If I have a force $F(x)$, can I assume it to be constant in any infinitesimal interval such as $Rd\theta$,$ dy \over cos\phi$, $dz$ etc. or can I assume it to only be constant in the interval $[x,x+dx)$? If yes, can you prove it mathematically and if not, which infinitesimal intervals can I assume a force to be constant in & why?

  • $\begingroup$ related: impulses and delta functions $\endgroup$ – user140606 Feb 19 '17 at 16:53
  • 2
    $\begingroup$ I think you can only do that, in general, if the element is infinitesimal in all dimensions. So $dx\,dy\,dz$ is fine, but $R\,d\theta$ is not. $\endgroup$ – lemon Feb 19 '17 at 16:57
  • 1
    $\begingroup$ @lemon : That looks like an answer, not a comment. $\endgroup$ – sammy gerbil Feb 19 '17 at 17:42
  • $\begingroup$ @lemon I disagree. If the force is finite, the change across an infinitesimal interval will always be infinitesimal, so the force can be treated as a constant. The triple infinitesimal is just a much smaller infinitesimal $\endgroup$ – user126422 Feb 19 '17 at 19:55
  • $\begingroup$ @AlbertAspect Consider a large $R$ and the force $f(R)=1/R$. The force is far from constant across $R\,d\theta$... $\endgroup$ – lemon Feb 19 '17 at 20:08

What you want to do is a Taylor Expansion. Assuming that $F\in C^1(\mathbb{R}^3)$ you have:

$$ F(\mathbf{x})=F(\mathbf{x}_0)+\nabla F(\mathbf{x})|_{\mathbf{x}_0}\cdot(\mathbf{x}-\mathbf{x}_0) + \mathcal{o}(\mathbf{x}-\mathbf{x}_0)$$

So, you can approximate $F$ to be constant only if the first (and higher) order terms are neglectable for your purposes. Here I've showed you a general expansion, but you can write the gradient in any coordinate system (for example spherical) you wish. If it's legitimate to consider only the zeroth order of the expansion usually depends on the explicit form of $F(\mathbf{x})$ and the particular application in which you want to make this approximation.

Let's now consider a one-variable function $F:\mathbb{R}\to\mathbb{R}$, $F\in C^N(\mathbb{R})$. It's then possible to write it using the Taylor theorem:

$$ F(x)=\sum_{k=0}^{N}\frac{d^kF}{dx^k}|_{x_0}\frac{(x-x_0)^k}{k!} +\mathcal{o}((x-x_0)^k)$$

If we put N=1, we get:

$$ F(x)=F(x_0)+\frac{dF}{dx}|_{x_0}\cdot(x-x_0)+\mathcal{o}(x-x_0)$$

That means that you can legitimately consider $F(x)\simeq F(x_0)=\text{constant}$ in the neighborhood of $x_0$ only if the first (and higher) order derivative of $F(x)$ evaluated in $x=x_0$ times $(x-x_0)$ is small enough to be considered neglectable for your purposes. Here I wrote $F:=F(x)$, but it's exactly the same if $F:=F(\theta)$. Just put $x\mapsto\theta$ (or any other variable you wish) in the above formulas.

Consider now the limit for $\mathbf{x}\to \mathbf{x_0}$, i.e. $\mathbf{x}-\mathbf{x_0}\simeq d\mathbf{x}$. The differential of the function $F(\mathbf{x})$ is its linear increase respect to its variables. It is defined as:

$$dF=\nabla F \cdot d\mathbf{x} =\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz$$

So F is not increasing (i.e. $F$ is constant) when moving along a direction only if the partial derivative in the same direction is zero. You can write this in any coordinate system you want. For example in spherical:

$$dF=\frac{\partial F}{\partial r}dr+\frac{1}{r}\frac{\partial F}{\partial \theta}rd\theta+\frac{1}{r\sin\theta}\frac{\partial F}{\partial \phi}r\sin\theta d\phi$$

So just evaluate the partial derivative and see if it's zero. If that's the case, then you can consider the function constant along that direction.

  • $\begingroup$ I only know single variable calculus(Calc I). Is there any way to do this using those basic tools? $\endgroup$ – xasthor Feb 20 '17 at 14:54
  • $\begingroup$ Yes, of course. I edited the answer to add the single variable case. Let me know if now it's more clear. $\endgroup$ – Alessandro Zunino Feb 20 '17 at 16:00
  • $\begingroup$ This is essentially telling me when/why I can assume $F(x)$ to be constant in the interval $[x_0, x_0+dx]$ and $F(\theta)$ to be constant in the interval $d\theta$, $f(z)$ to be constant in the interval $dz$ etc. but my question was whether I can assume $f(x)$ to be constant in any infinitesimal interval such as $d\theta$, $dz$, $dy/cos\phi$ etc., not just $dx$ because I've seen it done in my physics classes but I don't understand $why$ I can do it. $\endgroup$ – xasthor Feb 21 '17 at 9:57
  • $\begingroup$ Well, if you consider $F(\mathbf{x})$ to be a three-dimensional force (i.e. $F:\mathbb{R}^3\to\mathbb{R}^3$), you need to know multi-variable calculus, otherwise you can't be rigorous. Then the answer is the first one i wrote you. If you're thinking about $F(x)$ as having only a single variable (for example if you consider the line $R\theta$ with $R=constant$) the you can simply use the second answer I wrote. If you don't know enough math to understand the above formulas, I'm afraid the only thing you can do is evaluate $\frac{F(x+\Delta x)-F(x)}{F(x)}$ and check if is $\ll 1$. $\endgroup$ – Alessandro Zunino Feb 21 '17 at 10:17
  • $\begingroup$ I edited my answer again, trying to be even more clear and using explicitly infinitesimal quantities. I hope that now you can more easily understand the matter. Let me know. $\endgroup$ – Alessandro Zunino Feb 21 '17 at 10:46

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.