Why is relativity so hard to learn?[1]

[Dave_Oblad]

Hi Jorrie (or Mitch),

In your last diagram, we see Bob has a Blue dashed line at 0.8 that points at Alice's final Destination of Alpha Centauri and Alice also has a Red dashed line at 0.8 that points at Bob's Alpha Centauri.

Can you explain (in simple terms) the significance of these two dashed lines pointing at each others Alpha Centauri(s)?

Regards,
Dave :^)

[mitchellmckain]

Hi Jorrie (or Mitch),

In your last diagram, we see Bob has a Blue dashed line at 0.8 that points at Alice's final Destination of Alpha Centauri and Alice also has a Red dashed line at 0.8 that points at Bob's Alpha Centauri.

Can you explain (in simple terms) the significance of these two dashed lines pointing at each others Alpha Centauri(s)?

NewDiagram.png

Regards,
Dave :^)

Frankly the whole diagram is simply a way of representing the calculations of lorentz contraction and time dilation as trig functions. It is really nothing but mathematical artifice and the diagrams have no more basis in real physics than this. Accordingly those dashed lines represent those calculations of the the length and time as seen by the other observer in his inertial frame. If you are going to keep it in simple terms then that is the end of the matter.

On the other hand, in more abstract theoretical terms, representing a Lorentz transformation as a rotation is not a bad idea because they really are the same sort of thing. The Lorentz transformation can be taken as a Minkowsky 4 dimensional analog of a Euclidean 3 dimensional rotation.

[BurtJordaan]

Can you explain (in simple terms) the significance of these two dashed lines pointing at each others Alpha Centauri(s)?

Hi Dave, further to what Mitch said, remember that this diagram depicted the situation after the acceleration, when both Alice and Bob were inertial again. By the nature of relativistic spacetime, the situation is the same for both, i.e. it is reciprocal. Both observe Lorentz contraction in the other's distances, but we know there is no physical contraction. The Epstein diagram hammers that idea home much clearer then a Minkowski spacetime diagram. But, as Mitch implied, they both conform to the same math, just differently pictured.

I have used the analogy of the maps of Alice and Bob to be just tilted by an angle, which is determined through the Lorentz transformations. Each would then obviously see the other's distances as scaled down. The real point of the whole presentation is that it is acceleration relative to each other that causes the spacetime structures of each of the participants to look different to the other one. Whether you picture them as relatively tilted spacetimes or just distorted relative spacetimes (through relative motion) is immaterial.

The idea of the thread is not to give any proof of relativity, just to give another tool for visualizing it. And to show that you can predict real observable results by using this visualization (within limits, of course).

[hyksos]

I also happen to know that you still believe that you can detect motion (and speed) through it by comparing clocks that are moving inertially. Well, that idea has been falsified by many an experiment, so why you still cling to it, I don't understand.

..four years later...

[Dave_Oblad]

Hi Hyksos,

Good point Hyksos.

I also happen to know that you still believe that you can detect motion (and speed) through it by comparing clocks that are moving inertially. Well, that idea has been falsified by many an experiment, so why you still cling to it, I don't understand.

Frankly the whole diagram is simply a way of representing the calculations of lorentz contraction and time dilation as trig functions. It is really nothing but mathematical artifice and the diagrams have no more basis in real physics than this. Accordingly those dashed lines represent those calculations of the the length and time as seen by the other observer in his inertial frame.

Ok, so on one hand I'm told we can't detect motion or speed by comparing the clocks of two observers and then we are given diagrams that do exactly that. There is an elephant in the room that's pretty hard to miss.

I understand the concept of being reciprocal, but am I to conclude that at the 0.8 markers both presume the other as having completed the full travel distance? Extra confusing because I thought Bob was essentially not Moving at all.

By Relativity, I can maybe presume that Alice believes she has no Motion and Bob has the slower clock (reciprocal), which will ultimately be proven incorrect when Alice returns to Bob and finds Bob had aged more than her during her trip. Suddenly.. this reciprocal aspect vanishes and the truth is revealed.

But I'm getting ahead of us. So let's focus on my specific question regarding the Diagram:

I'm still confused. I thought the red marker Labeled 0.8 on Alice's distance line to Alpha Centauri meant she had completed 80% of the journey to her destination at a velocity of 60% Light Speed (V/c = 0.6) at the present moment of the diagram. But I still don't understand the relevance of, or significance of, Alice's Red Dashed Line pointing at Bob's Distance coordinate intersecting his Alpha Centauri.

Given the full trip span is 4.4 Light Years for the Distance axis of Start to Alpha Centauri:

Are we saying that at the 80% mark both Alice and Bob believe the other has completed the full distance (4.4 Light Years) between Start and Alpha Centauri? Obviously, Alice knows she still has 20% of her journey remaining (or at least she hasn't arrived at her destination yet), so no way can she perceive that Bob is already 4.4 Light Years behind her.. from her current 0.8 position.

So, obviously I still don't understand.. and need a much clearer English explanation to my Question in Bold just above.

Best wishes,
Dave :^)

[BurtJordaan]

Hi Dave, I'm afraid that you are misunderstanding the whole issue. You have either not read the whole thread, or your absolute-frame mindedness is blocking out the 'elephant in the room'. Maybe Mitch will give it another try, but I'm going to ask you respectfully to either reread the whole thread, or wait until I get to the end of the process (2 or 3 parts away). You may just spot the elephant when you have the whole picture.
--
Regards
Jorrie

[mitchellmckain]

Ok, so on one hand I'm told we can't detect motion or speed by comparing the clocks of two observers and then we are given diagrams that do exactly that. There is an elephant in the room that's pretty hard to miss.

You can only detect relative motion not absolute motion.

I understand the concept of being reciprocal, but am I to conclude that at the 0.8 markers both presume the other as having completed the full travel distance? Extra confusing because I thought Bob was essentially not Moving at all.

Your last statement has no meaning. Bob is not moving relative to some things like his ship and is moving relative to other things. The point is that only the motion relative to Alice effects the calculations we are talking about and any motion relative to something else is irrelevant.

By Relativity, I can maybe presume that Alice believes she has no Motion and Bob has the slower clock (reciprocal), which will ultimately be proven incorrect when Alice returns to Bob and finds Bob had aged more than her during her trip. Suddenly.. this reciprocal aspect vanishes and the truth is revealed.

The problem is that if Alice keeps going and instead Bob goes to Alice then the opposite will happen and it is Alice that will have aged more during the trip. This is not a matter of revealing any truth about anybody aging faster or slower but only about the structure of space-time. It is really about the relativity of simultaneity.

I'm still confused. I thought the red marker Labeled 0.8 on Alice's distance line to Alpha Centauri meant she had completed 80% of the journey to her destination at a velocity of 60% Light Speed (V/c = 0.6) at the present moment of the diagram.

Definitely not!!! These Epstein diagrams are only comparing space-time measurements of two inertial frames. These are not proper space versus time diagrams where distance over time would give you velocity -- as I explained before, it does not! For that you need a Minkowsky diagram with diagonal lines representing the speed of light.

But I still don't understand the relevance of, or significance of, Alice's Red Dashed Line pointing at Bob's Distance coordinate intersecting his Alpha Centauri.

The Alpha Centauri label simply marks the distance of 1 light year in Bob's coordinate system and the diagram shows this measures .8 light years in Alice's coordinate system, thus giving the lorentz contraction of this distance for Alice.

Given the full trip span is 4.4 Light Years for the Distance axis of Start to Alpha Centauri:

But the diagram is for when Alice is 1 light year away from Alpha Centauri according to Bob.

Are we saying that at the 80% mark both Alice and Bob believe the other has completed the full distance (4.4 Light Years) between Start and Alpha Centauri?

No.

Obviously, Alice knows she still has 20% of her journey remaining (or at least she hasn't arrived at her destination yet), so no way can she perceive that Bob is already 4.4 Light Years behind her.. from her current 0.8 position.

So, obviously I still don't understand.. and need a much clearer English explanation to my Question in Bold just above.

You are providing an excellent demonstration why I told BurtJordan that introducing Epstein diagrams in a beginner's explanation of relativity is a bad idea.

[Dave_Oblad]

Hi Jorrie,

I thought I was following perfectly up to a point.

The original Blue diagram, before you rotated it seemed perfectly clear. The Big Red Dot showed that Clock rate was the Vertical Axis marked +cT and the Horizontal Axis was Speed Marked as +X.

The Big Red Dot said that at 0.6c Speed (60% of Light Speed) that Alice's Clock was running at 0.8 Rate (80% normal time).

Then you rotated the diagram.

It seems that you redefined the Horizontal Axis from Speed to remaining Distance of 0.8 (80% complete) and even labeled the end of that Axis with Alpha Centauri.

When did +X stop representing Speed as 0.6c (60% of Light Speed) and became Distance instead.. with a Labeled tag of 0.8 (or 80%) of distance traveled thus far?

Did I misunderstand the original Diagram and the Pre-Rotation marked as 0.6 was Distance to +X as an Unknown Distance?

I'm kinda getting the impression that the Diagram is trying to show that:
1. At Speed = 0.. we get zero Distance with infinite time spent traveling at zero Speed.
2. At Speed = c (Speed of Light).. we get infinite Distance with zero time spent at Speed c (Speed of Light).
That perhaps Rotating the Graph is merely showing the in-betweens of the two extremes of (1) and (2)?

At least that is the persistent impression from the first time you showed me such a Diagram.

In this light, is the 0.8 dashed lines now showing real Distance traveled vs the Time spent traveling and the Time spent traveling is now modified by the Speed in a non-linear fashion?

We all know that Speed and Distance have a linear relationship (normally) or at 100 MPH we can say Distance = 100 Miles in 1 hour of Time or.. in 1 hour of Time we moved 100 miles. But Non-Linear means that at the 100 mile mark, the person in the vehicle will Time the trip taking less than 1 hour.

Again, my confusion was how we tagged the Horizontal Speed axis of +X to a Time Axis of +cT with a fixed distance labeled Alpha Centauri.

I suspect it's because my Brain doesn't want to see Speed and Distance defined on the same line perhaps. (Bob's line pointing to the Blue Labeled Alpha Centauri, with a 0.6 and a 0.8 sharing the same line but with the 0.6 and 0.8 having different definitions.

Again, this is not about Absolutism or any issues with Relativity.. it's only about trying to understand what the Graph Means and I'm starting to see my problem. It's the shortcut of having two different definitions sharing a single line on a Graph that is confusing to me (I think).

Sorry, my Brain is slowing down dramatically over the last few years and this stuff just doesn't process as well as it once did. It's the main reason I'm pushing to retire as soon as possible. It's taking me 3 times as long, and 1000% more errors, to produce something today.. than just 3 years ago (within my Field of Software Design).

For example, 2 years ago I discovered a Bug in one of my programs and located it within about 20 minutes. But before I could fix the bug.. my Boss shifted me to another Project. Last month I revisited the program and forgot where the Bug was at.. and it took me 3 days to track it down this time.

So.. please be patient with me.

Ever read "Flowers for Algernon"? I'm readily identifying with Charlie in that story. I'd put an "LOL" here but it's not that funny to me (it sucks).

Oh, I see I cross posted with Mitch. I'll submit this now (asis) and read what Mitch has to say.

Best Regards,
Dave :^)

[BurtJordaan]

You are providing an excellent demonstration why I told BurtJordan that introducing Epstein diagrams in a beginner's explanation of relativity is a bad idea.

Mitch, thanks for fielding Dave_O's 'curved balls' - we are at it for the last 5+ years and I have made some progress along the traditional line with Dave, but every so often, we are more or less back at square one.

You may be right on Epstein for beginners, and people have also told me that introducing acceleration in a beginners (popular) explanation is a relatively bad idea. But I will attempt to show that the combination of the two in the fashion I'm doing here, with a few numerical values on scenarios that beginners can cope with (and no math, since this is the philosophy side), can be superior to the traditional methods used in popular courses and books.

Yes, Dave is seemingly confused by Epstein, but then there are others that seem to be coping. The final proof of the pudding should come from the diners. If it fails, we may have to try another approach. My problem is that I progress too slowly and people may loose sight of the complete picture. It it was a paper, it would have had a comprehensive synopsis and length introduction/conclusions. Maybe we will get there, eventually...

[hyksos]

If I had to explain special relativity to a high school kid, or explain it to someone in a loud bar, or in a bowling alley. The first thing I would say, to keep it ultra-simple, is to tell the person :

Velocities do not add.

You travel 60% of the speed of light on the back of a rocket, and fire a bullet in your direction of your motion at 45% of the speed of light. If velocities add, the bullet should be going 105% the speed of light. But that does not happen, because velocities do not add.

{you educated turtleneck sweaters call this the "Galilean Transformation" i.e. velocities add under that transformation}.

Bringing laser light into the discussion could confuse a lay audience. But if you are sitting on a back porch, or around a warm fireplace late at night, you can give them the example of Bob riding a rocket wizzing by a motionless Alice on the ground. Bob's rockets carries him 80% of the speed of light, and he turns on a laser pointer in his direction of his motion. Nobody has a problem stomaching the fact that Bob sees the laser shoot out in front of him at c. But it is also a physical fact that Alice sees the laser light coming out at c.

Folk-mechanics tells us that if motionless Alice sees the laser light at c, Bob "must" see it moving slower than c as he "catches up" to it on his rocket.

(In our universe, comma) Bob's clocks are all running slower, at precisely the rate that causes him to measure the laser light at c. It's a conspiracy of the universe to perfectly time the 'rate of slowdown' of his clocks to match this.

If it helps to ease the medicine down : speed is distance divided by time. The light will travel a farther interval distance per each "Bob second" , which is now longer than a "regular second" because he is in motion.

{literate turtlenecks know what this is. It's the Lorentz Transformation. I just replaced the Galilean equation with the Lorentz factor. }

I am old enough to have taken calculus-based physics tract (all three quarters) at a major state university while a teenager. I took another two-semester dosage of calculus-based physics later in my life. In the first tract we did cover Special Relativity. I had to do homework in it, and took exam questions on it. The homework and exam problems were all more-or-less replacing the Galilean transformation of velocities with the Lorentz transformation. It was kind of plug-and-chug , I won't lie. We poor college kids had to crank the problems because they were "due next Thursday". If someone round these parts has had four years to engage with the material of Special Relativity. I mean, in that time such a person could have worked every single SR problem in the back of the chapter... in several textbooks.

[BurtJordaan]

Cool! Many thanks Mitchell. This shows that 60% of c is an easy to understand speed in many ways. As you have said, one can do a return trip and have a reserve fuel of 50% x 50% x 50% x 50% = 6.25% of the original fuel. That is if you have a 100% efficient mass-energy conversion, which may not be achievable, so practically one may have to settle for much slower trips.

I will continue with the 0.6c delta-v's, but just use a higher acceleration to reduce the time for Alice.

Alice's plan has changed somewhat since part [9] and she will now first stop in the region of Alpha Centauri (AC) for some observations. It is way too late to achieve the rocket braking at the original very low (0.075g) acceleration, so Alice asked her INS computer to get them into a 'parking' orbit, using a comfortable 1g retro-burn. The ship's INS and my trusted spreadsheet quickly determine that at -1g, it will take 0.69 years for the rocket to shrug off that 0.6c coasting velocity that it has relative to Bob.

The INS also knows that the braking must be initiated at a distance of 0.2 lyrs short of the intended 'parking' position, using Alice's map - it is 0.25 lyrs on Bob's map. (Dave_O, take notice, this comes from that rotated diagram that you asked about, just on a different scale). With this knowledge, the INS starts the retro burn at the right moment, and the ship slows down until it enters a slow orbit around AC. So slow, that we can consider Alice as stationary in Bob's structure now.

During the 0.69 years of retro deceleration, Alice's tilted structure has gradually rotated back to align with Bob's structure once again. While their structures were at an angle to each other, space and time were different for them, but with aligned structures it becomes identical again. Here is a summary of the space and time accounting along the way:

Stage/event _______ T_Bob yr __ D_Bob lyr __ T_Alice yr__ Alice g's __v_rel/c
Start burn 1 _________ 0 ________ 0 ________ 0 ________ 0.075 ___ 0
End burn 1 ________ 10.07 _____ 3.38 ______ 9.3 _______ 0 _______ 0.6
Duration coast 1 ____ 1.29 _____ 0.77 ______ 1.03 ______ 0 _______ 0.6
Start burn 2 _______ 11.36 _____ 4.15 _____ 10.33 ______ -1 _______ 0.6
Duration burn 2 _____ 0.75 ______0.25 _____ 0.69 _______ -1 _______ 0.6
End burn 2 ________ 12.11 _____ 4.40 _____ 11.02 _______ 0 _______ 0

(The spacing is a bit random between editor and post - does anyone know how to make a proper table using the 'table-tag' above in this editor?)

Alice and Bob are now separated by a distance of 4.4 lyrs. How do they test whether the final figures are correct? (Apart from asking an accountant to check the addition!) The simplest way is for Bob to have another observer, say Dot, stationed in slow orbit around AC. Bob and Dot are in the same inertial frame (spacetime structure), just separated by 4.4 lyrs in space. They have synchronized clocks, because they have done it long before Alice arrived. If Alice 'parked' her ship near Dot, the two of them can directly compare their clocks and confirm the bottom row of the table above. There are other methods as well, but let us only consider this one method for now, because it's simple and works superbly.

Alice and Dot can also confirm that their clocks now 'run the same', so we can safely assume that they are now in the same spacetime structure. Before the long journey, Alice and Bob were together and in the same spacetime structure. During the journey, Alice was in a spacetime structure of her own, which changed orientation almost continuously relative to Bob's structure. But now Alice's structure is again aligned with Bob's structure and the only observable difference between them is that less propertime has elapsed for Alice than for Bob and Dot.

The root cause of this difference is that Alice has actively changed her spacetime structure by expending a lot of energy in terms of fuel. We pictured it as a rotation of Alice's structure in relation to Bob's. The change in structure orientation caused Alice to have covered spatial distance at the cost of covering time, all in relation to Bob. This the nature of relativistic spacetime. Simple inertial Epstein diagrams illustrates the relationship beautifully, without looking at any math. It is obviously more complex to draw the diagram when there are portions of acceleration and portions of coasting.

I am working on such a space-propertime diagram to illustrate the complete situation, but that will have to stand over for the next part. In the meantime, I will leave you to ponder the following statement and decide if it is true: "it does not matter in which direction Alice did her one-way journey, or what Bob's speed relative to the 'universe at large' was, the outcome of the experiment would have been the same".

[bangstrom]

I
Folk-mechanics tells us that if motionless Alice sees the laser light at c, Bob "must" see it moving slower than c as he "catches up" to it on his rocket.

(In our universe, comma) Bob's clocks are all running slower, at precisely the rate that causes him to measure the laser light at c. It's a conspiracy of the universe to perfectly time the 'rate of slowdown' of his clocks to match this.

If it helps to ease the medicine down : speed is distance divided by time. The light will travel a farther interval distance per each "Bob second" , which is now longer than a "regular second" because he is in motion.

Folk-mechanics also tells us that Bob’s clocks must be moving faster if he is to measure the speed of a light beam moving to the rear of his rocket at the speed of c. This is where motion slowing time becomes counter intuitive.

[mitchellmckain]

I
Folk-mechanics tells us that if motionless Alice sees the laser light at c, Bob "must" see it moving slower than c as he "catches up" to it on his rocket.

(In our universe, comma) Bob's clocks are all running slower, at precisely the rate that causes him to measure the laser light at c. It's a conspiracy of the universe to perfectly time the 'rate of slowdown' of his clocks to match this.

If it helps to ease the medicine down : speed is distance divided by time. The light will travel a farther interval distance per each "Bob second" , which is now longer than a "regular second" because he is in motion.

Folk-mechanics also tells us that Bob’s clocks must be moving faster if he is to measure the speed of a light beam moving to the rear of his rocket at the speed of c. This is where motion slowing time becomes counter intuitive.

And that is why this explanation talking about clocks going faster or slower is wrong. Time is not slowing down OR speeding up. The real key to what is happening is the relativity of simultaneity.

If Alice fires two beams one towards the front of her ship and one towards the back with clocks lined up along the path, then Bob sees both of these beams of light traveling at the speed of light but the clocks all have different times on them. Assuming the same distance on Alice's ship both ways, the two beams of light arrive at the same time according to Alice so the clocks at the two arrival points will read that same time to both Bob and Alice, but for Bob these are happening at very different times. Alice's ship is moving in the direction of the forward light beam and so according to Bob that beam going forward has to reach an end of the ship that is moving way from the source, while the beam going backward has to reach and end of the ship that is moving towards the source. Thus according to Bob the backward beam arrives first and the forward beam arrives later. The two clocks at the ends of Alice's ship still give the same time but these are not synchronized according to Bob, for the clock at the back of Alice's ship gives this time much earlier than the clock at the front of Alice's ship.

This explains Lorentz contraction also. Alice's ship looks shorter, but in reality, what Bob is seeing is the parts of Alice's ship at different times according to Alice. He is seeing the front of Alice's ship at an earlier time when Alice's ship hasn't moved as far, and he is seeing the back of Alice's ship at a later time when it has moved farther. Thus Alice's ship looks shorter, at least it does in Bob's calculations after he has accounted for the speed of light and eliminated the aberration of light effect.

[mitchellmckain]

If I had to explain special relativity to a high school kid, or explain it to someone in a loud bar, or in a bowling alley. The first thing I would say, to keep it ultra-simple, is to tell the person :

Velocities do not add.

You travel 60% of the speed of light on the back of a rocket, and fire a bullet in your direction of your motion at 45% of the speed of light. If velocities add, the bullet should be going 105% the speed of light. But that does not happen, because velocities do not add.

{you educated turtleneck sweaters call this the "Galilean Transformation" i.e. velocities add under that transformation}.

Yeah you have to use a formula which keeps added velocities less than the speed of light.

vsum = (v1 + v2)/(1 + v1xv2/c²)

So for example v1 = v2 = .5 c gives vsum = (.5 c + .5 c)/(1 + .5x.5) = c/1.25 = .8 c

[vivian maxine]

(The spacing is a bit random between editor and post - does anyone know how to make a proper table using the 'table-tag' above in this editor?) (Burt)


By putting the exact same number of dots between the columns on each row. It doesn't always work to perfection because computers seem to measure the width of letters in their spacing. But it does fairly well. In fact, yours look fine.

That works from the second column onward. For the first, where some lines are shorter than others, count the letters in the longest line and use spaces on all the others to keep them all the same length.

[vivian maxine]

Velocities do not add.

Thank you! You have settled a problem and disagreement that I had months ago. Off topic here. So, I'll just say thank you.

[BurtJordaan]

I am working on such a space-propertime diagram to illustrate the complete situation, but that will have to stand over for the next part.

Here is the space-propertime diagram that can replace that poorly formatted data table. Much more informative than a thousand tables!

This is not an Epstein diagram, because it does not comprise solely of inertial frames. Therefore Alice's spatial axis cannot be depicted, but then Alice does not move along her own spatial axis, so it is somewhat superfluous for the scenario that has played out. However, it is a genuine space-propertime (SPT) diagram, embodying relativistic principles correctly.

To avoid confusion, here are some quick comments on the depicted diagram, specifically on those parts that may not be obvious and/or intuitive. The three blue arrow (Charlie, Joe and Dot) are inertial assistants with synchronized clocks, sitting at constant distances from Bob, as per the diagram. They are so placed that they are present at the end or start of Alice's acceleration phases, as applicable. So, they can read what is going on with Alice's clock as she flies by (or stops, as is the case at the end). Obviously, Alice can read their clocks as she flies by as well.

The solid red curve represents Alice's propertime as she progresses through Bob's space. The dotted blue curve is essentially Bob's propertime, but projected to Alice's position in Bob's space. This correlates with what Alice will read off Charlie, Joe and Dot's clocks as she are respectively collocated with them (look at the horizontal blue dashes). The horizontal red dashes logically indicate Alice's propertime, i.e. the readings on her clock as read by the assistants at flyby.

The difference in propertimes at the end of the test is obvious: Alice has recorded just over one year less than Bob. So, was Alice's clock simply slower than Bob's? No, Alice simply moved less through propertime. Actually the real length of the red curve is identical to the length of the blue arrows. It does not look like it, because I had to squash the diagram from top to bottom to about 1:3 in order not have an absurdly tall diagram. Trust me, the red curve is mathematically of identical length to the blue arrow lengths. IMO, this is really the clincher for this method of diagramming space and time for beginners.

Please ask if things are not clear, or if you spot errors... ;)

[BurtJordaan]

Actually the real length of the red curve is identical to the length of the blue arrows. It does not look like it, because I had to squash the diagram from top to bottom to about 1:3 in order not have an absurdly tall diagram. Trust me, the red curve is mathematically of identical length to the blue arrow lengths. IMO, this is really the clincher for this method of diagramming space and time for beginners.

Just to reemphasize this truth about space-propertime diagrams, it is not solely through the equations that one can test this. Attached is a simple, but rough graphical illustration of why this is so. The curvature of the space-propertime path, showing the accelerating Alice, is built up from small Epstein diagrams, as shown at an exaggerated scale. The individual blue and red arrows clearly are of equal length, but the curved segments clearly result in less gain in the vertical dimension than is the case for the vertical segments. When a lot of tiny segments are strung together, they form a smooth curve.

This in fact exactly how my spreadsheet does it, using nothing more than high school trig. This is in addition to to the appeal of the identically growing line lengths. It also takes some of the 'teeth' out of the treatment of accelerated observers. This is in contrast to the Minkowski format, where the line lengths of different observers grow differently over time and it needs more advance (hyperbolic) trigonometric functions.

I definitely see an advantage in utilizing this relative simplicity for introducing beginners to relativity. Added to that, relative time dilation, Lorentz contraction and even simultaneity are not needed in the explanations. They are obviously still there, but hidden from view, so to speak. IMO, once a beginner has a good feel for the nature of space and time, when thy are ready for more complexity, they could be gradually introduced to the more versatile and more complex Minkowski spacetime.

Finally, through the use of the equivalence principle, it is possible to introduce a beginner to general relativity (curved spacetime) in a similar, uncomplicated way.

What is your verdict?

PS: Dave_O, thanks for you patience, but the floor is now yours... ;)

[mitchellmckain]

Actually the real length of the red curve is identical to the length of the blue arrows. It does not look like it, because I had to squash the diagram from top to bottom to about 1:3 in order not have an absurdly tall diagram. Trust me, the red curve is mathematically of identical length to the blue arrow lengths. IMO, this is really the clincher for this method of diagramming space and time for beginners.

Just to reemphasize this truth about space-propertime diagrams, it is not solely through the equations that one can test this. Attached is a simple, but rough graphical illustration of why this is so. The curvature of the space-propertime path, showing the accelerating Alice, is built up from small Epstein diagrams, as shown at an exaggerated scale. The individual blue and red arrows clearly are of equal length, but the curved segments clearly result in less gain in the vertical dimension than is the case for the vertical segments. When a lot of tiny segments are strung together, they form a smooth curve.

But what you haven't demonstrated is the meaning or validity of making such a construction patching together small Epstein diagrams. Epstein's diagram is only for the purpose of comparing measurements in two inertial frames. What is it you think this patched/acceleration diagram showing anyway. It is not showing the person's progress through space time. You are making it look like these vector's in Epstein's diagram are displacement vectors -- which they are not. These vectors are fixed magnitude with an direction representing the relative velocity between two inertial frames. What can possibly be the meaning of adding such vectors like you have done? I do not see it.

To get the usual usage of the Epstein diagram you need to drop back to one of these composing diagrams, for I do not see how you are going to use the whole diagram to calculate anything meaningful. At first I thought, there are no obstacle to making this continuous and using the tangent vector to drop back to a real Epstein diagram. But then I had to ask myself how you construct such a continuous graph. I guess you would have to do it by taking the limit as you patch more and more of them together at smaller intervals. But then what is the mathematics for doing this, because that is where the real significance will be found.

[BurtJordaan]

You are making it look like these vector's in Epstein's diagram are displacement vectors -- which they are not. These vectors are fixed magnitude with an direction representing the relative velocity between two inertial frames. What can possibly be the meaning of adding such vectors like you have done? I do not see it.

The vectors are indeed displacement vectors in space-propertime, i.e. and . I see no problem in summing unit-vectors like this, provided one realize that you are not going to get the standard displacement in spacetime.

To get the usual usage of the Epstein diagram you need to drop back to one of these composing diagrams, for I do not see how you are going to use the whole diagram to calculate anything meaningful.

By vector summing the , displacements, I get the curve which gives me the values that I've shown in complete diagram of this post. They seem to be the correct values. One can check them using Rindler coordinates, which is outside the scope of this thread, for obvious reasons.

At first I thought, there are no obstacle to making this continuous and using the tangent vector to drop back to a real Epstein diagram. But then I had to ask myself how you construct such a continuous graph. I guess you would have to do it by taking the limit as you patch more and more of them together at smaller intervals. But then what is the mathematics for doing this, because that is where the real significance will be found.

Indeed, but I do not intend showing the math in this philosophy thread; I'll rather open a separate thread under physics for that. It is rather simple, as I have suggested earlier, but then the simplicity may be in the eye of the beholder. ;)

[mitchellmckain]

You are making it look like these vector's in Epstein's diagram are displacement vectors -- which they are not. These vectors are fixed magnitude with an direction representing the relative velocity between two inertial frames. What can possibly be the meaning of adding such vectors like you have done? I do not see it.

The vectors are indeed displacement vectors in space-propertime, i.e. and . I see no problem in summing unit-vectors like this, provided one realize that you are not going to get the standard displacement in spacetime.

Ok, checking this out seems to bear up BurtJordan's claim.
For everyone else, here is what I figured out.
Let phi be the angle between the two proper time axes A&B
Then cos(phi) = 1/gamma, and sin(phi) = v/c.
This means that the projection of an interval t' on the proper time axis of A gives the time dilated interval t=t'/gamma on the proper time axis of B, while the projection of this interval on the spatial axis of B gives the dispacement vt. To give the right units you have to multiply by c sin(phi) to get that projection onto the spatial axis.

Thus I was wrong and BurtJordan is correct and this path does indeed give the projected displacement in coordinate system in which it is moving (for Alice it would be in Bob's). And of course the successive displacements naturally add up just fine.

To get the usual usage of the Epstein diagram you need to drop back to one of these composing diagrams, for I do not see how you are going to use the whole diagram to calculate anything meaningful.

By vector summing the , displacements, I get the curve which gives me the values that I've shown in complete diagram of this post. They seem to be the correct values. One can check them using Rindler coordinates, which is outside the scope of this thread, for obvious reasons.

I guess you must be altering Epstein's method a bit by changing the magnitude of the vector from c as he has it to a variable proper time interval. I had this odd image of adding together his fixed length vectors which didn't make a lot of sense.

[BurtJordaan]

Thanks for confirming that it checks out, Mitchell,

I guess you must be altering Epstein's method a bit by changing the magnitude of the vector from c as he has it to a variable proper time interval. I had this odd image of adding together his fixed length vectors which didn't make a lot of sense.

I'm not quite sure what you mean here, but I'm only summing identical length space-propertime path vectors. With the years and light years units, c=1 and doesn't play any other role that I can see. This is the beauty (and perhaps the reason for the distrust in the physics community) of the Epstein diagrams. I will point out some other problems that Physicists may have with the method in the new thread in the physics subforum.

That notwithstanding, it could be a valuable tool in the quest to make relativity (even elements of the full theory) less intimidating to non-physicists of any level. It doesn't need to be that hard to learn! And AFAICS, there is really nothing that students will have to 'unlearn', should they carry on...

So what does the non-physicists think?

[dandelion]

So what does the non-physicists think?

Brilliant.

[mitchellmckain]

Thanks for confirming that it checks out, Mitchell,

I guess you must be altering Epstein's method a bit by changing the magnitude of the vector from c as he has it to a variable proper time interval. I had this odd image of adding together his fixed length vectors which didn't make a lot of sense.

I'm not quite sure what you mean here, but I'm only summing identical length space-propertime path vectors. With the years and light years units, c=1 and doesn't play any other role that I can see. This is the beauty (and perhaps the reason for the distrust in the physics community) of the Epstein diagrams. I will point out some other problems that Physicists may have with the method in the new thread in the physics subforum.

Giving you too much credit? Doing things in the easier units where c=1, is quite natural for me too. It is just easier. But to make sure everything is consistent you should check it out with real space-time units. And after all if the magnitude of the vector is really c then the projection on the x axis of the other coordinate system is not the displacement but the velocity: c sin(phi) = v. My observation that you can make the magnitude of the vector whatever proper time has passed made a lot of what you were saying fall into place. You even kept emphasizing that this was a proper time diagram. After all, there is no reason for a displacement to be a fixed quantity but should be a variable quantity. I am pretty sure Epstein making the magnitude fixed at c was because his intended use was quite different than what you are doing.

[hyksos]

And that is why this explanation talking about clocks going faster or slower is wrong. Time is not slowing down OR speeding up. The real key to what is happening is the relativity of simultaneity.

Wait a minute here. The real key is to replace the Galilean transformation with the Lorentz Transformation.

[BurtJordaan]

My observation that you can make the magnitude of the vector whatever proper time has passed made a lot of what you were saying fall into place.

I do not quite see it that way, but we must make sure we are talking about the same thing. The magnitude of the vectors that I was talking about represent the elapsed propertime of the reference frame, i.e. , not the of the 'moving frame' - the latter is the projection of vector onto the vertical () axis. So yes, if we work in unity coordinate time intervals, then the vector magnitude is c, but it really can be anything we choose, without changing any utility.

In my usage, the projection onto the coordinate space axis is really distance, not speed. I suppose one can interpret Epstein's "speedometer" as you did (v=c sin(phi)), so in that case it is correct that I modified the Epstein representation to suit the problem, but does that matter?

PS: Maybe it wasn't such a good idea to mention the Epstein diagram at all in the method that I use. It is really just an application of a space-propertime coordinate system for 'real-life' scenarios. Maybe the Epstein diagram distracts unnecessarily?

[mitchellmckain]

Maybe it wasn't such a good idea to mention the Epstein diagram at all in the method that I use. It is really just an application of a space-propertime coordinate system for 'real-life' scenarios. Maybe the Epstein diagram distracts unnecessarily?

I doubt it. Even if it isn't exactly what you are doing, it is a stepping stone to what you doing. You need something to make it clear you are changing the basic assumptions about what is done with those coordinate axis. Vectors on a space versus time graph are assumed to be velocity vectors. In fact this might be a good example of how the procedures we adopt in mathematics are far from inevitable -- so that mathematics may not be the universal language we often think it to be.

[BurtJordaan]

I was hoping to get the "myth" tag and possibly the negative connotations of Epstein diagrams out of the treatise. For lack of another name, I'm calling it simply a space-propertime diagram. I agree that one should then clearly state that a unit vector on this diagram is not a velocity vector like in normal spacetime diagrams. Maybe we should tag that with some other name, or perhaps just stick to its roots and call it a space-propetime vector (SPT vector for short).

I suppose the next step would be to clearly show how one can go back to the more versatile Minkowski domain.

I will slowly proceed with a summary of the main ideas in a new thread, while leaving this one open for further questions by the community.

[mitchellmckain]

I suppose the next step would be to clearly show how one can go back to the more versatile Minkowski domain.

It is not a 1-1 mapping, or you could say that it is a 1-1 mapping from the past and future light cones of the Minkowsky graph to the Epstein-BurtJordan graph (with a big singularity/discontinuity at the light speed of the Minkowsky graph). Mathematically it must have to with the change of trig functions to hyperbolic trig functions. I would guess that the circle in the Epstein graph becomes a hyperbola in the in the light cone. At least, that is what I have been visualizing all this time.

[BurtJordaan]

I have removed your accidental double post. The forum sometimes does that when you think it has not accepted your submit and you submit it again.

Mathematically it must have to with the change of trig functions to hyperbolic trig functions. I would guess that the circle in the Epstein graph becomes a hyperbola in the in the light cone. At least, that is what I have been visualizing all this time.

Yes, it would be, but since this is aimed mainly at the physicists, I was thinking about avoiding the math and just comparing things graphically. My idea is that once a student is comfortable with SPT diagrams, it should be easy for them to see the equivalence to the more versatile Minkowski. And why the Minkowski is the preferred spacetime diagram.

Unfortunately, it is still totally unclear as to whether non-physicists find the SPT diagrams useful at all. The silence could mean than they have no questions and have lost interest, or it makes no sense to them at all... ;)