transformations and experiments

General comments 11.24.15 In the "question barrage approach", you keep asking questions until there are no questions left that you cannot answer cleanly. At that point perhaps you understand the topic in question. Or perhaps you have not thought of the right questions to test what you understand. Think of this as a "stress test" or as "the scientific method". Transformations and Experiments Transformations of the type V'a = RabVb (Picture A) or [V(e)]a = Rab[V(u)]b (Picture E) are involved in "experiments" which involve an apparatus and a coordinate system. We image three different experiments : Experiment A : (active) Apparatus is transformed , basis vectors do not change. Experiment B : (passive) Basis vectors are transformed , apparatus does not change Experiment C : (both) Apparatus and basis are both transformed The reader might find it helpful to replace the word transformed with the word rotated, in order to obtain a simple physical picture for n = 3 dimensions. Critique: Later I add the direction of the transformation, forward or backward. Let the notation "a→b" mean "a is transformed into b for this experiment". Suppose the apparatus contains a vector v and a rank-2 tensor M and has basis is {ei}. Then Experiment A : (active) va → v'a = Rarvr (ei)a → (ei)a Mab → M'ab = RarRbsMrs Experiment B : (passive) va → va (ei)a → (e'i)a = Rar(ei)r Mab →Mab Experiment C : (both) va → v'a = Rarvr (ei)a → (e'i)a = Rar(ei)r Mab → M'ab = RarRbsMrs The above is all OK, directions will maybe get changed later. Question 1: What happens to the metric tensor in Experiment A? Is it associated with the apparatus, or is it associated with the basis? This is I think a very good question, I investigate in a separate doc called "an Experiment Question" and my conclusion there is clear: metric tensor gij is associated with the basis, not with the apparatus. Just as the ei are characteristics of the x-space, so is gij so it should not change just because you go and warp your apparatus in some way. Question 2: Does my entire discussion here have meaning for "general transformations", or only for rotations?? Yes, it has meaning for general transformations, see analysis in the same document. Answer 11.25.15. Since gij = (ei)n(ej)n and since in Experiment A we do not alter the basis vectors, I would say gij stays put. It is then NOT part of the apparatus which gets "rotated". The metric tensor is a characteristic of x-space correct Think of the example of curvilinear coordinates which has a non-linear general transformation. Does the above discussion have meaning in this situation? In that application we have g = 1 for x-space. Then "apparatus is transformed" would mean Va → Va = RabVb (Cartesian component → ei basis component) so at least that phrase has some meaning. Then "basis is transformed" means (ei)a → (ei)a = Rab (ei)b. Things seem to be OK here, except the notion of "apparatus" does not seem very useful. Maybe start over with this: Experiment A : (active) Tensors are transformed , basis vectors do not change. Experiment B : (passive) Basis vectors are transformed , tensors do not change Experiment C : (both) Tensors and basis are both transformed The tensors describe some apparatus perhaps, or are properties of some apparatus. This seems closer to what I mean. Then for Experiment A we have Va → Va = RabVb and the components of the tensor Va are transformed but they still describe some apparatus. So my tentative answer to Question 2 is yes. correct Back to Question 1. Well for polars I know that g = diag(1,1) → g' = diag(1,r2) or close to that. So under the transformation F I have g → g' which is to say gab → g'ab = RarRbsgrs . But which experiment is this? It is true that gab → g'ab just as en → e'n under a transformation, but this is not what happens in Exp A. Question 3: In my tensor doc treatment of curvilinear coordinates using Picture A, can I fit that treatment into one of my three experiment cubbyholes shown above? First consider, "tensors are transformed" Va → Va = RabVb This is certainly happening in tensor doc, so I could rule out Experiment B. correct But I also have going to for example, so it seems that the basis vectors are also being transformed. But this is like saying ei goes to e'i. That is true, but in Experiment A we stick with ei → ei. So it is then not correct to say we have → (or whatever) for Exp A. In tensor doc I write e'n , n = 1,2...N (e'n)i = δn,i e'1 = (1,0,0...) etc . (3.2.1) and en ≡ Se'n e'n = Ren So if I want, I can transform the basis vectors with R as shown above, and then they line up with the axes of x'-space. Now can I identify e'n with something like ? The answer is NO because e'n exists in x'-space and = e1 exists in x-space. STOP. Question 3 is answered in the same "an Experiment Question" doc. The cubbyhole is Experiment A. The machine is a skewed n-piped aligned with the en and the "warped machine" is a cube. A vector on the machine is the edge of the n-piped we call q and we have q' = Rq under the Exp A transformation Question 3a: In tensor doc, what does "basis vectors are transformed" mean? Answer #1 (3a): An example is un → en so for example is u1 and is e1. Note that both un and en exist in x-space. This is the wrong answer, see answer #2. agreed Answer #2 (3a): An example is en → e'n . This seems more like the correct answer. Yes, after doing the next question 3b, I conclude that this is the correct answer to question 3a. correct In tensor doc both un and en provide bases in x-space and have nothing to do with x'-space. Question 3b: What then is the meaning of writing un → en ? These are two different bases for x-space. We know from tensor/wedge that um = Rnm en em = Rmn un um = Rnm en em = Rmn un um = Rnm en em = Rmn un um = Rnm en em = Rmn un (2.5.5) so you could argue that un → en means em = Rmn un which is a linear combination of the un. But this is different from en→ e'n under a transformation F . Thus un → en would be a completely different meaning for the goes-to arrow →, so I would steer clear of ever saying un → en. Answer #2 (3b): If → means "transforms into", then un → en is an incorrect use of the → notation. So it has no meaning in the context of that notation. correct Now back to Question 3 about which experiment curvilinear coordinates might fit into. Consider the following "summary of all expansions" from tensor doc V = Vn un = Vn un = V'n en = V'n en = V'n n // x-space expansions, V'n = h'nV'n V' = V'n e'n = V'n e'n = Vn u'n = Vn u'n . // x'-space expansions (7.13.12) The second line suggests Experiment C. For a general tensor this would imply T = Σii....i Tii....i (ei ei ..... ei) . (5.2.1) T' = Σii....i T'ii....i (e'i e'i ..... e'i) . This second line was NOT what I was thinking of earlier. This second line is Experiment C. I am comparing these items from the above tensor doc quote, V = Σn V'n en V' = Σn V'n e'n Here is a repeat of the above four equations T = Σii....i Tii....i (ei ei ..... ei) . V = Σn V'n en T' = Σii....i T'ii....i (e'i e'i ..... e'i) . V' = Σn V'n e'n and here they are again in wedge/tensor Picture E notation T = Σii....i Tii....i (ei ei ..... ei) . V = Σn V'n en T(e) = Σii....i T(e)ii....i (ei(e) ei(e) ..... ei(e)) V(e) = Σn V(e)n en(e) Comment: These last two equation pairs are correct if you are doing Experiment C where both the tensors and the basis move, and they both move in the same direction (both forward). Doing Experiment C is the same as moving entirely from x-space to x'-space (from u-space to e-space). This is not normally what we mean when we "do a transformation". That usually means an active or a passive transformation, Exp A or B. I think the following claims are valid: 1. In Experiment A, the tensor (the machine) T = Σii....i T'ii....i (ei ei ..... ei) . V = Σn Tn en is transformed into (prime on both T's, no prime on e's) (this is the warped machine) T' = Σii....i T(e) ii....i (ei ei ..... ei) . V' = Σn T'n en Notice how strange the V = Σn Tn en equation looks. Consider [V]m = V em = [Σn Tn en] em = Tm so we are tempted to write V = Σn Vn en Question 4: Can you relate V' to V in a vector notation or perhaps some other notation? (V' )i = Σn V'n (e'n)i = Σn V'n [δni] from (7.18.1) tensor doc = V' n so the last line is self-consistent. I would guess that you can say V' = RV by which I mean (V' )i = Rij(V)j Question 4a: But what is the meaning of the above "components" in the sense of tensor/wedge clarification? Answer #1 (4a): Perhaps (V' )i = Rij(V)j is OK as long as you use the same coordinates on the two sides. After study below, I think this is the only possible correct answer. Let's test this conjecture: V'i = RijVj V'(u)i = RijV(u)j ?? V' ui = Rij (V uj) ?? Well here is another way to test the conjecture. Thing of the claim as being this V' = RV Then one should be able to analyze this in any basis one wants. So V' ui = (RV) ui V'(u)i = (RV)(u)i Question 4b : What is the meaning of (RV)(u)i ? Answer #1 (4b): (RV)(u)i ≡ (RV) ui = (RV)a(ui)a = valid in any basis you want But this is circular. In the u basis you just end up with (RV)(u)i = (RV)(u)i. Answer #2 (4b) The meaning of (RV)(u)i is W(u)i where W = RV is some vector, and so it has components in the u basis. There is no other meaning. But maybe you can evaluate (RV)(u)i = W(u)i = RijV(u)j Essay Question 5. What can you say about these two tensor doc equations?: V = Σn V'n en V' = Σn V'n e'n (a) I know that Ren = e'n from because this is obvious since en is a vector, and also here is a quote, e'n = R(x) en . tensor doc (3.3.2) Therefore I know that V' = RV which I prove by applying R to the first of the equation pair. (b) Corollary: V ≠ V' (c) One can write(V' )i = Rij(V)j where the components of the vectors must be taken in the same basis on the two sides, and the equation is then valid for any such components. (d) V is an object in x-space while V' is an object in x'-space. (e) To "transform" from the first line to the second line, you are transforming the basis vectors, but you are not transforming the tensor! This looks like Experiment B, Experiment B : (passive) va → va (ei)a → (e'i)a = Rar(ei)r Mab →Mab Notice that va → va implies v'a → v'a or V'n → V'n and this is what you are seeing going between the two equations. Therefore V = Σn V'n en → Σn V'n e'n is a transformation for Experiment B. Rephrase this with new names for the coefficients V = Σn Fn en → Σn Fn e'n is a transformation for Experiment B. And in Experiment B we have V' ≠ V. (f) Therefore, the two equations which are the topic of this essay question seem to be specific to Experiment B. Question 6. Can this pair of equations from tensor doc be connected to Experiment C ? V = Σn V'n en V = Σn Vn un In that experiment, the lower equation above would have to transform like this: Σn Vn un → Σn V'n u'n Now I see from tensor doc that (u'n)i = Rin = Sni (7.18.3) (en)i = Rni = Sin (7.18.1) Notice that the tilts are wrong, so we have u'n ≠ en . If however it were true that u'n = en. we would be able to say that Σn Vn un → Σn V'n u'n = Σn V'n en and then we could say our two equations are those of an Experiment C transformation. But clearly this is not the case, so the equation pair of Question 6 are not those of Experiment C. Maybe I should allow for "more experiments" as follows. Experiment A : Tensors are transformed forward , basis vectors do not change. Experiment B : Basis vectors are transformed forward , tensors do not change Experiment C : Tensors and basis are both transformed forward Experiment D : Tensors are transformed forward, basis is transformed backward Experiment E : Tensors are transformed backward, basis is transformed forward or to summarize tensors basis Experiment A : forward no change Experiment B : no change forward Experiment C : forward forward Experiment D : forward backward Experiment E : backward forward To supplement this list, we write tensors go forward: Mab → M'ab = RarRbsMrs A,C tensors go backward: Mab → M"ab = RraRsbsMrs E basis goes forward: (en)a → (e'n)a = Rab(en)b B basis goes backward: (en)a → (e"n)a = Rba(en)b D Detail for the above: Now in tensor doc (7.5.11) I show that (R-1)ca = Sca = Rac also using (7.5.13). So restate tensors go forward: Mab → M'ab = RarRbsMrs tensors go backward: Mab → M"ab = RraRsbsMrs So now I want to review: Essay Question 5 Reviewed. What can you say about these two tensor doc equations?: V = Σn V'n en starting position V' = Σn V'n e'n tensors not changed, basis vectors transform forward (a) I know that Ren = e'n from because this is obvious since en is a vector, and also here is a quote, e'n = R(x) en . tensor doc (3.3.2) Therefore I know that V' = RV which I prove by applying R to the first of the equation pair. (b) Corollary: V ≠ V' (c) One can write(V' )i = Rij(V)j where the components of the vectors must be taken in the same basis on the two sides, and the equation is then valid for any such components. (d) V is an object in x-space while V' is an object in x'-space. (e) To "transform" from the first line to the second line, you are transforming the basis vectors forward, but you are not transforming the tensor! This looks like Experiment B, Experiment B : (passive) va → va (ei)a → (e'i)a = Rar(ei)r Mab →Mab Notice that va → va implies v'a → v'a or V'n → V'n and this is what you are seeing going between the two equations. Therefore V = Σn V'n en → Σn V'n e'n is a transformation for Experiment B. Rephrase this with new names for the coefficients V = Σn Fn en → Σn Fn e'n is a transformation for Experiment B. And in Experiment B we have V' ≠ V. Notice that the coefficients do not change. (f) Therefore, the two equations which are the topic of this essay question seem to be specific to Experiment B. Question 6 Reviewed. Can this pair of equations from tensor doc be connected any experiment? V = Σn Vn un starting position V = Σn V'n en tensor goes forward, but basis goes neither forward nor backward If it were true that en = un we could say basis vector remained unchanged, but we know en ≠ un . Is this basis going forward or backward? Well, how are en and un related? (un)i = δni If it were true that en = u'n we could say that the basis vectors transformed forward. But I see from tensor doc that (u'n)i = Rin = Sni (7.18.3) (en)i = Rni = Sin (7.18.1) Notice that the tilts are wrong, so we have u'n ≠ en. Thus we cannot say basis goes forward STOP Question 7: In tensor doc displays like (7.18.1), we see various components like (en)i . Are these components the (en)(u)i ones or are they the (en)(e)i ones ?? Well in this case, we know (en)(e)i = δni so it is not that one, so it is going to be the other one. But how do we know that for sure? Answer: (as derived just below): All components are u-basis components! Well, the display says to look at (6.3.3) where I see (en)i = Sin which is from (3.2.6). There we see (en)i = Sin = ∂xi/∂x'n which goes back to (2.1.5). But that is just showing Sin = ∂xi/∂x'n. Let's go back instead to en ≡ Se'n in (3.2.4). This would be true in any basis. In (3.2.5) I say (en)i = Σj Sij (e'n)j = Σj Sij δn,j = Sin (3.2.5) which in turn is based on the claim that (e'n)j = δn,j. That in turn comes from e'n , n = 1,2...N (e'n)i = δn,i e'1 = (1,0,0...) etc . (3.2.1) So in the equation (e'n)i = δn,i, what component basis is implied? It must be u. Thus (e'n)i = (e'n(u))i = e'n ui = (e'n(u))r (ui)r = (e'n(u))r δir = (e'n(u))i So tracking forward again, we conclude that (e'n)i = (e'n(u))i (en)i = (en(u))i = Sin (e'n)i = (e'n(u))i (en)i = (en(u))i = Sin Fact: In the displays of (7.18.1) all "components" with specific values are components with respect to the u basis, not the e basis. I guess that is what I expected, but in tensor doc I never really thought about this issue. Only in tensor wedge do I bring out this issue, which I think is worth doing somewhere. But now consider en em = g'nm Essay Question 8: What can you say about the nature of the components in en em = g'nm ? Subquestion: what is the meaning of g'(u)nm and g'(e)nm ? Do these notations even make sense I suspect that they do. continue here -- OK : (a) One thing I know is that en em = e'n e'm . Can I verify this? en em = gab(en)a(em)b = gab(u)(en)(u)a(em)(u)b = gab(u)RnaRmb e'n e'm = g'ab (e'n)a (e'm)b = g'ab(u) (e'n(u))a (e'm(u))b = g'ab(u)δnaδmb = g'nm(u) So I end up with g'nm(u) = RnaRmbgab(u) and this is the statement that g is a rank-2 tensor written in the (u) basis, good. Question 9: I am used to un and en being vectors in x-space. What can be said about u'n and e'n ? Answer #1: We know that (e'n)i = Rij(en)j . This is not a lincom of vectors. But it is an equation which I presume can be evaluated in (u) coordinates or (e) coordinates. Try them out. (e'n(u))i = Rij(en(u))j ?? δni = Rij Rnj ?? yes, that is an orthog rule. (e'n(e))i = Rij(en(e))j ?? ?? = Rijδnj = Rin I can take this as a definition and I then find that (e'n(e))i = Rin e'n = Σi (e'n(e))i ei = Σi Rin ei Very good! Now I do have a lincom for e'n . This cross relation does not make my display tables in tensor doc. Does the above equation appear anywhere in tensor doc? NO! I hardly say anything about the e'n vector, and I have no equations like the above. But I am suspicious of saying that (e'n(e))i = Rij(en(e))j , because Rij is sitting there. OK, going to a whole new doc. Sec 3.2, no and not anywhere through 5.2 Question 10: What is the meaning of (e'n(e))i ? I am very confused about this and related matters, despite having written tensor doc! Question 10a: Can you think of e'n as being a vector in x-space? Could you draw it there? I am totally clueless. I think I know that e'n = Ren . In (u) components, this says (e'n)(u)i = Rij(en(u))j and I just showed above that this works right. Now try the same thing in the (e) basis: en em = gab(en)a(em)b = gab(e)(en)(e)a(em)(e)b = gab(e)δnaδmb = gnm(e) e'n e'm = g'ab (e'n)a (e'm)b = g'ab(e) (e'n(e))a (e'm(e))b = ?? I guess we know (e'n(e))a = e'n ea = Question 9: Is the quantity en em basis-independent, or not? (1) It is the same in both x-space and x'-space, I know that much (2) ********************************************* Note: In order to talk about what dot products "do", you have to know what metric tensors "do". Recall that metric tensor g'ij = ei ej = gab(ei)a(ej)b. In an experiment where the basis vectors do not change, it would seem that g → g and g'→g'. But in an experiment where the basis vectors transform Suppose the apparatus contains two vectors v and w, and the basis is {ei}. Let the arrow → mean "is transformed into for this experiment". Then consider for Experiment A: Experiment A: vw = gijviwj in x-space v'w' = g'ijv'iw'j in x'-space v'a = Rabvb v → v' w'a = Rabwb w → w' ei → ei Dot products: vw → v'w' = vw vei → v'ei ≠ vei v'(e)i ≠ v(e)i v'(e)i = Rij v(e)i wei → w'ei ≠ wei w'(e)i ≠ w(e)i w'(e)i = Rij w(e)i STOP. What happens to the metric tensor in the transformation? g → g' I guess