B Time dilation problem question

[IvicaPhysics]

So there is a problem I don't get. I special relativity, things moving relative to me, I see them as experiencing more time/ slower time. Ok...so then I should also see it's velocity decrease. If I see it's clock running slower, so should I it's velocity. V=...m/s
So if I see it's time slow down then V=...m/2s. And if the answer is, distance also decreases, if I saw something run past me at 0.5c, I should see the path before it shrink? Distance decreases only from the point of view of that thing (lenght contraction), but not from mine.
Thank you for help, I really don't get this.

 

[PeroK]

The velocity you measure is simply the spatial displacement you measure divided by the time you measure. SR has nothing to do with that.

 

[Maxila]

Realize it is not your clock that slows. If something were moving relative to you than it is the clock of that thing that is slow relative your clock; your measure of their velocity or distance doesn't change. Their measure of distance does in fact disagree with yours and would be shorter and they also would observe your clock to be running faster than theirs.

 

[Ibix]

You need to make measurements with your clocks and your rulers. They don't get contracted or dilated just because they're measuring someone who is moving. If you try to make measurements with someone else's clocks or rulers then you'll get strange answers - unless you are careful to account for simultaneity differences as well.

 

[PeroK]

Their measure of distance does in fact disagree with yours and would be shorter and they also would observe your clock to be running faster than theirs.

That's not correct. Something at rest relative to you will be length contracted to them. But, something at rest relative to them will be length contracted to you. The situation is symmetrical.

 

[Nugatory]

Their measure of distance does in fact disagree with yours and would be shorter and they also would observe your clock to be running faster than theirs.

That's not correct. Something at rest relative to you will be length contracted to them. But, something at rest relative to them will be length contracted to you. The situation is symmetrical.

And because the situation is symmetrical, they also find that your clock is running slower than yours, not faster. Yes, both observers find that the other clock is slower than their own, and they're both right.

(PeroK already knows this, of course - I'm adding this for others following the thread)

 

[Maxila]

That's not correct. Something at rest relative to you will be length contracted to them. But, something at rest relative to them will be length contracted to you. The situation is symmetrical.

I was talking about the distance between them as the OP inferred he should see an apparent velocity change and that should imply the distance from his position would decrease also. I was not discussing that the object would appear length contracted in its direction of motion. (like a muon traveling at near light speed would measure the distance to Earth as being much less than an observer on Earth would measure to the muon.)

 

[PeroK]

I was talking about the distance between them as the OP inferred he should see an apparent velocity change and that should imply the distance from his position would decrease also. I was not discussing that the object would appear length contracted in its direction of motion. (like a muon traveling at near light speed would measure the distance to Earth as being much less than an observer on Earth would measure to the muon.)

But, in the muon frame it's the Earth that is moving, so there is symmetry there too.

For both they are $vt$ apart $t$ seconds before they meet, where each measures $t$ on their clock.

 

[Maxila]

For both they are $vt$ apart $t$ seconds before they meet, where each measures $t$ on their clock.

Exactly and their clocks don't agree on the value of t.

 

[Herman Trivilino]

I was talking about the distance between them as the OP inferred he should see an apparent velocity change and that should imply the distance from his position would decrease also.

You meant to say his rate of change of distance, not the distance itself. To say the distance decreases is to say there is motion towards to the observer.

But anyway, to answer your original question ...

I special relativity, things moving relative to me, I see them as experiencing more time/ slower time. Ok...so then I should also see it's velocity decrease.

Not if you use your own clock to measure their velocity. Which is really the only way to do it that makes sense.

(If they use their own clocks to measure their own speed they get zero, since they're not moving relative to themselves. So in that sense you do see a decrease in velocity!)

 

[PeroK]

Exactly and their clocks don't agree on the value of t.

So, which one thinks they are "further apart"? Muon or Earth? Which one's clock is running slower?

 

[Maxila]

So, which one thinks they are "further apart"? Muon or Earth? Which one's clock is running slower?

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

 

[stevendaryl]

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

I think PeroK was asking you to explain how you would answer the question. He was being Socratic (teaching by asking leading questions).

 

[stevendaryl]

Realize it is not your clock that slows. If something were moving relative to you than it is the clock of that thing that is slow relative your clock; your measure of their velocity or distance doesn't change. Their measure of distance does in fact disagree with yours and would be shorter and they also would observe your clock to be running faster than theirs.

Everybody has pointed out that this is incorrect, but I don't think you have acknowledged it.

According to SR, if Alice and Bob are moving at constant velocity relative to one another, then

• According to an inertial coordinate system in which Alice is at rest, Bob's clock runs slower than Alice's.
• According to an inertial coordinate system in which Bob is at rest, Alice's clock runs slower than Bob.

 

[Maxila]

Everybody has pointed out that this is incorrect, but I don't think you have acknowledged it.

According to SR, if Alice and Bob are moving at constant velocity relative to one another, then

• According to an inertial coordinate system in which Alice is at rest, Bob's clock runs slower than Alice's.
• According to an inertial coordinate system in which Bob is at rest, Alice's clock runs slower than Bob.

Forget "Alice and Bob" use the muon example in the Georgia State University link above. It represents a real observed phenomenon.

 

[stevendaryl]

Forget "Alice and Bob" use the muon example in the Georgia State University link above. It represents a real observed phenomenon.

Geez. Okay:

• According to a frame in which the muon is at rest, clocks on the Earth are running slow.
• According to a frame in which the Earth is at rest, the muon is decaying slowly (that is, it has a longer lifetime than usual).

 

[stevendaryl]

Forget "Alice and Bob" use the muon example in the Georgia State University link above. It represents a real observed phenomenon.

But I don't want to forget "Alice" and "Bob". I want to know what your answer is. I think you're still confused, but I can only find that out by asking you about the details of your understanding of SR. A link to a paper does not tell me what your understanding is, unless you are the author of that paper.

 

[Maxila]

A link to a paper does not tell me what your understanding is...

It's not a paper, it's an example of the actual muon problem with a calculated solution as per SR. Just substitute Alice and Bob for muon and Earth observers to ask your question or make your point (I don't want misunderstandings between us do to syntax i.e. "Alice and Bob").

 

[Maxila]

Geez. Okay:

• According to a frame in which the muon is at rest, clocks on the Earth are running slow.

That appears to contradict this section of the example:

"In the muon experiment, the relativistic approach yields agreement with experiment and is greatly different from the non-relativistic result. Note that the muon and ground frames do not agree on the distance and time, but they agree on the final result. One observer sees time dilation, the other sees length contraction, but neither sees both.

These calculated results are consistent with historical experiments." link: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

Please explain?

According to a frame in which the Earth is at rest, the muon is decaying slowly (that is, it has a longer lifetime than usual).

Agreed.

 

[stevendaryl]

That appears to contradict this section of the example:

"In the muon experiment, the relativistic approach yields agreement with experiment and is greatly different from the non-relativistic result. Note that the muon and ground frames do not agree on the distance and time, but they agree on the final result. One observer sees time dilation, the other sees length contraction, but neither sees both.

The underlined sentence is incorrect, but irrelevant. What fraction of muons reach the ground before decaying has nothing to do with how fast Earth-clocks are running in a frame in which the muons are at rest.

Time dilation is seen from every frame, and length contraction is seen from every frame. But calculating how many muons make it to the surface of the Earth only involves time dilation in the Earth frame, and only involves length contraction in the muon frame.

 

[Maxila]

Time dilation is seen from every frame, and length contraction is seen from every frame. But calculating how many muons make it to the surface of the Earth only involves time dilation in the Earth frame, and only involves length contraction in the muon frame.

Not so, in the case of the twin paradox when the space traveling twin has turned around and accelerated back towards Earth they see a contracted distance to the Earth relative to the Earth twin (who does not); the space traveling twin also sees the Earth observers clock as running fast relative to their own while the Earth observers see the traveling twins clock running slow. Think of the traveling twin in this part as the Muon relative to the Earth.

The underlined sentence is incorrect, but irrelevant...

I disagree as it is also seen in the twin example above, and bear in mind, that site is hosted by the Department of Physic and Astronomy of Georgia State University while my only knowledge of your credentials is what I see as a forum poster here (and I don't know the criteria to get those acknowledgments).

 

[Ibix]

Not so, in the case of the twin paradox when the space traveling twin has turned around and accelerated back towards Earth they see a contracted distance to the Earth relative to the Earth twin (who does not); the space traveling twin also sees the Earth observers clock as running fast relative to their own while the Earth observers see the traveling twins clock running slow. Think of the traveling twin in this part as the Muon relative to the Earth.

No. Unless you are literally referring to what is seen and not correcting for the travel time of light. Both will determine that the other's clock ticks slowly if they do correct for that. Both will see the other's rulers as contracted. As @stevendaryl says, there is a minor and error in the paper that is irrelevant to its conclusions.

I disagree as it is also seen in the twin example above, and bear in mind, that site is hosted by the Department of Physic and Astronomy of Georgia State University while my only knowledge of your credentials is what I see as a forum poster here (and I don't know the criteria to get those acknowledgments).

Literally every textbook on relativity will show you that Steven is right. It's a trivial application of the Lorentz transforms, and you are making a naive mistake in applying the time dilation formula to the traveling twin's perspective and drawing the conclusion you do - namely ignoring the impact of the change of simultaneity convention at turnover. Ultimately, you are implying the existence of an absolute frame of reference, because you have a global sense in which one clock is moving faster than the other and hence ticking slowly.

 

[Nugatory]

and bear in mind, that site is hosted by the Department of Physic and Astronomy of Georgia State University while my only knowledge of your credentials is what I see as a forum poster here (and I don't know the criteria to get those acknowledgments).

Nonetheless, it is incorrect. It's clear that the author was not speaking precisely, because they used the word "see" - and we all understand (I hope) that neither time dilation nor length contraction are things that you see. I suspect that the author intended the word "see" to mean "use in our calculations of what's going on", in which case they're saying the same thing as Stevendaryl, except less precisely.

 

[stevendaryl]

Not so, in the case of the twin paradox when the space traveling twin has turned around and accelerated back towards Earth they see a contracted distance to the Earth relative to the Earth twin (who does not); the space traveling twin also sees the Earth observers clock as running fast relative to their own while the Earth observers see the traveling twins clock running slow. Think of the traveling twin in this part as the Muon relative to the Earth.

Okay, so I was right to suspect that you don't understand mutual time dilation and mutual length contraction. In the case of the twin paradox, there are three inertial frames that are relevant:

• F_1: The frame of the stay-at-home twin (on the Earth).
  
• F_2: The frame in which the traveling twin is at rest on his outbound journey.
• F_3: The frame in which the traveling twin is at rest on his return journey.

According to the inertial coordinate system associated with F_1, the traveling twin's clock runs slower than Earth clocks during both legs of his journey. According to F_2, the traveling twin's clock runs faster than the Earth clocks during the first leg of his journey, but slower during the return leg. According F_3, the traveling twin's clock runs slower during the outward journey, but faster on the return journey.

All three frames see time dilation and length contraction. They disagree about whose clocks are time-dilated when, but all three agree that the traveling twin is younger than the stay-at-home twin when the two get back together.

I disagree as it is also seen in the twin example above, and bear in mind, that site is hosted by the Department of Physic and Astronomy of Georgia State University while my only knowledge of your credentials is what I see as a forum poster here (and I don't know the criteria to get those acknowledgments).

I have never made any claims based on having particular credentials, but the particular line in question is either wrong or we're both misinterpreting it. Maybe what they mean is that there are two phenomena that can be used to explain the muon count reaching the Earth:

1. Time dilation of the muons.
2. Length contraction of the distance between the point of muon creation and the surface of the Earth.

Only the Earth frame sees 1, and only the muon frame sees 2.

 

[Maxila]

Do you see the contradiction for a case of the Muon and Earth observers if "seeing" each others clock running slowly"; the actual decay rate observations from either observer necessitate only the muon clock to be running slow relative to the Earth clock as calculated and explained in the Hyperphysics. The same is true for the traveling twin on their inbound leg, but muon decay is an actual observed phenomenon (they decay too slowly relative to clocks on Earth).

If we say both observers see the others clock running slow we break symmetry in the observations. The Muon decays at the expected rate according to its clock, but too quickly relative to how it sees the Earth clock (seen as running relatively slower). Meanwhile the Earth observer sees muon decay as being to slower than expected according to his own clock, but as expected relative to how Earth sees the muon clock. On the other hand the Hyperphysics explanation is symmetrical, both observers see the proper decay rate relative to how they see the muon clock and a slow decay rate relative to how they see the Earth clock.

 

[Ibix]

the actual decay rate observations from either observer necessitate only the muon clock to be running slow relative to the Earth clock as calculated and explained in the Hyperphysics.

Depends what you mean by "necessitate only" here. I can interpret this as "we only require the muon clock to be running slow as seen from the Earth frame and we don't care how the muon sees the Earth clock", in which case it is correct. But I suspect you mean "we require the muon clock to be running slow as seen from the Earth frame and the Earth clock to be running fast from the muon frame", in which case it is wrong.

If we say both observers see the others clock running slow we break symmetry in the observations. The Muon decays at the expected rate according to its clock, but too quickly relative to how it sees the Earth clock (seen as running relatively slower). Meanwhile the Earth observer sees muon decay as being to slower than expected according to his own clock, but as expected relative to how Earth sees the muon clock. On the other hand the Hyperphysics explanation is symmetrical, both observers see the proper decay rate relative to how they see the muon clock and a slow decay rate relative to how they see the Earth clock.

No, we do not break the symmetry of observations. You are doing so by saying that one must see the other's clocks tick slow while the other sees them tick fast. Symmetry is both observers saying the other's clock ticks slow.

The point is that the muon doesn't care about Earth's clocks. It sees the atmosphere length contracted, so has plenty of time to get to the (very close) ground in its proper lifetime. An Earth observer doesn't care about length contraction of the muon, only that it's clocks are time dilated so that it decays slowly enough to reach the (many-kilometer distant) ground. Both see length contraction and time dilation in the other, but both only care about one of the phenomena in this case.

 

[stevendaryl]

Do you see the contradiction for a case of the Muon and Earth observers if "seeing" each others clock running slowly";

Mutual time dilation is a prediction of SR, and so is the decay rate of high-speed muons. There is no contradiction. If you really don't understand this, you need to work through the equations yourself. They aren't difficult.

 

[Herman Trivilino]

The Muon decays at the expected rate according to its clock, but too quickly relative to how it sees the Earth clock (seen as running relatively slower). Meanwhile the Earth observer sees muon decay as being to slower than expected according to his own clock, but as expected relative to how Earth sees the muon clock.

You are looking at an example where the time measured in the muon's rest frame is a proper time.

Look also at an example where the time measured in the Earth's rest frame is a proper time. Then you will see that the time that elapses in the muon's rest frame is larger than the time elapsed in the Earth's rest frame. Here's one. A fire cracker at rest on Earth's surface explodes $6.8 \ \mathrm{\mu s}$ after it's lit. To an observer at rest relative to the muon you mentioned, the time elapsed between lighting and exploding will be $34 \ \mathrm{\mu s}$.

 

[vanhees71]

It's also a good exercise to show that the number of muons reaching the Earth is independent of any reference frame. It's a Lorentz scalar! For the very simple discussion from the point of view of the Earth and muon rest frames, see

http://th.physik.uni-frankfurt.de/~hees/art-ws16/lsg01.pdf

 

[Boing3000]

One observer sees time dilation, the other sees length contraction, but neither sees both

The last part of this sentence is very unfortunate, not so much because it is wrong, but because I think the author just meant to underline that there is two different explanation for the same fact, but one different in each frame.

But both frames will totally see both effect. For the muon, the Earth not only looks nearly flatten (along the motion axis) it is also look frozen, that is: the atmosphere atoms (also flatten) will move very slowly between them.
And the Earth would also see the muon flattened, if that means anything, its horizontal cross section is much smaller.

I made up those two other symmetrical observation, but I think they would both account for the same experimental result ... a muon is less scattered horizontally that is should have been. If I am wrong, the physicists here will bash my example to death... and rightly so

 

[Herman Trivilino]

And the Earth would also see the muon flattened, if that means anything, its horizontal cross section is much smaller.

Right. But the length contraction of the muon is not relevant, only the length contraction of the height of Earth's atmosphere.

Likewise, the dilation of any elapsed proper time measurements taken with an Earth clock is not relevant, only dilation of elapsed proper time measurements taken with the muon's "clock".

Perhaps the author's statement could have been better written as "One observer uses time dilation, the other uses length contraction, but neither uses both (to calculate the relative velocity of the other)."

 

[IvicaPhysics]

I found a solution by looking at the formula for time dialation. And the I realized that they are experiencing less time tham me, not more. That pretty much solves the problem

 

[Bartolomeo]

John, Tony, Jack and Bob are allocated at certain distance from each other. Let's say 1000 miles. Each of them possesses a light clock. They synchronize clocks by Einstein technique. Now these clocks oscillate synchronously in John – Tony - Jack – Bob reference frame.

Herb possesses a light clock too. He moves in John – Tony - Jack – Bob reference frame. His clock oscillates slower in the John – Tony - Jack – Bob reference frame because "in his clock light moves by hypotenuse". Everything is all right.

Herb passes by John, then Tony, then Jack and finally Bob. He compares clock readings i.e. how many oscillations his own clock has already done and any clock in reference frame John – Tony - Jack – Bob.

Look at this diagram. https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_dilation02.gif

How many oscillations Herbs clock did during time of travel? How many any synchronized one?

Does Herb see dilation or acceleration of time in the reference frame John – Tony - Jack – Bob?

What he has to do, so as to see that any clock (for example John's) dilates? What clock rate will measure John in Herb's reference frame then?

 

[stevendaryl]

John, Tony, Jack and Bob are allocated at certain distance from each other. Let's say 1000 miles. Each of them possesses a light clock. They synchronize clocks by Einstein technique. Now these clocks oscillate synchronously in John – Tony - Jack – Bob reference frame.

Herb possesses a light clock too. He moves in John – Tony - Jack – Bob reference frame. His clock oscillates slower in the John – Tony - Jack – Bob reference frame because "in his clock light moves by hypotenuse". Everything is all right.

Herb passes by John, then Tony, then Jack and finally Bob. He compares clock readings i.e. how many oscillations his own clock has already done and any clock in reference frame John – Tony - Jack – Bob.

Look at this diagram. https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_dilation02.gif

How many oscillations Herbs clock did during time of travel? How many any synchronized one?

Does Herb see dilation or acceleration of time in the reference frame John – Tony - Jack – Bob?

I can't tell from your post whether you are asking questions because you want to know the answers, or whether you are using the "Socratic method" of teaching others by asking them questions. That's why I never really liked Socrates. He was always asking questions that he perfectly well knew the answers to, like a manipulative lawyer.

In this scenario, there are certain things that everyone agrees on, in both reference frames. Let t_{1,john} be the time on John's clock when Herb passes him. Let t_{1, herb} be the time on Herb's clock when he passes John. Let t_{2,bob} be the time on Bob's clock when Herb passes him. Let t_{2, herb} be the time on Herb's clock when he passes Bob. Then both frames agree on the following values:

1. The elapsed proper time, \delta \tau_{herb} = t_{2,herb} - t_{1,herb} on Herb's clock between the time he passes John's clock and the time he passes Bob's clock.
2. The elapsed coordinate time, \delta t = t_{2,bob} - t_{1,john} in the John, Tony, Jack, Bob coordinate system.

They all agree that \delta \tau_{herb} < \delta t. But the two frames differ in how they explain this discrepancy:

• In Bob's frame, it is explained by the fact that Bob's clock is not synchronized with John's clock; it's ahead by a certain amount. So the time difference \delta t is comparing apples to oranges: times on two different unsynchronized clocks.
• In the John/Tony/Jack/Bob frame, it is explained by the fact that Bob's clock is running slower.

 

[Bartolomeo]

While moving clock makes 3 oscillations, any stationary makes 7. When Herb compares his own readings with John's readings, their clocks show 12 hours. When Herb compares his clock with Bob's, Herb's clock show 3 PM and Bob's 7 PM. And Bob's clock is perfectly synchronized with any other clock. What will be Herb's conclusion about stationary clocks? Simple comparison of clock readings show, that time in reference frame runs faster since his in motion in the reference frame John – Tony - Jack – Bob runs faster at gamma. So as to measure, that any single clock dilates, Herb has to change state of proper motion into proper rest (to change reference frame). Herb will introduce a new reference system then, in which he is at rest. Then every single clock, which moves in his reference frame will dilate.

 

[Bartolomeo]

We can think this way. When Herb passed by John, their clocks showed the same readings. Then, if Herb has another synchronized clock (of his reference frame) adjacent to Bob's clock, that Herb's - 2 clock shows different time than Bob's. But this method is equivalent to introducing Herb's rest frame. Then Bob approaches Herb from distant location. When they meet, Herb will make a conclusion, that Bob's clock dilated. However, If Bob will compare his time with time in Herb's reference frame, he will make a conclusion that time in Herb's reference frame accelerates.
The thing is that we change reference frames. John introduces his frame first and Herb moves in it. Herb dilates. Then Herb introduces his rest frame. John dilates. Bob is at rest and John is at rest either. If we will stay in one chosen frame, observations will not be reciprocal.
Observer "at proper rest" measures dilation. Observer "in proper motion" measures acceleration, since his own clock dilates in certain reference frame. Depending on "proper state" observations will be different.
If you are "at rest", you introduce your own frame. If you are "in motion", you don't introduce your own frame but use that you are "in motion" in.

 

[stevendaryl]

While moving clock makes 3 oscillations, any stationary makes 7. When Herb compares his own readings with John's readings, their clocks show 12 hours. When Herb compares his clock with Bob's, Herb's clock show 3 PM and Bob's 7 PM. And Bob's clock is perfectly synchronized with any other clock. What will be Herb's conclusion about stationary clocks? Simple comparison of clock readings show, that time in reference frame runs faster since his in motion in the reference frame John – Tony - Jack – Bob runs faster at gamma. So as to measure, that any single clock dilates, Herb has to change state of proper motion into proper rest (to change reference frame). Herb will introduce a new reference system then, in which he is at rest. Then every single clock, which moves in his reference frame will dilate.

I think you are missing the fact that in Herb's frame, Bob's clock is NOT synchronized with John's clock.

Here's a demo that I created that explains the time dilation from two different frames.

http://dee-mccullough.com/relativity/

It was created without reference to your exact problem, but you can make the connection by assuming that

• Herb's light clock corresponds to the right red clock.
• John's light clock corresponds to the left green clock.
• Bob's light clock corresponds to the right green clock.
• Tony and Jack were cut out of the demo, for budget reasons.

Click "Start" to see time dilation from the point of view of the John/Bob frame. In this frame:

• John's clock and Bob's clock are synchronized.
• John's clock and Herb's clock show the same time, initially: 12:00
• Herb's clock is advancing at half the rate of John's clock or Bob's clock.
• When Herb gets to Bob's clock, his clock only shows 12:30, while Bob's clock shows 1:00.

Click "Reset" and then "Red ship's frame" and then "Start" to see what things look like in Herb's frame. It's still true that

• Initially, John's clock and Herb's clock show the same time, 12:00.
• At the end, Bob's clock shows 1:00 while Herb's clock shows 12:30

But in Herb's frame:

• Bob's clock is ahead of John's by 45 minutes. It starts off showing time 12:45
  
• Bob's clock advances at half the rate of Herb's.
• So when Bob reaches Herb, Bob's clock has advanced only 15 minutes, to 1:00, while Herb's clock has advanced 30 minutes, to 12:30.

 

[Bartolomeo]

I think you are missing the fact that in Herb's frame, Bob's clock is NOT synchronized with John's clocks

I understand that very well. But when you say "in Herbs frame" that means that Herb (in Herb's mind) changes state of proper motion into proper rest. Since he introduces his own rest frame and places Einstein - synchronized clocks in different spatial positions. In this case he compares readings of a SINGLE moving clock, which moves in his reference frame.

 

[stevendaryl]

I understand that very well. But when you say "in Herbs frame" that means that Herb (in Herb's mind) changes state of proper motion into proper rest. Since he introduces his own rest frame.

Well, as far as the demo is concerned, Herb is always at rest in his frame, and John/Bob are always at rest in their frame. What's changing is our choice of whether to look at things from Herb's point of view or the John/Bob point of view. Herb isn't changing. (Well, I guess he could transform to John's frame as well as we can, but for the sake of the demo, assume that he always uses his own rest frame.)

 

[Bartolomeo]

Herb and John move relatively to each other at velocity v=0.9 c. There is a reference frame, in which Herb is at rest, and John moves with velocity 0.9 c. In this frame Herb sees dilation and John acceleration. There is a frame, in which John is at rest, and Herb is in motion at 0.9 c. In this frame John sees dilation and Herb acceleration. There is a frame, in which John and Herb move at velocity 0.45 and 0.45 respectively. They see the same clock rate. Is there a frame, in which John and Herb move with velocities 0 and 0 respectively?
Amount of time dilation depends on relative speed. But relative contributions of time dilation are frame dependant.

 

[Bartolomeo]

Whether an observer will see dilation or acceleration or the same clock rate purely depends on arbitrary choice of the reference frame.

 

[stevendaryl]

Herb and John move relatively to each other at velocity v=0.9 c. There is a reference frame, in which Herb is at rest, and John moves with velocity 0.9 c. In this frame Herb sees dilation and John acceleration.

Why do you bring up acceleration? Can't we, for the sake of simplicity, just assume that Herb and John have always been at rest in their respective rest frames?

There is a frame, in which John is at rest, and Herb is in motion at 0.9 c. In this frame John sees dilation and Herb acceleration.

I don't know what you mean. Whether Herb or John accelerates is something that Herb and John can determine on their own. It's not frame-dependent (well, the magnitude might be, but the fact that the acceleration is nonzero is frame-independent).

There is a frame, in which John and Herb move at velocity 0.45 and 0.45 respectively.

No, there isn't. Using the velocity addition formula, if John is moving at speed 0.45 in one direction, and Herb is moving at speed 0.45 in the other direction, then the speed of John relative to Herb is:

v_{rel} = \frac{v_{john} + v_{herb}}{1 + \frac{v_{john} v_{herb}}{c^2}} = \frac{.9 c}{1.2025} = 0.75 c

They see the same clock rate. Is there a frame, in which John and Herb move with velocities 0 and 0 respectively?
Amount of time dilation depends on relative speed. But relative contributions of time dilation are frame dependant.[/QUOTE]

To have a relative speed of 0.9 c, then in a frame where they are moving at the same speed, that speed would have to be around 0.63 c.

Is there a frame, in which John and Herb move with velocities 0 and 0 respectively?

No, that would mean that they would both be at rest in that frame, which would mean that they aren't moving, relative to one another.

 

[stevendaryl]

Whether an observer will see dilation or acceleration or the same clock rate purely depends on arbitrary choice of the reference frame.

I think you're using the word "acceleration" incorrectly here. But you're right--whether a clock is dilated depends on the choice of reference frame. On the other hand, the fact that (as shown in my demo http://dee-mccullough.com/relativity/) Herb's clock (the right red clock) is 30 minutes behind Bob's clock (the right green clock) is frame-independent.

 

[Ibix]

I think Bartolomeo is using "acceleration" to mean the opposite of time dilation. So he's either using frames in which the speed of light is not isotropic (so Herb's frame's synchronisation convention is equivalent to Einstein synchronisation performed by John) or a truly bizarre definition of "see". Or both.

 

[Bartolomeo]

I don't know what you mean. Whether Herb or John accelerates is something that Herb and John can determine on their own. It's not frame-dependent (well, the magnitude might be, but the fact that the acceleration is nonzero is frame-independent).

I mean acceleration of time in the reference frames he moves in. If all processes in Herb's body slow down, he will see that all processes around him run very fast.

To have a relative speed of 0.9 c, then in a frame where they are moving at the same speed, that speed would have to be around 0.63 c.

I simplify. Sure, there is relativistic velocities addition. That doesn't matter. There is a frame, in which they move with equal velocities. Herb dilates at gamma, and John dilates at gamma.

 

[Bartolomeo]

I think Bartolomeo is using "acceleration" to mean the opposite of time dilation. So he's either using frames in which the speed of light is not isotropic (so Herb's frame's synchronisation convention is equivalent to Einstein synchronisation performed by John) or a truly bizarre definition of "see". Or both.

I didn't ever mention another way of synchronization for moving observer. You have come to this conclusion yourself taking into account all the evidence. I think that you understand everything very well. Yes, if he will synchronize clocks in his frame not by Einstein, but will take into account his own velocity and will conduct measurements, all pieces of puzzle will take proper places. Everything leads straight to that, including transverse doppler effect.
Then even observations (measurements) by means of synchronized clocks would lead straight to the same outcome, that transverse Doppler effect and rate of time "from the point of view" of a single moving clock,

 

[stevendaryl]

I mean acceleration of time in the reference frames he moves in. If all processes in Herb's body slow down, he will see that all processes around him run very fast.

No, he won't. Herb sees nothing at all changed by his state of motion. Everything around him works as normal. What his motion does is:

• It makes it seem as if John's and Bob's clocks are running slow (in my demo, they advance at half the rate of Herb's clock)
  
• It makes it seem as if John's and Bob's clocks are out of synch (in my demo, Bob's clock is 45 minutes ahead of John's clock)
• It makes it seem as if the distance between John and Bob has shrunk (in my demo, it's half what it is in the John/Bob frame)

At no point does Herb see anything accelerated.

 

[Bartolomeo]

At no point does Herb see anything accelerated.

Sure as soon as he makes measurements by means of Einstein synchronized clocks. Because he arbitrarily synchronizes clock by Einstein. And everyone does the same.

 

[Bartolomeo]

I think Bartolomeo is using "acceleration" to mean the opposite of time dilation. So he's either using frames in which the speed of light is not isotropic (so Herb's frame's synchronisation convention is equivalent to Einstein synchronisation performed by John) or a truly bizarre definition of "see". Or both.

By the way. Imagine that red moving clock is the Aliens. Row of green synchronized clock are identical brothers Joes. Clockfaces of green clocks are higlighted in green monochromatic light. Aliens compare their own single clock rate with the time in the „Joes green“ reference frame. They see that the set of clocks runs faster at gamma. What color of clockfaces they will see? If they look straight down, it will be red. But since they move, they have to look into front. Due to aberration clockfaces will be blue. Frequency increases at gamma too. Now set of clock (time in reference frame) runs faster and every single clock too.

Please note that Einstein measuring technique is based on ASSUMPTION. It is not fact. You can make your own assumptions if you wish.

 

[Bartolomeo]

@Ibix, think about photocamera. Contracted or stretched? Hint: does film in moving photocamera Lorentz - contracts? Does it contracts, if the camera is "at rest"? If the film Lorentz contracts, will the square appear contracted or stretched on the photo?
How to put it into accordance with clock synchronization in different frames?