Can we differentiate with respect to a vector?

[KnightTheConqueror]

We know that a = vdv/dx
But is it applicable in only one dimensional components or is this actually a vector equation? If so then how do we exactly differentiate with respect to position 'vector'?

 

[KnightTheConqueror]

Thanks for answering, but I know that I can resolve it into components and solve. My problem is not that I am unable to use that equation in 3D. Rather, I am just curious whether we can write that equation as an vector equation directly without resolving.

Let me elaborate...

v = u + at here, although we can apply this equairon to x, y, z component separately and then add them but resolving into components is just something to make our work easier. v = u + at is still a vector equation and we can solve it using other methods of vector too for example geometry.
Likewise, I can apply a = vdv/dx to x, or vdv/dy to y
But
Can we just write that equation in vector form without needing to resolve into one dimensional components? Can we say that acceleration vector = velocity vector times d of velocity vector upon x of position vector.
If so then what exactly do we mean by differentiating with respect to a vector?

 

[TSny]

We know that a = vdv/dx
But is it applicable in only one dimensional components or is this actually a vector equation? If so then how do we exactly differentiate with respect to position 'vector'?

Let $s$ denote arc length along the trajectory of the particle. Consider the velocity of the particle as a function of arc length: $\vec v(s)$. Use the chain rule to express the acceleration as $\vec a = \dfrac{d\vec v}{ds} \dfrac{ds}{dt}$ and note that $\dfrac{ds}{dt}$ is the speed $v$ of the particle. So, $\vec a = v\dfrac{d\vec v}{ds}$.
(This expression is undefined at points of the trajectory where $v = 0$, just as the 1-D expression $a = v \dfrac{dv}{dx}$ is undefined when $v = 0$.)

 

[Orodruin]

What you are looking for is the gradient. Your derivatives wrt the spatial position need to be partial derivatives.
$$
\vec a = \frac{d\vec v}{dt} = \frac{d\vec x}{dt} \cdot\nabla\vec v = \vec v \cdot \nabla \vec v$$
Note that $\nabla \vec v$ is a rank 2 tensor.

 

[KnightTheConqueror]

Let $s$ denote arc length along the trajectory of the particle. Consider the velocity of the particle as a function of arc length: $\vec v(s)$. Use the chain rule to express the acceleration as $\vec a = \dfrac{d\vec v}{ds} \dfrac{ds}{dt}$ and note that $\dfrac{ds}{dt}$ is the speed $v$ of the particle. So, $\vec a = v\dfrac{d\vec v}{ds}$.
(This expression is undefined at points of the trajectory where $v = 0$, just as the 1-D expression $a = v \dfrac{dv}{dx}$ is undefined when $v = 0$.)

This is what I was looking for. Thank you very much!

 

[KnightTheConqueror]

What you are looking for is the gradient. Your derivatives wrt the spatial position need to be partial derivatives.
$$
\vec a = \frac{d\vec v}{dt} = \frac{d\vec x}{dt} \cdot\nabla\vec v = \vec v \cdot \nabla \vec v$$
Note that $\nabla \vec v$ is a rank 2 tensor.

I am still in high school and I haven't studied tensors... Can you suggest me some resource to get an idea of it?

 

[Orodruin]

(This expression is undefined at points of the trajectory where $v = 0$, just as the 1-D expression $a = v \dfrac{dv}{dx}$ is undefined when $v = 0$.)

This is not really correct. If the velocity field is such that it depends on position only (which needs to be the case for the derivative $\partial \vec v/\partial x^i$ to make sense), then the expression in #4 makes sense even if $\vec v = 0$. It is when you require the path length $s$ that it becomes a problem. It is not necessarily a problem of the one-dimensional case if it can be seen as the restriction of #4 to one dimension. It is just that if $\vec v = 0$, then in that case $\vec a = 0$ and consequently $\vec v$ is constantly zero. So there are certainly cases where the expression

 

[Orodruin]

I am still in high school and I haven't studied tensors... Can you suggest me some resource to get an idea of it?

If you are familiar with matrices, you may for practical purposes in Cartesian coordinates consider a rank 2 tensor a 3x3 matrix
$$
\begin{pmatrix}\partial v^x/\partial x & \partial v^y /\partial x & \partial v^z /\partial x \\
\partial v^x/\partial y & \partial v^y /\partial y & \partial v^z /\partial y \\
\partial v^x/\partial z & \partial v^y /\partial z & \partial v^z /\partial z \end{pmatrix}
$$
Multipyling with $\begin{pmatrix}v^x & v^y & v^z\end{pmatrix}$ from the left gives
$$
\begin{pmatrix}
v^x (\partial v^x/\partial x) +
v^y (\partial v^x/\partial y) +
v^z (\partial v^x/\partial z) &
v^x (\partial v^y/\partial x) +
v^y (\partial v^y/\partial y) +
v^z (\partial v^y/\partial z) &
v^x (\partial v^z/\partial x) +
v^y (\partial v^z/\partial y) +
v^z (\partial v^z/\partial z)
\end{pmatrix}
$$

 

[TSny]

This is not really correct.

If we try to evaluate $\vec a(s) = v(s) \dfrac{d\vec v(s)}{ds}$ at a point $s_0$ where $v(s_0) = 0$, then we would find $\left. \dfrac{d\vec v(s)}{ds} \right|_{s_0}$ is undefined. However, the acceleration at $s_0$ can be evaluated as a limit: $$\vec a(s_0) = \lim_{s \to s_0} \left[ v(s) \dfrac{d\vec v(s)}{ds} \right].$$

What you are looking for is the gradient. Your derivatives wrt the spatial position need to be partial derivatives.
$$
\vec a = \frac{d\vec v}{dt} = \frac{d\vec x}{dt} \cdot\nabla\vec v = \vec v \cdot \nabla \vec v$$
Note that $\nabla \vec v$ is a rank 2 tensor.

I don't understand how you are working with a velocity vector field to give meaning to the tensor $\nabla \vec v$.

For example, how would you evaluate the tensor component $\dfrac{\partial v^y}{\partial x}$ for a particle in projectile motion?

 

[Orodruin]

If we try to evaluate $\vec a(s) = v(s) \dfrac{d\vec v(s)}{ds}$ at a point $s_0$ where $v(s_0) = 0$, then we would find $\left. \dfrac{d\vec v(s)}{ds} \right|_{s_0}$ is undefined. However, the acceleration at $s_0$ can be evaluated as a limit: $$\vec a(s_0) = \lim_{s \to s_0} \left[ v(s) \dfrac{d\vec v(s)}{ds} \right].$$

With path length, no. With respect to a coordinate in a velocity field, yes.

I don't understand how you are working with a velocity vector field to give meaning to the tensor $\nabla \vec v$.

For example, how would you evaluate the tensor component $\dfrac{\partial v^y}{\partial x}$ for a particle in projectile motion?

A single particle in projectile motion is not in a velocity field. There are however many situations where velocity fields make perfect sense, such as for a fluid parcel. The above would apply to a fluid parcel in a stationary flow (for a non-stationary flow there would be additional terms due to the time derivative of the flow).

 

[TSny]

With path length, no. With respect to a coordinate in a velocity field, yes.

I’m not sure what the “no” and “yes” mean here.

A single particle in projectile motion is not in a velocity field. There are however many situations where velocity fields make perfect sense, such as for a fluid parcel. The above would apply to a fluid parcel in a stationary flow (for a non-stationary flow there would be additional terms due to the time derivative of the flow).

That makes sense. Thanks.