Special Relativity Question

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?

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.

This post has been edited by aintnuthin on Aug 3 2011, 03:48 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".

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)²−(Δx)² = 64 light-seconds²

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')²−(Δx')² = (cΔt')² = 64 light-seconds² and show that Δt' = 8 seconds exactly for Jill.

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?

This post has been edited by aintnuthin on Aug 4 2011, 01:40 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)²−(Δx)² = 100 light-seconds²

Now delta t' would be 10 seconds exactly for Jack.

This post has been edited by aintnuthin on Aug 4 2011, 01:42 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.

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.

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')². Jill is always stationary to Jill.

For Jack, not stationary to Jill, the only constraint for measuring the two events is (Δx)² < (cΔt)², which is to say Jill's speed relative to Jack is v = Δx/Δt, and v² = (Δx)² /(Δt)² < c².

So for Jack, who is in inertial motion relative to Jill, I = (cΔt)²−(Δx)² = (Δt)²(c²−v²) and for Jill I = (cΔt')² for the same two events (which both happen to Jill. So since I = I means (Δt')²c² = (Δt)²(c²−v²) we have (Δt')²/(Δt)² = (c²−v²)/c². So if 0 < v < c, then (Δt')² < (Δt)² and so Jill's measured time is always the shortest inertially measured time for two events that happen to Jill.

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.

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.

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.

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.

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')². 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)² < (cΔt)²..."

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?

This post has been edited by aintnuthin on Aug 4 2011, 08:10 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. 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?

This post has been edited by aintnuthin on Aug 4 2011, 08:38 AM

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?

This post has been edited by aintnuthin on Aug 4 2011, 09:48 AM

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?

This post has been edited by aintnuthin on Aug 4 2011, 10:31 AM

So, truth be told, I'm still confused about your original answer.

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.

This post has been edited by aintnuthin on Aug 4 2011, 12:41 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.

This post has been edited by aintnuthin on Aug 4 2011, 12:47 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.