# Homework Help: Tricky relativity problem

1. Nov 29, 2007

### mlazos

1.
I give the problem and the solution. Can someone tell me if and where im wrong? Thanks
The problem statement, all variables and given/known data

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.

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2. Nov 30, 2007

### neutrino

It may be strange, but it is not incorrect.

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?

3. Nov 30, 2007

### 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?

4. Nov 30, 2007

### 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.

5. Nov 30, 2007

### nrqed

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!

6. Nov 30, 2007

### Staff: Mentor

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).

7. Nov 30, 2007

### 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.

8. Nov 30, 2007

### Staff: Mentor

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

9. Nov 30, 2007

### nrqed

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?

10. Nov 30, 2007

### 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?

11. Nov 30, 2007

### Staff: Mentor

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.)

12. Nov 30, 2007

### nrqed

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)

13. Nov 30, 2007

### 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?

14. Nov 30, 2007

### nrqed

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)

15. Nov 30, 2007

### Staff: Mentor

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.

16. Nov 30, 2007

### 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?

17. Nov 30, 2007

### jambaugh

A side comment...

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.

Last edited: Nov 30, 2007
18. Nov 30, 2007

### 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?

19. Nov 30, 2007

### nrqed

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.

20. Nov 30, 2007

### 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.