# Two formulas for a particle's acceleration

While on a class my teacher was taking about particle's motion in space. At some point she said the following: Consider that the particle's path is described by a curve in space defined by the parametric equations $x^i=x^i(s)$, where $x^i$ are the coordinates in space and $s$ the length along the curve. The particles velocity is then given by $$\tag{1} v^i=\frac{dx^i}{dt}=\frac{ds}{dt}\frac{dx^i}{ds}=v~e^i,$$ where $v=ds/dt$ is the particle's speed and $e^i=dx^i/ds$ is a unit vector tangent to the curve. Then she wrote that the particle's acceleration would be given by: $$\tag{2} a^i=\frac{dv^i}{dt}=\frac{ds}{dt}\frac{d(v~e^i)}{ds}=v^2 \left(\frac{de^i}{ds}\right)+e^iv\frac{dv}{ds}.$$

Then I started wandering, what is the relation between this formula for the acceleration and then one using the covariant derivative: $$\tag{3} a^i=v^j~\nabla_j~v^i=v^j\left(\partial_j v^i+\Gamma^i_{k j}v^k \right)=\dot{v}^i+\Gamma^i_{k j}~v^k v^j~,$$ where $\Gamma^i_{k j}$ are the Chritoffel symbols of second kind? I know I should have asked her right away but she left pretty quickly from class. Can anyone help me? I can't seem to prove that the formulas are equivalent.

• I just have a question, how do you define the operator $\nabla _j$? – Ana S. H. Oct 19 '13 at 4:52
• @Anuar Oh sorry, I should have written it in the question. I define it as: $\nabla_i u^j=\partial_i u^j+\Gamma^j_{k i}u^k$. – PML Oct 19 '13 at 8:46

You're having trouble because the expressions aren't the same. On a general curved space (manifold), the definition of acceleration is the one you wrote down using the covariant derivative. Namely, for a given parameterization $x^i = x^i(t)$ of the path we define the velocity as $v^i = \dot x^i$ and the acceleration as \begin{align} a^i = \dot v^i + \Gamma^i_{jk}v^jv^k \end{align} where the overdot denotes differentiation with respect to the argument of a given function. Notice, in particular, that the acceleration is not in general equal to $\dot v$ as you indicate in your second string of equations. This is only true if \begin{align} \Gamma^i_{jk}v^iv^k = 0. \end{align} This would, for example, be the case for cartesian coordinates on $\mathbb R^3$ because then, the Christoffel symbols would all vanish.
• But why isn't Eq.$(2)$ equivalent to Eq.$(3)$? From what I was taught, Eq.$(3)$ is the most general but I don't see anything wrong with Eq.$(2)$. – PML Oct 19 '13 at 21:56
• @PML eq. ($2$) says $a^i = \dot v^i$ while equation ($3$) says $q^i = \dot v^i + \Gamma^i_{jk}v^jv^k$. They are the same only if $\Gamma^i_{jk}v^jv^k = 0$ as I remark in the answer. Think, for example, of polar coordinates on the plane. In these coordinates, one has $a_r = \ddot r - r\dot\phi^2\neq \ddot r$. – joshphysics Oct 19 '13 at 22:21
• @PML by $q^i$ I meant $a^i$. – joshphysics Oct 19 '13 at 22:31
• I'm feeling stupid. If I write the metric of $\mathbb{R}^2$ in polar coordinates as: $ds^2=dr^2 +r^2 d\theta ^2$ I find the same expression for $a^r$ as you, using Eq.$(3)$. But using Eq.$(2)$ isn't it the same as, for example, what I did in this answer? – PML Oct 19 '13 at 22:44
• @PML No need for the disclaimer. The equation $\vec a =d^2\vec r/dt^2$ in your other answer does not imply that $a^i = d^2 r^i/dt^2$ in any coordinate system. That vector equation assumes cartesian coordinates. But why do you get the correct polar coordinates result starting from the cartesian equation in your other answer? You're performing a coordinate transformation on an equation that holds in cartesian coordinates to obtain the equation in polar coordinates. The Christoffel symbols allow you to skip the step of transforming provided you know the metric in the desired coordinates. – joshphysics Oct 20 '13 at 1:02