- #1

gionole

- 281

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Would appreciate to let me know the flaw in my logic, no need to deviate from this approach.

We know that on small-scale, experiment such as dropping ball from some height to the earth tells us that space is non-isotropic, because of preferred direction. I was trying to mathematically detect non-isotropy of such case. Here is how.

Experiment 1:

in ##x, y## frame, we have: ##m\ddot y = -\frac{d}{dy}(mgy)## by which we get: ##\ddot y = -g##.

Experiment 2:

We rotate our own coordinate system, while everything stays in place. Some call this passive transformation. The rotation matrix is given by:

##y = x'sin\theta + y'cos\theta##

##x = x'cos\theta - y'sin\theta##

Since ##y = x'sin\theta + y'cos\theta## and earth stays the same place, potential energy of the ball doesn't really change, so we can replace ##y## by ##x'sin\theta + y'cos\theta## in the ##m\ddot y = -\frac{d}{dy}(mgy)## to get equation of motion in ##x', y'## frame.

##m\frac{d^2}{dt^2}(x'sin\theta + y'cos\theta) = -\frac{d}{dy'}(mg(x'sin\theta + y'cos\theta))##

##\ddot x'sin\theta + \ddot y'cos\theta = -gcos\theta##

##\ddot x'sin\theta = 0## since acceleration is non-present in that direction.

We're left with:

##\ddot y'= -g##.

It seems equation of motions of the ball is exactly the same in x,y and x',y' frame and this kind of suggests that space is isotropic, while we know that it's not isotropic. Where am I making a mistake ? I'm using general transformation matrix, so I shouldn't be needing to use specific ##\theta## insertions.