Continuing from Part 23, in the grand tradition of curious explorers, let’s keep digging..
We dig up diamonds by the score
A thousand rubies, sometimes more
Though we don't know what we dig 'em for
We dig dig dig a-dig dig
We are left with exp(tX) for yet unexplored candidates for X, namely e1, e2, ie1, ie2. Here the calculations can become cumbersome, and, in fact, this what they are. But they do not have to be such if we are clever enough. I was not that clever at the beginning, so I lost quite some time, until I remembered something vaguely, and an enlightenment suddenly occurred to me. After that I checked, and rechecked, and here comes you rabbit from the hat - the Eureka great discovery! The trick is extremely simple ..... once you know it!
We have these e1, e2, ie1, ie2 that we want to exponentiate. But why exactly THEM? They are part of the "Lie algebra" (you don't have to know what the Lie algebra is), they satisfy the fundamental requirement X+ ν(X) = 0, but so is any linear combination of them! So, let's take these linear combination, but in a clever way. And what is this "clever way" I am telling you right away, now.
Here are the linear combinations that work for us:
1. e1 + ie2
2. e2 - ie1
3. e1 - ie2
4. e2 + ie1.
We will take 1/2 of them. Today, we’ll focus on 1 and 2. Combinations 3 and 4? I leave those as a rewarding exercise for you, dear Reader.
Now, what is so special about these linear combinations? What is this "great discovery"? Take, for instance
X = (e1 + ie2)/2
We have that e1 and e2 anti-commute and their squares are 1, while i2 = -1. As the result X2 = 0. Check it! As the result the whole series for the exponential exp(tX) terminates after the linear term:
exp(tX) = 1 + tX. = 1 + t(e1 + ie2)/2
Then τ(X) = (e1 - ie2)/2,
and
τ(exp(tX)) = 1 + t(e1 - ie2)/2.
Thus we need to calculate
e'0 = (1 + t(e1 + ie2)/2) e0 (1 + t(e1 - ie2)/2),
e'1 = (1 + t(e1 + ie2)/2) e1 (1 + t(e1 - ie2)/2),
e'2 = (1 + t(e1 + ie2)/2) e2 (1 + t(e1 - ie2)/2),
e'3 = (1 + t(e1 + ie2)/2) e3 (1 + t(e1 - ie2)/2).
It’s a straightforward algebraic exercise—boring, yes, but manageable. I used the powerful and free computer algebra system "Reduce" to automate these tedious steps. If you’re not using tools like this, you’re missing out! Anyway, you tell the program the rules, and then it does the job. My rules were
Operator e$
Noncom e$
for all i let e(i)*e(i)=1$
for all i,j such that i>j let e(i)*e(j)=-e(j)*e(i)$
let e(1)*e(2)=i*e(3)$
let e(2)*e(3)=i*e(1)$
let e(1)*e(3)=-i*e(2)$
Note: Alternatively, one can play with the Pauli matrices instead.
In my mind I set e0=1, as it is an exception. Then you
give it expressions to calculate, and then, without even a blink of the
eye, it spits out the results. You tell the program to write them to a
file. Here they are, rewritten using our notation, where I manually
replaced the scalar part by e0:
e'0 = (1+t2/2) e0 + te1 + (t2/2)e3,
e'1 = te0 + e1 + te3,
e'2 = e2,
e'3 = -(t2/2)e0 - te1 + (1-t2/2)e3.
The transformation matrix, in Mathematica notation, is
{{1 + t^2/2, t, 0, t^2/2},
{t, 1, 0, t},
{0, 0, 1, 0},
{-t^2/2, -t, 0, 1 - t^2/2}}
The coordinates transformation matrix is the transposed matrix,
m={{1 + t^2/2, t, 0, -(t^2/2)},
{t, 1, 0, -t},
{0, 0, 1, 0},
{t^2/2, t, 0, 1-t^2/2}}
We are almost done. We recall from Part 18 the formula for embedding the x1=x, x2=y plane into the null cone of Minkowski space:
x ⟼ ζ={ζ0,ζ1,ζ2,ζ3}={1, 2 x/(1 + q), 2 y/(1 + q), (1 - q)/(1 + q)},
where I set x·x = x2 + y2 = q.
We act with the matrix m on vector ζ to get:
ζ'={(1 + q + q t^2 + 2 t x)/(1 + q), (2 (q t + x))/(1 + q), (2 y)/(1 + q), (1 + q (-1 + t^2) + 2 t x)/(1 + q)}
From this we can read x' = ζ'1/(ζ'0+ζ'3) and y' = ζ'2/(ζ'0+ζ'3). The result is:
x' = (x + q t)/(1 + q t^2 + 2 t x),
y' = y/(1 + q t^2 + 2 t x).
Now we can repeat the same procedure with
Y = (e2 - ie1)/2.
This time we get
x' = x/(1 + q t^2 + 2 t y),
y' = (y +qt)/(1 + q t^2 + 2 t y).
We can then check the very important property: X and Y commute! We replace t by "a" for tX, and by "b" in tY. Then
g = exp(aX)exp(bY)=exp(bY)exp(aX)=exp(aX+bY).
The transformation implemented by such a g is then
x' =(x + qa)/(1+ q (a2+b2) + 2(xa+yb)),
y' =(y + qb)/(1+ q (a2+b2) + 2(xa+yb)).
This
is known as a "special conformal transformation". Conformal
transformations are known to preserve angles, and to transform circles
into circles (straight lines are considered as infinite radius circles).
Exercise 1. Do the same with 3) and 4). This should be even easier. Why? You will see when you try.
P.S. 11-12-24 17:07 Differentiating the formulas
x' = x/(1 + q t^2 + 2 t y),
y' = (y +qt)/(1 + q t^2 + 2 t y).
with respect to t at t=0 we get the vector field {-x^2 + y^2, -2 x y} for our one-parameter group. Here is the image:
And, for comparison, here is the vector field of the one parameter group {exp(-2t)x,exp(-2t)y} calculated by Bjab in Part 23:
P.S. 13-12.24 13:02 There will be a delay with the following post. I am having problems with calculations for this post. They do not want to obey my expectations. Ends do not meet.
Is maybe in 3. e1 - ie1, instead of e1 - ie2? Since as they ste now, 1. and 3. differ only for a sign between e1 and ie2.
ReplyDelete3) was ok but 4) was messed up. Thanks.
ReplyDeletee'0 = (1+t2/2) e0 + te1 + (t2/2)e3 ->
ReplyDeleteHow could you get halves?
Fixed. Forgotten that I took 1/2 from the very beginning. Thanks
ReplyDelete"Fixed"
DeleteCan't see.
The updated version has:
DeleteTake, for instance
X = (e1 + ie2)/2
etc.
"And, for comparison, here is the vector field of the one parameter group {exp(2t)x,exp(2t)y} calculated by Bjab in Part 23..."
ReplyDeleteBut Bjab got:
x1' = x1 exp(-2t)
x2' = x2 exp(-2t)
that is the group {exp(-2t)*x,exp(-2t)*y} which when differentiated and evaluated for t=0 gives vector field {-2x,-2y}, i.e. pointing to the origin and not from the origin as the image shows.
Except if Bjab made a mistake in the calculation and there should be no minus sign in the exponent.
Fixed. Thanks
DeleteWhat would those vector fields represent (the images)?
DeleteDifferentiation WRT parameter t is basically looking at change of position (x,y) with change in rapidity, if t is rapidity, in which case t=0 would mean at rest, no boost and no rotation.
So, while being still, we look at the South Pole (origin) and the arrows are showing us what exactly? How the stars move when we start moving while still looking to the point at the South Pole, i. e. to the past?
The arrows represent the direction of flow at each point, and the color tells us the rate of flow.
DeleteBetter would be to show the animation of the flow of points. I will try to create such an animation.
OK, but I don't understand what kind of the flow you mean, as the differentiation is done WRT to rapidity, flow of what exactly?
DeleteIf you keep your hand on the "accelerate" gadget, the rapidity will be increasing proportionally to time. And so the stars will keep moving.
DeleteWith the conformal transformation the "accelerate" is coupled to the appropriate "rotate".
Maybe no need to spend time on producing animation, you can use the default setting, the arrow thickness for the vector magnitude, or perhaps arrow length, like in so called "brush" or "comb" diagram that is used to represent vector fields like sea currents or wind maps.
DeleteIf I read this right, red color is higher magnitude meaning that stars at the outer parts of (x,y) plane move faster than those in the center, and basically start to move as soon as we start boosting and/or rotating, which is exactly what would normally be expected as they have larger "paths" to travel to reach the convergence point in the center (in Bjab's case for example).
If I get this right, if we now plot differentiation WRT to t, in t=0, for combined cases of (1)+/-(3), or (2)+/-(4), we will get the "movement" of the stars in individual cases for boost in e1 and rotation about e2, or boost in e2 and rotation about e1, respectively, right?
DeleteSimilar to plotting {2y - 2x, -2x - 2y} that would visualize the "motion" or flow of stars for simultaneous boost in e3 and rotation about e3, which would mean that the end stereographic projection transformations are additive, i.e. linear?
Is that correct?
P.S. Which options for Mathematica's VectorDensityPlot do you use (for example ColorFunction, VectorScale, StreamPoints, etc.) to get such nice vector field plots?
DeleteP.P.S. The star flow lines, for combined boost in e3 and rotation about e3 for example, resemble quite to the lines on those tori you showed us in Part 2, or to Villarceau Circles or Penrose's twistors in
Deletehttps://ark-jadczyk.blogspot.com/2024/09/octagonal-complexigram.html
FWIW.
I used ColorFunction -> "ThermometerColors", StreamPoints -> Fine
DeleteThanks!
DeleteInterestingly, the end result turned up identical for the composition of boost e3 and then rotation ie3 (and rotation ie3 and then boost e3) and for just adding the results/transformations of both together, i.e. after d/dt at t=0, one gets what's written in previous comment, the vector field {2y-2x, -2x-2y} for all 3 cases.
DeleteI might guess the reason for that (like I already did in that comment), but maybe better is to ask, why do we get identical results at the end for just adding the transformations and applying them in succession/composition?
@Saša Can you rephrase your question, split it if possible, explain. I would like to answer, but there is too much for me in one sentence.
DeleteSo, when apply the composition of transformations for ie3 (rotation about e3 you derived in previous post Part 23) and then for e3 (the boost in e3 Bjab calculated in comment in Part 23), I get the same result like when applying composition in reverse order, i.e. first boost in e3 and then rotation about ie3, as expected and as you demonstrated for compositions in conformal cases. Normally, the result for the end vector field, after d/dt at t=0, is the same for both compositions, {-2x+2y,-2x-2y}.
DeleteThe surprising thing, for me at least at this point on my learning spiral, is that I get the same resulting vector field if I just add the transformation rules for e3 and ie3, and evaluate the d/dt at t=0 for these "sums" (like in that Mathematica notebook I sent you by mail yesterday evening).
The transformation rules for compositions and this ad hoc addition by hand are not even remotely similar. So I've been wondering why are the end results for vector field of stars' positions flow identical when the "origin" transformation rules are so evidently different? Coincidence?
Let me see if I understand it correctly. We have ie3, and generate rotations by taking exp( t ie3). We generate boosts by taking e3 and exp(s e3). Now, e3 and ie3 commute. Therefore exp(t ie3)exp(s e3) = exp(s e3)exp(t ie3) = exp(t ie3 + s e3). Now what it is about vector fields? I did not get it yet? Explain also this: "The transformation rules for compositions and this ad hoc addition by hand are not even remotely similar."
DeleteOK, for rotation about e3, i.e. for exp(t*ie3) we got:
Deletex' = cos(2t)x + sin(2t)y,
y' = cos(2t)y - sin(2t)x.
For boost in e3, i.e. for exp(s*e3) we got:
x' = x*exp(-2s),
y' = y*exp(-2s).
If we choose t=s and apply the composition, we get:
x' = cos(2t)*x*exp(-2t) + sin(2t)*y*exp(-2t) =
= (cos(2t)x + sin(2t)y)*exp(-2t),
y' = cos(2t)*y*exp(-2t) - sin(2t)*x*exp(-2t) =
= (cos(2t)y - sin(2t)x)*exp(-2t).
After differentiating WRT t and evaluating at t=0, we get the vector field {-2x+2y,-2x-2y}.
The identical vector field {-2x+2y,-2x-2y} is obtained from the transformation rules after their differentiation d/dt and evaluation at t=0, if the transformation rules for exp(t*e3) and exp(t*ie3) are just simply added:
x' = cos(2t)x + sin(2t)y + x*exp(-2t),
y' = cos(2t)y - sin(2t)x + y*exp(-2t).
Coincidence or there is some "deeper" physical or mathematical meaning behind this?
Thank you. Now I understand. The point is that it it is enough to differentiate at t=0 to get the "generating vector field. This is the case with your first calculation.
DeleteThen you have the second formulas. Changing t you draw some lines in the x,y plane. They will probably intersect for some valuse of t and different x,y, but you do not care. You just take tangeny vectors to these lines at t=0. This is the same vector field as in the first case. Can you recover from this vector field the formulas they came from? No. You can't. But you can recover from this vector fields the lines of the group case.
Or differently. If you draw the trajectories of you second formulas, they will not agree with your vector field except for one point: t=0.
Am I clear?
To simplify: take x(t) = e^t x. The vector field X(x)= 1x.
DeleteBut take x(t)= tx. The derivative at t=0 is also 1x.
But e^t is a solution of the differential equation
dx(t)/dt = 1 x(t)
while for x(t)=tx we have dx(t)/dt = (1/t) x(t).
The vector field 1 "generates" the exponential solution, but not the second formula.
"Or differently. If you draw the trajectories of you second formulas, they will not agree with your vector field except for one point: t=0.
DeleteAm I clear?"
You seem to be, but I don't quite get it, yet.
For the first transformation formulas, at t=0, we get:
x' = x and y' = y;
while for the second transformation formulas, at t=0, we get:
x' = 2x and y' = 2y;
which is obviously not the same point, i.e. those trajectories do not meet at t=0.
"But you can recover from this vector fields the lines of the group case."
What is "the group case" exactly?
So, from the vector field {-2x+2y,-2x-2y} I would be able to "recover" the correct transformation rules:
x' = (cos(2t)x + sin(2t)y)*exp(-2t),
y' = (cos(2t)y - sin(2t)x)*exp(-2t),
but not the wrong ones:
x' = cos(2t)x + sin(2t)y + x*exp(-2t),
y' = cos(2t)y - sin(2t)x + y*exp(-2t),
is that correct?
At the end, that those two set of formulas have the same tangent vectors for the same parameter t, at t=0, is just a mathematical quirk or the consequence that they are both the resulting actions of different subgroups of the same "origin" group G?
Quirk. Firs st of transformations has the group property. The second set describes a bunch of trajectories which at one point (t=0)are tangent to our vector field, but only at this one point.
DeleteWith the first set of you differentiate at t=1 instead of t=0, it will be the same if you take x(1) as the starting point of the trajectory.
Perhaps tomorrow I will explain it to you better, with a graphics.
Delete"To simplify: take x(t) = e^t x. The vector field X(x)= 1x.
DeleteBut take x(t)= tx. The derivative at t=0 is also 1x.
But e^t is a solution of the differential equation
dx(t)/dt = 1 x(t)
while for x(t)=tx we have dx(t)/dt = (1/t) x(t)."
I get it.
If x(t) = x*e^t, dx(t)/dt = x*e^t, which for t=0 gives x.
If x(t) = x*t, dx(t)/dt = x for any t and so also for t=0.
They both are the same value at t=0, but represent rather different things, first one is solution of diff. eq. dx(t)/dt - x*e^t = 0, while the other one is solution of diff. eq. dx(t)/dt - x = 0.
So, in principle, if I get this all right, that we get the same vector field {-2x+2y,-2x-2y} for both, correct and wrong transformation rules, is just a consequence of "appropriately writting" the wrong one to get the same result for d/dt at t=0 as for the correct one. Nice, mathematics at its best. :)
In other words, coincidence. Or, more precisely, as there are probably no coincidences in this wise and clever Universe of ours, 'mind' playing games with/on me, using my temporary ignorance. Luckily for me, you're right here to help me learn and eradicate a bit of my ignorance so to call the mind's tricks. Thanks! :)
"With the first set of you differentiate at t=1 instead of t=0, it will be the same if you take x(1) as the starting point of the trajectory."
DeleteAha, now I got it completely. Thank you very much!
this we cane ->
ReplyDeletethis we can
read x' = ζ'2/(ζ'1+ζ'3) ->
read x' = ζ'1/(ζ'0+ζ'3)
No we can ->
Oh, we can
We can, indeed. Thanks.
DeleteUprzednio włożyłem dwa grzyby w barszcz a tylko jeden został zjedzony, został:
Delete(ζ'1+ζ'3) ->
(ζ'0+ζ'3)
Barszcz z uszkami. Pychota! Thanks!
Delete"Differentiating the formulas
ReplyDeletex' = x/(1 + q t^2 + 2 t y),
y' = (y +qt)/(1 + q t^2 + 2 t y)
with respect to t at t=0 we get the vector field {-x^2 + y^2, -2 x y} for our one-parameter group."
Seems the X and Y cases got mixed, as the vector field {-x^2 + y^2, -2 x y} corresponds to d/dt of transformations for X:
x' = (x + q t)/(1 + q t^2 + 2 t x),
y' = y/(1 + q t^2 + 2 t x),
at t=0.
For the Y, i.e. the case in quotes, one gets vector field {-2 x y, x^2 - y^2}, the image of which is basically rotated for 90° WRT to the image in the P.S. 11-12-24 17:07.
FWIW.
P.S. Sent you the Mathematica notebook file with calculations and images by mail.
Would you consider a belated comment please?
ReplyDeleteI am trying to check explicitly the formulas for conformal tranformation by sequentially applying first
x' = (x + q a)/(1 + q a^2 + 2 a x),
y' = y/(1 + q a^2 + 2 a x)
and then
x'' = x'/(1 + q b^2 + 2 b y'),
y'' = (y' +qb)/(1 + q b^2 + 2 b y').
Expected to obtain those nice formulas
x'' =(x + qa)/(1+ q (a2+b2) + 2(xa+yb)),
y'' =(y + qb)/(1+ q (a2+b2) + 2(xa+yb)),
right?
But i cannot. Will try in the morning with fresh eyes.
No, i cannot obtain the formulas for conformal tranformation:
ReplyDeletex' =(x + qa)/(1+ q (a2+b2) + 2(xa+yb)),
y' =(y + qb)/(1+ q (a2+b2) + 2(xa+yb))
as a composition of
x' = (x + q t)/(1 + q t^2 + 2 t x),
y' = y/(1 + q t^2 + 2 t x)
and
x' = x/(1 + q t^2 + 2 t y),
y' = (y +qt)/(1 + q t^2 + 2 t y).
Can we can rather consider them as a generalization of those for coordinates? Like that:
x'_i = (x_i + q a_i)/(1 + 2(a_i x_i) + q a^2),
where x_i = (x, y), a_i = (a, b), and q = x^2 + y^2
This general formula can be found, e.g., at two last pages here http://theor.jinr.ru/~shnir/Group%20Theory/Groups_Lecture4.pdf
Thanks. I will check the formulas. During the night I was thinking of spinors. Now I can go back to conformal.
DeleteYou should consider
Deletex' = (x + q a)/(1 + q a^2 + 2 a x),
y' = y/(1 + q a^2 + 2 a x)
and
x' = x/(1 + q b^2 + 2 b y),
y' = (y +q b)/(1 + q b^2 + 2 b y).
to get
x' = (x +q a)/(1 + q (a^2+b^2) + 2 (a x + b y),
y' = (y +q b)/(1 + q b^2 + 2 b y).
I used Mathematica to verify the composition formula. Here is my code:
First the definition of the transformation. I think of a being a vector (alpha, beta) and x being a vector (x,y).
c[a_, x_] := (x + (x . x) a)/(1 + 2 a . x + (x . x) (a . a));
Now I check the composition formula:
Simplify[
c[{\[Alpha], 0}, c[{0, \[Beta]}, {x, y}]] -
c[{\[Alpha], \[Beta]}, {x, y}]]
The result is
{0,0}
Correction to the last equations
Deletex' = (x +q a)/(1 + q (a^2+b^2) + 2 (a x + b y)),
y' = (y +q b)/(1 + q (a^2+b^2) + 2 (a x = b y)).
It was my very bad idea to use t as a parameter. If you compose two transformations t in both does not to be the same. In the x case t translates x by some amount, in the y case we have translation of y by some other amount. I should have used a and b from the very beginning. Or a1 and a2 as in your linked lectures.
DeleteAnd one more comment: Very often when I write I mistype. Bjab knows it very well and every formula I write he takes with suspicion. You should never believe any formula I write. There may be a typo there with a very high probability!
DeleteSo, my writing is rather like an adventure. We are trying to find the truth together. My Reader is much a creator of the content as I the author. Of course the whole responsibility is on me. But the Reader is welcome to be sarcastic, make fun of me, sometimes be angry at me, when I deserve it.
DeleteI think you're maybe being a bit too harsh on yourself. And we the Readers are responsible for our emotional states, as well as for our presentation of them to the outside world.
DeleteOn the other hand, you extend your time and energy in effort of sharing your vast knowledge with us, while allowing us in the process to co-create something of value and at the same time learn quite a lot. For that you deserve all the credit, thank you!
Yes, this variant
ReplyDeletex' = (x +q a)/(1 + q (a^2+b^2) + 2 (a x + b y)),
y' = (y +q b)/(1 + q (a^2+b^2) + 2 (a x = b y))
looks more like my linked formula, indeed.
But now it differs from what we obtained using transformations based on (e1 + ie2) and other combinations of Cl elements :(
This is a general formula:
Deletex' = (x +q a)/(1 + q (a^2+b^2) + 2 (a x + b y)),
y' = (y +q b)/(1 + q (a^2+b^2) + 2 (a x = b y))
Setting a=t, b=0 we get the first formula from the blog, for X = (e1 + ie2)/2:
x' = (x + q t)/(1 + q t^2 + 2 t x),
y' = y/(1 + q t^2 + 2 t x).
Setting a=0, b=t we get the second formula from the blog, for Y = (e2 - ie1)/2.
x' = x/(1 + q t^2 + 2 t y),
y' = (y +qt)/(1 + q t^2 + 2 t y).
Oh, yes, agreed!
DeleteSo, not too bad then. Added P.S. at the end of the post.
Delete@Ark You should not blame "t", of course, i used vector a_i = (a, b) for calculation, just like i wrote in my first comment yesterday. Today i was in a hurry and copied the formulas with "t", sorry.
ReplyDelete"...takes with suspicion. You should never believe any formula..."
ReplyDeleteGenerally, if we took everything on trust and did not think beyond a suggested framework, we would not arrive here, at this Blog. So, Ark, you should not doubt in our critical approach, and thank you for the opportunity to participate in weaving the intricate net you conceived of.
Something top notice. Both rotations and dilations, if they are considered on the sphere instead on the sterographically projected plane, they have two fixed points: North and South poles. But the conformal transformations that we have just discussed have only one fixed point. The same concerns that other not yet identified transformation, left as an exercise.
DeleteTo be more precise: every particular conformal transformation has just one fixed point. If we change the transformation, the fixed point changes.
DeleteIn my Quantum Fractals book I classified these transformations with only one fixed point as "parabolic" - such is a terminology used by mathematicians.
The physical interpretaion of this one fixed point is "the spin direction measurement detector orientation". But tha is still in future.
DeleteOh, those stereographic projections, North and South poles... Too tough for me now. Hope to advance a bit with the help of your Quantum Fractals book
DeleteAnna, when I was writing this book, I understood very little. I still understand very little though little less little.
DeleteConcerning your private comment about surreal numbers and quantum theory: we do not understand quantum theory, and using new mathematics will not help us. It can only obscure the real problem. That is how I see it. Of course I may be totally wrong, and most probably I am.
"Too tough for me now."
DeleteDon't be shy, ask questions, ask questions, ask questions!!!!!!!!
Просите, и дано будет вам
Delete'ask questions!!!!!!!!' so far i am not able even to formulate clearly what is unclear to me. We did not study such projections at MSU (or i have forgotten it absolutely), thus, i must first take a course for beginners to learn the basics.
Delete"Просите, и дано будет вам". -- "Никогда ни о чем не просите; особенно у тех, кто сильнее"
DeleteС диким зверем это может быть правильным отношением. Но я не зверь.
Delete"we do not understand quantum theory, and using new mathematics will not help us. It can only obscure the real problem."
DeleteSome part of me agrees with the above, but another appeals to the known examples: Copernicus put things from head on feet, drastically simplified and clarified our description of the world. How can we say that it will not occur once again at the new turn of progress? The more so that the new mathematics is conceived as natural and intuitive, built in compliance with human deep conscious and preconscious mechanisms of interaction with reality.
"he more so that the new mathematics is conceived as natural and intuitive"
DeleteIf it appeals to quantum theory, I consider it un-natural and counter-intuitive.
But i can be totally wrong.
DeleteAs can any of us.
DeleteAs far as Copernicus goes, at least to my knowledge, he did not introduce new math to describe what he did, but by changing the "fixed point" he simplified extensively the already used mathematical descriptions on the market. FWIW.
'If it appeals to quantum theory, I consider it un-natural and counter-intuitive'.
DeleteThat is the position of the mainstream, but some scientists consider the undeterministic character of QM as physical correlate of human conscious experience. Our mental activity differs from AI by its unpredictable character, which can never be modeled regularly and completely. Similar ideas are developed, for example, by Lee Smolin in his works on the 'completion of quantum mechanics' (https://philpapers.org/archive/SMOTPO-3.pdf). Nonseparable spaces, unpredictability, entanglement - such things are typical of both QM and human mentality. But this is a hard problem and is very far from what we are doing here.
"Никогда ни о чем не просите; особенно у тех, кто сильнее". Cited from "Мастер и Маргарита" by M.Bulgakov
DeleteThanks for the link to Smolin's paper. Perhaps it will help me to see something new, even though I classify him, perhaps wrongly, as an "advanced mainstream".
DeleteAmd thanks for reminding me the quotation from Bulkhakov. But I again, it is the right attitude when you live among wild beasts, and our life is essentially such. But exceptions from rules are sometimes even more important than rules themselves.
It happens quite often that problems in our lives are being created by LACK OF COMMUNICATION. We do not communicate when communication is required. Sometimes we later say to someone we care about: "Then why didn't you tell me, why didn't you ask?" And we hear a lot of excuses. Because this, or that. So, the decision of whether telling or asking someone in some circumstances depends on details. The the devil hides in those details. There may be a situation when asking is a bad choice, but there may be other situations when not asking is a bad choice.
Delete"P.S. 14-12-24 18:21 I see that the exercise at the end of my post is too demanding. And no one got convinced enough to use the computer algebra software. Too bad. So, tomorrow I will have to do my exercise all by myself."
ReplyDeleteThat would be very bad, to waste your time on "student's" assignments or exercises instead of going on with the lectures. And since tomorrow (or today already) is Sunday, and there's a possibility for a Sunday Special blog post treat to grace us with, exercise's done.
Did not use the algebra software, as the installation on my Linux machine seemed to take more time than to practice the calculations by hand. At the end is seems that a shortcut could have been taken as for X = (e1 + ie2)/2 and X1 = (e1 - ie2)/2 (cases (1) and (3)), exp(tX) = τ(exp(tX1)) and τ(exp(tX)) = exp(tX1), i.e.
exp(tX1) τ(exp(tX1)) = τ(exp(tX)) exp(tX) = e0*(1+t^2/2) + e1*t + e2*0 + e3*(-t^2/2),
τ(exp(tX1)) exp(tX1) = exp(tX) τ(exp(tX)) = e0*(1+t^2/2) + e1*t + e2*0 + e3*(t^2/2),
and similar for cases (2), Y = (e2 - ie1)/2, and (4), Y1 = (e2 + ie1)/2),
exp(tY1) τ(exp(tY1)) = τ(exp(tY)) exp(tY) = e0*(1+t^2/2) + e1*t + e2*0 + e3*(-t^2/2),
τ(exp(tY1)) exp(tY1) = exp(tY) τ(exp(tY)) = e0*(1+t^2/2) + e1*t + e2*0 + e3*(t^2/2),
so certain commutation relations could have been used, for example for calculating the rules for e0.
From the above relations also turned out that transformation matrices for X and Y are now coordinates transformation matrices for X1 and Y1, which to my enormous surprise gave for X1,
x' = x + t,
y' = y,
and for Y1,
x' = x,
y' = y + t,
which seems to indicate that stars' positions for the cases X1 and Y1 just move along our x or y axes.
Since X1 and Y1 also commute, for a "combined" transformation in case of g=exp(aX1+bY1) we get simply,
x' = x + a,
y' = y + b.
Unfortunately, neither X and X1 nor Y and Y1 commute, so no easy way by using those "combinations" or compositions to get the transformation rules for only boosts in e1 and e2, or only rotations about e1 and e2, i.e. for exp(t*e1) and exp(t*e2) or for exp(t*ie1) or exp(t*ie2).
Indeed, x' = x + a, y' = y + b is the correct result. You are a very diligent student! I am nicely surprised. As Anna noticed in her comment human behavior is sometimes unpredictable. What is the source of this unpredictability - that I would like to know in details. Most of the time we are quite predictable, but sometimes, some of us, are not.
DeleteAnyway, I was planning yesterday to do the calculations leading to translations in today's post. Now I have to change my plan. By the butterfly effect this may change, in turn, the future of the universe. Which way - that we do not know.
'I am nicely surprised.'
DeleteAnd i am not surprised. It was obvious from the beginning that Sasha is digging deep and fast, and will not stop until find out something really precious. So, i've got a bit envious and also want to obtain that remarkable result. It looks like a special case of the general formula
x' = (x +q a)/(1 + q (a^2+b^2) + 2 (a x + b y)),
y' = (y +q b)/(1 + q (a^2+b^2) + 2 (a x + b y)),
but it is not.
Would somebody give a hint, please, which case should i consider:
2. e2 - ie1 or
3. e1 - ie2 or
4. e2 + ie1,
or all of them necessarily?
The post had 4 cases. The first two of them lead to special conformal transformations. 3) and 4) should lead to translations
Delete3. e1 - ie2
4. e2 + ie1.
Case 3) should lead to translation x+a, y, and 4) to x,y+b
if we consider exp(a(e2-ie1) ) and exp(b(e2+ie1)) respectively, with a,b real numbers.
Notice that special conformal transformations are non-linear on x,y plane, but translations (like boosts and rotations) are linear in x,y. The infinity is a stable point (the "absolute") for rotations, boosts, dilations and translations.
DeleteArk, thank you for the prompt. I hope i will manage to go through calculations now.
DeleteBesides, do you mean in the last note that infinity is a stable point for linear transformations and is not for nonlinear ones, right?
Right. If you take a particular conformal transformation, there is a denominator. You can easily calculated the particular point when this denominator vanishes. Ther should be one real point. So this point is being transformed into infinity - the South Pole on the sphere. So the inverse transformation (negative parameter) transforms infinity into this point. Thus infinity is not stable point for any nontrivial conformal transformation. On the other hand, given any particular conformal transformation you can calculate its stable point. You should find just one real solution on the plane. These are very good exercices.
DeleteA very clear explanation. Got it. Thank you very much!
DeleteSo late, but i have got that result x'=x+a and y'=y+b.
ReplyDeleteStill not happy, since it appeared as a rabbit from hat, i don't see the meaning of this result.
and have an entire list of questions:
1. Why for the transformations (1)-(4), which seem quite similar, the results differ so drastically?
2. Why Sasha spoke about boosts and rotations though we use their mixture when taking exp(e1+i e2), etc.
3. Conformal transform is inversion + translation + inversion. Hence, translation can be a special case of conformal transform if we eliminate the operation of inversion, i.e., take q=1.
These are vectors on a unit circle, and it seems to be the stable point of conformal transform. But what does it mean physically? Somewhat a mass shell if we work in the momentum space?
4. Finally, Ark, about your remark on a vanishing denominator. In our particular case, denominator is (1 + x a + y b)^2, and it vanishes when the scalar product of vectors (x,y) and (a,b) is minus 1, i.e. the stable point is (-a,-b), and somehow we must account for normalization, right?
@Anna
DeleteI will reply to your questions in a P.S. to the new Part 26. It is more convenient for me doing it within a post instead of a comment. Some of your questions are know how to answer, some I do not know yet.