How do you resolve seeming contradictions in SR?

[PeterDonis]

note that observers A and B, supposed to be at rest in the same inertial frame, share the same "now" sets in spacetime

As I said in the parenthetical note in the last part of what you quoted from my post.

 

[curiousburke]

I'm late for the OP question, but note that observers A and B, supposed to be at rest in the same inertial frame, share the same "now" sets in spacetime.

I am assuming this is in response to one of my latest posts about the distance between A and B in figure 1 (A, B perspective) and the distance between A and B in figure 4 (C's perspective), but I don't think I said the spatial separation effected their "now". Does something I said make that indirectly true?

 

[PeterDonis]

I don't think I said the spatial separation effected their "now".

See my post #16, which @cianfa72 quoted, and which was in response to a statement of yours (quoted in that post) that included "spatial separation".

 

[curiousburke]

See my post #16, which @cianfa72 quoted, and which was in response to a statement of yours (quoted in that post) that included "spatial separation".

Okay, I know I had said that near the beginning.

 

[robphy]

Here are a few spacetime diagrams. If you haven't come across them, they're simply plots of the position of an object over time, the custom being that time goes up the page. So a vertical line represents an object that isn't moving, and a line slanted to the left indicates one moving to the left. You may have come across these (usually with time horizontally) as "displacement-time graphs" in high school physics. We take them a little more seriously in relativity, since spacetime is a thing - these are maps of spacetime, and the lines are the 4d objects inhabiting it. You see one 3d slice at a time.

So here's your scenario, with my extra "D" observer. A and B are marked in red and are stationary, C and D are marked in blue and moving to the right.
View attachment 341321
Note that, in this frame, C and D are closer together than A and B - the horizontal distance between the lines ("the space between them") is shorter.

We could mark on the diagram the start and end of the experiment - when C passes A and then B. Let's do that with fine red lines:

[snip]

Here's @Ibix 's spacetime diagram supplemented with "light-clock diamonds"
to help visualize the ticks along segments in spacetime.

An inertial observer's light-clock diamond
has a timelike diagonal parallel to his worldline
and spacelike diagonal Minkowski-perpendicular to his worldline.
Thus the spacelike diagonal is simultaneous according to that inertial observer.
All light-clock diamonds
have lightlike edges (speed of light postulate: eigenvectors of a boost on this plane)
and have equal area (determinant of boost equals 1).

The scale is (1 on @Ibix 's scale)=(3 ticks).
Using @Ibix's labeling of worldlines,
B along x=1 is 3 red-spaceticks [red-"sticks"] away from A and
D is -3 blue-"sticks" from C.

• The undecorated parallel worldlines 5 ticks away from A and C
  help in counting diamonds to set up proportions (via similar triangles)
  when trying to count diamonds involving B and D.
  For instance:
  for v=(3/5)=(6/10)=$\frac{(PQ)}{(OP)}$, we have $\gamma$=(5/4) and k=2.
  
  Length contraction:
  A rod of proper-length 5 [red-]sticks carried by A
  is measured to be 4 [blue-]sticks according to C since $5/\gamma=4$. $(OM)=\frac{(OL)}{\gamma}$)
  A rod of proper-length 5 [blue-]sticks carried by C
  is measured to be 4 [blue-]sticks according to A since $5/\gamma=4$.
  
  So,
  A rod of proper-length 3 [red-]sticks carried by A [whose far end is traced out by B]
  is measured to be 2.4 [blue-]sticks according to C since $3/\gamma=3/(5/4)=(12/5) {\rm\ sticks}=(12/5) (1/3 {\rm\ Ibix})= (4/5 {\rm\ Ibix})$.
  ...and similarly for the rod carried by C [whose far end is traced out by D].

 

[curiousburke]

robphy, you're stuff is next level. One careful read, and I understand maybe 50%. Just another infinity-1 reads and I'll fully get it :)

Seeing you add features to this spacetime diagram gave me an idea. Just to make things more complicated, what about adding a 3rd dimension to represent the perspective? Basically, spacetime diagrams are slices through a block of any possible rotation.

 

[robphy]

robphy, you're stuff is next level. One careful read, and I understand maybe 50%. Just another infinity-1 reads and I'll fully get it :)

Seeing you add features to this spacetime diagram gave me an idea. Just to make things more complicated, what about adding a 3rd dimension to represent the perspective? Basically, spacetime diagrams are slices through a block of any possible rotation.

Thanks.
These approaches come from various attempts (some successful , some less-successful)
to convey the ideas of relativity and spacetime-geometry.

The light-clock diamonds are traced out by the light-signals in light-clocks.
They are already there, implicitly, in the spacetime diagram.
I highlighted them because they lead to this way of calculating by counting areas.

It's not clear what to "add" in a third dimension. There may be something... but I'm not sure.
One should take care to find representations that faithfully represent the physics.

 

[curiousburke]

It's not clear what to "add" in a third dimension. There may be something... but I'm not sure.
One should take care to find representations that faithfully represent the physics.

I thought the same thing, so left it ambiguous figuring you would know :)

I'm thinking two different things, one is the velocity of a third frame from which the other two are measured, so 3d could smoothly transition from one frames perspective to the other.

Second, I'm still trying to conceptualize what it means, but hand wavey, it's the choice of x=0 relative to the observers.

 

[robphy]

I thought the same thing, so left it ambiguous figuring you would know :)

I'm thinking two different things, one is the velocity of a third frame from which the other two are measured, so 3d could smoothly transition from one frames perspective to the other.

Second, I'm still trying to conceptualize what it means, but hand wavey, it's the choice of x=0 relative to the observers.

Here's the original diagram, followed by its boosted version.

If you look at these a little bit and learn how to interpret the geometry and physics encoded,
I think that you can see that they convey the same information...
(...Like a rotated photo has the same information as the original photo.)
by counting ticks and sticks.... and by interpreting the diagonals
of an inertial observer's light-clock diamonds as defining
"at the same place" and "at the same time" for that observer.

So, the second diagram isn't necessary... and, in fact, can be constructed by hand from the first diagram.
(Using v=(3/5)c or (4/5)c [or more generally velocities with a rational Doppler factor]
makes this easy to construct by hand on rotated graph paper.)

 

[robphy]

I still think he has the best explanations of the twin paradox on youtube.

Here's my clock-effect/twin-paradox spacetime diagram using the rotated graph paper.

This is for $\gamma=2$ [so, $v=\sqrt{3}/2$] (the video's situation),
which isn't ideal for handdrawing on rotated graph paper
since $k=2+\sqrt{3}$ (not rational).... so the piecewise-inertial traveler's diamonds don't fall nicely on the rotated graph paper.
I've drawn in the lines of simultaneity, which are parallel to the spacelike diagonal of the traveler's diamonds.
Note $2.5=\frac{10}{\gamma^2}$

With $v=4/5$, we have $\gamma=5/3$ and $k=3$.
Note $3.6=\frac{10}{\gamma^2}=10\left(\frac{3}{5}\right)^2$

 

[curiousburke]

@robphy this is going to take me some time to decipher.

 

[cianfa72]

View attachment 341329
If you ignore D for a moment, the only thing you can measure is the time between A and B reaching C - and here you can see that the time between those events as measured by C is shorter than the time measured by A and B, by comparing the vertical distance between the crossings.

Sorry, you mean draw another slanted red fine line that intersects the point (event) where the B red thin line meets the C blue thin line. Then evaluate the spacetime distance between the first red fine line through A/C (i.e. A meets C) and the last red fine line and compare with the vertical distance between those 2 events along the vertical C blue thin line.

 

[Ibix]

Sorry, you mean draw another slanted red fine line that intersects the point (event) where the B red thin line meets the C blue thin line. Then evaluate the spacetime distance between the first red fine line through A/C (i.e. A meets C) and the last red fine line and compare with the vertical distance between those 2 events along the vertical C blue thin line.

No, I just mean measure C's proper time between meeting A and B, which corresponds to coordinate time in this frame.

 

[cianfa72]

No, I just mean measure C's proper time between meeting A and B, which corresponds to coordinate time in this frame.

Yes, from the point of view of C along its worldline. But what about the difference of time between those two events (A meets C and B meets C) from the point of view of observers A and B (that share the same slanted red fine lines simultaneity convention) ?

 

[Ibix]

Yes, from the point of view of C along its worldline. But what about the difference of time between those two events (A meets C and B meets C) from the point of view of observers A and B (that share the same slanted red fine lines simultaneity convention) ?

You could add a red line of simultaneity through B reaching C if you want, but you can't read red times off that graph without a calculation or robphy's clock diamonds.

 

[Ibix]

I see I've missed a few posts in this thread.

FWIW, Ibix, it's probably very frustrating working with people like me, sorry.

No, it's fine. Nobody learns instantly and, as various other posters in this thread will know, I've needed things repeated a few times myself.

 

[cianfa72]

but you can't read red times off that graph without a calculation or robphy's clock diamonds.

Yes, that was my point: in that diagram we add the aforementioned slanted red fine line through the event (B meets C) and calculate the spacetime distance between this and the other slanted red fine line through event (A meets C) along a straight line Minkowski-orthogonal to them.