From complex analysis I know that we define the complex line integral as follows: $$\int_\gamma f(z)dz = \int_a^bf(\gamma(t))\cdot\gamma'(t)dt \tag 1$$ Assuming that $\gamma$ is continuously differentiable with endpoints $a$ and $b$. Back then I did not give this too much thought, as it seems like we are simply substituting $\gamma(t)$ for $z$.
Now we are defining the line integral over a real valued scalar field $\Omega$ as follows:$$\int_C \Omega(r) = \int_a^b\Omega(r(s))ds$$ Assuming that $r(s) = (x(s), y(s), z(s))$ is a curve parametrized by arc length. To generalise this definition for curves not parametrized by arc length we use the formula for the arc length $s$:$$s = \int \lvert r'(t)\rvert dt \implies \frac{ds}{dt}=\lvert r'(t)\rvert$$ $$\therefore \quad \int_C \Omega(r) = \int_a^b\Omega(r(s))ds \stackrel{?}{=} \int_a^b\Omega(r(t))\frac{ds}{dt}dt = \int_a^b\Omega(r(t))\lvert r'(t)\rvert dt \tag 2$$
My two questions are:
- Apart from the modulus, $(1)$ and $(2)$ look very similar. Is there any intuitive connection between the the complex and real line integral? Or is it just a coincidence. After all, the multiplication $\cdot$ in $(1)$ is the complex multiplication, so these definitions may be not as similar as I think.
- What exactly is happening at the penultimate step in $(2)$? My chain of equal signs seems very handwavy, so I am assuming there is some rigour missing in my explanation.
Best Answer
The objects $\int_\gamma f(z)\,dz$, where $f\in\mathbb{C}$ and $z\in\mathbb{C}$, and $\int_\gamma \Omega(\vec r)\,|d\vec r|$, where $\Omega(\vec r)\in\mathbb{R}$ and $\vec r\in \mathbb{R}^2$ are quite different.
In the complex plane, the arc length $s$ of the curve $\gamma$ parameterized by $z(t)$, $t\in [a,b]$ is witten
$$s=\int_\gamma |dz|=\int_a^b \left|\frac{dz}{dt}\right|\,dt$$
Therefore, one should compare the real part or imaginary part of $\int_\gamma f(z)\,|dz|$ with $\int_\gamma \Omega(\vec r(t))\,|\vec r'(t)|\,dt$.
Let me know if this helps.