# Difference between the two equations for acceleration

I came upon this while studying S.H.M.

Well,is there a difference between writing

$$a=\frac{dv}{dt}\;$$

and $$a=v\frac{dv}{dx}\;$$ do they differ on the basis of one being a vector and the other being a scalar equation?Please explain.

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$dv/dx$ is not a vector, it's a tensor of second order. When you multiply a vector, $v$, which is a first order tensor and the second order tensor, you get a first order tensor, which is a another vector – Pranav Hosangadi Jan 26 '14 at 6:17
Why do you say the second equation is a scalar equation? – John Rennie Jan 26 '14 at 8:02
Everything is formal without making explicit the physical context. Concerning the first equation you are dealing with a particle whose trajectory is given as a function of time $x=x(t)$, so $v=dx/dt$ and $a= d^2x/dt^2= dv/dt$. What about the second equation? It seems that you are given a function $v=v(x)$. What does this functional relation physically mean? – Valter Moretti Jan 27 '14 at 7:41
@johnrennie because my teacher told me. – soumyadeep Jan 29 '14 at 2:07

## 2 Answers

In the context of SHM they are probably meant to be interpreted as both scalar equations (ie $x$, $v$ and $a$ are all one-dimensional), especially since they use $x$ and not $r$, but they can also be written vector form. The first equation would be (using $r$ for the position vector)

$$\vec{a} = \frac{d\vec{r}}{dt}$$

no surprises here. The second would be

$$\vec{a} = \frac{d\vec{v}}{d\vec{r}} \vec{v}$$

this is strange - how can you differentiate a vector with respect to another vector? Well, let's say we're working in three dimensions. Then an expression like

$$\frac{d\vec{v}}{d\vec{r}}$$

would be something like, how does $\vec{v}$ change as I change $\vec{r}$? But you can change $\vec{r}$ by changing $x$, $y$ or $z$, then in each case there will be a change in $v_x$, $v_y$ and $v_z$, so you need 9 numbers to specify $\frac{d\vec{v}}{d\vec{r}}$. As @Pranav says, this is a second-order tensor; in cartesian coordinates this would be

$$\frac{d\vec{v}}{d\vec{r}} = \begin{bmatrix} \dfrac{\partial v_x}{\partial x} & \dfrac{\partial v_x}{\partial y} & \dfrac{\partial v_x}{\partial y}\\ \dfrac{\partial v_y}{\partial x} & \dfrac{\partial v_y}{\partial y} & \dfrac{\partial v_y}{\partial y}\\ \dfrac{\partial v_z}{\partial x} & \dfrac{\partial v_z}{\partial y} & \dfrac{\partial v_z}{\partial y}\\ \end{bmatrix}$$

so the equation does make sense in the end.

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We know that velocity is defined as $$v=\frac{dx}{dt}$$. Then using the chain rule for derivatives we can write acceleration as $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v.$$

For vectors lets take the index notation. Then $$v^i=\frac{dx^i}{dt}.$$ The expression for acceleration takes the following form $$a^i=\frac{dv^i}{dt}=\sum_{j=1}^{3}\frac{dv^i}{dx^j} \frac{dx^j}{dt}=\sum_{j=1}^{3}\frac{dv^i}{dx^j}v^j.$$

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Yes i know that.But what i am asking is whether there is a difference on the basis of one being a vector and the other being a scalar equation. – soumyadeep Jan 26 '14 at 6:05
They both are vector equations. In the second equation you do not do the scalar product, but just a summation of indices. – Edvard Jan 26 '14 at 6:07