# Very basic question- help?

1. Nov 27, 2011

### Lucy Yeats

Hello.

This is probably a stupid question, but I'm very confused, so any help would be much appreciated!

If there are two frames moving relative to each other, I cannot tell which is 'really moving'. If I have a time interval t in the frame S, the time interval in S' is t'=gamma x t. But if I label the frames the other way around, t=gamma x t', which isn't the same. So how do I know which frame is the prime frame?

2. Nov 28, 2011

### ghwellsjr

You really should start by defining some events in one frame and then use the Lorentz Transform to see what the coordinates of those events are in the other frame which is usually considered the prime frame. You can transform those events back to the first frame by negating the velocity factor that you used to get from the first frame to the second frame.

Do you know what we mean by "event" and Lorentz Transform?

3. Nov 28, 2011

### Lucy Yeats

Hi, thanks for replying so quickly!

Yes, I've just learnt what the Lorentz transforms are. I'm not really explaining my question well, so I'll use an example.

A spaceship is passing a planet with 0.6c. An interval of an hour is measured on the spaceship's clock.

To find the interval as measured on the planet, I find gamma is 5/4. Do I multiply or divide one hour by gamma? Why?

4. Nov 28, 2011

### ghwellsjr

You divide one hour by gamma to get 48 minutes.

To answer why will take some time which I'll put on another post, but in the meantime, let me just say that each observer sees the other one's clock as running slower.

5. Nov 28, 2011

### Lucy Yeats

Thanks! I look forward to finding out why in the next post, as I'm still quite confused.

:)

6. Nov 28, 2011

I'm new to this as well but I think the whole idea behind SR is there isn't really one that is moving. They are both moving relative to each other.

7. Nov 28, 2011

### Lucy Yeats

To expand on why I'm confused:
"A spaceship is passing a planet with 0.6c. An interval of an hour is measured on the spaceship's clock. To find the interval as measured on the planet, I find gamma is 5/4. Do I multiply or divide one hour by gamma? Why?"
The only information we have is that two frames have a relative velocity of 0.6c between them. How do we decide which is the frame with interval 48 minutes and which is the frame with interval one hour? They seem to be symmetrical.

8. Nov 28, 2011

### ghwellsjr

Now to answer why:

We start with one frame of reference, that of the spaceship and set up two events that are one hour apart. I'm assuming that your statement, "An interval of an hour is measured on the spaceship's clock" means that one hour has gone by on the spaceship's clock. This also means that one hour has gone by on all the coordinate clocks in the rest frame of the spaceship, no matter where they are located. So for our two events, we can pick different locations as long as the time interval is one hour. But in the rest frame of the planet, the clocks are running at different rates and so we cannot just pick any two events at different locations, we must make sure that the two events in the spaceship's frame are at the same location in the planet's frame.

Now in picking our two events, the easiest way to do this is to put the first event at the origin and the second event one hour later at some other location. We are going to use the standard configuration of events for the Lorentz Transform.

When doing problems where everything is happening along one line (and we can pretend that the spaceship and planet have no widths so they can pass right through each other), I like to shorten up the nomenclature for an event so that it includes just a time coordinate and an x-coordinate since the y- and z- coordinates are always zero. And I also like to use β for speed, as you did, so that the Lorentz Transform (LT) is much simpler, as follows:

t' = γ(t-βx)
x' = γ(x-βt)

The primed values are for the planet since we are starting with events defined in the spaceship's frame.

You already calculated γ as being equal to 1.25 and you are probably wondering why I said you divide by γ when you can see that the equation says you multiply by γ. Well, the answer to that question is that people like to take short-cuts and in the short-cut, you divide but you'll see when we get done using the LT that it works correctly the way it is written.

OK, so now we define two events in the spacecraft's frame. The first one will be, as I said, at the origin, with t=0 (in hours) and x=0 (in light-hours). So the coordinates of the first event are:

[0,0]

The second event will be one hour later which we naively might set up as [1,0] which would be an event on the spaceship that was one hour later but in order to do it correctly, as I stated earlier, we need to make sure that we use coordinates in which time is transformed at the same location on the planet, not on the spaceship. So we have to consider where the planet will be one hour later. We know that at time zero, the spacecraft and planet are at the same zero location, so one hour later, the planet will be 0.6 light-hours away since it is traveling at 0.6c, so this makes the event in the spaceship's frame of where the spaceship will be one hour later as:

[1,0.6]

Now we need to transform both these events from the spaceship's frame to the planet's frame but since we already know that the origins coincide (in other words, [0,0] transforms into [0,0] in any other frame), we don't have to do any work to transform the first event.

Now for the second event, [1,0.6], we have the following values:

t=1
x=0.6
β=0.6
γ=1.25

Solving for t' we have;

t' = γ(t-βx)
t' = 1.25(1-0.6*0.6)
t' = 1.25(1-0.36)
t' = 1.25(.64)
t' = 0.8

And 0.8 light-hours is 48 minutes which is what you get when you divide 1 hour by 1.25.

OK, now let's do x':

x' = γ(x-βt)
x' = 1.25(0.6-0.6*1)
x' = 1.25(0.6-0.6)
x' = 1.25(0)
x' = 0

Technically, when working with time intervals, we should subtract the two times that we calculated that define the starting and ending times of the interval, but since the starting time was at the origin (t=0) we don't have to explicitly do the subtraction.

Now to summarize, in the spaceship's frame, we have two events separated by one hour:
[0,0]
[1,0.6]

These transform into the planet's frame as:
[0,0]
[0.8,0]

We can see that in the planet's frame, the two events are at the same location, x=0, and we can see that the time has advanced by 0.8 hours.

9. Nov 28, 2011

### ghwellsjr

Hopefully, by the time you finish studying my previous long post, you will have discovered that the two frames are symmetrical and we could have started out with the problem stating that the clock on the planet progressed through an interval of one hour and that means that in the rest frame of the planet, all the clocks no matter where they are also went through an interval of one hour. Then we would pick out one clock in the spaceship's rest frame and see how much time transpired on it, remembering that it's location has to be the same for the two events. We make the location zero if we want a minimal amount of calculations to do. And we have to remember that β is now the negative of what it was when transforming in the other direction.

10. Nov 28, 2011

### Lucy Yeats

Wow, thanks so much for giving such a detailed answer, which I'll read through carefully now. Have a nice day!

11. Nov 28, 2011

### Lucy Yeats

So the spaceship time advances by 1 hr and the planet time advances by 0.8 hours. Isn't this asymmetrical?

Also, I'm confused by the shortcut. For my homework, I've just derived the expression t'=Yt, but you used the expression t'=t/Y. Which is right?

12. Nov 28, 2011

### ghwellsjr

When we say that the relationship is symmetrical, we mean that we can start with the rest frame of either observer, the spaceship or the planet, and we can calculate how the clocks that are moving in that rest frame are time dilated. It's just like I said in my previous post, the spaceship considers the planet to be moving at 0.6c and the planet considers the spacecraft to be moving at -0.6c. They each consider the other to be moving in the opposite direction. Does that make it symmetrical or asymmetrical? Well we could have just turned the whole x-axis around for the planet and then the spaceship would also be moving at 0.6c.
Can you show your derivation? And please note, the symbol for gamma is γ, not Y. You can get that symbol in the advanced editing mode where you'll see it in a box off to the right with lots of symbols in it.

13. Nov 28, 2011

### DrGreg

The Lorentz transformation is\begin{align} t' &= \gamma(t-vx/c^2)\\ x' &= \gamma(x-vt) \end{align}If you rearrange the equations, you get the inverse transform\begin{align} t &= \gamma(t'+vx'/c^2)\\ x &= \gamma(x'+vt') \end{align}From this, if follows that $t'=\gamma t$ applies only in the special case when x=0. And $t=\gamma t'$ applies only in the special case when x'=0.

So you need to think what does x=0 or x'=0 mean? They can't both be true (except at the origin where x=x'=t=t'=0). Which of them is true in the case you are interested in?