Coordinate transformation for isotropy

[gionole]

The following question struck me by accident.

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.

 

[renormalize]

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

This is wrong because $\ddot x'$ is a linear combination of both $\ddot x$, $\ddot y$ and $\ddot y \neq 0$.

 

[gionole]

This is wrong because $\ddot x'$ is a linear combination of both $\ddot x$, $\ddot y$ and $\ddot y \neq 0$.

Ah, true, thank you. so we got:

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

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

how do I continue from this to get to the equation that only contains $\ddot y'$

 

[renormalize]

Ah, true, thank you. so we got:

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

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

how do I continue from this to get to the equation that only contains $\ddot y'$

Remember that in the unprimed-frame you also have the second equation-of-motion (EOM) $m\ddot x=0$. You should transform this equation as well so you end up with two EOMs to solve in the primed-frame.

 

[gionole]

In x,y frame, system of equations:

$\ddot x = 0$ (1)
$\ddot y = -\frac{d}{dy}(gy)$ (2)

we know that:

$x = x'cos\theta - y'sin\theta$ (3)
$y = x'sin\theta + y'cos\theta$ (4)

so, plugging (3) into (1), we get:

$\ddot x' cos\theta - \ddot y' sin\theta = 0$
$\ddot x' = \frac{\ddot y' sin\theta}{cos\theta}$

plugging (4) into (2),

$\ddot x'sin\theta + \ddot y'cos\theta = -\frac{d}{dy}(g(x'sin\theta + y'cos\theta)$

My first question is: should I now do $\frac{d}{dy}$ or $\frac{d}{dy'}$ ? We only changed $y$ by $x'sin\theta + y'cos\theta$, so I guess, I can't change $\frac{d}{dy}$ by $\frac{d}{dy'}$, right ? If so, what's $-\frac{d}{dy}(x'sin\theta)$ ? I would need to substitute $x'$ by another transformation that I can get from (3) ?

 

[renormalize]

In x,y frame, system of equations:

$\ddot x = 0$ (1)
$\ddot y = -\frac{d}{dy}(gy)$ (2)

You can always apply a rotation-of-coordinates to the partial derivatives. But in this case it's easier to just perform the $y$-differentiation in eq.(2) to get $\ddot y = -g$ so that you only have to rotate $x$ and $y$.

 

[gionole]

You can always apply a rotation-of-coordinates to the partial derivatives. But in this case it's easier to just perform the $y$-differentiation in eq.(2) to get $\ddot y = -g$ so that you only have to rotate $x$ and $y$.

Yes, I know that, I asked the question in #5 for learning purposes. So, basically, this is how I do it: If you don't got time to go through, here is my 2 short questions:

--Short Version--

Q1: I end up with $\ddot y' = -gcos\theta$, would you say this is correct ?

Q2: When we have $\ddot y = -\frac{d}{dy}(gy)$, and when I replace $y$ by $x'sin\theta + y'cos\theta$, I still don't change wrt derivative (I leave $dy$ and not change it to $dy'$). Correct ?

-- Long Version--

$\ddot x = 0$ (1)
$\ddot y = -\frac{d}{dy}(gy)$ (2)

$\ddot x'cos\theta - \ddot y'sin\theta = 0$
$\ddot x' = \frac{\ddot y'sin\theta}{cos\theta}$

$\ddot x'sin\theta + \ddot y'cos\theta = -g\frac{d}{dy}(x'sin\theta + y'cos\theta)$ (Note that I still leave $\frac{d}{dy}$ here) (3)

Now, we know that $x'$ contains $y$ inside, so we can't really say that $\frac{d}{dy}(x'sin\theta) = 0$, so we need to get what $x'$ and $y'$ are.

After some math, I end up with:

$x' = xcos\theta + ysin\theta$
$y' = -xsin\theta + ycos\theta$

we put them inside (3).

$\ddot x'sin\theta + \ddot y'cos\theta = -g\frac{d}{dy}(x'sin\theta + y'cos\theta)$
$\ddot x'sin\theta + \ddot y'cos\theta = -g\frac{d}{dy}((xcos\theta + ysin\theta)sin\theta + (-xsin\theta + ycos\theta)cos\theta)$

$\ddot x'sin\theta + \ddot y'cos\theta = = -g(sin^2\theta + cos^2\theta)$
$\frac{\ddot y'sin\theta}{cos\theta}sin\theta + \ddot y'cos\theta = -g$
$\frac{\ddot y'sin^2\theta}{cos\theta} + \ddot y'cos\theta = -g$
$\frac{\ddot y'sin^2\theta + \ddot y'cos^2\theta}{cos\theta} = -g$
$\ddot y' = -gcos\theta$

Would you say this is all correct ? @renormalize

 

[renormalize]

$\ddot y' = -gcos\theta$

Would you say this is all correct ? @renormalize

Yes, you have the right expression for $\ddot y'$, but don't forget to similarly display the second EOM giving $\ddot x'$ in terms of $g$ and $\theta$.