Gradient of spherical co-ords/ differentiation help

I think the problem lies with what [itex]\theta[/itex] and [itex]\phi[/itex] are. In physics the azimuth and zenith are often reversed. So what convention does your book use? In this case differentiation with respect to [itex]\phi[/itex] does not yield 0 where as differentiation with respect to [itex]\theta[/itex] does yield 0. So I am pretty sure you're using the wrong convention.

Secondly why do you even bother to transform that function into spherical coordinates?

 

There are two different conventions.

1) The mathematics convention where the azimuth, the angle between the x-axis and the radius in the x-y plane, is called [itex]\theta[/itex] and the zenith, the angle between the z axis and the radius is called [itex]\phi[/itex]

2)The physics convention where the azimuth, the angle between the x-axis and the radius in the x-y plane, is called [itex]\phi[/itex] and the zenith, the angle between the z axis and the radius is called [itex]\theta[/itex]

Read this carefully, draw a picture perhaps so that you fully understand the difference.

The definition in your book has [itex]\theta[/itex] as the zenith. Where as seeing as maple gives 0, has [itex]\phi[/itex] as the zenith. Do you understand why maple gets zero and you get a non-zero value now?

Write down the definition of the gradient in spherical coordinates (on the forum) and label each term with, radius, zenith, azimuth. Since you want to find the [itex]\phi[/itex] term you will need to use which term in the definition of the gradient given the convention used in the book? Since maple uses the other convention which term did maple use?

And no you can't just apply spherical coordinates straight after taking the Cartesian gradient.

 

Even though this may be for maths the convention your book uses is the convention I labeled the physics convention. It doesn't really matter though feel free to call it anything you want as long as you always pay attention to how [itex]\phi[/itex] and [itex]\theta[/itex] are defined. This way you can use the appropriate convention.

So the gradient your book gives you is:

[tex]\nabla f=\frac{\partial f}{\partial r}e_r+\frac{1}{r} \frac{\partial f}{\partial \theta}e_\theta+\frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi}e_\phi[/tex]

maple:
[tex]\nabla f=\frac{\partial f}{\partial r}e_r+\frac{1}{r} \frac{\partial f}{\partial \phi}e_\phi+\frac{1}{r \sin \phi} \frac{\partial f}{\partial \theta}e_\theta[/tex]

So if you want to know the [itex]\phi[/itex] component then you have to use the the [itex]\theta[/itex] component in maple while also switching all [itex]\theta s[/itex] and [itex]\phi s[/itex] in the definitions of x,y,z in spherical coordinates. Nice and confusing isn't it?

Just calculate all three terms and I will check if you got the correct answer or not.

I would really suggest you do it by hand instead of using some function in a program you don't now its definition of.

 

I made an error in the last sentence of my previous post. I will fix it. If you include the exponent your answer is the correct one. You can simplify it a bit if you like.

 

In the third one the cosine should read [itex]\cos \phi[/itex]. The others seem correct. Try to simplify the second one by using [itex]\sin 2x=2 \sin x \cos x[/itex] and [itex]\sin^2x+\cos^2x=1[/itex].