How Ebeb's Diagram Reveals Time Dilation

[sweet springs]

Thanks. Again my misuse of words, I am afraid. How about Only time component and No space component, instead of "time-like" and Both time and space components, instead of "time-space like" in my previous post ? All lines here expressing motion of body are time-like as you taught. Best.

 

[jbriggs444]

Thanks. Again my misuse of words, I am afraid. How about No space component, instead of "time-like" and Both time and space components, instead of "time-space like" in my previous post ? Best.

The correct terms are "time-like", "space-like" and "null". Those are invariant properties of an interval. That is to say that they do not depend on choice of coordinate system.

The terms "time-space like" and "space-time like" are never used. Those would not refer to invariant properties.

 

[sweet springs]

Hi.

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...

Your chart seems like 2d radar charts as in the back of video . I have no idea of your "spherical time coordinate".

 

[jbriggs444]

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.

@Ibix proposed parallel strands. That translates to a set of unaccellerated clocks all at rest with respect to one another.

Straight strands = unaccelerated.
Parallel = at rest with respect to one another.

Let's not think about knots until we have parallel strands sorted.

The distance along any particular strand of spaghetti is analogous to proper time.
If you lay the spaghetti on a sheet of graph paper, the grid lines in the horizontal direction mark out coordinate time.

Note that the analogy is imperfect because spaghetti can be rotated past 45 degrees on a sheet of Euclidean graph paper. Clocks cannot do a rotation past 45 degrees on a space-time diagram.

 

[Dale]

So in my amateurish way of thinking I imagine drawing time as a spherical coordinate

Please, don't do that. You really need to stop just making things up on your own. Please just try to learn and use the standard techniques.

For how can one assign a direction to proper time?

Proper time is not a direction. It is the spacetime equivalent of length.

If you have two points on a piece of paper and you connect them with a line then you can measure the length along that line from the first point to the second. Similarly, if you have two events in spacetime and you connect them with a worldline then you can measure the proper time along that worldline from the first event to the second.

This one way I visualize the difference between proper and coordinate times; ...

The difference between proper time and coordinate time is the difference between coordinates and lengths. Please stop making stuff up

 

[Grimble]

I am sorry if you think that, but I am not trying to make things up but to explore what they mean.
I now understand that propertime is one dimensional; which means it is the time measured between events on a worldline. (It would be wrong to describe it as the propertime measured on a worldline because we are defining propertime).
Propertime in invariant. It is the same whoever is measuring it. That can only be because it is measured between points on a worldline, in that dimension alone. It can have no location, nor direction. For the 'clock' whose time is being measured it happens at a single location - that of the clock; For a different observer it may be between different physical locations, but only the duration of the interval measured on the clock is relevant.

So, if we liken the single dimension of proper time to spaghetti, the length of the spaghetti is the proper time, its direction, whether it is straight, parallel or twisted in knots is meaningless.

 

[Ibix]

I now understand that propertime is one dimensional; which means it is the time measured between events on a worldline. (It would be wrong to describe it as the propertime measured on a worldline because we are defining propertime).

It's the time measured between two events along a worldline that passes through those events. Thus two observers may experience two different times between a pair of events if they follow different worldlines. That's the resolution of the twin paradox, in one line.

That can only be because it is measured between points on a worldline, in that dimension alone.

Careful with "dimension" here. Better to say that the proper time for a worldline is defined only on that worldline and leave "dimension" out of it.

It can have no location, nor direction

Depends a bit what you mean. I'd agree that direction in spacetime isn't related to proper time. However there is a clear meaning to "along the worldline in the direction of increasing proper time".

So, if we liken the single dimension of proper time to spaghetti, the length of the spaghetti is the proper time, its direction, whether it is straight, parallel or twisted in knots is meaningless.

Well, it's not meaningless. It's irrelevant to the proper time, though, which is what I suspect you meant.

Note that proper time is only defined for time-like worldlines and these cannot curve back on themselves in flat spacetime. So no knots. The rest is fine.

 

[jbriggs444]

Propertime in invariant. It is the same whoever is measuring it

This piece needs some additional details. Indeed, proper time is invariant. It is the same regardless of what reference frame is used to measure it. But "measure" means that there has to be some physical procedure to come up with the measurement. How does a fellow using reference frame B "measure" the proper time between when clock a at rest in frame A starts and stops ticking?

A fellow at rest in frame B could measure it by using a telescope and copying down the readings of clock a when it starts and stops. He could subtract one from the other and say "there -- that's my measurement of proper time". That would be a valid measurement procedure. It is obvious that such a measurement procedure would produce an invariant result. No matter what frame of reference is used, the difference between the readings of clock a when it starts and stops will be the same. This measurement procedure is uninteresting. Let us not use it. Instead...

Let the fellow using frame B lay out rulers at rest in his frame to establish a three dimensional coordinate grid. He could then use clocks at rest in his frame and synchronized in his frame to extend this to a four dimensional coordinate grid. He could then note the coordinate position and time (x_i, y_i, z_i, t_i) at which clock a starts ticking. He could note the coordinate position and time (x_f, y_f, z_f, t_f) at which clock a stops ticking.

The fellow in frame B could then calculate $\sqrt{(t_f-t_i)^2 - (x_f-x_i)^2 - (y_f-y_i)^2 - (z_f-z_i)^2}$ (all using coordinates from frame B).

Now we have an interesting question: Is this number equal to the difference between the starting and stopping times read directly from the face of clock a?

The claim is that yes, this calculated figure will be the same regardless of what frame of reference is used to do the measurement and will match the difference between the starting and stopping times read directly from the face of clock a.

For example, the fellow in frame B could measure a five second elapsed time and a three light-second displacement for clock a. He would then calculate a four second proper time.

 

[sweet springs]

Hi. World interval between the events is invariant. Everybody in any coordinate can observe it using clocks and measures and gets the same value. Among them one observer is privileged to observe world interval only by clock without measure. The events happen at the same place for him so he does not have to measure space between the events. Zero. In respect to him invariant world interval is called proper time. Proper time is time or coordinate time for him but square root of time^2 - space^2 for all the others.

Analogy in 2d geometry. Unit circle in xy plane
$x=cos\theta$
$y=sin\theta$
By rotation we can live in the coordinate where X=1,Y=0.
In respect to this simplicity let us call the radius of circle Proper x.
$Proper \ x = \sqrt{X^2} =\sqrt{x^2+y^2}=1$
Best.

 

[Ibix]

Among them one observer is privileged to observe world interval only by clock without measure.

...assuming the events are time-like separated. If they're null- or space-like separated there is no observer who can do this.

 

[PeterDonis]

Among them one observer is privileged to observe world interval only by clock without measure.

Somewhat pedantic note: this is only true if you restrict to observers traveling on inertial worldlines (i.e., in free fall). If you allow accelerated worldlines there are an infinite number of possible observers whose worldlines pass through the same two timelike separated events.

(For completeness: it's also only true in flat spacetime, even if you restrict to inertial worldlines.)

 

[Herman Trivilino]

I am sorry if you think that, but I am not trying to make things up but to explore what they mean.

You are attempting to create descriptions, no? That's what I see you doing. But your descriptions never focus on the phenomena, just on the terms that physicists use to describe phenomena.

Physics is usually taught by asserting generalizations and then showing some specific cases of how each generalization applies. It is better, in my opinion, to start with the specifics, which are the phenomena, and then present the generalizations. You seem to be skipping over the specifics and going right to the generalizations.

 

[Dale]

I am sorry if you think that, but I am not trying to make things up

Then please stop trying to make nonstandard types of diagrams. You seem to go out of your way to draw everything possible besides a standard spacetime diagram. You should spend that effort learning and drawing standard spacetime diagrams.

 

[Grimble]

Hi. World interval between the events is invariant. Everybody in any coordinate can observe it using clocks and measures and gets the same value. Among them one observer is privileged to observe world interval only by clock without measure. The events happen at the same place for him so he does not have to measure space between the events. Zero. In respect to him invariant world interval is called proper time. Proper time is time or coordinate time for him but square root of time^2 - space^2 for all the others.

Thank you @sweet springs, this is exactly what I have been trying to say!

 

[Grimble]

Note that proper time is only defined for time-like worldlines and these cannot curve back on themselves in flat spacetime. So no knots. The rest is fine.

But being a single dimension it cannot curve or have direction, it is only the measurement along the length of the spaghetti. Flat spacetime has no relevance to the experiment.
That is why (it seems to me) that proper time is invariant, because it is only measure along the time coordinate, otherwise the location and motion of other observers would alter the measurement...

 

[jbriggs444]

But being a single dimension it cannot curve or have direction, it is only the measurement along the length of the spaghetti. Flat spacetime has no relevance to the experiment.

A one dimensional trajectory cannot have intrinsic curvature. It can have extrinsic curvature and direction. Whether that is relevant to an experiment depends on the experiment.

Extrinsic curvature is the curvature of the trajectory in the context of the larger multi-dimensional space within which it is embedded. For instance, the world-line of a clock in our four-dimensional universe. Or a strand of cooked spaghetti on your plate.

 

[Grimble]

It's the time measured between two events along a worldline that passes through those events. Thus two observers may experience two different times between a pair of events if they follow different worldlines. That's the resolution of the twin paradox, in one line.

Yes, that is a good way to express it, time measured along a worldline; yet the time measured along a worldline depends on who is measuring the world line: if it is the observer holding the 'clock' the time is the measure of the time difference at a single point - the elapsed time at that location for the observer present at that location. Whereas, measured by another observer for whom there is distance between the events, that must surely be coordinate time(?) as it involves differences in the other (spatial) dimensions. Now it takes time to traverse the extra distance, so there there must be a greater time measured by the distant observer between the same two points (events?) on the longer worldline.
The proper time measured along a worldline by the present observer, between two events at the same location will be the time difference displayed on that clock = τ, (measured in a single dimension); while the timelike Spacetime interval, measured by a distant observer relative to whom the clock is moving, t', is:
t'² = (ct)² - x²
t'² = (ct)² - (vt)²
t' = t√(1 - v²/c²)

There cannot be different proper time measurements on different worldlines, for the propertime difference between two events is invariant; in the same way that the spacetime interval between two events is invariant. They are both measurements solely of the time for an observer to pass through both events and there will be a frame in which that observer will be at rest. That is why, in any frame in which the observer travels between the events, the distance traveled is subtracted from the time measured by that observer, in order to take it out of the invariant measure.

 

[Grimble]

This piece needs some additional details. Indeed, proper time is invariant. It is the same regardless of what reference frame is used to measure it. But "measure" means that there has to be some physical procedure to come up with the measurement. How does a fellow using reference frame B "measure" the proper time between when clock a at rest in frame A starts and stops ticking?

A fellow at rest in frame B could measure it by using a telescope and copying down the readings of clock a when it starts and stops. He could subtract one from the other and say "there -- that's my measurement of proper time". That would be a valid measurement procedure. It is obvious that such a measurement procedure would produce an invariant result. No matter what frame of reference is used, the difference between the readings of clock a when it starts and stops will be the same. This measurement procedure is uninteresting. Let us not use it. Instead...

Let the fellow using frame B lay out rulers at rest in his frame to establish a three dimensional coordinate grid. He could then use clocks at rest in his frame and synchronized in his frame to extend this to a four dimensional coordinate grid. He could then note the coordinate position and time (x_i, y_i, z_i, t_i) at which clock a starts ticking. He could note the coordinate position and time (x_f, y_f, z_f, t_f) at which clock a stops ticking.

The fellow in frame B could then calculate $\sqrt{(t_f-t_i)^2 - (x_f-x_i)^2 - (y_f-y_i)^2 - (z_f-z_i)^2}$ (all using coordinates from frame B).

Now we have an interesting question: Is this number equal to the difference between the starting and stopping times read directly from the face of clock a?

The claim is that yes, this calculated figure will be the same regardless of what frame of reference is used to do the measurement and will match the difference between the starting and stopping times read directly from the face of clock a.

For example, the fellow in frame B could measure a five second elapsed time and a three light-second displacement for clock a. He would then calculate a four second proper time.

Yes that is what I am saying. Thank you.

 

[sweet springs]

Hi.

t'2 = (ct)2 - x2
t'2 = (ct)2 - (vt)2
t' = t√(1 - v2/c2)

Square of world intervals is
c^2\tau^2=c^2t^2-x^2=c^2t'^2-x'^2=c^2t''^2-x''^2=...
\tau=t\sqrt{1-\frac{v^2}{c^2}}=t'\sqrt{1-\frac{v'^2}{c^2}}=t''\sqrt{1-\frac{v''^2}{c^2}}=...
where $\tau$ is proper time between the events, (t,x),(t',x'),(t'',x''), are coordinates of inertial frames of reference..v,v',v'',... are velocity of the clock ticking $\tau$ in these frames of reference. Obviously
\tau < t,t',t'',...Best.

 

[jbriggs444]

yet the time measured along a worldline depends on who is measuring the world line

Time measured along a worldline is invariant. Delta time assessed between the endpoints depends on a coordinate system and need not match the time measured along any particular worldline between those endpoints.

 

[sweet springs]

Along a world line all the observers measures time coordinate and space coordinate.

I would rather say world interval of two events is invariant. In frame of reference where two events happen at the same place, in another words the worldline call is timeline, invariant world interval is time interval.

 

[Dale]

yet the time measured along a worldline depends on who is measuring the world line:

It would help if you did not say "the time" but were specific and said either "the proper time" or "the coordinate time".

Coordinate time, $\int_P dx^0$, does depend on who is measuring but proper time, $\int_P \sqrt{dx_{\mu} dx^{\nu}}$, does not.

 

[Herman Trivilino]

There cannot be different proper time measurements on different worldlines, for the propertime difference between two events is invariant;

If we look at the twin paradox as an example, different amounts of proper time elapse between the same two events. Each twin's worldline determines the amount of proper time experienced by that twin, and hence his age. Each twin will agree about how much both twins have aged, even though they aged different amounts. That is the meaning of invariant. Being invariant does not mean that each twin ages the same amount.

 

[Grimble]

Hi.

Square of world intervals is
c^2\tau^2=c^2t^2-x^2=c^2t'^2-x'^2=c^2t''^2-x''^2=...
\tau=t\sqrt{1-\frac{v^2}{c^2}}=t'\sqrt{1-\frac{v'^2}{c^2}}=t''\sqrt{1-\frac{v''^2}{c^2}}=...
where $\tau$ is proper time between the events, (t,x),(t',x'),(t'',x''), are coordinates of inertial frames of reference..v,v',v'',... are velocity of the clock ticking $\tau$ in these frames of reference. Obviously
\tau < t,t',t'',...Best.

Like this?

where it shews how your calculations give, the same proper time for each scenario.

 

[Dale]

The problem with plotting proper time as the vertical axis and x as the horizontal axis is that x is a coordinate and proper time is not. The resulting plot is not a spacetime plot and the points on that plot are not events.

I again recommend that you learn to make and understand the standard spacetime diagrams instead of trying to make up your own stuff.

 

[Grimble]

OK. There have been many occasions where I have been asked just what my problem is and after much thought I think the simplest way to elucidate it is like this:
Everything about relativity and time dilation, proper time, coordinate time, world lines etc. makes perfect sense; however ...

Take a moving clock with a two second tick - (e.g. a light clock with the mirrors 1 light second apart)
It travels for 10 seconds at 0.6c, traveling 6 light seconds, from event 1 to event 2.
This inertial clock will tick 5 times in that journey. The light will be reflected from the distant mirror 5 times and will arrive back at the clock's base, for the fifth time at event B. So we know that, measured from the frame of the clock, 10 seconds elapses between event 1 and event 2; that the clock travels inertially for 10 seconds covering a distance of 6 light seconds at 0.6c.
However, as shewn in this diagram from Wikipedia, where in our scenario
Δt = 2 seconds;
Δt' = 2.5 seconds
1/2 v Δt' = .75 light seconds which gives us 7.5 light seconds between events 1 & 2 in 12.5 seconds.
Measured from a stationary frame the traveling clock takes longer and travels further.

Now the problem that I have is that it is the same journey between events 1 & 2.
10 seconds measured for the inertial clock by the observer traveling with the clock is proper time - time measured between two events on the clock traveling inertially between those events.
12.5 seconds measured by the stationary observer measuring the time passing for the moving clock in the stationary frame.
In the Lecture in which he introduced his Spacetime theory, when discussing length contraction, Minkowski said

Space and Time[/B] (1920)]This hypothesis sounds rather fantastical. For the contraction is not to be thought of as a consequence of resistances in the ether, but purely as a gift from above, as a condition accompanying the state of motion.

The distance the clock travels is the same = 6 light seconds.
The proper time, measured by/recorded on, the clock between the events is 10 seconds. (specified in the description)
The coordinate time for the moving clock is longer = 12.5 seconds measured by the observer at rest in the stationary frame. This is the total of the time measured internally by the clock, the proper elapsed time, and the time taken to travel the distance from event 1 to event 2. (Added as vectors - simple pythagorean triangles)

 

[Vitro]

OK. There have been many occasions where I have been asked just what my problem is and after much thought I think the simplest way to elucidate it is like this:
Everything about relativity and time dilation, proper time, coordinate time, world lines etc. makes perfect sense; however ...

Take a moving clock with a two second tick - (e.g. a light clock with the mirrors 1 light second apart)
It travels for 10 seconds at 0.6c, traveling 6 light seconds, from event 1 to event 2.
This inertial clock will tick 5 times in that journey. The light will be reflected from the distant mirror 5 times and will arrive back at the clock's base, for the fifth time at event B. So we know that, measured from the frame of the clock, 10 seconds elapses between event 1 and event 2; that the clock travels inertially for 10 seconds covering a distance of 6 light seconds at 0.6c.

That's all wrong. The clock above is STATIONARY, it doesn't travel anywhere. It just stays in one place and counts 10 seconds (proper time). It has no speed. Both events 1 and 2 happen at the same place, same x coordinate.

However, as shewn in this diagram from Wikipedia, where in our scenario
Δt = 2 seconds;
Δt' = 2.5 seconds
1/2 v Δt' = .75 light seconds which gives us 7.5 light seconds between events 1 & 2 in 12.5 seconds.
Measured from a stationary frame the traveling clock takes longer and travels further.

Travels farther? Farther than what?

Now the problem that I have is that it is the same journey between events 1 & 2.
10 seconds measured for the inertial clock by the observer traveling with the clock is proper time - time measured between two events on the clock traveling inertially between those events.

Again, from the clock's perspective there is no travel, it just counts a time interval between two events happening at the same place.

12.5 seconds measured by the stationary observer measuring the time passing for the moving clock in the stationary frame.
In the Lecture in which he introduced his Spacetime theory, when discussing length contraction, Minkowski said
The distance the clock travels is the same = 6 light seconds.

There is no frame in your setup where the clock travels 6 light seconds. There's only one travel distance of 7.5 light seconds.

The proper time, measured by/recorded on, the clock between the events is 10 seconds. (specified in the description)
The coordinate time for the moving clock is longer = 12.5 seconds measured by the observer at rest in the stationary frame. This is the total of the time measured internally by the clock, the proper elapsed time, and the time taken to travel the distance from event 1 to event 2. (Added as vectors - simple pythagorean triangles)

 

[sweet springs]

Hi.

Like this?

No like the first figure in your OP.　Like　

in https://en.wikipedia.org/wiki/Minkowski_diagram　　Best.

 

[sweet springs]

Hi.

However, as shewn in this diagram from Wikipedia, where in our scenario

\delta t'=\frac{2D}{c}=\frac{2\sqrt{L^2+\frac{1}{4}v^2 (\delta t')^2}}{c}
solving this
(\delta t')^2= \frac{\frac{4L^2}{D^2}}{1-\frac{v^2}{c^2}}
\delta t'= \frac{\delta t}{\sqrt{1-\frac{v^2}{c^2}}}
time dilation was derived.

 

[PeterDonis]

Take a moving clock with a two second tick

You've already created two problems here.

First, you said "a moving clock", but "moving" is frame-dependent. In which frame is the clock moving?

Second, you didn't properly specify what a "two second tick" means. There are two ways of doing that, and your basic problem might be that you are failing to keep them conceptually distinct:

(1) You can define a "two second tick" as two seconds of proper time along the clock's worldline. (Strictly speaking, it would be the worldline of mirror A, since that's the mirror where you are counting ticks.)

(2) You can define a "two second tick" as two seconds of coordinate time in an inertial frame in which the clock is at rest. (This assumes that the clock is moving inertially, which you also should have specified, for completeness and clarity.)

The reason these two things have to be kept conceptually distinct is that (1) is an invariant but (2) is not. The coordinate time elapsed for one tick of the clock is frame-dependent. But the proper time along the clock's worldline is an invariant number, the same in all frames--although the mathematical expression for this invariant number looks different in different frames when you fill in the details.

It travels for 10 seconds at 0.6c, traveling 6 light seconds, from event 1 to event 2.
This inertial clock will tick 5 times in that journey.

No, it won't. It will only tick 4 times. Here's why:

Since you said the clock travels for 10 seconds and covers 6 light seconds at 0.6c, you must mean 10 seconds of coordinate time in the frame in which the clock is moving at 0.6c. But the clock does not tick once every two seconds of coordinate time in that frame. To figure out how many times the clock ticks in 10 seconds of coordinate time in this frame, you need to figure out the proper time along the clock's worldline between the two events you've specified. That proper time will be $\sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$ seconds, or 4 ticks.

I suggest that, before you post anything else at all on this subject, you read the above again and again and think about it until you understand why it is true. The rest of your post just repeats this same basic error, as have your previous posts in this thread. This thread has already gone on for far too long trying to explain something this basic to you.