How does time dilation work in relativity?

[mlazos]

1.
I give the problem and the solution. Can someone tell me if and where I am wrong? Thanks
Homework Statement

We have an observer $$O'$$, who travels with constant speed $$U=0.8c$$, and another observer $$O$$ with speed $$0$$.
The Observer $$O'$$ is traveling from point $$A$$ to point $$B$$ $$(1.2 \cdot 10^8m)$$.

Since we know the speed of $$O'$$, and the distance from $$A \rightarrow B$$, we can calculate the needed time $$(t')$$ for the $$O'$$ to arrive to the point $$B$$.The observer $$O$$ counts the elapsed time, $$t_1$$ so the $$O'$$ to arrive to the point B.
$$U=\frac{d}{t'} \Rightarrow t'= \frac{d}{U} \Rightarrow t' = \frac{1.2 \cdot 10^8}{0.8c}$$
$$t'=\frac{1.2 \cdot 10^8}{0.8 \cdot 3 \cdot 10^8} \Rightarrow t'=0.5s$$
The time $$t'$$ is the dilated time for the Observer $$O'$$.Therefore, we can calculate the elapsed proper time $$t_0$$ for the Observer $$O'$$
$$t'= \frac{t_0}{\sqrt{1-(\frac{u}{c})^2}} \Rightarrow t_0 =t\sqrt{1-(\frac{u}{c})^2}$$
$$t_0=0.5 \sqrt{1-0.8^2} \Rightarrow t_0 = 0.3s$$

So when the Observer $$O'$$ is at point B reads $$t_0=0.3s$$.
The first Observer who has zero speed is also counting the time needed for the Observer $$O'$$ to be to the point B. This means two things

1. When $$O'$$ is to the point B the proper time for $$O'$$ is $$0.3s$$
2. The time needed to go from point $$A$$ to point $$B$$ is $$0.5s$$. The Observer $$O$$, with a pocket calculator, calculated the time (later) because his reading was strange.

So the observer $$O$$ is reading $$0.5s$$ plus the time needed for the light (information) to get from point $$B$$ to point $$A$$! This is very logical since we observe by using electromagnetic waves which they have speed $$c$$. So in order the information to get from point $$B$$ to point $$A$$ needs time $$T=\frac{d}{c}=0.25s$$. Therefore the total time for the Observer $$O$$ is $$t_1= t' +T= 0.5+0.25 \Rightarrow t_1=0.75s$$.
So when the Observer $$O'$$ reads $$0.3s$$ the Observer $$O$$ reads $$0.75s$$, which is the time needed for the $$O'$$ to get from point $$A$$ to point $$B$$, plus the time needed for the information (light) to arrive to us.

 

[neutrino]

1. When $$O'$$ is to the point B the proper time for $$O'$$ is $$0.3s$$
2. The time needed to go from point $$A$$ to point $$B$$ is $$0.5s$$. The Observer $$O$$, with a pocket calculator, calculated the time (later) because his reading was strange.

It may be strange, but it is not incorrect.

So the observer $$O$$ is reading $$0.5s$$ plus the time needed for the light (information) to get from point $$B$$ to point $$A$$! This is very logical since we observe by using electromagnetic waves which they have speed $$c$$. So in order the information to get from point $$B$$ to point $$A$$ needs time $$T=\frac{d}{c}=0.25s$$. Therefore the total time for the Observer $$O$$ is $$t_1= t' +T= 0.5+0.25 \Rightarrow t_1=0.75s$$.
So when the Observer $$O'$$ reads $$0.3s$$ the Observer $$O$$ reads $$0.75s$$, which is the time needed for the $$O'$$ to get from point $$A$$ to point $$B$$, plus the time needed for the information (light) to arrive to us.

When one usually makes calculations in relativity, the light-travel-time from event to observer is taken into account. The results you have obtained (O' 0.3s and O 0.5s) are correct. To confirm your answer, think of A and B being marked by two asteroids, say, and answer the following questions:
With respect to O'...
What is the velocity with which the asteroids travel?
What is the distance between the asteroids?
How long does it take for B to reach the origin of O' since A passed the origin?

 

[mlazos]

Yes but in order to know that the asteroid is at point B we have somehow to observe him. Therefore we count also the delay of light since the speed is finite.Thats why when the observer is at the point B we count the time it took to get there plus the delay of light from B to our instruments. Is this not logical?

 

[mlazos]

Now the tricky question is:
When the Observer O' reads 0.3 sec what do we read to our clock? This is the answer I need actually. I calculated that to be 0.75s and I want to verify the answer. Thank you for reading me.

 

[nrqed]

1.
I give the problem and the solution. Can someone tell me if and where I am wrong? Thanks
Homework Statement

We have an observer $$O'$$, who travels with constant speed $$U=0.8c$$, and another observer $$O$$ with speed $$0$$.
The Observer $$O'$$ is traveling from point $$A$$ to point $$B$$ $$(1.2 \cdot 10^8m)$$.

Since we know the speed of $$O'$$, and the distance from $$A \rightarrow B$$, we can calculate the needed time $$(t')$$ for the $$O'$$ to arrive to the point $$B$$.The observer $$O$$ counts the elapsed time, $$t_1$$ so the $$O'$$ to arrive to the point B.
$$U=\frac{d}{t'} \Rightarrow t'= \frac{d}{U} \Rightarrow t' = \frac{1.2 \cdot 10^8}{0.8c}$$
$$t'=\frac{1.2 \cdot 10^8}{0.8 \cdot 3 \cdot 10^8} \Rightarrow t'=0.5s$$
The time $$t'$$ is the dilated time for the Observer $$O'$$.Therefore, we can calculate the elapsed proper time $$t_0$$ for the Observer $$O'$$
$$t'= \frac{t_0}{\sqrt{1-(\frac{u}{c})^2}} \Rightarrow t_0 =t\sqrt{1-(\frac{u}{c})^2}$$
$$t_0=0.5 \sqrt{1-0.8^2} \Rightarrow t_0 = 0.3s$$

So when the Observer $$O'$$ is at point B reads $$t_0=0.3s$$.
The first Observer who has zero speed is also counting the time needed for the Observer $$O'$$ to be to the point B. This means two things

1. When $$O'$$ is to the point B the proper time for $$O'$$ is $$0.3s$$
2. The time needed to go from point $$A$$ to point $$B$$ is $$0.5s$$. The Observer $$O$$, with a pocket calculator, calculated the time (later) because his reading was strange.

So the observer $$O$$ is reading $$0.5s$$ plus the time needed for the light (information) to get from point $$B$$ to point $$A$$! This is very logical since we observe by using electromagnetic waves which they have speed $$c$$. So in order the information to get from point $$B$$ to point $$A$$ needs time $$T=\frac{d}{c}=0.25s$$. Therefore the total time for the Observer $$O$$ is $$t_1= t' +T= 0.5+0.25 \Rightarrow t_1=0.75s$$.
So when the Observer $$O'$$ reads $$0.3s$$ the Observer $$O$$ reads $$0.75s$$, which is the time needed for the $$O'$$ to get from point $$A$$ to point $$B$$, plus the time needed for the information (light) to arrive to us.

One comment: it's much better to use a prime on t to indicate values measured in O prime and t for times measured in O. Your notation is a bit confusing here.

Your final result is not correct. 0.3 second is the time in O' when O' reaches point B. But it takes some time before the signal reaches O so O's will say that the signal reaches A a bit after 0.3 second!

 

[Doc Al]

Now the tricky question is:
When the Observer O' reads 0.3 sec what do we read to our clock? This is the answer I need actually. I calculated that to be 0.75s and I want to verify the answer. Thank you for reading me.

No, that answer is incorrect. 0.75 s is the answer to a different question, which is equivalent to "If O' turned on a light when he reached point B, at what time--according to O observers--would that light be seen at point A?"

But that's not the time (according to observer O) when observer O' reads 0.3 sec (that time you already calculated to be 0.5 s).

 

[mlazos]

0.3 seconds is the clock inside the object which moves. According to the stationary observer, the moving object needs at least 0.5 seconds from A to be plus the time the signal needs to get from B to A which is 0.25s.Thats why i said 0.75s. I have a drawing so you could check there too. In the book of young and freedman the solution is...strange since the time for the stationary observer is 0.12s.They set 0.3 as the dilated time so the proper time is even smaller and makes no sense at all.

 

[Doc Al]

0.3 seconds is the clock inside the object which moves. According to the stationary observer, the moving object needs at least 0.5 seconds from A to be plus the time the signal needs to get from B to A which is 0.25s.Thats why i said 0.75s. I have a drawing so you could check there too. In the book of young and freedman the solution is...strange since the time for the stationary observer is 0.12s.They set 0.3 as the dilated time so the proper time is even smaller and makes no sense at all.

Can you post (or scan in) the original problem exactly as it's given?

 

[nrqed]

Now the tricky question is:
When the Observer O' reads 0.3 sec what do we read to our clock? This is the answer I need actually. I calculated that to be 0.75s and I want to verify the answer. Thank you for reading me.

As Doc al said, this is quite a different question from what you discuss in your initial post!

Could you write down the entire question so that we know what we are discussing?

 

[mlazos]

The exact problem says
As you pilot your space utility vehicle at a constant speed towards the moon, a race pilot flies past you in her spaceracer with a constant speed of 0.8c relative to you.At the instant the spaceracer passes you,both of you start timers at 0. a) At the instant when you measure that the spaceracer has traveled $$1.2 \cdot 10^8m$$ past you, what does the race pilot reads on her timer? b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be her distance from you? c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

 

[Doc Al]

The exact problem says
As you pilot your space utility vehicle at a constant speed towards the moon, a race pilot flies past you in her spaceracer with a constant speed of 0.8c relative to you.At the instant the spaceracer passes you,both of you start timers at 0. a) At the instant when you measure that the spaceracer has traveled $$1.2 \cdot 10^8m$$ past you, what does the race pilot reads on her timer? b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be her distance from you? c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Good. That's perfectly clear. As we've stated already, the answer to (a) is not 0.75s, but what you calculated earlier to be 0.3 s. (There is no mention of a signal traveling back to you.)

 

[nrqed]

The exact problem says
As you pilot your space utility vehicle at a constant speed towards the moon, a race pilot flies past you in her spaceracer with a constant speed of 0.8c relative to you.At the instant the spaceracer passes you,both of you start timers at 0. a) At the instant when you measure that the spaceracer has traveled $$1.2 \cdot 10^8m$$ past you, what does the race pilot reads on her timer? b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be her distance from you? c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Ok, so there is nothing about a signal being sent here.

And I missed this in your initial post, but you got the wrong time in O's (I missed that because you switched the prime and unprime in your time). The time in O's should be larger than 0.5 second, so you must multiply 0.5 second by gamma. (You had written the correct equation but you got the wrong result because you changed the meaning of t' and t unprime)

 

[mlazos]

Yes but how do we know someone arrived to the destination?We have to observe it somehow. So when we see someone is at the point B then we count the time to go there plus the light to get to our eyes. Since the distance is big we have to count also the time that light traveled from B to A. right?

 

[nrqed]

Yes but how do we know someone arrived to the destination?We have to observe it somehow. So when we see someone is at the point B then we count the time to go there plus the light to get to our eyes. Since the distance is big we have to count also the time that light traveled from B to A. right?

This is a very good question!

One thing that your prof should have made clear is the following (most profs do not explain this so it's understandable if he/she didn't): in those problems, you have to imagine that there are an infinite number of observers at all possible points in the frame O. Now assume that all measurements are made by the observers who happen to be exactly at the location of the event (thsi is why they are called "local observers"). What they are asking is the time recorded by the local observer in O who happens to be right next to the spaceship O' when that spaceship arrives at destination. so that way no signal is needed, so no time delay for observation.
(That local observer records the time on his watch and then later calls you at the origin of O an dtells you what the result was)

 

[Doc Al]

Yes but how do we know someone arrived to the destination?We have to observe it somehow. So when we see someone is at the point B then we count the time to go there plus the light to get to our eyes. Since the distance is big we have to count also the time that light traveled from B to A. right?

If you assume that the information given in the problem is correct, that the speed racer really is going at that speed and really does travel the stated distance, then you can certainly calculate all the needed answers as to clock times.

But even if you insist on observing O' looking at his watch as it reads 0.3 seconds, you would have to subtract the signal travel time of the light signal from your observed time to get the needed answer. So the process is a bit circular. You "know" that when O' reaches point B that his clock reads 0.3s. The signal takes 0.4 seconds to get back to observer O, so O "sees" O' reading his clock at a time of 0.5 + 0.4 = 0.9 seconds. So at a time of 0.9 s on the O clock, O "sees" the O' clock read 0.3 seconds. Of course, he knows that it took 0.4 s for the light to reach him, so he knows that the O' clock was actually reading 0.3 s when his cock read 0.9s - 0.4s = 0.5s.

You're making the problem more complicated than necessary.

 

[mlazos]

Yes but in order all those observers to synchronize their clocks still use information and this takes time.So there is always a delay this way. Why in the book of Young and Freedman the time measured by the stationary observer is 0.3/gamma ?They substitute the 0.3 in the space vehicle as the proper time, so the stationary observer reads less than the moving one. Makes sense?

 

[jambaugh]

A side comment...

Now the tricky question is:
When the Observer O' reads 0.3 sec what do we read to our clock? This is the answer I need actually. I calculated that to be 0.75s and I want to verify the answer. Thank you for reading me.

In a way this question is ill posed as stated. Remember simultaneity is relative so the "When" of "When the observer O' reads..." is not well defined except in the special case where both observers are at the same spatial position. This is why the twin "paradox" isn't really one. The question should be one of either:

"When observer O' reads 0.3 sec, what does he see our clock reading? (at the "same instant" as he defines simultaneous events)"

or distinctly:

"When we see observer (O')'s clock read 0.3 sec (as we see simultaneous events) what is our clock reading?"

The two questions will have very different answers. It is the difference between two events lying on the t=const. hyperplane vs lying on the t'=const. hyperplane.

 

[mlazos]

I see, makes sense. Though for the last question, the part c) is the answer 0.5 seconds? Again i say that in the solutions they say "When the racer reads 0.3 the stationary observer reads 0.3/gamma".Is there any logical explanation for this or its just a mistake?

 

[nrqed]

Yes but in order all those observers to synchronize their clocks still use information and this takes time.So there is always a delay this way. Why in the book of Young and Freedman the time measured by the stationary observer is 0.3/gamma ?They substitute the 0.3 in the space vehicle as the proper time, so the stationary observer reads less than the moving one. Makes sense?

This is the whole question of synchronization of clocks. It is assumed in those problems that the clocks have been synchronized. This can be done. If you are in my frame and you are 3 x 10^8 m away from me, I will send you a signal when my clock reads noon. when you receive my signal, you adjust your clock to noon plus one second. The guy who is 6 x 10^8 meters away will set his clock at noon plus 2 seconds and so on. That way, we all have our clocks synchronized.

About 0.3 / gamma, i sthat their answer for part a or for part c? I lose tracj of which question you are discussing.

 

[mlazos]

is the thread number 10 of this topic and question c. This is really strange. The problem itself is not difficult its just the solution strange for me. I will be back in 2:30h. thank you for your time.

 

[Doc Al]

I see, makes sense. Though for the last question, the part c) is the answer 0.5 seconds? Again i say that in the solutions they say "When the racer reads 0.3 the stationary observer reads 0.3/gamma".Is there any logical explanation for this or its just a mistake?

The word "when" is ambiguous unless you specify according to whom. They mean: When the racer reads 0.3s on her clock, what time does she say that the O clock reads. Of course, you will disagree because simultaneity is frame-dependent.

 

[nrqed]

The word "when" is ambiguous unless you specify according to whom. They mean: When the racer reads 0.3s on her clock, what time does she say that the O clock reads. Of course, you will disagree because simultaneity is frame-dependent.

If we mean what is the time on a local observer in O, here there is no ambiguity. What is the time read by a local observer in O (who is therefore aligned with the clock in 0' when the clock in O' reads 0.3 second) is non-ambiguous. Both her and the local observer in O will see at that instant the clock in O reading the same value. There is no ambiguity here because we are not talking about time interval between two events but we are talking about the time indicated in O corresponding to a single event.

 

[mlazos]

So what is this value? how much? Is it 0.5sfor the O while is 0.3 for the O'? Or its 0.3/gamma for the O while its 0.3 for the O'?

 

[Doc Al]

If we mean what is the time on a local observer in O, here there is no ambiguity. What is the time read by a local observer in O (who is therefore aligned with the clock in 0' when the clock in O' reads 0.3 second) is non-ambiguous. Both her and the local observer in O will see at that instant the clock in O reading the same value. There is no ambiguity here because we are not talking about time interval between two events but we are talking about the time indicated in O corresponding to a single event.

But in this problem there is no local observer in O aligned with the racer when her clock reads 0.3 seconds (except in our imagination). Only two clocks are mentioned: hers (race pilot) and ours (utility vehicle). Part c of the problem asked:

c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?​

This is describing two spatially separated events: (1) me in the utility vehicle looking at my clock and (2) she in her racer looking at her clock when it reads 0.3 seconds. Unless I know who's "instant" we're talking about, it seems ambiguous to me.

But you are of course correct: If there was a local observer in O aligned with the racer at the moment in question, his clock would read 0.5 seconds. Everyone would agree with that. And the O observer in the utility vehicle would agree that "at that instant" his clock also reads 0.5 seconds. But the race driver would disagree: Since only 0.3 seconds elapsed during the trip--according to her--she'd say that the O clock on the utility vehicle would only have advanced by $0.3/\gamma = 0.18$ seconds. So, according to her, at the instant her clock reads 0.3 s, my clock reads 0.18 s.

 

[nrqed]

But in this problem there is no local observer in O aligned with the racer when her clock reads 0.3 seconds (except in our imagination). Only two clocks are mentioned: hers (race pilot) and ours (utility vehicle). Part c of the problem asked:

c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?​

This is describing two spatially separated events: (1) me in the utility vehicle looking at my clock and (2) she in her racer looking at her clock when it reads 0.3 seconds. Unless I know who's "instant" we're talking about, it seems ambiguous to me.

But you are of course correct: If there was a local observer in O aligned with the racer at the moment in question, his clock would read 0.5 seconds. Everyone would agree with that. And the O observer in the utility vehicle would agree that "at that instant" his clock also reads 0.5 seconds. But the race driver would disagree: Since only 0.3 seconds elapsed during the trip--according to her--she'd say that the O clock on the utility vehicle would only have advanced by $0.3/\gamma = 0.18$ seconds. So, according to her, at the instant her clock reads 0.3 s, my clock reads 0.18 s.

Yes, you are completely right (I had not paid enough attention to the phrasing of the question).

But the notion of local observesr is still key. Now, the question becomes: at the instant she has reached her destination (at that instant, in her frame), what does a local observer in her frame will read on your clock (that local observer is in O' but happens to be aligned with you at the same instant she has reached the destination, as measured in O').

Then teh answer is clearly given by a Lorentz transformation.

 

[mlazos]

So how is possible the moving object to have a higher reading that the stationary. Since time slows down then we should have a higher reading that her. Right?

 

[jambaugh]

So how is possible the moving object to have a higher reading that the stationary. Since time slows down then we should have a higher reading that her. Right?

It is not good conceptually to say "time slows down". Time is how far we move through space-time. We each see a time axis in the direction we're moving (we each see ourselves as stationary through space and "moving" through our time at a rate of 1 second per second.). If you and I are not moving in parallel (i.e. we're not spatially stationary w.r.t. each other) then my time direction and yours don't agree. When you compare what my watch reads to what yours does at a given event we see distinct numbers because we are measuring temporal distances along distinct paths.

It is not different in principle than the fact that you walking North will see me, walking North-East, take more steps to cross a given latitude line. (Just looking at footprints and do not invoking time out of context.) Likewise I defining N' latitude' lines so I'm moving North' and you're moving North'-West' will see you take more footsteps than I to cross my latitude' lines.

The distinction, obvious here, is the analogous principle to time dilation and relativity of simultaneity but the actual effect is reversed with time due to the peculiar hyperbolic geometry of space-time. Relatively moving observers will each see the other taking less clock-steps to pass our own temporal "latitude" lines.

If you understand this then you can correctly parse the problems you are given and read in them the implicit assumptions (e.g. one observer O is said to be stationary and so it is relative to his frame that statements of simultaneity are intended even though they are stated in absolute terms.)

BTW I find it much clearer to express Lorentz transformations in terms of hyperbolic trig:
For some boost parameter b we have:

$$\beta = \tanh(b),\quad \gamma = \cosh(b),\quad \beta\cdot\gamma = \sinh(b)$$
thence
$$\Delta x' = \cosh(b) \Delta x + \sinh(b) c\Delta t$$
$$c\Delta t' = \sinh(b)\Delta x + \cosh(b) c\Delta t$$
where you solve for the boost parameter b using known quantities. Note here the sign of beta and b matter. Be sure beta refers to the velocity of the unprimed (pre-transform) observer relative to the primed (post-transform) observer.

Note also the problem you were given makes use of the Pythagorean triple:
$$3^2 + 4^2 = 5^2$$
but in hyperbolic form:
$$5^2 - 4^2 = 3^2$$
Thence $$\cosh(b) = 5/3,\quad \sinh(b) = 4/3,\quad \tanh(b) = 4/5 = 0.8$$