Recent content by ianhoolihan

OK, after some time away, and some different perspectives, I think I have it sorted. Firstly, the short answer is that the answer to my original question is that method A. is correct. Secondly, active and passive transformations are equivalent --- there is no physical difference. (Sorry...

If I have a vector ##u## and a vector ##v## (maybe a basis vector) then doing a passive transformation on one, and an active on the other will change their relative displacements/orientation. 'Physically'. Anyway, in response to you question, if we have a coordinate tuple, we must know the basis...

Frederik, we still disagree I think. Active = observable, passive = unobservable. If you mean former, as in the former paragraph, then yes, that's what I mean. An active transformation is not the inverse of a passive one (which, I admit, is what is usually bandied around). See this post...

For now this: why then is a passive transformation not physically observable, while an active one is? Now I need just about an odit of sleep :wink: !

I disagree. Active transformations lead to physically observable effects. Passive ones do not --- they are trivial changes of coordinates. In a passive transformation, both component and basis are changed (inversely). In an active transformation, only one of those is changed --- either the...

I'll reply to this quickly, as I've got to shoot off. The point is, I disagree, and that's what I've been trying to say. In an active transformation, the coordinate system does not change, only the function --- as before ##y(p) \to y(p)##. In a passive one, ##y(p)\to z(p)## and the corresponding...

I do not think we change coordinate systems in an active transformation, in the sense that the basis (or axis) does not change. And I also explained that, for an active transformation, ##x(p)\to x(p)##, i.e. does not change. Maybe 'change coordinate system' means more in this context. Anyway...

Your notation etc is unfamiliar. However, to clarify, by "the same coordinate system", I mean that the basis vectors do not change. You'll note that the derivative is ##\partial /\partial x'## in the last bit.

OK, reading the wiki again, the example makes it clear what is active and passive. By passive, the geometric thing of the vector does not change, only it's coordinate representation, and trivially so. The basis is transformed, and the coordinate representation of the vector transformed by th...

OK, we are trying to prove, not "recall" the rule. Also, I'm not sure we agree on what is "active". Passive: $$\quad\phi'(x')=\phi(x)=\phi(\Lambda^{-1} x'), \quad x'=\Lambda x \; \Leftrightarrow \; x=\Lambda^{-1} x'.$$ This is a trivial change of coordinates. Active...

OK, I still don't like this. In the linked post, you state the equality between $$\frac{\partial (f\circ g)(x)}{\partial x_i}=\sum_{j=1}^m\frac{\partial f(g(x))}{\partial g_j}\frac{\partial g_j(x)}{\partial x_i},$$ and $$(f\circ g)_{,i}(x)=f_{,j}(g(x))g_{j,i}(x)$$ which I disagree with, as...

I just skimmed through this before lectures, and to clarify, I was confused by notation: I see ##(\Lambda^{-1})^\nu{},_\mu = 0## whereas ##(\Lambda^{-1}x)^\nu{},_\mu = (\Lambda^{-1})^\nu{}_\mu##. I'll read everything else later.

OK, things got busy, so my apologies for the delay. After reading the Wiki on active and passive transformations, I am happy to say that a passive transformation is just a trivial change of coordinates. That is, the basis vectors are changed to different ones, or the axis are changed, if you...

Thanks Frederik. I'll have to do some reading on this today, and get back to you. Specifically, what is meant by "active" and "passive". Also, we were disagreeing (I think) on the derivative point. Your contention was that ##\partial_\mu \to \partial_{\mu'} =...

I do know some differential geometry, and your explanation gets a big thumbs up! I guess that was what I was going for with the ##\phi \circ \Lambda^{-1}## comment, just not as thorough. That said, I disagree with the very last bit --- according to those notes, ##x\to x' = \Lambda x##...