# Relative positions in time - How does SRT define them.

1. Jan 8, 2005

### Haemon

How does SRT define simultaneousness?
Therefore how does SRT define relative simultaneousness?

Depending on the answers I would like to ask how relative can simultaneousness [not tick rates] get and how would you calculate it?

My reasoning:
What I am trying to consider here is that we can readily accept that tick rates are relative to velocity. But do we consider how this relative tick rate effects an objects "NOW location rest frame"

When we deem an object at Rest we also deem it to be existing it in it's own Now which according to SRT is relative.

So when we look at another object from our perspective we deem the now of that other object as simultaneous with our Rest frame.

How ever that Now of our rest frame extrapolated onto our moving object is not the NOW [being it's past or future] of the object but the Now of the RF.So when we swap Rest frames where does our new Rest Frame object exist in its own Now location?
This location I assume can only be arrived at by knowing the time [past and future]separation between our objects

I understand my question may be a little confusing but maybe someone out there can get the gist of what I am endeavoring to explore here.

However first thing to do is clearly define what relative time is regards to the future and the past of both our objects I feel.

Last edited: Jan 8, 2005
2. Jan 9, 2005

### pervect

Staff Emeritus
"Now" is unambiguous only for a suffiiently small object. Thus, for a 6 foot tall human being, there is an inherent 6 nanosecond fuzziness in "now" between one's head and one's feet.

Actual human biological perceptions of time are inherently a lot more fuzzy than this small number. The brain does sophisticated processing on signals that arrive via nerves that transmit information at a speed that's a lot lower than 'c' to in order to order events into a sequence. It is quite possible to fool the brain into placing events "out of sequence" - this has been investigated experimentally, but unfortunately I don't recall exactly where I read about these experiments to be able to provide a link / URL.

As far as synchronizing clocks goes, the standard method applies to clocks that are stationary relative to one another. An observer says that the clocks are synchronized when a light pulse emitted from the midpoint of the clocks (a point which is the same distance from both of them) arrives at both clocks at the same time.

To set up a complete coordinate system for space-time, one can imagine an imaginary array of stationary clocks neatly organized into a cube, filling all of space-time. One can then synchronize these clocks via the method described (by ensuring that clocks at the same distance from the origin read the same time when a short light pulse is emitted from the origin). Conceptually, this array of clocks then allows one to determine both the position and the time of any event that may occur. A real implementation will have only a finite number of clocks, of course, but the thought experiment sets up the basic principle one uses to think of coordinate time.

3. Jan 9, 2005

Staff Emeritus
This definition seems to imply a reliable spacelike rigidity between the spacelike separated clocks. But I thought there was no spacelike rigidity in SR? cf. the pole and barn, rotating disc, etc.

4. Jan 9, 2005

### jdavel

I'm not sure what you're asking, but SR does not exclude point-like objects (e.g., clocks) being rigidly placed at fixed distances from eachother. It only excludes rigid objects whch extend over some distance in space. But that's different.

5. Jan 9, 2005

### pervect

Staff Emeritus
Given that we are talking about special relativity, the stationkeeping problem of keeping the clocks a constant distance is simple. Ideally, if we just place them a constant distance apart initally, they will stay a constant distance apart. Because the net force required to keep the clock in place is zero in the flat space-time of SR, there would be no stresses placed on a rigid body used to keep the clocks the correct distance apart. So in the simple case of SR, one doesn't have to worry about any such "rigid" bodies deforming, because there is no force to deform them.

The notion of a coordinate system built of clocks and rods, and even the notion of clocks & rods themselves, can get more involved when one moves from SR to GR. MTW on and around pg 397 talks about how one can create clocks and rods built out of geodesic world lines, for instance. This mainly answers some conceptual criticisms about how we measure time and distance when our clocks and rulers are accelerating or being torn apart by tidal forces.

The geodesic clocks decribed by MTW are basically like the "light clocks", which time things by watching a light pulse bounce back and forth "the same distance".

The "same distance" is constructed from geodesics by the mathematical mechanism of "Schild's ladder". One starts out with a reference geodesic path, and winds up with a second, parallel path (not necessarily a geodesic) that's a constant distance from the reference path.

6. Jan 10, 2005

### Haemon

Thanks for that Guys,
I'll pose a simple scenario, just to clear the point I am concerned about.

WE have two ships
Ship A and Ship B.

The velocity of ship B according to ship A is 0.8c. They are separating.

At a given point in time ship A measures the distance between them as 100,000 kms.
Question:
At that moment that ship A makes his measurement how far is Ship B according to his perspective?

Ship A records 100,000 kms
Ship B records ????

7. Jan 10, 2005

### Janus

Staff Emeritus
It depends on what you mean by "a given point in time". As measured by who? Ship A?
And it depends on what you mean by "at that moment". Do you mean when ship B determines that a clock on ship A reads the same time as the Clock reads according to Ship A when ship A determines their distance as being 100,000 km? Or do you mean that Ship B is a distance from Ship A such that it would determine that Ship A would measure the same distance as 100,000 km?. These are not the "same moment" according to ship B.

8. Jan 10, 2005

### pervect

Staff Emeritus

Two events occur "at the same time" according to some specific observer O if observer O assigns the same time-coordinate to both events.

Because of the relativity of simultaneity, different observers will describe a different set of points in space-time as occuring "at the same time".

So, in general, it is necessary to specify an observer when one says that two events occur "at the same time", something that you didn't do in this case.

Here's another way of saying the same thing

spaceship 1 has a coordiante system which assigns coordiantes (x1,t1) to events.

Spaceship 2 has a coordiante system which also assigns coordinates (x2,t2) to events.

You can transform from (x1,t1) to (x2,t2) via use of the Lorentz transforms, In units where 'c' = 1, we can write the Lorentz transforms as follows

x2 = (x1-v*t1)/sqrt(1-v^2)
t2 = (t1 - v*x1)/sqrt(1-v^2)

"At the same time" means to an observer on spaceship 1 that two events have the same t1 coordinate. To an observer on spaceship 2, "at the same time" means that two events have the same t2 coordinates.

It's easy to see mathematically that events that have the same coordiante t1 and different coordinates x1 will have different coordinates t2, i.e. points that occur "at the same time" in the coordinate system of spaceship 1 will occur at different times on spaceship 2.

9. Jan 10, 2005

### Haemon

From this am I correct in understanding that Ship B's perspective on distance is unavailable unless Ship B's clocks are read.

In other words knowing B's velocity and distance according to A is not enough to determine what B sees?

for example:

If ship A is stationary next to Jupiter and according to Ship A, ship B passes ship A at v=0.8c heading for a collision with Jupiter, for all intents and purposes when Ship A records the distance as 100,000 kms. Ship B may have already collided with Jupiter from it's own perspective. However this is impossible to deduce from A's perspective as all Ship A sees is a ship passing at v = 0.8c at a distance of 100000kms. [if we assume that knowledge of Ship B's perspective is unavailable to us]

Last edited: Jan 10, 2005
10. Jan 10, 2005

### JesseM

No, you can just use the Lorentz transforms which pervect gave in his previous post to figure out what space and time coordinates B assigns to an event, if you know what space and time coordinates A assigned to them.
Your question seems ill-defined--what do you mean by "when Ship A records the distance"? When you ask a question like "when event 1 happens, has event 2 already happened", what you really mean is something like "is the time-coordinate of event 1 greater or smaller than the time-coordinate of event 2", and that type of question can only be answered if you specify whose coordinate system you're using.

11. Jan 10, 2005

### Haemon

Thanks for that.....

12. Jan 10, 2005

### Haemon

OK, I think I see what you mean....I shall think on it.....Thanks