1 "David Seppala" |
SR Length Contraction - how do physicists explain this | maandag 8 november 2004 1:21 |
2 "Tom Roberts" |
Re: SR Length Contraction - how do physicists explain this | dinsdag 9 november 2004 1:04 |
3 "David Seppala" |
Re: SR Length Contraction - how do physicists explain this | dinsdag 9 november 2004 3:35 |
4 "Paul Cardinale" |
Re: SR Length Contraction - how do physicists explain this | woensdag 10 november 2004 0:49 |
5 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | donderdag 11 november 2004 16:29 |
6 "David Seppala" |
Re: SR Length Contraction - how do physicists explain this | donderdag 11 november 2004 17:11 |
7 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | vrijdag 12 november 2004 13:57 |
8 "David Seppala" |
Re: SR Length Contraction - how do physicists explain this | vrijdag 12 november 2004 18:03 |
9 "Tom Roberts" |
Re: SR Length Contraction - how do physicists explain this | zaterdag 13 november 2004 19:58 |
10 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | zondag 14 november 2004 12:17 |
11 "shuba" |
Re: SR Length Contraction - how do physicists explain this | zondag 14 november 2004 17:09 |
12 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | dinsdag 16 november 2004 13:45 |
13 "shuba" |
Re: SR Length Contraction - how do physicists explain this | woensdag 17 november 2004 5:40 |
14 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | woensdag 17 november 2004 12:27 |
15 "shuba" |
Re: SR Length Contraction - how do physicists explain this | donderdag 18 november 2004 0:17 |
16 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | zaterdag 20 november 2004 10:50 |
17 "shuba" |
Re: SR Length Contraction - how do physicists explain this | zaterdag 20 november 2004 17:08 |
18 "Bill Hobba" |
Re: SR Length Contraction - how do physicists explain this | zondag 21 november 2004 0:20 |
19 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | zondag 21 november 2004 10:44 |
20 "Nicolaas Vroom" |
Re: SR Length Contraction - how do physicists explain this | zondag 21 november 2004 10:46 |
21 "shuba" |
Re: SR Length Contraction - how do physicists explain this | zondag 21 november 2004 21:48 |
22 "Bill Hobba" |
Re: SR Length Contraction - how do physicists explain this | maandag 22 november 2004 0:21 |
23 | Re: SR Length Contraction - how do physicists explain this |
Consider two identical steel rods, side by side, positioned along the
x-axis in the rest frame. At time t0, the two rods are accelerated
along the x-axis into another inertial frame which has a velocity V
along the x-axis. Assume each point of the rods is accelerated in an
identical fashion. When the acceleration stops, let these rods have
zero relative velocity wrt to this moving frame. From previous posts,
most responders agreed that Einstein's length contraction notion
doesn't occur instantaneously. (This is easy to see when you consider
extremely long rods - 100's of light years in length which are
accelerated at familiar velocities).
If this notion is indeed true, then the following problem occurs.
When the two rods stop accelerating and have zero velocity wrt to the
moving frame, the length contraction process hasn't stopped. If the
length contraction hasn't stopped, then what happens if in the moving
frame we attach a weight to one or both ends of one of the rods? In
this frame, the rods and weight have zero relative velocity. If the
length contraction were to continue, one rod has to drag this weight
as it contracts, or it must end up at a different position than the
other rod. This implies that the two rods end up with different
energies. When the acceleration stopped, both rods must have had
identical energies. But when the length contraction stopped, some of
the energy in one rod was used to drag a weight, or used to stretch
the rod, or used to change the position of the rod or a combination
thereof. When you consider springs, and stretching, it appears that
the rod with no weight attached ends up with less energy than the one
that is attached to a weight. Where did the energy go to during the
length contraction process?
Thanks,
David Seppala
> | Consider two identical steel rods, side by side, positioned along the x-axis in the rest frame. At time t0, the two rods are accelerated along the x-axis into another inertial frame which has a velocity V along the x-axis. Assume each point of the rods is accelerated in an identical fashion. |
I assume you mean d^2x/dt^2 measured in the initial inertial frame is the same for every point along the rods. Note this is not Born rigid motion, and this will induce internal stresses within the rods. This is also quite difficult to achieve in practice, for rods. But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically (ignore the effects of the exhaust from a rocket on its neighbors).
> | When the acceleration stops, let these rods have zero relative velocity wrt to this moving frame. |
OK. And while there are stresses internal to the rods, we'll ignore them.
> | From previous posts, most responders agreed that Einstein's length contraction notion doesn't occur instantaneously. |
I have no idea where you get this. I'd be careful about believing what people write in this newsgroup. And be even more careful about what you write yourself -- this is rather imprecise in that "instantaneously" can be taken to mean various things here. In particular, all portions of the rod stop accelerating simultaneously in the initial inertial frame, but they do NOT stop accelerating simultaneously in the final inertial frame (the ultimate rest frame of the rod). So what does "instantaneously" mean?? -- in which frame is the term applied?
For any small region of the rod, its "length contraction" wrt some specific inertial frame is determined only by its velocity wrt that frame. So if this small region stops accelerating wrt that frame, then its "length contraction" instantly ceases to change.
This is analogous to the clock hypotheses: the tick rate of a clock measured by a given inertial frame does not depend upon the clock's acceleration, it depends only on its speed wrt that frame. Length contraction is similar, but is also direction dependent.
> | (This is easy to see when you consider extremely long rods - 100's of light years in length which are accelerated at familiar velocities). |
The length of the rod does not matter. What matters is how imprecise your use of language is, and how that permits you to confuse yourself.
> | If this notion is indeed true, then the following problem occurs. When the two rods stop accelerating and have zero velocity wrt to the moving frame, the length contraction process hasn't stopped. |
It has stopped in the initial ienrtial frame. There is no meaning to the phrase "when the two rods stop accelerating" in the final inertial frame, unless you wait until every region of the rods has stopped. Do that and you description is invalid.
> | If the length contraction hasn't stopped, then what happens if in the moving frame we attach a weight to one or both ends of one of the rods? |
In the final inertial frame, the trailing end of the rod stops before the leading end. If you attach the weight to the trailing end there is no effect. If you attach the weight to the leading end before the leading end stops moving wrt this frame, then in order to keep the "identical acceleration" claimed, then you need to increase the force on that end of the rod (plus its weight) -- for the rockets I mentioned the leading rocket must increase its thrust.
> | [... further nonsense based on the above confusion...] |
Tom Roberts tjroberts@lucent.com
On Mon, 08 Nov 2004 18:04:26 -0600, Tom Roberts
> | dseppala@austin.rr.com wrote: |
>> | Consider two identical steel rods, side by side, positioned along the x-axis in the rest frame. At time t0, the two rods are accelerated along the x-axis into another inertial frame which has a velocity V along the x-axis. Assume each point of the rods is accelerated in an identical fashion. |
> |
I assume you mean d^2x/dt^2 measured in the initial inertial frame is the same for every point along the rods. Note this is not Born rigid motion, and this will induce internal stresses within the rods. This is also quite difficult to achieve in practice, for rods. But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically (ignore the effects of the exhaust from a rocket on its neighbors). |
>> |
When the acceleration stops, let these rods have zero relative velocity wrt to this moving frame. |
> |
OK. And while there are stresses internal to the rods, we'll ignore them. |
>> |
From previous posts, most responders agreed that Einstein's length contraction notion doesn't occur instantaneously. |
> |
I have no idea where you get this. I'd be careful about believing what people write in this newsgroup. And be even more careful about what you write yourself -- this is rather imprecise in that "instantaneously" can be taken to mean various things here. In particular, all portions of the rod stop accelerating simultaneously in the initial inertial frame, but they do NOT stop accelerating simultaneously in the final inertial frame (the ultimate rest frame of the rod). So what does "instantaneously" mean?? -- in which frame is the term applied? For any small region of the rod, its "length contraction" wrt some specific inertial frame is determined only by its velocity wrt that frame. So if this small region stops accelerating wrt that frame, then its "length contraction" instantly ceases to change. |
No. Consider a long row of people holding a long rod. Let the length of the row of people and the rod be equal to 10**17 meters (or about 10.6 light-years). Simultaneously, as measured in their rest frame, let them step onto a conveyer belt that is moving at 3 meters/sec. When I use Einstein's equations, this rod while in the rest frame spans about 5 meters more than an identical rod that's sitting on the conveyer belt. Thus when the row of people step on to the conveyer belt, the rod must undergo a length contraction of about 5 meters (out of 10*17 meters). The difference in time between one end starting to accelerate and the other end starting to accelerate is about 3 seconds (according to Einstein's equations). Let's say the acceleration lasts about 0.1 seconds (the time necessary to accelerate each point from zero to 3 meters/second). The person holding the end point of the rod must move 2.5 to 5 meters from his original position where he stepped on to the conveyer belt due to length contraction. This length contraction cannot occur in 3 seconds. That would imply that some sort of physical effect traveled the 10*17 meters in 3 seconds, since the amount of movement is dependent on the length of the rod. This is far greater than the speed of light. So here we are left with a situation in which the acceleration has stopped (in both frames), but the length contraction hasn't completed. That's the situation in the problem I posted. What happens when a weight is now attached to either or both ends of that rod. That rod ends up differently than an identical rod in which no weight was attached. Both rods before the weight is attached have the same energy. But after the length contraction stops, there is a difference in energy. I don't see how to explain the difference. David
> |
This is analogous to the clock hypotheses: the tick rate of a clock measured by a given inertial frame does not depend upon the clock's acceleration, it depends only on its speed wrt that frame. Length contraction is similar, but is also direction dependent. |
>> |
(This is easy to see when you consider extremely long rods - 100's of light years in length which are accelerated at familiar velocities). |
> |
The length of the rod does not matter. What matters is how imprecise your use of language is, and how that permits you to confuse yourself. |
>> |
If this notion is indeed true, then the following problem occurs. When the two rods stop accelerating and have zero velocity wrt to the moving frame, the length contraction process hasn't stopped. |
> |
It has stopped in the initial ienrtial frame. There is no meaning to the phrase "when the two rods stop accelerating" in the final inertial frame, unless you wait until every region of the rods has stopped. Do that and you description is invalid. |
>> |
If the length contraction hasn't stopped, then what happens if in the moving frame we attach a weight to one or both ends of one of the rods? |
> |
In the final inertial frame, the trailing end of the rod stops before the leading end. If you attach the weight to the trailing end there is no effect. If you attach the weight to the leading end before the leading end stops moving wrt this frame, then in order to keep the "identical acceleration" claimed, then you need to increase the force on that end of the rod (plus its weight) -- for the rockets I mentioned the leading rocket must increase its thrust. |
>> |
[... further nonsense based on the above confusion...] |
> |
Tom Roberts tjroberts@lucent.com |
dseppala@austin.rr.com wrote in message news:<4190262d.127453199@news-server.austin.rr.com>...
> |
On Mon, 08 Nov 2004 18:04:26 -0600, Tom Roberts |
> > |
dseppala@austin.rr.com wrote: |
> > > | Consider two identical steel rods, side by side, positioned along the x-axis in the rest frame. At time t0, the two rods are accelerated along the x-axis into another inertial frame which has a velocity V along the x-axis. Assume each point of the rods is accelerated in an identical fashion. |
> > |
I assume you mean d^2x/dt^2 measured in the initial inertial frame is the same for every point along the rods. Note this is not Born rigid motion, and this will induce internal stresses within the rods. This is also quite difficult to achieve in practice, for rods. But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically (ignore the effects of the exhaust from a rocket on its neighbors). |
> > > |
When the acceleration stops, let these rods have zero relative velocity wrt to this moving frame. |
> > |
OK. And while there are stresses internal to the rods, we'll ignore them. |
> > > |
From previous posts, most responders agreed that Einstein's length contraction notion doesn't occur instantaneously. |
> > |
I have no idea where you get this. I'd be careful about believing what people write in this newsgroup. And be even more careful about what you write yourself -- this is rather imprecise in that "instantaneously" can be taken to mean various things here. In particular, all portions of the rod stop accelerating simultaneously in the initial inertial frame, but they do NOT stop accelerating simultaneously in the final inertial frame (the ultimate rest frame of the rod). So what does "instantaneously" mean?? -- in which frame is the term applied? For any small region of the rod, its "length contraction" wrt some specific inertial frame is determined only by its velocity wrt that frame. So if this small region stops accelerating wrt that frame, then its "length contraction" instantly ceases to change. |
> | No. Consider a long row of people holding a long rod. Let the length of the row of people and the rod be equal to 10**17 meters (or about 10.6 light-years). Simultaneously, as measured in their rest frame, let them step onto a conveyer belt that is moving at 3 meters/sec. |
If there is no slipping of their hands gripping the rod, nor any slipping of their feet on the belt, and if the conveyer belt is not deformed as they step on, then as the rod is accelerated, it will be stretched in its rest frame (i.e. its proper length will increase), and its length in the original rest will be unchanged. In other words, in the original rest frame, length contraction will cancel out the stretching of the proper length.
Paul Cardinale
"Tom Roberts"
> | dseppala@austin.rr.com wrote: |
> | But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically |
Strange suggestion. But I expect my understanding is wrong.
I agree that it is easy to build all those rockets identical. I agree that it is easy to program all the motors identical. The problem lies in the fact what is the object of this program (what is it that you try to perform, mimic) i.e. what are the mathematics involved and how do you prove that the program is correct.
This writing is solely based on the initial rest frame. In the initial rest frame there are n equal spaced (distance = l, length rod = l) clocks all synchronised Clock 1 is at the end of the rod and clock 2 at beginning. I agree with David Seppala that length contraction is hindered with an instantaneous issue.
Suppose I start the motor at the end at 12.00 (clock 1) Does the front also move at 12.00 (clock 2) ? IMO, In reality not, because internal within the rod there are time delays (propagation delays). Now let us suppose (1) that the front starts instantaneous.
Suppose that the end of the rod reaches clock 2 at 12.01 Does the front meets clock 3 at 12.01?
There are two answers Yes or No.
(A lawyer will say: it depends)
IMO taking care of SR length contraction and if you assume
supposition (1) i.e. that the forces within the rod act
instantaneous the answer is NO
i.e. the front end will meet clock 3 later.
This is the general rule. When the back is at clock n the front will meet clock n+1 later. The later, the faster the rod moves.
This implies that the speed within the rod is not everywhere
the same:
The front moves the slowest (based on the initial rest frame!)
If your answer is Yes than the speed within the rod is everywhere the same (based on the rest frame) and there is no SR length contraction.
If the answer is No, than I expect what the program in each motor tries to perform is to mimic length contraction such that the whole rod behaves in agreement with SR. This implies that each motor has a different speed (ofcourse during acceleration) as measured in the initial rest frame.
However the whole issue is more complex, but that is for later. (For example all motors have to stop simultaneous in the initial rest frame)
Nicolaas Vroom
https://www.nicvroom.be/
On 9 Nov 2004 15:49:15 -0800, pcardinale@volcanomail.com (Paul Cardinale) wrote:
> | dseppala@austin.rr.com wrote in message news:<4190262d.127453199@news-server.austin.rr.com>... |
>> |
On Mon, 08 Nov 2004 18:04:26 -0600, Tom Roberts |
>> > |
dseppala@austin.rr.com wrote: |
>> > > | Consider two identical steel rods, side by side, positioned along the x-axis in the rest frame. At time t0, the two rods are accelerated along the x-axis into another inertial frame which has a velocity V along the x-axis. Assume each point of the rods is accelerated in an identical fashion. |
>> > |
I assume you mean d^2x/dt^2 measured in the initial inertial frame is the same for every point along the rods. Note this is not Born rigid motion, and this will induce internal stresses within the rods. This is also quite difficult to achieve in practice, for rods. But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically (ignore the effects of the exhaust from a rocket on its neighbors). |
>> > > |
When the acceleration stops, let these rods have zero relative velocity wrt to this moving frame. |
>> > |
OK. And while there are stresses internal to the rods, we'll ignore them. |
>> > > |
From previous posts, most responders agreed that Einstein's length contraction notion doesn't occur instantaneously. |
>> > |
I have no idea where you get this. I'd be careful about believing what people write in this newsgroup. And be even more careful about what you write yourself -- this is rather imprecise in that "instantaneously" can be taken to mean various things here. In particular, all portions of the rod stop accelerating simultaneously in the initial inertial frame, but they do NOT stop accelerating simultaneously in the final inertial frame (the ultimate rest frame of the rod). So what does "instantaneously" mean?? -- in which frame is the term applied? For any small region of the rod, its "length contraction" wrt some specific inertial frame is determined only by its velocity wrt that frame. So if this small region stops accelerating wrt that frame, then its "length contraction" instantly ceases to change. |
>> | No. Consider a long row of people holding a long rod. Let the length of the row of people and the rod be equal to 10**17 meters (or about 10.6 light-years). Simultaneously, as measured in their rest frame, let them step onto a conveyer belt that is moving at 3 meters/sec. |
> |
If there is no slipping of their hands gripping the rod, nor any slipping of their feet on the belt, and if the conveyer belt is not deformed as they step on, then as the rod is accelerated, it will be stretched in its rest frame (i.e. its proper length will increase), and its length in the original rest will be unchanged. In other words, in the original rest frame, length contraction will cancel out the stretching of the proper length. Paul Cardinale |
Instead of stepping onto a conveyer belt, let's use the equally spaced
rockets approach to accelerate the rod. Let the rod be 10*17 meters
in length. Let's accelerate it from zero to 3 meters/second in 0.1
seconds (about 3 g's). In the original rest frame, when the
acceleration stops, each point of the rod has traveled about 15
centimeters. Einstein's length contraction formula says that the
original 10**17 meter rod spans 10**17 + 5 meters of length in the
moving frame (the 3 meter/second frame). When the acceleration stops,
the length contraction hasn't yet completed (this violates Einstein's
notion that nothing can travel faster than c). The end points of the
rod are still about 10**17 + 5 meters apart.
As viewed in the moving frame (the 3 meter/second frame), all points
of the rod do not start accelerating at the same time. According to
Einstein's equations, there is a 3 to 4 second delay between one end
point starting its acceleartion wrt to the other end in this example.
In the moving frame the 10**17 meter rod is measured to be about
10**17 - 5 meters in length before the acceleration starts. Since in
both frames each rocket fires for only about 0.1 seconds, the moving
frame observers see one end of the rod travel about 10 meters before
the other end finishes accelerating (3 m/s times 3 to 4 seconds).
They see the rod grow from 10**17 - 5 meters to a length of 10**17 + 5
meters. In both frames, the length contraction has not completed
when there is no more acceleration. Assuming the 3 g acceleration did
not cause a permanent deformation of the rod, the rod will be 10**17
meters when the length contraction process stops (its proper length).
This cannot occur in under 4 seconds with a rod 10**17 meters in
length. This conflicts with Einstein's notion that physical effects
cannot travel faster than the speed of light in a vacuum. Therefore
in both frames the length contraction has not completed when the
acceleration goes to zero.
So once you see that the length contraction can't be completed
when the acceleration has stopped (as measured in either frame), then
we must answer the question in my original post - namely, what happens
to the energy in a rod in which weights are now attached to the end
points or elsewhere compared to an identical rod that is free to
continue to return to its propert length. When I consider spring
shaped rods, rods dragging weights, rods perpendicular to the motion,
etc. it seems that the rod with no weights attached has less energy
when it reaches its proper length compared to a rod that is stretched.
I couldn't find any reference explaining where this energy goes in the
length contraction process.
David
"David Seppala"
Keep Things simple. Try to raise one question at a time.
> | Instead of stepping onto a conveyer belt, |
> | let's use the equally spaced rockets approach to accelerate the rod. Let the rod be 10*17 meters in length. Let's accelerate it from zero to 3 meters/second in 0.1 seconds (about 3 g's). In the original rest frame, when the acceleration stops, each point of the rod has traveled about 15 centimeters. |
Using rockets is asking for problems. See my posting dd 11/11/04. You have to specify what the mathematics is, used, in order to control the motors of the rockets. The last sentence implies, assuming that my understanding is correct, that all the motors are stopped simultaneous in the initial reference frame. The last sentence also implies that each point of the rod has travelled exactly the same distance (about 15 cm) This is very important if you read my posting dd 11/11/04
That means your answer on my question is Yes, i.e. there is No length contraction in the rod.
This has to be clarified before to continue.
(The use of rockets is tricky, because you are using something "artificial" in order to unravel the laws of physics. This is also true if you use rigid rods or Born rigid rods, without any clear difinition what they are.)
Nicolaas Vroom
https://www.nicvroom.be
On Fri, 12 Nov 2004 12:57:15 GMT, "Nicolaas Vroom"
> |
"David Seppala" Keep Things simple. Try to raise one question at a time. |
>> |
Instead of stepping onto a conveyer belt, |
> | conveyer belts are very tricky issues to test SR. They resemble very much the rotating disc problem or better the rotating wheel problem. The question is: does the radius change (decrease) when the speed increases. |
> > |
let's use the equally spaced rockets approach to accelerate the rod. Let the rod be 10*17 meters in length. Let's accelerate it from zero to 3 meters/second in 0.1 seconds (about 3 g's). In the original rest frame, when the acceleration stops, each point of the rod has traveled about 15 centimeters. |
> |
Using rockets is asking for problems. See my posting dd 11/11/04. You have to specify what the mathematics is, used, in order to control the motors of the rockets. The last sentence implies, assuming that my understanding is correct, that all the motors are stopped simultaneous in the initial reference frame. The last sentence also implies that each point of the rod has travelled exactly the same distance (about 15 cm) This is very important if you read my posting dd 11/11/04 That means your answer on my question is Yes, i.e. there is No length contraction in the rod. |
> |
This has to be clarified before to continue.
(The use of rockets is tricky, because you are using something "artificial" in order to unravel the laws of physics. This is also true if you use rigid rods or Born rigid rods, without any clear difinition what they are.) Nicolaas Vroom https://www.nicvroom.be |
> |
"Tom Roberts" |
> > | But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically |
> |
Strange suggestion. But I expect my understanding is wrong. |
Yes, it is. You get confused by making ambiguous and imprecise statements. This is SR, and you must ALWAYS state wrt which frame you are measuring, and wrt which frame things are simultaneous. This includes such innocuous words as "when".
Start with the rockets lined up along X in the initial frame, let all motors operate identically in the original frame and accelerate each rocket to 3 m/s wrt the initial frame, and then turn off. Let them all do this simultaneously in the initial frame. The first and last rocket will behave identically wrt this initial frame, and the distance between them will remain 10^17 meters before their motors turn on, while their motors are firing, and after their motors turn off.
In the final frame, the front rocket starts accelerating about 10/3 second before the last rocket motor starts. The result of this, of course, is that in the final frame the distance between the front and last rocket increases; after all motors turn off, the distance between the first and last rocket is 10^17+5 meters.
Tom Roberts tjroberts@lucent.com
"Tom Roberts"
> | Nicolaas Vroom wrote: |
> > |
"Tom Roberts" |
> > > | But consider instead a series of rockets lined up along the x-axis and it becomes easy to program all their motors identically |
> > |
Strange suggestion. But I expect my understanding is wrong. |
> |
Yes, it is. You get confused by making ambiguous and imprecise statements. This is SR, and you must ALWAYS state wrt which frame you are measuring, and wrt which frame things are simultaneous. This includes such innocuous words as "when". |
See Comment 1.
> | Start with the rockets lined up along X in the initial frame, let all motors operate identically in the original frame and accelerate each rocket to 3 m/s wrt the initial frame, and then turn off. Let them all do this simultaneously in the initial frame. The first and last rocket will behave identically wrt this initial frame, and the distance between them will remain 10^17 meters before their motors turn on, while their motors are firing, and after their motors turn off. |
See Comment 2.
> | In the final frame, the front rocket starts accelerating about 10/3 second before the last rocket motor starts. The result of this, of course, is that in the final frame the distance between the front and last rocket increases; after all motors turn off, the distance between the first and last rocket is 10^17+5 meters. |
See Comment 3
After reading this reply and the same with the latest reply from David Dd 12/11/2004 it becomes clear what you both want.
Comment 1.
I do not understand why you write this.
In my previous posting Dd 11/11/2004 I clearly state:
"This writing is solely based on the initial rest frame."
i.e. no where I mentioned a moving frame or the final frame.
Comment 2.
Your startegy is based as if the answer on my question is Yes
that means the speed within the rod is everywhere the same
(based on the rest frame) and there is no SR length contraction.
i.e. physical
To be more precise:
You start all the rockets at the same time and you stop all
the rockets at the same time (rest frame).
Comment 3.
You write: "In the final frame, the front rocket starts accelerating
about 10/3 second before the last rocket motor starts."
But you should also write:
"In the final frame, the front rocket stops accelerating about 10/3
second before the last rocket motor stops".
In short each motor runs the same duration.
What is more also in the final frame there is no physical
SR length contraction.
(In no frame what so ever there is SR lc using rockets as described)
On the other hand an Observer in the final (moving) frame using a moving clock he (she) will claim that the length has increased. That means this Observer is able to detect relative motion and that is in disagreement with EEP i.e. that all laws of physics are the same in each relative moving frame.
The reason of course is that you perform this experiment in an artificial way with (multiple) rockets , such that, in the rest frame there is no SR length contraction.
(for a different example see: https://www.nicvroom.be/calc1.htm
IMO the best way to perform this experiment is with only one rocket in the center and observe what happens only in the rest frame with no moving clocks. If there is no lenght contraction Newton wins. If there is length contraction (5m) Einstein wins.
The problem of course is (the sad part) that in reality no one can perform (has performed) such an experiment to decide who is right.
Nicolaas Vroom
https://www.nicvroom.be/
Datum: zondag 14 november 2004 17:09
> | The reason of course is that you perform this experiment in an artificial way with (multiple) rockets , such that, in the rest frame there is no SR length contraction. |
The length contraction of special relativity is always associated with *two* different frames. It is a phenomenon of geometric perspective, quite different from the length of a rod being increased by induced stresses e.g. by a rocket.
---Tim Shuba---
"shuba"
> | Nicolaas Vroom wrote: |
> > |
The reason of course is that you perform this experiment in an artificial way with (multiple) rockets , such that, in the rest frame there is no SR length contraction. |
> |
The length contraction of special relativity is always associated with *two* different frames. It is a phenomenon of geometric perspective, quite different from the length of a rod being increased by induced stresses e.g. by a rocket. |
I do not fully agree with you.
IMO there are two different situations. First you can study different experiments in the same frame Secondly you can study the same experiment in different frames. In the first case different experiments imply different physical conditions. In the second case the physical conditions are identical. (except for the speed of the clocks) In the first case in each experiment you should try to change only one parameter. IMO the first case (different experiments) is the most important and the results require physical explanations. The second case is much less important and is much more a mathematical exercise.
An example of the first case is to study the length of a rod in the rest frame using clocks at rest with different speeds i.e. v = 0 and v > 0 (only one rocket !) The result can be that the length is different i.e. physical different inside the rod.
An example of the second case is to study the length of a with v >0 (only one rocket !) in the rest frame (clock at rest) and in a moving frame (moving clock). The result can be that the length is different, but the cause is not physical (rod based) but is caused as a result of the measuring process i.e. the condition of the clock, at rest versus moving.
A second example of the first case is to study the length in the rest frame with v > 0, with one rocket at different positions.
A third example of the first case is to study the length in the rest frame with v > 0, with different number of rockets.
All those examples of the first case imply different physical conditions inside the rod. For examples of the second case this is not the case.
As I already mentioned examples of the first case are the most important i.e. study only one reference frame. This same physlosophy is used in the discussion "How important is SR (GR) in order to calculate the precision of Mercury" in sci.physics.research, sci.physics and sci.astro.research.
> |
"shuba" |
> > | The length contraction of special relativity is always associated with *two* different frames. It is a phenomenon of geometric perspective, quite different from the length of a rod being increased by induced stresses e.g. by a rocket. |
> |
I do not fully agree with you. |
I assumed you wouldn't. Nevertheless this is what will be meant by a physicist discussing length contraction within the context of special relativity, so it might be worth your while to learn the standard meaning of the term.
[..]
> | As I already mentioned examples of the first case are the most important i.e. study only one reference frame. |
Then you are not talking about relativistic length contraction.
---Tim Shuba---
"shuba"
> | Nicolaas Vroom wrote: |
> > |
"shuba" |
> |
> > > |
The length contraction of special relativity is always associated with *two* different frames. It is a phenomenon of geometric perspective, quite different from the length of a rod being increased by induced stresses e.g. by a rocket. |
> > |
I do not fully agree with you. |
> |
I assumed you wouldn't. Nevertheless this is what will be meant by a physicist discussing length contraction within the context of special relativity, so it might be worth your while to learn the standard meaning of the term. [..] |
> > |
As I already mentioned examples of the first case are the most important i.e. study only one reference frame. |
> |
Then you are not talking about relativistic length contraction. |
IMO this last sentence also implies that the length of the body in the rest frame will be smaller than the rest length or proper length for v>0 as measured by clocks at rest.
That means I make a "prediction" by only considering one frame
This is in agreement with what further is written: "This (MMX) is rather different from the length contraction of SR, which is not to be regarded as illusory but is a very real effect"
If you study the length of a moving (relative to rest frame) rod
relative to a moving frame the speed becomes zero (relative to
the moving frame).
But this is a more complex exercise because moving clocks
have to be introduced.
For details see https://www.nicvroom.be/calc1.htm
Nicolaas Vroom
https://www.nicvroom.be/
> |
In the book "Introducing Einstein's Relativity" by Ray d'Inverno
at page 33 I read:
"Since etc the result shows that the length of a body in the
direction of its motion with uniform velocity v is reduced
by a factor SQRT (1-v*v/c*c). This phenomena is called length
contraction. Clearly the body will have greatest length in
its rest frame, in which case it is called the rest length
or proper length".
IMO this last sentence also implies that the length of the body in the rest frame will be smaller than the rest length or proper length for v>0 as measured by clocks at rest. |
Proper length is an invariant; you are thouroughly confused if you think that two frames disagree on its value.
> |
That means I make a "prediction" by only considering one frame
This is in agreement with what further is written: "This (MMX) is rather different from the length contraction of SR, which is not to be regarded as illusory but is a very real effect" |
I expect this is way out of context. In special relativity, there is no "real" physical contraction of the arms of the MMX apparatus, so I have no idea what d'Inverno is referring to here.
> | If you study the length of a moving (relative to rest frame) rod relative to a moving frame the speed becomes zero (relative to the moving frame). But this is a more complex exercise because moving clocks have to be introduced. |
I don't see why it should take more than 10 seconds thought to realize that the speed of an object in its (momentarily comoving) rest frame is zero. Or do you have a meaning for "at rest" that no one else uses?
> | For details see https://www.nicvroom.be/calc1.htm |
No thanks. I looked at your site some years ago, and I would not recommend it to anyone. Except maybe for one thing. You are a member of one of strangest group of characters on this newsgroup, that is the qbasic crowd. It seems that none of you guys who are into qbasic have any kind of a decent grasp on the fundamentals of reativity. It's just a bizarre fact I've noticed.
---Tim Shuba---
"shuba"
> | Nicolaas Vroom wrote: |
> > |
If you study the length of a moving (relative to rest frame) rod
relative to a moving frame the speed becomes zero (relative to
the moving frame). |
> |
I don't see why it should take more than 10 seconds thought to realize that the speed of an object in its (momentarily comoving) rest frame is zero. Or do you have a meaning for "at rest" that no one else uses? |
You are right, but the importance is not in the speed but
in the length, or better in the question:
is the proper length of a rod at rest in a rest frame the same
as the the proper length of a rod at rest in a moving frame.
Accordingly to EEP it should.
For the answer see the following link where that problem is
explained in a sequence of steps.
In each step only ONE PARAMETER is changed.
> > | For details see https://www.nicvroom.be/calc1.htm |
> |
No thanks. I looked at your site some years ago, and I would not recommend it to anyone. Except maybe for one thing. You are a member of one of strangest group of characters on this newsgroup, that is the qbasic crowd. It seems that none of you guys who are into qbasic have any kind of a decent grasp on the fundamentals of reativity. It's just a bizarre fact I've noticed. |
The program in calc1.htm is written in Java Script.
IMO the language is completely unimportant.
The importance is the mathematics behind the program
which is explained in the text.
As I already mentioned above the program explains in a
sequence of steps that the proper length in the restframe
is the same as the proper length in a moving frame.
The program performs some steps in the rest frame
and some steps in moving frame.
Each time only ONE PARAMETER is changed.
In fact to switch rest frame to moving frame is easy, because
this means switching from a clock at rest into a moving clock,
which runs slower.
Mathematically this means dividing by gamma.
The program also explains what happens if no length
contraction is involved.
In fact this is what david seppala does in his example
when he introduces multiple rockets in order to keep
the length constant, as such there is no (physical)
length contraction involved in his example.
The length 10**17 + 5 he mentions is a result of the
measuring process when you perform the same
measurement "in two frames" one with clocks at rest,
versus one with moving clocks (the factor gamma).
Nicolaas Vroom
https://www.nicvroom.be/
> | the question: is the proper length of a rod at rest in a rest frame the same as the the proper length of a rod at rest in a moving frame. Accordingly to EEP it should. |
Wrong. The equivalence principle deals with gravity. Proper length is an invariant. All frames agree on whether the proper length of a rod is different before and after the rod has been accelerated and is at rest in a new frame. A change in an object's proper length is NOT what is known as relativistic length contraction.
---Tim Shuba---
"Nicolaas Vroom"
> |
"shuba" |
> > | Nicolaas Vroom wrote: |
> > > |
If you study the length of a moving (relative to rest frame) rod relative to a moving frame the speed becomes zero (relative to the moving frame). But this is a more complex exercise because moving clocks have to be introduced. |
> > |
I don't see why it should take more than 10 seconds thought to realize that the speed of an object in its (momentarily comoving) rest frame is zero. Or do you have a meaning for "at rest" that no one else uses? |
> |
You are right, but the importance is not in the speed but
in the length, or better in the question:
is the proper length of a rod at rest in a rest frame the same
as the the proper length of a rod at rest in a moving frame. |
Your grasp of the basics is very poor indeed. The POR says the laws of physics are the same in an inertial frame of a frame moving at constant velocity relative to an inertial frame (the last bit is strictly not necessary with regard to the usual definition of an inertial frame - but I use a less usual one based on symmetries - but that is another issue). Thus if by moving frame you man one moving at constant velocity wrt to an inertial frame it must also have the same length. The EEP principle has to do with gravity - not with inertial frames.
Bill
> | For the answer see the following link where that problem is explained in a sequence of steps. In each step only ONE PARAMETER is changed. |
> > > |
For details see https://www.nicvroom.be/calc1.htm |
> > |
No thanks. I looked at your site some years ago, and I would not recommend it to anyone. Except maybe for one thing. You are a member of one of strangest group of characters on this newsgroup, that is the qbasic crowd. It seems that none of you guys who are into qbasic have any kind of a decent grasp on the fundamentals of reativity. It's just a bizarre fact I've noticed. |
> |
The program in calc1.htm is written in Java Script. IMO the language is completely unimportant. The importance is the mathematics behind the program which is explained in the text. As I already mentioned above the program explains in a sequence of steps that the proper length in the restframe is the same as the proper length in a moving frame. The program performs some steps in the rest frame and some steps in moving frame. Each time only ONE PARAMETER is changed. In fact to switch rest frame to moving frame is easy, because this means switching from a clock at rest into a moving clock, which runs slower. Mathematically this means dividing by gamma. The program also explains what happens if no length contraction is involved. In fact this is what david seppala does in his example when he introduces multiple rockets in order to keep the length constant, as such there is no (physical) length contraction involved in his example. The length 10**17 + 5 he mentions is a result of the measuring process when you perform the same measurement "in two frames" one with clocks at rest, versus one with moving clocks (the factor gamma).
Nicolaas Vroom |
"Bill Hobba"
> |
"Nicolaas Vroom" |
> > |
You are right, but the importance is not in the speed but in the length, or better in the question: is the proper length of a rod at rest in a rest frame the same as the the proper length of a rod at rest in a moving frame. Accordingly to EEP it should. |
> |
Your grasp of the basics is very poor indeed. The POR says the laws of physics are the same in an inertial frame of a frame moving at constant velocity relative to an inertial frame (the last bit is strictly not necessary with regard to the usual definition of an inertial frame - but I use a less usual one based on symmetries - but that is another issue). Thus if by moving frame you man one moving at constant velocity wrt to an inertial frame it must also have the same length. The EEP principle has to do with gravity - not with inertial frames. |
What we are discussing here is Postulate 1 "Principle of Special Relativity"
All inertial observers are equivalent as defined at page 19 of the
book "introducing Einstein's Relativity" by Ray d'Inverno.
A different question is if the Principle of SR is a sub set of, is part of,
is incorporated in the EEP or SEP.
For an explanation see the following:
http://www.mazepath.com/uncleal/eotvos.htm#b22
Specific point 8:
The Weak Equivalence Principle extends to the
Strong (Einsteinian) Equivalence Principle :
8 "The Strong Equivalence Principle embraces all laws of nature;
all reference frames accelerated or not,
in a gravitational field or not, rotating or not,
anywhere at any time"
You can also raise a different question: Is SR an Integral part of GR, Is SR incorporated in GR.
But those questions are not relevent to the subject of this thread.
( For a discussion of that subject please go to: "Is Einstein's Principle of Equivalence true?" in sci.physics.relativity )
The subject of this thread is:
Is the program CALC1.htm correct.
Is the text within this program correct.
Is the strategy within this program, namely only to change
one parameter at a time correct.
Is it correct to state that there is no (physical) length contraction
involved in the experiment by David.
Is it correct to state that the difference in length mentioned by
David is caused by the measuring process used i.e. different
observers and different clocks. Different meaning at rest versus
moving.
Nicolaas Vroom
http://user.pandora.be/nicvroom/
"shuba"
> | Nicolaas Vroom wrote: |
> > |
the question: is the proper length of a rod at rest in a rest frame the same as the the proper length of a rod at rest in a moving frame. Accordingly to EEP it should. |
> |
Wrong. The equivalence principle deals with gravity. Proper length is an invariant. All frames agree on whether the proper length of a rod is different before and after the rod has been accelerated and is at rest in a new frame. A change in an object's proper length is NOT what is known as relativistic length contraction. |
See my reply to Bill Hoppa
Nicolaas Vroom
> | See my reply to Bill Hoppa |
I did. It misses the point. I'm really not interested in arguing with another kook who refuses to learn the basics of a subject. If you disagree with the simple things I've said about proper length and length contraction, fine.
---Tim Shuba---
"Nicolaas Vroom"
> |
"Bill Hobba" |
> > |
"Nicolaas Vroom" |
> > > |
You are right, but the importance is not in the speed but in the length, or better in the question: is the proper length of a rod at rest in a rest frame the same as the the proper length of a rod at rest in a moving frame. Accordingly to EEP it should. |
> > |
Your grasp of the basics is very poor indeed. The POR says the laws of physics are the same in an inertial frame of a frame moving at constant velocity relative to an inertial frame (the last bit is strictly not necessary with regard to the usual definition of an inertial frame - but I use a less usual one based on symmetries - but that is another issue). Thus if by moving frame you man one moving at constant velocity wrt to an inertial frame it must also have the same length. The EEP principle has to do with gravity - not with inertial frames. |
> |
What we are discussing here is Postulate 1 "Principle of Special Relativity" |
The POR is about inertial frames so before addressing that you need to be sure you understand what an inertial frame is.
> | All inertial observers are equivalent as defined at page 19 of the book "introducing Einstein's Relativity" by Ray d'Inverno. |
So?
> | A different question is if the Principle of SR is a sub set of, is part of, is incorporated in the EEP or SEP. |
Of course the POR is incorporated into the EEP because it says, locally, freely falling frames of reference are indistinguishable from inertial frames. The SEP says all the laws of physics are equivalent in such freely falling frames, even the laws of gravity, the WEP says such may not apply to al the laws of nature - gravity may be excluded. We also have the strong-weak version etc.
> | For an explanation see the following: http://www.mazepath.com/uncleal/eotvos.htm#b22 Specific point 8: The Weak Equivalence Principle extends to the Strong (Einsteinian) Equivalence Principle : 8 "The Strong Equivalence Principle embraces all laws of nature; all reference frames accelerated or not, in a gravitational field or not, rotating or not, anywhere at any time" |
I have found the discussion in the following very useful - http://www.pupress.princeton.edu/sample_chapters/ciufolini/chapter2.pdf
> |
You can also raise a different question: Is SR an Integral part of GR, Is SR incorporated in GR. |
The answer is well known and obvious ie when gravity is not present and we consider inertial frames GR reduces to SR because GR is based on the SEP.
Bill
> |
But those questions are not relevent to the subject of this thread. ( For a discussion of that subject please go to: "Is Einstein's Principle of Equivalence true?" in sci.physics.relativity )
The subject of this thread is:
Is the program CALC1.htm correct.
Nicolaas Vroom |
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