# Am I thinking SR correctly?

1. Sep 8, 2004

### JasonRox

This is a question in David W. Hoggs SR Book, so it is not homework.

Problem 3-4

A rocket ship passes the Earth at speed B = 0.8 (B=v/c). Observers on the ship and on Earth agree that it is noon. Answer the following questions:

a)At 12:30pm, as read by a rocket ship clock(assuming it is the same rocket ship as above), the ship passes an interplanetary navigational station that is fixed relative to the Earth and whose clocks read Earth time. What time is it at the station?

I simply just used:
$$\frac{1}{\gamma}\Delta t = t_e$$

I used this because if I take the rocket ships frame, than the Earth is moving, therefore Earth time is going slower relative to the rocket ship.

My answer is 12:18pm, this is the time at the station seen by the rocket ship. Also, if we go with the rest frame of Earth, than the station reads 12:18pm in the rocket ship, while the time on Earth/station would be 12:30pm.

If we assume that it is a different rocket ship clock, which the question does not clearly state, than time changes relative to v of the other rocket ship, which v is not given, so we can only assume v, or come up with a partial equation. I didn't bother doing this. I'm only concerned with my train of thought, and not my algebra skills.

That is only a), but I'm sure I'll get the rest if I get my thinking right.

This is what I am thinking...

If a rocket going v passes by us, we see the rocket's time as going slow. Also, for the rocket passing us, the rocket will see our time as slow.

I think that because if we take the rocket's frame, than we are moving at speed v and the rocket at 0, therefore the rocket will see our time as slow.

Last edited: Sep 8, 2004
2. Sep 8, 2004

### pervect

Staff Emeritus
I think that the Earth clock is going to read a longer time - you have it backwards.

The rocket will say that the Earth-clocks are not synchronized properly. So if you want to avoid the clock synch problem, use the Earth frame, it's easier.

So the Earth people will say that the rocket's clocks are slow, and the rocket will say that the distance is length contracted, and that the clock synchronization is off. The Earth clock will read $$\gamma$$ * 30 min.

Here's another way to confirm this. What is invariant here? The Lorentz interval between the two points in space time (rocket leaves earth) -- (rocket arrives at station) is what's invariant.

In the rocket frame, the lorentz interval is rocket_distance^2- c^2*rocket_time^2 but the rocket_distance is zero, so the answer is just c^2*rocket_time^2

In the earth frame, the lorentz interval is earth_distance^2 - c^2*earth_time^2.

So earth_distance^2 - c^2*earth_time^2 = -c^2*rocket_time^2, or
earth_time^2 = rocket_time^2 + earth_distance^2/c^2, so it's clear that earth_time > rocket_time.

3. Sep 9, 2004

### Fredrik

Staff Emeritus
I agree with pervect. The answer should be 12:50, not 12:18.

In the rocket ship's frame, the time on Earth will be 12:18, but that's the right answer to the wrong question.

These two results may seem to contradict each other, but they really don't. In the rocket ship's frame, the clock on the station and the clock on Earth will show two different times (12:50 and 12:18 respectively).

Last edited: Sep 9, 2004
4. Sep 9, 2004

### JasonRox

You didn't understand what I said.

We are assuming Earth as ZERO all the time.

If someone who is born on a spaceship, will think the rocketship ship as ZERO all the time. When that person views the Earth fly right past him at xc, so shouldn't Earth time be going slower in time relative to him because the Earth is going at xc, and not the rocket. We always assume the rocket is flying past because of life experience, but this is not the case, and the rocket is goign ZERO.

Another note, I don't know what the Lorentz interval is yet.

The time dilation equation can be deduced from the pythagorean theorem, which is where it comes from. The book used the example:

A beam of light goes up and down, to complete 1m, and we will say the time for that round trip to take 1 second for simplicities sake.

If something is moving relative to you with this beam clock, than you will see something like this:

----/\
---/--\
--/----\
-/------\
/--------\
S------->F

S=Start, F=Finish
The height is .5m.

If you are standing on the ground with your own clock, you see the clock go straight up and down. If you see a train fly past you at xc, you will see a similiar diagram as above because the train is moving. Now the beam of light, in the train, has travelled a larger distance, therefore a second takes longer to complete.

BUT...

Someone on the train will view her/his clock as going up and down in a straight fashion. We know this because if you are on a train, you can throw a ball straight up and it will come right back down with your hand WITHOUT having to move your hand because the train is "moving". I know you know this, but I need to make sure you are on the same page as me on this thought, so you know what to correct. So, when the person on the train views you past them at xc, because they are going ZERO, they will see your beam clock like the diagram I made above.

See where I'm getting at?

We see their clock as slow, but they also see our clock as slow relative to them.

5. Sep 9, 2004

### Staff: Mentor

"time dilation" formula

Right. Every observer views themselves as being at rest (speed zero) in their own frame.

It is certainly true that a moving clock will be observed to run slow. If that moving clock shows a time interval of $\Delta t'$, the time interval measured by the other frame will be $\Delta t = \gamma \Delta t'$. Of course, this works both ways. The rocket observer will see Earth-frame clocks going slow, and Earth-frame observers will see the rocket clock as going slow.

Note that this "time dilation" rule applies to measurements made by a single moving clock. In Hogg's problem, the Earth-frame observes a single moving clock (the rocket clock)... so the rule applies from the Earth frame with no ambiguity. The Earth observers see a moving rocket clock measure 30 minutes, thus they will measure this interval as taking 1.67 x 30 = 50 Earth minutes.

But from the rocket frame, things aren't so simple. There are two Earth-frame clocks involved: one on Earth, the other on the station. As pervect explained, according to the rocket frame these clocks are not synchronized. Rest assured, the rocket observer will see both the Earth clock and the station clock as running slow. But to apply the "time dilation" rule from the rocket frame (without having to worry about synchronization) you need a time interval measured by a single Earth clock.

Going back to Hogg's problem, check out part (c) which asks you to figure out when the Earth receives a signal from the rocket. Now you'll have a time interval as measured by a single Earth-frame clock (the interval from 12:00 to the time the signal was received) so you can apply the time dilation rule to find out how long that interval is measured according to the rocket observer.

6. Sep 9, 2004

### pervect

Staff Emeritus
Eeek! You should find out, right away. It's easy enough for me to tell you that the Lorentz interval is the difference of the squared of the space interval and the speed of light times the time interval, i.e. it's x^2+y^2+z^2-c^2*t^2, as measured between two events in space-time, but you'll probably need some convincing before you believe that it's constant in all inertial frames if you haven't heard of it before. It may help a little to point out that the Lorentz interval of a light beam is always zero.

Wikipedia calls this the "space-time" interval
http://en.wikipedia.org/wiki/Space-time_interval
(I just looked, they chose the opposite sign than I did).

It's about 10x as easy to get the correct results in relativity when you focus on what's constant (the Lorentz interval) rather than what varies (time and space). In my opinion, anyway.

7. Sep 9, 2004

### Staff: Mentor

Hogg's book covers the invariant interval and the Lorentz transformation in Chap 4---but this problem was given in Chap 3.

8. Sep 9, 2004

### JasonRox

I would just prefer to wait till Chapter 4, like Doc Al mentionned.

Now I'm picking up with the synchronization thing now, which will probably be covered in detail.

To make sure I'm on the right track all the time, is there a answer book for there questions. If I have a good understanding of the material and go through the questions without so much difficulty, I move on.

I'll come back with all my answers, and hopefully someone can confirm whether or not it is wrong.

Thanks for the help.

9. Sep 9, 2004

### Fredrik

Staff Emeritus
Actually, we did.

Correct. Now imagine what things will look like to the people on that ship. When their clock shows 12:50, the clock on the station (which has been slow compared to the ship's clock) will only have reached 12:30.

10. Sep 10, 2004

### JasonRox

The question said the rocket ship read 12:30pm, and now you are saying 12:50pm?

If Earth time is slower seen by the rocketship, which you showed yourself above, than 12:18pm should be correct. It might be the right answer to the wrong question, but the analogy should be right.

If you came up with an earlier time, why can't I. This exactly what I did above. I made it clear that if the rocketship read 12:30pm, what time will they read at the station? Assuming they are synchronized, I came up with 12:18pm, which is no different than what you said. Instead you started with 12:50pm, and it makes no difference because you gave an earlier time anyways, in essence, if I'm wrong, you are too.

Anyways, I didn't post the answers yet, I have been working on my homework; real homework. :)

Note: There may difficulties, but I still appreciate the help.

11. Sep 10, 2004

### pervect

Staff Emeritus
Once more into the breach.....

The earth clock at the outpost is synchronized in the earth frame. Therefore what we are measuring is how much the rocket clock slows down.

We know that the time dilation factor is
$$\gamma$$ = 1/ sqrt(1 - .8^2) = 1.6666

We know that the rocket clock is running slow, so rocket_clock / outpost_clock < 1, and outpost_clock / rocket_clock > 1.

So outpost_clock / rocket_clock = 1.666666

since the rocket clock reads 30 min, the outpost clock reads 50 minutes.

12. Sep 10, 2004

### Staff: Mentor

It's not clear what you are talking about. If you are talking about what the rocket observes/deduces, then you are mistaken. We have established that when the rocket passes the station, the rocket clock reads 12:30 (given) and the station clock reads 12:50 (deduced from time dilation). When the rocket clock reads 12:50, 20 rocket-minutes will have elapsed since passing the station, so the rocket frame will deduce that the station clock will show only 20/1.67 = 12 station-minutes to have passed: Thus according to the rocket frame, when the rocket clock reads 12:50, the station clock will read 1:02. (Of course, the station frame will disagree.)

Or did you mean something else?

For JasonRox: Here's where 12:18 comes in. According to the rocket observer, what will the Earth clock read at the moment the rocket passes the station? The rocket sees 30 minutes pass on the rocket clock. And, of course, the rocket sees the Earth clock as running slow, so he claims that only 30/1.67 = 18 Earth minutes will have elapsed on the Earth clock. So, according to the rocket observer, the Earth clock reads 12:18 at the moment he passes the station. (So the rocket observer thinks that the Earth clock reads 12:18 when the station clock reads 12:50---they are obviously not synchronized in the rocket frame.)

13. Sep 10, 2004

### pervect

Staff Emeritus
I just had a thought. I think it might be easier for JasonRox to grasp the problem if we modify it a bit. Jason, assume that the earth_outpost clock reads 50 minutes when the rocket passes by, and calculate the reading on the rocket clock.

14. Sep 10, 2004

### Fredrik

Staff Emeritus
Oops. That was a mistake! My first answer was correct. The second one (the one talking about the rocket clock showing 12:50) was just nonsense that I wrote several hours after I should have been asleep already. You should ignore it completely.

This is the correct solution:

The rocket ship passes Earth at noon. Let's say that this event has coordinates (0,0) in both frames, and that the rocket ship's coordinates are (t,x) in the rocket's frame and (t',x') in the station's frame. We are looking for the value of t' at the event specified by (t,x)=(30,0).

All observers will agree about the value of the Lorentz invariant quantity $-t^2+x^2$. Therefore we have

$$-t^2+x^2=-t'^2+x'^2$$

But x=0, so the equation simplifies to

$$-t^2=-t'^2+x'^2=-t'^2\Big(1-\frac{x'^2}{t'^2}\Big)=-t'^2(1-v'^2)=-t'^2\frac{1}{\gamma^2}$$

where v' is the speed of the rocket in the station's frame (which is of course equal to v, the speed of the station in the rocket's frame). Now drop the minus sign and take the square root of both sides.

$$t=t'\frac{1}{\gamma}$$

$$t'=\gamma t=\frac{1}{\sqrt{1-0.8^2}}\times 30=\frac{1}{0.6}\times 30=50$$

This means that 50 minutes have passed in the station's frame, so the clock on the station must be showing 12:50.

I'm using units in which c=1.

15. Sep 10, 2004

### JasonRox

Sorry frederick, but I am ignoring the Lorentz invariance for now. I will be sure to come back on this post next week, when in Chapter 4.

I think understand now.

Doc Al, you are saying that if the rocketship captain looked through his telescope at Earth, he would see 12:18pm (ignoring the time it takes light to get there) ?
What you are saying is that the rocketship captain thinks 18 minutes has passed for Earth, but reading the "synchronized" clock at the station, he realizes that 50 minutes has actually passed. Because of this, the captain should know that HE is the one going at velocity xc, and not Earth.

If a new born on the rocketship new special relativity, he would be aware that even though he views Earth people aging slowly, Earth people are actually aging quicker.

Is this right?

Last edited: Sep 10, 2004
16. Sep 11, 2004

### Chronos

Imagine this. Two video cameras, one at NASA and one on your shuttle. Both have satellite links to each other. Camera 1 at NASA is pointed at an atomic clock on the ground and a monitor image of the clock on the shuttle. Camera 2 on the shuttle is pointed at an atomic clock on the shuttle and a monitor image of the clock on the ground. Both clocks are synchronized while the shuttle is still on the launch pad and a video recording is initiated. You then launch the shuttle with a constant acceleration until it reaches a velocity of 0.5c in ten seconds, NASA time [which will be very painful]. After completing the mission, you bring your tape back and compare it to the NASA tape. Will the tapes be identical?

Last edited: Sep 11, 2004
17. Sep 11, 2004

### Fredrik

Staff Emeritus
That's what we're saying, yes.

Actually, no. In the rocket's frame, the time on Earth is 12:18 when the rocket gets to the station, and at the same time, the time on the station is 12:50.

You probably won't believe me the first time you read this.

But this is a very deep, and in my opinion, a very cool, thing about special relativity. Events that are simultaneous to one inertial observer are not simultaneous to another, unless their relative velocity is zero. For example, the station's clock displaying 12:50, and Earth's clock displaying 12:50, those are two events that are simultaneous in the station's frame (and in Earth's frame), but they are not simultaneous in the rocket's frame. In the rocket's frame, the event on Earth that is simultanous with the station's clock displaying 12:50 is Earth's clock displaying 12:18.

He is, in Earth's frame, but in his own frame Earth is moving away from him. Neither of the two frames is "special" in any sense. Both descriptions are equally valid.

No. People on Earth appear to be aging slowly in the rocket's frame, because they are aging slowly (still in the rocket's frame). The funny thing is that if the ship turns around and goes back to Earth at the same speed, the people on Earth will have aged more than the people on the ship. It's hard to explain why without drawing a spacetime diagram, but the reason is that when the rocket ship has reversed its velocity, the observers on the ship will be in another inertial system than they were before. In the new frame, the Earth clock will be ahead of the station clock. If the captain is watching Earth through a telescope as the rocket is changing its velocity, he will se people aging very, very quicky. This is not just what appears to happen. In his frame, it is what actually happens.

18. Sep 11, 2004

### pervect

Staff Emeritus
What would have to happen for the rocket ship captain to know the earth time would be more like this.

At t=6 minutes (by his watch, of course), the rocketship captain sends a signal to the earth's time server asking "what time is it".

At t=54 minutes, the rocketship captain gets the answer back, "It's now 12:18 here on earth. Have a nice day."

So, muses the captain, it took 48 minutes for the signal to get there and back, so the time it took light to reach the earth was 24 minutes, and it took another 24 minutes to get back.

I sent the signal at 12:06, so it arrived at earth at 12:30pm, and arrived back to the ship at 12:54.

But why did the earth people reported that their clock only read 12:18? Oh yeah, right, because of relativity. Of course, their clocks are slow. I can tell that, when I come back from a trip everyone is a lot older, just like they warned me.

He also thinks - hmmm, at 12:30 pm, where was I? Oh yes, I was right by that earth oupost. The outpost clock read 12:50 pm. That clock must not be synchronized right. Oh wait, I remember now. That clock synchronization difference business is part of relativity too, it's what makes it impossible for anyone to determine whose clock is actually running slower, I remember that now from those courses on relativity I had to take in at the academy. Boy, that was a long time ago.

19. Sep 11, 2004

### Staff: Mentor

I think Fredrik and pervect gave excellent answers to your questions. But here is my two cents anyway, for the record.
In a sense, yes. But obviously you cannot ignore the time it takes light to travel, so you must be very careful to interpret what you see. If the rocket ship captain looked through his telescope he would see a certain clock reading on the Earth clock. But since he knows how far he is from Earth (24 light-minutes), and since he knows how moving clocks run slow, he will be able to deduce what time the clock must be reading at the moment he passes the station. And his observations will force him to conclude that the Earth clock must read 12:18 when his rocket passes the station.

Another way to think of it is this. Image not just one rocket, but two rockets moving at exactly the same speed following each other at a distance. They maintain a 24 light-minute distance so that (from the rocket frame) the second rocket passes earth exactly as our first rocket passes the station. The clocks in both rockets are synchronized of course. Both rocket observers write down what the Earth/station clocks read just as they pass them. What do they see? We know that the first rocket will record that when his clock read 12:30 the station clock showed 12:50. The second rocket will record that when his clock read 12:30 the Earth clock read 12:18.

Allow me to rephrase it. The rocket captain will say that only 18 Earth minutes have passed during his 30 minute trip, but he will also say that the Earth frame will calculate that the trip took 50 minutes since the Earth/station clocks are not in synch.

The rocket captain realizes that the Earth clock and station clock are not in synch. The station clock is 32 minutes ahead of the Earth clock! So when he passes Earth at 12:00, he knows that the station clock is reading 12:32 at that moment (according to the rocket frame). So when he gets to the station (30 rocket minutes later) at 12:30, he is not surprised to see the station clock reading 12:50. According to the rocket captain, only 18 Earth/station minutes have passed. This makes sense, since he knows that moving clocks will be observed to go slow. And remember, from the rocket viewpoint, the Earth and station clocks are moving at 0.8c.
Not at all! In relativity there is no way to say that one frame is the one really moving. The rocket sees the Earth moving at 0.8c past him, and the Earth sees the rocket moving at 0.8c past Earth. Both frames are equally entitled to view themselves as being at rest.

No! The "time dilation" effect works the same for both frames. The rocket observer would say that Earth clocks (and a person is a clock) tick slowly, but the Earth observers will say the same thing about the rocket clocks. (While everyone agrees that the rocket trip took 30 rocket minutes, each frame will disagree as to the relative aging. According to Earth, the trip took 50 minutes--so the rocket aged less. But according to the rocket, only 18 Earth minutes passed during his 30 minute flight--so Earth people aged less!)

If you introduce acceleration by having the rocket make a round trip journey returning to Earth--then you will find that the rocket folks will have aged less than their friends who stayed on Earth. When one frame accelerates, things are no longer symmetric. (See Section 4.5 in Hogg's book.)

These ideas are subtle. When you get a chance, do all of the problems in Hogg. Learn how to draw a spacetime diagram. Take your time.

20. Sep 11, 2004

### JasonRox

Now that you say two rockets, than what is the difference in 2 stations. You wrote this later:

Now you are assuming that the station clock and earth clock start at different times, it is no wonder it is not synchronized. How am I suppose to know that it started at 12:32am? You imagined two rockets that started their clocks at the SAME time, and why not start both Earth and station clock at the same time, since they are moving at the same speed relative to each other right?

Another thing:

This is EXACTLY what I wanted to know. Remember when I said:

You did in fact say this earlier, but I wanted to confirm that time dilation has nothing to do with the aging paradox (well it might, but it's not the only thing.) I will learn this later.

I was typing something out now, and I GET IT. Finally! :surprised

Thanks Fredrik, pervect, and Doc Al.