Post by: MikeFontenot on 23/04/2023 22:31:35
More than 20 years ago, I plotted a chart showing two separated objects undergoing the same constant acceleration "A". (That chart still hangs on the wall above my desk, and I've never questioned it before). The plot supposedly shows the view of things according to an inertial reference frame (the IRF) that is stationary wrt the two objects immediately before the acceleration begins. One curve starts from the origin with slope zero at the origin, but then curves upward with a curvature that monotonically decreases as time increases, and asymptotically approaching a slope of "c", the speed of light. I use units where "c" equals 1.0, so the curve approaches a slope of 1.0 on the chart.
The other curve has exactly the same shape, but starts at some distance "D" above the origin. The two curves are always separated by a vertical distance of "D".
The idea, I think, was that the two curves must have exactly the same shape because of "the Principle of Relativity" ... i.e., it shouldn't matter where in space you start the curve, the curves should always have the same shape.
But here's the quandary: An observer in the inertial frame IRF is told by the chart that the two objects always have the same distance apart. But the length contraction equation (LCE) of special relativity says that an inertial observer should conclude that a moving yardstick should get shorter and shorter as its speed wrt the inertial observer increases. That seems to contradict what the chart says, and it seems to contradict the Principle of Relativity. The LCE seems to require that the two curves get closer together as time increases. Does the upper curve slowly get closer to the lower curve? Or does the lower curve approach the upper curve? Or is there some combination of those two movements? Any of those movements contradicts what the chart says, and it thus seems to contradict the Principle of Relativity.
Any ideas? I'm really stuck ... I don't know the answer.
Post by: Halc on 23/04/2023 23:24:21
More than 20 years ago, I plotted a chart showing two separated objects undergoing the same constant acceleration "A". (That chart still hangs on the wall above my desk, and I've never questioned it before).We will assume constant proper acceleration since constant coordinate acceleration isn't possible after a while. The chart sounds legit and seems to have no reason to question it.
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The plot supposedly shows the view of things according to an inertial reference frame (the IRF) that is stationary wrt the two objects immediately before the acceleration begins. One curve starts from the origin with slope zero at the origin, but then curves upward with a curvature that monotonically decreases as time increases, and asymptotically approaching a slope of "c", the speed of light. I use units where "c" equals 1.0, so the curve approaches a slope of 1.0 on the chart.All good so far. It seems that vertical is the D axis and horizontal is T, kind of opposite of what I'm used to, but not wrong. OK, so these objects start simultaneously (relative to the frame of the chart) and thus stay at constant separation relative to that frame exactly as they should.
The other curve has exactly the same shape, but starts at some distance "D" above the origin. The two curves are always separated by a vertical distance of "D".
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The idea, I think, was that the two curves must have exactly the same shape because of "the Principle of Relativity" ... i.e., it shouldn't matter where in space you start the curve, the curves should always have the same shape.That principle doesn't say that, but yes, relative to that IRF, those curves will be identical.
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But here's the quandary: An observer in the inertial frame IRF is told by the chart that the two objects always have the same distance apart.Relative to the IRF, they do. The observer doesn't need to be told this if he already has the description above.
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But the length contraction equation (LCE) of special relativity says that an inertial observer should conclude that a moving yardstick should get shorter and shorter as its speed wrt the inertial observer increases.Again, so far so good.
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That seems to contradict what the chart says, and it seems to contradict the Principle of Relativity.I don't think the chart shows contraction of moving things. Maybe it does. You didn't post an image. Principle of relativity seems unreferenced here. There's no specific contradiction specified. Yes, a ruler moving with the objects contracts in the D direction, but it doesn't sound like your chart shows moving rulers. You can fit more of them between the objects over time, as many as you want if you wait long enough.
Principle of relativity just says that physics is the same relative any inertial frame. It sounds like you have only one inertial frame specified here, so there's nothing to compare, hence no particular violation of PoR.
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The LCE seems to require that the two curves get closer together as time increases.It says no such thing. It says that the moving rulers get shorter, meaning more fit between, meaning that in the accelerating frame of one of the rulers, the objects are getting further apart, not closer. The chart doesn't show that since it shows the IRF, not the frame of any of the objects.
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Does the upper curve slowly get closer to the lower curve?Frame dependent question. Relative to the IRF in which the two objects always have identical velocity, the separation remains constant. In the accelerating frame of either object, the two get further apart, a consequence of relativity of simultaneity. All this is covered in Bell's spaceship scenario, something with which you really should familiarize yourself since so many of your topics seem to run amok on this. https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
Post by: MikeFontenot on 23/04/2023 23:47:23
"The LCE seems to require that the two curves get closer together as time increases. Does the upper curve slowly get closer to the lower curve? Or does the lower curve approach the upper curve?"
I've realized that the bottom curve doesn't move upward, because it already has speeds that approach the speed of light "c", and so it's speeds can't be increased any. So all of the decrease in their separation has to come from a lowering of the upper curve.
So I suppose that is enough information to allow the correct upper curve to be plotted ... just subtract the amount of length contraction (using the LCE) from each point of the upper curve.
A new question: Do the two observers who are doing the accelerating agree that their separation is decreasing?
(Inertial observers don't ever think the yardsticks between them contract, so maybe accelerating observers don't think the yardsticks between them contract either.)
Post by: MikeFontenot on 24/04/2023 00:06:23
Post by: Halc on 24/04/2023 00:19:57
In my original post, I said:The LCE applies only to rigid inertial objects. 'The distance between two things' is not a rigid object, nor is it inertial in this case. Read my post above. Your chart is correct and the curves stay equally separated in that IRF.
"The LCE seems to require that the two curves get closer together as time increases. Does the upper curve slowly get closer to the lower curve? Or does the lower curve approach the upper curve?"
Quote
I've realized that the bottom curve doesn't move upward, because it already has speeds that approach the speed of light "c", and so it's speeds can't be increased any. So all of the decrease in their separation has to come from a lowering of the upper curve.You persist in using the mathematics of a rigid object. That's fine, but not the scenario depicted on your chart.
So I suppose that is enough information to allow the correct upper curve to be plotted ... just subtract the amount of length contraction (using the LCE) from each point of the upper curve.
Suppose you had a long rigid object of length D, a rocket say, stretching the distance between the two points on your chart. The rear of it accelerates per the curve shown in the chart. Now the LCE comes into play as you describe here. All of the contraction of the rocket has to come from, as you say, a lowering of the upper curve, but this also has a consequence of lower proper acceleration of the upper curve since the full proper acceleration is the not-lowered curve that your chart shows. Yes, that is enough information to allow the alternate upper curve to be plotted, albeit a somewhat complicated way to do so. The plot of the full proper acceleration curve remains unchanged as your chart correctly shows.
Anyway, it means that accelerometers at either end of a rocket read different values.
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Do the two observers who are doing the accelerating agree that their separation is decreasing?They'd be wrong if they decided that. In the rocket (with the front guy under less proper acceleration), they'd agree that the rigid rocket remains the same proper length at all times. In the identical proper acceleration case that your chart depicts (and the Bell's scenario discusses, and you still haven't read), they'd agree that their separation is increasing as evidenced by the string between them breaking.
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(Inertial observers don't ever think the yardsticks between them contract, so maybe accelerating observers don't think the yardsticks between them contract either.)That's right, so in the long rocket case, the rocket always remains a constant number of yardsticks in length. The marking are in fact painted along the length of the rigid rocket so it really isn't possible for them to measure a different length.
I've scanned the chart into a jpeg. How do I post that?A bit complicated. Apologies.
http://www.thenakedscientists.com/forum/index.php?topic=45718.msg397740#msg397740
It all works through the 'Attachments and other options' link just below the edit window
Post by: Jaaanosik on 24/04/2023 13:23:02
Here is a figure, is it similar to your picture above your desk?
More than 20 years ago, I plotted a chart showing two separated objects undergoing the same constant acceleration "A". (That chart still hangs on the wall above my desk, and I've never questioned it before). The plot supposedly shows the view of things according to an inertial reference frame (the IRF) that is stationary wrt the two objects immediately before the acceleration begins. One curve starts from the origin with slope zero at the origin, but then curves upward with a curvature that monotonically decreases as time increases, and asymptotically approaching a slope of "c", the speed of light. I use units where "c" equals 1.0, so the curve approaches a slope of 1.0 on the chart.
The other curve has exactly the same shape, but starts at some distance "D" above the origin. The two curves are always separated by a vertical distance of "D".
The idea, I think, was that the two curves must have exactly the same shape because of "the Principle of Relativity" ... i.e., it shouldn't matter where in space you start the curve, the curves should always have the same shape.
But here's the quandary: An observer in the inertial frame IRF is told by the chart that the two objects always have the same distance apart. But the length contraction equation (LCE) of special relativity says that an inertial observer should conclude that a moving yardstick should get shorter and shorter as its speed wrt the inertial observer increases. That seems to contradict what the chart says, and it seems to contradict the Principle of Relativity. The LCE seems to require that the two curves get closer together as time increases. Does the upper curve slowly get closer to the lower curve? Or does the lower curve approach the upper curve? Or is there some combination of those two movements? Any of those movements contradicts what the chart says, and it thus seems to contradict the Principle of Relativity.
Any ideas? I'm really stuck ... I don't know the answer.
(https://i.imgur.com/knUfyTr.png)
As Halc says, Bell's paradox is important to read and analyze:
https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox#Constant_proper_acceleration
... specifically the section: "Constant proper acceleration".
Edit: from the wiki page, these two scenarios are different, the Bell's paradox explains it.
(https://i.imgur.com/j5Ub62G.png)
Post by: MikeFontenot on 24/04/2023 14:18:05
In my original post, I said:The LCE applies to rigid objects. 'The distance between two things' is not a rigid object. Read my post above. Your chart is correct and the curves stay equally separated in that IRF.
"The LCE seems to require that the two curves get closer together as time increases. Does the upper curve slowly get closer to the lower curve? Or does the lower curve approach the upper curve?"QuoteI've realized that the bottom curve doesn't move upward, because it already has speeds that approach the speed of light "c", and so it's speeds can't be increased any. So all of the decrease in their separation has to come from a lowering of the upper curve.You persist in using the mathematics of a rigid object. That's fine, but not the scenario depicted on your chart.
So I suppose that is enough information to allow the correct upper curve to be plotted ... just subtract the amount of length contraction (using the LCE) from each point of the upper curve.
Suppose you had a long rigid object of length D, a rocket say, stretching the distance between the two points on your chart. The rear of it accelerates per the curve shown in the chart. Now the LCE comes into play as you describe here. All of the contraction of the rocket has to come from, as you say, a lowering of the upper curve, but this also has a consequence of lower proper acceleration of the upper curve since the full proper acceleration is the not-lowered curve that your chart shows. Yes, that is enough information to allow the alternate upper curve to be plotted, albeit a somewhat complicated way to do so. The plot of the full proper acceleration curve remains unchanged as your chart correctly shows.
Anyway, it means that accelerometers at either end of a rocket read different values.QuoteDo the two observers who are doing the accelerating agree that their separation is decreasing?They'd be wrong if they decided that. In the rocket (with the front guy under less proper acceleration), they'd agree that the rigid rocket remains the same proper length at all times. In the identical proper acceleration case that your chart depicts (and the Bell's scenario discusses, and you still haven't read), they'd agree that their separation is increasing as evidenced by the string between them breaking.Quote(Inertial observers don't ever think the yardsticks between them contract, so maybe accelerating observers don't think the yardsticks between them contract either.)That's right, so in the long rocket case, the rocket always remains a constant number of yardsticks in length. The marking are in fact painted along the length of the rigid rocket so it really isn't possible for them to measure a different length.I've scanned the chart into a jpeg. How do I post that?A bit complicated. Apologies.
http://www.thenakedscientists.com/forum/index.php?topic=45718.msg397740#msg397740
It all works through the 'Attachments and other options' link just below the edit window
Post by: Eternal Student on 24/04/2023 17:01:22
OK, we can see the diagrams you ( @MikeFontenot ) have posted. We can also see that you have drawn some diagonal lines between the two worldlines.
So I'm going to agree with previous comments from others. The situation is very much like Bells's spaceship paradox and it is probably best explained just by looking through a good explanation of that situation ("paradox" - although it isn't really a paradox, a perfectly fine explanation does exist).
I'm, not sure that the link to Wikipedia's article provides a great explanation. Try this explanation:
https://math.ucr.edu/home/baez/physics/Relativity/SR/BellSpaceships/spaceship_puzzle.html
(That's a physics FAQ webpage from the University of California and should be safe enough).
I've noticed that their diagram and explanation has the same approach of drawing diagonal lines from one worldline to another (lines of constant time or of "simultaneity" in a different frames of reference). I think it will help to remind you of what was actually done in your own diagram and answer many of your questions. Obviously they have chosen to put time on the y-axis and not the x-axis but other than that switch it's showing exactly the same as your diagram.
The phrasing of these situations is always complicated and English Language is rarely the best tool to use, diagrams really help. The phrasing used on their ( ucr) webpage isn't without some problems but it's a good effort and I doubt I could do much better. Notice that they do end the discussion with a digram showing a very different pair of worldlines the objects could have traced out in the lab frame (the one you called IRF in your posts), where the people in the rockets now would find that the distance between the rockets remains constant - but the person in the lab frame no longer does.
A new question: Do the two observers who are doing the accelerating agree that their separation is decreasing?If you manage to follow the explanation of Bell's spaceship paradox you'll see that the answer is "no", not if the objects have the worldlines with the shape you've given them. The distance also doesn't decrease as you stated, they would notice the distance has increased.
Best Wishes.
Post by: MikeFontenot on 24/04/2023 19:48:59
(I couldn't get the attachment to work ... you'll just have to view it in one of the other posts.)
The above diagram (without the diagonal straight lines) shows the perspective of the two accelerating observers.
One thing that diagram DOESN'T show is how the ages of those two observers compare, as time progresses. (The horizontal axis is the age of the TRAILING accelerating person). The two accelerating observers do NOT age at the same rate. Einstein (in his 1907 paper) said the leading person ages exp(D * A) times faster than the trailing person, but I showed in an earlier paper [] that that exponential equation is incorrect.
Post by: Jaaanosik on 24/04/2023 20:38:16
Hi.There is a problem though.
OK, we can see the diagrams you ( @MikeFontenot ) have posted. We can also see that you have drawn some diagonal lines between the two worldlines.
So I'm going to agree with previous comments from others. The situation is very much like Bells's spaceship paradox and it is probably best explained just by looking through a good explanation of that situation ("paradox" - although it isn't really a paradox, a perfectly fine explanation does exist).
...
Best Wishes.
The initial lab frame grid of inertial observers predicts the rockets separation increase therefore the string breaks.
The formation of the uniformly accelerated frame, a Rindler frame, can be done only when one origin from the initial lab frame is chosen as a preferred origin.
That means not all observers within the original lab frame grid of inertial observers are equal.
Post by: Jaaanosik on 24/04/2023 22:48:45
...I can confirm that. :D ;)
Statements to the effect of 'Einstein was wrong' will get this topic moved like all the others.
Post by: MikeFontenot on 25/04/2023 15:23:08
There is a problem though.
The initial lab frame grid of inertial observers predicts the rockets separation increase therefore the string breaks.
What exactly are the "initial lab frame grid of inertial observers"? The inertial observers who are stationary wrt the rockets immediately before the acceleration begins will say that the rockets get closer together as the acceleration progresses. So they will conclude that the string DOESN'T break.
Post by: MikeFontenot on 25/04/2023 15:34:44
Special relativity theory (and not some frame) predicts that the proper separation (which is not frame dependent) increases, and for that reason the string breaks (an objective fact, not a frame dependent one).
I don't believe that SR predicts that.
Post by: Jaaanosik on 25/04/2023 16:45:13
From the link provided by Eternal Student:
There is a problem though.
The initial lab frame grid of inertial observers predicts the rockets separation increase therefore the string breaks.
What exactly are the "initial lab frame grid of inertial observers"? The inertial observers who are stationary wrt the rockets immediately before the acceleration begins will say that the rockets get closer together as the acceleration progresses. So they will conclude that the string DOESN'T break.
Quote
The distance that the blue rocket measures from A to B is approximately γL (in fact, it's somewhat more than γL). But if we ask the blue rocket to re-measure the distance from itself to the red rocket at a later time marked by event P, then the line of simultaneity will have changed: it will be the upper dotted blue line. This is not parallel to the lower dotted blue line, and this line crosses the red world line at event Q. The distance PQ will be larger than distance AB, and so the blue rocket will conclude that the red rocket is actually pulling away from it.
(https://i.imgur.com/XSqqikx.png)
Post by: MikeFontenot on 24/05/2023 19:25:44
The diagram (two identical attachments) shows worldlines of two objects with constant proper acceleration, and depicts the original IRF, thus the perspective of an inertial observer.
The diagram is SUPPOSED to show the perspective of two sets of inertial observers. The first set is the inertial observers who are stationary with respect to the rocket passengers immediately before the rockets are ignited. For those inertial observers, a given instant of time is a vertical line on the diagram, and a given spacial location is a horizontal line on the diagram. (The other set of inertial observers in the diagram are stationary wrt the rocket passengers at a later instant of time. Since both sets of inertial observers must agree about whether or not the string breaks, I will focus on the first set, and ignore the second set.)
The important thing to understand is that the diagram, as shown, is INCORRECT. It does NOT show the correct viewpoint of that first set of inertial observers. The well-known length contraction equation (LCE) says that for ANY inertial observer (HE), a line of end-to-end yardsticks that are moving at a constant speed relative to him will be shorter than his own yardsticks, by the gamma factor
1 / sqrt( 1 - v * v ).
So that means that the given inertial observers (who are stationary with the rockets immediately before the rockets are turned on) MUST conclude that the two rockets get closer together during their acceleration. So the diagram, as drawn, is incorrect ... it does NOT show the correct conclusions of those given inertial observers.
To obtain the correct diagram, at each instant of the given inertial observers' time, it is necessary to compute the gamma factor (where "v" is the speed of the rockets at that instant), and divide the constant separation "L" of the rockets (according to the observers on the rockets) by gamma. The result is then added to the location of the trailing rocket, to get the location of the leading rocket.
That correct diagram shows that, according to the given inertial observers, the two rockets get closer together during the acceleration, and therefore the string does NOT break.
Q.E.D.
Post by: MikeFontenot on 24/05/2023 23:58:14
Quote from: Halc
[...]
Halc, you've got things exactly backwards:
The diagram that you like, and which you contend is standard special relativity, is wrong, because it violates one of the most important laws of special relativity: the length contraction equation.
The diagram that you hate, and which you contend ISN'T special relativity, is correct, because it obeys the length contraction equation of special relativity.
Post by: Halc on 25/05/2023 01:39:32
The important thing to understand is that the diagram, as shown, is INCORRECT. It does NOT show the correct viewpoint of that first set of inertial observers. The well-known length contraction equation (LCE) says that for ANY inertial observer (HE), a line of end-to-end yardsticks that are moving at a constant speed relative to him will be shorter than his own yardsticks, by the gamma factor 1 / sqrt( 1 - v * v ).Translation: The mathematics (the picture) and the entire physics community (Einstein included) contradict your intuitions, so the mathematics and the physicists must be wrong. Another conclusion is inconceivable.
Moving this accordingly to New Theories as this is no longer a question, but an assertion of alternate physics.
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To obtain the correct diagram, at each instant of the given inertial observers' time, it is necessary to compute the gamma factor (where "v" is the speed of the rockets at that instant), and divide the constant separation "L" of the rockets (according to the observers on the rockets) by gamma. The result is then added to the location of the trailing rocket, to get the location of the leading rocket.Excellent! Now do exactly that for the lead rocket when it is 10 ly ahead of the trailing one instead of 0.5. Compute the gamma factor and compute where the trailing rocket needs to be after a year (rocket time or inertial frame time, your choice), of acceleration at 1 ly/y2 (a smidge over 1g), in order for the string not to break.
That correct diagram shows that, according to the given inertial observers, the two rockets get closer together during the acceleration, and therefore the string does NOT break.
The diagram that you like, and which you contend is standard special relativity, is wrong, because it violates one of the most important laws of special relativity: the length contraction equation.Jano posted some diagrams in post 5, which I'll call (P and QL and QR). You posted one (twice) in post 6 (R). ES put one in post 14 (S), and it's hard to tell but acceleration seems to cease after a certain amount of time in that one.
The diagram that you hate, and which you contend ISN'T special relativity, is correct, because it obeys the length contraction equation of special relativity.
Don't remember hating any of them, but they don't all depict the same thing. QR, R & S all depict identical proper acceleration profiles. P & QL depict constant acceleration of an extended rigid object. Only those two show a ruler to be contracted.
None are inconsistent with SR. What you describe is inconsistent, which would become super apparent if you actually did this:
Now do exactly that for the lead rocket when it is 10 ly ahead of the trailing one instead of 0.5.I notice you decline this. Now why is that? Could it be that your assertions can trivially be driven to contradiction?
I don't need a plot. I just need to know where the other end is after one end accelerates at about 1g for a year. What are the coordinates (relative to the initial inertial frame) of both ends? A simple 2 digits of precision will do.
Oh wait, you can't do that. You won't do that. The length contraction which Einstein describes must be wrong as well, and relativity of simultaneity along with it. All bunk.
Post by: MikeFontenot on 25/05/2023 03:42:06
"If you're loosing an argument, change the subject."
Post by: Eternal Student on 25/05/2023 10:19:41
If you're loosing an argument, change the subject.That seems to be what you ( @MikeFontenot ) have done. You CAN have a situation where the two rockets get closer together in the lab frame and the string between them does not break. However, that wasn't the situation you originally described or what was shown in your original diagrams.
Change the original situation and you will change the final consequences. I think we're all in agreement with that. It doesn't make the original situation an impossible situation to have, just one that you didn't really want to be examining.
Best Wishes.
Post by: MikeFontenot on 25/05/2023 14:43:18
Hi.If you're loosing an argument, change the subject.That seems to be what you ( @MikeFontenot ) have done. You CAN have a situation where the two rockets get closer together in the lab frame and the string between them does not break. However, that wasn't the situation you originally described or what was shown in your original diagrams.
Change the original situation and you will change the final consequences. I think we're all in agreement with that. It doesn't make the original situation an impossible situation to have, just one that you didn't really want to be examining.
Best Wishes.
Here is my post that you are referring to:
____________________________________________________
The diagram is SUPPOSED to show the perspective of two sets of inertial observers. The first set is the inertial observers who are stationary with respect to the rocket passengers immediately before the rockets are ignited. For those inertial observers, a given instant of time is a vertical line on the diagram, and a given spacial location is a horizontal line on the diagram. (The other set of inertial observers in the diagram are stationary wrt the rocket passengers at a later instant of time. Since both sets of inertial observers must agree about whether or not the string breaks, I will focus on the first set, and ignore the second set.)
The important thing to understand is that the diagram, as shown, is INCORRECT. It does NOT show the correct viewpoint of that first set of inertial observers. The well-known length contraction equation (LCE) says that for ANY inertial observer (HE), a line of end-to-end yardsticks that are moving at a constant speed relative to him will be shorter than his own yardsticks, by the gamma factor
1 / sqrt( 1 - v * v ).
So that means that the given inertial observers (who are stationary with the rockets immediately before the rockets are turned on) MUST conclude that the two rockets get closer together during their acceleration. So the diagram, as drawn, is incorrect ... it does NOT show the correct conclusions of those given inertial observers.
To obtain the correct diagram, at each instant of the given inertial observers' time, it is necessary to compute the gamma factor (where "v" is the speed of the rockets at that instant), and divide the constant separation "L" of the rockets (according to the observers on the rockets) by gamma. The result is then added to the location of the trailing rocket, to get the location of the leading rocket.
That correct diagram shows that, according to the given inertial observers, the two rockets get closer together during the acceleration, and therefore the string does NOT break.
Q.E.D.
__________________________________________________
Show me where in the above I have changed anything. The diagram that I am referring to is given early in this thread (the one showing the viewpoint of the inertial observers who are stationary with the rockets immediately before they are fired). I stated why that diagram is incorrect (because those inertial observers say that the rocket separation doesn't change, and that violates the length contraction equation (LCE) of special relativity). I then described how that diagram must be changed in order to be consistent with special relativity.
Post by: Eternal Student on 25/05/2023 19:28:57
Show me where in the above I have changed anything.
You said this:
The important thing to understand is that the diagram, as shown, is INCORRECT. It does NOT show the correct viewpoint of that first set of inertial observers.If it doesn't show the situation you wanted to discuss, then don't show it. Show the diagram that does show the situation you wanted to discuss.
There is a diagram that corresponds to the situation you wanted to discuss.
A diagram like this one, for example:
(https://math.ucr.edu/home/baez/physics/Relativity/SR/BellSpaceships/rindler.png)
Diagram from https://math.ucr.edu/home/baez/physics/Relativity/SR/BellSpaceships/spaceship_puzzle.html
If you had shown a diagram like that and had the rockets experience the appropriate accelerations to create that, then exactly as you have stated:
That correct diagram shows that, according to the given inertial observers, the two rockets get closer together during the acceleration, and therefore the string does NOT break.
We're all in agreement with that.
You seem to be suggesting that the original situation, with the original diagram you did supply was just outright impossible - it isn't, it is a possible situation but just a different one to the situation you are describing later.
1. There is a situation (where the rockets experience a certain pattern of acceleration) where the distance between the rockets stays constant in the lab frame AND then the string must break.
2. There is another situation (a different pattern of acceleration) where the string does not break but then the distance between the rockets would be seen to reduce in the lab frame as time progressed.
This comment was made in a much earlier post from me:
Notice that they (ucr.edu website on Bell's spaceship paradox) do end the discussion with a diagram showing a very different pair of worldlines the objects could have traced out in the lab frame (the one you called IRF in your posts), where the people in the rockets now would find that the distance between the rockets remains constant - but the person in the lab frame no longer does.
Just to be clear then, both situations are possible. They require a different pattern of acceleration for the rockets. The first situation is not made "impossible" just by talking about the second situation.
Best Wishes.
Post by: MikeFontenot on 25/05/2023 21:27:46
Post by: MikeFontenot on 25/05/2023 22:00:12
Post by: Eternal Student on 26/05/2023 19:31:55
WHEN you do that, then the original diagram (that hung over my desk for 20 to 30 years) violates special relativity ... because the inertial observers who are stationary with the rockets immediately before the rockets are fired, claim that the separation between rockets is constant.It doesn't violate special relativity and it's not impossible for that to be the motion and corresponding worldlines of the rockets. Are the observers who were "stationary with the rockets immediately before the rockets were fired" going to observe those worldlines for the rockets or not? That is a choice you have. Decide how the rockets will move in this frame (which I'll call the lab frame). You can make this choice - but then it will determine what happens to a piece of string that connected the two rockets.
Special relativity (via the length contraction equation) says that any inertial observer will conclude that yardsticks that are moving (in the direction of their length) wrt himself are shorter than his own yardsticks (by the gamma factor). I.e., if gamma = 2.0, the yardsticks are only half as long as they would be if they weren't moving relative to the inertial observer. So the inertial observers who are stationary with the rockets immediately before the rockets are fired MUST (according to special relativity) say that the two rockets get closer together as their speed increases.There weren't any rigid connections like yard sticks between the rockets. If you do put a rigid connection rod between them (e.g. the rope or piece of string in Bell's spaceship paradox) then you do find there is a problem, exactly as you have outlined.
If the motion of the rockets was as originally described (constant separation in the lab frame), then the string must break.
Alternatively you can start the problem the other way round. Stipulate that a string of fixed length was attached between the two rockets and it did not break as the rockets accelerated. Then the motion of the rockets (in the lab frame) could not have been as previously described, it would have been different. The motion of the rockets in the lab frame would be exactly as you have described later in the post - the observers who stay in the lab frame would see the rockets getting closer together as time progresses. One consequence of this is that, in the lab frame, the rockets did not have the same acceleration at every moment of time.
I hope that makes some sense. You can choose how the rockets move in the lab frame OR if the string will break. However, you can't choose to have the rockets accelerate equally at every moment of time in the lab frame AND ALSO avoid the string breaking ---> That's a combination that is not possible.
Best Wishes.
Post by: MikeFontenot on 26/05/2023 21:21:36
Each rocket has an attached accelerometer, and those two accelerometers always show exactly the same acceleration. (That is part of the initial specification of the scenario). The INITIAL diagram says that the inertial observers who are stationary wrt the rockets immediately before the rockets fire, say that the separation of the rockets doesn't vary. But THAT violates special relativity: special relativity says (via the length contraction equation) that an inertial observer MUST conclude that a yardstick moving away from himself (in the direction of its length) is shorter than his own yardsticks. (And the faster the yardstick moves, the shorter it becomes, according to that inertial observer). So those inertial observers must conclude that the separation of the rockets MUST decrease as the rockets accelerate away. And the faster the rockets go, the more the separation decreases. So that initial diagram is incorrect, because it violates special relativity.
Post by: Eternal Student on 27/05/2023 02:31:36
I'm just not able to follow you, Eternal.That's ok. It's probably my fault. Also the entire Bells spaceship paradox is quite tricky.
Each rocket has an attached accelerometer, and those two accelerometers always show exactly the same acceleration. (That is part of the initial specification of the scenario).This part is ok. It's best to assume the accelerometers show a constant acceleration at all times. Have the acceleration shown on the accelerometer = a(t) = a = a constant, independent of how long the rocket has been in flight.
The INITIAL diagram says that the inertial observers who are stationary wrt the rockets immediately before the rockets fire, say that the separation of the rockets doesn't vary.Yes. That is the space-time diagram you would have for the rockets if they follow the prescribed behaviour (both rockets have the same acceleration a, showing on their accelerometers). In the frame of reference of those observers (which I will call the lab frame from here onwards), the worldlines of the two rockets are exactly as you've shown in your first diagram.
In the lab frame the separation of the two rockets would not vary with time.
But THAT violates special relativity: special relativity says (via the length contraction equation) that an inertial observer MUST conclude that a yardstick moving away from himself (in the direction of its length) is shorter than his own yardsticks.It doesn't violate special relativity. Yardsticks with some non-zero velocity in the lab frame would show contraction - but there weren't any yardsticks hanging between the two rockets, so it's not an issue.
Hypothetically you can imagine there were yardsticks between the rockets and indeed, if those yardsticks were to move with the rockets, then as the rockets accelerate, those yardsticks have to take up less space (contract) in the lab frame. That doesn't change the space between the rockets in the lab frame, that will be whatever is shown on the diagram (it is always measured along a straight line parallel to the x-axis and in our diagram that will be a constant). However, it does mean that using a frame of reference that moves with the rockets will disagree about the distance that exists between the rockets. An (instantaneous) rest frame for the rocket (i.e. a frame that moved with the rocket) shows that the space between the two rockets is bigger than that reported in the lab frame.
I know that was complicated so let's say the same thing again but in a different way: We have a lab frame, that's where all the rockets were at rest before the engines were started. Yardsticks that remain at rest in the lab frame, will remain 1 yard long in the lab frame for ever. So if you already had these yardsticks nailed up and held in the background of space, then as the two rockets race past them you could still see that at every moment of time, there were (say) 100 yard sticks between the rockets. Meanwhile, yardsticks that aren't at rest in the lab frame do show length contraction in the lab frame. Since these yardsticks are moving with the rockets you can't just have them nailed up on the background of what you're considering static space in the lab frame. Instead you need some other system where they do move along with the rockets. Do that by whatever system you care for, for example have a million other mini-rockets at your disposal, each mini-rocket precisely 1 yard long (proper length, or length measured when that rocket was stationary in the lab frame) and have them race after the two main rockets we are interested in and try to get themselves lined up nicely head-to-toe in-between the those two main rockets. You would find that you do need more than 100 of those yard-sticks (the moving ones, the 1-yard mini-rockets) to bridge the gap between the two main rockets. Indeed the faster the two main rockets go, the more that a yardstick or 1-yard mini-rocket undergoes contraction in the lab frame - so you'd find that more of the mini-rockets need to move in and join the line of mini-rockets bridging the gap between the two main rockets.
I hope that makes sense.
There has been no violation of special relativity required so far. The observers who remained in the lab frame were not "forced" to conclude the distance between the two main rockets was reducing. Instead all they notice is that while the distance stays the same in their frame of reference (which they can see by looking at the stationary yard sticks nailed up in the background), it can't be staying the same in a frame of reference that moved with the rockets (because more 1-yard mini-rockets had to be used to bridge the gap).
Best Wishes.
Post by: MikeFontenot on 27/05/2023 17:21:05
In order to be consistent with special relativity, that original diagram must be modified as I have previously described, with the lower curve remaining unchanged, but with the original upper curve (at each instant "t") being lowered so that the vertical distance between the lower curve and the new upper curve is the original constant vertical distance, divided by the factor "gamma". "gamma" is equal to
1 / (1 - v * v),
where "v" is the velocity of the trailing rocket, and is given by the slope of the lower curve. "v" increases from zero at "t" = 0, and asymptotically approaches the speed of light (1.0 lightyear per year) as "t" goes to infinity.
If you disagree with any of my above statements, identify the first such statement that you disagree with, and tell me exactly why you disagree with it.
Post by: Eternal Student on 28/05/2023 00:51:22
If you disagree with any of my above statements, identify the first such statement that you disagree with
This one:
But that contradicts special relativity
and tell me exactly why you disagree with it.Because I think it's wrong.
You seem to have said much the same thing in your last 3 posts and I know I've been trying to say much the same thing, just in some slightly different ways, in a few of my last posts. I'm not sure it's going to be very productive for me to say the same stuff again or vice versa.
You said, in one of your very earliest posts, that the diagram had been hanging on your wall for the last 20 years. I was guessing that you had studied Special Relativity at that time. If that's not correct, that's fine - I know I learn a lot of stuff and certainly wasn't born with much knowledge of anything. However, it suggests an alternative course of action would be sensible - perhaps we should cover some basic principles of SR before trying to tackle the tricky example of Bell's spaceship paradox. On the other hand, you may just have reached an alternative understanding or point of view and remain quite firmly set in that view. That's also fine, there's no law against it. I don't think I will easily change your opinion and equally, I don't think you'll change my opinion. If you bring something new then I may read it BUT on the short term I've already spent a lot of time here and I am way behind on other tasks.
What-ever happens I wish you well.
Post by: MikeFontenot on 28/05/2023 14:31:59
Hi.If you disagree with any of my above statements, identify the first such statement that you disagree with
This one:But that contradicts special relativityand tell me exactly why you disagree with it.Because I think it's wrong.
What a pathetic cop-out!
Post by: Origin on 28/05/2023 15:34:22
What a cop-out!It of course it is not a 'cop out' since ES has patiently pointed out your errors in multiple posts. I think it's disingenuous for you to imply that was their only reply was that you are wrong.
It appears to me that you came up with an idea that you think shows a flaw in Einstein's relativity and this is such an exciting prospect that you are refusing to consider any of the errors pointed out to you.
Post by: MikeFontenot on 28/05/2023 17:37:26
[...]
It appears to me that you came up with an idea that you think shows a flaw in Einstein's relativity [...]
[...]
You've got that exactly backwards! I think Einstein is the greatest physicist that ever lived. And I am the strongest of believers in the truth (and MEANINGFULNESS) of special relativity. But Einstein himself said he got a few things wrong: he regretted adding a cosmological constant to his theory of general relativity. And the fact that his exponential equation in gravitational time dilation was wrong (as I've proven) WOULD have been accepted by him if he had known about it while he was alive ... it had gotten past him only because he never applied gravitational time dilation (in its special relativity equivalent) to the case of near-instantaneous speed changes. Apparently no one else (besides me) has ever done that either.
You have also misunderstood what I've said about the original diagram in this thread. That original diagram (which is the basis of the contention that the thread breaks in Bell's spaceship scenario) is INCONSISTENT with special relativity ... i.e., that diagram VIOLATES special relativity. I've shown how to correct that diagram so that the new diagram IS consistent with special relativity. And that new diagram says that the inertial observers who are stationary with the rockets immediately before the rockets are fired, will conclude that the rockets get CLOSER TOGETHER during the acceleration. Therefore the thread does NOT break in Bell's spaceship scenario.
Post by: Origin on 28/05/2023 18:03:59
And the fact that his exponential equation in gravitational time dilation was wrong (as I've proven)No physicists believe that you have proven time dilation as expressed in relativity as wrong.
This is evident since time dilation is still taught in universities and I have not noticed any headlines proclaiming that "Mike has fixed Relativity.
Like I said before you seem to be caught up in the excitement in the belief that you have found a flaw in relativity and that is preventing you from seeing your own errors.
Post by: Halc on 28/05/2023 18:08:13
If you disagree with any of my above statements, identify the first such statement that you disagree with, and tell me exactly why you disagree with it.It doesn't work that way. We disagree with several of your statements, and the first one is perhaps not the source of your confusion.
I thought at first (years ago when the whole CADO thing was going on) that you had a basic grasp of at least special relativity, which is simple enough to be taught in high school. This seems to have degraded over time, and you've now descended into full purveyor of, how did you put it? Oh yea:
crackpot theoriesas evidenced by a total refusal to consider the possibility that all the other posters have a point and that these assertions you hold so tight might in fact be mistaken. We're giving up because you refuse to actually listen to having your errors pointed out. I'll give it one more try, and then let you 'win the argument' as you put it, if that's what your goal is.
So lets try to point out the core errors. I'll put the important ones in bold. First I'm going to name some frames. We have the initial inertial frame in which all ships are initially stationary and relative to which they all simultaneously commence identical (and unending) proper acceleration. That is frame F.
Then there is the inertial frame in which a given ship is momentarily stationary after one year as measured on the ship clock. This is S0 for the ship staring at F's origin, S0.5 for the ship starting at x=0.5, S3 for the ship starting at 3, and S-5 for the ship staring at x=-5. One can have an entire array of ships, each with its own frame. For the record, your diagram shows the line of simultaneity for S0.5.
The idea, I think, was that the two curves must have exactly the same shape because of "the Principle of Relativity" ... i.e., it shouldn't matter where in space you start the curve, the curves should always have the same shape.This is correct, but indirectly so.
The PoR just says that physics is the same for everybody doing the same thing, making local measurements. But this conclusion (that the curves in Minkowski spacetime are identical relative to the initial IRF)
So suppose we populate F with stationary mile posts, 100 per light year. Given that, each ship can look out the window and see his progress relative to F. Per the PoR, each ship is going to see the exact same number of posts go by during that year. From that one can demonstrate that indeed, the curves are identical relative to F, and that the separation between the ships relative to F must remain constant. This latter point is one you deny, but we'll get to that. Your chart from 20 years ago (presumably the one entered twice in post 6) is entirely correct. It depicts the scenario with identical proper acceleration, one of the two scenarios carefully distinguished by ES. The other one is rigid motion where the string doesn't break, which doesn't involve identical proper acceleration. This all was carefully explained by ES, but then dismissed because you 'know better'.
But the length contraction equation (LCE) of special relativity says that an inertial observer should conclude that a moving yardstick should get shorter and shorter as its speed wrt the inertial observer increases.This is also correct, but all your objects are point objects (the ships). You've not depicted any rigid extended objects like rulers. The string represents such a ruler. You could color the string red and green, switching every thousandth of a light year and write numbers on it. If it's pre-stressed, it won't stretch further and it would make a wonderful ruler. We can tow it behind the lead ship and not attach it at all to the trailing one so it can pull away from the trailing ship as it accelerates.
The above diagram (without the diagonal straight lines) shows the perspective of the two accelerating observers.This is wrong. The one chart you've posted shows F, the perspective of somebody who doesn't accelerate at all. You put out no pictures other than that one.
One thing that diagram DOESN'T show is how the ages of those two observers compare, as time progresses.It does show it, but since it depicts F, it shows their ages relative to F. Both curves have little age marks on them, showing their ages to always be identical relative to each other in frame F, even if they're both younger than the non-accelerating clocks.
The inertial observers who are stationary wrt the rockets immediately before the acceleration begins ...There doesn't need to be a mess of them. One observer (or none) will do. The frame defines the coordinates of all the events, not the observers. Any observer with any motion can still use frame F
Quote
... will say that the rockets get closer together as the acceleration progresses.This is very wrong. SR does not posit this nor does it conclude this. I think this is one of the most important assertions you erroneously believe. Unlearn this. It cannot be true.
Take the rocket starting at x=0. After a year it moves to F coordinate of about x=0.54 and is moving at ~.76c relative to F, for a dilation factor of 1.56 or so.
That means that a ship starting at x=-3 would have to move to x = 0.54 - (3/1.56) = 1.38 which is moving 1.62 in a year, which is over light speed.
A ship starting at x=5 would need to be at x=0.54 + (5/1.56) = 3.74, or moving backwards at over light speed despite accelerating forward. This is what results from the assertion you make. It cannot be. It would be evident if you ran some example numbers, but you repeatedly refuse to do so. Feel free to correct my arithmetic if you find it in error.
Looking at it from the S0 frame makes it even worse since the ship at x=-3 hasn't even pulled out of the parking lot yet when the ship defining S0 is a year into its trip. Or do you also deny relativity of simultaneity?
I don't believe that SR predicts that.Science isn't a religion. Belief hasn't a role to play. It's all about the mathematics working out or not, and your 'beliefs' mathematically lead to direct contradictions with the premises of SR.
After this you mostly just keep repeating the same assertions over and over without addressing any of the corrections provided. You do say this:
To obtain the correct diagram, at each instant of the given inertial observers' time, it is necessary to compute the gamma factor (where "v" is the speed of the rockets at that instant), and divide the constant separation "L" of the rockets (according to the observers on the rockets) by gamma. The result is then added to the location of the trailing rocket, to get the location of the leading rocket.But you never do this. You don't run any numbers. You don't provide a 'corrected diagram'. If you did, it would run into the contradiction demonstrated just above with things needing to move faster than light to get to where you insist they should be.
"If you're loosing an argument, change the subject."You feel the need to quote some anonymous crybaby. Are you implying that you channel this sentiment. You may notice that I'm not nearly so polite in my dealing with a stubborn crank as is ES, who seems to have not made a single mistake in his posts. I tend to walk away from conversations such as this.
Quote from: MarkTwain
Never argue with a fool; onlookers may not be able to tell the difference.That quote is not so anonymous.
That's ironic, because my postings take the existing diagram (which I've shown VIOLATES special relativity) and replace it with a new diagram (never before defined) which OBEYS special relativity!This is wrong, but also a lie. There has been no replacement diagram.
If you reject all this due to it contradicting your personal belief system about what SR says, then as ES says, I wish you all the best. If you're actually interested in correcting your knowledge of the theory, then I'll respond to questions.
And the fact that his exponential equation in gravitational time dilation was wrong (as I've proven)This is another error. You've yet to demonstrate even the beginnings of an understanding of what the equation in question calculates, let alone whether it is correct or not.
The rest is repetition. From other threads:
NOTE: My use of the phrase "Proper Separation" in the title of this submission means that it is the separation of the two people undergoing the acceleration, ACCORDING TO THOSE TWO PEOPLE THEMSELVES.Proper separation only applies between two objects that are relatively stationary. This is the same scenario in this topic, and at no time after commencement of acceleration (assuming we never cease acceleration) is either observer stationary relative to the other.
I also seem to remember that you've incorrectly asserted that opposite ends of an accelerating rigid object (a ship, a ruler, whatever) must experience identical proper acceleration. If I remember that incorrectly, I apologize. If you still assert that, that would be another important thing that SR neither posits nor concludes.
Post by: MikeFontenot on 28/05/2023 19:22:34
Einstein's exponential time dilation equation has only been tested for very small values of its argument L*A, where the exponential is essentially linear. It works fine in that linear range. It fails miserably for large values of "A" ... in particular, it disagrees with the outcome of the twin paradox, as I show in https://vixra.org/abs/2109.0076 .And the fact that his exponential equation in gravitational time dilation was wrong (as I've proven)No physicists believe that you have proven time dilation as expressed in relativity as wrong.
Post by: Origin on 28/05/2023 22:19:27
Einstein's exponential time dilation equation has only been tested for very small values of its argument L*A, where the exponential is essentially linear. It works fine in that linear range. It fails miserably for large values of "A" ... in particular, it disagrees with the outcome of the twin paradox,Millions of physicists over the past 100+ never noticed this obvious error? Does that make any sort of sense to you?
I know that you can't, for some reason, allow yourself to admit you are wrong on this. I personally think this is a sad situation since you appear intelligent, but in reality if this makes you happy then have fun fighting this battle.
Post by: MikeFontenot on 28/05/2023 22:47:04
Post by: MikeFontenot on 28/05/2023 22:53:10
Millions of physicists over the past 100+ never noticed this obvious error?
Have YOU ever applied the (special relativity version of the) gravitational time dilation equation to the twin paradox turnaround? Apparently no one has but me. It's not hard to do. See what you get when YOU do it. I do it in
https://vixra.org/abs/2109.0076
Your silence is deafening! Did you do the calculation? Did you get the same answer I got, or a different answer? Why the silence?
Post by: MikeFontenot on 10/06/2023 17:05:31
Scan 2023-4-23 17.01.18.jpg (243.13 kB . 1700x2338 - viewed 2764 times)
The above diagram (which I produced many years ago, and until very recently believed to be correct) is INCONSISTENT with special relativity. That fact is still being ignored by people who should now know better.
In the above diagram, the vertical and horizontal axes refer to the viewpoint of the initial inertial observers who are stationary wrt the rockets immediately before they ignite. The two curves show the progress of each of the two rockets.
The two sloped straight lines refer to inertial observers momentarily stationary wrt the leading rocket at two times later in the acceleration. They say that the two rockets get farther apart during the acceleration, and so they say that the thread will break in Bell's Spaceship Paradox.
The diagram says that, according to those initial inertial observers, the distance between the rockets is CONSTANT during the entire acceleration. BUT THAT CONTRADICTS SPECIAL RELATIVITY: special relativity says that ANY inertial frame that is moving wrt the initial inertial observers will have SHORTER yardsticks than the inertial observers' yardsticks, according to the initial inertial observers. Therefore, according to the initial inertial observers, the distance between the rockets continually DECREASES during the acceleration. Therefore the above diagram (which says otherwise) is INCONSISTENT with special relativity.
In order to be consistent with special relativity, the above diagram must be modified, with the lower curve remaining unchanged, but with the upper curve (at each instant "t") being lowered so that the vertical distance between the lower curve and the new upper curve is the original constant vertical distance, divided by the factor "gamma". "gamma" is equal to
1 / sqrt(1 - v * v),
where "v" is the velocity of the trailing rocket, and is given by the slope of the lower curve. "v" increases from zero at "t" = 0, and asymptotically approaches the speed of light (1.0 lightyear per year) as "t" goes to infinity.
For example, when v = 0.866 ly/y, gamma = 2.0 . So when v = 0.866 ly/y, the distance between the rockets is half of what the above diagram incorrectly says it is.
Post by: Origin on 11/06/2023 14:48:36
The above diagram (which I produced many years ago, and until very recently believed to be correct) is INCONSISTENT with special relativity.Then stop bringing it up since wrong and move on. ::)
Post by: MikeFontenot on 11/06/2023 17:40:34
The above diagram (which I produced many years ago, and until very recently believed to be correct) is INCONSISTENT with special relativity.Then stop bringing it up since wrong and move on. ::)
I bring it up because most people still believe that the thread will break in Bell's Paradox (which is what that original diagram implies). The corrected diagram, which I describe, shows that the thread in Bell's Paradox does NOT break. Most people still believe I'm wrong about that.
And, after I had shown that the original diagram is incorrect and violates special relativity (and after I had described the corrected diagram which agrees with special relativity), Halc moved this discussion from the main section (which is supposed to be "the special relativity section") to the "basement" non-relativity-theories section. Halc has it exactly backwards!
Post by: MikeFontenot on 27/06/2023 19:56:44
Post by: MikeFontenot on 27/06/2023 20:04:37
[ Invalid Attachment ]
Post by: MikeFontenot on 28/06/2023 18:21:41
two rockets (immediately after they are ignited) always have the same
constant acceleration, as reported by accelerometers attached to the two
rockets.
In that scenario, the separation of the rockets, according to the accelerating traveler in the trailing rocket, is constant, and the string doesn't break.
According to the initial inertial observers who are stationary wrt the rockets immediately before they are fired, the separation of the two rockets decreases as the acceleration proceeds, as required by the famous length contraction equation (LCE) of special relativity. So they also conclude that the string doesn't break.
The scenario, as given in Bell's Paradox, may be a completely different scenario from the above scenario. As far as I know, there is no mention of rocket accelerometer readings in that Wiki article on Bell's Paradox. If, in Bell's Paradox, the initial inertial observers correctly conclude that the rocket separation doesn't decrease, then the rockets CAN'T be accelerating at the same rate (according to accelerometers attached to the rockets) ... the leading rocket must have a greater acceleration than the trailing rocket. If so, I have no interest in that different scenario.
Post by: paul cotter on 28/06/2023 21:16:35
Post by: pzkpfw on 28/06/2023 23:53:47
But two accelerometers on the _same spaceship_ (one at the front, one at the back) won't even agree with each other.
Post by: MikeFontenot on 01/07/2023 19:00:58
It is possible to show that the string doesn't break in Bell's Paradox (assuming that the accelerations of the two spaceships are confirmed with accelerometers to be equal), without referring to ANY inertial observers.
In 1907, Einstein derived the gravitational time dilation equation (GTD), which says that if two clocks are separated by a fixed distance "L", in a constant gravitational field (with the separation along the direction of the field), such that the field strength doesn't vary with distance from the source of the field, then the clock farther from the source of the field will run faster than the clock closer to source of the field, by the factor f(L*g).
From the equivalence principle, that means that two spaceships (with no gravitational fields anywhere), each accelerating with a constant acceleration "A" (as confirmed by accelerometers), and separated by the distance "L" in the direction of the acceleration at some instant, will always maintain that separation (for as long as that acceleration equals "A"). And clocks in the leading spaceship will run faster than clocks in the trailing spaceship, by the factor f(L*A).
Since the distance between the two spaceships is constant, a thread stretched between them never breaks.
Post by: paul cotter on 01/07/2023 21:12:58
Post by: Halc on 01/07/2023 22:59:05
Your 'corrected' graph is interesting. I copied it and added some lines for ships at 1, 1.5, and 2. I did my best to keep the separation of successive ships identical. I don't have numbers, so I did it by eye. I didn't take as many data points so my curve isn't as nice as the original ones. I get this:
FiveShips.jpg (539.02 kB . 1613x1457 - viewed 1114 times)Note that all the new ships actually, which accelerating forward at first, move backwards initially.
This makes for an interesting way to communicate faster than light.
The superbowl is played at x=0. If the Bills win, the ship there takes off (at exactly kickoff time + 4 hours) at fixed proper acceleration of 1 ly/yr2. If they lose, they stay put.
There is another ship at x=1.5 ly. It takes off unconditionally at kickoff time + 4 hours and accelerate at a fixed proper acceleration of 1 ly/yr2. If the ship moves backwards (or for that matter, doesn't reach the 20 meter post in 2 seconds), the Bills have won. If not, they've lost. Presto, faster than light communication.
Concerning your plot:
The ship starting at 0 does seem to be accelerating at the correct rate. Assuming rigid motion (no string breakage), the other ship seems to do ok for at least a year, but looking at year 2, ship0 is moving at 0.9c (hard to measure slope to that precision), location 1.32 (should be 1.25), and gamma of 2.3. 0.5/2.3 is ~0.22 but your contraction is more than that. They seem about 1.4 apart at t=2 Your lower ship is accelerating harder than 1 ly/yr2.
All sort of other silliness results from your assertions. You don't care of course. Your reaction to any critique since the beginning is just to repeat the opening argument with the same mistakes, as you have done yet again in what was your new topic.
In the scenario that I am interested in, and which I have analyzed, the two rockets (immediately after they are ignited) always have the same constant acceleration, as reported by accelerometers attached to the two rockets.Then the accelerometers are broken, because if they look out the window and watch the graph lines go by (one every 0.1 ly), one ship notices a lot more of them going by than the other. Your assertion contradicts your graph.
In 1907, Einstein derived the gravitational time dilation equation (GTD), which says that if two clocks are separated by a fixed distance "L", in a constant gravitational field (with the separation along the direction of the field)The paper says no such thing. You're making things up. You've become a crank, which is too bad.
Anyway, citation needed. Where does Einstein associate gravity with the equation mentioned? What coordinate system is Einstein using? Where does he make any suggestion of the possibility of a uniform gravitational field? Sure, they exist in Newtonian mechanics, but we're not talking about that anymore.
Post by: MikeFontenot on 03/07/2023 18:22:25
Note that all the new ships actually, which accelerating forward at first, move backwards initially.
No they don't. You are apparently not following my description of how the curves are to be determined. Try again.
Post by: MikeFontenot on 04/07/2023 17:47:16
Note that all the new ships actually, which accelerating forward at first, move backwards initially.
That doesn't happen. You have apparently misinterpreted my description of how to produce the new chart. I'll explain it again.
(In this scenario, two spaceships are initially separated at time t = 0 by distance "D", with their rockets off. People on the spaceships will say that the separation between the spaceships is constant, before and after the rockets are fired. After the rockets are fired, the constant acceleration is confirmed by accelerometers on the two spaceships.)
To get a chart that shows the conclusions of INERTIAL OBSERVERS who are stationary wrt the spaceships immediately before the rockets are ignited:
Start out with the two identical curves separated by some fixed distance "D", one starting at the origin, and the other starting a distance "D" up the vertical axis.
The bottom curve is obtained this way:
The rapidity "theta" is
theta(t) = A * t,
where "A" is the constant acceleration that begins at t = 0.
The velocity is
v(t) = tanh(theta[t]);
The distance traveled is
d(t) = ln(cosh[v{t}]) .
That should allow you to plot the lower curve (which starts at the origin). And you can plot the upper (incorrect) curve by just moving that lower curve vertically up by the amount "D".
That chart is wrong, because the famous length contraction equation (LCE) of special relativity says that an inertial observer (stationary with the two spaceships immediately before their rockets are fired) will conclude that the spaceships get closer together as their speed increases, by the factor
gamma = 1 / sqrt(1 - v * v) .
Gamma is equal to 1.0 when v = 0, and increases monotonically as "t" increases.
To make the two curves get closer together as "t" increases, we can't raise the lower curve, because the lower curve already asymptotically approaches the speed of light as "t" goes to infinity. So we need to lower that upper curve. We do that by, at each time "t", dividing the original separation between the two curves "D" by gamma(v[t]). Gamma starts out equal to 1.0 at t = 0, but then increases (slowly at first, but increasingly fast) as "t" and "v" increase. For example, when v = 0.866 ly/y, gamma = 2.0 .
If you do the above correctly, you will get the revised diagram that I gave. Neither of the two curves ever decrease (or stay constant) as "t" increases.
Post by: Origin on 05/07/2023 21:03:01
That chart is wrong, because the famous length contraction equation (LCE) of special relativity says that an inertial observer (stationary with the two spaceships immediately before their rockets are fired) will conclude that the spaceships get closer together as their speed increasesWhy do you think this? Clearly each ship will be length contracted, but why would space be length contracted, certainly the space is not moving relative to the stationary observer right?
Post by: MikeFontenot on 05/07/2023 22:35:10
It's understandable that reproducing my diagrams may be difficult for you ... you may not have access to a computer that allows you to compute the tanh, cosh, and log functions. I can scan a page that gives v, gamma, and d1 and d2 for t = 0 to t = 3 in increments of 0.1, and I will try to post it.
But in the meantime, let me try again to qualitatively explain what's going on. Regardless of what initial separation you choose (0.5 like I did, or 1.0, or 2.0 ...), the process is the same: for a given value of "t", you determine "v" on the bottom curve, and then you compute the corresponding gamma for that velocity. (gamma = 1/sqrt{1-v*v}). You then take the distance between the two curves at t = 0 (call it D), and divide that by gamma ... call that "d". Then add "d" to the value of the lower curve at that time "t". The result is the value of the upper curve at that time "t". The important point to understand is that that process is the same regardless of how large you chose "D". The curves you get for the various choices of "D" are all qualitatively the same ... they are just scaled up.
Just FYI: Here's how the bottom curve is determined:
For each value of "t", determine "theta" (the rapidity) at that time from theta = A * t. (I had chosen A = 1 ly/y/y.) Then compute v = tanh(theta). Then compute d = integral of v. It's not necessary to do the numerical integration: the closed form solution is d = ln(cosh[theta(t)]).
Post by: MikeFontenot on 05/07/2023 22:43:51
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Post by: MikeFontenot on 05/07/2023 22:45:53
Post by: MikeFontenot on 05/07/2023 22:52:49
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Post by: MikeFontenot on 06/07/2023 00:14:46
To get a chart that shows the conclusions of INERTIAL OBSERVERS who are stationary wrt the spaceships immediately before the rockets are ignited:I have no beef with the chart in post 6 that shows this. It seems totally accurate.
In the above, you seem to be OK with the original diagram (the one showing a constant separation), which I now know is incorrect. The axis labeling uses terms from a long time ago, and don't need to be discussed now. Basically, the horizontal axis gives the time (or age) of the initial inertial observers who are stationary wrt the two rockets before they are fired, and I would today label that "t". And similarly for the distance "d" being the distance from the starting locations to the current locations, according to those initial inertial observers (although when I drew that sketch, I used "x" for the vertical axis, not "d"). But my intent was to use the same axes on the revised diagram as I used on the original diagram. They both show the viewpoint of the initial inertial observers, and I don't discuss the viewpoints of the people in the rockets at all.
Then, you discuss (I think) my hand-drawn revised diagram, and there, you criticize my labeling, etc. As I look at that hand-drawn diagram now, I agree that I didn't take the time to define the x and t axes as being the viewpoints of the initial inertial observers ... probably because I thought it was obvious that I wanted the same perspective as in the original diagram.
You then said that I use the same symbol t for both Earth (IRF) time and ship time. That's not true: I don't discuss the viewpoint of the people in the rockets at all in that entire discussion.
Post by: MikeFontenot on 06/07/2023 01:09:23
Post by: MikeFontenot on 06/07/2023 17:42:56
Post by: MikeFontenot on 07/07/2023 03:19:04
Your chart says at t=1, d1 is 0.4338, v=.7616, gamma=1.543
If you take D at 2 (the top line of the edited picture I posted), it starts at x=2 at time 0.
2/gamma is 1.2962 which we add to d1 0.4338 to get 1.73 which is exactly where I drew the data point.
In the above, you use the value of gamma at t = 1 to divide the distance between the curves at t = 0 to get the new curve at t = 1. That's completely incoherent!
I think any further discourse would waste both our times.
Post by: MikeFontenot on 07/07/2023 16:58:08
I found your post extremely hard to follow. But I THINK I see a (really bizarre) mistake you're making here:
Your chart says at t=1, d1 is 0.4338, v=.7616, gamma=1.543
If you take D at 2 (the top line of the edited picture I posted), it starts at x=2 at time 0.
2/gamma is 1.2962 which we add to d1 0.4338 to get 1.73 which is exactly where I drew the data point.
In the above, you use the value of gamma at t = 1 to divide the distance between the curves at t = 0 to get the new curve at t = 1. That's completely incoherent!
I think any further discourse would waste both our times.
I said:
"In the above, you use the value of gamma at t = 1 to divide the distance between the curves at t = 0 to get the new curve at t = 1. That's completely incoherent!"
A better way for me to say that is:
"In the above, you use the value of gamma at t = 1 to divide the distance "D" between the curves at t = 0 to get the portion of the new curve for 0 < t < 1." THAT violates the principle of causality: effects can't precede causes." In other words, the value of gamma at t = 1 has NO effect on anything that happens before t = 1. The value of gamma at any instant can only affect things on or after that instant.
So, what you need to do is, for any t = T where you want a data point for the upper curve, determine gamma(T) for the lower curve, and then divide the original separation "D" by THAT value gamma(T), and add that result to the value of the lower curve at time T, and plot that point.
The result is that no matter what the chosen value of "D" is, the resulting curve will be monotonically increasing everywhere, as time "t" increases. The curve will not anywhere DECREASE as time increases. No matter what the choice of "D" is, the shape of the resulting curve will be qualitatively similar to the curve I gave for D = 0.5. Specifically, it just gets scaled up by the ratio of the two choices of "D".
For example, once you've got the D = 0.5 curve (which we DO have), if you want the D = 1.0 curve, just divide 1.0 by 0.5 to get 2.0, and then,
for each value of "t" you want, multiply the vertical distance from the 0.5 curve down to the bottom curve by 2.0, and THAT gives the vertical distance from the D=1.0 curve down to the bottom curve.
And, to get the D = 2.0 curve, just divide 2.0 by 0.5 to get 4.0, and then,
for each value of "t" you want, multiply the vertical distance from the 0.5 curve down to the bottom curve by 4.0, and THAT gives the vertical distance from the D=2.0 curve down to the bottom curve.
Don't don't lose sight of the original plan. You start with the original (incorrect) diagram, where the two curves were identical except for their starting point on the vertical axis ... the upper curve was just the lower curve, shifted up by the original distance "D" between the spaceships. That corresponds to the contention that, ACCORDING TO THE INITIAL INERTIAL OBSERVERS WHO ARE STATIONARY WRT THE SPACESHIPS BEFORE THE ROCKETS ARE IGNITED, the two spaceships maintain a constant distance apart during the acceleration. That contention is wrong.
So, you use the fact that those initial inertial observers MUST instead conclude that that original diagram is incorrect, because the two spaceships MUST get closer together, by the factor gamma, as their speed increases ... that is required by the Length Contraction Equation (LCE) of special relativity. So those initial inertial observers then conclude that the two spaceships must get closer together, by the gamma factor, as their speed increases ... i.e., the distance "d" between the spaceships must be d = D/gamma, where gamma is a function of the velocity, and the velocity is a function of the time "t".
But HOW do the two curves change, to get the correct separation "d"? The lower curve can't be moved upward, toward the upper curve (because it already gets arbitrarily close to the speed of light as the duration of the acceleration approaches infinity). It is the upper curve which must be modified ... it must not curve upward as much as the lower curve curves upward. Its curvature must change so that, for each time "t", the vertical distance between the two curves equals the original separation "D" divided by the value that gamma has for the speed at that moment "t".
Post by: MikeFontenot on 09/07/2023 20:50:29
Post by: MikeFontenot on 09/07/2023 20:55:28
Here is the graph for D = 0.5 and 1.0, and the data from the program.
I wanted full size. I'll try again.
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Post by: MikeFontenot on 09/07/2023 21:00:39
Post by: MikeFontenot on 09/07/2023 21:03:58
Trying again to get full size
Post by: MikeFontenot on 10/07/2023 17:46:12
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Post by: MikeFontenot on 10/07/2023 17:48:29
And here is the computer output for the D = 3 case:
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Post by: Halc on 10/07/2023 18:49:22
Your own graph contradicts your own denial. Ships at 2 and 3 are both moving backwards. A ship at 4 would do so faster than c. You seem to not find any of this problematic. Faster than light communication, and even faster than light ships, accelerating in the opposite direction of where their accelerometers indicate. Ship at 1 accelerating at 1g but not actually going anywhere for weeks.Note that all the [ships with greater initial separation] actually, which accelerating forward at first, move backwards initially.No they don't. You are apparently not following my description of how the curves are to be determined. Try again.
So what's up with all this nonsense? Why are you sticking with it?
Post by: MikeFontenot on 10/07/2023 19:11:09
Post by: Halc on 10/07/2023 21:17:25
at least that's what my computer program says, and I haven't been able to find any errors in it, so far.I've pointed out about 6 of them so far. If you paste your code, I can be more specific. As I said, excepting the d1 numbers, I reproduced all your numbers (every one of which is wrong).. Code would show how you got those. None of it is consistent with SR.
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The more accurate smooth curves are consistent with your straight-line approximations.My approximations took only 4-5 data points each, not 30, and they worked with only 1.5 digits of precision since you had not yet put out the number table when I drew that. I simply did it by eye.
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I need to look back at that analysis, and try to see where it goes wrong.Doesn't occur to you to just look at the posts in this thread, which says pretty much what goes wrong? Even if you got the d1, v, and gamma numbers correct, if you still do the same trick to get d2 numbers, you'll still get the backwards, faster than light motion. You're completely disregarding relativity of simultaneity in the whole analysis. Under SR, if a rigid accelerating object is stationary (has everywhere identical velocity) for a moment in an inertial frame, it cannot have identical velocity anywhere in any other frame, so the gamma cannot apply along its entire length like you are attempting. It would be a contradiction if this were not so.
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At least, the latest results still DO say that the thread doesn't break!Your program shows that the thread would have to move faster than light in order to be that length in the original inertial frame. I notice you don't compute its length in any other frame. Guess what? It breaks (or bunches up, depending on frame of choice). The SR solution has none of these problems.
Post by: MikeFontenot on 10/07/2023 21:57:42
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Post by: Halc on 10/07/2023 22:38:42
I'll try to print out my program:One programming nit bug: d2 is assigned twice, the first one being immediately discarded. It doesn't affect the results, but that line can be deleted.
Thanks for the code. I updated one line of mine and it now reproduces yours.
I don't have any input, just a couple of defines at the top for D0 and MF, so a recompile is needed to change these. You can add the scanf to make that easier.
The MF variable is boolean, 1 to do it your way, and 0 to do it the SR way.
I don't specify A. Since it is always 1, the math is a little simpler. It can be added in to get a different curve than this sort of unit acceleration.
And my compiler will not compile yours without stdio.h. Yours is apparently fine with referencing undeclared functions, or maybe your math.h includes stdio. Mine doesn't.
Code: [Select]
#include <stdio.h>
#include <math.h>
#define DD 0.5
#define MF 1 // 1 to print Mike's number, 0 to print mine.
int main()
{
double ctime; // Coordinate time of inertial frame
double ptime; // Proper time of ship
double md1; // Computed distance per Mike
double srd1; // Computed distance per SR
double vm, vsr; // Speed per Mike and SR
double gim, giSR; // Gamma factor inverted, per Mike & SR
int mf = MF;
printf("ctim%s d1 d2 v gamma\n", mf ? "" : " ptime ");
for (ctime = 0.1; ctime <= 3.05; ctime += 0.1)
{
ptime = asinh(ctime); // proper time
// Compute speeds as a function of time
vm = tanh(ctime);
vsr = tanh(ptime);
// Compute gamma as a function of those speeds
gim = sqrt(1 - vm * vm);
giSR = sqrt(1 - vsr * vsr);
// Compute distance traveled by lower ship
md1 = log(cosh(ctime));
srd1 = cosh(ptime) - 1.;
// I think we're good. Print results
if (mf)
printf("%.1f %.5f %.5f %.6f %7.4f\n",
ctime, md1, md1 + gim*DD, vm, 1./gim);
else
printf("%.1f %.5f %.5f %.5f %.6f %8.4f\n",
ctime, ptime, srd1, srd1 + DD, vsr, 1./giSR);
}
}Note the corrections in computing
time: ptime vs ctime,
velocity: vm and vsr,
gamma gim and giSR,
md1 and srd1 (distance of rear ship).
In my code d2 is just d1 + D. Any other curve and the 2nd ship simply doesn't have 1G of acceleration.
Your have d2 being a function of what d1 does, which is instant cause/effect at a distance, which is, as I've pointed out, faster than light communication, and every retrocausality in different frames. This (plus the faster than light travel, backwards motion when accelerating forward, etc) all don't seem to strike you as contradictory.
Post by: MikeFontenot on 11/07/2023 00:38:12
In my code d2 is just d1 + D.
The whole point of this exercise is that the length contraction equation (LCE) of special relativity REQUIRES that the initial inertial observers say the distance between the spaceships MUST decrease by the gamma factor as the speed increases. That's one of the two most important equations in all of special relativity (the other being the time dilation equation (TDE) ). And this diagram IS from the point of view of the initial inertial observers. So the appropriate equation is
d2 = d1 + (D / gamma).
You aren't working the appropriate problem.
Post by: MikeFontenot on 11/07/2023 16:54:27
Post by: MikeFontenot on 15/07/2023 01:02:34
Post by: MikeFontenot on 15/07/2023 18:11:09
But you're already positing going FTL, [...]
Why do you say that?
Update: OK, I see why you said that. I ran the case where D = 6. It shows a maximum velocity for that leading rocket to be about 2.3 ly/y between t = 0.8 and t = 0.9.
But I'm not sure that actually violates any rules. Lets just start with the length contraction equation. It says an inertial observer (he) at some time t1 will say a yardstick moving at speed v1 wrt himself is only 1/gamma_1 yards long. Suppose that observer then instantaneously changes his speed with respect to that yardstick to v2 (but then becoming inertial again), so that the yardstick is instantly 1/gamma_2 yards long. So in an infinitesimal time, the yardstick has instantaneously gotten shorter, which means that one or both ends of that yardstick just moved at an infinite speed (drastically greater than the speed of light).
My question to you (Halc) is, how would YOU apply the length contraction equation in this case? And, a more general question for you, is what does the SR theory that you mentioned have to say about this scenario? I.e., what is your alternative solution?
New addition:
My GUESS is that your alternative solution is that the upper curve should look just like the bottom curve, just shifted up by "D". (i.e., like the original chart that started this whole discussion). If I'm right about that, how do you square that chart with the length contraction equation, which says the inertial observer MUST conclude that the separation of the two rockets must get smaller by the factor gamma?
Still newer addition:
That original chart was claimed to be the perspective of the initial inertial observers, who were stationary wrt the rockets before the rockets were fired. For that to be true, it means that the people on the rockets would say that the separation of the two rockets was INCREASING with time, which means that the leading rocket was producing more thrust that the trailing rocket. So that is a different scenario from what we've been talking about.
What that original chart ACTUALLY showed was the perspective of the people on the trailing rocket. THEY said the separation of the rockets is constant, and that accelerometers on the two rockets show the same acceleration. And for that chart, the viewpoint of the initial inertial observers is the one we've been discussing (where the upper curve is above the lower curve by the distance D / gamma).
Post by: MikeFontenot on 15/07/2023 21:26:41
Post by: MikeFontenot on 15/07/2023 21:45:58
Post by: MikeFontenot on 15/07/2023 23:18:52
Post by: Halc on 16/07/2023 14:45:52
But I'm not sure that actually violates any rules.Yea, I noticed that violations of causality and locality don't seem to bother you. You don't seem to know the difference between an abstract coordinate choice and physical causation. You don't mind an accelerometer on a ship that lies and says the ship is accelerating forward when in fact it is accelerating the other way.
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how would YOU apply the length contraction equation in this case?I wouldn't of course. The equation is only applicable to describe the coordinate separation between two parallel straight (unaccelerating) worldlines. There are none of those in this case.
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I.e., what is your alternative solution?It's not my solution. My program prints the numbers per SR, as did your chart from 20 years ago. Read any website on Bell's spaceships, since that is exactly this scenario. None of it is anything I'm personally speculating.
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If I'm right about that, how do you square that chart with the length contraction equationSince the equation is entirely inapplicable to accelerating worldlines, there's no conflict with it.
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which says the inertial observer MUST conclude that the separation of the two rockets must get smaller by the factor gamma?This is your assertion, completely unbacked, and driven to obvious self contradiction, which doesn't seem to bother you.
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For that to be true, it means that the people on the rockets would say that the separation of the two rockets was INCREASING with timeWhich is why the 'string' breaks, yes. The rest of your addition to this post is just more trolling.
I don't really think you care what anybody says. You seem entirely closed to actually learning something (that you've apparently forgotten over time). I see little point in going on. Post your nonsense. Put it online and sell it on Amazon, so everybody can have a good laugh. Post on another forum and ignore their corrections. But remember: by definition, you cannot be wrong about this. It's the rest of the world that has gotten it all wrong for over a century.
Post by: MikeFontenot on 19/07/2023 00:48:53
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Post by: MikeFontenot on 19/07/2023 20:53:57
The original (incorrect) curve (which shows a constant distance between the two rockets, according to the initial inertial observers who are stationary wrt the rockets before the rockets are fired), violates the length contraction equation (LCE) of Special Relativity. The curve below shows what the initial inertial observers conclusions must be, according to the LCE.
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Post by: MikeFontenot on 20/07/2023 02:15:25
Post by: MikeFontenot on 20/07/2023 15:48:05
https://vixra.org/abs/2307.0151
* But the answer it gives when there are only two rockets is still incorrect, even though it doesn't produce an obvious inconsistency. The only correct method is the previous one, where the trajectory of a sufficiently distant rocket has a negative slope on the chart initially, before curving upward eventually .
Post by: MikeFontenot on 10/08/2023 15:04:06
Title: A Proof that the Separation Between Accelerating Rockets is Constant
Author: Michael Leon Fontenot
email: [removed by mod]
____________________________________________________________________________________
Abstract:
For rockets whose accelerometers show identical, constant readings, their separation is constant. The proof of that fact makes use of the limit of a sequence of accelerations "A", lasting for a time "delta_t", such that the total change in rapidity "A*delta_t", and therefore the total change in the velocity, don't vary for each iteration of the sequence of accelerations. In the limit, as "A" goes to infinity, and "delta_t" goes to zero, the velocity of the rockets changes instantaneously, and their separation doesn't change. The result is analogous to the CoMoving Inertial Frame (CMIF) simultaneity method of Special Relativity, which says that, according to the traveling twin (him), the home twin (she) instantaneously gets older during his instantaneous turnaround. Likewise, the ages of the people on the leading rocket instantaneously get older during their instantaneous velocity change.
____________________________________________________________________________________
The above abstract really says all that needs to be said about the proof that the separation between accelerating rockets is constant. The only thing that would be useful to add, is to elaborate a bit about the CoMoving Inertial Frame (CMIF) simultaneity method used to resolve the twin paradox, and to give the "delta_CADO" equation that makes the CMIF method especially easy and quick to use.
The CMIF simultaneity method says that the accelerating person (the "AP") must agree with the inertial person ("IP") who is momentarily stationary with respect to the accelerating person at any given instant. In the case of the instantaneous turnaround, there is an IP1 immediately before the turnaround, and an IP2 immediately after the turnaround. For each of those IP's, their line of simultaneity ("LOS") can be plotted on a Minkowski diagram. Where those two LOS's intersect the home twin's world line then give her age, according to the AP, immediately before and after the turnaround. I.e., that gives the amount by which she instantaneously ages during his turnaround, according to him.
It's even easier to get that instantaneous age change by using the "delta_CADO" equation:
delta_CADO = - L * delta(v),
where
delta(v) = v_after_turnaround - v_before_turnaround,
and "L" is their separation, according to HER. Velocities are positive when directed away from her, and negative when directed toward her.
Post by: MikeFontenot on 14/08/2023 23:58:07
There are two DIFFERENT scenarios:
Scenario 1: The two accelerometers always display the same value during the trip, and the thread DOESN'T break.
Scenario 2: The initial inertial observers REQUIRE that the separation THEY measure is constant, and therefore those inertial observers conclude that, in the rocket frame, the separation must be increasing, and so the thread WILL break.
In the WIKI article on the Bell's Paradox, it says (for the initial inertial frame "S")
"The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S, it is effectively DEFINED to remain the same, due to the equal and simultaneous acceleration of both spaceships in S."
So they didn't specify that the rockets had accelerometers that showed equal constant readings. And, most important, they specified that the separation was constant, ACCORDING TO THE INITIAL INERTIAL FRAME. That rules out the separation being constant in the trailing rocket's frame ... it REQUIRES that the separation increases in the rocket's frame and it REQUIRES that the accelerations are different, as measured by the accelerometers.
The two scenarios are DIFFERENT, and it's not surprising that their conclusions about the survival of the thread are different.
Acknowledgement: Neddy Bate was the first one to recognize that the two scenarios are different, and I am grateful for his insight.
Post by: pzkpfw on 15/08/2023 02:01:31
Post by: MikeFontenot on 15/08/2023 16:39:23
If you have accelerometers at the front and back of a single spaceship, will they show the same acceleration?
That's not a question that interests me.
Post by: MikeFontenot on 15/08/2023 17:43:03
What if a rigid rod connects the two separated spaceships. Perhaps with a rigid connection to the leading spaceship, but with a sliding connection to the trailing spaceship. And with distance markings printed on the rod, like a tape measure.