# Finding a field line of a vector field

1. Aug 5, 2015

### S. Moger

1. The problem statement, all variables and given/known data

Find the field line of

$$\vec{E}(\vec{r}) = \frac{m}{4 \pi r^3} (2 \cos\theta, \sin\theta, 0)$$

through the point (a, b, c)

(Spherical coordinates).

m is a constant

I know the answer, but I don't see what I do wrong.

3. The attempt at a solution

$$\frac{d\vec{r}}{d \tau} = C \cdot \vec{E}(\vec{r}(\tau))$$

$$\frac{dr}{d \tau} = C \cdot \frac{m}{4 \pi r^3} 2 \cos\theta$$
$$\frac{d\theta}{d \tau} = C \cdot \frac{m}{4 \pi r^3} \sin\theta$$
$$\frac{d\phi}{d \tau} = 0$$

By setting $C = \frac{4 \pi}{m}$ I get

$$\frac{dr}{d \tau} = \frac{ 2 \cos\theta}{r^3}$$
$$\frac{d\theta}{d \tau} = \frac{\sin\theta}{r^3}$$
$$\frac{d\phi}{d \tau} = 0$$

To get rid of $r^3$ I divide $\frac{d\theta}{d \tau}$ by $\frac{dr}{d \tau}$ (must not be zero and so on) and get

$$2 \tan^{-1}\theta \frac{d\theta}{d \tau} = \frac{dr}{d \tau}$$

Then I multiply both sides with $d\tau$ (which is a somewhat mysterious operation to me).

After integration I obtain $2 \log (\sin\theta) = r + const$. I could of course determine the const and so on, but this isn't the answer anyway.

Where's the error and why?

Scale factors? But if so, why?

2. Aug 5, 2015

### DEvens

$\text{tan}^{-1} \theta$ is the inverse tan function. That is, it is the angle that has tan equal to $\theta$.

I think you want $\text{tan} \theta \frac{dr}{d\tau} = 2 \frac{d\theta}{d\tau}$ meaning $\frac{dr}{d\theta} = \frac{1}{2} \text{tan} \theta$

3. Aug 5, 2015

### Zondrina

EDIT: See my post below.

Last edited: Aug 5, 2015
4. Aug 5, 2015

### Zondrina

After staring at this for a bit, I think the whole problem statement is wrong. Could you please provide the original problem statement as it was given with no modifications?

I say this because you should have started with something like $\vec E(x, y, z, t) = E_x \space \hat i + E_y \space \hat j + E_z \space \hat k$.

I'm guessing this would reduce to $\vec E = E_x \space \hat i + E_y \space \hat j$ due to the zero component.

Then you would simply solve $\frac{dy}{dx} = \frac{E_y}{E_x}$.

5. Aug 5, 2015

### S. Moger

Ok, the notation might not be the best, $\tan^{-1} \theta$ above is meant to be $1 / \tan\theta$

The problem statement is unmodified, with the exception of the coordinates of interest that I generalized. It's given in spherical coordinates (or practically in polar as you said)

I get the right answer if I use the scale factor $$r$$ (used in your expression):

$$\frac{dr}{d \theta} = r \frac{E_r}{E_{\theta}}$$

So I think I made my error here

$$\frac{d\vec{r}}{d \tau} = C \cdot \vec{E}(\vec{r}(\tau))$$

which seems to hold for cartesian coordinates only.

To generalize I seem to require something like this, but I can't convince myself of it.

$$\frac{d\vec{r}}{d \tau} = C \cdot \frac{1}{h_i} \vec{E}(\vec{r}(\tau))$$

With
$$h_{\theta} = | \frac{ \partial{\vec{r}} }{ \partial{\theta} }| = r$$

Last edited: Aug 5, 2015
6. Aug 5, 2015

### gleem

Or is the field expressed in spherical coordinates?

7. Aug 5, 2015

### S. Moger

Yes my bad if it's unclear, everything is in spherical coordinates.

theta would be the angle measured from the z axis

Ok, in a general case a change in cartesian coordinates can be described like this with other (normalized, orthogonal) base vectors $\vec{e_i}$

$$d\vec{r} = \sum_{i=1}^{3} h_i \vec{e_i} du_i$$

So I guess

$$\frac{d\vec{u}}{d \tau} = C \cdot \frac{1}{h_i} \vec{E}(\vec{u}(\tau))$$

could make sense.

Last edited: Aug 5, 2015