# The physical interpretation of the Minkowski spacetime diagram

Discussion in 'Physics & Math' started by geordief, Dec 21, 2017.

1. ### Mike_FontenotRegistered Member

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I'm not following your description of what you're reading on my webpage. It might help if you copied/pasted a few short specific statements in my webpage that are confusing you. For now, though, it might help for you to know that the basic results I give for the traveler's conclusions about the home twin's ageing during various portions of his trip are identical to the results of the analysis most commonly known as the "co-moving inertial frames" determination of the traveler's conclusions . So what is different about the description on my webpage primarily has to do with terminology, not results. The reason I have developed different terminology is that I've witnessed over many years a lot of confusion and mistakes being made by people trying to apply the conventional terminology. So my goal (undertaken many years ago) was to come up with terminology and variable names which reduce that confusion and those commonly made errors. And before that development of different terminology, I had derived an equation (now called "the CADO equation") which gets the same traveler's perspective normally obtained via the traveler's "line of simultaneity" on a Minkowski diagram, but in a much easier and less error-prone way. The CADO equation is also now expressed using the new terminology, again to minimize misunderstandings and errors.

Last edited: Dec 25, 2017

3. ### Confused2Registered Senior Member

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506
Imagine bouncing a radio signal off the moon - it takes about 2.5 seconds to come back. This is of no help determining how much younger an astronaut (Alice) travelling to the moon would be if compared to her stay-at-home twin Bob.

5. ### Confused2Registered Senior Member

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506
It may be helpful to know that the Lorenz Transforms predated any understanding of why they worked. It may not be easy (or even possible) to reverse engineer the why by looking at the transforms. If you looked at them as something for people who know what they're doing and why they're doing it and you concentrated on the what and why you may find you progress more rapidly.

7. ### geordiefRegistered Senior Member

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941

I have not got involved in the twin paradox scenario at all in this thread.I have simply been trying to tie down the simplest geometric relationship on the chart with a pair of physical events.

I know you referenced me that muon link earlier in the thread but it was above my pay grade (and also not really what I was asking I think)

8. ### Confused2Registered Senior Member

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My pay grade is pretty darn low - but I have been looking at exactly this type of problem for (more than) several years. The first time you get numbers that match an experiment is pure delight - well worth the effort.

9. ### Neddy BateValued Senior Member

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1,767
Yes, but the ct axis is, by convention, charted perpendicular to the x axis.

That means that while "time" might be charted going north, the "displacement" might be charted going east.

Meanwhile, there is a real "NORTH" in three dimensions, and that dimension is not ct, but y.

I think that is why it is confusing for people to look at Minkowski diagrams before trying the Lorentz transform equations.

10. ### Neddy BateValued Senior Member

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1,767
Yes, I understand your motivation was to make it easier. My main question is why t=20+ became t=20- as found here:

"We can denote the instant in the traveler's life, immediately before the turnaround, as t = 20-, and the instant immediately after the turnaround as t = 20+."

Why are the signs after the numbers? Is this different than +20 and -20?

11. ### Confused2Registered Senior Member

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506

Last edited: Dec 25, 2017
12. ### Neddy BateValued Senior Member

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1,767
In light of the above post, this one will probably lose a lot of readers. Sorry about that, but I use equations to calculate these things.

This post will use the same scenario as the one on Mike_Fontenot's webpage, with a basic "Twins" scenario, and "Idealized Instantaneous Acceleration." However, I will use the conventional variables, and the standard Lorentz transformation equations. This will probably be more complicated than his method, but I just wanted to try it, to see if I could do it.

Standard Variables:
t = time as measured by the Stay-At-Home-Twin
x = location of the Traveling-Twin as measured by the Stay-At-Home-Twin
t' = time as measured by the Traveling-Twin
x' = location of the Stay-At-Home-Twin as measured by the Traveling-Twin
v = relative velocity between the two twins
γ = gamma = Lorentz factor = 1 / √(1 - (v²/c²))

FIRST LET'S SEE HOW IT IS DONE FROM THE EASIEST REFERENCE FRAME:

From Start to Turnaround, (from the reference frame of the Stay-At-Home-Twin):
v = 0.866c
t = 20.00
x = vt = 17.32
t' = γ(t - (vx / c²)) = 10.00

From Turnaround to End, (from the reference frame of the Stay-At-Home-Twin):
v = -0.866c
t = 20.00
x = vt = -17.32
t' = γ(t - (vx / c²)) = 10.00

Totals, (from the reference frame of the Stay-At-Home-Twin):
t = 20.00 + 20.00 = 40.00 = Final Age of Stay-At-Home-Twin
t' = 10.00 + 10.00 = 20.00 = Final Age of Traveling-Twin
x = 17.32 + (-17.32) = 0.00 = Final Position of Traveling-Twin

AND NOW LET'S SEE HOW IT IS DONE IT FROM THE OTHER REFERENCE FRAME:

From Start to Turnaround, (from the reference frame of the Traveling-Twin):
v = 0.866c
t' = 10.00
x' = -vt' = -8.66
t = γ(t' + (vx' / c²)) = 5.00

From Turnaround to End, (from the reference frame of the Traveling-Twin):
v = -0.866c
t' = 10.00
x' = vt' = -8.66
t = γ(t' + (vx' / c²)) = 35.00

Totals, (from the reference frame of the Traveling-Twin):
t = 5.00 + 35.00 = 40.00 = Final Age of Stay-At-Home-Twin
t' = 10.00 + 10.00 = 20.00 = Final Age of Traveling-Twin
x' = -8.66 - (-8.66) = 0.00 = Final Position of Stay-At-Home-Twin

Last edited: Dec 25, 2017
13. ### Confused2Registered Senior Member

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506
Just to confirm...
Assuming all clocks start at zero...
When the ship arrives at the space station
The ship clock shows 10 seconds elapsed time and the space station shows 20 seconds elapsed time.

This way round the ship clock again shows 10 seconds elapsed time but time in the Earth/space station frame has only advanced by 5 seconds. So, when the ship arrives does the space station clock show 5 seconds or 20 seconds?

Edit ... trivial point but your x and x' distances need a c factor added. I have occasionally spent hours trying to work why a typo makes sense when it doesn't.

Last edited: Dec 25, 2017
14. ### Neddy BateValued Senior Member

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1,767
The two twins start out in the same place as one another, with both of their personal wristwatches at zero. The stay-at-home twin does nothing special, but the traveling-twin instantly accelerates away at v=0.866c or 86.6% of the speed of light. Because of time dilation, the stay-at-home-twin finds that the traveling-twin's watch is running slow, so when the stay-at-home-twin's own watch accumulates 20 seconds, the stay-at-home-twin says the traveling-twin's watch has accumulated only 10 seconds.

But by this time, the stay-at-home-twin would say that the two are a distance of 17.32 lightseconds apart, so the stay-at-home-twin can't instantly "see" the distant watch displaying 10 seconds at that instant of time. He will have to wait 17.32 seconds for the light from the distant watch face to reach him, and then he will see it.

That is from the other reference frame, so it goes like this: The two twins start out in the same place as one another, with both of their personal wristwatches at zero. The stay-at-home twin does nothing special, but the traveling-twin instantly accelerates away at v=0.866c or 86.6% of the speed of light. Because of time dilation being reciprocal, the traveling-twin finds that the stay-at-home twin's watch is running slow, so when the traveling-twin's own watch accumulates 10 seconds, the traveling-twin says the stay-at-home-twin's watch has accumulated only 5 seconds.

But by this time, the traveling-twin would say that the two are a distance of 8.66 lightseconds apart, so the traveling-twin can't instantly "see" the distant watch displaying 5 seconds. He would have to wait for the light from the distant watch face to reach him, but before that even happens, the traveling-twin instantly reverses direction and heads back at v=-0.866c where the negative sign represents the opposite direction. In that instant the traveling-twin changes reference frames, and finds the distant watch must be displaying 35. That is how the stay-at-home-twin's clock can end up being 40 in the end, while the traveling-twin's clock ends up being 20.

Sorry, I should have stated that if the units of time (t and t') are seconds, then the units of length (on the x and x' axes) would have to be lightseconds. Likewise, if the units of time (t and t') were years, then the units of length (on the x and x' axes) would have to be lightyears.

Similarly, since the units of v are given as a percentage of c, (0.866c or 86.6% of c), the units of c would have to be the same units, 1.000c or 100% of c, which is why c=1 in all those calculations.

Last edited: Dec 26, 2017
15. ### Confused2Registered Senior Member

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506
Duplicate post, sorry.

Last edited: Dec 26, 2017
16. ### Confused2Registered Senior Member

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506
Sorry, duplicate post ... see next (with edits)

Last edited: Dec 26, 2017
17. ### Confused2Registered Senior Member

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506
So can you say what time a clock at the turnaround point would show as the ship arrives? There would be no light travel delay time so this would show the time of flight of the ship in the frame of the stay-at-home twin (yes?). Assume clocks in the stay-at-home frame are Einstein synchronised.
Edit...
I don't have the original question so I'm guessing the start and turn points are 17.32 light seconds apart. When the ship launches (start_0) seen from the turn point the turn point clock reads turn_17.32. When the ship arrives (after 20 seconds in the start/turn frame) the turn clock reads turn_37.32 and the start clock reads start_20 seen from the turn point. As the ship arrives at the turn both ship and turn should see the same value on the start clock (yes?).

Last edited: Dec 26, 2017
18. ### Neddy BateValued Senior Member

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1,767
Yes, in the stay-at-home-twin's reference frame, 17.32 lightseconds was the distance that I was using. But please note that in the traveling-twin's reference frame, the start and turn points would be 8.66 lightseconds apart, due to length contraction. Remember, the stay-at-home-twin would say that the one-way journey took twice as much time as the traveling-twin would say, and yet they agree on the relative velocity of v=0.866c. So naturally they disagree on the distance between the start and turn points, but only while they are in relative motion.

If there is an Einstein-synchronised clock at rest in the stay-at-home-twin's reference frame, and located at the turn around point, then it must display t=20 at the turn-around event, while the traveling-twin is right next to it, with the traveling twin's own wristwatch displaying t'=10 at that same event.

But that does not prevent the traveling-twin from saying the time on the distant stay-at-home-twin's watch would be t=5 in the instant before the turn-around, nor does it prevent the traveling-twin from saying the time on the distant stay-at-home-twin's watch would be t=35 in the instant after the turn-around. Those two Einstein-synchronised clocks would be synchronous with each other in their own frame, but not in the traveling-twin's frame. That is one of the weird things about "relativity of simultaneity".

Last edited: Dec 26, 2017
19. ### Confused2Registered Senior Member

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506
Yes, it has to be right despite looking weird to me. Thanks for your patience.

20. ### geordiefRegistered Senior Member

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941
I thought acceleration did (also) cause time dilation. What is the point in modeling the outward and return journey as if acceleration was not a factor?

Introducing a third clock seems to me a bit like imaging an object approaching the observer at a relativistic speed and instantly reversing course at the same speed ( an instantaneous change of speed of up to 2c).

Am I seeing that right?

21. ### Neddy BateValued Senior Member

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1,767
Yes, that is exactly right.

One might be tempted to say something like, "Since both reference frames agree that the two clocks at the turn-around point display t'=10 and t=20 at the turn-around event, doesn't this prove that the clocks at rest in the stay-at-home-twin's frame are running at TWICE the rate of the clocks at rest in the traveling-twin's frame?" But the answer would be, "No, because the traveling-twin would say that the added third clock displayed t=15 when the journey first began, and therefore it only had to accumulate 5 more ticks to reach t=20, demonstrating that the traveling-twin would still say that all of the clocks at rest in the stay-at-home-twin's frame are running at HALF the rate of the clocks at rest in the traveling-twin's frame."

I can show you the calculation for the t=15 if you want, but I didn't want to clutter the post if you don't need it.

Last edited: Dec 27, 2017
22. ### Neddy BateValued Senior Member

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1,767
No, the acceleration itself does not cause the time dilation.

But what breaks the symmetry of time dilation being perfectly reciprocal is when one of the twins changes reference frames, (especially when that happens when they are far apart). So, with the outward-and-return-journey model, the symmetry is broken when the one twin turns around at the turning point, thus changing reference frames, and changing his own notion of simultaneity.

That change in his own notion of simultaneity is why he can say the distant watch on his twin's wrist displays t=5 just before turning around, but then just after turning around he can say the distant watch on his twin's wrist displays t=35. His notion of simultaneity changes in that turnaround.

No, I don't see how you can see it that way. The third clock introduced by Confused2 was just to try to eliminate the potential problem of the one twin having to predict a time on a watch face that was too far away from himself to be able to see it in real time.

Furthermore, if we consider the "speed limit" to be 1.000c and the traveling twin only travels at 0.866c in one direction, and then he travels at -0.866c in the opposite direction, surely he has not broken the speed limit?

Last edited: Dec 27, 2017
23. ### arfa branecall me arfValued Senior Member

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6,190
I think the idea is to make space and time mathematically equivalent, both axes have the same units of distance. So actually time is 'transformed' into a distance; of course another mathematical trick which we have to undo at some point to recover physics, which has that darn time thing.