# Absolute Motion?

1. Jul 14, 2006

### actionintegral

In another thread, born2perform asked an interesting question. I would like to re-ask that question in the most concise way I know how.

I stand ,at rest, shoulder-to-shoulder with other observers in an infinitely long line.Our wristwatches are all syncronized.

My line faces another line of observers, also at rest, wristwatches syncronized.

So everybody is at rest and and all wristwatches are syncronized.

My group simultaneously begins moving to the right with some velocity.

I look the wristwatches of the people who pass by me. How will the time on their wristwatches compare to the time on my wristwatch?

2. Jul 14, 2006

### Born2Perform

They answered me something like this:
since the sistem is not linear, in order that you must accelerate to gain speed, your moving line sees the other line's time speed up, at least during the acceleration; and that line sees your time slow down.
If your line is alredy in move, each one sees the other's time slow down, until the simmetry is broken, respecting the relativity principle.

but i think the usual problem is this: how do you synchronize the two lines? at least pervect said that in the previous thread, as i understood

Last edited: Jul 14, 2006
3. Jul 14, 2006

### actionintegral

Thank you for your reply. However, I am not interested in the short period of time during which my line is accelerating. I am interested in the period of time after we have all reached given final velocity and we are all coasting along.

4. Jul 14, 2006

### MeJennifer

It has nothing to do with some symmetry being broken or so. Both people's elapsed time will be the same when the meet, it is simply measured to go slower from the other frame of reference.

Many people seem to have a particular problem with time. For instance this issue rarely comes up with length.
Both see the other's length in the direction of travel shorter, but in fact none of them gets any shorter.

5. Jul 14, 2006

### Born2Perform

in 10 lines you have upsetted all i learned about relativity

"Both see the other's length in the direction of travel shorter, but in fact none of them gets any shorter"
"Both people's elapsed time will be the same when the meet"

so you are saying that nobody gets younger than the other??

6. Jul 14, 2006

### pervect

Staff Emeritus
What happens:

As soon as you start moving to the right, you re-check the synchronization on your writwatches, and find that they are no longer synchronized.

How do you synchronize watches? You use the Einstein convention - you put a light source at the midpoint of the watches to synchronize, and because the speed of light is constant, you know that light hits the two clocks to be synchronized "at the same time".

The following space-time diagram might help

$$\] \unitlength 1mm \begin{picture}(60,90)(0,0) \linethickness{0.3mm} \put(10,10){\line(0,1){80}} \linethickness{0.3mm} \put(20,10){\line(0,1){80}} \linethickness{0.3mm} \put(30,10){\line(0,1){80}} \linethickness{0.3mm} \put(40,10){\line(0,1){80}} \linethickness{0.3mm} \put(50,10){\line(0,1){80}} \linethickness{0.3mm} \put(60,10){\line(0,1){80}} \linethickness{0.3mm} \put(15,10){\line(0,1){20}} \linethickness{0.3mm} \multiput(15,30)(0.12,0.24){250}{\line(0,1){0.24}} \linethickness{0.3mm} \put(25,10){\line(0,1){20}} \linethickness{0.3mm} \multiput(25,30)(0.12,0.24){250}{\line(0,1){0.24}} {\color{red} \linethickness{0.3mm} \multiput(10,22.5)(0.24,0.12){208}{\line(1,0){0.24}} \linethickness{0.3mm} \multiput(10,27.5)(0.24,0.12){208}{\line(1,0){0.24}} } {\color{blue} \linethickness{0.3mm} \put(10,15){\line(1,0){50}} \linethickness{0.3mm} \put(10,20){\line(1,0){50}} } \end{picture} $$$ Vertical lines are the clocks that "stay still". The lines that start out vertical and bend represent the lines that start moving. The blue lines represent events that are simultaneous in the "non-moving" frame. The red lines represent events that are simultaneous in the "moving" frame. As you can see, the idea of your line "simultaneously" starting to move is ambiguous. Which notion of simultaneity do you use, the notion "before" you started moving, or the notion "after"? The diagram is drawn using the "before" notion. There is a simple law for the "line of simultaneity" - if you are moving at a velocity v, the slope of the line of simultaneity is c^2 / v. Thus the product of the slope of a line and the slope of the line of simultaneity is always c^2. This is most conveneint when c=1, then the product is always 1. This rule can be derived from the Lorentz transform. Last edited: Jul 14, 2006 7. Jul 14, 2006 ### MeJennifer Well for sure nobody is getting younger, that is a fact. Now if traveler A travels on a path in space-time that is longer than that of person B (which will happen if he accelerates) and then when they meet later it will the case that the elapsed time for A is less than B so he will actually have aged less. But that does not mean that his clock went slower, clocks don't slow down due to relativity! What it means is that he took a longer path in space-time with a result that less time was spent compared to the time spent for person B. If you were to accelerate almost instantly to almost light speed and then for us after a year you would come back then for you almost no time has elapsed. So we are talking about two different things here. Relativistic measurements (clocks slowing down, lengths contracting, masses increasing) and paths in space-time. For the twin experiment the essense is the paths in space-time not the relativistic measurements. So I think it is important to remember the following things in relativity: • Clocks never slow down • Masses never increase • Lengths never contract • Each object can have its own age due to the paths it took in space-time • Clocks seem to go slower, lengths seem shortened and masses seem increased if measured in relative motion to us. Last edited: Jul 14, 2006 8. Jul 14, 2006 ### actionintegral Excellent. So a delta-v will cause us to lose simultaneity. 9. Jul 14, 2006 ### pervect Staff Emeritus Yep. I've been busy editing my diagram to make it look better and cleaning up the text - you might take another look at it. 10. Jul 14, 2006 ### pervect Staff Emeritus Here's another diagram, showing in more detail how the Einstein synchronization method yields the results I mentioned previously. The slanted lines are all moving clocks. The "middle" moving clock (middle slanted line) emits a light pulse, which arives "at the same time" at the other two clocks (slanted lines). Therfore, the red line through the two points at which the light pulses hit the inner and outer slanted lines must represent the moving frames notion of "simultaneous events" - as long as the distance between the middle clock and each of the outer clocks is the same (which it is in this diagram). The light beams in this diagram are all drawn to scale at a 1:1 slope, i.e. at 45 degrees. $$$ \unitlength 1mm \begin{picture}(90,80)(0,0) \linethickness{0.3mm} \multiput(20,20)(0.12,0.24){250}{\line(0,1){0.24}} \linethickness{0.3mm} \multiput(40,20)(0.12,0.24){250}{\line(0,1){0.24}} \linethickness{0.3mm} \multiput(60,20)(0.12,0.24){250}{\line(0,1){0.24}} \linethickness{0.3mm} \multiput(40,20)(0.12,0.12){333}{\line(1,0){0.12}} \put(80,60){\vector(1,1){0.12}} \linethickness{0.3mm} \multiput(26.88,33.12)(0.12,-0.12){109}{\line(1,0){0.12}} \put(26.88,33.12){\vector(-1,1){0.12}} {\color{red} \linethickness{0.3mm} \multiput(10,25)(0.24,0.12){333}{\line(1,0){0.24}} } \end{picture} \[$$

Note that the Einstein clock synchronization method is equivalent to "slow clock transport" in relativity, and is the only synchronization scheme that allows rapidities to be proportional to velocities.

I'm going to digress a bit, and attempt to motivate Einstein clock synchronization a bit more - i.e explain why it is a good idea, and other schemes for synchronizing clocks are not good (unless they are equivalent).

A velocity measurement requires two syncronized clocks - a rapidity measurement only requires one clock, on the moving object itself.

To measure velocity, one synchronizes the clocks at the start and end of the "course". One then measures the time at which the object crosses the "starting line", subtracts it from the time that the object arives at the "finish line" and defines the velocity as the distance between the start and finish lines divided by the elapsed time.

To measure rapidity, one uses one clock on the object itself. One notes the time at which the clock crosses the starting line, and one notes the time at which the clock crosses the finish line on the same clock.

Rapidities can only be measured for physical objects - one cannot make a clock go at lightspeed.

If clocks are synchronized fairly, one expects rapidities, measured using one clock, to be proportional to velocities, using a "fair" synchronization scheme, as described above.

[add]This wasn't very clearly explained - see later post for more detail and discussion. The issue is really isotropy, not proportionality. Given a set of particles with constant and isotropic rapidity - i.e. the rapidity is independent of direction, we say that an "fair" clock synchronization is one which gives a set of velocities which are also isotropic (independent of direction).

This isotropy is an important motivator of Einstein's theory. In Einstein's original paper, this is just referred to as "isotropy" or "isotropy of space-time". I've always found this a bit murky, hence the longer explanation.

Last edited: Jul 14, 2006
11. Jul 14, 2006

### robphy

Can you clarify your definition of "rapidity"?

It does not agree with the more common usage as the "Minkowski angle", i.e., the additive-parameter associated with a boost. With this more common definition, in SR, the velocity is not proportional to the rapidity... instead it is proportional to the hyperbolic-tangent of the rapidity.

12. Jul 14, 2006

### pervect

Staff Emeritus
Rapidity is just distance divided by proper time - which should be equivalent to the Minkowski angle.

I probably have failed to make my point clear, so let me explain at further length what I was attempting to say. (I've also taken the liberty of fixing up my earlier post significantly).

What I'm concerned with is isotropy. Consider an object with a rapidity of 'n' going east-west, where n is some number. Measure its velocity using two synchronized clocks and some suitable clock synchronizaton scheme.

Now measure the velocity of an object with the same value of rapidity, 'n', going west-east. Measure its velocity again, using the same two clocks, and not changing their synchronization for the second measurement.

If the clocks are synchronized fairly, the velocity east-west will be the same as the velocity west-east.

The exact value of the ratio of rapidity / velocity will vary with the rapidity 'n' chosen. But in order to make the ratio of velocity / rapidity isotropic (the same ratio E-W as W-E), one must chose Einstein clock synchronization, or some equivalent.

Last edited: Jul 14, 2006
13. Jul 15, 2006

### CarlB

I think it is also important to remember that the alternative viewpoint is also tenable. That is,

• The rate at which clocks tick depends on their speed and the gravitational potential
• Masses depend on speed and gravitational potential
• Each object can have its own age due to the paths it took in spacetime, but there can also be a universal time.

Carl

14. Jul 15, 2006

### MeJennifer

Well I am not sure who makes those claims of tenability, I see several problems with the points listed.

Apart from the fact that speed is relative, it must be clear that the laws of physics cannot depend on the speed of something as Einstein correctly postulated in his theory of relativity. So clocks must run at the same speed in all inertial frames, and all lengths must be the same and masses must also be the same.
It is only the measurements between frames that show differences.
And lastly, the existence of an absolute time is simply inconsistent with Einstein's theory of relativity.

So, I would not consider them tenable at all.

15. Jul 19, 2006

### yogi

MeJennifer ...I think your post 7 was very well put