His initial one dimensional derivation of Newton's Second Law using the Principle of Least Action, I believe is fairly concise and easy to read. However, I did get hung up on his use of the Taylor series expansion, and I noticed this quote from his lecture:

I have written $V′$ for the derivative of $V$ with respect to $x$ in order to save writing.

This is how I come about the same expansion for the potential function, $V$:


Where x is the approximated path taken by a classical particle, $\underline{x}$ is the actual path, and $\eta$ is the error (which is zero at the start and finish point). Expanding the first two terms of the taylor series for V,

$V(\eta)_{centered\ on\ \underline{x}} \approx V(\underline{x}) + V'(\underline{x})(\eta - \underline{x})$

$V(\eta + \underline{x}) \approx V(\underline{x}) + V'(\underline{x})\eta$

Which is, no surprise, the same thing noted in the lecture. I won't bother with following the math through to the end result (it's simple), but I will note that the derivative of the potential function, V, evaluated at $\underline{x}$ shows up in the final result:

$m\ddot{\underline{x}} = -V'(\underline{x})$

And I have been scratching my head at this for many days (far too many days). The original potential function should have been with argument $\eta$, centered on the actual path, $\underline{x}$. This is counter to my notion that the potential functional should be dependent only on a position in real space. This implies greatly that,

$m\ddot{\underline{x}} = -\frac{dV}{d\eta}\bigg{|}_{\eta=\underline{x}}$

Revealing a force function as follows,

$F = -\frac{dV}{d\eta}\bigg{|}_{\eta=\underline{x}}$

Can someone explain what this means? Why would the force function be proportional to the derivative of the potential function with respect to the error in the path compared to the actual path taken?

  • $\begingroup$ $\uparrow$ Link? $\endgroup$ – Qmechanic Nov 29 '18 at 2:52
  • $\begingroup$ I think your notation may be backward - typically one says that the approximate path $x$ is the sum of the true path $\underline{x}$ and an error term $\eta$. $\endgroup$ – J. Murray Nov 29 '18 at 2:55
  • $\begingroup$ feynmanlectures.caltech.edu/II_19.html $\endgroup$ – Michael Burt Nov 29 '18 at 2:55
  • $\begingroup$ J. Murray- you're right, I've got the notation backwards. I'm checking this, but I don't think it changes the math. The potential function is still required to be a function of the error in order for the Taylor expansion to turn out correct $\endgroup$ – Michael Burt Nov 29 '18 at 2:57

This is one of those places where almost-universally helpful notation can trip you up.

$V$ is a function. It eats a single real number and then spits out another real number. As a result, you can differentiate it with respect to one thing and one thing only - its argument. The result is another function, which we give the related name $V'$.

$V(\underline x + \eta)$ is a number - specifically, the number which $V$ spits out when you feed it the number $\underline x + \eta$. If $V$ is sufficiently well-behaved and $\eta$ is sufficiently small, Taylor's theorem tells us that $V(\underline x + \eta)$ can be approximated as follows:

$$V(\underline x + \eta) \approx V(\underline x) + V'(\underline x)\cdot \eta $$

In words,

The function $V$ evaluated at the number $\underline x + \eta$ is approximately equal to the function $V$ evaluated at the number $\underline x$, plus $\eta$ times the function $V'$ evaluated at the number $\underline x$.

Hopefully this doesn't come across as overly mathematical and pedantic - the point I'm trying to convey is that you are confusing yourself by writing $\left.\frac{dV}{d\eta}\right|_{\eta = \underline x}$. It doesn't make sense to talk about differentiating $V$ with respect to anything other than its argument. In that expression, $\eta$ is essentially a dummy variable - it means exactly the same thing as $\left.\frac{dV}{dy}\right|_{y=\underline x}$ or $\left.\frac{dV}{d\star}\right|_{\star = \underline x}$.

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  • $\begingroup$ No apologies need about being pedantic. My original question is pedantic, mathematically, as well. How is $\eta$ a dummy variable? It's clear from the expansion that V is a power function who's principal argument is $\eta$, centering the function near $\underline{x}$. $V(\eta + \underline{x})$ is just that function with a shifted input. No? $\endgroup$ – Michael Burt Nov 29 '18 at 3:53
  • $\begingroup$ Edit: formatting. Sorry about that $\endgroup$ – Michael Burt Nov 29 '18 at 3:58
  • $\begingroup$ My remark about $\eta$ being a dummy variable was referring to the notation $\left.\frac{dV}{d\eta}\right|_{\eta = \underline x}$. I believe you're getting confused by the $\frac{d\bullet}{d\bullet}$ notation for derivatives; if you use only the "primed" notation, the heart of your question evaporates. You find that $m\underline{\ddot x} = -V'(\underline x)$, which is perfectly clear because we define $V$ in the first place to be a function such that $-V' = F$ (force is the negative derivative of the potential energy). $\endgroup$ – J. Murray Nov 29 '18 at 4:03
  • $\begingroup$ In my understanding of the Taylor expansion, $V'(\underline{x})$ is not "the derivative of V which results in a function of $\underline{x}$", but rather "the derivative of V (with respect to $\textit{something}$) evaluated at argument = $\underline{x}$. This is the crux of my issue here. I'm not trying to abuse notation, by the way, I'm trying to understand what variable we take the derivative of V with. Is it x? Is it $\underline{x}$? $\eta$? Intuition tells us the answer, but I would like it to be borne out in the mathematics. $\endgroup$ – Michael Burt Nov 29 '18 at 4:09
  • $\begingroup$ Let's continue this discussion in chat $\endgroup$ – J. Murray Nov 29 '18 at 4:16

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