Epstein Spacetime Diagrams

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Epstein Spacetime Diagrams

Postby BurtJordaan on July 21st, 2013, 6:13 am 

In his book "Relativity Visualized", Lewis Carroll Epstein introduced a kind of spacetime diagram called a Space-Proper-time diagram. It has a few issues and is hence not often used in the literature, but for a quick grasp of special relativity theory, there is none better. In fact, it does just the thing that Dave_Oblad was searching for in this reply on another thread. It actually makes it possible to use Pythagoras' theorem for triangles directly in many special relativity problems. The Epstein diagram looks like this:

EpsteinDiagramBasic.jpg

Where the standard Minkowski spacetime diagram has a decidedly square "look and feel" to it, the Epstein diagram looks somewhat circular, or rather like a conversion from rectangular Cartesian coordinates to polar coordinates. The circle has a radius of cT, i.e. the speed of light multiplied by the time read on a clock that moves radially from the center. It is usual to make the radius 1 spacetime unit, as shown on the diagram. Further, we only show two dimensions (the radius is a two-vector, one of space and one of time), but it can be done for more dimensions. With two dimensions of space and one of time, the circle will be a sphere and for full 3-D space plus time, it will become what is known as a 4-D hyper-sphere.

The blue and red arrows from the origin are rotated relative to each other by an angle determined by the relative speed between two inertial observers. In the diagram above it is V=0.6c, so let us use this in a worked example. The grid shown is arbitrarily chosen to coincide with the blue axes. In one second of blue coordinate time, the blue observer has moved 1 light-second straight up, just like in any spacetime diagram. In the same time, the red observer has moved 0.6 light-seconds to the right and 0.8 light-seconds up, so the red unit vector points to 0.6,0.8 on the grid.

Using Pythagoras, we know that 0.62 +0.82 = 12, so what Dave wanted to happen now actually makes sense: add a space displacement vector to a time displacement vector and you get an invariant spacetime vector. What is more, it can be shown that if we take the origin and the two arrow-head positions each as a separate event in spacetime, the values conform to the Lorentz transformations (LTs) for converting space and time intervals observed in one inertial frame to another inertial frame. If the red frame's clock reads 1s when reaching the position (0.6, 0.8), a clock synchronized in the blue frame will observe the same event at 0.8 seconds. More about this later.

If we flash a light at the origin (when the two observers were collocated there), the light-pulse will go along the the blue +x axis, i.e. 90 degrees. The present red frame will however not observe that light flash to progress along the blue +x axis, but rather along its own red +x axis. This is a peculiar feature of the Epstein diagram - the normal common lightcone is replaced by a light-sphere that "looks the same" for all inertial observers, but they observe different portions of it. Lewis Carroll Epstein has cautioned against taking this concept too literally - he actually called it "a myth" in his book, albeit a very useful one. It helps to visualize special relativity more naturally.

Before I leave you with the above to sink in a little, just one more tantalizing bit of Epstein. Look in the bottom part of the diagram at the dotted line from the red +x, drawn normal to the blue +x axis. Can you see the Lorentz contraction there? It is just the projection of the length of 1 light-second on the red x-axis onto the blue x-axis. Can it really be that simple?

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Re: Epstein Spacetime Diagrams

Postby owleye on July 21st, 2013, 9:59 am 

Given, this, I don't exactly see how Dave's comments would conflict with SR as it is interpreted within the Minkowski diagram. Is there something special about this diagram that explains the twin paradox differently or the notion of a time axis that one travels through? I would agree that time travel is perhaps a confusing term, but that clocks run slower or faster than other clocks in relative motion seems to me to be bedrock. And that 4D space-time is a solution to the perplexing problem of having the speed of light constant regardless of that relative motion.

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 21st, 2013, 11:32 am 

James, I think some of Dave's arguments are due to the imprecision of language - things like "movement through time" and hence "speed of time" are just not well defined. We all look at spacetime diagrams and say something like I did above: "In one second of blue coordinate time, the blue observer has moved 1 light-second straight up, just like in any spacetime diagram. In the same time, the red observer has moved 0.6 light-seconds to the right and 0.8 light-seconds up, so the red unit vector points to 0.6,0.8 on the grid."

Once we have implicitly drawn axes in which the scale for space and time are the same, it becomes almost natural to use these terms. The technically correct terms would have been space interval, time interval and space-time interval, implying the existence of events, not just continuous lines.

You asked: "Is there something special about this diagram that explains the twin paradox differently or the notion of a time axis that one travels through?"

I will get to the twin paradox in a later reply, because the diagram lends itself to a correction of a misconception about time dilation in relativity, but we must first discuss the diagram more fully. Otherwise the essence may get lost in the argument.

Clocks "running slower and faster" perhaps reinforce the idea that there is a "speed of time", while relativity avoids that and goes to great lengths to teach us that time is relative and different observers measures different time intervals between the same two events, while all will agree on the space-time interval between the same two events. If someone wants to interpret this as a constant progress (or even speed) through time, so be it, but I have no idea how to define that speed...

Be this as it may, the Epstein spacetime diagram is an excellent tool for visualizing these principles.
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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 22nd, 2013, 4:20 am 

BurtJordaan wrote:I will get to the twin paradox in a later reply, because the diagram lends itself to a correction of a misconception about time dilation in relativity, but we must first discuss the diagram more fully. Otherwise the essence may get lost in the argument.

The Epstein diagram only works intuitively for unaccelerated (inertial) observers in flat space (no gravity).

Image

Two purely inertial observers are completely equivalent in the eyes of relativity and hence the woldlines of the observers are shown as equal in length and with the same scale in the original diagram above. It becomes very tempting to declare that this shows that the "spacetime movement" for all inertial observers are the same in the same time, or alternatively that "time progresses at the same rate" for all inertial observers in free space.

However, as soon as we take any one of the observers out of the "purely inertial state", the diagram is no longer so intuitive in that respect. An example is the "twin-paradox", where one of the two inertial observers must change inertial frame at least once (at the turnaround point).

EpsteinTwinParadox.jpg

Red must accelerate in order to turn around and head back. By choosing very large spacetime scales for the diagram, one can make the turnaround period negligible in terms of the total picture, so it is essentially just a switch of inertial frames by Red. If Red returns at the same speed, the naive impression would be that Red's clock will read one second when it passes Blue in space. SR and the Epstein diagram tells us a different story - the Red clock will read 0.8 sec when it passes the position of Blue. It is irrelevant whether Red "stops" at blue or just fly past. They are momentarily in the same location in space and may read each others clocks directly.

It is rather intuitive that the two 0.5 unit spacetime vectors are now adding up at an angle and that this may have an influence on what time they represent. A rather difficult question to answer is: what did the Red clock read when that turnaround was made? The fairly obvious answer is that it must have been 0.4 seconds, half the final 0.8 seconds. But, what if that turnaround was never made? Assuming there is some sort of beacon there, what would the red clock have read when it passed that beacon? Here I mean a reading by Red herself, or by another observer sitting at the beacon, reading Red's clock as she passed.

It is a different experiment than the original one and the answer is that we cannot know the answer, unless we specify a number of extra things. If Red was originally stationary with blue and then accelerated quickly to the constant V=0.6c, the answer will be 0.4 seconds. Red has then experienced more total action over the time and will have recorded less elapsed time than what Blue did.

If Red happened to be just an inertial observer that flashed by Blue and they both zeroed their stopwatches at that instant, SR cannot tell,** because all inertial observers are treated identically. Who is then to say which one experienced more action? There are variants of this situation that become even more difficult to answer. More about that later.

The simple lesson here is that we must be careful in not drawing conclusions on elapsed times, unless the conditions are very carefully defined.

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** Edit: This is somewhat controversial. Way back in 2009 we had a thread that ran for 16 pages without reaching complete resolution. I do not want to divert this thread into that direction right now.
Last edited by BurtJordaan on July 23rd, 2013, 2:36 am, edited 1 time in total.
Reason: Added endnote
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Re: Epstein Spacetime Diagrams

Postby owleye on July 22nd, 2013, 8:36 am 

I personally could not agree more with what you say here. And, as you've emphasized many times, aging cannot be understood merely from the arbitrariness of the coordinate systems being used. (I'm paraphrasing here, so I can be mistaken.)

In any case, my most recent thought on this is that there is a distinction between SR in its abstract formulation and one that is concrete in nature where actual experiments can be carried out that are capable of testing the theory. In its abstract formulation certain principles become evident that help in understanding and explaining evidence that was heretofore unexplained. In other words it amounts to a general explanatory framework. In concrete situations, what is needed is to pin down or anchor that framework in such a way that actual results can be obtained that confirm or disconfirm what the theory predicts. It isn't so much about objects in relation to each other, though this can be a by-product, if you like, but rather in figuring out causal relations, in so far as they become specific explanations of events. Basically the general framework gives us that part of the explanation that becomes the context in which the experiments are conducted, that part which is true for all observers in relative motion, so as to get down to those things which are exclusive to the individual objects under discussion (or the set of events in need of an explanation).

(As usual the above is a work in progress for me, and I'm working toward one that will eventually become (I think) a co-variant formulation, one that I have in mind from what my earlier cited Michael Friedman's book is informing me of. I welcome all criticism.)

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Re: Epstein Spacetime Diagrams

Postby Dave_Oblad on July 22nd, 2013, 6:29 pm 

Hi Jorrie,

While I'm trying to digest this.. a funny sidebar: My Quote was from "Lewis Carroll", the Author of "Alice through the looking Glass".. etc. His real name was "Charles Dodgson".

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 22nd, 2013, 11:33 pm 

Lol! I thought you just used half the name...

I was obviously just watching future time and did not look around in space ;-)

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 23rd, 2013, 3:03 am 

owleye wrote:In any case, my most recent thought on this is that there is a distinction between SR in its abstract formulation and one that is concrete in nature where actual experiments can be carried out that are capable of testing the theory. ...


Yes, it is not always possible to construct a test that measures exactly what we can calculate in SR. In real experiments I guess we decide on what is technologically achievable and then use the theory to predict what we should measure in such a test (and calculate the precision that can be achieved).

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 24th, 2013, 12:37 pm 

BurtJordaan wrote:In his book "Relativity Visualized", Lewis Carroll Epstein introduced a kind of spacetime diagram called a Space-Proper-time diagram.

In http://www.sciencechatforum.com/viewtop ... 40#p237240, SSDZ replied:
SSDZ wrote:There is nothing new in this "Epstein Spacetime" practically, the approach is well known as "Euclidian relativity" – see, e.g.:
Newburgh, R. G., Phipps, T. E.: A space-proper time formulation of relativistic geometry. Air Force Cambridge Res. Lab. Physical research paper No. 401, Nov. 1969.


I just want to make it absolutely clear that the "Euclidian relativity" that SSDZ referred to is not what Epstein diagrams support. There are aspects of similarity that may date back to the 1960's, but AFAIK, Epstein first published the graphical utilization for educational purposes in his 1981 book (fully supporting SR, not some variant of it). There is an excellent website covering virtually everything in his book (plus some, if I remember correctly).

In the rest of this thread, I will not cover what is already written there, but rather attempt to show some of the pros and cons of the Epstein when compared to Minkowski diagrams.

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Re: Epstein Spacetime Diagrams

Postby Dave_Oblad on July 24th, 2013, 11:40 pm 

Hi Jorrie,

Ok.. digested..lol. You have already pointed out that this diagram has limitations, but just for giggles, if Red was a Photon then Reds Dot would be at the 3:00 mark on the circular grid, meaning that, from Red's point of view (Red's hypothetical clock) zero time has lapsed.

You can see my original confusion in that, by the first diagram on this thread, that there is a 0.2 second gap between Blue and Red. I took that to mean Blue was 0.2 seconds ahead of Red (on some Temporal Axis) and thus Red would never be able to catch up with Blue (due to some kind of Temporal Distance). Now I know what Time-Dilation is.. I'm ready to move on. (with sanity intact)

Thanks again Jorrie.. for straightening me out on the "Absolute Time Vs Special Relativity" thread: http://sciencechatforum.com/viewtopic.php?f=2&t=24945

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 25th, 2013, 12:46 am 

Yes Dave, you have spotted one of limitations of the Epstein: it actually works correctly only for time-like intervals (measured by material clocks), but excludes photons (null or light-like intervals) and all space-like intervals, where not even photons can be present a both events.

To keep our sanity, we should let Red be a particle that happens to pass the origin at a speed approaching c (say 0.999c) relative to Blue. That particle will for all practical purposes be at the 3 o'clock position on the circle with Blue at 12 -o'clock. So per Blue, the particle's clock will read zero time interval, but per Red, its own clock reads one unit and it is Blue that reads zero time interval. The whole diagram has then just been rotated 90 degrees anti-clockwise.

Now you can see why we cannot put photons on the diagram - they do not carry clocks that tick, while massive particles have them built-in (decay times). We can show photon paths (different for each frame), but we cannot attach time and distance scales to those paths. This is also true for Minkowski diagrams, except that the lightcone provides a common light path for all frames (but still no scale).

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 25th, 2013, 7:27 am 

I have left one point hanging in the air in the opening post.
BurtJordaan wrote:Before I leave you with the above to sink in a little, just one more tantalizing bit of Epstein. Look in the bottom part of the diagram at the dotted line from the red +x, drawn normal to the blue +x axis. Can you see the Lorentz contraction there? It is just the projection of the length of 1 light-second on the red x-axis onto the blue x-axis. Can it really be that simple?


As this now had a little time to "sink in", on with interpreting Epstein diagrams.

Image

Can it really be that simple? Yes, it can if we tread with caution. What is 1 unit along the +x axis for Red will be observed as 0.8 units along the +x axis by Blue. This is the standard Lorentz transformation for lengths.

Physically, it may be interpreted as follows: the events at x = 0.8, cT = 0 and at x = 1.0, cT = 0 are spatially collocated in the reference frame, with the red event happening before the blue event, at time x = 0.8, cT = -0.6. Had we chosen Red as the reference frame and Blue was moving to the left, exactly the opposite would have been true - essentially, the vectors would have been rotated so that red lines up with the grid. Not that the grid has any physical meaning. We have to guard against viewing it as some absolute spacetime; it represents nothing of that sort - it is just convenient 'paper' for drawing charts.

This seems to indicate that Lorentz contraction is no more than a projection from one inertial frame onto another and not something physical happening to any of the objects. This interpretation is often controversial, because it does not define what "something physical" means and different people may have different definitions.

More in a later installment.

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Re: Epstein Spacetime Diagrams

Postby owleye on July 25th, 2013, 8:56 am 

It took me a bit to see your last point (and I may not be getting it), however, some of my early confusion came from events written in colors in the same colors as observers are drawn in the diagram. I'm hoping this wasn't supposed to be intentional, but it may be, and that there is a connection between the color representations that I'm missing. Could you clarify this?

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 25th, 2013, 9:26 am 

Any confusion it may have been caused was not intentional, but the use of corresponding colors was. Red coordinate values are as measured in the red coordinate system, meaning there is a second (invisible) rotated set of grid lines that line up with the red vectors. The red x-axis represents all points that are at cT = 0 for red - it is a line of simultaneity for red (and so are any line drawn parallel to the red x-axis, just for a different time). Hence x = 1.0, cT = 0 sits where the red +x axis intersects the unit circle.

I should have made these points clearer by labeling the diagram better - sorry about that.

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Re: Epstein Spacetime Diagrams

Postby owleye on July 26th, 2013, 8:17 am 

Ok. I should be able to get a handle on this, but for some reason I'm having trouble. Here you say "... the events at x = 0.8, cT = 0 and at x = 1.0, cT = 0 are spatially collocated in the reference frame, with the red event happening before the blue event, at time x = 0.8, cT = -0.6."

I should understand this to mean that from Blue's perspective, there is some event at x = 0.8, when his clock reads 0, and there is another event, co-located with it, from Red's perspective, at x = 1.0, when her clock reads 0. To make this more comprehensible, I'm going to assume they are one and the same event, say E. So, when we start our clocks, at the origin of the drawing in which the two observers cross paths, the event, E, can be said to have a difference of value for that event's spatial separation of 0.2 between the two observers, Blue's perspective would have it shorter than Red's. However, neither Blue nor Red are able to make that determination at their joint cT = 0, since at that time the interval is space-like. However, we can determine when Blue and Red observed E, by the projection onto the other's frame. Blue will have earlier observed E when his clock had reached -0.6, while Red will observe E when her clock reads something else (looks like it would be around +0.6). However, I made this latter determination by mentally projecting a line orthogonal to Red's cT axis from where event E takes place on the grid. However, it appears that what I'm supposed to do is take the projection from Red's perspective when her time is at cT = 1.0 onto the Blue spatial axis, which might yield the same value. In that case, clearly, Blue will have observed E before Red does, Blue before they cross paths while Red afterwards. This will probably be the case even if I switch reference frames, though I'm not sure. I have the strong feeling I'm not following the rules for the drawing, since you conclude the opposite.

My apologies for not coloring the perspectives. I hope I'm clear enough any way. In any case, I'm giving the diagram a fair amount of thought.

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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 26th, 2013, 9:34 am 

owleye wrote:I should understand this to mean that from Blue's perspective, there is some event at x = 0.8, when his clock reads 0, and there is another event, co-located with it, from Red's perspective, at x = 1.0, when her clock reads 0. To make this more comprehensible, I'm going to assume they are one and the same event, say E.

No, I'm afraid you can't assume that they are the same spacetime event; they are collocated for Blue, but they happen at different times. For Red, they are neither collocated, nor simultaneous, so we should call them events E1 and E2. This is throwing your following discussion a bit off track here and there.

It is correct to project each event's position at right angles to the other coordinate system and so determine the Lorentz transformation to the other frame. The temporal order of the two events will remain the same, irrespective of which frame you consider the events in. I may have confused the issue when I remarked:

Had we chosen Red as the reference frame and Blue was moving to the left, exactly the opposite would have been true - essentially, the vectors would have been rotated so that red lines up with the grid.


I had in mind a mirror image of the diagram, with the two colors swapped, but unfortunately that would have meant two different events, so the remark is not true for the temporal order. Sorry about that. The idea was simply to demonstrate the Lorentz contraction and the absence of a preferred frame. It is normally clearer when we rotate both axes relative to the grid, rather than drawing another (sloped) grid on the same diagram.

So, maybe you should reconsider your post with the two distinct events in mind.

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Re: Epstein Spacetime Diagrams

Postby owleye on July 26th, 2013, 10:23 am 

Ok. Thanks. However, I'll have to hold off on re-thinking it, as I'm about to post a bit more on the principle of equivalence in my topic. I think there may be a semantical issue that I'd like to clear up. And it hampers my ability to understand Greene's sequence of steps that lead to the warping of space-time. Indeed, I don't think he's all that clear about it. Rather instead, he seems to be mixing several strands of thought to arrive at the conclusion. It is not a seamless transition, the way I read it.

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Re: Epstein Spacetime Diagrams

Postby phyti on July 26th, 2013, 3:13 pm 

Others have proposed the idea of motion through time, i.e. a literal 'time dimension'.
The Epstein drawing doesn't appear to provide a method of light interaction, unless the pre-relativity notion of instantaneous light speed is used. Obviously it falls way short of the Minkowski drawing for purposes of education and analysis.
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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on July 26th, 2013, 3:27 pm 

phyti wrote:The Epstein drawing doesn't appear to provide a method of light interaction, unless the pre-relativity notion of instantaneous light speed is used.


You misinterpret the diagram. It is not called a "space-proper-time" diagram for no reason - the proper-time along a photon path is zero. Any time-like interval analysis can be done as well on an Epstein diagram as it can be done on a Minkowski diagram; sometimes it is much more intuitive than Minkowski. This web-site should convince you if I cannot.

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Re: Epstein Spacetime Diagrams

Postby phyti on July 29th, 2013, 2:48 pm 

BurtJordaan wrote:
phyti wrote:The Epstein drawing doesn't appear to provide a method of light interaction, unless the pre-relativity notion of instantaneous light speed is used.


You misinterpret the diagram. It is not called a "space-proper-time" diagram for no reason - the proper-time along a photon path is zero. Any time-like interval analysis can be done as well on an Epstein diagram as it can be done on a Minkowski diagram; sometimes it is much more intuitive than Minkowski. This web-site should convince you if I cannot.

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The Epstein site is quite extensive, it will take a while to read it.

[img]
light%20clock.gif
[/img]

Moving in time is a literal interpretation of the space and time merger proposed by Minkowski. It’s also another perspective of the moving light clock.
Light is emitted from a source in a direction p, perpendicular to x, the direction of motion, and reflects from a mirror a distance d=1, to a detector/counter. For the clock to function, the photon path must have an x and p component. The x component compensates for the motion of the clock at speed v. The p component becomes the active part of the clock. Since the photon speed is constant, its path in any direction generates a circular arc for the 90º between the p axis and x axis. This means the relative photon speed along p = c*sqrt(1-(v/c)^2) = c/γ, i.e. the clock ticks slower, the faster it moves past an observer.
With vt the x component and pt the p component, it can be rephrased as
1. (vt)^2 + (pt)^2 = (ct)^2, or
2. (object motion)^2 + (light motion)^2 = (light motion)^2, or
3. (object motion)^2 + (object time)^2 = (light motion)^2
Conclusion:
Line 3 is the misconception, equating ‘object time’ to ‘object motion’. The clock moves in a 1-dimensional space, while (simultaneously) the photon moves in a 2-dimensional space. The clock is counting spatial increments of (2γd) which are labeled as ‘time’.
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Re: Epstein Spacetime Diagrams

Postby TinkTank on September 6th, 2013, 4:38 pm 

Here are some applets and animations that use Epstein diagrams (or space-propertime diagrams):

The basic idea how you can show coordinate time, proper time, coordiante length, proper length in one diagram:
http://www.adamtoons.de/physics/relativity.swf

Interactive comparison of Minkowski & Epstein diagrams for the twins scenario:
http://www.adamtoons.de/physics/twins.swf

How gravitational time dilation and the apparent gravitational pull are connected:
http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_spacetime.html

Animated version of the link just above explaining local gravity:



Interactive applet on gravity on a more global scale:
http://www.adamtoons.de/physics/gravitation.swf
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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on September 8th, 2013, 11:51 am 

Welcome to the forum, TinkTank!

These are really cool applets and animations; thanks for posting.

Having read the Epstein book, they make sense to me, but I will look a little deeper and give some more feedback later.

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Re: Epstein Spacetime Diagrams

Postby GrayGhost on October 9th, 2013, 5:55 pm 

Hello Jorrie,

I was looking thru this interesting thread here. Regarding this illustration you had posted, and related comment ...
BurtJordaan wrote:The Epstein diagram (see hyperlink below) ...

http://www.sciencechatforum.com/download/file.php?id=2994&mode=view

Red must accelerate in order to turn around and head back. By choosing very large spacetime scales for the diagram, one can make the turnaround period negligible in terms of the total picture, so it is essentially just a switch of inertial frames by Red. If Red returns at the same speed, the naive impression would be that Red's clock will read one second when it passes Blue in space. SR and the Epstein diagram tells us a different story - the Red clock will read 0.8 sec when it passes the position of Blue. It is irrelevant whether Red "stops" at blue or just fly past. They are momentarily in the same location in space and may read each others clocks directly.


I must disagree, respectfully. Regarding the clock readings at intersection on reuinion ...

SR tells us that RED reads 0.8s-B x γ = 0.8s-B x 0.8 = 0.64s-R if RED changed frames, or that RED reads 0.8s-B / γ = 0.8s-B / 0.8 = 1.0 sec if BLUE changed frames. We assume here that one properly accelerated (changed frames) while the other did not. Or, did I misunderstand you maybe?

BurtJordaan wrote:** Edit: This is somewhat controversial. Way back in 2009 we had a thread that ran for 16 pages without reaching complete resolution. I do not want to divert this thread into that direction right now.


I would like to talk about it further, and see if I can understand your reasoning there. In your leisure, if necessary. Keep in mind that the excepted result of the classic twins scenario, has long been that the non-always-inertial twin will age less than the always-inertial twin on reunion. You made the comment regarding "very large spacetime scales" being used. Are you suggesting that flat spacetime in its grandier (which is curved) will precisely counter the typical predicted twins result of one aging less than the other? By grandier, I mean a reasonable percentage of the cosmos if journeyed, by twin RED ... where any locally flat spacetime is spherical in the collective grander scope of huge roundtrip journeys. Or, did you mean something else?
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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on October 10th, 2013, 1:24 am 

Hello Gray, nice to see you back. :)

GrayGhost wrote:I must disagree, respectfully. Regarding the clock readings at intersection on reuinion ...

I said around that post that the Epstein diagram is not recommended for more than one inertial frame, but have just shown how one can in principle just bend the red worldline back to find the Lorentz transformation for the time component of the away-twin. The problem is that I should then have had two scales on the vertical axis, blue and red. Remember that the speed is 0.6c, so the blue time would be 1.0 and the red time 0.8 units at reunion. Just badly indicated on the diagram.

On the "controversy", recall the context:

Jorrie wrote:If Red happened to be just an inertial observer that flashed by Blue and they both zeroed their stopwatches at that instant, SR cannot tell,** because all inertial observers are treated identically. Who is then to say which one experienced more action? There are variants of this situation that become even more difficult to answer. More about that later.

** Edit: This is somewhat controversial. Way back in 2009 we had a thread that ran for 16 pages without reaching complete resolution.


The reference to "very large spacetime scales" was just to justify making the turnaround to seem instantaneous, by comparison to trip time. No other meaning...

Maybe we should start another thread for the "ABC" scenario with a "time-hand-off" at the turnaround?

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Re: Epstein Spacetime Diagrams

Postby GrayGhost on October 10th, 2013, 6:08 am 

OK Jorrie,

Do you such an ABC handoff scenario graphic available?
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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on October 10th, 2013, 6:39 am 

GrayGhost wrote:Do you [have] such an ABC handoff scenario graphic available?


I just have those that we discussed during the 2010 thread, e.g. here: http://www.sciencechatforum.com/viewtopic.php?f=2&t=13635&start=390#p145115

I think you have also done quite a few of them, which you could either trace on your computer, or I think you may get hold of them again from your posts on the forum, but maybe only if you can see them as attachments.

I recently had a sad loss (break-in theft) of my laptop with one external drive used for backups attached to the damn thing. I have only some old diagrams still available...

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Re: Epstein Spacetime Diagrams

Postby GrayGhost on October 11th, 2013, 2:51 am 

Very sorry to hear about your break-in/loss, Jorrie. That really stinks, as I know how much work you put into your studies, and in support of teaching the theory. I figure your archives are rather large and organised. I wonder if the web can be searched for your laptop's MAC ADDR, to find the house it now sits in, and recover it? Odds are, no thief is going to go thru the trouble to replace the NIC card. You'd think ISP providers have a copy of all active MAC ADDRs, and that they could compare those against a "reported stolen" list.

I'll look thru that prior thread you referenced first, make sure I understand your points well. Refresh a bit. Thanx.
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Re: Epstein Spacetime Diagrams

Postby GrayGhost on October 11th, 2013, 3:22 am 

GrayGhost wrote:Hello Jorrie,

I was looking thru this interesting thread here. Regarding this illustration you had posted, and related comment ...
BurtJordaan wrote:The Epstein diagram (see hyperlink below) ...

http://www.sciencechatforum.com/download/file.php?id=2994&mode=view

Red must accelerate in order to turn around and head back. By choosing very large spacetime scales for the diagram, one can make the turnaround period negligible in terms of the total picture, so it is essentially just a switch of inertial frames by Red. If Red returns at the same speed, the naive impression would be that Red's clock will read one second when it passes Blue in space. SR and the Epstein diagram tells us a different story - the Red clock will read 0.8 sec when it passes the position of Blue. It is irrelevant whether Red "stops" at blue or just fly past. They are momentarily in the same location in space and may read each others clocks directly.


I must disagree, respectfully. Regarding the clock readings at intersection on reuinion ...

SR tells us that RED reads 0.8s-B x γ = 0.8s-B x 0.8 = 0.64s-R if RED changed frames, or that RED reads 0.8s-B / γ = 0.8s-B / 0.8 = 1.0 sec if BLUE changed frames. We assume here that one properly accelerated (changed frames) while the other did not. Or, did I misunderstand you maybe?


Just noticed that I had my gamma and contraction factors backwards there. I meant (of course) to say ...

SR tells us that RED reads 0.8s-B / γ = 0.8s-B / 1.25 = 0.64s-R if RED changed frames, or that RED reads 0.8s-B x γ = 0.8s-B x 1.25 = 1.0s-R if BLUE changed frames. We assume here that one properly accelerated (changed frames) while the other did not. Or, did I misunderstand you maybe?

Anyway, I just wanted to correct that type-O there. That's in relation to this post, about 5 posts back ...

http://www.sciencechatforum.com/viewtopic.php?p=240540#p240540
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Re: Epstein Spacetime Diagrams

Postby BurtJordaan on October 11th, 2013, 4:47 am 

I think there is a miscommunication somewhere. I do not understand your:

SR tells us that RED reads 0.8s-B / γ = 0.8s-B / 0.8 = 0.64s-R if RED changed frames, or that RED reads 0.8s-B x γ = 0.8s-B x 0.8 = 1.0 sec if BLUE changed frames. We assume here that one properly accelerated (changed frames) while the other did not.


At v=0.6c both ways, the time dilation factor is simply 0.8 (γ=1.25), so Blue will record 1.0 units of time at reunion, and Red 0.8 units.

The equivalent LT is:

TBlue = 1s, XBlue = 0.6ls, so TRed = (1s - 0.6ls * 0.6)/0.8 = 0.8s.

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Last edited by BurtJordaan on October 12th, 2013, 10:48 pm, edited 1 time in total.
Reason: Corrected color reference as Gray pointed out.
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Re: Epstein Spacetime Diagrams

Postby GrayGhost on October 12th, 2013, 9:46 pm 

BurtJordaan wrote:I think there is a miscommunication somewhere. I do not understand your:

GrayGhost wrote:SR tells us that RED reads 0.8s-B / γ = 0.8s-B / 0.8 = 0.64s-R if RED changed frames, or that RED reads 0.8s-B x γ = 0.8s-B x 0.8 = 1.0 sec if BLUE changed frames. We assume here that one properly accelerated (changed frames) while the other did not.

At v=0.6c both ways, the time dilation factor is simply 0.8 (γ=1.25), so Blue will record 1.0 units of time at reunion, and Blue 0.8 units.

The equivalent LT is:

TBlue = 1s, XBlue = 0.6ls, so TRed = (1s - 0.6ls * 0.6)/0.8 = 0.8s.

Did you maybe mean RED 0.8 units? Either way ...

It seems that your are assuming apriori that RED arrives at BLUE on reunion with a RED clock reading of 0.8s-R. However, I was responding in reference to this Epstein illustration you had depicted prior here ...

http://www.sciencechatforum.com/download/file.php?id=2994&mode=view

which shows that the BLUE clock definitively reads 0.8s-B on reunion, and its generally presumed they all began at time 0 from the co-located origins. I therefore considered that carved into tablets, and then considered what the RED clock would have to read on reunion (which the figure did not depict). So,

Per the traditionally accepted analysis of the classic twins scenario, the twin that changed frames is the twin that ages the least over the common interval. As such, RED must age less than BLUE over that defined interval. At reunion, the RED clock must read ...

tRed = tBlue / γ = 0.8s-B / 1.25 = 0.64s-R.

I have a feeling you are going to disagree with this, and I am interested in knowing how and why.
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