Brake Orbits Fill the N-Body Hill Region

Abstract

A brake orbit for the N-body problem is a solution for which, at some instant, all velocities of all bodies are zero. We reprove two “lost theorems” regarding brake orbits and use them to establish some surprising properties of the completion of the Jacobi – Maupertuis metric for the N-body problem at negative energies.

APPENDIX A. BOUNDS ON POTENTIAL DERIVATIVES IN AN ANNULUS

Here we prove the bounds relating the potential and its 1st and 2nd derivatives, (inequality (4.2)), used in the proof. We repeat this inequality now. It asserts the existence of positive constants $c,c_{1},C_{1},C_{2}$ such that for all $q,v$ we have:

$$1\leqslant U(q)\leqslant 1+c\implies c_{1}\leqslant\|\nabla U(q)\|\leqslant C_{1},\qquad|d^{2}U(q)(v,v)|\leqslant C_{2}\|v\|^{2}.$$

(A.1)

We begin by showing that it suffices to prove the following modified bound. There exist positive constants $a_{1},A_{1},A_{2}$ such that

$$U(q)=1\implies a_{1}\leqslant\|\nabla U(q)\|\leqslant A_{1},\quad|d^{2}U(q)(v,v)|\leqslant A_{2}\|v\|^{2}.$$

(A.2)

To go from (A.2) to (A.1) use that $U$ is homogeneous of degree $-1$ in $q$ and thus $\nabla U$ is homogeneous of degree $-2$ and $d^{2}U$ is homogeneous of degree $-3$. It follows that inequality (A.2) implies

$$U(q)=\frac{1}{\lambda}\implies\frac{a_{1}}{\lambda^{2}}\leqslant\|\nabla U(q)\|\leqslant\frac{A_{1}}{\lambda^{2}},\qquad|d^{2}U(q)(v,v)|\leqslant\frac{A_{2}}{\lambda^{3}}\|v\|^{2}.$$

(A.3)

Now write the upper bound $1+c$ of inequality (A.1) as $1+c=1/\lambda$. Bounds on $\nabla U$ and $d^{2}U(q)(v,v)$ immediately follow with the upper bounds $C_{1},C_{2}$ now depending on $c$ through the scaling. (We do not need to keep track of the precise dependence.)

To establish the upper bounds of inequalities (A.2), observe that each summand $m_{a}m_{b}/r_{ab}$ occurring in $U$ is positive. Thus, if $U=1$ then, for each pair $a\neq b$, we have $m_{a}m_{b}/r_{ab}<1$. It follows that $m_{a}m_{b}/r_{ab}^{2}\leqslant 1/m_{a}m_{b}$ and $m_{a}m_{b}/r_{ab}^{3}\leqslant 1/(m_{a}m_{b})^{2}$. Let’s call these type 1 and type 2 functions. Each component of $\nabla U$ consists of a sum of terms of the form $m_{a}m_{b}/r_{ab}^{2}$ times a unit vector $\hat{q}_{a}-\hat{q}_{b}=q_{ab}/r_{ab}$. The Hessian $d^{2}U$ can be written out as an $N\times N$ block symmetric matrix whose $d\times d$ entries consist of the function $m_{a}m_{b}/r_{ab}^{3}$ times quadratic forms $H_{ab}$ made out of a fixed linear combination of the identity and $\hat{q}_{ab}\otimes\hat{q}_{ab}$. The $H_{ab}$ satisfy a uniform bound independent of $a,b$ and the masses. Thus, we get $|\nabla U(q)|\leqslant C$ and $|d^{2}U(q)(v,v)|\leqslant C^{\prime}|v|^{2}$ where $C,C^{\prime}$ only depend on the masses and the number $N$ of bodies $N$. Essentially the same algebra and logic works for any power law potential, with the exponents $1/r^{2}$ and $1/r^{3}$ replaced by $1/r^{\alpha+1}$ and $1/r^{\alpha+2}$.

To establish the lower bound for $\nabla U$ observe that $\nabla U(q)\to 0$ if and only if all $r_{ab}\to\infty$, in which case $U\to 0$. Consequently, if we fix $U=1$ there must be a positive lower bound for $\|\nabla U(q)\|$.

APPENDIX B. VARIATIONS ALONG THE RAY

Here we provide the proof promised within the step “Step two: variation along the rays” within the part of the proof labeled “Showing the collar map is a local diffeomorphism”. We are to find a $\tau_{0}$ small enough so that the variation $X(q_{0},\tau)={{\partial}\over{\partial\tau}}\Psi(q_{0},\tau)$ of $\Psi$ along the ray directions is transverse to its variation $\delta q(0)\mapsto D\Psi(q_{0},\tau)(\delta q(0),0)$ along the brake initial conditions surface $\partial\Omega$. To this end we will use a variant of the following lemma:

Lemma 3

Let $\alpha$ be a nonvanishing one-form defined on an open set of an inner product space $\mathbb{E}$ . Suppose that $X$ is a vector field and $D$ is a hyperplane field on $W$ . If there is a constant $c>0$ such that

$$|\alpha(X)|>c\|X\|$$

(B.1)

while

$$|\alpha(V)|<c\|V\|$$

(B.2)

for all $V$ tangent to $D$ , then $X$ is transverse to $D$ on the open set.

Proof

The conditions $|\alpha(X)|>c\|X\|$ and $\alpha(X)<c\|X\|$ are mutually exclusive, showing that $X\notin D$.     $\square$

We need a slight variant of the lemma for use in our situation: allow the vector field and hyperplane field be along a smooth map into $\mathbb{E}$. Thus, we are given a smooth map $\Psi:M\to U\subset\mathbb{E}$ defined on some manifold $M(=\partial\Omega\times[0,\tau_{0}])$ and smooth maps $X:M\to\mathbb{E}$ and $D:V\to Gr(k-1,\mathbb{E})$, the Grassmannian of hyperplanes in $\mathbb{E}$. (So $k={\rm dim}(\mathbb{E})$.) The variant of the lemma asserts that if, for all $m\in M$, we have $\alpha\big{(}\psi(m)\big{)}\big{(}X(m)\big{)}>c\|X(m)\|$ while $|\alpha\big{(}\psi(m)\big{)}(V)|<c\|V\|$ for all nonzero $V\in D(m)$, then $D(m)$ and $X(m)$ are everywhere transverse. The proof is the same.

To apply the variant of the lemma to our situation, take $\psi=\Psi:\partial\Omega\times[0,\tau_{0}]\to\mathbb{E}$ — the collar map, as before. Take the one-form to be $\alpha=dU$. Take the vector field $X$ along $\Psi$ to be $X(\tau,q)={{\partial}\over{\partial\tau}}\Psi(\tau,q)$. Take the hyperplane field along $\Psi$ to be the image of the differential at $q$ of $q\mapsto\Psi(q,\tau)$. For the open set $W$ we can take an energy annulus $\{1<U<1+c^{\prime}\}$ where $c^{\prime}<c$ is such that the image of $\Psi$ covers all of $W$. We are guaranteed such a $c^{\prime}$ exists by inequality (4.14). Indeed, that inequality says we can take $c^{\prime}=k_{1}t_{0}^{2}$.

We will have established transversality of $X$ and $D$ once we have established the two inequalities (B.1), (B.2) of the lemma.

Establishing inequality (B.1). We begin by establishing

$$X(q,\tau)=\nabla U\big{(}\Phi(q,\tau)\big{)}+O(\tau)$$

(B.3)

where here and throughout the rest of this section we use “$O(\tau)$” to indicate a vector or scalar variable whose dependence on $\tau$ is uniformly bounded by a positive constant times $\tau$ up to $\tau=\tau_{0}$. Thus, inequality (B.3) asserts the existence of a positive constant $M$ independent of $q_{0}$ such that $\|X(q,\tau)-\nabla U\big{(}q(\tau)\big{)}\|\leqslant M\tau$ for all $q_{0}$ and for $0\leqslant\tau\leqslant\tau_{0}$. From the chain rule and the definition of $\Phi$ we get

$$X(q,\tau)=\frac{1}{2t}\dot{q}(t,q_{0});\quad t^{2}=\tau.$$

Inequality (4.5) can be differentiated term by term, yielding

$$\dot{q}(t,q_{0})=t\nabla U(q_{0})+O(t^{3}),$$

so that $X(q,\tau)=\nabla U(q_{0})+O(\tau)$. Here we use the obvious analogue of the “$O$” notation for powers of $t$ so that $O(t^{3})$ means bounded by some $Ct^{3}$ over $[0,t_{0}]$. We have already seen that $\nabla U\big{(}q(t,q_{0})\big{)}=\nabla U(q_{0})+O(t^{2})=\nabla U(q_{0})+O(\tau)$. Combining these inequalities and using $q(t,q_{0})=\Phi(q_{0},\tau)$ yields the result, inequality (B.3).

Thus, $\alpha\big{(}X(q_{0},\tau)\big{)}=dU(\nabla U\big{(}\Phi(q_{0},\tau)\big{)}+O(\tau)=|\nabla U\big{(}\Phi(q_{0},\tau)\big{)}\|\|X(q,\tau)\|+O(\tau)$. Now recall the bound (4.2), which asserts that $c_{1}\leqslant\|\nabla U(q)\|$ on our energy annulus. Set $c=c_{1}/2$ and choose $\tau_{0}$ so small that the error term $O(\tau)$ satisfies $|O(\tau)|\leqslant c_{1}/2$ provided $\tau\leqslant\tau_{1}$. With these choices, we get inequality (B.1) with $c=c_{1}/2$.

Establishing inequality (B.2). Any $V\in D(q,\tau)$ has the form

$$V=\delta q(t),\quad t=\sqrt{\tau}$$

for a Jacobi field $\delta q(t)$ as before. (Recall $\tau=t^{2}$ as per usual.) Recall the boundary conditions $\delta q(0)\in T_{q_{0}}\partial\Omega$ and $\dot{\delta}q(0)=0$. We will need the following two bounds relating $\delta q(0)$ and $\delta q(t)$:

$$\delta q(t)=\delta q(0)+R(t),\text{ where }\|R(t)\|\leqslant M\|\delta q(t)\|t^{2},$$

(B.4)

and

$$\frac{1}{2}\|\delta q(0)\|\leqslant\|\delta q(t)\|,$$

(B.5)

which we will derive below. We can write the bound of inequality (delta bd 0) as $R(t)=O(\tau)\|\delta q(t)\|$. We have that $\alpha(q)\big{(}\delta q(0)\big{)}=0$ since $\alpha=dU$ annihilates the tangent space to $\partial\Omega=\{U=1\}$. Moreover, $dU(q(t,q_{0})=dU(q_{0})+O(t^{2})$ since we have uniform Lipschitz bounds on $U$ and since $q(t,q_{0})=q_{0}+O(t^{2})$. Thus, at points $q=\Phi(q_{0},\tau)$ we have that $\alpha(q)=\alpha(q_{0})+O(\tau)$, so that

$$\displaystyle \begin{aligned} \alpha(q) \big(\delta q(t)\big) &= \big(\alpha (q_0) + O(\tau)\big)\big(\delta q(0)\big) + O(\tau)| \delta q(0)| \\ &= 0 + O(\tau) \| \delta q (0)\| + O(\tau) \| \delta q (0)\| + O(\tau^2) \|\delta q (0) \| \\ &= O(\tau) \| \delta q(t) \| \\ &= O(\tau) \| V \|, \end{aligned}$$

where to go from the second to the third line we invoke (B.5). This result asserts the existence of a constant, independent of $q_{0}$, such that $\alpha(q)(V)\leqslant K\tau\|V\|$ for all $\tau\leqslant\tau_{0}$. Take $\tau_{1}$ so small that $K|\tau_{1}|<c_{1}/2:=c$ to get the desired $|\alpha(q)(V)|<c\|V\|,$ inequality (B.2).

We proceed to establishing inequalities (B.4) and (B.5). Inequality (B.4) is implied by (B.5) and

$$\|\delta q(t)-\delta q(0)\|\leqslant K\|\delta q(0)\|t^{2}.$$

(B.6)

(Take $M=2K$ in (B.4).) It now remains to establish inequalities (B.5), (B.6) and for this we will need the third inequality:

$$\|\delta q(t)\|\leqslant 2\|\delta q(0)\|.$$

(B.7)

We proceed to establishing inequalities (B.7), (B.6) and (B.5) in that order. For this purpose, fix the brake orbit $q(t)=q(t,q_{0})$ and set

$$A(t)=d^{2}U\big{(}q(t)\big{)}\text{ and }J(t)=\delta q(t)$$

so that the Jacobi equation (4.16) becomes

$$\ddot{J}(t)=A(t)J(t)\text{ with }\dot{J}(0)=0.$$

We will need the bound

$$\|A(t)\|\leqslant C_{2},$$

(B.8)

which is exactly the second derivative bound of inequality (A.3).

Establishing inequality (B.7). Set $\xi(t)=\big{(}J(t),\dot{J}(t)\big{)}\in\mathbb{E}\times\mathbb{E}$ and

$$M(t)=\left(\begin{array}[]{cc}0&I\\ A(t)&0\end{array}\right),$$

$M(t):\mathbb{E}\times\mathbb{E}\to\mathbb{E}\times\mathbb{E}$ is a time-dependent linear operator with which we can rewrite our Jacobi equation in the equivalent first-order form as

$$\dot{\xi}(t)^{T}=M(t)\xi(t)^{T};\quad\xi(0)=\big{(}J(0),0\big{)}.$$

If we use the norm $\|(q,v)\|=\|q\|+\|v\|$ on $\mathbb{E}\times\mathbb{E}$, then the operator norm of $M(t)$ satisfies

$$\|M(t)\|\leqslant C*=\min\{1,C_{2}\}$$

with $C_{2}$ as in inequality (B.8). It follows that

$$\|\xi(t)\|\leqslant e^{C_{*}t}\|\xi(0)\|.$$

But $\|\xi(0)\|=\|J(0)\|$ and $\|J(t)\|\leqslant\|\xi(t)\|$, so we get

$$\|J(t)\|\leqslant C_{M}\|J(0)\|$$

with $C_{M}=e^{C*t_{0}}$ for all $|t|\leqslant t_{0}$. To insure that $C_{M}\leqslant 2$, choose $t_{0}\leqslant\log_{e}(2)/C_{*}$.

Establishing inequality (B.6). Since $\dot{J}(0)=0$ we have that $\dot{J}(t)=\int_{0}^{t}\ddot{J}(s)ds=\int_{0}^{t}A(s)J(s)ds$. Thus,

$$\displaystyle J(t)-J(0)=\int_{0}^{t}\dot{J}(s)ds$$

(B.9)

$$\displaystyle=\int_{0}^{t}\int_{0}^{s}\ddot{J}(u)ds$$

(B.10)

$$\displaystyle=\int_{0}^{t}\int_{0}^{s}A(u)J(u)ds.$$

(B.11)

Now combine $\|A(t)\|\leqslant C_{2}$ and $\|J(t)\|\leqslant 2\|J(0)\|$ for $|t|\leqslant t_{0}$ to get $\|A(u)J(u)\|\leqslant 2\alpha\|J(0)\|$ for $u$ in the range of integration when $t\leqslant t_{0}$ and so

$$\displaystyle\|J(t)-J(0)\|\leqslant\int_{0}^{t}\int_{0}^{s}2C_{2}\|J(0)\|ds$$

(B.12)

$$\displaystyle=2C_{2}\|J(0)\|\int_{0}^{t}\int_{0}^{s}ds$$

(B.13)

$$\displaystyle=K\|J(0)\|t^{2}$$

(B.14)

with $K=2C_{2}$.

Establishing inequality (B.5). From $J(t)=J(0)+J(t)-J(0)$ we get $\|J(t)\|\geqslant\|J(0)\|-\|J(t)-J(0)\|$. But inequality (B.6) says $\|J(t)-J(0)\|\leqslant Kt^{2}\|J(0)\|$. By taking $|t_{0}|$ so that $Kt_{0}^{2}\leqslant(1/2)$ we get $\|J(t)\|\geqslant\|J(0)\|-\frac{1}{2}\|J(0)\|=\frac{1}{2}\|J(0)\|$ as desired.

Finally, replace the old $\tau_{0}=t_{0}^{2}$ by the minimum of all the upper bounds $\tau_{0}=t_{0}^{2}$, appearing for each of these finite numbers of inequalities to insure that all of them hold on the same interval.

APPENDIX C. A GAUSS LEMMA TYPE ARGUMENT: PROOF OF LEMMA 2

We prove the claim by the standard “calibration” argument of Hamilton – Jacobi theory. To begin let us observe that we can reparameterize the collar using JM arclength $s$ rather than time $t$. Thus, we write

$$F(s,q)=\Phi\big{(}t(s,q),q\big{)},\quad q\in\partial\Omega$$

where $t(s,q)$ relates Newtonian time $t$ to JM arclength along the brake orbit starting at $q\in\partial\Omega$. Set

$$T(s,q)=\frac{d}{ds}F(s,q)$$

for the tangent vectors along the brake orbits. For simplicity of notation, in the remainder of the argument, inner products, as in $\langle V,W\rangle$ , and norms $\|V\|$ will mean JM inner products and norms. The calibration (or Hamilton – Jacobi) argument relies on establishing

$$ds(v)\leqslant\|v\|$$

(C.1)

and

$$ds(v)=\|v\|\iff v\text{ is parallel to }T.$$

(C.2)

We have actually already established the implication “$\Leftarrow$” of Eq. (C.2). For when we parameterize a brake curve by its JM arclength, then its tangent vector is $T$, which, by construction, has JM length $1$.

Assume for the moment that the calibration conditions (C.1) and (C.2) hold. We show how they imply that $s$ is equal to the JM distance $S(A)$ in the collar. Fix a point $A$ in the collar and consider any smooth path $c:[0,T]$ in the collar connecting the Hill boundary to $A$. Then $s(A)-s(0)=s(A)=\int_{c}ds$. But $|ds(\dot{c})|\leqslant\|\dot{c}\|_{JM}$, from which it follows that $s(A)\leqslant\int_{c}\|\dot{c}\|_{JM}dt=\ell(c)$. This shows that $s(A)\leqslant{\rm dist}(A,\partial\Omega):=S(A)$. On the other hand, if $c$ is the brake orbit passing through $A$ we get equality, proving that the brake orbits in the collar are JM minimizers. (The astute reader will be able to work out how to exclude paths that leave the collar neighborhood on their way to the boundary as well, using the fact that at some point they must reenter the collar to arrive at the boundary.)

We now verify the two calibration conditions (C.1) and (C.2). For inequality (C.1) and the implication “$\Rightarrow$” of Eq. (C.2) it suffices to show that $T$ is orthogonal to the level sets of $s$. For if this is the case, then any vector $v$ in the collar neighborhood can be decomposed as an orthogonal direct sum:

$$v=v^{T}+cT,\ c\in\mathbb{R},\ v^{T}\perp T,\ ds(v^{T})=0.$$

(C.3)

We then get $ds(v)=ds(cT)=c$ and, because $\langle v^{T},T\rangle=0$, we find $\|v\|=\sqrt{\|v^{\perp}\|_{JM}^{2}+c^{2}}\geqslant|c|$.

We establish the required perpendicularity, Eq. (C.3), by following the standard proof of what is known as the “Gauss lemma”, recorded in [7] as Lemma 10.5 on p. 60, but without that name. Any vector $W$ tangent to the level set of $s=s_{0}$ at some point $F(s_{0},q_{0})$ can be expressed in the form

$$W=\frac{d}{d\lambda}|_{\lambda=0}F\big{(}s_{0},q(\lambda)\big{)}$$

where $\lambda\mapsto q(\lambda)$ is some embedded curve passing through $q_{0}$ when $\lambda=0$. We can make $W$ part of a vector field along a parameterized surface containing our “base” brake orbit $s\mapsto F(s,q_{0})$ by forming the surface

$$f(s,\lambda)=F\big{(}s,q(\lambda)\big{)}.$$

So set

$$W={{\partial}\over{\partial\lambda}}f(s,\lambda).$$

Along the surface the two vector fields $T$ and $W$ commute and the vector field $T$ is a field of (JM) geodesics parameterized by arclength. The perpendicularity condition (C.3) is equivalent to $\langle T,W\rangle=0$ Differentiating this expression with respect to $s$ and using the Levi-Civita connection $\nabla$ for the JM metric, the fact that the JM brake orbits are characterized by $\nabla_{T}T=0$, and the fact that $0=[T,W]=\nabla_{T}W-\nabla_{W}T$, we get

$$\displaystyle\begin{aligned} \frac{d}{ds} \langle T, W \rangle & = T \langle T, W \rangle \\ & = \langle \nabla_T T , W \rangle + \langle T , \nabla_T W \rangle\\ & = \langle T , \nabla_T W \rangle \\ & = \langle T , \nabla_W T \rangle \\ & = \frac{1}{2} W \langle T , T \rangle \\ & = {{\partial}\over{\partial\lambda}} \langle T , T \rangle \\ & = 0, \end{aligned}$$

where in arriving at the final line we used that $s$ is arclength, so that $\langle T,T\rangle=\text{const}=1$. Thus, the inner product $\langle T,W\rangle$ is constant along brake orbits. To finish the proof, we show that $\lim_{s\to 0}\langle T,W\rangle=0$. Indeed, we already know this! $T$ points in the direction $\nabla U(q_{0})$ which is orthogonal to $\partial\Omega$.

But let us be careful! The orthogonality $\nabla U(q_{0})\perp\partial\Omega$ is true in the Euclidean inner product, so, in a formal sense, also in the JM metric which is conformal to the Euclidean one. But maybe $W$ or $T$ blow up as $s\to 0$ in such a way as to make this obvious fact not true. (Indeed, $T$ must blow up!) So we roll up our sleeves and take the pain to make sure the limit $\lim\limits_{s\to 0}\langle T,W\rangle$ actually exists and is zero. This grungy computation will finish off the proof.

In the remaining paragraphs all estimates of the form $O(t^{2}),O(\tau)$ etc refer to estimates made using Euclidean norms. When we need to differentiate from the Euclidean and JM inner product, we use $\langle\cdot,\cdot\rangle_{\mathbb{E}}$ for the Euclidean norm. From our earlier expressions we have, along a brake orbit, that $T=\dot{q}\frac{dt}{ds}$ and $\dot{q}=\frac{1}{2}t\nabla U(q_{0})+O(t^{3})$. But $ds/dt=t^{2}w(q,t)$, so $dt/ds=\frac{1}{t^{2}w(q,t)}$ by Eq. (4.19) with $w$ smooth and bounded away from zero as $t\to 0$. It follows that

$$T=\frac{1}{2}\frac{1}{tw(q_{0},t^{2})}\nabla U(q_{0})+O(t)\text{ as }t\to 0.$$

To estimate $W$ as $s\to 0$ we write $F(s,q)=\Psi(\tau(s,q),q)$ where $\Psi$ was our original collar parameterization and $\tau(s,q)$ is the inverse of the arclength parameterization $s=s(\tau,q)$ along the brake orbit corresponding to $q\in\partial\Omega$. By the chain rule

$$W={{\partial\Psi}\over{\partial q}}\delta q+{{\partial\Psi}\over{\partial\tau}}{{\partial\tau}\over{\partial q}}\delta q$$

where $\delta q=\frac{d}{d\lambda}q(\lambda)$. The first term ${{\partial\Psi}\over{\partial q}}\delta q$ is one of our Jacobi vectors $J(t)=\delta q(t)$ evaluated at $t=\sqrt{\tau}$ and whose growth with $t$ we investigated earlier, so that $J(t)=J(0)+O(t^{2})$ with $\langle\nabla U(q_{0}),J(0)\rangle_{\mathbb{E}}=0$. We also need to estimate the second term ${{\partial\Psi}\over{\partial\tau}}{{\partial\tau}\over{\partial q}}\delta q$. ${{\partial\Psi}\over{\partial\tau}}=\nabla U(q_{0})+O(\tau)$ as verified near the beginning of this section. We will also need to understand the decay of the scalar multiple ${{\partial\tau}\over{\partial q}}\delta q:=\frac{d}{d\lambda}\tau\big{(}s,q(\lambda)\big{)}$ which multiplies it as $s\to 0$.

We have seen that

$$\tau=t^{2},\quad s=t^{3}a(q,t^{2}),\quad u:=U-1=t^{2}b(q,t^{2})$$

near $t=0$ where $a,b$ are smooth and bounded away from zero. The parameter $\tau$ is a smooth normal coordinate for $\partial\Omega$, meaning that as a one form $d\tau\neq 0$ along the hypersurface $\partial\Omega=\{\tau=0\}$. Let us define a new function

$$v=s^{2/3}.$$

We have that $v=\tau a(q,\tau)^{2/3}$ is also a smooth normal coordinate for $\partial\Omega$ since the function $a(q,\tau)^{2/3}=c(q,\tau)$ is smooth and bounded away from zero since $a$ is smooth and bounded away from zero as $\tau\to 0$. Thus, $(q,\tau)\mapsto\big{(}q,v(q,\tau)\big{)}$ is a diffeomorphism near $\tau=0$ and we can write its inverse as $(q,v)\mapsto\big{(}q,\tau(q,v)\big{)}$ with

$$\tau=vh(q,v),\quad h(q,0)>0.$$

Plugging in the definition of $v$, we get that

$$\tau(s,q)=s^{2/3}h(q,s^{2/3}),$$

from which it follows that ${{\partial\tau}\over{\partial q}}\delta q=s^{2/3}{{\partial h}\over{\partial q}}\delta q=O(s^{2/3})=O(\tau)$. Putting these pieces together, we get that

$$W=\nabla U(q_{0})+O(\tau).$$

Consequently, $\langle T,W\rangle=\langle T,W\rangle_{JM}=u\langle T,W\rangle_{\mathbb{E}}$ has the following structure: $\langle T,W\rangle_{JM}= u\langle\frac{1}{2}\frac{1}{tw(q_{0},t^{2})}\nabla U(q_{0})+O(t),J(0)+O(t^{2})\rangle_{\mathbb{E}}$. The leading order term of this expression, namely, $u\langle\frac{1}{2}\frac{1}{tw(q_{0},t^{2})}\nabla U(q_{0}),J(0)\rangle_{\mathbb{E}}$ is zero since $\langle\nabla U(q_{0}),J(0)\rangle_{\mathbb{E}}=0$. Also, $u=O(t^{2})$. Thus $\langle T,W\rangle_{JM}=uO(t)=O(t^{3})\to 0$.     $\square$