# Time dilation problem question

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

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

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.

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.

Stephanus
Maxila said:
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.

Stephanus
Maxila said:
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.
PeroK said:
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.

Stephanus
PeroK said:
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.)

Maxila said:
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.

PeroK said:
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.

Maxila said:
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.

IvicaPhysics said:
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!)

Maxila said:
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 said:

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

PeroK
Maxila said:
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.

stevendaryl said:
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.

Maxila said:
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).

Maxila said:
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.

stevendaryl said:
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").

stevendaryl said:
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

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.

Maxila said:
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.

stevendaryl said:
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).

Maxila said:
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.
Maxila said:
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.

Maxila said:
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.

Maxila said:
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.

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.

Maxila said:
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.

Maxila said:
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.

Maxila said:
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.

Maxila said:
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}##.

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

Maxila said:
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

Boing3000 said:
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)."

PeroK
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

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?

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Bartolomeo said:
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.

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.

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