next up previous
Next: Scaling potentials and the Up: Appendix D: How Many Previous: Requirement on time-integral of

General potential case

We start with the general potential case (with constant isotropic mass). The deformation field ${\mathbf D}{({\mathbf r})}$ must be defined everywhere in space (note that it does not change in time). The deformation parameter $x$ controls the hamiltonian

{\mathcal{H}}({\mathbf r},{\mathbf p};x) =
...}{2m} + U\bm{(}{\mathbf r} - x{\mathbf D}{({\mathbf r})}\bm{)}
\end{displaymath} (D.4)

by distorting the potential in space. Note that there is already an infinite class of `trivially special' deformation fields such that ${\mathbf D} \bot \nabla U$ everywhere: these cause no deformation. Therefore a scalar field $D{({\mathbf r})}\equiv \nabla U \cdot {\mathbf D}$ is sufficient to uniquely define the deformation. However working with the vector field ${\mathbf D}{({\mathbf r})}$ is more convenient because every function ${\mathbf D}{({\mathbf r})}$ is a valid deformation. (Not every function $D{({\mathbf r})}$ is valid; consider regions of constant potential). At a frozen (fixed) $x$ value, the fluctuating force on the $x$ parameter, defined by (2.6), can be written
{\mathcal{F}}(t) \; = \; -m\ddot{{\mathbf r}} \cdot {\mathbf D}{({\mathbf r})}- F(x),
\end{displaymath} (D.5)

where ${\mathbf r} = {\mathbf r}(t)$ is the (time-dependent) particle position, and $m\ddot{{\mathbf r}} = \dot{{\mathbf p}} = -\nabla U{({\mathbf r})}$ is the force on the particle. Therefore $\mathcal{F}$ is a function of the particle position alone. The constant $F(x)$ is chosen such that the time-average of ${\mathcal{F}}$ is zero; this is necessary for the asymptotic value $C(\tau\rightarrow\pm\infty)$ to be zero, to allow finite moments of $C(\tau)$. We will not restrict ourselves to phase-space-volume-preserving deformations (i.e. ones which have $F(x) = 0$).

The integral of (D.5) gives the general expression

{\mathcal{G}}(t) \; = \; -m \int_0^t dt' \: \ddot{{\mathbf ...
...athbf a} + B{\mathbf r} + O(r^2) \cdots \right] \; - \; F(x)t.
\end{displaymath} (D.6)

where the square brackets enclose the expansion of ${\mathbf D}{({\mathbf r})}$ (each vector component is assumed to be a multi-dimensional Taylor series about ${\mathbf r}={\mathbf 0}$), and ${\mathbf r}$ is the position at time $t'$. The coefficients ${\mathbf a}, B, \cdots$ are rank $1, 2, \cdots$ objects. The linear growth rate $F(x)$ is determined uniquely for a given ${\mathbf D}$. The linear growth (final term) is capable of cancelling deterministic linear growth in the time-integral term; however, if the time-integral grows diffusively, then no value of $F(x)$ can prevent $\mathcal{G}$ from being non-stationary. Therefore a choice of ${\mathbf D}{({\mathbf r})}$ which is `special' will have no terms in the time-integral which grow diffusively. Note that, since ${\mathbf r}(t)$ decorrelates on a timescale of $t_{{\mbox{\tiny erg}}}$, diffusive growth (random walk) is the generic case.

We treat the time-integral terms in (D.6) in increasing powers of $r$, (writing with Einstein summation notation):

$\displaystyle \int_0^t dt' \ddot{r_i} a_i$ $\textstyle =$ $\displaystyle a_i [\dot{r_i}]_0^t$ (D.7)
$\displaystyle \int_0^t dt' \ddot{r_i} B_{ij} r_j$ $\textstyle =$ $\displaystyle - B_{ij} \int_0^t dt' \dot{r_i} \dot{r_j}
+ B_{ij} [ \dot{r_i} r_j ]_0^t$ (D.8)
$\displaystyle \int_0^t dt' \ddot{r_i} C_{ijk} r_j r_k$ $\textstyle =$ $\displaystyle - C_{ijk} \int_0^t dt' \dot{r_i} (\dot{r_j}r_k + r_j\dot{r_k})
+ C_{ijk} [ \cdots ]_0^t ,$ (D.9)

and so on. What constraints arise on the coefficients for there to be no diffusive growth terms? The first result shows that since the integrand is an exact time derivative, there is no diffusive growth for any ${\mathbf a}$. In general, any boundary-type terms (those coming from exact time derivatives, shown with square brackets) cannot cause diffusion since all functions of ${\mathbf r}(t)$ are stationary and bounded (since the energy-surface is bounded in phase-space); so these terms can be ignored. The time-integral term in (D.8) will vanish if $B$ is any antisymmetric tensor. Any symmetric component in $B$ will give a quadratic form in the momentum, and therefore a fluctuating function whose time-integral grows diffusively. The one important exception to this is when this quadratic form is proportional (up to an exact time derivative) to ${\mathcal{H}}$, the Hamiltonian. ${\mathcal{H}}$ is a constant of the motion, so gives uniform linear (non-diffusive) growth of the time-integral. This will correspond to the special nature of dilation in the hard-walled case discussed in Sec. D.3.

We can show that the higher terms (D.9), etc cause diffusive growth, unless either all the coefficients $C_{ijk},
D_{ijkl}$, etc are zero, or unless the coefficients fall into the `trivially special' subspace corresponding to ${\mathbf D} \cdot \ddot{{\mathbf r}} = 0$ at all accessible ${\mathbf r}$. (This latter case imposes linear conditions on the coefficients; we exclude the resulting deformation-less subspace). For any given $k^{th}$-order term, for instance (D.9) with $k=3$, the coefficient tensor must be totally symmetric with respect to interchange of its last $k{-}1$ indices (nonsymmetric parts cancel in the summation because of this symmetry in the $r$ factors). Note that this symmetry forbids the type of argument that allowed an antisymmetric $B$ to give no diffusive growth for $p=2$. Manipulation by parts conserves the symmetry of the $r$ terms, the power of $r$ (namely $k$), and the number of time-derivatives, or `dots' (namely 2). (It also generates ignorable boundary-type terms). There are only two cases:

  1. Both the dots are on the same $r$, in which case the integrand is a (non-zero) function of position alone (since $\ddot{r_i}$ is a function of position alone).
  2. The dots are on different $r$'s, giving a quadratic form in ${\mathbf p}$ multiplied by a (non-zero) function of ${\mathbf r}$.
It is clear that by such manipulation, no exact time-derivative can be produced.

The only other way to prevent diffusive growth is by making the integrand a multiple of $\mathcal{H}$ (plus exact time- derivatives). Case 2) cannot give the required $p^2$ term so it is no help. (The only $p^2$ term can come from an isotropic $B$ tensor). It is unknown whether case 1) can give a term proportional to $U{({\mathbf r})}$, which could add to an existing $p^2$ term to give $\mathcal{H}$. It seems this latter occurrence could only happen for particular forms of $U{({\mathbf r})}$, for instance the scaling potentials given below D.1. If it happens, it would give a `dilation-type' special deformation corresponding to conservation of $\mathcal{H}$.

In conclusion this gives the form of a `special' deformation (in a general potential) as

{\mathbf D}{({\mathbf r})}\; = \; {\mathbf a} + B^a {\mathbf r}, \hspace{1in}
\mbox{$B^a$\ an antisymmetric tensor.}
\end{displaymath} (D.10)

The first term accounts for all possible translations, and the second all possible rotations, in $d$ dimensions. The existence of a `dilation-type' special deformation for a general potential is still open.

next up previous
Next: Scaling potentials and the Up: Appendix D: How Many Previous: Requirement on time-integral of
Alex Barnett 2001-10-03