Background:
Let $M$ be a Riemannian manifold. Let $p \in M$ and $\epsilon \gt 0$.
For sufficiently small $\epsilon$, the standard definition (correct me if I'm wrong) for the 'geodesic ball of radius $\epsilon$ centered at $p$' is:
B = \{ \gamma(1) \mid \gamma \text{ is a geodesic}, \gamma(0)=p, \text{ and } \langle \gamma'(0), \gamma'(0) \rangle \, \lt \epsilon \} .
I am wondering: suppose we instead consider the set
B_{\text{alt}} = \{ \gamma(\epsilon) \mid \gamma \text{ is a geodesic}, \gamma(0)=p, \text{ and } \langle \gamma'(0), \gamma'(0) \rangle \lt 1 \} .
(Roughly speaking, $B$ takes an $\epsilon$-ball in $T_pM$ and runs the geodesics forward by 1, while $B_{\text{alt}}$ takes the unit ball in $T_pM$ and runs the geodesics forward by $\epsilon$. $\langle \cdot , \cdot \rangle$ is the metric inner product.)
Question:
Do we have $B = B_{\text{alt}}$? If not, is there a nice counter-example? Or an intuitive reason why we should not expect the equality to hold?
Note:
I have tried to compute the volume of $B_{\text{alt}}$ (to next-to-leading order in $\epsilon$) in the special case of a two-dimensional manifold with a diagonal metric. Despite many careful checks, my answer does not match the standard answer for the volume of a geodesic ball -- so I suspect that $B \ne B_{\text{alt}}$.
Thanks for any help!
Appendix - calculation of $Vol(B_{alt})$ (added 12/23)
I'm going to redo my calculation of the volume of $B_{alt}$, taking account of joriki's comment below. The problem I was having (which I don't think is resolved by joriki's suggestion to replace $\epsilon$ by $\epsilon^{1/2}$, but will have to check to be sure) is described in the next two paragraphs.
I'm doing the calculation in a particular coordinate chart, assuming the metric is diagonal: $ds^2 = g_{11}(x^1,x^2) (dx^1)^2 + g_{22}(x^1,x^2) (dx^2)^2$. I find that $Vol(B_{alt})$ includes (at next-to-leading order in $\epsilon$) the terms $(\partial_1)^2 g_{11}|_p$ and $(\partial_2)^2 g_{22}|_p$. In a diagonal metric (and using the Levi-Civita connection), the only Christoffel symbol in which $\partial_1 g_{11}$ appears is $\Gamma^1_{11} = \frac{1}{2} g^{11} \partial_1 g_{11}$. However, the formula $R^a_{bcd} = \partial_c \Gamma^a_{db} - \partial_d \Gamma^a_{cb} + \Gamma^a_{ce} \Gamma^e_{db} - \Gamma^a_{de} \Gamma^e_{cb}$ shows that $\partial_1 \Gamma^1_{11}$ doesn't appear in the scalar curvature - the first two terms in the preceeding formula just cancel when all indices are equal to 1.
Now I will explain how these bad terms appear in my calculation. My approach is to calculate $Vol(B_{alt}) = \int_{B_{alt}} dx^1 dx^2 \sqrt{g_{11}(x^1,x^2) g_{22}(x^1,x^2)}$ by Taylor-expanding the integrand about the point $p \leftrightarrow (x^1_0,x^2_0)$. Given $a\ \partial_1 |_p + b\ \partial_2 |_p$ in the unit ball in $T_pM$, the corresponding point in $B_{alt}$ is $(x^1_0 + a\ \epsilon + C(a,b)\ \epsilon^2 + O(\epsilon^3), x^2_0 + b\ \epsilon + D(a,b)\ \epsilon^2 + O(\epsilon^3))$, where $C,D$ are constants we can get from the geodesic equation.(**) There's a Jacobian factor to change variables from $(x^1, x^2)$ to $(a,b)$. But the only way (it seems to me) to get the second derivatives of the metric that appear in $R$ is to Taylor-expand $\sqrt{det\quad g}$ to second order. That's exactly the order at which the bad terms $(\partial_1)^2 g_{11}|_p$ and $(\partial_2)^2 g_{22}|_p$ appear (even if we replace $\epsilon$ by $\epsilon^{1/2}$).
(**) $C(a,b) = -\frac{1}{2}(a^2 \Gamma^1_{11} + b^2 \Gamma^1_{22} + a\ b\ \Gamma^1_{12})|_p$.
Thank you Srivatsan for your TeX edits... I am still fairly new to TeX.