From 'the fabric of the cosmos'

[jainabhs]

Hi
I am reading 'The Fabric Of the Cosmos' by Brian Greene.
I am not able to understand a particular part of it.I have a doubt as below.

In one of the chapters he says that suppose there are two observers. One on the Earth and the other one on a planet (lets call it planetA ) which is 10 lightyears far from the earth. He assumes that there is no relative motion between the two observers. As there is no relative motion between them, their current time slices are same. There is no time dilation.
Now observer on the planetA starts walking slowly toward earth.
At this point the author says that the observer will be able to see in the past of the observer on the earth.In other word the toime dilation would be of years in magnitude even due to slow walk of the planetA observer.

As a whole, the author's basic idea is that the time dilation is observable if the relative speed between the observers is approaching light speed.
He says, the same time dilation can be made observable with slow speeds but with the distance between the two observers is very large (like 10 light years...).
I don't understand this. I think time dilation depends only on relative velocity.
Please explain...

Thanks in anticipation.

Abhishek Jain

 

[robphy]

$$\] \begin{picture}(200,200)(0,0) \unitlength 2mm \qbezier(0,50)(0,0)(0,0)\put(0,50){O} \qbezier(0,50)(30,0)(30,0)\put(0,0){\circle*{2} \put(-5,15){\gamma T}} \qbezier(0,0)(30,0)(30,0)\put(30,0){\put(-8,15){ T }} \end{picture} \[$$

In the "time-dilation formula", one is essentially comparing the length of the [timelike] hypotenuse with one [the timelike] side of a right triangle. Although the time-dilation factor $$\gamma$$ (which could be thought of as a "[hyperbolic] cosine" factor... think "adjacent") depends only on the relative speed $$v$$ (which could be though of as a "[hyperbolic] tangent"), the observable difference-in-time (from comparing their previously-synchronized clocks upon meeting) $$\gamma T-T=(\gamma-1)T$$ depends on both $$v$$ and $$T$$, the duration of the return trip according to the traveller.

So, this observable time-difference is easier to measure if $$v$$ is large (as you say), or if $$T$$ is large (as Greene says), or if you had an accurate enough clock [which may not need a large $$v$$ or a large $$T$$].

 

[Chris Hillman]

Time dilation is relative to a family of observers and a method of clock comparison

Hi, jainabhs,

Looks like Robphy has answered your question, but I must add a crucial warning:

As there is no relative motion between them, their current time slices are same. There is no time dilation.

Don't talk or think like that, or you will be bitten! (By a misconception which impedes your understanding, I mean.)

"Time dilation" always involves some method for comparing timekeeping by ideal clocks carried by some family of observers, often just one pair of observers, either of whom might be accelerating, or perhaps an entire congruence (a family of observers whose world lines fill up some region of the spacetime model under investigation). In the case of such a pair of observers, the comparison is often carried out, following Einstein, using light or radio signals sent between the two observers, so "time dilation" is closely related to "frequency shift" (red or blue shift). It is never admissible to refer to "frequency shift" or "time dilation" without mentioning a specific pair (or a larger family) of observers.

When you (or Greene?) refer to "time slices", you are actually referring to an entire congruence of static observers. Their world lines forms an irrotational congruence, so it is hypersurface orthogonal, i.e. there is a unique slicing into three-dimensional spatial hyperslices such that all the world lines in the congruence are orthogonal to each hyperslice. These hyperslices can be taken to comprise constant coordinate time surfaces, and as such they give a basis for constructing a coordinate system; the same construction works in the FRW models in cosmology, where the congruence still (neccessarily) has vanishing vorticity, but now as nonvanishing expansion.

 

[MeJennifer]

Actually, Greene is not talking about time dilation, although of course it is related. He is talking about the effect that distance has on a rotation of the plane of simultaneity.

For two observers, very far removed and at relative rest with each other we could say that they are on a plane of simultaneity. Now as soon as there is even a small motion with respect to each other, the plane of simultaneity will rotate in space-time. The rotation is very small for non-relativistic speeds, but since the distances are large the effect will be large.
So, when the distance is decreasing, each observer will regard the other observer's future on his plane of simultaneity, and if the distance is increasing, each observer will regard the other observer's past on his plane of simultaneity.

That's what Greene was talking about.

 

[quantum123]

I see!
The rotating binary star must be in my past and future back and forth.

 

[jainabhs]

Hi
Thanks a lot MeJennifer.
I got the idea of what you are saying, perhaps.
Please explain following
You say that two far removed relatively at rest observers are on the plane of simultaneity.But if some event occurs in the close vicinity of observer1, he will see it immediately whereas it will take years for oberserver2 to see that event; and vice versa. So the events are not simultaneous for them (unless they happen at the middle of the line joining both obeservers ). are they still on the plane of simultaneity?
OR being on the plane of simultaneity just means that time of the event that has occurred and time that both relatively at rest observers note is same, but they may not see the event??

Please tell me where I can read more on this plane of simultaneity and effect of distance on time dilation??

Thanks a lot
Abhishek