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Official Leibniz notation for double derivative is:

$$\frac{\mathrm d^2s}{\mathrm dt^2}$$

This term seems inconsistent. Two considerations:

  1. We have infinitesimal change in distance $\mathrm ds$ per infinitesimal change in time $\mathrm dt$: $\mathrm ds/\mathrm dt$. Both terms are a tiny value/interval. Because the $\mathrm d$ symbolizes difference, I would as a change of the change of the distance to time intuitively write: $$\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm ds^2/\mathrm dt^2)}{\mathrm dt}=\frac{\mathrm ds^2}{\mathrm dt^3}$$ where the extra $\mathrm d$ says that both terms are now "double" infinitesimal differences.

  2. Maybe more properly following mathematical logic and not my intuition, the $\mathrm d$ could be considered a "free" variable in itself that can be multiplied onto this $\mathrm ds/\mathrm dt$ fraction numerator: $$\frac{\mathrm d(\mathrm ds/\mathrm dt)}{\mathrm dt}=\frac{(\mathrm d^2s/\mathrm dt)}{\mathrm dt}=\frac{\mathrm d^2 s}{\mathrm dt^2}$$ That agrees with the actual notation but doesn't really make physical sense now. $\mathrm d$ means (infinitesimal) difference, so that $\mathrm ds=s_{final}-s_{start}$, and therefore it makes no physical sense to consider the $\mathrm d$ and the $s$ separate. The $\mathrm ds$ is physically just a "name"/"symbol" for one term, which could just as well have been called $x$ or $a$ or anything else.

Now, while searching for an explanation, the answers always tend to consider $\frac{\mathrm d}{\mathrm dt}$ as one symbol in itself, so that a double derivative is $\frac{\mathrm d}{\mathrm dt}\frac{\mathrm d}{\mathrm dt}s=\frac{\mathrm d^2}{\mathrm dt^2}s=\frac{\mathrm d^2s}{\mathrm dt^2}$ - which makes even less physical sense, since the $\mathrm dt$ term has to be a separable term before we can treat $\frac{\mathrm ds}{\mathrm dt}$ as a normal fraction (as done in integration e.g.). $\frac{\mathrm d}{\mathrm dt}$ can't possibly be just "a symbol".

Why is $\frac{\mathrm d^2s}{\mathrm dt^2}$ the correct one in a physical context, where $\mathrm ds$ actually means the infinitesimal difference in $s$? Are my considerations in point 2 correct, and I just can't figure out that splitting $\mathrm d$ and $s$ is allowed?


Update

The answers already given at this time both indicate the use of $\mathrm d/\mathrm dt$ as merely a symbol. So, neither of my two suggestions mentioned above are the case. Sure, I can accept that. But the question still remains of why as well of how come we still treat them as variables then, e.g. in integration?

Let me clarify those two points:

Firstly, if it indeed is the case that $\mathrm d/\mathrm dt$ is merely a symbol and should be thought of as just a symbol, then I do not understand the motivation for this symbol.

  • Why did Leibniz choose $\mathrm d/\mathrm dt$ as a symbol, which causes the confusion and inconsistency described in the question above?
  • Why not, say, $\mathrm d/\mathrm d$, in which case we would get a writing-style that at least looks a bit more "consistent": $$\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm ds}{\mathrm dt}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^2s}{\mathrm d^2t}\qquad \frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac{\mathrm d}{\mathrm d}\frac st=\frac{\mathrm d^3s}{\mathrm d^3t}\qquad \cdots$$
  • Or even better yet, if the two $\mathrm d$ involved in this $\mathrm d/\mathrm dt$ symbol have no meaning as neither a variable nor an indicator of a change in the parameter, then why use this letter at all? Why not stick to, say, the prime-notation throughout and never jump into the Leibniz notation: $$s_t'\qquad s_t''\qquad s_t'''\qquad \cdots$$

And secondly, if the $\mathrm d/\mathrm dt$ really just is a symbol, and that's it, then how come we suddenly can treat it as a fraction again containing a set of variables $\mathrm ds$ and $\mathrm dt$ that we can split apart during for instance integration? Such as here: $$\frac{\mathrm ds}{\mathrm dt}=v\quad\Leftrightarrow\quad \mathrm ds=v\,\mathrm dt\quad\Leftrightarrow\quad \int 1 \,\mathrm ds=\int v\,\mathrm dt \quad\Leftrightarrow\quad s=\int v\,\mathrm dt$$

I hope to get this notion cleared out and appreciate all comments and answers that can help.

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  • $\begingroup$ I think the last one makes the most sense $\endgroup$ – Karl Feb 22 '17 at 20:48
  • $\begingroup$ @Karl I think "sense" then has a different meaning for me when thinking in physical context. How can $\mathrm d$ and $s$ be seperated and treated like were they two distinct variables? $\endgroup$ – Steeven Feb 22 '17 at 20:51
  • $\begingroup$ I'm not totally sure they are. Are you referring to solving variable separable equations? The notations conveniently hide the behind the scenes maths. $\endgroup$ – Karl Feb 22 '17 at 20:55
  • $\begingroup$ $\dfrac d {dt}$ is a symbol: it is the name of the operation of "differentiation". See Second derivative : Notation. $\endgroup$ – Mauro ALLEGRANZA Feb 22 '17 at 21:01
  • $\begingroup$ Thank for the answers and comments at this point. It has helped my insight into this topic, but I unfortunately have not gotten the question fully cleared out yet. I have made an update to clarify with the new information I have gotten. $\endgroup$ – Steeven Aug 7 '19 at 6:11
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I think the answer to the first question is quite straitforward: the symbol $\frac{ds}{dt}$ is short for the limit $\delta t\to 0$ of $$ \frac{\delta s}{\delta t} = \frac{s(t+\delta t)-s(t)}{\delta t}. $$

Whenever you see $d$ think of the change $\delta$ with that in mind that the limit of $\delta$ to zero will be taken.

Now, second derivate is the limit $\delta t\to 0$ of $$ \frac{\frac{s(t+2\delta t)-s(t+\delta t)}{\delta t}-\frac{s(t+\delta t)-s(t)}{\delta t}}{\delta t} = \frac{\overbrace{(s(t+2\delta t)-s(t+\delta t))}^{\delta s(t+\delta t)}-\overbrace{(s(t+\delta t)-s(t))}^{\delta s(t)}}{\delta t^2} = \frac{\delta(\delta s)}{\delta t^2}. $$ and hence the symbol $\frac{d^2s}{dt^2}$.

For the second question, there are two answers. The official answer is that $\frac{ds}{dt}$ is just a symbol for derivative, and you shouldn't separate $ds$ and $dt$, and the fact that it works is just a coincidence, ...

I don't believe in coincidences, so let's dig deeper to see if there is a reason for that to work. Let us start with $$\frac{ds}{dt} = v$$ This means at every point $t$, $\delta s$ is approximately $v\,\delta t$. And this approximation is more accurate the smaller the $\delta t$. Now, if you want to know how much $s$ changes over some time interval $[0,t]$, you just add up all of the small changes in $\delta s$ for all the smaller time intervals $\delta t$. Now, we have $$ s \approx \sum \delta s \approx \sum v\, \delta t $$ where approximations become exact when $\delta t$ goes to zero. The limit $\delta t \to 0$ of $$ \sum v\, \delta t$$ is defined as the integral $$\int v\, dt$$

So basically, when you separate $ds$ and $dt$, you are postponing taking the limit $\delta t\to 0$.

Also, see my answer to a relevant question.

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$\frac {d^2s}{dt^2}$ is the rate of change of the rate of change of distance. It is a differential operator acting on something that is already a derivative.

While Leibnitz originally thought of $\frac {ds}{dt}$ as a ratio of infinitesimals that was effectively a fraction in all senses, that is no longer what the notation means today.

$\frac {d}{dt}$ is indeed one symbol!

$\frac {ds}{dt}$ is the differential operator, $\frac {d}{dt}$ applied to the function $s(t)$ i.e. $\frac {ds}{dt} = \frac {d}{dt} s(t)$

$\frac {d^2s}{dt^2}$ is the differential operator applied to $\frac {ds}{dt}$ or $\frac {d}{dt}\frac{ds}{dt}$

so $\frac {d^2s}{dt^2} = (\frac {d}{dt})(\frac {d}{dt}) s(t)$

Futhermore, $\frac {ds^2}{dt^2}$ could be interpereted to be $(\frac {ds}{dt})^2$

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  • $\begingroup$ Aha, thank you for this answer. A follow-up: So $\frac{ds}{dt}$ has become intepretted as $\frac{d}{dt}s$ - why not something like $\frac{d}{d}\frac{s}{t}$, which would then cause the second derivative to be $\frac{d}{d}\frac{d}{d}\frac{s}{t}=\frac{d^2s}{d^2t}$? Such notation would at least look consistent allover. $\endgroup$ – Steeven Feb 22 '17 at 21:05
  • $\begingroup$ It would not be any more a linear operator, which by definition operates on functions $\endgroup$ – Bernard Feb 22 '17 at 21:09
  • $\begingroup$ Hi @Bernard, thanks for the comment. I'd like to come back to this question and ask if you wouldn't mind elaborating on the issue you mention with this not being a "linear operator" any more. If $d/dt$ is merely a symbol, because we have chosen it to be a symbol, and it has no meaning as variables $d$ and $dt$, then why couldn't we just as well have chosen $d/d$ as a symbol as per my first comment? Is there a "rule" when choosing to consider this term as a symbol that must to considered? $\endgroup$ – Steeven Aug 6 '19 at 19:27
  • $\begingroup$ @Steeven: in $d/df$, the $t$ stipulates that the variable in the function is $t$ (just in case it depends on other variables/parameters. It is the equivalent of Lagrange's notation $f'_t$ or Arbogast's $D_t $. $\endgroup$ – Bernard Aug 7 '19 at 10:44
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The mathematical answer, as has been noted by other in comments and answers, is that $d/dt$ is a single symbol in modern usage, and so it makes sense that the operator applied twice would be $(d/dt)^2 = \frac{d^2}{dt^2}$.

The physicist does often treat something like $ds/dt$ as a ratio between small quantities - at least intuitively. Strictly speaking mathematically this is somewhere between sloppy and wrong. In a lot of cases, however, it works functionally and the physicist goes along happily. This seems to be more along the lines of what you're asking.

In this case, (again noting that this is not mathematically rigorous at all), you could look at it this way. Maybe you have

$v(t) = \frac{ds}{dt} \\ a(t) = \frac{dv}{dt}$

So far so good since each one has the form that your intuition demanded - The first "looks" like a ratio between small changes in position with small changes in time. The second "looks" like a ratio between small changes in velocity with small changes in time. If you substitute, then you get

$a(t) = \frac{d}{dt} \frac{ds}{dt} = \frac{d^2 s}{dt^2}$

I'll again emphasize that thinking of these as separate symbols is mathematically sloppy and/or wrong, but it is consistent with what you were calling "physical intuition" in your question. The $d^2$ in the numerator came because you "simplified" the fraction - If you want to see it in terms of infinitesimals, you need run that last equation right-to-left to put it in a different form.

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I will summarize a few ideas presented in this paper: http://online.watsci.org/abstract_pdf/2019v26/v26n3a-pdf/4.pdf

The best way of thinking of $\frac{d}{dt}$ is as two separate things. (1) the top $d$ is differential (linear) operator on whatever is to the right and (2) the $dt$ is an infinitesimal, i.e., the same differential operator applied to a single variable, in this case $t$ (e.g., $d(x^2) = 2xdx$ and dividing by $dx$ yields $2x$, i.e., the derivative of $x^2$). Infinitesimals are supported by the hyperreal numbers (https://en.wikipedia.org/wiki/Hyperreal_number) of non-standard analysis that are given an introductory treatment in https://www.math.wisc.edu/~keisler/calc.html.

In the original paper I referenced the notation $\frac{d^2s}{dt^2}$ is explained using the description I gave for the first order version. Barlett's argument leads to the actual mathematical second order equivalent of $\frac{d}{dt}$ being (using the more concise Arbogast notation on the left, also favored by Euler)

$$ D_t^2 = \frac{d^2s}{dt^2} - \frac{ds}{dt}\frac{d^2t}{dt^2} $$

where usually $t$ is taken to be an independent variable and so the right half of the equation becomes zero. This explanation is more general than cases were the independent variable is $t$ and allows the second order Leibniz notation to be manipulated algebraically. In summary, $dt$ is not "just a symbol," it's a infinitesimal and $\frac{d}{dt}$ is a differential operator on top with the $dt$ being an algebraic division by an infinitesimal.

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The notation came from considering first the finite differences before going to infinitesimal differences.

Thus, the first difference in $s=s(t)$ is $\Delta s =s(t+\Delta t)-s(t),$ which gives $$\frac{\Delta s}{\Delta t}=\frac{s(t+\Delta t)-s(t)}{\Delta t},$$ and the second difference is $$\Delta \Delta s =\Delta (s(t+\Delta t)-s(t))=\Delta s(t+\Delta t)-\Delta s(t)=s(t+2\Delta t)-2s(t+\Delta t)+s(t).$$ Thus the second difference quotient is $$\frac{\Delta \Delta s}{\Delta t\Delta t}\frac{s(t+2\Delta t)-2s(t+\Delta t)+s(t)}{\Delta t\Delta t}.$$

Note that we usually write $\Delta \Delta s=\Delta^2 s$ and $\Delta t \Delta t={\Delta t}^2.$

This shows why that choice of notation was made by Leibniz.

And now a word on differentials. A differential is an infinitesimal change in a function, and an infinitesimal is an infinitely small quantity, as it were. One may semi-formally think of it as a quantity approaching zero. Not the quantity alone, or the zero limit it approaches, but the quantity and its approach to zero. Thus this semiformal definition looks like that of the vector quantities of physics, which capture a notion of directed quantities. That's what the notion of infinitesimal captures -- our intuition of instants of time, points on a line, etc.

Having said that, then it's clear that we may handle differentials separately, and perform operations on them -- and that's how they were used from the beginning (hence, the differential calculus, differential equations, etc.). In particular, if we're thinking of quantities depending on a single independent variable, then we may add, subtract, multiply the differentials of such quantities to get similar differentials. However, with division, even when the differential in the denominator is not that of a constant function, we have to be careful, because the result is no longer always a differential. In many cases it is a function, and this is why the differential calculus is about calculating this type of ratio of differentials.

When we come to integrating, what we're doing is gathering together a continuous stream of differentials, to give a quantity.

The upshot of this all is that you're right. Even those who would banish talk of differentials (even after their justification in the so-called nonstandard analysis) cannot avoid using them when calculating integrals. And I can't imagine them still talking like that if they want to integrate over arbitrary manifolds, for example. So, with care and understanding, one can calculate with differentials, once one understands their provenance and rules of behavior -- they're just infinitesimal differences, that is we may think of $\rm d$ as the pair $(\Delta, \Delta\to 0),$ and they obey many of the usual rules of a ring (once you're dealing with only one independent variable). With more than one independent variable, things become complicated a little and one may need to run to multilinear algebra for some help. But that's another story.

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