Direction of ##d/dx##

[laser1]

Homework Statement

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Relevant Equations

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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