# OEMB Section 3, confused

1. Apr 11, 2010

### JulianM

In the section shown here

is the right hand side of the equation intended to refer to light in the moving frame, or to light in the stationary frame ?

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2. Apr 11, 2010

### starthaus

Frame k

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3. Apr 12, 2010

### JulianM

Thank you, but I already understood that.

I thought that (k) was the moving frame and (K) the stationary one. I am trying to follow the math.
From that should I understand that both sides of the equation refer to the moving frame ?

If so then the equation really confuses me.

4. Apr 12, 2010

### starthaus

Correct.

Yes. They two sides express $$\tau$$ as a general function of the coordinates in frame K.

Why?

5. Apr 12, 2010

### JulianM

What I think he says is

- in the moving frame,
the time at which the light ray reaches the far position (x') is 1/2 of the round trip time.

This is the meaning of 1/2(T0 + T2) = T1

then the equation I am studying means:

taking into account the movement of the system

- the time to make the round trip is the sum of the time out and back

- it will take longer for the light to travel outbound (c - v), than it takes to return (c+v) because of the movement (v) of the system

So then we get the equation which says

1/2(round trip time) = Outbound time

Do I understand this correctly ?

6. Apr 12, 2010

### JesseM

Yes, in the coordinates of frame k this will be true, though not in the coordinates of frame K.
In frame K, not in frame k. Remember, the $$\tau$$ coordinate is the time in frame k, but it's expressed as a function of the x and t coordinates in frame K (with x' defined as x-vt...if this seems confusing, take a look at my explanation of Einstein's derivation and terms in post #4 here).
In frame k, yes.

7. Apr 12, 2010

### JulianM

This is hard without a picture. Do you know of anyone who has made sketch of what this all looks like ? I think that might help me.

I am presuming that both x and x' are in the moving system (k), right, because in the paper the coordinates of the stationary system are given by greek letters (which I don't know how to type in this text box ;-(
However maybe my presumption is wrong.
Actually I do find the choice of symbology very difficult to follow.

8. Apr 12, 2010

### JesseM

The problem is you couldn't really draw a diagram with both frames together without assuming the Lorentz transformation, which we aren't supposed to have derived yet at this point. You could draw a diagram showing how things look just from the perspective of the stationary system alone, or just from the perspective of the moving system alone, though.
No, he defines the "stationary system" as K which uses normal letters, and the moving system k as the one which uses greek letters (read the third paragraph in section 3 of the original paper). And x' is not actually the space coordinate of the stationary system K, it's defined as x'=x-vt, where x and t are the coordinates in K. You may find that section a bit easier to follow if you just substitute x-vt in for x'. And again, you may want to look at my post #4 here for help following the derivation as a whole.

9. Apr 12, 2010

### JulianM

Thanks Jesse,

I have OEMB in front of me, and I am working through line by line, looks like I have go back and re-read it

10. Apr 12, 2010

### JulianM

Well I read Jesse's post, and it's much clearer. Jesse, you are a better writer than old Al ;-)

One point, which being clearer than Einstein, slowed me up, and I have read it at least 10 times now is
So, the postulate that all observers must
measure the speed of light to be c in their own rest frame doesn't apply
to Kg.

Jesse, my mental picture of Kg was essentially a set of axes placed at the position x'
If I got this right why doesn't Kg see light at speed c ?

11. Apr 12, 2010

### starthaus

ALL observers (including the one if Jesse's frame Kg) measure the light propagating at c.

12. Apr 13, 2010

### JulianM

Starthaus,

Kg is really the position x' of a body in Einstein's moving frame k, which seems to detect it's difference in speed from c.

13. Apr 13, 2010

### JesseM

Kg as I defined it is not an "inertial frame" as defined in relativity. Rather it is a coordinate system that you get by doing a Galilei transformation on an SR inertial frame (Einstein didn't use the notation Kg in his paper but the use of this sort of non-inertial coordinate system seems to have been sort of implicit in some of his equations). So, if a ruler is at rest in Kg the coordinate distance between its ends will not be equal to the ruler's rest length as measured by inertial observers, likewise the coordinate time between two events on the worldline of a clock at rest in Kg will not be equal to the proper time as measured by the clock itself, and if you have multiple clocks at rest in Kg which are synchronized according to the Einstein synchronization convention (which is what Einstein is discussing in part 1 of OEOMB) then they will not be synchronized in the coordinates of Kg. Since all measurements of speed are really just measurements of distance/time, it's not surprising that since Kg differs from a standard SR inertial frame in this way, it does not measure light to move at c.

14. Apr 13, 2010

### JulianM

From the origin of system k let a ray be emitted at the time T0 along the X-axis to x', and at the T1 time be reflected thence

From this my mental picture was of a mirror or something like that located at distance x' from the origin of the moving frame k

15. Apr 13, 2010

### JesseM

Yes, that's what I was talking about in my explanation on the other thread in this paragraph:
Since I was using x' to refer to the general position variable in the non-inertial coordinate system Kg, I thought it would be more clear to define a constant xm' that represented the position of the mirror in Kg rather than using x' as both a general position variable and the specific position of the mirror.

16. Apr 24, 2010

### JulianM

I have read and re-read the first 3 sections of OEMB and spent a couple of weeks studying the replies in this thread.
I am still lost and confused.

What I find is that Frame k (the moving frame) contains nothing but a reflecting "device" and a ray of light, yet the reflecting device does not experience the light as moving at a speed c.

I'm sure this is a real "newbie" question, but how should I understand this so I can get on to the more meatier parts.

17. Apr 24, 2010

### JulianM

As an additional question how should we understand "Synchronized Clocks"

a) as clocks which are ticking at the same rate
b) as clocks which are set at the same time.

Section 1 basically talks about the two-way trip made by a light ray and, if we use real place names instead of letters can be expressed as:

- if it takes light the same time to go from London to New York and back as it takes for light to go to New York to London and back then the clocks are synchronized.

There doesn't seem to be anything in here which results in them showing the same time, it just reads as if they tick at the same rate.
This gets confusing because the term "synchronized clocks" gets used in so many discussions and experiments where it looks like it should be interpreted as - clocks set to the same time.

What is Einstein really saying ?

18. Apr 24, 2010

### matheinste

Ideal clocks at rest with respect to each other in the same inertial frame of reference tick at the same rate as each other. Synchronization sets these all to show the same time as each other as seen by an observer at rest in the same frame of reference as the clocks.

The speed of light being the same in both directions between two locations is usually assumed. It is not possible to measure the one way speed of light. It does not mean that clocks at these locations are synched. I have not recently looked at OEMB but will when I have time to see if this gives a different interpretation, but that is unlikely.

Matheinste.

19. Apr 24, 2010

### JesseM

Clocks at rest relative to each other in an inertial frame are automatically ticking at the same rate--"synchronized" refers to b, that they are set in such a way that they read the same time at any time according to the frame's definition of simultaneity (which is chosen so that light moves at c in both directions in the frame)
No, that is a condition that guarantees they show the same time. For example, if a signal leaves clock A when it reads 3:00, bounces off clock B when it reads 3:30, and returns to 4:00, then since the light had the same distance to travel both ways, and since we want to assume the light travels at c in all directions, it should have taken equal times to go from A to B and from B back to A. Therefore if it took 1 hour according to A to make the round trip, each half of the trip must have taken a half an hour, so it must have hit B when A read 3:30. So if B also reads 3:30 when the light hits it, that means the clocks are synchronized according to this frame's definition of simultaneity.

There would be other equivalent ways of synchronizing clocks using light--for example, you could set off a flash of light at the exact midpoint of A and B, and set both clocks to read the same time at the instant the light reaches them. Again, this is all based on the idea that we choose a simultaneity convention to assume light moves at the same speed regardless of what direction it's moving.