time dilation between infinite clocks

[514void]

I am not sure what would happen if there were multiple lines of infinite clocks where each line of clocks was at rest in an inertial frame.

 

 

each line of clocks is syncronised so that you can tell what time it is in each inertial frame from any position.

I realise that in SR each inertial frame that you would see the others going slower, but wouldn't that create a paradox?

a clock in inertial frame A would read the year 3000 and see a clock in inertial frame B as having the year 2000, and the clock in frame B would see the clock in frame A as having the year 1000

3000 = 1000? thats a paradox right?

[swansont]

It's not a paradox, because each clock will behave exactly as relativity predicts it will behave. It seems like one only because your naive expectation differs from relativity.

[DimaMazin]

t'=gamma(t - xv/c²)

gamma = 1/(1-v²/c²)^1/2

 

You can try to use the formula.

[md65536]

each line of clocks is syncronised so that you can tell what time it is in each inertial frame from any position.

[...]

a clock in inertial frame A would read the year 3000 and see a clock in inertial frame B as having the year 2000, and the clock in frame B would see the clock in frame A as having the year 1000

The clocks can be synchronized in their own rest frame, but another frame's observer would not see the clocks as synchronized.

 

At any instant in A's frame, each of the clocks in B's frame generally has a different time. There's no paradox with what you wrote, but it doesn't apply to all the clocks.

[514void]

so frame B's clocks would not look syncronised? like all of frame B's clocks would show different times?

that seems weird since you could check all of frame B's time from all of frame A. I would of thought that each frame would see each other frame as being syncronised.

Unless there was a preferred point in space, I'm not sure if there is any mention of that in SR.

[md65536]

so frame B's clocks would not look syncronised? like all of frame B's clocks would show different times?

that seems weird since you could check all of frame B's time from all of frame A. I would of thought that each frame would see each other frame as being syncronised.

Unless there was a preferred point in space, I'm not sure if there is any mention of that in SR.

Simultaneity is not shared across different inertial frames.

 

Say you have an event at each clock in B's frame that happens at time 0 according to that clock. In B's frame, with all of its clocks synchronized, all of those events are simultaneous.

 

In another inertial frame, those events would not be simultaneous. The clocks would mark a particular time, at different times according to a clock in A's frame.

 

Anyone (A, B, C, etc) can agree that B's clocks are synchronized in B's frame, but they're not synchronized in A's frame.

No preferred frame or point, of course.

Edited February 19, 2014 by md65536

[514void]

but all clocks in frame A would be syncronised, and all clocks in frame B would be syncronised

so for each clock in frame A, the nearest clock in frame B would show a certain time.

how would there be any difference in any B's clocks observed from frame A?

[md65536]

but all clocks in frame A would be syncronised, and all clocks in frame B would be syncronised

so for each clock in frame A, the nearest clock in frame B would show a certain time.

how would there be any difference in any B's clocks observed from frame A?

It's a basic consequence of special relativity, called relativity of simultaneity.

The clocks in B's frame, synchronized in B's frame, are not synchronized in A's frame.

 

The flashing bulbs can be considered to be ticking clocks, synchronized in the cart's frame. Note that in the ground's frame they don't flash at the same time.

 

 

According to A, all of B's clocks will tick at the same rate (which will be slower than A's "normal rate" clocks), but they'll be set to different times at any given moment (where the moment is a set of events that are simultaneous according to A).

Edited February 19, 2014 by md65536

[Janus]

so frame B's clocks would not look syncronised? like all of frame B's clocks would show different times?

that seems weird since you could check all of frame B's time from all of frame A. I would of thought that each frame would see each other frame as being syncronised.

Unless there was a preferred point in space, I'm not sure if there is any mention of that in SR.

 

No preferred frame, and each frame sees its own clocks as synchronized and the other frames clocks as not synchronized,

 

Let's consider four clocks, two in each frame.

 

We will start with the frame that you show as stationary in the animation and call it frame A.

We will call the frame above it frame B.

 

Now the first thing we must note that in your animation which shows things according to frame A, you show the distances between the clocks in both frames as being the same for both lines of clocks.

 

But we also know that the distances in frame B will be length contracted as measured from frame A.

 

So for example, if the relative speed is 0.5c and the distances between clocks is 1 light sec as measured from frame A, then the distance between its own clocks in frame B is going to be 1.155 light sec.

 

So let's use the above numbers to figure out what happens in each frame with our clocks.

 

Starting with in the A frame, we will assume that all its clocks read zero when clock 1A is next to clock 1B. We will also assume that clock 1B also reads zero at this moment.

Going with your animation this places clock 2A next to clock 2B, and since 2A is in sync with 1A, it will also read zero at this moment.

 

But as stated earlier, even though the clocks is frame B are in sync as measured in frame B, they are not so as measured from frame A. Every clock will be 0.578 sec ahead of the clock to its left ( as shown in the animation.) This means that B2 will read 0.578 sec when it is next to A2.

 

It will take 2 secs according to the A frame for A1 and B2 to meet. B2 will be time dilated by a factor of 0.866, so it will tick off 1.732 sec during this time. Added to the 0.578 sec it already read, and it will read 2.31 sec when A1 and B2 meet.

 

Now consider things from the B frame:

 

We start start again with A1 and B1 next to each other and reading zero. The distance between B1 and B2 will be 1.16 light sec, and B2, being in sync with B1 reads zero.

Since the distance between A1 and A2, which was 1 light sec in frame A, is length contracted to 0.866 light sec. A1 and A2 will be out of sync with each other by the same factor as B2 and B1 were as measured from frame A (0.578 sec per 1.16 light sec) this gives a difference of 0.5 sec between the clocks, with A2 being 0.5 sec behind A1.

 

So to recap; When A1 is next to B1, both clocks read zero, clock B2 is 1.16 light sec away and also reads 0, while A2 is 0.866 light sec away in the same direction and reads -0.5 sec.

 

A2 is 0.289 light sec from B2 and moving towards it at 0.5c, so it will take .578 sec in the B frame for it to reach B2, so B2 will read 0.578 sec when it gets there. A2 is time dilated so it will run slow by a factor of 0.866 and will tick away 0.5 sec during this time, Thus it will read zero when it reaches B2. This is exactly what happened according to frame A; A1 and B1 reading zero as they pass each other and B1 reading 0.578 sec and A2 reading zero when they meet.

 

You can do the same between frames A&C, and between frame C&B.

 

One thing to note when working between A&C. Since B is moving in one direction at 0.5c in one direction and C at 0.5c in the other relative to frame A, you have to use relativistic velocity addition to get the relative velocity between B&C. This works out to 0.8c, this also gives a time dilation and length contraction factor of 0.6. The difference in simultaneity will be 1.333 sec per light sec separation. Thus if the distance between clocks in their own frame is the 1.16 light sec, the distance as measured from the other frame will be .696 light sec as measured from the other.

 

Each frame sees exactly the same type of relativistic effects in the other frames, and all frames agree what times are on each others clocks at the moment they pass each other. All without invoking a preferred frame.

[514void]

so in frame A all of B's clocks would go slower...

and in frame B all of A's clocks would go slower...

so the paradox would grow as time elapsed?

o, i didn't realise that moving clocks changed the length of space between them.

[Janus]

so in frame A all of B's clocks would go slower...

and in frame B all of A's clocks would go slower...

so the paradox would grow as time elapsed?

o, i didn't realise that moving clocks changed the length of space between them.

Are you not listening to what anyone is telling you? There is NO paradox.

Edited February 19, 2014 by Janus

[514void]

yer, i started writing that before you answered, then i added the bit where i said that i didn't realise that the length of space changed