Recall that I've asserted that one cannot move their now "faster" or "slower" than others - even without "turning around".
You discussed stopping and reversing your time direction in the same breath in post 74. I'm going to come back to that in a bit, but in the above you are incorrect as well.
Let's say you're looking at an ordinary plane (Euclidean), and you want to describe directionality of something, so you draw a bunch of vectors of the same length Δs, say some standard meter-stick. The Pythagorean theorem/distance formula then says that from their point of origin, the vectors satisfy (Δs)² = (Δx)² + (Δy)², like so:
Code:
/| Δs² = (Δx)² + (Δy)²
Δs/ | Δy = (Δs)²cos²θ + (Δs)²sin²θ
/ | ^y
/___| | Euclidean case
Δx +->x (angle θ from pos. x-axis)
As you vary your angle θ, the operation traces out the curve s² = x²+y² if the vectors are from the origin: a circle. That seems like a lot of work just to justify what everyone intuitively already knows--that in space, you can turn any which way you like. But this is going to be important in contrasting what happens when time is involved.
Suppose you're recording events, when and where they occur, and you plot them in four-dimensional spacetime. There's yet no claim about any physicality/'reality' or any other significance of events in this spacetime. For now, it's just 4-tuples (t,x,y,z), so it makes sense in both Newtonian and relativistic physics. You can draw vectors in this spacetime; physically, they represent a change of position (Δx,Δy,Δz) during some change of your time time measurement (Δt), i.e., a
velocity. What does rotation mean? Well, it's changing the 4-vector, which changes the velocity, so rotation in spacetime =
acceleration. I'm going to supress two spatial dimensions.
(BTW, note that there is no mathematical problem in the clock going backwards in time; just make Δt < 0 in your 4-vector.)
But the velocity does not determine the length of that vector, since (Δt,Δx) represents the same velocity as, say, (2Δt,2Δx), as (2Δx)/(2Δt) = Δx/Δt. So just as we had a standard meter-stick before, let's take a standard clock-tick Δτ as the length, representing the time some standard clock would give when it travels through the events along the vector. The Newtonian case is kind of boring, though:
Code:
Δx
+---
| / ^t (Δτ)² = (Δt)²
| /Δτ | Galilean/Newtonian case
|/ +-->x
It just says that the clock's measurement is the same as your measurement. Your time is everybody's time. No matter which way you turn in spacetime, you can't change that. No one goes "faster" or "slower" in time: |Δτ/Δt| = 1, always. Hyperplanes of simultaneity are also same for everybody:
Code:
<-------->
<--------> ^t Simultaneity in Newtonian physics
<--------> |
<--------> +-->x
So if you wanted to, you could consistently claim that 'now' is in some way
more special than the past or future. You don't have to, of course; "they're all equally real" is also consistent with Newtonian physics. But there's nothing in the physics that forces you to choose one way or another. It's a metaphysical question at best.
If we describe spacetime as four dimensions, "now" is a special hyperplane sweeping through the time dimension - at the exact same pace and time location for everybody.
I don't agree that we're forced to admit that in Newtonian physics, but I agree that it's consistent. But it's definitely
not consistent with relativistic physics. There, things a bit different there:
Code:
Δx (Δτ)² = (Δt)² - (Δx)²
+--- = (Δτ)²cosh²α - (Δτ)²sinh²α ]
| / ^t Lorentzian/special-relativistic case
| /Δτ | (angle α from pos. t-axis)
|/ +-->x
Already there is clear contradiction with your claim that "one cannot move their now 'faster' or 'slower' than others", since if (Δτ)² = (Δt)² - (Δx)², then the moving clock's measurement Δτ will not be the same as your measurement Δt. With a bit of algebra, one can rearrange it to:
|Δt/Δτ| = 1/sqrt[ 1 - (Δx/Δt)² ] = 1/sqrt[ 1 - v² ] = γ ("Lorentz gamma"),
which is the usual form of relativistic time dilation.
Your 'now' and the moving clock's 'now' are "advancing to the future" at different rates!.
So you can indeed affect this just by accelerating. Can you turn around completely, i.e., the go backwards in time? Well, no; just as in Euclidean geometry, rotation traces out a circle, here it traces out the curve τ² = t² - x², which is a hyperbola. Hyperbolas have two asymptotes; no matter how hard you accelerate, you won't cross them. Your hyperplanes of simultaneity and the clock's hyperplanes of simultaneity will be different as well. Unfortunately, I can't draw good diagrams in ASCII, but I can try to explain it if you like.
I'm not in all honestly understanding how the analogy of "turning around in a tight space" applies here.
Suppose there were two time dimensions, t and t'. Say the spacetime metric was (Δτ)² = (Δt)² + (Δt')² - (Δx)² - (Δy)² - (Δz)² or something. Then the time parts together would act like an Euclidean subspace and you could turn around any way you liked in time. "Why is there just one time dimension?" is not any more asburd than "why are there three spatial dimensions?".
... it sounds like hand waving - electrons could back up in time without turning around, but humans would have to turn around first. Up to what size is it possible for particles to reverse themselves in time without "turning around"?
No, the first point was that you can't turn around because there only is one time dimension (if there were more, then you could), and time does not mix in the same way with space (see above).
The second point was that if you can't turn around by actual rotations, then the only other way available to you is some kind of discrete "flip" in your time orientation: one moment you're going forward, the next backward without any any intermediate turning. That's the "tight space" analogy. Electrons can do this in some sense. You can't.
Is there experimental evidence of that? If electrons can reverse in time (with or without turning around) can electrons stop in time, or move faster or slower?
Sure: time dilation, as above (though it can't
stop for massive objects). As for reversing, also yes, again in the sense that there are phenomena that can be interpreted that way in a logically consistent manner. Suppose you have an electron bouncing in time several times, sometimes going forward an sometimes going backward
Code:
/\ /^ / = diagonal-up (can't draw diagonal
/\ / \ / t^ \ = diagonal-down arrows very well)
/ \/ \ / |
/ \/ +-->x
It needs something to bounce
from, and besides gravity, electrons interact only with the electromagnetic field (and to other charges
via the electromagnetic field). So imagine than in every vertex, there is also a photon likewise bouncing into the other time direction (of energy E = γmc², so it's nothing to sneeze at for larger masses!).
What would that look like to you? Well, an electron going backward in time would react oppositely to electromagnetic fields, so you would measure opposite charge. Every down (V) vertex in the above diagram would look like two photons coming in and an electron and its mirror twin coming out, while every up (^) vertex would look like an electron and its twin coming in and two photons coming out. In other words, electron-positron pair creation and annihilation, respectively.
At least we're all in a similar stew of trying to discuss metatime, and lacking good words for it, so I'm not picking on the wording.
