Undergrad How Ebeb's Diagram Reveals Time Dilation

[Grimble]

because they are in relative motion their clocks may have been synchronised at some time (e.g. both set to zero by some convention) but will not tick at the same rate.

Hmmm.
You see, this is the sort of thing I am referring to; a completely sensible and rational summary that fits perfectly with what we know. It is fine taken on its own; but I have a problem making it fit with Einstein's first Postulate:

The laws of physics are invariant (i.e. identical) in all inertial systems (non-accelerating frames of reference).

Take any number of identical, inertial, light clocks; each and every one will read 1 second after the light in that clock has traveled 1 second to its mirror 1 light second away. These are identical light clocks in identical frames that happen to be traveling at different speeds relative to one another, but each is at rest in its own frame, subject to identical scientific laws. The proper time for each clock on its worldline is invariant being the same wherever it is measured from, as it is the time for light to travel 1 light second. That, the proper time of one second is what that clock will read after 1 second has passed.

i.e. each and every clock in this scenario will read the same time = 1second (How could they not?)

Note: this is not absolute time, as each observer will measure the other clocks to be time dilated.

That interval measured by the clock in one frame for a clock(A) in another, moving frame(B), can be found by applying the Lorentz Transformation Equation in order to transform that measurement of 1 second coordinate time; to calculate what that interval would be relative to the observer's frame. At 0.6c it would be 1.25 seconds.

Taking A as the stationary system K, and B as the moving system K',
then

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but

seconds, i.e. a somewhat larger time.

t' = γt = 1.25seconds.

In the special case of an inertial observer in special relativity, by convention the coordinate time at an event is the same as the proper time measured by a clock that is at the same location as the event, that is stationary relative to the observer and that has been synchronised to the observer's clock using the Einstein synchronisation convention.

Therefore if we place a clock adjacent to B's mirror at the moment B's light hits the mirror (at the same location as the event), in frame A(that is stationary relative to the observer) synchronised to A's clock, then the coordinate time of that clock must be the same as the proper time (i.e. 1 second).
the coordinate time is 1 second, the clock calculated to read 1.25 seconds is the moving clock in frame B.
The time 1.25 is the time for the light in B to travel the distance to the clock plus the distance traveled by the clock. (by vectorial addition)

 

[Dale]

Take any number of identical, inertial, light clocks; each and every one will read 1 second after the light in that clock has traveled 1 second to its mirror 1 light second away

I am not sure that what you are saying is correct. Again, the semantics are off. To avoid getting bogged down in semantics, can you write mathematically what you mean?

Be sure to use different symbols for coordinate time and proper time. Make sure that all coordinate times are associated with a specific reference frame and that all proper times are associated with a specific clock. Label any important events or times that you are comparing, particularly whatever you think is in conflict with SR.

For instance I might use $A$, $B$, and $C$ to label reference frames, $t_A$ to label the coordinate time in frame $A$, and $\tau_a$ to label the proper time along the worldline of the clock which is at rest at the spatial origin of $A$. I might use labels like $a_0$ to designate the event along the worldline of clock $a$ where the clock reads 0, and $a_1$ to designate the event on the worldline of $a$ where it reads 1. And I would then use labels like $t_{Ba1}$ to indicate the coordinate time in frame B at which clock $a$ reads 1. I would use units where c=1 and I would write coordinates as (t,x,y,z) and I would use the convention where spacelike intervals $ds^2>0$ so $d\tau^2=-ds^2$. You can do it differently as long as you are clear.

 

[PeterDonis]

That interval measured by the clock in one frame for a clock(A) in another, moving frame(B), can be found by applying the Lorentz Transformation Equation

No, it can't. The interval--the actual elapsed time on the clock--is an invariant, it doesn't change when you change frames.

The coordinate time difference between the two events that define the interval--i.e., the two events on the clock's worldline that define the "elapsed time"--will change when you change frames; but the coordinate time difference, by itself, has no physical meaning.

 

[sdkfz]

The point of post #26 was to try to boil down to where you and everyone else differs, simply, and away from the minutia. It's disappointing that you ignored it and instead tried again to add more sloppy complexity.

Hmmm.
You see, this is the sort of thing I am referring to; a completely sensible and rational summary that fits perfectly with what we know. It is fine taken on its own; but I have a problem making it fit with Einstein's first Postulate:

That's because you read that postulate (whether you realize it or not) as meaning "absolute time". In fact that postulate is well covered by the results in post #26: both A and B consider their own mirror to be 1 light second away from themselves, and their own clock ticks 1 second when the light strikes it. Yes, the laws of physics were invariant for A and B.

Take any number of identical, inertial, light clocks; each and every one will read 1 second after the light in that clock has traveled 1 second to its mirror 1 light second away. These are identical light clocks in identical frames that happen to be traveling at different speeds relative to one another, but each is at rest in its own frame, subject to identical scientific laws. The proper time for each clock on its worldline is invariant being the same wherever it is measured from, as it is the time for light to travel 1 light second. That, the proper time of one second is what that clock will read after 1 second has passed.

i.e. each and every clock in this scenario will read the same time = 1second (How could they not?)

Yes. As noted in post #26.

Note: this is not absolute time, as each observer will measure the other clocks to be time dilated.

Yes. As noted in post #26.

That interval measured by the clock in one frame for a clock(A) in another, moving frame(B), can be found by applying the Lorentz Transformation Equation in order to transform that measurement of 1 second coordinate time; to calculate what that interval would be relative to the observer's frame. At 0.6c it would be 1.25 seconds.

Taking A as the stationary system K, and B as the moving system K',
then
t' = γt = 1.25seconds.

Therefore if we place a clock adjacent to B's mirror at the moment B's light hits the mirror (at the same location as the event), in frame A(that is stationary relative to the observer) synchronised to A's clock, then the coordinate time of that clock must be the same as the proper time (i.e. 1 second).
the coordinate time is 1 second, the clock calculated to read 1.25 seconds is the moving clock in frame B.
The time 1.25 is the time for the light in B to travel the distance to the clock plus the distance traveled by the clock. (by vectorial addition)

This is where your thinking gets sloppy (I took great pains to be very clear about who was measuring what in post #26) and you end up off the rails. Simply placing a clock at some location doesn't make it read the same as some other clock at the location. If that clock is "in frame A", i.e. stationary with respect to A, then it's in a frame where the B light moves more than 1 light second to hit the B mirror. The 1.25 seconds is B's light-hit-mirror event, as measured by A. You can't claim it's the proper time for B, as according to B the light only moved 1 light second to hit the mirror.
Just above, you wrote "as each observer will measure the other clocks to be time dilated", now you seem to argue the opposite. Why is your scenario so self-contradictory?
You seem to be trying hard to again make absolute time, and demote relativity to mere illusion.

 

[Grimble]

No, it can't. The interval--the actual elapsed time on the clock--is an invariant, it doesn't change when you change frames.

The coordinate time difference between the two events that define the interval--i.e., the two events on the clock's worldline that define the "elapsed time"--will change when you change frames; but the coordinate time difference, by itself, has no physical meaning.

Sorry... I was using the word 'interval' literally rather than scientifically...

The proper time displayed by clock B, (that is the coordinate time in frame B?), transformed by LTE gives the measurement relative to the observer A.

 

[PeterDonis]

The proper time displayed by clock B, (that is the coordinate time in frame B?), transformed by LTE gives the measurement relative to the observer A.

No. You don't transform proper time; it's an invariant, the same in all frames. Proper time is just the length along a specific curve (in this case, B's worldline) between two chosen points (whichever events on B's worldline you want to know B's elapsed time between).

IMO you should forget about coordinates altogether until you understand the underlying geometry that the coordinates are representing, which consists of the invariant lengths along particular curves, and the angles between the curves when they intersect. The "angle between the curves" in SR corresponds to relative velocity.

 

[Dale]

The proper time displayed by clock B, (that is the coordinate time in frame B?).

What have I told you before about this? In my notation above, what is the difference between $\tau_b$ and $t_B$? Where is each defined?

 

[Grimble]

The proper time displayed by clock B, (that is the coordinate time in frame B?), transformed by LTE gives the measurement relative to the observer A.

Let me rewrite what I meant here...
The coordinate time in frame B, which is synchronised to the proper time of frame B's clock (is that the right way to say it?), transformed by LTE gives the measurement relative to the observer A.[/QUOTE]

 

[Grimble]

What have I told you before about this? In my notation above, what is the difference between τ_b and t_B? Where is each defined?

τ_b is the proper time on the worldline of clock B.
t_B is the coordinate time at an event in clock B

In the special case of an inertial observer in special relativity, by convention the coordinate time at an event is the same as the proper time measured by a clock that is at the same location as the event, that is stationary relative to the observer and that has been synchronised to the observer's clock using the Einstein synchronisation convention.

So, I read this as meaning that τ_ba1 is equal to t_Ba1 is that right?

 

[Dale]

τ_b is the proper time on the worldline of clock B.
t_B is the coordinate time at an event in clock B

Yes.

So, I read this as meaning that τ_ba1 is equal to t_Ba1 is that right?

No. If I understand what you are trying to denote here then $\tau_{ba1}$ doesn't exist. That would be the proper time of clock b at event a1. But event a1 is not on the worldline of clock b. So it is undefined.

 

[Grimble]

I have tried to take in all I have been told and apply it to the scenario of moving clocks.
I now understand that proper time involves measurement along a single dimension between events that occur on a clock's worldline.
In Fig. A I have drawn two clocks, A & B.
Event τ_a0.0 and Event τ_b0.0 are a single event that occurs when the two clocks are co-located. The clocks then move apart. The lights emitted at the common event labelled a0.0/b0.0 strike their respective mirrors after 1 second, having each traveled 1 light second.

​

 

[Dale]

@Grimble I have looked fairly carefully at your diagram and although it is pretty complicated, I think I understand it. Most of them seem correct to me, but Fig A and Fig 5 have some problems. You have labeled several spatial locations with $\tau_a$ or $\tau_b$ but in this case the proper time should only label the worldline of the end of the clock. So all of the $\tau$ labels should be on the horizontal axis.

 

[Grimble]

@Grimble I have looked fairly carefully at your diagram and although it is pretty complicated, I think I understand it. Most of them seem correct to me, but Fig A and Fig 5 have some problems. You have labeled several spatial locations with $\tau_a$ or $\tau_b$ but in this case the proper time should only label the worldline of the end of the clock. So all of the $\tau$ labels should be on the horizontal axis.

Thank you Dale, I can see that, I am finding it difficult to draw what I was trying to shew. Would it be better to label Fig. A as the diagram of the lights within the clocks, as it is the lights that travel to the mirrors whereas the clocks are stationary (within their individual frames)?

How should one draw a diagram shewing proper time?

 

[Dale]

Would it be better to label Fig. A as the diagram of the lights within the clocks, as it is the lights that travel to the mirrors whereas the clocks are stationary (within their individual frames)?

There is no proper time along the worldline of a pulse of light. Proper time is defined along timelike worldlines only.

 

[Grimble]

There is no proper time along the worldline of a pulse of light. Proper time is defined along timelike worldlines only.

Thank you, I wondered if that would be the case...
I think using a light clock is causing complications, and is unnecessary for what I am trying to do. I suppose it was a way to help me visualize time by the movement of the light. I can now see how it does that but at the same time how it muddies the waters.
So I have changed my diagram to represent two clocks and their proper times plotted on their worldlines.

The purpose of this diagram is to highlight that proper time is the same for every perfect clock that ticks at the same rate of 1 second per second. Which brings me to another realisation: coordinate time is individual to each reference frame, being relative to that frame. Proper time is invariant because it is only measured relative to its own frame, in this case to the frame of each clock.
Proper time is the measurement of time by a clock and in that clock's frame the clock is at rest and its worldline moves in one dimension, that of time. A worldline only moves through space when viewed by an observer outside the clock's frame.

So I hope I am right if I say that clock A and clock B will each record (and display) time at the same rate, that in Spacetime an observer can read the invariant proper time from either clock, but will measure the clock to run slow as a function of its relative motion?

 

[Dale]

I have changed my diagram to represent two clocks and their proper times plotted on their worldlines.

That is better, but I am not sure what the vertical axis represents now. Previously it was a second direction in space. If it were now representing time then you would need to show them as non-parallel since they are moving away from each other.

 

[Grimble]

If proper time is seen as one dimensional, how does that work? For it seems to me that the 1D must be time, but if that is the case then how can two 1D paths, for the two clocks be non-parallel for being 1D direction would have no meaning?
Anyway I have redrawn it as two separate diagrams Fig. A and Fig. B; what I am trying to convey is the idea that proper time works in exactly the same way for any clock, or other object.
In its own reference frame an object is at rest and only moves through time?
And time progresses at the same rate everywhere in every frame - c.
It seems to me there is a dichotomy between proper time, which, is invariant because it is measured directly along an objects worldline and coordinate time which is measured remotely from a particular observer's perspective and is therefore specific to each observer.

 

[Ibix]

A piece of spaghetti is one dimensional - or close enough for this example - but if you lay a dozen pieces out parallel they form the lines of a grid.

Coordinate time is just what happens when you agree on a direction to call "time" and a procedure for defining "time zero".

 

[Grimble]

But when you add more pieces of spaghetti and form a grid are you not adding more constraints that are nothing to do with our one dimensional proper time? Even when two worldlines interact are they not still separate 1 dimensional lines?

 

[Ibix]

But when you add more pieces of spaghetti and form a grid are you not adding more constraints that are nothing to do with our one dimensional proper time?

Yes. That's what a simultaneity convention is - an arbitrary choice out of many possible ways to define "the same time not at the same place".

 

[sweet springs]

Hi.

I would like to return to the question of how this diagram, courtesy of 'Ebeb' from #124 in thread "Proper (and coordinate)

In A system, velocities of A,O,B are
A: 0
O: V
B: about 2V, exactly $\frac{2V}{1+\frac{V^2}{c^2}}$

In O system, velocities of A,O,B are
A: -V
O: 0
B: V

In B system, velocities of A,O,B are
A: about -2V, exactly $-\frac{2V}{1+\frac{V^2}{c^2}}$
O: -V
B: 0

The three systems have their proper times and time of other systems are slow according to their velocities. Diagram of O is as you showed. In diagram of A, Lines of O and B lie right side. In diagram B, Lines of A and B lie left side. A and B seem to have same proper time only for O not for other systems.
Best.

 

[Dale]

Anyway I have redrawn it as two separate diagrams Fig. A and Fig. B; what I am trying to convey is the idea that proper time works in exactly the same way for any clock, or other object.

Ok, as two separate figures that makes sense.

It seems to me there is a dichotomy between proper time ... and coordinate time

Most definitely, yes. That is why I was being such a stickler about using them correctly. They are fundamentally different things.

 

[sweet springs]

PS to #51

the proper time = τ, the distance to the mirror is cτ
the coordinate time = t the distance to B's mirror is ct
the distance between A and B is vt
and (cτ)2 = (ct)2 - (vt)2
Yes the proper time squared (which is another way of saying the Spacetime interval) is equal to the coordinate time squared minus the distance squared.

World interval between A clock ticks 0:00 and A clock ticks 1:00 is $ct_A$. This is minimum time of all the systems because t_A < t =\frac{\sqrt{c^2t_A^2+x^2}}{c}
World interval between O clock ticks 0:00 and O clock ticks 1:00 is $ct_O$. This is minimum time of all the systems because $t_O < t =\frac{\sqrt{ct^2_O+x^2}}{c}$
World interval between B clock ticks 0:00 and B clock ticks 1:00 is $ct_B$. This is minimum time of all the systems because $t_B < t =\frac{\sqrt{ct^2_B+x^2}}{c}$

So the system in which two events occur at the same place has minimum time interval among all the systems. In your mirror setting, light path is clock at rest 0:00, mirror 0:30 and clock at rest again 1:00. Proper time of things in motion is known and shared with all the systems calculating $\frac{\sqrt{c^2t^2-x^2}}{c}$ respectively. Best.

 

[jbriggs444]

But when you add more pieces of spaghetti and form a grid are you not adding more constraints that are nothing to do with our one dimensional proper time? Even when two worldlines interact are they not still separate 1 dimensional lines?

Let's be clear. @Ibix called for strands of spaghetti laid out parallel. He did not call for any cross-wise strands. As I understand the analogy:

The strands are clocks. One end is when they are started. The other end is when they are stopped.

You can lay out strands in parallel (a bunch of clocks at rest with respect to one another). You can line up their ends (you can start them all "at the same time" in some fashion). You can do this with the start times at right angles to the strands. Or you can do it on a slant.

But if the start points are "all at the same time" and are not all at the same place, you are not allowed to have a strand of spaghetti that goes through all of them. They are space-like separated. A clock cannot go from anyone to any other. A cross-wise line of synchronization is not a strand of spaghetti.

 

[sweet springs]

Hi. For whom spaghetti is vertical in his tx diagram, it is time spaghetti, e.g. 8:00 car and 9:00 car. However for whom same spaghetti is perpendicular it is time-space mixed spaghetti. It has both time interval length and space interval length, e.g. 8:00 home and 9:01 office.

Time spaghetti for A is time-space spaghetti for B and vice versa. Length of spaghetti is defined as time interval of A for whom spaghetti is time-one not time-space one.

 

[Ibix]

However for whom same spaghetti is perpendicular it is time-space mixed spaghetti.

You can't have perpendicular timelike vectors. You can certainly have non-parallel ones, but never perpendicular. For any time-like vector there exists a frame in which it is parallel to (1,0,0,0). The inner product of any time-like vector with this is clearly non-zero.

 

[sweet springs]

Thanks. Oblique, not perpendicular, due to my poor English. Best.

 

[Grimble]

Now it seems you are trying to befuddle me with spaghetti!
I have just come to accept that 1D spaghetti is a unique representation of each clock; and that as soon as one refers to multiple strands, whether parallel, orthogonal, oblique, or tied in knots, you have to be referring to coordinate times, not proper time.

I can understand that when one reduces spacetime diagrams to 2 or 3 dimensions, time is drawn vertically for a frame, so a resting object does not move as time increases. And to extend that to 4 dimensions one could leave space to the 3 cartesian coordinates, but then time would have to be somehow drawn so that while extending it did not move in any physical direction.

So in my amateurish way of thinking I imagine drawing time as a spherical coordinate; it would be like measuring the time like a flash of light - moving at c in every direction...
And that drawing it vertically would then be merely a convention.
For how can one assign a direction to proper time? To assign it a direction one must add extra dimensions.

This one way I visualize the difference between proper and coordinate times; for proper time, Fig 58A there is only one dimension and it has no direction; for coordinate time we add coordinates for the spatial dimensions: we know that when Clock B has traveled 0.6 light seconds along the x axis, the light in clock B will have traveled 1 light second from the null point. Therefore the point where the light is after 1 second measured by the observer is fixed by the movement of the clock. It has, therefore a full set of coordinates. It is coordinate time.

(Please note I did not intend these diagrams to conform to any specific diagram 'type', but merely to shew how I visualize it.

 

[sweet springs]

I have just come to accept that 1D spaghetti is a unique representation of each clock; and that as soon as one refers to multiple strands, whether parallel, orthogonal, oblique, or tied in knots, you have to be referring to coordinate times, not proper time

More appropriate name for spaghetti is world-line, straight lines as drawn in your figure in OP. World line is space-time like but if events to be considered are given we can choose special coordinate that world line through the events is time-like. Events on this world-line has common place to occur in that coordinate. The coordinate time of this special coordinate is called proper time. Propertime is one of coordinate times.

 

[jbriggs444]

World line is space-time like but we can choose special coordinate that world line is time-like.

A time-like world line is always time-like regardless of coordinates. There is no such thing as a "space-time like" world line.

What is true is that in a coordinate system where a clock is at rest, one second of coordinate time is traversed for every second of proper time that the clock ticks off.