both dl1 and dl2 are measured in the rest frame.

>	When you measure the length of the rod, the different parts of the rod will be Lorentz contracted according to the speed that part of the rod has, and note that it is the deformed length that is Lorentz contracted.

The front rod has a deformed length of l0-dl1
The back rod has a deformed length of l0+dl2 with v being constant.
Both dl1 and dl2 are measured in the rest frame.

>	So the part of the rod that is physically stretched will still be Lorentz contracted, e.g. measured to be a little shorter than the stretched length measured in the frame of the rod.

I can measure the length moving rod in two ways:
In the rest frame using clocks in the rest frame.
In the moving frame using clocks in the moving frame.

I prefer to measure everything in the rest frame. In the rest frame a moving rod l0 has a length l (l = l0/ gamma) ie it has contracted with a distance dl0

When you push a rod (v) and you measure the length of the rod in the rest frame the rod will be contracted with dlx.
The question is now how much of this dlx is related to dl0 (Lorentz contraction) and how much to dl1 (deformation)

When you pull a rod (v) and you measure the length of the rod in the rest frame the length of rod will be changed with dlx.
The question is now how much of this dlx is related to dl0 (Lorentz contraction) and how much to dl2 (deformation/expansion)

Because both effects have an opposite sign the result can be zero.

>	Thus the Lorentz contracted rod may still be longer than the rest length of the not deformed rod.

I agree If the rod is pulled.

However this also could imply that dl0 part is very small and even could be zero.

Nick.

"Bennett Standeven" schreef in bericht news:24c3076b.0203142123.11f7b174@posting.google.com...

>	"Nicolaas Vroom" wrote in message news:..

> >

I want to study two rods with in the rest frame make contact at point m rod2 m rod 1 <----------------><----------------> -----> pull/push direction. At point m (back of rod1, front of rod 2) "I" push against rod 1 and "I" pull rod 2 When I push rod 1 is compressed and rod 2 expands. (This is the case when a<>0) In the first dt point m moves a distance dl In the second dt point m moves a distance 2dl In the third dt point m moves a distance 3dl Starting from the fourth dt we keep v>0 and constant ie distance is 3dl In first, second and third dt a>0 During that whole period (a>0) front rod is contracted and back rod is expanded. When v > 0 and constant (a=0) both rods should contract. How is this explained ?

>

When the shock waves hit the far ends of the rods, the one hitting the front end will be more severe, because of the Doppler effect. Thus, the front rod will expand more rapidly than the back rod contracts. This will tend to cancel out the initial differences in length, so that when the two rods have settled down, they will be the same length.

You must consider each rod separately. You should also consider the length of the rod when both rods have the same speed v (of point m) IMO the front rod is physical contracted and the back rod is physical expanded (when each rod has the same constant speed v) The amount of contraction expansion are respectivily dl1 and dl2. It is possible that if you add both effects they could cancell each other but that is NOT the issue of this posting.

Nick

ande452@attglobal.net schreef in bericht news:3C918910.576D@attglobal.net...

>	Nicolaas Vroom wrote:

> >

I want to study two rods with in the rest frame make contact at point m rod2 m rod 1 <----------------><----------------> -----> pull/push direction. At point m (back of rod1, front of rod 2) "I" push against rod 1 and "I" pull rod 2 When I push rod 1 is compressed and rod 2 expands. (This is the case when a<>0) In the first dt point m moves a distance dl In the second dt point m moves a distance 2dl In the third dt point m moves a distance 3dl Starting from the fourth dt we keep v>0 and constant ie distance is 3dl In first, second and third dt a>0 During that whole period (a>0) front rod is contracted and back rod is expanded. When v > 0 and constant (a=0) both rods should contract. How is this explained ? A whole different issue is what happens if "you" accelerate all points of one rod at the same time (simultaneous) in the rest frame

> > >

Again, this is discussed in the FAQ entry on Bell's "paradox".

> >	Which FAQ ?

>	The FAQ for this newsgroup.

> >	Bell's Spaceship Paradox?

>	Yes, I sort of remember that this is what it's called there. The idea is to look at two spaceships attached by a cable and accelerated "identically" in some particular frame. The cable is the same length in the original rest frame, but stretched in the locally comoving frame.

This has nothing to do with this.
I want to study the difference between when you push or pull a rod with a constant speed v. The idea is that the the rods are all the time touching each other and that the point of contact has this constant speed v in the rest frame. (I call this the point m ie the point at the middle of the two rods)

My point is that the rod that is pushed is contracted while the rod that is pulled expands.
Accordingly to SR both rods should contract. How do you explain this.

>

That means that the acceleration process has changed the length of this system in its local rest frame. It's a different object as you acclerate it. The internal stresses in it change. Also, as I and others have pointed out, this has nothing to do with length contraction in SR which considers the length of something in the same state of motion wrt different frames of reference. The object itself is not accelerated. John Anderson

Nicolaas Vroom wrote:

>	ande452@attglobal.net schreef in bericht news:3C918910.576D@attglobal.net..

> >	Nicolaas Vroom wrote:

> > >

I want to study two rods with in the rest frame make contact at point m rod2 m rod 1 <----------------><----------------> -----> pull/push direction. At point m (back of rod1, front of rod 2) "I" push against rod 1 and "I" pull rod 2 When I push rod 1 is compressed and rod 2 expands. (This is the case when a<>0) In the first dt point m moves a distance dl In the second dt point m moves a distance 2dl In the third dt point m moves a distance 3dl Starting from the fourth dt we keep v>0 and constant ie distance is 3dl In first, second and third dt a>0 During that whole period (a>0) front rod is contracted and back rod is expanded. When v > 0 and constant (a=0) both rods should contract. How is this explained ? A whole different issue is what happens if "you" accelerate all points of one rod at the same time (simultaneous) in the rest frame

> > > >

Again, this is discussed in the FAQ entry on Bell's "paradox".

> > >

Which FAQ ?

> >	The FAQ for this newsgroup.

> > >

Bell's Spaceship Paradox?

> >	Yes, I sort of remember that this is what it's called there. The idea is to look at two spaceships attached by a cable and accelerated "identically" in some particular frame. The cable is the same length in the original rest frame, but stretched in the locally comoving frame.

>

This has nothing to do with this. I want to study the difference between when you push or pull a rod with a constant speed v. The idea is that the the rods are all the time touching each other and that the point of contact has this constant speed v in the rest frame. (I call this the point m ie the point at the middle of the two rods) My point is that the rod that is pushed is contracted while the rod that is pulled expands. Accordingly to SR both rods should contract.

SR doesn't say this. This is your fantasy. SR describes how different observers measure the length of the same object in the same state of motion moving wrt them.

It doesn't say anything about how ONE observer measures the length of an object before and after acceleration. That is the whole point of Bell's "paradox" as explained in the FAQ. Resolving it involves analyzing the scenario that you're describing according to SR not according to half-baked word arguments.

SR tells you that an accelerated object can expand or shrink depending on how it's accelerated. It also tells you that the same object (unaccelerated, but observed in different frames in relative motion) will have different measured lengths with the longest being in the frame in which the object is at rest.

Your argument is based on your misconceptions, not on a problem in relativity. If you think otherwise, post a counterargument based on using Lorentz transformations and not pseudo relativity crap like trying to pretend that accelerating an object is the same as observing the unaccelerated object in another frame. This is not so in relatvity for an extended object as the FAQ entry demonstrates.

>	How do you explain this.

I don't have to explain it. SR doesn't predict it.

John Anderson

ande452@attglobal.net schreef in bericht news:3C92DBCA.4BF0@attglobal.net...

>	Nicolaas Vroom wrote:

> >	My point is that the rod that is pushed is contracted while the rod that is pulled expands. Accordingly to SR both rods should contract.

>	SR doesn't say this. This is your fantasy. SR describes how different observers measure the length of the same object in the same state of motion moving wrt them.

I expect what you are saying that if an observer, ,at rest in rest frame, measures the length of a rod at rest versus an moving observer which wants to measure the length of the same rod, but now moving rod, that both observers will measure the same. (Equivalence principle) To do that the observer at rest, with the rod at rest, uses a clock at rest. To do that the moving observer, with the moving rod, uses a moving clock.

I do not want to do that. I want to study how a moving rod "behaves" as measured by an observer at rest with a clock at rest.

For an example see: https://www.nicvroom.be/calc1.htm

>	It doesn't say anything about how ONE observer measures the length of an object before and after acceleration.

But that is basicly what I want to study:

>	That is the whole point of Bell's "paradox" as explained in the FAQ. Resolving it involves analyzing the scenario that you're describing according to SR not according to half-baked word arguments.

Bells space ships paradox discusses two rods that are both accelerated in the same frame in the same manner. The paradox is that the distance between the two increases while you are expecting Lorentz contraction. The Faq mentions: "But after a time they will acquire a large velocity, and so the distance between them should suffer Lorentz contraction. Which is it?" http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html

In the Bell space ship paradox you accelarate both space ships in the same manner For example you push both from the back.

I do not want to do that.
I want to push one rod and to pull the other rod for a certain period of time such that they always touch each other.
This is what I called the point m.
The speed v of point m increases lineair first (a>0) and then v becomes constant (a=0)
The question is now what is the length of each rod when a=0 measured in the rest frame.

>	SR tells you that an accelerated object can expand or shrink depending on how it's accelerated.

How is that measured ?
In the rest frame ?

>	on how it's accelerated.

What "means" how.
Is this a distinction between increase versus decrease ?

>	It also tells you that the same object (unaccelerated, but observed in different frames in relative motion) will have different measured lengths with the longest being in the frame in which the object is at rest.

As I mentioned before I want to study the behaviour of the rod(s) only in the rest frame. (Using clocks at rest)

Nick

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