Multiocular O (ꙮ) is the most rare and exotic glyph variant of Cyrillic letter O. This glyph variant can be found in certain manuscripts in the phrase «серафими многоꙮчитїи» ("many-eyed seraphim"). It was documented by Yefim Karskiy from a copy of Psalms from around 1429, now found in the collection of the Trinity Lavra of St. Sergius, and subsequently incorporated into Unicode as character U+A66E.
== See also ==
Monocular O
Binocular O
Dual Monocular O
Cyrillic characters in Unicode
== References... ==
@xnor I mean that the light from A and the light from the end of the minus pole reach B simultaneously. Subtracting yields zero time from A to the end of the minus pole. Instantaneous.
Quantum calculus, sometimes called calculus without limits, is equivalent to traditional infinitesimal calculus without the notion of limits. It defines "q-calculus" and "h-calculus", where h ostensibly stands for Planck's constant while q stands for quantum. The two parameters are related by the formula
where is the reduced Planck constant.
== Differentiation ==
In the q-calculus and h-calculus, differentials of functions are defined as
and
respectively. Derivatives of functions are then defined as fractions by the q-derivative
and by
In the limit, as h goes to 0, or equivalently as q goes...
@GlenTheUdderboat I don't follow. A is moving with v. A generates a light flash in its origin when it overlaps with B. In A both ends of the 1 Ls long poles are reached after 1s. In B my calculations don't match what I'd expect, but it certainly is not the same for both ends
@xnor I believe this is the first time that you mention that A and B are overlapping at some point. I assumed A was quite a bit away (more that the length of the pole).
@GlenTheUdderboat it could be away... My example changed to this (including technical details): A has a photon source, detector and timer in its origin and from there extending a 1 Ls long pole in positive x direction with a mirror at the end. B also has a detector and timer. B sees A moving with v=0.866c. When both timers line up they are reset and A generates a photon in positive x direction.
A detects the photon coming back after 2*1s. B detects it when and why?
A explodes. The light of the explosion triggers a secondary explosion at the end of the "minus" pole. B is somewhat behind (so the end of the pole is between A and B). B sees both explosions simultaneously. That was my point.
In B A moves. The initial light takes some time to travel to both ends, it takes a lot longer to travel to the positive x direction end, because that is where A moves with 0.866c
@xnor I'm only talking about the "minus" pole. The one of which I assumed the end to be between A and B. I'm currently not concerned with the other pole (which I think you might have gotten right).
@xnor Well, if you're going to insist on A and B being in the same spot at the time of the flash, then my example doesn't work. Too bad, because I liked it.
Well if they don't overlap at the instant of the flash then how would you synchronize them? You'd have no clue in B when A is flashing. You'd just receive some flashes after some random time.
@0celo7 "The h-calculus is just the calculus of finite differences, which had been studied by George Boole and others, and has proven useful in a number of fields, among them combinatorics and fluid mechanics. The q-calculus, while dating in a sense back to Leonhard Euler and Carl Gustav Jacobi, is only recently beginning to see more usefulness in quantum mechanics, having an intimate connection with commutativity relations and Lie algebra."
@GlenTheUdderboat no, it simplifies things, because then you have an instant both agree on, can reset their timers on and agree on when the initial flash is created
So I still don't know why the lorentz factor doesn't match or what exactly is going on
@xnor : maybe you need to ask a question and get some carefully explained answers that do help. Meanwhile try to imagine what kind of world it would be if light didn't move at all.
Imagine you want to simulate a cubic meter down to the particle. By following the Standard Model and other basic physical equations, how much computing power would be required to do this, in say, a day?
Would a quantum computer help you in this task? Could you somehow directly simulate the parti...
@ChrisWhite So the graduate algebra class is starting soon. I'm really curious about discrete subgroups of lie groups and that kind of thing, and I found undergraduate algebra to be pretty entertaining, but I'm not so sure it would be the class for me... any thoughts? I had actually already bought Hungerford's book (for being Rudin-like and fun) for cheap, but I'm not sure if I should focus on general relativity instead...
@punkerplunk Haha well I mean, groups are so general they come up everywhere, physics is mostly about calculus, so groups+calculus=lie groups=lotsa physics!
It's like in classical mechanics. If you're in a rotating frame of reference Newton's laws look different and there's no galilean invariance and all that stuff, straight linesl ook like circles, it's ugly.
@0celo7 as it's introduced normalllyyyyyy dudeeee, I know you can do a transformation on the metric, and I know Landau defines the special relativistic metric as one which can be transformed to be the usual one everywhere.
I was attempting to make a statement that, because this (not) paradox holds true, it would be possible to travel to a location, even if at nearly the speed of light, and return OLDER than your twin
@0celo7 I just mean to say that the laws of physics need to be modified. Like going into a rotating frame in CM causes the addition of fictitious forces
@punkerplunk Right, sidetracked, sorry. The point I'm making is that regular, introductory special relativity assumes you're working in a frame that is inertial throughout all time. But in Alice's frame, objects without force acting on them would move in a line with a sharp corner.
Because blanch returns to Alice in the illistration, and is OLDER than alice when she returns - because the entire question was framed as the earth as the inertal frame
I attempted to make the case that the choice of the earth is arbitrary. That I could have chosen nearly any frame of reference as the 'inertial' frame . . .
@punkerplunk Yeah, but if you consider a short enough period of time in its path around the sun, objects in motion will continue moving in a straight line, etc.
so it will be inertial.
It's like the issue of angular velocity being absolute in Newtonian mechanics.
@punkerplunk The Earth's frame, and any frame that moves with respect to the Earth's frame with a constant velocity, is inertial. Inertial frames are the simplest frames in special relativity, so yes, in that sense, special relativity prefers inertial frames.
But we're not talking about one frame, we're talking about "Earth's [momentary rest] frame and any frame that moves with respect to that frame with a constant velocity"
-sigh- I'm making a bunch of tiny mistakes in my wording, sorry.
No, a journey in a spaceship is a line in spacetime. If $e_i$ are your basis vectors and $x_i(s)$ are your positions as functions of time, a journey in a spaceship is a line that can be parameterized as $\sum e_i x_i(s)$. It's independent of the basis you work in.
A change of reference transformation is a change of basis plus a translation.
but a path is not an action on the basis. It's just a line/set of points.
There's another thing I think you might be getting at: Are you concerned whether proper time (the time elapsed for an observer) is an axis of spacetime or a scalar?
So, would that suggest it would be possible for an object to leave the earth and nearly the speed of light, go to 'some location X' and return OLDER - the opposite effect of the normal twin paradox
for instance, we have the following general expression for the Lorentz factor $$\gamma=\sqrt{(1+\vec a\cdot\vec{OM})^2-(\vec V+\vec \omega\times_u \vec{OM})^2}^{-1}$$
@punkerplunk $v$ is a vector that is one leg of the trip, $w$ returns to Earth. $v+w$ is just a line going in a straight path. The given inequality implies that the time elapsed for Earth, $\|v+w\|$, is always greater than the sum of the time elapsed on the two legs, $\|v\|+\|w\|$.
@punkerplunk No, because that inequality applies to every velocity vector [velocity vectors are easier to phrase stuff with than line segments here] in Minkowski space, which holds for any inertial frame of reference. (And to appease 0celo7, yes, non-inertial ones too)