1
$\begingroup$

I am new to stack exchange, the only other place I have asked questions on thus far is stackoverflow, and though technically I am a student of computer science (a late bloomer, started my degree two years ago when I was 26), my sole hobby for the past three to four years has been mathematics, pure mathematics. I'm sorry to say but I am horrible at applying it to real life, but boy do I love crunching the numbers. My knowledge ranges up to Vector Analysis, right now I am just completing the exercise section on the Fundamental Theorem of Line Inegrals (Larson's Essential Calculus ETF). My question is quite simple actually:

Is there a specific notation to denote the potential function $\mathcal f$?

It is known that for a conservative vector field: $$\overrightarrow A = \nabla\mathcal f = M\hat{i}+N\hat{k}$$ where $M$ and $N$ are continous functions of $x$ and $y$ and, $$\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}$$ There is also the case that for a vector field in space to be conservative, $\mathbf{curl \overrightarrow A} = \mathbf{\overrightarrow 0}$, but there exist potential functions for the plane and for space.

I'm sort of a freak when it comes to notation, so though it may be a dumb question for my first, I'd love know. Thank you in advance!

$\endgroup$
0
$\begingroup$

While we have all sorts of simple notation for derived functions or fields ($f'$, $\nabla f$, ${\rm curl}\,{\bf F}$, etc.) there is no such thing for antiderivative. Note that even in the one-dimensional case we resort to "let $F$ be a primitive of $f$", or need the clumsy $\int f(x)\>dx$ to denote the infinite set of all primitives. A reason for this lack might be that the potential of a conservative field ${\bf F}$ is defined only up to an additive constant, while the various derived objects are uniquely determined by ${\bf F}$.

$\endgroup$
  • $\begingroup$ I have noticed that there really isnt a specific notation for antiderivatives. $\int f(x)\>dx$ really is sort of clumsy, but I guess it will do. Though it becomes a bother when there are aleady a lot of calculations done. $\endgroup$ – SSBASE Feb 9 '18 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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