Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was looking at the color wheel graph of $e^{1/z}$, and my girlfriend commented that it looked just like a dipole. Does anyone have an explanation for that, why the geometry would be so similar? I guess as we follow e.g. the red color from the left side as it goes up, and then back down to the right side, we are tracing a ray outward from the origin in $\mathbb{C}$. enter image description here enter image description here

share|improve this question
1  
+1 to question and answer -- Stunningly beautiful picture and answer. –  Domi Dec 7 '13 at 10:18
add comment

3 Answers

up vote 3 down vote accepted

This answer is in response to Marvis's (a.k.a. "Please delete account") answer, which suggests that the dipole-like behavior of $e^{1/z}$ and $e^{1/z^2}$ is due to resemblance to $1/z^2$, but looking at the color graph of $1/z^2$ itself suggests that this is not the case.

The gradient field of a dipole is $$E=\left(\frac{x^2-y^2}{(x^2+y^2)^2},\frac{2xy}{(x^2+y^2)^2}\right)=\frac1{z^2}$$ (expressing the two real variables as real and imaginary part of a single complex variable). We are interested in a function which increases in magnitude along the field lines, i.e. $f'(z)/f(z)$ is proportional to the gradient field. (Note that the original observation was that the color changes as you pass from one field line to another, and the color bands follow the field lines. The only way for an analytic function to do this is for the magnitude, but not the phase, to change as you follow the field lines.) Thus, we have the differential equation:

$$\frac{f'(z)}{f(z)}=\frac d{dx}\log f(x)=\frac c{z^2}\Rightarrow\log f(x)=a-\frac cz\Rightarrow f(x)=Ae^{-c/z},$$

and the original post had $A=-c=1$. Thus we see that it is not just any function, but precisely the function $e^{1/z}$ that looks like a dipole in the color plot. (Indeed, $e^{1/z^2}$ looks much more like a quadrupole, for the same reason.)

share|improve this answer
add comment

Try $e^{1/z^2}$, you will get an even better match. This is due to the following:

The potential due to the dipole you have in the figure is given by $\phi(x,y) = -\dfrac{x}{x^2+y^2}$. Hence, the electric field is given as $$\vec{E} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, \dfrac{2xy}{(x^2+y^2)^2}\right) \tag{$\star$}$$ If we expand $e^{1/z}-1$, we get $$\dfrac{\bar{z}}{r^2} + \dfrac{\bar{z}^2}{2r^4} + \mathcal{O}(1/r^3)$$ If we look at the first two terms, we get $$\dfrac1z + \dfrac1{2z^2} = \dfrac{x-iy}{x^2+y^2} + \dfrac{(x-iy)^2}{2(x^2+y^2)^2}$$ which gives us $$\left(\dfrac{x^2-y^2+2x(x^2+y^2)}{2(x^2+y^2)^2},- \dfrac{xy+y(x^2+y^2)}{(x^2+y^2)^2}\right)$$ If you look at $e^{1/z^2}$, and truncate with the term $\dfrac1{z^2}$, you will get the exact thing as $\star$. We have $$e^{1/z^2}-1 = \dfrac1{z^2} + \mathcal{O}(1/z^4)$$ And $$\dfrac1{z^2} = \dfrac{\bar{z}^2}{r^4} = \dfrac{(x^2-y^2) -2ixy}{(x^2+y^2)^2} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, -\dfrac{2xy}{(x^2+y^2)^2}\right)$$ which agrees with $\star$ but for the sign.

share|improve this answer
    
This answer is incorrect; see my post. If this were true, $e^{1/z^2}$ and $1/z^2$ would look like dipoles, and they don't. Also, why is this looking at $e^{1/z}-1$, which looks quite different from $e^{1/z}$? –  Mario Carneiro Dec 11 '13 at 7:15
add comment

Here is the color wheel chart for the exponential function from Wikipedia. Notice that the horizontal lines have "constant color", Each of these lines can be written as $t + bi$, where $b$ is fixed and $t$ goes from $-\infty$ to $+\infty$. You notice that the constant color lines for $e^\frac{1}{z}$ are little arcs. This is just a consequence of the fact that the map $z \mapsto \frac{1}{z}$ twists the horizontal lines into the arcs. You can even see in the picture how the "left side" of the plane is mapped to the left size of 0, and the "right size" of the plane is mapped to the right side of zero. The point at $\infty$ is mapped to 0. So you should really think of this as being a picture of what $\frac{1}{z}$ is doing.

share|improve this answer
1  
most probably it is just a co-incidence that it looks like the magnetic field produced by the dipole . Anyways the dipole doesn't produce magnetic fields that are perfect circles . –  Zoro Dec 7 '13 at 6:17
    
Yep. The two just happen to look fairly similar. The answer was just trying to give a little insight to what made the little circles appear in the color wheel graph. Hopefully it's a somewhat helpful to the OP. –  Zach L. Dec 7 '13 at 6:21
    
Actually I loved the insight you gave on this answer . But still I find it amazing that how the magnetic field produced by a dipole can be approximated by such an elegant function ( in terms of shape ) . –  Zoro Dec 7 '13 at 6:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.