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## Length contraction

I am not against the mainstream beliefs (I support them). I am starting this thread because several mainstream supporters seem to be claiming that a moving object does not contract in the direction of motion. I believe this actually goes against mainstream beliefs.

So the point of this thread is: For mainstream supporters on this site to come to a consensus on the predictions of SR regarding length contraction so that there is no confusion in the future. (The discussion may also serve as a learning tool for those watching that are still having trouble with relativity.)

As such, please do not post in this thread unless you feel you support special relativity.

I have already been accused of "strawman arguements" by clj4. So let me show why I feel there is non consensus.

Tim Thompson wrote:
As for the misleading part, in real, 3D space, there is never a contraction of any real 3D object, nor has the mainstream ever held that there was. The contraction you are thinking of is a teaching tool meant to simplify the problem and reveal fundamental physics, and is valid in one and only one dimension. (link to post)

When discussing the predictions according to an observer moving relative to a sub, MacM stated that this observer would see the sub contracted and therefore denser.

clj4 wrote:
No, this is not correct. Objects moving at very high speeds do not contract
and then contradicted himself in the same post saying
...they appear contracted when viewed from other frames of reference...

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

To help make this discussion on the predictions straight forward, let me attempt to focus the discussion as much as possible.

We are discussing: does a moving object contract in the direction of motion according to special relativity?
And here are three comments on the matter in case there are questions of ambiguity:

1] By definition an object is not moving in its rest frame. So of course an observer in its rest frame does not see it contracted, because it is not a moving object in this frame.

2] The "terrell rotation" is what "appears" to an observer in the direct sense of what he sees. In other words, before correcting for the fact that light from different parts of the object take different amounts of time to reach the observer. So this is just an optical effect, and we need to account for the finite speed of light to obtain the object's orientation and size in this frame.

3] A well known relativity thought experiment to help us focus this discussion:

Let's take two small rockets (I'll just treat them as points, if you really want to be picky the points I am referring to are their center of masses) and an observer. All three are initially at rest with respect to each other.

Initially have the rockets laying horizontally some distance d apart. To be even more unambiguous, have the observer set up a coordinate system according to Einsteins conventions. Now, program the rockets to launch simultaneously according to the observer and use constant thrust for a set amount of time. Will the distance between the rockets decrease according to the observer? Do the calculations, blah blah, and the answer is no.

Okay, now let's take this farther. Put an elastic string between the two rockets (initially straight / no slack but also no tension). Give them extra strong thrusters so that they can maintain their correct path no matter what (basically, make sure the string's effect on the rockets' trajectory is negligible). Same thing, launch simultaneously. Now again the distance between the rockets does not change according to the observer. But, will there be tension in the string? Do the calculations, blah blah, and the answer is yes.

At what length would there be no tension in the string? It's at the normal length contracted distance.

In short, if an object WASN'T length contracted in the direction of motion, there would be tension trying to bring it into the now shorter equilibrium length.

Some (not all) mainstream followers at this site seem to be claiming otherwise.

Maybe we are actually in agreement but using different meanings in our phrases (which is fine, I would just like to make sure this is the case and then we can be aware of it).

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Originally Posted by Jonny
I am not against the mainstream beliefs (I support them). I am starting this thread because several mainstream supporters seem to be claiming that a moving object does not contract in the direction of motion. I believe this actually goes against mainstream beliefs.

So the point of this thread is: For mainstream supporters on this site to come to a consensus on the predictions of SR regarding length contraction so that there is no confusion in the future. (The discussion may also serve as a learning tool for those watching that are still having trouble with relativity.)

As such, please do not post in this thread unless you feel you support special relativity.

I have already been accused of "strawman arguements" by clj4. So let me show why I feel there is non consensus.

Tim Thompson wrote:
As for the misleading part, in real, 3D space, there is never a contraction of any real 3D object, nor has the mainstream ever held that there was. The contraction you are thinking of is a teaching tool meant to simplify the problem and reveal fundamental physics, and is valid in one and only one dimension. (link to post)

When discussing the predictions according to an observer moving relative to a sub, MacM stated that this observer would see the sub contracted and therefore denser.

clj4 wrote:
No, this is not correct. Objects moving at very high speeds do not contract
and then contradicted himself in the same post saying
...they appear contracted when viewed from other frames of reference...

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

To help make this discussion on the predictions straight forward, let me attempt to focus the discussion as much as possible.

We are discussing: does a moving object contract in the direction of motion according to special relativity?
And here are three comments on the matter in case there are questions of ambiguity:

1] By definition an object is not moving in its rest frame. So of course an observer in its rest frame does not see it contracted, because it is not a moving object in this frame.

2] The "terrell rotation" is what "appears" to an observer in the direct sense of what he sees. In other words, before correcting for the fact that light from different parts of the object take different amounts of time to reach the observer. So this is just an optical effect, and we need to account for the finite speed of light to obtain the object's orientation and size in this frame.

3] A well known relativity thought experiment to help us focus this discussion:

Let's take two small rockets (I'll just treat them as points, if you really want to be picky the points I am referring to are their center of masses) and an observer. All three are initially at rest with respect to each other.

Initially have the rockets laying horizontally some distance d apart. To be even more unambiguous, have the observer set up a coordinate system according to Einsteins conventions. Now, program the rockets to launch simultaneously according to the observer and use constant thrust for a set amount of time. Will the distance between the rockets decrease according to the observer? Do the calculations, blah blah, and the answer is no.

Okay, now let's take this farther. Put an elastic string between the two rockets (initially straight / no slack but also no tension). Give them extra strong thrusters so that they can maintain their correct path no matter what (basically, make sure the string's effect on the rockets' trajectory is negligible). Same thing, launch simultaneously. Now again the distance between the rockets does not change according to the observer. But, will there be tension in the string? Do the calculations, blah blah, and the answer is yes.

At what length would there be no tension in the string? It's at the normal length contracted distance.

In short, if an object WASN'T length contracted in the direction of motion, there would be tension trying to bring it into the now shorter equilibrium length.

Some (not all) mainstream followers at this site seem to be claiming otherwise.

Maybe we are actually in agreement but using different meanings in our phrases (which is fine, I would just like to make sure this is the case and then we can be aware of it).

If you edit things out of context you can prove anything. The complete quote was:

Objects moving at very high speeds do not contract (this was the old FitzGerald hypothesis, long refuted, they appear contracted when viewed from other frames of reference (this is the Lorentz, actually Einstein acception of the term).
In the context of the discussion with MacM, who thought that different observers will see different things (some will see the sub sink, some not, hence the apparent SR "contradiction' that he was trying to prove), a more complete statement would have been:

Objects moving at very high speeds do not contract IN THEIR PROPER FRAME OF REFERENCE (this was the old FitzGerald hypothesis, long refuted), they appear contracted when viewed from other frames of reference (this is the Lorentz, actually Einstein-Poincaire acception of the term).
Everyone else in the thread seemed to get that. Now, that we are done with this strawman argument can we get on with life?
Show us please your calculations for the "well known thought experiment", it looks interesting and non-trivial.

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I'm not saying that all of MacM's statements were correct. But he stated that in a frame in which the sub is moving that the sub would be more dense due to length contraction. You stated this was incorrect.

If the sub is moving in an inertial frame, then it is contracted compared to its rest frame and it is more dense compared to its rest frame. Do you disagree with this?

If not, and the problem is that you assume the phrase "according to its proper frame" is inserted after everything, then it is just a terminology issue and we have no disagreement and are done. (Although we still need to hear from Tim.)

Originally Posted by clj4
This is a common thought experiment well held my mainstream. If you disagree with it, let me know other wise it is not worth wasting my time on something we agree on. And since if you disagree you'd be disagreeing with the mainstream, I would be interested in seeing your calculations as well.

Just as there are several variations to the thought experiment "a length contracted car is shut in a garage with a shorter proper length". This is just one of the variations of the Bell's spaceship "paradox".

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Originally Posted by Jonny
I'm not saying that all of MacM's statements were correct. But he stated that the sub would be more dense due to length contraction if it was moving according to an observer. You stated this was incorrect.

If the sub is moving in an inertial frame, then it is contracted compared to its rest frame and it is more dense compared to its rest frame. Do you disagree with this?

If not, and the problem is that you assume the phrase "according to its proper frame" is inserted after everything, then it is just a terminology issue and we have no disagreement and are done. (Although we still need to hear from Tim.)
we are done.

This is a common thought experiment well held my mainstream. If you disagree with it, let me know other wise it is not worth wasting my time on something we agree on. And since if you disagree you'd be disagreeing with the mainstream, I would be interested in seeing your calculations as well.

Just as there are several variations to the thought experiment "a length contracted car is shut in a garage with a shorter proper length". This is just one of the variations of the Bell's spaceship "paradox".
Looks like ithere is a lot of controversy on this item that appears to be non-trivilal. You seem to imply that you have done the calculations yourself ("blah-blah"). Could you show them to us?.

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Originally Posted by clj4
we are done.
Thank you, I am glad we cleared that up.

Originally Posted by clj4
Looks like ithere is a lot of controversy on this item that appears to be non-trivilal. You seem to imply that you have done the calculations yourself ("blah-blah"). Could you show them to us?.
If you really wish. To calculate the actual trajectories of the rockets, more information is needed, but that is an inconsequential aside. The important part is that the length between the rockets stays constant according to the observer. We can show this without any more details.

Common starting point that we can agree on:
- momentum in a given frame is conserved in special relativity

Thus:
- the result of an experiment is translationally invarient (invoking Noether's theorem)

Therefore:
- The velocity of the spaceship as a function of time cannot depend on the rocket's initial location.

Using the fundemental theorem of calculus, $x(t) = x(0) + \int_0^t \frac{dx}{dt'} dt'$ where x(t) is the position of the rocket according to the observer. What is the distance between the rockets as a function of time?

$d(t) = x_1(t) - x_2(t) = x_1(0) + \int_0^t v_1(t') dt' - x_2(0) - \int_0^t v_2(t') dt' = (x_1(0) - x_2(0)) + \int_0^t (v_1(t')-v_2(t')) dt' = x_1(0) - x_2(0) = constant$

Because the end point of the string is the rockets, its length is also d.

After the rockets have finished their programmed thrust, they will be moving at some constant velocity V. I will define the rest frame of the rockets as the primed (') frame. Let's see what the length of the string is in this frame.

Let b = V/c, g = 1/Sqrt[1-b^2].

In observer frame, events defining length of string:
event 1 (ct,x,y,z) = (0,0,0,0)
event 2 (ct,x,y,z) = (0,d,0,0)

In rocket frame
event 1 (ct',x',y',z') = (0,0,0,0)
event 2 (ct',x',y',z') = (- b g d , g d,0,0)

Thus, in the rocket frame L' = g d > d.

The string is stretched in the rest frame of the string. Therefore there is tension on the string.

At what length would there be no tension on the string? The string must not be stretched in its rest frame, so L'=d. I won't bother writing out the transforms this time, as they are analogous. We find L=d/g in this case.

There is tension on the string until it contracts to the "length contracted" value according to the observer.

If you want even more details than this, it is best if you read the papers on this subject.

==========

6. It was never my intention for someone to think that I did not know what the Lorentz transform was, nor that I was unaware of the fairly obvious conclusion that the Lorentz transform predicts a Lorentz contraction. Is that not simply common knowledge? Why even bother to talk about it? My intention was to point out that when you look at something, that's not what you see. You don't see a contraction, you see a rotation, and as far as I am concerened, what you see is what you get. All of this is neatly described in the 2nd edition of Spacetime Physics, 2nd ed (1992), page 92, exercise 3-17. I agree with Taylor & Wheeler.

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Originally Posted by Jonny
Thank you, I am glad we cleared that up.

If you really wish. To calculate the actual trajectories of the rockets, more information is needed, but that is an inconsequential aside. The important part is that the length between the rockets stays constant according to the observer. We can show this without any more details.

Common starting point that we can agree on:
- momentum in a given frame is conserved in special relativity

Thus:
- the result of an experiment is translationally invarient (invoking Noether's theorem)

Therefore:
- The velocity of the spaceship as a function of time cannot depend on the rocket's initial location.

Using the fundemental theorem of calculus, $x(t) = x(0) + \int_0^t \frac{dx}{dt'} dt'$ where x(t) is the position of the rocket according to the observer. What is the distance between the rockets as a function of time?

$d(t) = x_1(t) - x_2(t) = x_1(0) + \int_0^t v_1(t') dt' - x_2(0) - \int_0^t v_2(t') dt' = (x_1(0) - x_2(0)) + \int_0^t (v_1(t')-v_2(t')) dt' = x_1(0) - x_2(0) = constant$

Because the end point of the string is the rockets, its length is also d.

After the rockets have finished their programmed thrust, they will be moving at some constant velocity V. I will define the rest frame of the rockets as the primed (') frame. Let's see what the length of the string is in this frame.

Let b = V/c, g = 1/Sqrt[1-b^2].

In observer frame, events defining length of string:
event 1 (ct,x,y,z) = (0,0,0,0)
event 2 (ct,x,y,z) = (0,d,0,0)

In rocket frame
event 1 (ct',x',y',z') = (0,0,0,0)
event 2 (ct',x',y',z') = (- b g d , g d,0,0)

Thus, in the rocket frame L' = g d > d.

The string is stretched in the rest frame of the string. Therefore there is tension on the string.

At what length would there be no tension on the string? The string must not be stretched in its rest frame, so L'=d. I won't bother writing out the transforms this time, as they are analogous. We find L=d/g in this case.

There is tension on the string until it contracts to the "length contracted" value according to the observer.

If you want even more details than this, it is best if you read the papers on this subject.

==========

thank you,

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Originally Posted by Tim Thompson
...as far as I am concerened, what you see is what you get.
Okay. Most people refer to the coordinates in an inertial frame when referring to an event in special relativity. You are referring to events/measurements before correcting them for the finite speed of light. When you state something "is" blah, you are then referring to these uncorrected measurements.

Stated that way, I now see what you meant in your previous statements. (Although your statement that there is contraction in 1-d while there is not contraction in 3-d still seems strange.)

Regardless, we have now resolved the issue of "differing terminology / viewpoints". I was hoping this was the case, and I thank you for your help. (I had a feeling this would be quick without the side conversations, thank you Nereid for moving the discussion.)

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Originally Posted by clj4
thank you,
This seems unnecessary since we agree on the results. I even tried the temporary solution of using mimetex links to forkosh.com, but this site doesn't allow image tags either.

I assumed most people here can read tex. Until this board supports it natively, I'll just have to leave it as such. Sorry.

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Originally Posted by Jonny
This seems unnecessary since we agree on the results. I even tried the temporary solution of using mimetex links to forkosh.com, but this site doesn't allow image tags either.

I assumed most people here can read tex. Until this board supports it natively, I'll just have to leave it as such. Sorry.
Never mind, your solution seems incomplete. But thank you for the link to wiki.
Here is one that is rigurous and complete:

http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf

A non trivial problem, none of the "blah-blah" , "do your calculations". quite a bit of heavier lifting. Shows that you can't simply apply length contraction, you need to go thru the complete SR calculations.

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Originally Posted by clj4
Here is one that is rigurous and complete:

http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf

The reason my solution is simplier is that I attached the string at the center of mass of the two rockets. It simplifies the problem greatly and makes it easier to discuss and use as an example. That is also how I've normally seen it introduced in classes.

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Originally Posted by Jonny

The reason my solution is simplier is that I attached the string at the center of mass of the two rockets. It simplifies the problem greatly and makes it easier to discuss and use as an example. That is also how I've normally seen it introduced in classes.
you are welcome

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Well, Tim beat me to it, and that is a neat link clj4, thanks to both of you.

14. (just as an aside)
Interestingly, this whole discussion is very reminiscent on the discussion about falling bodies in Huygens' time.
There it was "obvious" that heavy bodies fall faster than light bodies. That is just common sense.
Then Galileo reasoned the following way:
1. I have two bodies, a heavy one and a light one, and according to Aristotle the heavy bodie would fall faster than the light one
2. Now we connect the two bodies with a wire that itself is massless. Then the heavy body must drag the light body making it fall faster, and vice versa the light body will hinder the heavy body making it fall slower.
3. Naturally we can observe the connected bodies as a new one with a mass the sum of both, which will be heavier than the two separate bodies. And therefore, this new body should fall even faster than the two separate components.
4. The only solution that Galileo could find to this paradox was to prove Aristotle wrong, and conclude that all bodies fall at the same rate.

In some way this little step aside in history is relavant to the discussion of length contraction and rockets and....
I will let the educated reader make her/his own conclusions.

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Originally Posted by Tensor
Well, Tim beat me to it, and that is a neat link clj4, thanks to both of you.
You are welcome , Tensor. The link show the importance of doing your calculations from base principles, Bell's "paradox" stems from inappropratelly applying length contraction. As we can see from the proper solution, length contraction is never used.

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Originally Posted by clj4
As we can see from the proper solution, length contraction is never used.
Are you still trying to imply that my solution is incorrect?

As I already explained, the reason my solution is simplier is that the variation of the "paradox" that I used was simpler. I attached the string at the center of mass of the two rockets. This simplifies the problem greatly and makes it easier to discuss and use as an example. That is also how I've normally seen "Bell's paradox" introduced in classes.

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Originally Posted by Jonny
Are you still trying to imply that my solution is incorrect?

As I already explained, the reason my solution is simplier is that the variation of the "paradox" that I used was simpler. I attached the string at the center of mass of the two rockets. This simplifies the problem greatly and makes it easier to discuss and use as an example. That is also how I've normally seen "Bell's paradox" introduced in classes.
Yes, it is incomplete and incorrect. The "paradox" originates in the fact that one has to apply length contraction to the objects (rockets, string) but apparently not to the spacing between rockets. Your solution resorts to applying length contraction to the string. The complete solution included in the link does not.

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Originally Posted by clj4
Yes, it is incomplete and incorrect. The "paradox" originates in the fact that one has to apply length contraction to the objects (rockets, string) but apparently not to the spacing between rockets. Your solution resorts to applying length contraction to the string. The complete solution included in the link does not.
As already explained several times (but you seem to keep ignoring), the variation of the "paradox" that I worked out is simplier than the variation of the "paradox" worked out in your link. It was purposely chosen to be simplier for the sake of discussion (as I also explained, for this reason this is usually how the "paradox" is introduced in a classroom setting). You are right in that the main "paradox" here is the issue of the spacing between the rockets. My solution may look simplier (only because the problem was simplier in setup as well as only asking about the final state of the string) but that does not make the calculations wrong.

Let me put it this way. I assume you agree with the results of my calculations, but just disagree on my methods. Is this correct?

If not: please explain what you believe the results should be.
If so: please explain where you believe I have made several mistakes that seem to cancel each other out.

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I assume you agree with the results of my calculations, but just disagree on my methods. Is this correct?
For a correct approach see the link I have provided.
Getting the right result through questionable methods is a bad way of teaching physics.
Last edited by clj4; 2006-Mar-23 at 04:23 PM.

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Originally Posted by clj4
For a correct approach see the link I have provided.
Getting the right result through questionable methods is a bad way of teaching physics.
Again. My solution may look simplier (only because the problem was simplier in setup as well as only asking about the final state of the string) but that does not make the calculations wrong.

If you insist on repeating that my calculation is wrong, please point out where it is wrong. (It does not "reinforce" the paradox. It clearly shows that the length between the rockets does not contract, but that the string has tension on it at the end ... there is no ambiguity.)

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In case you didn't read or you refuse to understand:

The "paradox" originates in the fact that one has to apply length contraction to the objects (rockets, string) but apparently not to the spacing between rockets. Your solution resorts to applying length contraction to the string. The complete solution included in the link does not. Using an incorrect approach in the context of knowing the result apriori is not the right way to teach physics.

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My application of SR is correct. I can apply length contraction there (actually, I used the lorentz transformations in my solution instead of directly applying length contraction, but the result would have been the same if you wished to do it that way).

If you insist on repeating that my calculation is wrong, please point out where it is wrong. I have asked this of you several times and you have not responded with specifics.

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It is plain for everyone (but you) to see:

Why did you apply length contraction to the object(s) (string, you reduced the rockets to points) and not to the distance between them? (this is Bell's paradox). By bringing in length contraction you ruin the proof.

24. I thought that thought experiment was just another classic example of the relativity of simultaniety.

In the inertial observer's frame the two rockets accelerate identically so the distance between them stays the same but the string contracts as successive Galilean boosts (or acceleration) put it into an ever faster moving instantaneous inertial frame relative to the observer.

From the point of view of the rockets each successive Galilean boost makes clocks in front of them leap forward while clocks behind leap backwards (in the limit this becomes time dilation identical to that due to the pseudo gravitational field they both experience.) This asymmetry causing the distance between them to increase thus breaking the string joining them.

The rockets aren't even necessary for this thought experiment, all you need to do is consider a rigid body uniformly accelerating in the inertial observers frame. So I don't see the problem with reducing the rockets to points.

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Originally Posted by clj4
Why did you apply length contraction to the object(s) (string, you reduced the rockets to points) and not to the distance between them? (this is Bell's paradox). By bringing in length contraction you ruin the proof.
Again, you have failed to point out what step is incorrect in my calculations. Why?

I don't apply length contraction. I applied the lorentz transformations. And I only applied them where they are applicable (between two inertial frames). You can't apply this logic to the rockets' frame while they are accelerating because that is not an inertial frame. I used a simple mathematical arguement to work out the distance between the rockets according to the observer so that I wouldn't have to resort to accelerating frames.

My calculations are fine. I am not doing anything wrong, and I don't understand why you keep stating I am without pointing out which step you believe is incorrect.

We agree my calculation gives the correct answer. We don't agree on whether I applied SR correctly.
Which step in my calculations do you believe is incorrect and why?

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Lifted straight from your initial post (also note the name of the thread)

Originally Posted by Jonny

At what length would there be no tension in the string? It's at the normal length contracted distance.

In short, if an object WASN'T length contracted in the direction of motion, there would be tension trying to bring it into the now shorter equilibrium length.

Some (not all) mainstream followers at this site seem to be claiming otherwise.

Originally Posted by Jonny
Thus, in the rocket frame L' = g d > d.

The string is stretched in the rest frame of the string. Therefore there is tension on the string.

At what length would there be no tension on the string? The string must not be stretched in its rest frame, so L'=d. I won't bother writing out the transforms this time, as they are analogous. We find L=d/g in this case.

There is tension on the string until it contracts to the "length contracted" value according to the observer.
By contrast to your solution, the correct solution calculates the equations of the trajectories of the string points.
Yes, you get the same result BECAUSE you KNOW the correct answer apriori.
So the issue is one of approach, the paper pointed by the link uses a uniform approach, your solution does not. Please stop telling us that this is because the rockets accelerate. The string between them accelerates just the same.
Last edited by clj4; 2006-Mar-24 at 07:15 AM.

27. There was a reason why I posted the free fall experiment by Galileo and the reasoning behind the "all bodies fall at the same rate" even though it got ignored.

Consider that the string is unbreakable then what happens? Then you can consider the whole system as 1 object, that is what it reduces to. Indeed, then Worzel's limit is there. Everything then contracts.

Now then, consider the system, but the string is now a thin tube through which astronauts can move from one ship to the other. So does the tube contract and the space inside the tube not? That would be troublesome, because that would mean we would not be able to build a spaceship anyway, or at least not one that will be able to sustain any relativistic velocities.

I guess I would come here to the conclusion that the whole space contracts, at least in SR. I would have to take a closer look at acceleration, and maybe read up again in "The philosophy of space and time" by Reichenbach (an excellent book).

well, just tell me if I am wrongly philosiphizing here.

Martin

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Originally Posted by clj4
By contrast to your solution, the correct solution calculates the equations of the trajectories of the string points.
They claculate the trajectories of each point separately (true).
To make this much simplier, I just calculated the distance between the points (as that is the relevant quantity in this problem).

Yes, this is simplier. That does not make my calculations incorrect.
Again, stop making these accusations unless you are willing to point out which step is mathematically incorrect and why.

Originally Posted by clj4
Yes, you get the same result BECAUSE you KNOW the correct answer apriori.
No, I get the correct result because I did the calculations correct.

Originally Posted by clj4
Please stop telling us that this is because the rockets accelerate. The string between them accelerates just the same.
You seem to be missing the entire point here. We are treating the rockets as separate points (instead of as one object) because we consider the effect of the string on the trajectory as negligible (even the solution you linked does).

If we attached a strong rod between the two, then the distance between the rockets would indeed contract. That is why I included that problem in this discussion.

Do you disagree with that statement? If so, that is the root of our disagreement. And I'd like to see you back up this disagreement with mainstream with a calculation of your own.
Last edited by Jonny; 2006-Mar-25 at 08:12 AM.

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