# Spatial dependence of time

1. Feb 21, 2012

### ShayanJ

I heard that the lorentz transformation for time means that time depends on the space.(I'm only talking about special relativity)I wanna know is there a explicit relationship between time and space in special relativity?
thanks

2. Feb 21, 2012

### Staff: Mentor

Yes the Spacetime interval dist^2 - (ict)^2 a pythagorean distance for Spacetime measurement notice the ict term that coverts time in seconds to imaginary meters.

3. Feb 21, 2012

### Chronos

Under GR, time and space are interchangeable properties.

4. Feb 21, 2012

### Staff: Mentor

That should be $(\Delta x)^2 + (ic\Delta t)^2 = (\Delta x)^2 - (c\Delta t)^2$. Actually, by convention we usually use the negative of this, $(c \Delta t)^2 - (\Delta x)^2$.

The version with "i" is the original one that Minkowski set forth in the early 1900s, but it's considered old-fashioned notation now. Practically all textbooks and researchers have used the version with the - sign and without the i, for fifty years or more.

5. Feb 21, 2012

### ShayanJ

I mean sth like t=t(x).

6. Feb 21, 2012

### Staff: Mentor

It sounds like you want the Lorentz transform equation:
http://en.wikipedia.org/wiki/Lorentz_transformation
$t'=\gamma \; (t-vx/c^2)$

The spacetime interval and the Lorentz transform are very closely related so you can go from one to the other pretty easily.

7. Feb 21, 2012

### BruceW

You can do that in non-relativistic models anyway. For example, x = f(t), so therefore t = f-1(x) (as long as we can define the inverse of the function).

In pre-Einstein physics, the time difference between two events would be the same for all observers, but the position difference would depend on the relative velocity of the observer. (This is the 'Newtonian relativity').

In special relativity, the time difference between two events also depends on the relative velocity of the observer. So this is why special relativity is said to bring time on an equal footing with the spatial dimensions.

In your first post, you talked about the Lorentz transformation for time. Here are the Lorentz transformations (using only 1 spatial dimension, for simplicity):
$$ct' = \gamma ( ct - \beta x)$$
$$x' = \gamma ( x - \beta ct)$$
where t and x are the spacetime coordinates of an event in one frame of reference and t' and x' are the spacetime coordinates according to another frame.

You heard that "the lorentz transformation for time means that time depends on the space" But as you can see from the transformations, time depends on time and space of the same event in another frame. So this isn't giving an explicit relationship between t' and x'. (It just tells us how the coordinates transform when we choose a different frame of reference).

8. Feb 21, 2012

### ShayanJ

No guys.I know these things.let me clarify more.
In some books,it is said that we should place a clock in every point of space.then we send light signals from one clock to others to synchronise them.but because it takes time for the light to reach other clocks,there would be difference between the time shown by clocks in different places.It suggests that the time in one reference frame is different in different places and if you think it explains why time dilation occurs.So Time in one frame should depend on the space in the same frame.But space-time interval is just an invariant quantity not such a relationship.

This is an example of mathematical actions with no physical meaning,

9. Feb 21, 2012

### jadrian

interchangable or the same thing?

10. Feb 21, 2012

### ShayanJ

Not interchangeable nor the same thing.They are different dimensions of the manifold of our world.

11. Feb 21, 2012

### BruceW

Yes, but this is not how time is defined in a frame of reference. When we imagine a frame of reference, we can think of an instantaneous moment in that frame of reference. And there are clocks with dials at each point in space. So the time at each of those locations is the instantaneous reading on each of the dials. It doesn't matter that an observer could not in practise know what all of these dials are saying at that instantaneous moment.

This is the idea in special relativity that we have a rigid spacetime coordinate system that extends throughout space. And we define far-away events by a time and position, even though we were not there when the event happened. How is this theory useful? Well, for example, if there was a solar flare on the sun, and we already knew the sun-earth distance, then we can calculate the time difference between when the solar flare happened and when we received the light from it. (By using the fact that light travels at c). So in this case, we have 2 events, event 1 is when the solar flare happened on the sun at time t1, and event 2 is when the light reaches us at time t2. So you can see that we define the time at which the solar flare happened on the sun, even though we were not actually there.

12. Feb 21, 2012

### jadrian

i have a deep problem with this. i believe the solar flare happened when you saw it.

13. Feb 21, 2012

### Staff: Mentor

The time that it takes for the light to reach the other clock is accounted for in the synchronization process.

14. Feb 21, 2012

### Staff: Mentor

This synchronization process can be used, but it has several problems. First, using this approach the one-way speed of light is not isotropic. Second, it violates the principle of relativity. Third, it is not transitive, meaning that if A uses this method to synchronize with B and B uses this method to synchronize with C then A is not necessarily synchronized with C.

15. Feb 21, 2012

### ShayanJ

You mean with the process I mentioned?!I guess yes,that's a problem!But just the first two.
Why do we need it to be transitive?

And to you BruceW.
If you study the train thought experiment(You sit somewhere near a railroad and two light signals are emmited simoultaneity from two windows of the train each one on a different side of yours and having equal distances to you.because the train is moving one reaches you sooner and so you don't say the distances are equal)I know this is not a proof or so but it suggests that It does matter that an observer could not in practise know what all of these dials are saying at that instantaneous moment.

16. Feb 22, 2012

### BruceW

If the events were simultaneous and the distances were equal (according to the person near the railroad), then he would receive the light signals at the same time.

17. Feb 22, 2012

### Staff: Mentor

You don't need it to be transitive, nor to obey the priniciple or relativity, nor to have an isotropic one-way speed of light. But they are convenient properties to have, and usually it is smarter to adopt convenient conventions rather than inconvenient conventions when you have a choice.

18. Feb 22, 2012

### ShayanJ

Oh.I guess I should study about SR again.Then maybe I can answer.I'm not sure about what you guys say nor what I remember now.

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