Direction of ##d/dx##

[laser1]

Homework Statement

N/A

Relevant Equations

N/A

I read that $d/dx$ is the direction that x increases (paraphrased from Griffiths e-mag). Why is this? I can't find any information online when I search this up.

Original text:

 

[PeroK]

That's essentially the definition of a unit vector associated with a coordinate. For curvilinear coordinates, of course, it's only exact in the differential limit. A small increase in $\theta$, say, with no change in $r$ or $\phi$ means a small change in the $\hat \theta$ direction.

 

[laser1]

That's essentially the definition of a unit vector associated with a coordinate. For curvilinear coordinates, of course, it's only exact in the differential limit. A small increase in $\theta$, say, with no change in $r$ or $\phi$ means a small change in the $\hat \theta$ direction.

How do we know that it is the direction that x increases, as opposed to x decreasing?

 

[PeroK]

How do we know that it is the direction that x increases, as opposed to x decreasing?

By definition.

 

[Mark44]

I read that d/dx is the direction that x increases (paraphrased from Griffiths e-mag).

This doesn't make sense to me. d/dx is an operator, indicating that a derivative or rate of change of something is to be calculated with respect to changes in x. Everything in the text that you copied has to do with vectors and unit vectors.

 

[laser1]

A small increase in θ, say, with no change in r or ϕ means a small change in the θ^ direction.

I am confused on what the definition is.

Let's say a small increase in $\theta$ means a small positive change in $\hat{\theta}$ by definition. But $\hat{\theta}$ is made up of $\partial / \partial \theta$ etc.

 

[Orodruin]

This doesn't make sense to me. d/dx is an operator, indicating that a derivative or rate of change of something is to be calculated with respect to changes in x. Everything in the text that you copied has to do with vectors and unit vectors.

That is what the tangent vector space is. A vector space of differential operators.

This is usually not considered until one starts learning differential geometry though, so I’m a bit confused that it would appear in Griffith’s EM.

Alternatively, in Euclidean space the curvilinear base is defined by using partial derivatives of the position vector wrt the coordinates. This just tells us the basis vectors are the tangents to the coordinate lines.

 

[Mark44]

That is what the tangent vector space is. A vector space of differential operators.

No problem with that. What I was disputing was the statement that d/dx and others are directions rather than operators that are applied to vectors.

 

[laser1]

(From Griffiths).

What is the vector? I take it the vector is $\frac{\partial \bf{r}}{\partial r}$. Also isn't it $$d\mathbf{r} = \frac{\partial \mathbf{r}}{\partial r} dr + \frac{\partial \mathbf{r}}{\partial \theta} d\theta + \frac{\partial \mathbf{r}}{\partial \phi} d\phi$$

 

[PeroK]

View attachment 354797
(From Griffiths).

What is the vector? I take it the vector is $\frac{\partial \bf{r}}{\partial r}$. Also isn't it $$d\mathbf{r} = \frac{\partial \mathbf{r}}{\partial r} dr + \frac{\partial \mathbf{r}}{\partial \theta} d\theta + \frac{\partial \mathbf{r}}{\partial \phi} d\phi$$

Why does that not make sense? The only confusion is that $\mathbf r$ is not to be confused with the coordinate $r$. In any case:
$$\frac{\partial \bf{r}}{\partial r} = \sin \theta \cos \phi \bf{\hat x} + \sin \theta \sin \phi \bf{\hat y} + \cos \theta \bf{\hat z}$$Which is how we would define $\bf{\hat r}$ geometrically as the increasing radial direction. Likewise:
$$\frac{\partial \bf{r}}{\partial \theta} = r(\cos \theta \cos \phi \bf{\hat x} + \cos \theta \sin \phi \bf{\hat y} - \sin \theta \bf{\hat z})$$Which, you can check is the direction of increasing $\theta$.

If you are still stuck with this, you'd be better to look at plane polar coordinates first.

 

[Orodruin]

No problem with that. What I was disputing was the statement that d/dx and others are directions rather than operators that are applied to vectors.

That’s just the point. These differential operators indeed represent directions as the tangent vector of an equivalence class of curves.

 

[laser1]

$$\frac{\partial \bf{r}}{\partial \theta} = r(\cos \theta \cos \phi \bf{\hat x} + \cos \theta \sin \phi \bf{\hat y} - \sin \theta \bf{\hat z})$$Which, you can check is the direction of increasing $\theta$.

How can I check that it is the direction of increasing theta?

 

[PeroK]

How can I check that it is the direction of increasing theta?

You need to be able to visualise things in 3D! And imagine your orthonormal basis of $\bf{\hat r, \hat \theta, \hat \phi}$ at an arbitrary point on a sphere.

 

[laser1]

You need to be able to visualise things in 3D! And imagine your orthonormal basis of $\bf{\hat r, \hat \theta, \hat \phi}$ at an arbitrary point on a sphere.

Ok, what about if vector r is <x, y, z>. So $\partial \mathbf{r} / \partial x = \hat{i}$ which is in direction of increasing x! Does the same logic apply to spherical and polar?

 

[PeroK]

Ok, what about if vector r is <x, y, z>. So $\partial \mathbf{r} / \partial x = \hat{i}$ which is in direction of increasing x! Does the same logic apply to spherical and polar?

Yes.

 

[laser1]

Yes.

alright so I'd say it's just my lack of imagination. I'll try again tomorrow, thanks

 

[laser1]

If you are still stuck with this, you'd be better to look at plane polar coordinates first.

Ok, I tried checking this and I have kind of satisfied myself. Taking the partial derivative with respect to e.g. $\theta$ does point in the direction of increasing $\theta$, i.e. anticlockwise. What I now want to understand is why does taking the derivative do this? I understand that, with a partial, you are only changing one thing, but why does taking the derivative point in the increasing variable direction as opposed to decreasing?

Earlier you answered "by definition", but I am not sure what you meant exactly.

 

[vela]

I understand that, with a partial, you are only changing one thing, but why does taking the derivative point in the increasing variable direction as opposed to decreasing?

Go back to the limit definition of the derivative and consider the difference quotient. In what direction does the numerator point when you increase $\theta$ by a small amount?

 

[Orodruin]

The definition of the partial derivative (of anything , call it $f$) wrt $\theta$ is
$$
\frac{\partial f}{\partial \theta} = \lim_{h\to 0}\left(
\frac{f(r,\theta+h,\varphi) -f(r,\theta,\varphi)}{h}
\right)
$$
In particular, if you let $f$ be the position vector then the numerator is the difference of the position vectors given by coordinates $(r,\theta+h,\varphi)$ and $(r,\theta,\varphi)$. For positive $h$ this therefore points from $\theta$ to $\theta+h > \theta$, ie, towards larger values of $\theta$. For negative $h$ the numerator points towards smaller $\theta$, but the denominator is also negative.

 

[laser1]

Ah cheers, in my head that was the definition of the "right-sided limit", but I guess not lol

 

[laser1]

Ah cheers, in my head that was the definition of the "right-sided limit", but I guess not lol

actually it doesn't even matter because h can be negative like you said.

 

[PeroK]

Ok, I tried checking this and I have kind of satisfied myself. Taking the partial derivative with respect to e.g. $\theta$ does point in the direction of increasing $\theta$, i.e. anticlockwise. What I now want to understand is why does taking the derivative do this? I understand that, with a partial, you are only changing one thing, but why does taking the derivative point in the increasing variable direction as opposed to decreasing?

Earlier you answered "by definition", but I am not sure what you meant exactly.

What we have, in general, is the definition of a unit vector based on the position vector $\vec r$. Let's use two general, orthogonal curvilinear coordinate $u, v$. We define the unit vector by:
$$\hat u = \frac 1 {h_u} \frac{\partial \vec r}{\partial u}$$Where $h_u = \big |\frac{\partial \vec r}{\partial u}\big |$. Using this definition, we can express the gradient of a function in these coordinates:
$$\vec \nabla f = \frac 1 {h_u}\frac{\partial f}{\partial u}\hat u + \frac 1 {h_v}\frac{\partial f}{\partial v}\hat v$$In particular, we can take the gradient of the coordinate $u$:
$$\vec \nabla u = \frac 1 {h_u}\hat u = \frac 1 {h_u^2} \frac{\partial \vec r}{\partial u}$$The gradient is the direction of greatest increase of a function, hence the greatest increase in $u$ is in the direction of the unit vector so defined.

This won't apply in non-orthogonal (skew) curvilinear coordinates.

 

[Orodruin]

This won't apply in non-orthogonal (skew) curvilinear coordinates.

Generally, that is why we have the metric tensor. No need to normalise the basis vectors and a formalism that covers both orthogonal and skew coordinates as well as generalises easily to curved manifolds.

Now if only I could find a good textbook that introduces all of that on a basic level …

 

[Orodruin]

actually it doesn't even matter because h can be negative like you said.

Indeed, the right-side limit would be denoted $\lim_{h\to 0^+}$