Second triplets paradox (muons)

[esbo]

I don’t usually post to these forums because I don’t have time to do a back and forth dialog. But I have to reply to this thread. The “paradox” presented by the author that esbo referenced is so basic it could be given as a homework problem to a student taking “Introduction to Special Relativity”. The key to the resolution, as has already been mentioned, is Relativity of Simultaneity.

 

Let me walk thru the homework problem.

 

Given:

You are sitting on the surface of the Earth monitoring a box of decaying muons with a clock. I am monitoring an identical box of decaying muons with my own clock. But I am at an altitude of 10 light-seconds above hurtling toward you at .866c. (10 light-seconds is the distance light travels in 10 seconds). We both start monitoring at the same time according to you.

 

As far as you are concerned, when I reach you, 11.55 seconds will have elapsed on your clock (10c/.866c). But only 5.77 seconds will have elapsed on my clock since my clock is running at 50% of yours. Since less time has elapsed for me, you expect that I will have more muons in my box than you have in your box.

 

Assignment:

Analyze the problem from my reference frame. What do I expect?

 

Solution.

From my point of view, the distance to you is only 5 light seconds (50% length contraction). So I expect to meet you when 5.77 seconds have elapsed on my clock (5c/.866c) and I expect to have an amount of muons in my box consistent with that time. That’s the same time (and the same amount of muons) as you expected for my clock and my box. So far, no contradiction.

 

Now for the hard part.

I believe your clock is running at 50% the rate of my clock. So if 5.77 seconds elapsed on my clock, only 2.89 seconds will have elapsed on your clock between the time I started monitoring and the time we met. But it was given that 11.55 seconds elapsed on your clock when we met. I’m standing right next to you. I have to see the same number on your clock or there is a paradox. How do I resolve that? Well, you must have started monitoring 17.33 seconds before I started monitoring. Since your clock was running at 50%, 17.33 /2 means 8.67 seconds already elapsed on your clock before I even started my clock. Add 2.89 more seconds while both of us were monitoring and I will expect 11.55 seconds elapsed on your clock when we meet. That eliminates the contradiction. If you’re thinking I invented my answer just for this problem, I didn’t. It’s part of the theory.

 

To Summarize:

From your point of view we both started monitoring our boxes of muons at the same time. From my point of view you started monitoring your muons 17.33 seconds before I started monitoring mine. What is simultaneous to you is not simultaneous to me. That’s how Length Contraction, Time Dilation and Relativity of Simultaneity work together to eliminate any contradiction.

 

Can I show that using actual calculations? Sure. I could post some equations with the right numbers on the other side of the “equals” sign. Would you then be convinced? I doubt it.

 

 

 

 

 

Thanks at least you appear to given a comprehensive answer, I don't have time to go through it right now,, but I will later.

[uncool]

I have now but it basically boils do to "Do the simultaneity calculation." and I don't believe that is a good enough answer it is too vague.

You need to highlight the error.

I mean you are basically say there is an error, go find it.

However, my basic question is "where is the error in this?"

So it's your job to find it!! Not mine!!

Uh. No, it doesn't. I highlighted the error - that the supposed "contradiction" was not a contradiction. I pointed out that the reasoning behind the claimed contradiction was that the three observers would see different numbers of the two types of muons at the same time, but that "at the same time" meant different things to the different observers, meaning that there was no such contradiction. Pointing out the logical error is highlighting the error.

=Uncool-

[esbo]

I don’t usually post to these forums because I don’t have time to do a back and forth dialog. But I have to reply to this thread. The “paradox” presented by the author that esbo referenced is so basic it could be given as a homework problem to a student taking “Introduction to Special Relativity”. The key to the resolution, as has already been mentioned, is Relativity of Simultaneity.

 

Let me walk thru the homework problem.

 

Given:

You are sitting on the surface of the Earth monitoring a box of decaying muons with a clock. I am monitoring an identical box of decaying muons with my own clock. But I am at an altitude of 10 light-seconds above hurtling toward you at .866c. (10 light-seconds is the distance light travels in 10 seconds). We both start monitoring at the same time according to you.

 

As far as you are concerned, when I reach you, 11.55 seconds will have elapsed on your clock (10c/.866c). But only 5.77 seconds will have elapsed on my clock since my clock is running at 50% of yours. Since less time has elapsed for me, you expect that I will have more muons in my box than you have in your box.

 

Assignment:

Analyze the problem from my reference frame. What do I expect?

 

Solution.

From my point of view, the distance to you is only 5 light seconds (50% length contraction). So I expect to meet you when 5.77 seconds have elapsed on my clock (5c/.866c) and I expect to have an amount of muons in my box consistent with that time. That’s the same time (and the same amount of muons) as you expected for my clock and my box. So far, no contradiction.

 

Now for the hard part.

I believe your clock is running at 50% the rate of my clock. So if 5.77 seconds elapsed on my clock, only 2.89 seconds will have elapsed on your clock between the time I started monitoring and the time we met. But it was given that 11.55 seconds elapsed on your clock when we met. I’m standing right next to you. I have to see the same number on your clock or there is a paradox. How do I resolve that? Well, you must have started monitoring 17.33 seconds before I started monitoring. Since your clock was running at 50%, 17.33 /2 means 8.67 seconds already elapsed on your clock before I even started my clock. Add 2.89 more seconds while both of us were monitoring and I will expect 11.55 seconds elapsed on your clock when we meet. That eliminates the contradiction. If you’re thinking I invented my answer just for this problem, I didn’t. It’s part of the theory.

 

To Summarize:

From your point of view we both started monitoring our boxes of muons at the same time. From my point of view you started monitoring your muons 17.33 seconds before I started monitoring mine. What is simultaneous to you is not simultaneous to me. That’s how Length Contraction, Time Dilation and Relativity of Simultaneity work together to eliminate any contradiction.

 

Can I show that using actual calculations? Sure. I could post some equations with the right numbers on the other side of the “equals” sign. Would you then be convinced? I doubt it.

 

 

 

 

 

 

OK I have had a bit more time on this, and it sort of sound plausible to some extent apart from the fact I am not too happy with length contraction.

It's perhaps better if that is addressed first as follows.

 

Now,as I understand it we will always measure the speed of light as a constant.

So say you have two people on opposite side of the earth measuring the speed, one spinning towards the light and one away from it.

Because of the earth's spin on it's axis they are both doing the same speed but in opposite direction (we can ignore orbit speed to keep it simple).

So doing the same speed they will the same (rate of) time, correct??

And also the same length contraction? Correct?

 

So that is the problem, how can they measure the same speed when light has to travel further over the identical metre ruler they had when they

met up?

 

You see the problem is one ruler is travelling towards the light and one away so light will have to go further to cross the ruler moving away from the light.

 

So I do not quite see how someone can answer this with "oh it's due to simultaneousness".

 

Or maybe you can?

 

I suppose you can say one twin could be considered stationary and the other travelling at twice the rate of spin???

 

So either way it is a bit confusing to think about when things we take for granted change.

 

So what is the explanation to that problem, is it as I said?

 

So you could say for the stationary one there is no contraction, but there is for the one moving towards the sun so his length will contract

*and* his time will slow down. AT first though that seems to give the opposite of the what I want to explain it, but I am not sure.

 

I need to think about it a bit more, I think. It might give the answer possibly.

 

I suppose if the ruler length is the distance light travels in a second and his time is slower then it will not have travelled as far

consistent with his shorter ruler?

 

But I am not OK with that yet, need to give it more though.

[swansont]

The spinnin earth is not actually an inertial frame. If you do an experiment where the spin is important, e.g. send light one way vs the other, around the equator, you will notice the effect of the spin.

 

But in an inertial set of frames, moving toward or away from the light will not affect the speed you measure.

[J.C.MacSwell]

The proposer has in effect provided his proof, you are saying it is wrong, I think the onus is you to prove he is wrong by

say specifically where he is wrong.

 

It's like in a maths question and you saying he has done his maths wrong, you need to provide a correction to the detail

ie say in which specific line of maths there is an error.

 

Just saying "you need to check your maths' is a bit of a cop out, I because anyone could say that irregardless of whether

they knew were the error was.

 

So if anyone can provide that specific detail I would be very grateful grateful.

 

If they can't I guess I will be forced to look elsewhere.

This is basically about math. Regardless of how you view, or believe in, the physics,

SR is consistent mathematically. So if you use SR, as an underlying assumption, in a thought

experiment and come up with a conflict in the results, you have either made a mistake as to how to use the theory, or made a mathematical error, or have proven the math of SR itself incorrect. So if you are convinced you have not made an error...where exactly is the math of SR inconsistent or wrong?

 

If you do not believe SR is mathematically consistent, point out the inconsistency.

[xyzt]

The spinnin earth is not actually an inertial frame. If you do an experiment where the spin is important, e.g. send light one way vs the other, around the equator, you will notice the effect of the spin.

 

But in an inertial set of frames, moving toward or away from the light will not affect the speed you measure.

The local speed of light is "c" even in rotating frames. There are preciously few papers on this subject but there are a few very good ones. So, esbo claim is false, he mixes the speed of light with closing speed. Closing speed is indeed [math]c \pm \omega r[/math], as evidenced by the class of experiments known as "Sagnac experiments".

[DimaMazin]

This is basically about math. Regardless of how you view, or believe in, the physics,

SR is consistent mathematically. So if you use SR, as an underlying assumption, in a thought

experiment and come up with a conflict in the results, you have either made a mistake as to how to use the theory, or made a mathematical error, or have proven the math of SR itself incorrect. So if you are convinced you have not made an error...where exactly is the math of SR inconsistent or wrong?

 

If you do not believe SR is mathematically consistent, point out the inconsistency.

Simultaneousness was useful always.The non-simultaneousness is useless.

Edited June 21, 2013 by DimaMazin

[swansont]

The local speed of light is "c" even in rotating frames. There are preciously few papers on this subject but there are a few very good ones. So, esbo claim is false, he mixes the speed of light with closing speed. Closing speed is indeed [math]c \pm \omega r[/math], as evidenced by the class of experiments known as "Sagnac experiments".

 

IOW, you will notice the effect of the spin.

[xyzt]

IOW, you will notice the effect of the spin.

Yes, this is well known, as the "Sagnac effect".

Edited June 21, 2013 by xyzt

[esbo]

This is basically about math. Regardless of how you view, or believe in, the physics,

SR is consistent mathematically. So if you use SR, as an underlying assumption, in a thought

experiment and come up with a conflict in the results, you have either made a mistake as to how to use the theory, or made a mathematical error, or have proven the math of SR itself incorrect. So if you are convinced you have not made an error...where exactly is the math of SR inconsistent or wrong?

 

If you do not believe SR is mathematically consistent, point out the inconsistency.

 

 

Well there are examples which can be used and have been used to point out inconsistency.

What I am saying is look at this example, it does not seem to fit with SR, where is the error.

 

Now I think it is reasonable to expect someone to point out the error rather than simply parrot out some line

about SR or maths or simultaneous need. I am finding those kind of non answers rather tiresome.

 

And the "this is about maths bit" is a bit silly, it is about the question asked and the solution, the examples point out the inconsistencies

that is the whole point of them.

 

The error is pointed out in the examples provided,.

 

the best way of to replying to a response like yours would be to simply repeat the question, however I won't do that, but you can

read it again if you so wish, and that is your answer.

 

The spinnin earth is not actually an inertial frame. If you do an experiment where the spin is important, e.g. send light one way vs the other, around the equator, you will notice the effect of the spin.

 

But in an inertial set of frames, moving toward or away from the light will not affect the speed you measure.

 

 

The spinning of the earth is pretty irrelevant to the question, the point is they are moving in opposite directions and that was just a

way of illustrating that point.

[swansont]

The spinning of the earth is pretty irrelevant to the question, the point is they are moving in opposite directions and that was just a

way of illustrating that point.

 

OK, then: closing speed is always measured to be c in your frame. Relative motion of the light source only changes the frequency of the light.

[esbo]

OK, then: closing speed is always measured to be c in your frame. Relative motion of the light source only changes the frequency of the light.

 

 

I know c is always measured as a constant, I just want the apparent anomalies explain but in a specific rather than generic way.

The sun is not moving in the example, or at least not considered to be.

[Delta1212]

 

 

I know c is always measured as a constant, I just want the apparent anomalies explain but in a specific rather than generic way.

The sun is not moving in the example, or at least not considered to be.

From the perspective of the sun. If you have two observers moving in opposite directions with respect to the sun, the sun is most definitely moving as seen by one or both observers.

[md65536]

You see the problem is one ruler is travelling towards the light and one away so light will have to go further to cross the ruler moving away from the light.

This is true for certain observers, for example someone at the north pole, or others similarly "stationary relative to the center of the Earth". For those observers, yes, light crosses the length of the ruler in a different amount of time depending on the direction the ruler is traveling. But that's not a measure of the speed of light, that's a measure of the change in difference of position of the ruler and the light (aka "closing speed"). Speed of light is measured relative to a stationary, non-contracted ruler.

 

Consider the same thing from either of the observers on the equator. The observer has its own ruler, which is stationary and not contracted. Light takes 1m/c seconds to cross the ruler. Consider the other observer and you get the same result. Each considers its ruler as stationary, neither considers the ruler to be traveling toward or away from the photons.

[esbo]

The local speed of light is "c" even in rotating frames. There are preciously few papers on this subject but there are a few very good ones. So, esbo claim is false, he mixes the speed of light with closing speed. Closing speed is indeed [math]c \pm \omega r[/math], as evidenced by the class of experiments known as "Sagnac experiments".

 

 

Why claim is false, I am not aware of making a claim, I am basically say, ""explain this".

So what specifically are you referring to when you say my claim is false.

I am not sure so perhaps you could refresh me on the claim you say i have made.

 

From the perspective of the sun. If you have two observers moving in opposite directions with respect to the sun, the sun is most definitely moving as seen by one or both observers.

 

 

Well not really the sun is stationary and they are the ones moving.

[Mike-from-the-Bronx]

Oh, I didn’t read the whole thread. Now I see why it is called a triplets paradox. There’s one doing the observing and two being observed. This could be an extra credit homework problem in an introductory course in SR.

 

Let me restate the problem in more precise terms. One of a set of triplets, the observer, is at rest with respect to the source of a distant light pulse which is far off in the x-direction. The second triplet has a measuring stick 1 light-second long with a clock attached at either end and is moving at a speed of .866c toward the source of the light pulse. The third triplet, also with a measuring stick and two clocks is moving at a speed of .866c away from the source of the light pulse. How can the observer triplet believe that the other two, moving in opposite directions, will measure the same value for the speed of the light pulse?

 

Time Dilation, Length Contraction and Relativity of Simultaneity all play a role again. But the key to understanding what is going on is this: The leading clock has the trailing time. The trailing clock has the leading time. (That’s just another version of Relativity of Simultaneity.)

 

Start with the triplet moving toward the light pulse. The right side clock is its leading clock and the left is its trailing clock. The right side clock records the passage of the light pulse first. Say it records the passage at time t=10.000sec. The left side clock reads .866sec ahead of the right side clock. Familiar number? So, according to the observer triplet, at the moment the light pulse passes the leading clock (10.000sec) the left side clock already reads 10.866sec. (I’ll leave it for another day to show how that’s calculated) How long will it take the light pulse to reach the left side clock? Because of length contraction the distance between clocks is only half a light second and the left side clock and light pulse are closing on each other at 1.866c. So the observer triplet calculates that it will take .5c/1.866c = .2679sec for the light pulse and left side clock to meet. Since the clocks moving with the second triplet are ticking at 50%, only .2679 / 2 = .134sec will elapse on the left side clock. So, starting time of 10.866 +.134 gives 11.000sec elapsed. The second triplet will calculate 11.000 -10.000 or 1sec for the time for the pulse to traverse the 1-light second long measuring stick and the second triplet will conclude the light pulse is traveling at speed “c”.

 

Now, for the third triplet, the one moving away from the light pulse. (I’m getting tired) Again the passage of the light pulse will be recorded by the right side clock first. But now the right side clock is the trailing clock and has the leading time. So this time according to the observer triplet, when the right side clock records the passage of the light pulse at, say t=100.000sec, the left side clock will only read 99.134sec. How long for the light pulse and left side clock to meet? Well, according to the observer triplet, the light pulse is closing on the left side clock at a speed of only (1c -.866c) = .134c. (.5c/.134c) = 3.731sec for the light pulse and this left side clock to meet. Dividing by 2 again to account for 50% time dilation gives 1.866sec elapsed on the left side clock. (99.134 + 1.866) = 101.000sec at the meeting. The third triplet will calculate 101.000 -100.000 or 1sec for the time for the pulse to traverse the 1-light second long measuring stick and the third triplet will also conclude the light pulse is traveling at speed “c”.

Edited June 26, 2013 by Mike-from-the-Bronx

[DimaMazin]

 

The burden of that detail is on the proposer who claims relativity is wrong. You don't get a pass because you didn't follow the proper rigor in presenting the problem. Do the simultaneity calculation.

We can do it

 

Link removed

[hypervalent_iodine]

!

Moderator Note

DimaMazin, please stop using other threads to advertise your pet theory.