craigthone said:
If ##x_a \rightarrow \lambda x_a##, then ##\phi(x_a) \rightarrow \phi'(\lambda x_a)## rather than ##\phi'(x_a)##.
So the second transformation has some problem.
Am I right?
No, there is no problem because the symbol \bar{\varphi}(x) stands for \bar{\varphi}(x ; \lambda ), i.e. it depends on the coordinate
as well as the transformation parameter, with \bar{\varphi}( x ; 1) = \varphi (x) representing the identity transformation. So, the statement \varphi (x) \to \bar{\varphi}(\bar{x}) is completely
equivalent to the statement \varphi (x) \to \bar{\varphi}(x), provided (of course) one knows the meaning of \varphi (x) , \bar{\varphi} (\bar{x}) and \bar{\varphi} (x).
Let us consider a simple example (which you should consider doing whenever you get confused about something). We take our field to be a vector field V on \mathbb{R}^{2}, i.e., for every point p \in \mathbb{R}^{2}, we have V(p) = \begin{pmatrix} V_{1}(p) \\ V_{2}(p) . \end{pmatrix} Let S be a system in which the point p has coordinate value x = (x_{1} , x_{2}), and the field V(p) has the following simple form V(x) \equiv \begin{pmatrix} V_{1}(x_{1} , x_{2}) \\ V_{2} (x_{1} , x_{2}) \end{pmatrix} = \begin{pmatrix} x_{2} \\ 0 \end{pmatrix} . So, in the S-system, the components of the field at p are given by V_{j}(x) = \delta_{j1} x_{2} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
Notice that our vector field has length given by \left(V_{i}(x)\right)^{2} = (x_{2})^{2}
Now, let \bar{S} be another system in which the
same point p \in \mathbb{R}^{2} has coordinates \bar{x} = (\bar{x}_{1} , \bar{x}_{2}) and the vector field at the
same p is represented by \bar{V}(p) = \bar{V}(\bar{x}). Let us assume that the system S is related to \bar{S} by infinitesimal SO(2) rotation. This means that the coordinates of the point p in the two systems are related by \bar{x}_{i} = x_{i} + \theta \ \epsilon_{ij} x_{j} , \ \ \ \ \ \ \ \ (2a) where | \theta | \ll 1 is the infinitesimal rotation angle (infinitesimal means you put \theta^{2} = \theta^{3} = \cdots = 0), and \epsilon_{ij} = - \epsilon_{ji} \ , \epsilon_{12} = 1 is the antisymmetric SO(2)-invariant tensor. The inverse transformation is obtained by letting \theta \to - \theta x_{i} = ( \delta_{ij} - \theta \ \epsilon_{ij}) \bar{x}_{j} . \ \ \ \ \ \ \ (2b)
Being a vector, our field V_{i}(x) must transform in the vector representation of SO(2), i.e., it obeys the same transformation law of the coordinates: \bar{V}_{i}(\bar{x}) = V_{i}(x) + \theta \ \epsilon_{ij} V_{j}(x) . Substituting eq(1), we get \bar{V}_{i}(\bar{x}) = \delta_{i1} x_{2} + \theta \ \epsilon_{i1} x_{2} . \ \ \ \ \ \ (3) This equation looks very ugly because the RHS is written in terms of
untransformed coordinate x_{2} only. To give eq(3) a face-left, we use eq(2b) to write x_{2} = \bar{x}_{2} + \theta \bar{x}_{1} and substitute it in the RHS of (3). Thus, to first order in \theta we get \bar{V}_{i}(\bar{x}) = \delta_{i1} \left( \bar{x}_{2} + \theta \bar{x}_{1} \right) + \theta \ \epsilon_{i1} \bar{x}_{2} \ . \ \ \ \ (4)
Now, we check that eq(4) gives us the
expected actions of rotations on a vector field, these are
A) No rotation corresponds to \theta = 0. Indeed putting \theta = 0 in eq(4) gives us \bar{V}_{i}(\bar{x}) = \delta_{i1} \bar{x}_{2} = \delta_{i1}x_{2} = V_{i}(x) .
B) Rotation mixes the components of the field. Indeed, while V_{i}(x)|_{x \neq 0} has only one non-zero component (see eq(1)), the transformed field in eq(4) \bar{V}_{i}(\bar{x})|_{\bar{x} \neq 0} has two non-zero components.
C) More importantly, rotation
preserves the length of our vector field. Indeed, from eq(4) it follows \left( \bar{V}_{i}(\bar{x})\right)^{2} = \left( \bar{x}_{2} + \theta \ \bar{x}_{1}\right)^{2} = \left(x_{2}\right)^{2} = \left( V_{i}(x) \right)^{2} .
Now, I don’t like those
barred coordinates in eq(4)! The question then: Do I loose any of the above 3 properties, if I evaluate function \bar{V}_{i}(\bar{x}) at \bar{x} = x? As you can check for yourself, you lose
absolutely nothing if you write \bar{V}_{i}(x) = \delta_{i1} \left( x_{2} + \theta \ x_{1}\right) + \theta \ \epsilon_{i1} x_{2} \ . \ \ \ \ (5)
Not just you don’t loose any property, in fact you
gain more insight when you consider the transformation law eq(5): Since \bar{V}_{i}(x) and V_{i}(x) are evaluated at the same coordinate
value x, the difference \bar{V}_{i}(x) - V_{i}(x) = \theta \left( \delta_{i1} x_{1} + \epsilon_{i1} x_{2}\right) \equiv \delta V_{i}(x) \ , gives you the infinitesimal change in the functional form of the field.
Why did you post this in the Homework section? Can some one move this thread to the Relativity or QM forums?
