
#19
Jun611, 10:51 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500





#20
Jun611, 10:51 PM

Sci Advisor
P: 1,563

[tex]\varphi(x) \rightarrow \varphi(x(y))[/tex] where y are the new coordinates. By contrast, a vector field [itex]A^\mu(x)[/itex] would transform as [tex]A^\mu(x) \rightarrow \frac{\partial y^\mu}{\partial x^\nu} A^\nu(x(y))[/tex] where you see there is an additional change of basis. But note that linear coordinate transformations are subsumed under the first equation above. And if you do a nonlinear coordinate change (such as changing from Cartesian to spherical coordinates), you will find that only the transformation law for a scalar field makes sense. 



#21
Jun611, 11:06 PM

Emeritus
Sci Advisor
PF Gold
P: 5,500

I think the whole issue here revolves around different interpretations of the OP's notation [itex]\tau[/itex]. As far as I can tell, we're all in agreement that if [itex]d/d\tau[/itex] just means some kind of differentiation with respect to an arbitrary real parameter (which is a function of Godknowswhat), then the manipulation in #1 is invalid. On the other hand, if it represents differentiation with respect to proper time along a worldline, then I'm pretty sure that the manipulation is valid. 



#22
Jun611, 11:35 PM

P: 2,071

[tex]F_\mu=\frac{dp_\mu}{d\tau}[/tex] ? What I mean is, is the lowered index on the F so simple as just lowering the index on the p? 



#23
Jun711, 05:16 AM

P: 2,896

It is remarkable how often this misuse of SR notions confuses differential geometry issues. 



#24
Jun711, 08:36 AM

P: 95

This point is often confused because there are four coordinates, and hence they come with an index. But it's really just a collection of scalars. I'm not too dogmatic about the "geometric" versus "coordinatebased" definitions of tensor fields, but in this case I think the geometric approach makes things much clearer. EDIT: In reading quickly I didn't notice you're thinking of F as the fourforce and p as the fourmomentum. In this case your expression should indeed be using [itex]F^\nu = u^\mu \nabla_\mu p^\nu = D p^\mu / d\tau[/itex], not the ordinary derivative. This is the standard definition of fourforce, so that F^\mu = 0 is the geodesic equation. 


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