aintnuthin
My friend and I have a disagreement about the solution to a rather simple relativity problem, as follows:

Jack and Jill are in relative motion at the rate of .6c, with Jill being "stationary" at Point A, with Jack about to pass her at that point. Jack has previously established two synchronized clocks in his frame, which are 6 light seconds apart. One (clock 2) he has with him and the other "lead" clock (clock 1) encounters Jill first. After clock 1 passes by Jill, Jack, with clock 2, passes her.

The question is about how Jill will see these two events in her frame.

We agree that Jack will the the elapsed time as being 10 seconds and, as established, the distance to be 6 light seconds. The question is about Jill's time and distance. Two alternative answers have been proposed:

Answer A: Jill's time and distance will be 8 seconds and 4.84 light seconds

Answer B: Jill's time and distance will be 12.5 seconds and 7.5 light seconds

Which, if either, do you think is correct?

aintnuthin
A brief summary of the reasoning behind each proposed answer:

Answer A: Since the clocks are co-moving with Jack, Jill's distance must be less.

Answer B: Since, by stipulation, Jill is stationary, her distance must be greater.

We agree that the rate of change for both time and distance will be 80%, it's mainly just a question of whether this factor would result in more or less distance (and time) for Jill.
rpenner
Right from the beginning, neither you nor your friend understand thing one about relativity if you can call something stationary and expect that designation to matter, or attribute meaning to place-names like "A".

There are two "events":
E: Jill and the first clock meet
F: Jill and Jack meet.

In Jack's coordinate frame F happens at (x=0,t=0) so E happens 6 light-seconds to the left and 10 seconds in the past. E happens at (x=-6, t=-10).
F - E = (Δx=6 light seconds,Δt=10 seconds) and the invariant interval is
I = (cΔt)˛−(Δx)˛ = 64 light-seconds˛

In Jill's coordinate frame, we could set up a Lorentz transform, plug in the velocity, double-check that we got the signs right, convert events E and F to x' and t' and then find out what Δx' and Δt' where. Or we could just note that by definition, Δx' = 0 since both events E and F happen to Jill. so I = (cΔt')˛−(Δx')˛ = (cΔt')˛ = 64 light-seconds˛ and show that Δt' = 8 seconds exactly for Jill.
aintnuthin
QUOTE (rpenner+Aug 3 2011, 04:40 PM)
Right from the beginning, neither you nor your friend understand thing one about relativity if you can call something stationary and expect that designation to matter, or attribute meaning to place-names like "A".

Thanks for your response, Rpenner. The term "stationary" here is basically derived from the proposition that "moving clocks run slow," and may be misleading in that sense.

Here's the question: Is there any way to interpret the information given in this problem is such a way that Jill ends up being the one with the greater amount of time on her clock (and therefore measures the greater distance)?

As a matter of convention, established for the purpose of getting a starting point for constructing a spacetime diagram, the rule is to start with the frame where the clocks are co-moving.

But is that mathematical convention a universal law of phsyics? Is there any physical law that prevents a person with clocks which are co-moving in his frame from having a lesser time and distance than something travelling at .6c with respect to it?
aintnuthin
Conventions aside, would this scenario be physically possible?

Establish a point from Jill's frame (constructed from the premise that it is a stipulation that her time and distance must be longer):

In Jill's coordinate frame F happens at (x=0,t=0) so E happens 7.5 light-seconds to the left and 12.5 seconds in the past. E happens at (x=-7.5, t=-12.5).

F - E = (Δx=7.5 light seconds,Δt=12.5 seconds) and the invariant interval is
I = (cΔt)˛−(Δx)˛ = 100 light-seconds˛

Now delta t' would be 10 seconds exactly for Jack.
aintnuthin
Just in case I haven't sufficiently cleaned up the mess I seem to have created, let me add this.

The disagreement with my friend was not, ultimately, about what answer a standard approach would give you.

It was about the difference, if any, between (1) rules and conventions which are established to govern the approach to problem-solving in formal mathematical systems and (2) physical possibility.

My friend says it is "impossible" for the answer to be 12.5 seconds for Jill.

I say there is no law of physics which says if a party happens to have established synchronized clocks in his own frame, then all other objects moving relative to him MUST experience shorter times and distances.
rpenner
QUOTE (aintnuthin+Aug 4 2011, 12:53 AM)
Here's the question:  Is there any way to interpret the information given in this problem is such a way that Jill ends up being the one with the greater amount of time on her clock (and therefore measures the greater distance)?
Jill can't be the one who measure's the greater distance because Jill's distance is always 0.

For Jill measuring, with Jill's Clocks and Jill's rulers, two different events that happen to Jill, Δx' = 0 and I = (cΔt')˛. Jill is always stationary to Jill.

For Jack, not stationary to Jill, the only constraint for measuring the two events is (Δx)˛ < (cΔt)˛, which is to say Jill's speed relative to Jack is v = Δx/Δt, and v˛ = (Δx)˛ /(Δt)˛ < c˛.

So for Jack, who is in inertial motion relative to Jill, I = (cΔt)˛−(Δx)˛ = (Δt)˛(c˛−v˛) and for Jill I = (cΔt')˛ for the same two events (which both happen to Jill. So since I = I means (Δt')˛c˛ = (Δt)˛(c˛−v˛) we have (Δt')˛/(Δt)˛ = (c˛−v˛)/c˛. So if 0 < v < c, then (Δt')˛ < (Δt)˛ and so Jill's measured time is always the shortest inertially measured time for two events that happen to Jill.

QUOTE (aintnuthin+Aug 4 2011, 12:53 AM)
As a matter of convention, established for the purpose of getting a starting point for constructing a spacetime diagram, the rule is to start with the frame where the clocks are co-moving.

But is that mathematical convention a universal law of phsyics?  Is there any physical law that prevents a person with clocks which are co-moving in his frame from having a lesser time and distance than something travelling at .6c with respect to it?
No -- a frame in special relativity is an inertial coordinate system (x,y,z and t) where the coordinates tell you what comoving rulers and clocks would read if any such rulers and clocks exist, and which faithfully reproduce the laws of physics. It doesn't matter which one you start with -- all frames are exactly like every other frame. The Lorentz transform lets you convert between any two coordinate systems even if they are both moving relative to the one you call stationary.

Since all you have in this example are bodies in inertial motion, and both events happen at the same place in one of the inertial frames, that frame has the special shortest time between those two events.

In Euclidean geometry, the shortest distance between two points is a straight line. For any side of any triangle, the sum of the other two sides is LONGER than the chosen side, because the other two sides form a non-straight line between those two points.

In the twin paradox, the time of the twin outbound is the shortest time between those two events, and the time of the twin inbound is the shortest time between those two events, and the sum of those two times can be (it is) less than the time of the stay-at-home twin. But the traveling twin is not always in the same state of inertial motion. So of all the ways the same two events can happen to the same observer, the observer who takes an inertial trajectory between the two events is the one with the longest elapsed time. The straight line in space-time has the longest elapsed time. But this doesn't reflect the situation you described.
QUOTE (aintnuthin+Aug 4 2011, 01:12 AM)
Conventions aside, would this scenario be physically possible?

Establish a point from Jill's frame (constructed from the premise that it is a stipulation that her time and distance must be longer):

In Jill's coordinate frame F happens at (x=0,t=0) so E happens 7.5 light-seconds to the left and 12.5 seconds in the past. E happens at (x=-7.5, t=-12.5).

F - E = (Δx=7.5 light seconds,Δt=12.5 seconds) and the invariant interval is
I = (cΔt)˛−(Δx)˛ = 100  light-seconds˛

Now delta t' would be 10 seconds exactly for Jack.
F and E aren't frames, they are events which in this setup don't both happen to Jill. If Δx' = 0 for Jack, then both events E and F happen in the same place in Jack's coordinate system. You have basically swapped Jack for Jill from your earlier setup, for purposes unknown. So (Δt' = 10 seconds, Δx' = 0) is consistent with the value of I in this setup but not with the earlier setup.
QUOTE (aintnuthin+Aug 4 2011, 03:40 AM)
Just in case I haven't sufficiently cleaned up the mess I seem to have created, let me add this.

The disagreement with my friend was not, ultimately, about what answer a standard approach would give you.

It was about the difference, if any, between (1) rules and conventions which are established to govern the approach to problem-solving in formal mathematical systems and (2) physical possibility.

My friend says it is "impossible" for the answer to be 12.5 seconds for Jill.
Yes. But it is also impossible for it to be 4.84 light-seconds for Jill when both events happen to Jill. You had to change the problem completely for it to be 12.5 seconds for Jill. That's cheating.

QUOTE (aintnuthin+Aug 4 2011, 03:40 AM)
I say there is no law of physics which says if a party happens to have established synchronized clocks in his own frame, then all other objects moving relative to him MUST experience shorter times and distances.
An object in inertial motion always measures (experiences) the shortest measurable elapsed time for two events (time-like separated) on its world-line and by definite the distance is zero, the shortest possible distance. That is a law of physics derived from special relativity called the invariant interval. Another way to say this is that any two time-like-separated events defines a unique state of inertial motion where the two events are perceived to happen in the same "place" -- and thus defines the frames where the time between the events is the shortest.
aintnuthin
Thanks for your gracious reply, rpenner. I'm afraid I'm still somewhat confused. You say, for example:

"Jill can't be the one who measure's the greater distance because Jill's distance is always 0."

But isn't it generally true that the party who has 0 distance ends up having the greater distance (and greatest elapsed time)? You then say:

"For Jill measuring, with Jill's Clocks and Jill's rulers, two different events that happen to Jill, Δx' = 0 and I = (cΔt')˛. Jill is always stationary to Jill."

But by the same token, Jack is always stationary to Jack, right? If you construct the graph I proposed, wouldn't this still be true? You also say:

"For Jack, not stationary to Jill, the only constraint for measuring the two events is (Δx)˛ < (cΔt)˛..."

Here again, isn't it also true that jill is not stationary to Jack?

Your analysis makes sense, and is a valid way to look at it, but isn't it still just a conventional way to do an analysis? Couldn't we turn that around, and still reach the same result? I mean wouldn't it be possible to just adopt different rules for determining "proper length" versus "proper time" and still get an invariant spacetime interval?

For example, I have looked at an online exposition which indicates (as I read) that derivations from the lorentz transformations can be used to get either time dilation or time contraction between two parties, just depending on the way the measurment is made. Unfortunately, the formatting is such that I can't cut and paste it here, and it's too much to re-type. Unfortunately, I am not, as a newcomer, allowed to post links.

As I read it, you must therefore be careful with which formula you use, but it seems you could use either, and would do so according to how you thereafter defined the measurment process. Another website I have looked at says: "the observer who measures the proper length will not measure the proper time, and vice versa" (Again, can't link, sorry). This suggests to me that it could be somewhat arbitrary when it comes to deciding who measures the so-called "proper" time and length, but that you would then have to be careful how you chose one of each for each observer.

I don't mean this as a literal example, but, for example, say the following rules applied:

1. The "proper" distance between two points is the distance measured by the party travels between those two points, and

2. The "proper" time between two events separated in time (such as a person going from A to B) is the time measured by a person who remains stationary with respect to A and B.

Given those assumptions (which I intend to be the opposite of the rules you are applying), couldn't it still be consistently worked out that the "stationary" party still ended up with the longest time and the greatest distance? In such a case the "proper distance" would simply be assigned to the "moving" party, and the "proper time" would simply be assigned to the "stationary" party, and vice versa.

This might not make much sense, but you could achieve the same "end results," couldn't you? If, for example, you knew the distance for the "stationary" party, and if you then consistently applied the rules, you could derive the time and distance for both parties (given a stated relative speed between them) in such a way that the "stationary" party always ended up with the longest time and distance, and the "moving" party always ended up with the shortest time and difference

I may have messed up with my "non-conventional" rules, but, particulars aside, it could be done in theory, couldn't it?
aintnuthin
Elsewhere in your post you say: "You had to change the problem completely for it to be 12.5 seconds for Jill. That's cheating."

Well, maybe it's a matter of semantics, I don't know. Let me tell you how the discussion arose.

I said that if the motion of the parties were reversed, then Jill would have the greater elapsed time and longer distance. So I was, in effect, asking that the problem be changed.

His response was twofold. On the one hand he said that if you changed your assumptions about who was moving (the parties were reversed from the example I gave here) you would get exactly the same answer because it made no difference. I countered with the argument that I didn't believe that an inertial frame where one had a velocity of zero could possibly be the "same as" one where that same party was presumed to have a relative speed of .6c.

His second response was to effect that, as you said at the beginning, all motion was relative, so it therefore made no difference who was moving.

But all I really intended by "moving" was the party who ended up with the shorter time and difference.

So, since you have accused me of being a cheater (just kidding) I'm curious to know if you would still consider it "cheating" if I proposed to change the problem, but was told it would make no difference, and furthermore, that it couldn't be done.

I would, without cheating, at least be right in contending that with different postulated inertial states, you wouldn't still get the "same answer," wouldn't I?
aintnuthin
Rpenner, I can tell you really know your stuff when it comes to SR (and physics in general), so If you don't mind, let me ask you a new question.

Your first response was to say that terms such as "point A" and "stationary" were meaningless. As a matter of "philosophy" that may be true, but it's not really true (in substance as opposed to semantics) as a matter of physics, is it?

For example, in the course of your explanation, you resort to such concepts as the "same place" (what I called point A) and you yourself use the term "stationary" in several places. I'm certainly not complaining or condemning you for that. I know exactly what you mean and, as a layman without an attachment to a particular philosophical view of motion, I understand you better that way.

As a particular instance, at one point you say: "Since all you have in this example are bodies in inertial motion, and both events happen at the same place in one of the inertial frames, that frame has the special shortest time between those two events."

By referring to events happening at the "same place" aren't you implicitly just saying (if put in layman's terms), that the person in the "same place" is stationary, i. e., not moving? I don't mean absolutely stationary, just relative to the other.

Can't the explanation you offered in more technical terms be translated to say something like: "the party who is relatively at rest will experience the greater amount of time and measure the longer distance, while the party who is relatively moving with respect to that party will experience the lesser amount of time and will measure the shorter distance?"

In other words, what you referred to "a unique state of inertial motion where the two events are perceived to happen in the same "place"" is just a fancy way of saying "at relative rest," isn't it?
aintnuthin
You may not have the least bit of interest in discussing issues of philosophy or semantics, rpenner, and if that's the case, I certainly understand. But I want to point out a few things that cause me, as a layman, confusion when discussing SR. I will use this excerpt from your posts to try to elucidate the questions I am raising:

"Jill can't be the one who measure's the greater distance because Jill's distance is always 0."

Now, reading this, I form the following conclusions (probably mistakenly):

1. I read "Jill's distance is always 0" as an acceptance of the stipulation that Jill is "not moving"

2. "measure the greater distance" is somewhat ambiguous. Does this mean measure the distance between the two clocks? Granted that she cannot directly "measure" the distance in another frame, couldn't she nonethless "calculate it, using the lorentz transform? And if she did calculate it wouldn't she conclude that the distance separating the clock was greater in her frame than Jack perceives it to be in his?

Based on those impressions, I end up reading your statement as saying "Jill will (indirectly) measure (calculate) his distance to be less than hers, and will and therefore conclude that her distance is "greater."

See what I'm getting at? And that would comport with my general understanding that the "stationary" party will end up measuring the greater distance.

And yet, you are presumably trying to say the opposite. So I get confused. Any comment?
aintnuthin

By stipulation, we have said there that Jill is stationary. Assuming that she will measure his distance to be shorter than he measures it to be, what does that say about her instruments of measurement (yardsticks, whatever)? If his lengths are shorter than hers, doesn't that imply that her yardsticks are longer than his? So, from that perspective, her lengths are longer than his.

Looking at it from a time perspective: If time is faster for her, won't her clock show more time elapsed between the two events than his time does? If his clock ticks off 10 seconds between two events, won't her clock tick off more time than that? And, because of that, she will conclude, in her frame, that the distance between his clocks is more than he perceives it to be?

If you say that, because the distance in his frame is 6 light seconds, hers must be shorter, that's fine, but isn't it kind of arbitrary?, a matter of mere convention? As soon as you say hers MUST be shorter, haven't you simply, by fiat, made him the "stationary" party? Suppose Jack and Jill are together, then he accelerates to .6c relative to her, then turns back toward her at that same relative speed. Then he sets up clocks 6 light seconds apart in his frame and passes her. If he did that, how would that automatically make him the "stationary" party?

I just keep going around and around with this, it seems.
AlexG
QUOTE
By stipulation, we have said there that Jill is stationary.

Relative to what?

From Jack's POV, he is stationary and Jill is moving.
aintnuthin
QUOTE (AlexG+Aug 4 2011, 12:38 PM)

Relative to what?

From Jack's POV, he is stationary and Jill is moving.

Hi, Alex. Well, yeah, they can both "perceive" themselves to be stationary, but if there is relative movement between them, they can't BOTH be stationary, so at least one is in relatve motion. Which one? Well, that's the reason for the stipulation. Jill is, we stipulate, stationary relative to Jack, rather than vice versa.
AlexG
QUOTE (aintnuthin+Aug 4 2011, 07:47 AM)

Hi, Alex. Well, yeah, they can both "perceive" themselves to be stationary, but if there is relative movement between them, they can't BOTH be stationary, so at least one is in relatve motion. Which one? Well, that's the reason for the stipulation. Jill is, we stipulate, stationary relative to Jack, rather than vice versa.

It's not a matter of perception, it's a matter of physics. If you are declaring Jill to be stationary, then what you are saying is that All measurements and observations will be taken in Jill's frame of reference. You are establishing a preferred frame of reference. If you are observing from Jack's reference frame, then Jill's
is the moving frame.
aintnuthin
QUOTE (AlexG+Aug 4 2011, 12:58 PM)

It's not a matter of perception, it's a matter of physics.  If you are declaring Jill to be stationary, then what you are saying is that All measurements and observations will be taken in Jill's frame of reference.  You are establishing a preferred frame of reference.  If you are observing from Jack's reference frame, then Jill's
is the moving frame.

"All measurements and observations will be taken in Jill's frame of reference."

Well, yes, but not exactly as I see it. In essence you are saying that all measurements, wherever taken, will be standardized in accordance with her frame. At least that's the way I would put it.

"You are establishing a preferred frame of reference."

Yes, in effect you are. Of course this must always be done if you're going to try solve any problem. You can't get anywhere if you simply say that because each can perceive himself as stationary, each must be treated as stationary. That approach simply gives you a universe in which nothing ever moves and we know (or at least think we know) better.
AlexG
QUOTE
each must be treated as stationary.

No, each must be treated as being in relative motion with the other.
aintnuthin
QUOTE (AlexG+Aug 4 2011, 01:08 PM)

No, each must be treated as being in relative motion with the other.

Which puts us back where we started, doesn't it? One of them is moving. Are you suggesting that stipulations as to which one that is are forbidden?
AlexG
They're both moving, relative to each other.

synthsin75
QUOTE (AlexG+Aug 4 2011, 06:58 AM)

It's not a matter of perception, it's a matter of physics. If you are declaring Jill to be stationary, then what you are saying is that All measurements and observations will be taken in Jill's frame of reference. You are establishing a preferred frame of reference. If you are observing from Jack's reference frame, then Jill's
is the moving frame.

If he stipulated that Jack's frame had been previously accelerated to the relative velocity with Jill's, he would be perfectly fine stating Jill's to be relatively at rest, as Jack's would have had to overcome it's own inertia.
Sithdarth
QUOTE
If he stipulated that Jack's frame had been previously accelerated to the relative velocity with Jill's, he would be perfectly fine stating Jill's to be relatively at rest, as Jack's would have had to overcome it's own inertia.

No he wouldn't. At best he could stipulate that Jack changed frames but that doesn't change the fact that once Jack has arrived at his new inertial frame the situation becomes symmetric once again. To put it another way since we aren't worrying about any measurements made by Jack before he accelerated into his new frame the acceleration is a non-issue to the problem at hand. If we were comparing Jack's measurements before and after the frame change things would be different. But again all we could say is that Jack changed frames not that Jill is relatively at rest.

To put it yet another way they are both equally able to be called relatively at rest regardless of Jack's acceleration. Specifically in this case because all of Jack's measurements are happening in one frame. But even if that were not true the only distinction that can be made is that Jack changed frames and Jill did not which does not make it ok to say Jill is relatively at rest. Well at least not if you are using the terms correctly in the context of special relativity were terms like rest have very specific and precise meanings.
synthsin75
It would make a difference to any third non-accelerating observer.

http://en.wikipedia.org/wiki/Rest_%28physics%29
QUOTE (^+)
By Albert Einstein's celebrated definition, two observers measure having been at rest relative to each other in a particular trial if they succeed to identify a third observer as middle between each other.

"At rest", relatively (and no, there is no absolute rest frame), is defined utilizing a third observer.

Note: I didn't say it'd have an impact on the situation, merely that there was a valid way to define one frame at relative rest.
Sithdarth
QUOTE
It would make a difference to any third non-accelerating observer.

Except there isn't one and the only difference would be that both inertial observers would agree that Jack moved in a non-inertial manner at some point.

QUOTE (->
 QUOTE It would make a difference to any third non-accelerating observer.

Except there isn't one and the only difference would be that both inertial observers would agree that Jack moved in a non-inertial manner at some point.

"At rest", relatively (and no, there is no absolute rest frame), is defined utilizing a third observer.

QUOTE
By Albert Einstein's celebrated definition, two observers measure having been at rest relative to each other in a particular trial if they succeed to identify a third observer as middle between each other.

Is a means by which two observes can use a third observer to check if they are relatively at rest with each other. This allows you to say only that the two observers are relatively at rest with respect to each other. It does not allow you to tag a specific observer or frame as simply "At rest" or even "Relatively at Rest".

To put it another way, in order to be relatively at rest with respect to something there must be no relative velocity between you and that something. There is clearly a relative velocity between Jack and Jill therefore the usage of the term "at rest" (relatively or not) is inappropriate. "At rest" as a very specific meaning in special relativity and when talking about special relativity it is imperative to keep to that meaning. The term that best describes Jill's motion is not "at rest" but rather "inertial".

QUOTE (->
 QUOTE By Albert Einstein's celebrated definition, two observers measure having been at rest relative to each other in a particular trial if they succeed to identify a third observer as middle between each other.

Is a means by which two observes can use a third observer to check if they are relatively at rest with each other. This allows you to say only that the two observers are relatively at rest with respect to each other. It does not allow you to tag a specific observer or frame as simply "At rest" or even "Relatively at Rest".

To put it another way, in order to be relatively at rest with respect to something there must be no relative velocity between you and that something. There is clearly a relative velocity between Jack and Jill therefore the usage of the term "at rest" (relatively or not) is inappropriate. "At rest" as a very specific meaning in special relativity and when talking about special relativity it is imperative to keep to that meaning. The term that best describes Jill's motion is not "at rest" but rather "inertial".

Note: I didn't say it'd have an impact on the situation, merely that there was a valid way to define one frame at relative rest.

It most certainly is not. As shown above it is a misuse of the term "at rest" which has a very specific meaning in special relativity.
synthsin75
I'm not doing this BS quibbling over your poor comprehension.
rpenner
QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
Thanks for your gracious reply, rpenner.  I'm afraid I'm still somewhat confused.  You say, for example:

"Jill can't be the one who measure's the greater distance because Jill's distance is always 0."

But isn't it generally true that the party who has 0 distance ends up having the greater distance (and greatest elapsed time)?
No. That's a wholly inadequate summary of special relativity. If you want a general statement about how the same space-time is measured by two different observers in two different motions you need to use the Lorentz transform. Here, we don't need a completely general statement because both events happen to Jill (i.e. both happen on Jill's world-line), and we can prove that the invariant interval is a quantity unchanged by the Lorentz transform. Since Δx' = 0 in Jill's coordinates, solving for Δt' is a snap with just knowledge of the invariant interval.

The "distance" of the 6 light-second-long "ruler" is not a distance measured by Jill, since Jill doesn't care where both ends of the rulers are at the same time (in Jill's frame). Jill only cares about the events of when the front and back of the ruler meet her. Thus this problem is about events and not rulers. That would be a question of where Jill measures Jack to be when the first bit of the ruler meets her -- a question you didn't ask.
QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
You then say:

"For Jill measuring, with Jill's Clocks and Jill's rulers, two different events that happen to Jill, Δx' = 0 and I = (cΔt')˛. Jill is always stationary to Jill."

But by the same token, Jack is always stationary to Jack, right?  If you construct the graph I proposed, wouldn't this still be true?
That would be a different pair of events to consider and has nothing to do with the setup you and your friend disagree on.
QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
You also say:

"For Jack, not stationary to Jill, the only constraint for measuring the two events is (Δx)˛ < (cΔt)˛..."

Here again, isn't it also true that jill is not stationary to Jack?
Yes, Jack and Jill are not stationary to each other. But the pair of events you describe happen both at Jill's position and not both at Jack's position. So the math is very simple for Jill. The laws of physics don't prefer Jill's situation to Jack's -- only the problem setup does.
QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
Your analysis makes sense, and is a valid way to look at it, but isn't it still just a conventional way to do an analysis? Couldn't we turn that around, and still reach the same result?  I mean wouldn't it be possible to just adopt different rules for determining "proper length" versus "proper time" and still get an invariant spacetime interval?

For example, I have looked at an online exposition which indicates (as I read) that derivations from the lorentz transformations can be used to get either time dilation or time contraction between two parties, just depending on the way the measurment is made.  Unfortunately, the formatting is such that I can't cut and paste it here, and it's too much to re-type. Unfortunately, I am not, as a newcomer, allowed to post links.
There's a lot of junk on the Internet. I would not trust someone's personal website over a good textbook, approved by a scientifically minded publisher and adopted by school boards or universities with a vested interest in teaching good science in a useful manner.

QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
As I read it, you must therefore be careful with which formula you use, but it seems you could use either, and would do so according to how you thereafter defined the measurment process.  Another website I have looked at says: "the observer who measures the proper length will not measure the proper time, and vice versa"  (Again, can't link, sorry). This suggests to me that it could be somewhat arbitrary when it comes to deciding who measures the so-called "proper" time and length, but that you would then have to be careful how you chose one of each for each observer.
Just as switching frames only relabels the coordinates of events without actually changing the events, so "choosing which formula" cannot change reality. Either your understanding of that person's website is very confused or you have been hanging out at the website of a child-predator, promising candy and an easy road to enlightenment when all he has to give is darkness, despair, misery and pain.

QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
I don't mean this as a literal example, but, for example, say the following rules applied:

1.  The "proper" distance between two points is the distance measured by the party travels between those two points
Then all proper distances would be zero. Wholly wrong.

The actual distance between two co-moving world lines is measured between two simultaneous event and the proper distance is the longest such measurement which happens to be in the frame where the co-moving world lines are stationary. Since the events are simultaneous (Δx)˛ > (cΔt)˛ and the Invariant Interval is negative, which is the definition of space-like separation of the events. But just like the distance between parallel lines can be different than the distance between two points chosen on those lines, so the "proper length" between two co-moving world lines can be different than the "proper distance" between two events.
QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
, and

2.  The "proper" time between two events separated in time (such as a person going from A to is the time measured by a person who remains stationary with respect to A and B.
Correct, and already accepted in physics and as part of the notion of the Invariant Interval.

QUOTE (aintnuthin+Aug 4 2011, 08:04 AM)
Given those assumptions ...
I may have messed up with my "non-conventional" rules, but, particulars aside, it could be done in theory, couldn't it?
I try to only use "theory" to describe that which is a communicable, precise and useful description of phenomena of the universe. Your proposal fails on more than one criteria.

How you must hate your so-called friend that you would go to such dishonest lengths to try and demonstrate that your understanding of special relativity is not deeply flawed. There was nothing hard about the problem, and I told you two ways to solve it. Naturally both ways get the same answer because we are describing two events which happen on Jill's world-line.
QUOTE (aintnuthin+Aug 4 2011, 08:33 AM)
Elsewhere in your post you say: "You had to change the problem completely for it to be 12.5 seconds for Jill. That's cheating."

Well, maybe it's a matter of semantics, I don't know.
No, its not semantics. You described a fundamentally different physical setup.
QUOTE (aintnuthin+Aug 4 2011, 08:33 AM)
I would, without cheating, at least be right in contending that with different postulated inertial states, you wouldn't still get the "same answer," wouldn't I?
If you swap Jack for Jill, then both events happen on Jack's worldline and the distance between events in Jack's frame is zero. That is the "same answer" swapping Jack for Jill. You have just relabeled the frames -- the same thing as if you relabeled them Josephine and Jacques.

If you do anything more than relabeling, then you are changing the physical situation, which is cheating if you are trying to demonstrate that you were right about a setup you already described to your friend.
Sithdarth
QUOTE
I'm not doing this BS quibbling over your poor comprehension.

And yet again you prove you are incapable of having a reasoned discussion without resorting to insults and personal attacks.
Montec
Hello aintnuthin

You could always transform the times and distances of the Jack and Jill frames to a third stationary (agreed on by both Jack and Jill) frame. I like the CMBR frame as a third frame myself. You can then compare Jack's data with Jill's data and see what is what.

When you remove the third frame then either Jack or Jill must be in a stationary frame (V = 0). You cannot have both Jack and Jill with a velocity (actually a speed) since that would imply a third frame. To imply that both Jack and Jill are both moving and stationary, at the same time, with respect to each other brings up the question of reciprocity. Which I believe is your "sticking" point. This "uncertainty" of position, stationary frame or moving frame, and relative speed (who is moving) lies at the heart of the reciprocity debate.

The common answer is "The frame that undergoes acceleration will have a slower time rate". ie Acceleration implies a moving frame.

My observation is that the frame that travels farther will have aged less. The implication of this is that there is no reciprocity in SR. Just false assumptions on who traveled farther.

Lorentz transformations allow the conversion between frames but said transformations don't determine which frame is stationary. This is left up to the user.

rpenner
QUOTE (Montec+Aug 4 2011, 07:22 PM)
You could always transform the times and distances of the Jack and Jill frames to a third stationary (agreed on by both Jack and Jill) frame. I like the CMBR frame as a third frame myself. You can then compare Jack's data with Jill's data and see what is what.
CMBR has nothing to do with special relativity, nor is it a preferred frame.

QUOTE (Montec+Aug 4 2011, 07:22 PM)
When you remove the third frame then either Jack or Jill must be in a stationary frame (V = 0). You cannot have both Jack and Jill with a velocity (actually a speed) since that would imply a third frame.
Incorrect. Both Jack and Jill are already stationary with respect to themselves. Even without a third observer it completely possible to describe a frame (a center-of-mass frame, for example) where both Jack and Jill are in motion. A frame isn't "real" -- it's just a coordinate system where the laws of physics look like any other inertial coordinate system. You can't add and remove frames, because they aren't "real" -- you just use what every frames are useful for the situation as described.

QUOTE (Montec+Aug 4 2011, 07:22 PM)
To imply that both Jack and Jill are both moving and stationary, at the same time, with respect to each other brings up the question of reciprocity. Which I believe is your "sticking" point. This "uncertainty" of position, stationary frame or moving frame, and relative speed (who is moving) lies at the heart of the reciprocity debate.
Wrong again. Jack an Jill are only stationary with respect to themselves. Jack and Jill are inertial observers in relative motion, which means relative to each other, they each have a non-zero velocity in the natural coordinate frame of the other. So you label each event in each frame. So E has a value of x and a value of t, and a value of x' and a value of t'. Further, event F also has the same number of coordinates in every frame. We can sensibly talk about the difference from E to F as F-E = (Δx,Δt) in one frame and F-E = (Δx',Δt') in another, and use the Lorentz transform to flip-flop between them. Since any Lorentz transform preserves I, we have (cΔt)˛−(Δx)˛ = I = (cΔt')˛−(Δx')˛.

QUOTE (Montec+Aug 4 2011, 07:22 PM)
The common answer is "The frame that undergoes acceleration will have a slower time rate". ie Acceleration implies a moving frame.
No, acceleration implies an object has changed its state of motion, which is a statement about real things. Frames are just man-made coordinate systems used to label events. They are not "real."

QUOTE (Montec+Aug 4 2011, 07:22 PM)
My observation is that the frame that travels farther will have aged less. The implication of this is that there is no reciprocity in SR. Just false assumptions on who traveled farther.
You are back to the twin paradox, which is already explained as the geometry of space-time. The longest time-like path between two events is the inertial trajectory. That's true and important even if two clocks traveled the same distance. So your principle of "farthest distance" is not a physical law, but an imprecise statement which will lead one into error precisely as the OP's original summary of length contraction.

QUOTE (Montec+Aug 4 2011, 07:22 PM)
Lorentz transformations allow the conversion between frames but said transformations don't determine which frame is stationary. This is left up to the user.
Finally you have said something true. Jill's rulers and clocks say she is stationary, Jack's rulers and clocks say he is stationary, and they are both right because comparing their rulers and clocks is a case of apples and oranges which is why we distinguish Δx from Δx' even when we are talking about the coordinate difference between the same two events.
Montec
Hello rpenner
Isn't a "center of mass frame" a type of third reference frame? I am thinking of the GPS Earth centered frame.

There is nothing special about the CMBR frame. It is just a frame that can be used anywhere by anyone and have all users agree to its usage.

Any object/particle is always stationary with respect to its self. This is a no brain-er. However, if you have two or more particles with different velocities then there is a common reference/lab frame as well as velocities as measured between said particles. Think of a shotgun blast and pellet spread.

Frames, as I understand them, are associated with particles/mass and encompass the physics affecting said particle. Our understandings may differ on this.

As I see it, the "twins paradox" comes about because the assumed stationary twin is switched when looking at the other twin. ie The moving twin will always see the stationary twin as aging faster and observe a shrunken stationary ruler. This is because the moving twin's clock-rate is slower than the stationary twin. If you do not switch between the moving and stationary twin then there is no paradox. The moving twin is assumed to have undergone an acceleration to alter it's state of motion.

aintnuthin
QUOTE (Montec @ Aug 4 2011, 07:22 PM)
"Lorentz transformations allow the conversion between frames but said transformations don't determine which frame is stationary. This is left up to the user."

rpenner responded: "Finally you have said something true. Jill's rulers and clocks say she is stationary, Jack's rulers and clocks say he is stationary, and they are both right because comparing their rulers and clocks is a case of apples and oranges which is why we distinguish Δx from Δx' even when we are talking about the coordinate difference between the same two events."

rpenner, you say Montec is right that the question of who is stationary is "left up to the user." Who are you referring to by the term "user." Would that be you (and me) or anyone else trying to resolve a problem? Are you suggesting that question can only be answered by arbitrary stipulation?

You also say both observers are right. Yet, on the other hand, you have also said, as I understand you, that one observer is right and that only that one user is right in the original (Jack/Jill) problem. On what non-arbitrary grounds do you make that determination?

1. Jack sees his own time and speed as 10 seconds, 6 LS. This is tautologically true, and will never change, I agree with that. It's a simple matter of definition.

2. Using the LT Jack therefore imputes a time of speed of 8 seconds and 4.84 LS to Jill.

If I'm understanding you correctly, you are saying Jack is right, as a matter of fact, about Jill's state of relative motion, and that nothing imaginable could keep him from being right. Why is that?

3. If Jill, assuming she is stationary, uses the LT to calculate Jack's time and distance, she will come up with 6.4 seconds and about 3.8 LS.

If I'm understanding you correctly, you are saying that Jill is wrong, as a matter of fact, and that nothing imaginable could keep her from being wrong. Why is that?

Phrased in a different way: When you conclude that, from the information given, one must necessarily impute a time and distance of 8 seconds, 4.84 LS to Jill, haven't you, in effect, said that Jack must be the stationary party in this hypothetical situation? That is, that his time and distance MUST both be greater than Jill's?
synthsin75
aintnuthin,

Separate spacetime events that occur in the same space have the largest separation in time, whereas those which occur at the same time have the largest separation in space.

Since both events occur at the same place in Jill's frame, she will measure the largest separation in time between Jack's two clock events. In Jack's frame, the clock events maintain a separation in space, so the time between events in his frame will be shorter.
aintnuthin
QUOTE (synthsin75+Aug 4 2011, 10:10 PM)
aintnuthin,

Separate spacetime events that occur in the same space have the largest separation in time, whereas those which occur at the same time have the largest separation in space.

Since both events occur at the same place in Jill's frame, she will measure the largest separation in time between Jack's two clock events. In Jack's frame, the clock events maintain a separation in space, so the time between events in his frame will be shorter.

Thanks for the input, Synth. I understand what you are saying here, easily enough, I think. But my question is more about the underlying basis for that assertion. In your opinion is that conclusion based upon (arbitrary) convention, or is it necessarily true as a matter of physical law?

If the latter, then it seems there is, after all, a sure-fire way to determine and know who is "really" moving (faster).

That "sure-fire" way, would naturally presuppose that you have a way of knowing whether Jack or Jill (in the original problem) remains in the "same place," of course.

aintnuthin
QUOTE (synthsin75+Aug 4 2011, 10:10 PM)
aintnuthin,

Since both events occur at the same place in Jill's frame, she will measure the largest separation in time between Jack's two clock events. In Jack's frame, the clock events maintain a separation in space, so the time between events in his frame will be shorter.

Applying these rules, Synth, what would your answer be to the original problem?

If I'm reading you correctly, Jill would have the greater time, i.e., more than Jack's, which is 10 seconds.
aintnuthin
QUOTE (Sithdarth+Aug 4 2011, 03:13 PM)

But even if that were not true the only distinction that can be made is that Jack changed frames and Jill did not which does not make it ok to say Jill is relatively at rest.

Sithdarth, what would your answer to the original problem be? Would it be that there is no possible answer?
aintnuthin
QUOTE (aintnuthin+Aug 4 2011, 12:20 PM)
So, truth be told, I'm still confused about your original answer. If you say that, because the distance in his frame is 6 light seconds, hers must be shorter, that's fine, but isn't it kind of arbitrary, a matter of mere convention?  As soon as you say hers MUST be shorter, haven't you simply, by fiat, made him the "stationary" party?  Suppose Jack and Jill are together, then he accelerates to .6c relative to her, then turns back toward her at that same relative speed.  Then he sets up clocks 6 light seconds apart in his frame and passes her.  If he did that, how would that automatically make him the "stationary" party?

rpenner, if you have already answered this question, which I asked back on page 1, then I'm sorry, I must have missed it. That's why I more or less asked it again in the last post I addressed to you.
aintnuthin
General forum question addressed to anyone who may know:

Sometimes I am given the option to "edit" my own posts, and sometimes that option disappears. Anyone know why that might be?
synthsin75
QUOTE (aintnuthin+Aug 4 2011, 04:33 PM)
Thanks for the input, Synth.  I understand what you are saying here, easily enough, I think.  But my question is more about the underlying basis for that assertion.  In your opinion is that conclusion based upon (arbitrary) convention, or is it necessarily true as a matter of physical law?

If the latter, then it seems there is, after all, a sure-fire way to determine and know who is "really" moving (faster).

That "sure-fire" way, would naturally presuppose that you have a way of knowing whether Jack or Jill (in the original problem) remains in the "same place," of course.

Nothing to do with convention. For any spacetime interval (separation between events in both time and space) there is only the interval. What ratio of space and time is observed depends upon the relative motion of the observer.

Assuming both observers pass through both events, as they do in this case, simultaneity of events determines which is slower than the other. Remaining stationary is not required though. It is only that Jill observes both events to happen at her position, regardless of how that may change in between events, and Jack observes one to be at a set distance from his position, even though stationary relative to his frame.

Since Jill can observe no space between events, the entire spacetime interval must be a separation solely in time for her, and thus the greatest time separation between events.

QUOTE
Applying these rules, Synth, what would your answer be to the original problem?

If I'm reading you correctly, Jill would have the greater time, i.e., more than Jack's, which is 10 seconds.

Yes, you are reading me correctly.

QUOTE (->
 QUOTE Applying these rules, Synth, what would your answer be to the original problem?If I'm reading you correctly, Jill would have the greater time, i.e., more than Jack's, which is 10 seconds.

Yes, you are reading me correctly.

Sometimes I am given the option to "edit" my own posts, and sometimes that option disappears. Anyone know why that might be?

There's a timer on editing until someone else posts.
rpenner
QUOTE (aintnuthin+Aug 4 2011, 11:51 PM)
rpenner, if you have already answered this question, which I asked back on page 1, then I'm sorry, I must have missed it.

QUOTE (aintnuthin+Aug 3 2011, 03:36 PM)
Jack and Jill are in relative motion at the rate of .6c, with Jill being "stationary" at Point A, with Jack about to pass her at that point.  Jack has previously established two synchronized clocks in his frame, which are 6 light seconds apart.  One (clock 2) he has with him and the other "lead" clock (clock 1) encounters Jill first.  After clock 1 passes by Jill, Jack, with clock 2, passes her.

The question is about how Jill will see these two events in her frame.

We agree that Jack will the the elapsed time as being 10 seconds and, as established, the distance to be 6 light seconds.  The question is about Jill's time and distance.  Two alternative answers have been proposed:

Answer A:  Jill's time and distance will be 8 seconds and 4.84 light seconds

Answer B:  Jill's time and distance will be 12.5 seconds and 7.5 light seconds

Which, if either, do you think is correct?

QUOTE (rpenner+Aug 3 2011, 04:40 PM)
There are two "events":
E: Jill and the first clock meet
F: Jill and Jack meet.

In Jack's coordinate frame F happens at (x=0,t=0) so E happens 6 light-seconds to the left and 10 seconds in the past. E happens at (x=-6, t=-10).
F - E = (Δx=6 light seconds,Δt=10 seconds) and the invariant interval is
I = (cΔt)˛−(Δx)˛ = 64  light-seconds˛

In Jill's coordinate frame, we could set up a Lorentz transform, plug in the velocity, double-check that we got the signs right, convert events E and F to x' and t' and then find out what Δx' and Δt' where. Or we could just note that by definition, Δx' = 0 since both events E and F happen to Jill. so I = (cΔt')˛−(Δx')˛ = (cΔt')˛ = 64 light-seconds˛  and show that Δt' = 8 seconds exactly for Jill.

Answer: Jill's time and distance will be Δt' = 8 seconds and Δx' = 0
aintnuthin
Thanks for your response, Synth. I may be wrong, but I agee with your analysis insofar as it seems to be an experimentally verified objective fact that time "really" does dilate with increased speed and contract with decreased speed. For that reason, I would see it as a "law of physics."

Others seem to disagree and assert that there is no objective fact about the matter.
aintnuthin
QUOTE (rpenner+Aug 5 2011, 12:58 AM)

Answer: Jill's time and distance will be Δt' = 8 seconds  and Δx' = 0

Well, thanks, rpenner, but that's not the question I was referring to.
Sithdarth
QUOTE
Sithdarth, what would your answer to the original problem be? Would it be that there is no possible answer?

The same as rpenner with one difference. It seems what you are trying to ask for is not the distance between events for Jill but rather the distance between the two clocks (which is not the same as the distance between the two events). The answer is that Jill sees the distance between the two clocks (which are 6 light seconds apart in Jack's frame) as 4.84 light seconds apart.

QUOTE (->
 QUOTE Sithdarth, what would your answer to the original problem be? Would it be that there is no possible answer?

The same as rpenner with one difference. It seems what you are trying to ask for is not the distance between events for Jill but rather the distance between the two clocks (which is not the same as the distance between the two events). The answer is that Jill sees the distance between the two clocks (which are 6 light seconds apart in Jack's frame) as 4.84 light seconds apart.

Separate spacetime events that occur in the same space have the largest separation in time, whereas those which occur at the same time have the largest separation in space.

Since both events occur at the same place in Jill's frame, she will measure the largest separation in time between Jack's two clock events. In Jack's frame, the clock events maintain a separation in space, so the time between events in his frame will be shorter.

Did you not pay attention at all to rpenner's posts? To quote him:

QUOTE
here are two "events":
E: Jill and the first clock meet
F: Jill and Jack meet.

In Jack's coordinate frame F happens at (x=0,t=0) so E happens 6 light-seconds to the left and 10 seconds in the past. E happens at (x=-6, t=-10).
F - E = (Δx=6 light seconds,Δt=10 seconds) and the invariant interval is
I = (cΔt)˛−(Δx)˛ = 64  light-seconds˛

In Jill's coordinate frame, we could set up a Lorentz transform, plug in the velocity, double-check that we got the signs right, convert events E and F to x' and t' and then find out what Δx' and Δt' where. Or we could just note that by definition, Δx' = 0 since both events E and F happen to Jill. so I = (cΔt')˛−(Δx')˛ = (cΔt')˛ = 64 light-seconds˛  and show that Δt' = 8 seconds exactly for Jill.

You'll notice that in the equation for the invariant interval Δx is subtracted from cΔt. In other words if the interval is to remain invariant when Δx decreases to zero it is required that Δt decrease as well. For example, X - 4 = 1 means that x = 5 and X - 1 = 1 means x = 2. In other words Jill's time = 8 seconds and Jack's time = 10 seconds.

At this point I will also point out that Δx' is entirely different then Jill's observation of the separation of the clocks. Which is the point rpenner is attempting to make.
aintnuthin
QUOTE (Sithdarth+Aug 5 2011, 04:53 AM)

The same as rpenner...

Well, it seems that, for all your insistence that nobody can say who's moving, you're easily able to give a definite answer to the question of who's clock is running faster in this case, Sithdarth.

Let me give you two examples, and the ask you a question, if you don't mind:

Example 1: Jack and Jill are together. Jack acclerates to a speed of .6c, then turns around and head back toward Jill at that same speed. Before he reaches her, he places a clock 6 LS in front of him, then passes Jill. Who will have the greater time and distance.

Your answer: Jack will have the greater time and distance (is "stationary").

Why? Because the clocks are co-moving with him.

Example 2: Same facts with one exception: This time, while Jack is heading back to Jill she sets up a clock 6 LS away from her, and then Jack passes both clocks.

Who will have the greater time and distance this time?

Your answer: Jill will have the greater time and distance (is "stationary").

Why? Because the clocks are co-moving with her, that's why.

Apart from the matter of which one beat the other to the task of setting up a clock, the two situations are identical.

Is that a difference dictated by the "laws of phsyics?" Does the answer to the fundamental question of whose clock is running slower change when one guy sets up a clock in his frame first?
Sithdarth
QUOTE
Well, it seems that, for all your insistence that nobody can say who's moving, you're easily able to give a definite answer to the question of who's clock is running faster in this case, Sithdarth.

Only because we've fixed the frame we are measuring from. If we switched to a frame where both events happened at the same place for Jack then it would be Jack with the shortest time not Jill. It has nothing to do with who is moving.

QUOTE (->
 QUOTE Well, it seems that, for all your insistence that nobody can say who's moving, you're easily able to give a definite answer to the question of who's clock is running faster in this case, Sithdarth.

Only because we've fixed the frame we are measuring from. If we switched to a frame where both events happened at the same place for Jack then it would be Jack with the shortest time not Jill. It has nothing to do with who is moving.

Example 1: Jack and Jill are together. Jack acclerates to a speed of .6c, then turns around and head back toward Jill at that same speed. Before he reaches her, he places a clock 6 LS in front of him, then passes Jill. Who will have the greater time and distance.

Why? Because the clocks are co-moving with him.

This question isn't well enough defined to give an answer.

QUOTE
Example 2: Same facts with one exception: This time, while Jack is heading back to Jill she sets up a clock 6 LS away from her, and then Jack passes both clocks.

Who will have the greater time and distance this time?

Why? Because the clocks are co-moving with her, that's why.

Once again the question isn't defined well enough for a good answer.

QUOTE (->
 QUOTE Example 2: Same facts with one exception: This time, while Jack is heading back to Jill she sets up a clock 6 LS away from her, and then Jack passes both clocks.Who will have the greater time and distance this time?Your answer: Jill will have the greater time and distance.Why? Because the clocks are co-moving with her, that's why.

Once again the question isn't defined well enough for a good answer.

Apart from the matter of which one beat the other to the task of setting up a clock, the two situations are identical.

No they aren't and you said so yourself when you changed who the clock was co-moving with respect to.

QUOTE
Is that a difference dictated by the "laws of phsyics?" Does the answer to the fundamental question of whose clock is running slower change when one guy sets up a clock in his frame first?

In a perfectly symmetric inertial setup neither clock can be said to be running slower. Rather they both run slow when looked at from the other frame. The only thing the changes the answer is changing the setup which is exactly what you did above. You changed the situation and therefore changed the answer.

synthsin75
SD and RPenner are correct. I misled you there. I didn't take into account the length contraction. Jill will see the length between Jacks clocks (6cs as observed by Jack) to be contracted to 4.8cs. This will indeed make her time between events to be 8secs, thus shorter than Jack's.

So yes, the frame co-moving with both clocks will have the greater time between events.
aintnuthin
QUOTE (Sithdarth+Aug 5 2011, 06:01 AM)

This question isn't well enough defined to give an answer.

What information is missing here that was provided in the original question?
aintnuthin
QUOTE (synthsin75+Aug 5 2011, 06:11 AM)
SD and RPenner are correct. I misled you there. I didn't take into account the length contraction. Jill will see the length between Jacks clocks (6cs as observed by Jack) to be contracted to 4.8cs. This will indeed make her time between events to be 8secs, thus shorter than Jack's.

So yes, the frame co-moving with both clocks will have the greater time between events.

Why is what Jill "sees" (I would say calculate) determinative of the entire situation, Synth? Everybody "sees" everybody's length as contracted, don't they? Other than observers in his own frame, an infinite number of observers, all going at different speeds, would also "see" his length contracted, whatever speed they were going, right? If what others saw my lengths to be determined my state of motion, I would have to be going millions of different speeds, all at the same time, it seems. What makes Jill's view so unique?

The problem stipulates that Jill is stationary. If she were stationary, her length (and time) would be longer, and she would still "see" his as contracted. The difference is that it would then be contracted from 7.5 LS down to 6LS (in Jack's frame).

In that case Jack would "see" Jill's length contracted to 8 seconds, just as he does now, so he wouldn't see anything different.

To say that Jill's length must be 4.84 is equivalent to saying she must be moving, while Jack is stationary. But, again, the stipulation was that she was the one moving.
synthsin75
The simple answer is that, since Jill already observes the two events to occur in the same place, there is just no space between events to be contracted. And both events co-moving with Jack, Jack cannot observe a length contraction between events.
aintnuthin
QUOTE (synthsin75+Aug 5 2011, 08:13 AM)
The simple answer is that, since Jill already observes the two events to occur in the same place, there is just no space between events to be contracted. And both events co-moving with Jack, Jack cannot observe a length contraction between events.

Well, maybe I'm not following you, but if Jill is "already" in the same place, that sounds like you've already presumed that she is stationary. But as soon as you say Jack's length must be the longest, that presupposes that he is stationary, doesn't it? If he is already stationary, then wouldn't that put Jill in different places, and Jack in the "same" place?

I asked Sithdarth a question, which he says is incomplete. Have you looked at that? There I'm suggesting that hinging things on co-moving clocks seems to lead to a contradiction.

There may well be some contradiction here between Jill being stationary and Jack's length being 6LS, but I'm not sure where the source of that contradiction lies. I do agree that Jack's distance is set, by definition, at 6 LS, but that, standing alone, is not incompatible with Jill's length being 7.5 LS
synthsin75
QUOTE (aintnuthin+Aug 5 2011, 04:02 AM)
Well, maybe I'm not following you, but if Jill is "already" in the same place, that sounds like you've already presumed that she is stationary. But as soon as you say Jack's length must be the longest, that presupposes that he is stationary, doesn't it? If he is already stationary, then wouldn't that put Jill in different places, and Jack in the "same" place?

Jill is only in the same place relative to the two events (not absolute), they both are defined as happening at her position. This doesn't require her position to be stationary, as her and Jack could both, or neither, be moving faster than the relative velocity between them.

Jill will always observe things co-moving with her to be normal, un-contracted length, and the same for Jack. So the clock being co-moving with Jack precludes him observing length contraction. If he did, that would be like you seeing the hood of your car to be shorter even though it is at rest and has no velocity relative to yourself.

QUOTE
I asked Sithdarth a question, which he says is incomplete.  Have you looked at that?  There I'm suggesting that hinging things on co-moving clocks seems to lead to a contradiction.

Co-moving determines longest length, as that is the only frame in which everything is relatively at rest and cannot be length contracted.

QUOTE (->
 QUOTE I asked Sithdarth a question, which he says is incomplete.  Have you looked at that?  There I'm suggesting that hinging things on co-moving clocks seems to lead to a contradiction.

Co-moving determines longest length, as that is the only frame in which everything is relatively at rest and cannot be length contracted.

There may well be some contradiction here between Jill being stationary and Jack's length being 6LS, but I'm not sure where the source of that contradiction lies.  I do agree that Jack's distance is set, by definition, at 6 LS, but that, standing alone, is not incompatible with Jill's length being 7.5 LS

Sorry if I've confused you. Jill will never observe a longer length than Jack, like SD and rpenner said. I had concatenated this example with one in which both observers have two co-moving point that align for two events, i.e. the barn/pole paradox.

But perhaps that will help though. The barn/pole paradox is equivalent to both Jill and Jack having two clocks with the two events being 1) when Jack's front clock and Jill's back clock meet and 2) when Jack's back clock and Jill's front clock meet. Even if the length between clocks is 6cs in both frames, both clocks will not meet at the same time.

Each observer will see their front clock to meet before their back clock, since both sees the length between the others clocks to be contracted. This is the symmetry SD was referring to.
aintnuthin
QUOTE (synthsin75+Aug 5 2011, 05:32 PM)
Co-moving determines longest length, as that is the only frame in which everything is relatively at rest and cannot be length contracted.

Well, but indirectly it determines everything. As soon as you assign one party, arbitrarily or not, the longest length, you have also, in effect, assigned the maximal time to them. Of course, once you have done that, they have to be the "stationary" party.

I am just defining "stationary" as "the party with the greater time on his clock and the greatest distance." The other is, of course, "moving" in this context.

One limitation of possible "movement" in time or length comes from the fact that, in SR, one can only go a certain way. At the subjective level, every observer must "see" contraction, even if they are, in fact, dealing with relative expansion by the other party. Neither party is allowed to "see" the other has elongated lengths and contracted time, even though one of them must, in fact, experience it relative to the other.

I guess the "relativity of simultaniety" formula is designed to balance that out, but it really seems very superficial, to me, at least.

I'm just trying to figure out which "constraints" are due to physical law, and which are just artifacts of the need for a coherent mathematical system. Not easy for me to tell.
synthsin75
Co-moving only determines "stationary" relative to that particular co-moving observer, nothing else. Both can have the same two clock setup. In which case, you have no way to determine a "stationary" frame at all. Both would be equally co-moving with two clocks which determine the events.

In the symmetrical setup, both observers are "the party with the greater time on his clock and the greatest distance", and both would observe a completely symmetrical length contraction of the other.

All of this is physical law, as what each observes, even though seemingly contradictory, will effect the relativistic energy and mass that each measures.
aintnuthin
QUOTE (synthsin75+Aug 5 2011, 11:35 PM)
Co-moving only determines "stationary" relative to that particular co-moving observer, nothing else. Both can have the same two clock setup. In which case, you have no way to determine a "stationary" frame at all. Both would be equally co-moving with two clocks which determine the events.

Which, I guess, is just another way of saying that clocks are NOT determinative, right, Synth? One is going have the faster time, either way, it's just that this method won't "tell" us which one. It seems to me that the "co-moving clock" criterion is merely conventional, and designed to fit our pattern of mathematical analysis. If one clock is going faster, and one slower, it shouldn't, in theory, matter which one you read.

"In the symmetrical setup, both observers are "the party with the greater time on his clock and the greatest distance", and both would observe a completely symmetrical length contraction of the other."

I agree that SR it is not a truly symmetrical (relational) theory of motion. And yet most of it's adherents want to insist that it is, for some reason. Why's that, do you think?

"All of this is physical law, as what each observes, even though seemingly contradictory, will effect the relativistic energy and mass that each measures."

You can have all that without the strong emphasis on subjective "observer perceptions" being assigned a major role, though, can't you?
aintnuthin
QUOTE (synthsin75+Aug 5 2011, 08:13 AM)
The simple answer is that, since Jill already observes the two events to occur in the same place, there is just no space between events to be contracted. And both events co-moving with Jack, Jack cannot observe a length contraction between events.

Coming back to this, Synth, if Jack were "allowed" to see length elongation, he would, using the LT, calculate her length as 7.5 LS (and 12.5 seconds), right?

The LT itself doesn't say which one is contracted. It simply says that one of the two is, i.e, the moving one. There is nothing inherent in the LT which prevents a party (say one on a train) from acknowleding that they are the moving party, is there?

boit
My practise has been to accept SR as a fact. Starting from this premise I struggle to bring my understanding to per. Relativity for dummies agreed very well with my cheap scenarios but I get a headache when I try to apply it at the level of the big boys. I quickly lock myself in a loop and only by giving up am I able to retain my sanity. Perhaps I'll come to master it one day. . .
synthsin75
QUOTE (aintnuthin+Aug 6 2011, 08:21 AM)
Coming back to this, Synth, if Jack were "allowed" to see length elongation, he would, using the LT, calculate her length as 7.5 LS (and 12.5 seconds), right?

The LT itself doesn't say which one is contracted. It simply says that one of the two is, i.e, the moving one. There is nothing inherent in the LT which prevents a party (say one on a train) from acknowleding that they are the moving party, is there?

1. I've already said I was wrong there.
2. There is no such thing as a physical "length elongation".
http://en.wikipedia.org/wiki/Length_contra...#Visual_effects

Special relativity says there is no way to determine what any absolute velocities may be. You only know the relative velocities, and can only "call" one stationary relative another. With only two observers, there is no third with which to assign rest to either one.

It is only the idea of "stationary" that is an artifact of frame observed from, as all observers see their local frame as stationary relative to themselves.

QUOTE
Which, I guess, is just another way of saying that clocks are NOT determinative, right, Synth? One is going have the faster time, either way, it's just that this method won't "tell" us which one. It seems to me that the "co-moving clock" criterion is merely conventional, and designed to fit our pattern of mathematical analysis. If one clock is going faster, and one slower, it shouldn't, in theory, matter which one you read.

There is no absolute "faster", only relative velocity between them. The relative velocity can be composed of any two velocities with that difference. IOW, the relative velocity is .6c between objects moving at .1c and .7c (assuming aligned vectors), but it is also .6c between objects moving at .3c and .9c. And there is no way to tell which may be moving at which velocity.

The notions of "stationary" and "absolute velocity" are the only conventions, and they have nothing to do with the math, just our intuitive assumptions based solely on what we experience locally.

QUOTE (->
 QUOTE Which, I guess, is just another way of saying that clocks are NOT determinative, right, Synth? One is going have the faster time, either way, it's just that this method won't "tell" us which one. It seems to me that the "co-moving clock" criterion is merely conventional, and designed to fit our pattern of mathematical analysis. If one clock is going faster, and one slower, it shouldn't, in theory, matter which one you read.

There is no absolute "faster", only relative velocity between them. The relative velocity can be composed of any two velocities with that difference. IOW, the relative velocity is .6c between objects moving at .1c and .7c (assuming aligned vectors), but it is also .6c between objects moving at .3c and .9c. And there is no way to tell which may be moving at which velocity.

The notions of "stationary" and "absolute velocity" are the only conventions, and they have nothing to do with the math, just our intuitive assumptions based solely on what we experience locally.

I agree that SR it is not a truly symmetrical (relational) theory of motion. And yet most of it's adherents want to insist that it is, for some reason. Why's that, do you think?

SR is completely relative. Absolute values are an artifact of your intuitive experience where you "call" something at rest relative to the Earth, for example, "stationary". It is only at rest relative to the Earth, as it shares the same velocity with that surface of the Earth.

QUOTE
You can have all that without the strong emphasis on subjective "observer perceptions" being assigned a major role, though, can't you?

No, you can't. As the effective energy and mass has a different actual and physical effect based on which frame it is acting upon.
aintnuthin
QUOTE (synthsin75+Aug 6 2011, 04:38 PM)
1. I've already said I was wrong there.
2. There is no such thing as a physical "length elongation".

I think you may have understood me, Synth. Nothing I said had any thing to do with absolute motion, that I know of. By "length elongation" I didn't mean a physical process, just a relative one. If your lengths are contracted, then relative to you, mine are elongated, that's all. That said, the theory seems to imply that length will physically elongate as one decreases his speed, but that's not what I'm talking about here.

I'm just noting that the Lorentz tranform indicates that only one of the two objects in question can be (relatively) contracted, leaving the other one (relatively) elongated.

But it must be one or other, not "both," and it could be either as far as the LT goes--it doesn't tell you that.

So even though both observers "see" length contraction, only one of them can be seeing it correctly.

It's not the math (conventions aside), or the physics, which has the frame in which the clocks rest "determine" the person with the greater length. If any part of SR does that, it's the "philosophical" part.
aintnuthin
QUOTE (synthsin75+Aug 6 2011, 04:38 PM)
1It is only the idea of "stationary" that is an artifact of frame observed from, as all observers see their local frame as stationary relative to themselves.

No, you can't. As the effective energy and mass has a different actual and physical effect based on which frame it is acting upon.

Well, as I said, I'm am using "stationary" to mean which (between two) has the longer lengths and time. The theory says all observers (MUST) see their frame as stationary, but it's not true in practice, is it? Who, when riding a train, "really" thinks they're stationary and ask the conductor if Chicago stops here?

"No, you can't. As the effective energy and mass has a different actual and physical effect based on which frame it is acting upon."

Well, sure, but that doesn't require that all observers insist they are stationary when, by hypothesis, one of them is not, does it?
synthsin75
QUOTE (aintnuthin+Aug 6 2011, 11:00 AM)
I think you may have understood me, Synth. Nothing I said had any thing to do with absolute motion, that I know of. By "length elongation" I didn't mean a physical process, just a relative one. If your lengths are contracted, then relative to you, mine are elongated, that's all. That said, the theory seems to imply that length will physically elongate as one decreases his speed, but that's not what I'm talking about here.

Your length, to you, is never contracted.

QUOTE
I'm just noting that the Lorentz tranform indicates that only one of the two objects in question can be (relatively) contracted, leaving the other one (relatively) elongated.

But it must be one or other, not "both," and it could be either as far as the LT goes--it doesn't tell you that.

So even though both observers "see" length contraction, only one of them can be seeing it correctly.

Both can indeed be contracted, and actually are regardless of what events or setup you use. The laws of physics are consistent because they are the same in every frame. One is just as valid as the other.

QUOTE (->
 QUOTE I'm just noting that the Lorentz tranform indicates that only one of the two objects in question can be (relatively) contracted, leaving the other one (relatively) elongated.But it must be one or other, not "both," and it could be either as far as the LT goes--it doesn't tell you that.So even though both observers "see" length contraction, only one of them can be seeing it correctly.

Both can indeed be contracted, and actually are regardless of what events or setup you use. The laws of physics are consistent because they are the same in every frame. One is just as valid as the other.

It's not the math (conventions aside), or the physics, which has the frame in which the clocks rest "determine" the person with the greater length.  If any part of SR does that, it's the "philosophical" part.

There is no "philosophical part" of SR. "At rest" is merely a human convention that doesn't have any absolute meaning in reality.
aintnuthin
QUOTE (synthsin75+Aug 6 2011, 05:46 PM)
Both can indeed be contracted, and actually are regardless of what events or setup you use.

"Actually are?" You mean each is actually shorter than the other? You get that from the math of the LT? If not where does that come from, if not philosophy?
synthsin75
QUOTE (aintnuthin+Aug 6 2011, 12:44 PM)
"Actually are?" You mean each is actually shorter than the other? You get that from the math of the LT? If not where does that come from, if not philosophy?

Anything of any extent in space (size) with be length contracted as viewed from an observer at a relative velocity. Both observations and LT verify this.
aintnuthin
QUOTE (synthsin75+Aug 6 2011, 06:52 PM)
Anything of any extent in space (size) with be length contracted as viewed from an observer at a relative velocity. Both observations and LT verify this.

Could be. Although I've been led to believe that length contraction has never experimentally (as opposed to theoretically) verfied.

Either way, I don't see how or where the LT indicates that "both" are contracted, Synth.
brucep
QUOTE (aintnuthin+Aug 6 2011, 07:02 PM)
Could be.  Although I've been led to believe that length contraction has never experimentally (as opposed to theoretically) verfied.

Either way, I don't see how or where the LT indicates that "both" are contracted, Synth.

It's been experimentally verified. Why not look it up before you flap your jaw. Then I wouldn't have to say anything.

http://en.wikipedia.org/wiki/Length_contra...l_verifications

aintnuthin
QUOTE (brucep+Aug 6 2011, 07:20 PM)
It's been experimentally verified. Why not look it up before you flap your jaw. Then I wouldn't have to say anything.

Hmmm, well, I see this:

"A direct experimental confirmation of Lorentz contraction is of course hard to achieve, as such an effect can only be observed at particles that nearly travel at the speed of light, and which spatial dimensions are vanishingly small. In addition, it could only be measured by an observer not at rest in the same inertial frame, as the observer would undergo a contraction as well, that is, he can judge himself and the object as at rest in the same inertial frame in accordance with the relativity principle (as it was demonstrated by the Trouton-Rankine experiment)."

And I see this: "In addition, there are other indirect confirmations (besides the Michelson-Morley experiment) of this effect. For example, heavy ions that are spherical when at rest should assume the form of "pancakes" or flat disks when traveling nearly at the speed."

But I don't see where it says it has been "observed," as you suggested. Not a big deal, anyway, Synth. But I kinda sense a bit of hostility creeping into your tone. Am I angering you?
brucep
QUOTE (synthsin75+Aug 4 2011, 04:09 PM)
I'm not doing this BS quibbling over your poor comprehension.

You're pretty confused for the most part but you've improved somewhat. At least you're no longer claiming the proper frame is a preferred frame or that light is at rest in some frame of reference. You should appreciate SD since he has the patience to help you out. I noticed that you actually admiited to error which is a step in the right direction.
brucep
QUOTE (aintnuthin+Aug 6 2011, 07:33 PM)
Hmmm, well, I see this:

"A direct experimental confirmation of Lorentz contraction is of course hard to achieve, as such an effect can only be observed at particles that nearly travel at the speed of light, and which spatial dimensions are vanishingly small. In addition, it could only be measured by an observer not at rest in the same inertial frame, as the observer would undergo a contraction as well, that is, he can judge himself and the object as at rest in the same inertial frame in accordance with the relativity principle (as it was demonstrated by the Trouton-Rankine experiment)."

And I see this: "In addition, there are other indirect confirmations (besides the Michelson-Morley experiment) of this effect. For example, heavy ions that are spherical when at rest should assume the form of "pancakes" or flat disks when traveling nearly at the speed."

But I don't see where it says it has been "observed," as you suggested.  Not a big deal, anyway, Synth.  But I kinda sense a bit of hostility creeping into your tone.  Am I angering you?

That's because you don't know what the term 'observed' means in physics. You want to claim the natural phenomena we refer to as length contraction hasn't been observed in nature go for it.
aintnuthin
QUOTE (brucep+Aug 6 2011, 07:44 PM)
That's because you don't know what the term 'observed' means in physics. You want to claim the natural phenomena we refer to as length contraction hasn't been observed in nature go for it.

Enlighten me. What does the term 'observed' mean in physics?

On second thought, never mind. I really don't care for semantical quibbles.
brucep
QUOTE (aintnuthin+Aug 6 2011, 07:47 PM)
Enlighten me.  What does the term 'observed' mean in physics?

On second thought, never mind. I really don't care for semantical quibbles.

You figure it out. There were examples in the link such as the path of the muons. Magnetism as a consequence of length contraction. Without length contraction there would be no observation of magnetic fields.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

It's called terminology not semantics. You say length contraction hasn't been observed then folks versed on what observation means know you don't understand what it means.
synthsin75
QUOTE (brucep+Aug 6 2011, 01:37 PM)
You're pretty confused for the most part but you've improved somewhat. At least you're no longer claiming the proper frame is a preferred frame or that light is at rest in some frame of reference. You should appreciate SD since he has the patience to help you out. I noticed that you actually admiited to error which is a step in the right direction.

I never claimed that any frame was preferred, so why don't you pull your head out of your *** and quit trolling me.
synthsin75
aintnuthin,

http://en.wikipedia.org/wiki/Length_contra...l_verifications
QUOTE (^+)
On the other side, "real" is also used in connection with "absolute", and "apparent" is thus "relative".

We know that there are no absolutes, so the first definition of "real" must be true:

QUOTE
On one side, the word "real" is used for things that we can measure without considerable observational errors

So length contraction is unequivocally real, as it can be measured without considerable error. IOW, we can measure it consistently enough to fully verify it.
brucep
QUOTE (synthsin75+Aug 6 2011, 08:34 PM)
I never claimed that any frame was preferred, so why don't you pull your head out of your *** and quit trolling me.

Sure you did in that unwanted petulant email you sent me. The email where you called me an idiot because you thought the proper frame I was discussing was a preferred frame and you were pissed off because I blocked your harassing PM's. Bet somebody else knows about your petulant harangues outside the forum. As far as relativity goes you think you can read a few wiki pages and you're an expert For the most part you're clueless. You should learn some math so you could figure out what it's all about. You should kiss SD's behind for having been so patient with a twit like you.
synthsin75
QUOTE (brucep+Aug 6 2011, 04:35 PM)
Sure you did in that unwanted petulant email you sent me. The email where you called me an idiot because you thought the proper frame I was discussing was a preferred frame and you were pissed off because I blocked your harassing PM's. Bet somebody else knows about your petulant harangues outside the forum. As far as relativity goes you think you can read a few wiki pages and you're an expert For the most part you're clueless. You should learn some math so you could figure out what it's all about. You should kiss SD's behind for having been so patient with a twit like you.

Bruce, you're the fool who kept claiming that the pole could never fit in the barn specifically because of its length in its own frame. That directly implies that you were the ***** assuming a preferred frame, as I clearly stated the correct answer to that question with a very large array of references backing up exactly what I said.

You just like to keep talking out of you hat, even when you are completely refuted by a host of very reputable references. So quit blaming me for your embarrassment.

And all this was what, three, four, five months ago? You've been stewing on it all this time waiting for you opportunity to troll me over it. Pathetic. Go lick your wounds elsewhere.
brucep
QUOTE (synthsin75+Aug 6 2011, 11:11 PM)
Bruce, you're the fool who kept claiming that the pole could never fit in the barn specifically because of its length in its own frame. That directly implies that you were the ***** assuming a preferred frame, as I clearly stated the correct answer to that question with a very large array of references backing up exactly what I said.

You just like to keep talking out of you hat, even when you are completely refuted by a host of very reputable references. So quit blaming me for your embarrassment.

And all this was what, three, four, five months ago? You've been stewing on it all this time waiting for you opportunity to troll me over it. Pathetic. Go lick your wounds elsewhere.

You couldn't understand what I was talking about. That's why you thought it was a preferred frame. For the most part you're pretty clueless. You should be thanking SD instead of telling him he has a comprehension problem. You're the one with the comprehension problem. Know nothing sexual intellectual internet punk.I'm not stewing over anything related to you but I do like to call you on something like what you said to SD. Petulant whining bullshit because someone needs to show how sloppy your responses are.
synthsin75
QUOTE (brucep+Aug 6 2011, 06:02 PM)
You couldn't understand what I was talking about. That's why you thought it was a preferred frame. For the most part you're pretty clueless. You should be thanking SD instead of telling him he has a comprehension problem. You're the one with the comprehension problem. Know nothing sexual intellectual internet punk.I'm not stewing over anything related to you but I do like to call you on something like what you said to SD. Petulant whining bullshit because someone needs to show how sloppy your responses are.

You are the one who said the pole wouldn't fit, as viewed from the barn frame, specifically because you claimed a, what, .8c relative velocity magically became zero at the passing point. If there were any "sloppy responses", they were yours. I couldn't understand what you were talking about specifically because you were talking complete rubbish.

And your string of vitriol does nothing to change the facts. You are simply old and impotent (in body and mind), and like to feel better about yourself by berating hapless people online.

The facts, which I provided in many references, prove you were talking pure bollocks. End of story. So sorry, no redemption for Brucie.