Inertia tensor around principal Axes

[Lambda96]

Homework Statement

Show that if $\rho$ is invariant (i.e. stays the same) under a $\pi$-rotation around the 1 axis, then

Relevant Equations

none

Hi,

unfortunately, I am not getting anywhere with the following task

The inertia tensor is as follows

$$\left( \begin{array}{rrr}
I_{11} & I_{12} & I_{13} \\
I_{21} & I_{22} & I_{23} \\
I_{31} & I_{32} & I_{33} \\
\end{array}\right)$$

I had now thought that I could simply rotate the inertia tensor around the angle $\pi$ using the rotation matrix for x, i.e.

$$\left( \begin{array}{rrr}
1 & 0 & 0 \\
0 & cos & -sin \\
0 & sin & cos \\
\end{array}\right)$$

I have received the following

$$\left( \begin{array}{rrr}
I_{11} & I_{12} & I_{13} \\
I_{21} & I_{22} & I_{23} \\
I_{31} & I_{32} & I_{33} \\
\end{array}\right) \left( \begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{array}\right)=\left( \begin{array}{rrr}
I_{11} & -I_{12} & -I_{13} \\
I_{21} & -I_{22} & -I_{23} \\
I_{31} & -I_{32} & -I_{33} \\
\end{array}\right)$$

And as you can unfortunately see, that is not the right solution at all. Unfortunately, I don't know what exactly I should calculate here to get the result.

 

[Filip Larsen]

First of all, the inertia tensor is a second order tensor so it doesn't rotate like that (how does it rotate?)

Secondly, you may want to take a look at the defining integral/sum for each of the components, assuming your lecture material contains such. What does the hint given in the problem mean in that context?

 

[PhDeezNutz]

$RMR^T$

Is how you rotate a tensor I believe.

 

[Lambda96]

Thank you for your help,

@Filip Larsen do you mean the following?

$$ I_{xx}=\int_{}^{} \rho(r)(y^2+z^2)\ d^3r $$
$$ I_{xy}=\int_{}^{} -\rho(r)xy \ d^3r $$
$$ I_{xz}=\int_{}^{} -\rho(r)xz \ d^3r $$
$$ I_{yx}=\int_{}^{} -\rho(r)yx \ d^3r $$
$$ I_{yy}=\int_{}^{} \rho(r)(x^2+z^2) \ d^3r $$
$$ I_{yz}=\int_{}^{} -\rho(r)yz \ d^3r $$
$$ I_{zx}=\int_{}^{} -\rho(r)zx \ d^3r $$
$$ I_{zy}=\int_{}^{} -\rho(r)zy \ d^3r $$
$$ I_{zz}=\int_{}^{} \rho(r)(x^2+y^2) \ d^3r $$

I have now replaced the index of $11$ by $xx$ etc., as I find this notation easier to write in Latex.

The integrals integrate over the volume, which means that the integral can be represented as follows.

$$ I_{xx}=\int_{}^{}\int_{}^{}\int_{}^{} \rho(r)(y^2+z^2)\ dxdydz $$

Unfortunately, I am now just unsure about the interval of integration, so I thought of the following

$$ I_{xx}=\int_{-z}^{z}\int_{-y}^{y}\int_{x}^{x} \rho(r)(y^2+z^2)\ dxdydz $$

This would make all terms with a $x$ disappear, but unfortunately this also makes all other integrals 0, so I assume that these are the wrong intervals of integration.

 

[FactChecker]

$RMR^T$

Is how you rotate a tensor I believe.

With the rotor, $R$, being for half the angle of the rotation.
CORRECTION: This is wrong. Thanks, @vanhees71

 

[Filip Larsen]

Look at the integrals for the components that you have to show becomes zero, e.g. $I_{xy}$. What happens if you split this integral into two parts, one integral for $y \gt 0$, the other for $y \lt 0$?

 

[Lambda96]

If I split the y-integral, it would look like this:

$$ \int_{-y}^{0} -\rho(r)xy \ dy + \int_{0}^{y} -\rho(r)xy \ dy $$

Then the left integral $\int_{-y}^{0} -\rho(r)xy \ dy=-A$ and the right integral $\int_{-y}^{0} -\rho(r)xy \ dy=A$ gives $A-A=0$.

 

[vanhees71]

With the rotor, $R$, being for half the angle of the rotation.

No it's just the rotation matrix you also use for vector components. The point is that a 2nd-rank tensor is a bilinear form, i.e., the value $\Theta(\vec{x},\vec{y})$ is independent of the choice of coordinate systems. Especially when using cartesian coordinates the change from one to the other system is given by an orthogonal matrix $\hat{R}$ with $\hat{R}^{-1}=\hat{R}^{T}$. Thus you have
$$\vec{x}'=\hat{R} \vec{x}, \quad \vec{x}=\hat{R}^T \vec{x}',$$
where $\vec{x}$ and $\vec{x}'$ are now the components of the vector written as a column of numbers with respect to the "old" and the "new" basis.

In order to have the outcome of the tensor be the same its representing matrices $\hat{Theta}$ and $\hat{\Theta}'$ must fulfill
$$\vec{x}^{T} \hat{\Theta} \vec{y} = (\hat{R}^T \vec{x}')^T \hat{\Theta} \hat{R}^T \vec{y}' = \vec{x}^{\prime T} \hat{R} \hat{\Theta} \hat{R}^T \vec{y}'=\hat{x}^{\prime T} \hat{\Theta}' \hat{y}'.$$
Since this must hold for any vectors $\vec{x}$ and $\vec{y}$ you necessarily get the transformation law for cartesian components of a 2nd-rank tensor,
$$\hat{\Theta}'=\hat{R} \hat{\Theta} \hat{R}^T.$$

 

[Lambda96]

Thank you all for your help

However, I still have two questions

My professor's script says the following about the rotation of a second order tensor

$$I_{ab} \rightarrow I'_{ab}=R_{ac}R_{bd}I_{cd}$$

Alternatively in vector notation

$$I \rightarrow I'=RIR^T$$

Unfortunately, I still don't quite understand how I could have solved the task with it. Would I have to do the calculation as follows?

$$\left( \begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{array}\right)\left( \begin{array}{rrr}
I{11} & I{12} & I{13} \\
I{21} & I{22} & I{23} \\
I{31} & I{32} & I{33} \\
\end{array}\right)\left( \begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{array}\right)=\left( \begin{array}{rrr}
I{11} & -I{12} & -I{13} \\
-I{21} & I{22} & I{23} \\
-I{31} & I{32} & I{33} \\
\end{array}\right)$$

Where $R$ is the rotation matrix around the x-axis with angle $\theta=\pi$.

To my second question, I now have to calculate another problem, namely the inertia tensor of a cuboid about a principal axis, can I post this discussion here too or should I open a new post for it?

 

[vela]

You're assuming the rotation leaves the tensor invariant, i.e., $I=I'$. So what does that imply about the individual elements of $I$?

 

[PhDeezNutz]

I would set up the expressions for the entries in Cartesian and then convert to spherical.

With $\phi$ and then $\phi + \pi$

Once you brutally trudge through each the integrals in each case I think they will evaluate to the same.

After that you can reason which ones are zero. I’ll take a look at it more closely later.

 

[pasmith]

I would set up the expressions for the entries in Cartesian and then convert to spherical.

Cylindrical seems more relevant.

 

[PhDeezNutz]

Cylindrical seems more relevant.

I agree.

 

[Lambda96]

@vela that means that the objects must all be the same after the rotation, i.e. as before the rotation. In my calculation in post 9 I see that all objects except $I_{12},I_{13},I_{21},I_{31}$ remain the same. Does this mean that for an invariant rotation, I can take these objects as zero?

Thanks PhDeezNutz and pasmith for the tip, would $I_{xy}=\int_{}^{} -\rho(r) xy \ d^3r$ then have to be as follows, with $x=rcos(\theta)$, $y=rsin(\theta)$ and $z=z$?

$$I_{xy}=\int_{0}^{z}\int_{0}^{2\pi}\int_{0}^{r} -\rho(r) r^3cos{\theta}sin{\theta} \ dr d\theta dz$$

 

[vela]

If $I=I'$, that means the corresponding entries are equal, so for example, you have $I_{12} = -I_{12}$. What does that tell you about $I_{12}$?

 

[pasmith]

@vela that means that the objects must all be the same after the rotation, i.e. as before the rotation. In my calculation in post 9 I see that all objects except $I_{12},I_{13},I_{21},I_{31}$ remain the same. Does this mean that for an invariant rotation, I can take these objects as zero?

Thanks PhDeezNutz and pasmith for the tip, would $I_{xy}=\int_{}^{} -\rho(r) xy \ d^3r$ then have to be as follows, with $x=rcos(\theta)$, $y=rsin(\theta)$ and $z=z$?

$$I_{xy}=\int_{0}^{z}\int_{0}^{2\pi}\int_{0}^{r} -\rho(r) r^3cos{\theta}sin{\theta} \ dr d\theta dz$$

Technically $x_1 = z$, $x_2 = r\cos\theta$ and $x_3 = r\sin\theta$.

You can make the domain of integration all of $\mathbb{R}^3$ by taking $\rho = 0$ outside the object. The assumption is that $\rho(r,\theta,z) = \rho(r,\theta + \pi, z)$ for every $(r, \theta, z)$.

 

[PhDeezNutz]

On second thought I think you should use Cartesian.

I think you can reason that

1) The diagonal entries must be nonzero and positive by virtue of the integrand of the integral is a categorically positive quantity; density is positive and you’re multiplying by the sum of two squared quantities.

2) The remaining entries in the first row (and column) must be zero due to @vela ’s reasoning. When you use the hint and compare initial vs. rotated integrals you’ll get something like $A = - A \Rightarrow A = 0$.

Which leaves entries 23 and 32. I don’t know how to tackle those.

Edit: in post 9 you completed the problem. Just use vela’s hint and you’re done.

 

[Lambda96]

Thanks to all of you for your help

@vela, if the following holds, $I_{12}=I_{12}$ this condition can only be valid if $I_{12}=-I_{12}=0$.

So then I can assume the following:

$$\left( \begin{array}{rrr}

I_{11} & -I_{12} & -I_{13} \\

-I_{21} & I_{22} & I_{23} \\

-I_{31} & I_{32} & I_{33} \\

\end{array}\right) \rightarrow I_{12}-I_{12}=0 \rightarrow \left( \begin{array}{rrr}

I_{11} & 0 & 0 \\

0 & I_{22} & I_{23} \\

0 & I_{32} & I_{33} \\

\end{array}\right)$$I have one more question, if $\rho$ were not invariant, would the following hold for the above rotation $I \neq I'$ then the following would also hold $I_{12} \neq -I_{12}$, right?

 

[PhDeezNutz]

If I’m not mistaken each entry of the moment of inertia tensor must be $\geq 0$.

If $I21 \geq 0$ and $-I21 \geq 0$ then the the only way for this to happen is if $I21 = 0$.

Edit: that’s wrong

 

[vela]

I have one more question, if $\rho$ were not invariant, would the following hold for the above rotation $I \neq I'$ then the following would also hold $I_{12} \neq -I_{12}$, right?

I wouldn't go that far. It could be that $I \ne I'$ because a different pair are not equal.

 

[Filip Larsen]

The original question was that, given $\rho$ is invariant under the given rotation, show that specific components of the inertia tensor is zero. Unless you also know or show the invariance of the density function implies invariance of the inertia tensor, i.e. $I = R I R^T$, you cannot use that to prove the components are zero as this would be kind of circular logic. And showing the inertia tensor is invariant requires you to look at the defining integrals anyway so why not just use those integrals directly to see the end up as zero?

 

[vanhees71]

If $\rho$ is invariant under the rotation by $\pi$ around a given axis, then the components of the tensor of inertia transform as discussed above $\hat{\Theta}'=\hat{R} \hat{\Theta} \hat{R}^T=\hat{\Theta}$. As was calculated in some postings above, this implies that $\Theta_{12}'=\Theta_{12}=-\Theta_{12}$ and $\Theta_{23}'=\Theta_{23}=-\Theta_{23}$, which immediately shows that $\Theta_{12}=\Theta_{23}=0$, and thus that the rotation axis is a principle axis of the tensor of inertia.

 

[Filip Larsen]

If is invariant under the rotation by around a given axis, then the components of the tensor of inertia transform as discussed above

I guess it all depends on whether or not the material quoted from already has introduced relationships which will allow the OP to conclude, by other means than from the integrals directly, that invariance of the inertia tensor under the rotation follows from the invariance of density function. The way I read the problem text it suggests a proof that involves the integrals directly.

 

[vanhees71]

But this is really easy to see:
$$\Theta_{jk} = \int_{\mathbb{R}^3} \mathrm{d}^3 x \rho(\vec{x}) [\vec{x}^2 \delta_{jk}-x_j x_k].$$
Then you get
$$\Theta_{ab}'=\int_{\mathbb{R}^3} \mathrm{d}^3 x \rho(\vec{x}) [\vec{x}^2 \delta_{ab} -R_{aj} R_{ak} x_j x_k].$$
Now substitute $\vec{x}'=\hat{R} \vec{x}$, \quad $\mathrm{d}^3 x'=\mathrm{d}^3 x$, and you get
$$\Theta_{ab}'=\int_{\mathbb{R}^3} \mathrm{d}^3 x \rho(\hat{R}^{-1} \vec{x}') [\vec{x}^{\prime 2} \delta_{ab} \vec{x}_a' \vec{x}_b'].$$
If the mass distribution is invariant under the rotation $\hat{R}$ (and thus also $\hat{R}^{-1}$), you have
$$\rho(\hat{R}^{-1} \vec{x}')=\rho(\vec{x}').$$
That shows
$$\Theta_{ab}'=\Theta_{ab}.$$

 

[PhDeezNutz]

I did it a little bit different than @vanhees71 in #24 but underneath it all is probably the same thing. I did along the lines of @Filip Larsen 's suggestion.

Take $I_{12}$ for instance.

$I_{12} = - \int_{z_1}^{z_2} \int_{y_1}^{y_2} \int_{x_1}^{x_2} \rho \left(x,y,z\right) xy \, dx \, dy \, dz$

make change of variables

$x \rightarrow x$ and $dx \rightarrow dx$
$y \rightarrow -y$ and $dy \rightarrow -dy$
$z \rightarrow -z$ and $dz \rightarrow -dz$

$I_{12} = - \int_{z_2}^{z_1} \int_{y_2}^{y_1} \int_{x_1}^{x_2} \rho \left(x,-y,-z\right) x \left(-y\right) \, dx \, \left(-dy\right) \, \left(-dz\right)$

the negatives on the differentials multiply to 1:

$= - \int_{z_2}^{z_1} \int_{y_2}^{y_1} \int_{x_1}^{x_2} \rho \left(x,-y,-z\right) x \left(-y\right) \, dx \, dy \, dz$

now un-reverse the limits of integration with the minus sign up front on $z$

$= \int_{z_1}^{z_2} \int_{y_2}^{y_1} \int_{x_1}^{x_2} \rho \left(x,-y,-z\right) x \left(-y\right) \, dx \, dy \, dz$

now un-reverse the limits of integration with the minus sign on $y$

$= \int_{z_1}^{z_2} \int_{y_1}^{y_2} \int_{x_1}^{x_2} \rho \left(x,-y,-z\right) x y \, dx \, dy \, dz$

Note that $\rho\left(x,-y,-z\right) = \rho\left(x,y,z\right)$

$= \int_{z_1}^{z_2} \int_{y_1}^{y_2} \int_{x_1}^{x_2} \rho \left(x,y,z\right) x y \, dx \, dy \, dz$

Which is $-I_{12}$

$I_{12} = - I_{12} \Rightarrow I_{12} = 0$

 

[Filip Larsen]

But this is really easy to see:

My point was that you need a proof similar to yours in order to establish the invariance of the inertia tensor in the first place, you can't just assume it. If the OP is really able to establish (and not just copy) such a proof in his solution, then that is all fine and good. But I will still suspect most students new to this topic will prefer a more simple and direct proof that just show the integrals in question evaluates to zero under the given assumptions, similar to what PhDeezNutz showed.

 

[vanhees71]

Well, first of all, it's not more simple and second it's a very nice example for the application of symmetry principles in physics, and one cannot overestimate the importance of symmetry! Imho the most underrepresented subject in the standard physics curriculum is group and representation theory!

 

[Filip Larsen]

Well, first of all, it's not more simple and second it's a very nice example for the application of symmetry principles in physics

So you assume that every student who has just started to learn about moment of inertia and the inertia tensor automatically knows all the magic required to understand the tensor notation in your proof?

And now we are on this topic I suddenly recall my recurrent puzzlement when you and others here provide help to homework, which does not by itself introduce use of tensor notation, by including derivations in tensor notation, often accompanied with a remark about how simple something is to show. That may be nice and elegant if the student actually has a clue what the notation mean, but pretty useless if he doesn't.

 

[vanhees71]

So you assume that every student who has just started to learn about moment of inertia and the inertia tensor automatically knows all the magic required to understand the tensor notation in your proof?

No, I don't assume this, but I advertize to teach it early on!

And now we are on this topic I suddenly recall my recurrent puzzlement when you and others here provide help to homework, which does not by itself introduce use of tensor notation, by including derivations in tensor notation, often accompanied with a remark about how simple something is to show. That may be nice and elegant if the student actually has a clue what the notation mean, but pretty useless if he doesn't.

Then he or she should learn it as early as possible, and here in the forum they can ask, if something is not clear!