dropping a particle into a vector field

Question

I'm independently studying Colley's Vector Calculus and am on the section on line integrals. I understand that the line integral gives the amount of work done on a vector field for a predetermined path through the field, much like me walking across a river to the other side in a straight line, how much work do I need to do against the current. But what about a particle dropped into a vector field? I think this would be easy to compute if say the field were given by $$\mathbf{F}(x,y)=\mathbf{i}$$ But for something more dynamic like $$\mathbf{F}(x,y)=y\mathbf{i}-x\mathbf{j}$$ How does one go about finding the path $\mathbf{r}(t)$ that the particle traces out after say $t=n$ seconds? We can get the velocity vector and the acceleration vector, but would this be a prolbem for systems of differential equations and gradient? Intuitively it seems like the particle will follow the path of least resistance so perhaps the $-\nabla$ operator would help take care of that?

Answer 1

Indeed, to find such $\mathbf{r}$ you have to solve the system of differential equations $\mathbf{r}'(t) = \mathbf{F}(\mathbf{r}(t))$. In general this can be hard to do in closed form, but there are many numerical methods. Euler's method is perhaps the simplest.

Answer 2

I'm still in high school so my answer is a little "sus" (so don't destroy me if some of it is not mathematically rigorous) but for a vector field of the form (,), one way is as following:

The path taken by a particle "dropped into a vector field" is intuitively going to be the path that maximises the work done by the field on the particle. A neat way to determine this is by considering a "perpendicular vector field". What I mean by this is a vector field that is, at all points, perpendicular to (,). Let's call this perpendicular vector field, P(,) $$ F(x,y)\bullet P(x,y)=0 $$ If F(x,y) = $<f_1(x,y),f_2(x,y)>$, then a suitable P(x,y) is $<-f_2(x,y),f_1(x,y)>$. (There is an infinite amount of suitable P(x,y), this one however is just an elegant one to use). Play around with a vector field grapher if need be, to convince yourself that a path taken in F(x,y) of maximal work done by the field is a path of minimal work done by the field in P(x,y).

So to find the path r(t) taken by a particle dropped in a field F(x,y), r(t) must satisfy the following equation: $$\int_cP(x,y)\bullet dr=0$$ $$\int_{t_0}^{t_1}P(x,y)\bullet \frac{dr}{dt}dt = 0$$ This holds when: $$P(x,y)\bullet \frac{dr}{dt}=0$$ $$<-f_2(x,y),f_1(x,y)> \bullet <\frac{dx}{dt},\frac{dy}{dt}>=0$$ Hence the path r(t) taken by the particle must satisfy the following differential equation: $$-f_2(x,y)\frac{dx}{dt}+f_1(x,y)\frac{dy}{dt}=0$$ Interestingly if we consider the curve traced by the particle in the vector field, it yields a nice result.

$$-f_2(x(t),y(t))\frac{dx}{dt}+f_1(x(t),y(t))\frac{dy}{dt} \Rightarrow \frac{dy}{dt} = \frac{f_2}{f_1}\frac{dx}{dt}$$

Using the chain rule, $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$, $$\frac{dy}{dx}\frac{dx}{dt} = \frac{f_2}{f_1}\frac{dx}{dt}$$ $$\frac{dy}{dx}=\frac{f_2(x,y)}{f_1(x,y)}$$ An Example: If F(x,y)= <-y,x> (This vector field is "circular" in nature), Find the (general solution) curve traced by a particle dropped in the vector field. $$\frac{dy}{dx}=-\frac{x}{y}$$ $$ydy=-xdx$$ $$\frac{y^{2}}{2}=-\frac{x^{2}}{2}+k$$ $$y^{2}=c-x^{2}$$ Thus: $$x^{2}+y^{2}=c$$ As expected, this is the cartesian equation of a circle.

Conclusion: Thus we have it, a particle dropped in a vector field F(x,y),will travel in a path r(t)=<x(t),y(t)> which satisfies:$$\frac{dy}{dt} = \frac{f_2}{f_1}\frac{dx}{dt}$$ and trace out a curve that satisfies the differential equation: $$\frac{dy}{dx}=\frac{f_2(x,y)}{f_1(x,y)}$$