# Coordinate transformation for isotropy

• I
• gionole
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.

gionole said:
##\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 and Dale
renormalize said:
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'##

gionole said:
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) ?

gionole said:
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
renormalize said:
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

gionole said:
##\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##.

gionole

## What is coordinate transformation for isotropy?

Coordinate transformation for isotropy refers to the mathematical process of changing the coordinate system in such a way that the properties of a material or physical system remain unchanged in all directions. This is particularly important in fields like physics and engineering where isotropic properties, meaning uniform properties in all directions, are assumed or required.

## Why is coordinate transformation important for isotropic materials?

Coordinate transformation is crucial for isotropic materials because it allows scientists and engineers to analyze and model the behavior of these materials under different conditions and orientations. By transforming coordinates, one can simplify complex problems, making it easier to apply theoretical models and computational methods to predict material behavior accurately.

## How do you perform a coordinate transformation to achieve isotropy?

To perform a coordinate transformation to achieve isotropy, you typically use rotation matrices or tensor transformations. These mathematical tools help you rotate or transform the coordinate system such that the material properties appear the same in all directions. The specific steps involve defining the initial coordinate system, applying the transformation matrix, and then recalculating the material properties in the new coordinate system.

## What are the common applications of coordinate transformation for isotropy?

Common applications include stress analysis in mechanical engineering, where isotropic materials are often assumed for simplicity; fluid dynamics, where isotropic turbulence models are used; and materials science, where understanding the isotropic properties of composites or crystalline structures is essential. Additionally, it is used in computer graphics for rendering isotropic textures and in geophysics for modeling isotropic layers of the Earth.

## Can coordinate transformation be applied to anisotropic materials?

Yes, coordinate transformation can be applied to anisotropic materials, but the process is more complex. Anisotropic materials have direction-dependent properties, so the transformation must account for these variations. In such cases, the transformation involves more advanced tensor mathematics to ensure that the directional dependencies are accurately represented in the new coordinate system.

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