The meaning of V-bar

The meaning of PhL 4.6.15 I have created a monster and now I have to make it go away! In (2.5.1) I define how contravariant and covariant vectors transform. V' = R V contravariant Rik(x) ≡ (∂x'i/∂xk) R = S-1 ' = ST covariant Sik(x') ≡ (∂xi/∂x'k) = STki(x') (2.5.1) I can hardly say that the vector does not exist! So how do I avoid the notation which I want to claim is somehow meaningless! Maybe I need some picture like this: V'(x') = R(x) V(x) '(x') = ST(x) (x) (x) = g(x)V(x) Update 4.17.15. A. I have concluded that the partner vectors = V really do exist and this is true for all the basis vectors as well. For example, V = V1 u1 + V2 u2 +... = ΣnVnun where Un V = Vn . (6.6.1) and then apply from the left to get = V1 1 + V2 2 +... = ΣnVnn where n = Vn . (6.6.1) where I am guessing here about the dot product on the right. Just as I could plot basis vectors en in x-space, so too could I plot n in x-space. In general, they are of course different vectors. They really do exist, so stop trying to make them go away. Plan A. Now here is my way out of the jam for the dot product. Consider these steps: 1. Define the dot product in this way: A B ≡ Aii // a combination of contra and covariant components. 2. It follows then that A B ≡ Aii = Ai(ijBj) = ijAiBj = (jiAi)Bj = jBj = iBi 3. My idea at this point was to say that = A and therefore = ii = iBi = .... = Aii = A B But this Plan A is no go! What is the meaning of ? You first do ≡ A to get the partner vector. But then if you want to go the other way, you have to do A = (-1) = g . In fact = = A so this object is not just A as I was thinking. So Plan A is wrong. Plan B. Comment: Consider again the definition and demonstrated fact from above, A B ≡ Aii = iBi . The dot product involves the contravariant components of one vector and the covariant components of the other vector. In the dot product notation A B , we have indicated each vector by its contravariant name just as a convention. We could just as well have indicated one or both vectors by its covariant name, but the dot product indicated by whatever name would be the same: contravariant components of one vector and the covariant components of the other vector. Thus, A B ≡ Aii = iBi = B = A = We shall always use the first notation A B since it is the simplest. We then have the interesting fact that, if A and B are tensorial vectors, A ≠ but A B = B for any B If B = A, this says A ≠ but A A = A = |A|2 = Aii = gijAiAj Consistency Check with Series Go back now to V = V1 u1 + V2 u2 +... = ΣnVnun where Un V = Vn . (6.6.1) Apply just to the series to get = V1 1 + V2 2 +... = ΣnVnn where Un V = Vn . (6.6.1) where we have not altered the dot product. Is this really valid. = ΣnVnn ? To find out, I want to do a dot product of this with something, but that will involve . So Um = ΣnVnn Um = ΣnVn (n Um) = ΣnVn (un Um) = ΣnVn δnm = Vm So here is a case where I am forced to use the notation Um where one vector is barred, but then I use my Plan B claim and I end up with this consistent result = V1 1 + V2 2 +... = ΣnVnn where Un = Un V = Vn . (6.6.1) So the interesting fact is that ≠ V in general, but A = V A . Although the two vectors V and are different vectors, they have the same covariant dot product with any third vector A. In Standard Notation we have V = V1 u1 + V2 u2 +... = ΣnVnun where Un V = Vn . (6.6.1) Vi = V1 (u1)i + V2 (u2)i +... = ΣnVn(un)i where Un V = Vn . contra Vi = V1 (u1)i + V2 (u2)i +... = ΣnVn(un)i where Un V = Vn cov Compare with V = V1 u1 + V2 u2 +... = ΣnVnun where Un V = Vn . (6.6.1) = V1 1 + V2 2 +... = ΣnVnn where Un = Un V = Vn . (6.6.1)