Wikipedia:Reference desk/Archives/Mathematics/2012 May 16

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May 16

The norm of a multivariate normally-distributed vector

Consider a k-dimensional random vector x = [X₁, X₂, …, X_k] which has a multivariate normal distribution with mean zero

$\mathbf {x} \ \sim \ {\mathcal {N}}(0,\,{\boldsymbol {\Sigma }}),$

what is the distribution of its 2-norm, as defined below?

$\|\mathbf {x} \|_{2}:={\bigg (}\sum _{i=1}^{n}|x_{i}|^{2}{\bigg )}^{1/2}$

We can simplify this question by considering ${\boldsymbol {\Sigma }}$ is a diagonal matrix with diagonal entries $\sigma _{i}$, then we know that the distribution of

$Y={\sqrt {\sum _{i=1}^{k}\left({\frac {X_{i}}{\sigma _{i}}}\right)^{2}}}$

is the chi distribution. But what is the distribution of

$Z={\sqrt {\sum _{i=1}^{k}\left(X_{i}\right)^{2}}}\;?$

Yymao (talk) 06:37, 16 May 2012 (UTC)

Inertia tensor question

Yes, I'm aware that this is the mathematics reference desk and not the physics one; however, given the mathematical nature of the question I figured that this desk was the more appropriate.

Suppose I have a body composed of an infinite number of infinitesimal point masses. I am trying to find the value of the following integral.

$\int dV\rho (\mathbf {x} ){\sqrt {{\ddot {\mathbf {x} }}\cdot {\ddot {\mathbf {x} }}}}$

where $\mathbf {x}$ is the location of a point in the body in world space and the dot signifies differentiation with respect to time.

However, I do not know everything about the body. All I know is

1. The position of the body for all time.
2. The orientation of the body for all time.
3. The body's inertia tensor.
4. (addition) The body's mass.

So, my question: do I have enough information to evaluate the integral above, and if so, how do I do it? I've been trying to do this using a homogeneous geometric algebra, but I seem to be missing some trick.--Leon (talk) 20:37, 16 May 2012 (UTC)

It doesn't seem like the integral is uniquely determined by these values. To take a simple example: If the body isn't rotating at all, the integral is just the mass times the magnitude of the acceleration, but the mass cannot be deduced (even if you have the mass I still think it's impossible). -- Meni Rosenfeld (talk) 22:05, 16 May 2012 (UTC)

I've made an addition to include the body's mass as a given. In the homogeneous geometric algebra that I use, mass is a part of the inertia tensor, but you've reminded me that most objects called the inertia tensor don't include the mass.--Leon (talk) 09:23, 17 May 2012 (UTC)

Ok. I think the answer is still the same though. If that's true it shouldn't be hard to find an example with some linear and rotational accelerations, and two bodies with the same tensor but different values of the integral.

If you do the integral $\int dV\rho (\mathbf {x} )({\ddot {\mathbf {x} }}\cdot {\ddot {\mathbf {x} }})$ instead, and what you refer to as the moment of inertia tensor also includes $\int e\times \mathbf {x} \ dM$ for every basis vector e, then I think this is solvable. You should use ${\ddot {\mathbf {x} }}={\ddot {c}}+{\ddot {A}}v$ where c is the center of mass, A is the rotation matrix and v is the original location vector of the point mass. -- Meni Rosenfeld (talk) 13:25, 17 May 2012 (UTC)

Maybe a dumb question, but what is meant by knowing the position and orientation at all times? I'm thinking that if this is a body on which no external forces are acting, then the only acceleration would be due to rotation of the body. But I'm not sure I understand the question. Ssscienccce (talk) 06:24, 17 May 2012 (UTC)

The OP didn't say there are no external forces acting. An arbitrary external force and torque are applied, and we are given the object's center of mass and rotation matrix as a function of time that results from these forces. -- Meni Rosenfeld (talk) 07:20, 17 May 2012 (UTC)