I often see in physics that, we say that we can multiply infinitesimals to use chain rule. For example,

$$ \frac{dv}{dt} = \frac{dv}{dx} \cdot v(t)$$

But, what bothers me about this is that it raises some serious existence questions force me, when we say that we take derivative of $v$ velocity with respect to distance, that means we can write velocity as a function of distance. But, how do we know that this is always possible? As in like when we do these multiplication of differentials we are implicitly assuming that $v$ can be change from a function of time into a function of displacement.

I seen this used ubiquitously, and I some crazier variations I've seen of literally swapping differentials like $ dv \frac{dm}{dt} = dm \frac{dv}{dt}$ , as shown by the answer of user "Fakemod" in this stack post.

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    $\begingroup$ Welcome to life as a physicist :-) $\endgroup$ – John Rennie Aug 13 at 8:46
  • $\begingroup$ If velocity=v(x(t),t), means velocity is an explicit function of both 'x' and 't' - then for same x , at two different times - it will have same veocity. It v=v(x(t))- then same as in v(x(t),x). If v=v(t), then v can have different velocity at same x at two different times.(since v is no longer a function of x- v has to have a unique value corresponding to a x,to be a function).- in this case, we can't write v as a function of x. $\endgroup$ – Syed Jaffri Aug 13 at 11:11
  • $\begingroup$ dv/dx might mean the implicit differentiation of v and x, both as functions of time. Look up 3Blue1Brown's Essence Series on Implicit Differentiation, it might be helpful :-) $\endgroup$ – Koustubh Jain Aug 14 at 14:42
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    $\begingroup$ @JohnRennie Mathematicians hate this one weird trick! $\endgroup$ – Mark H Aug 14 at 19:37

You are correct that you cannot (globally) write velocity as a function of distance. For example, as one commenter has already mentioned, throw a ball directly up in the air and wait for it to come down. When the ball is at height $h$ on the way up, it has a positive (upward directed) velocity. When it is at the same height $h$ on the way down, it has a negative (downward directed) velocity. So velocity is definitely not a (global) function of distance.

But this much is true: For any height $h$ except for the maximum height the ball ever reaches, there is some open interval around $h$ --- some range of heights from $h-\epsilon$ to $h+\epsilon$ --- in which you can treat velocity as a well-defined function of height while the ball is on its way up, and another well-defined function of height while the ball is on its way back down. And moreover that function is differentiable and obeys the chain rule. All of this is part of the content of the implicit function theorem, which you can google for.

If you just write velocity as a function of height, you do have to be careful to make it clear from context which of the two functions --- the "on the way up" function and the "on the way down" function --- you're referring to. You also have to make sure you don't try to pull this stunt when the ball is at the very top of its trajectory (or more generally, at points where its velocity is zero). Many books take it for granted that you're being careful about this, so they don't have to worry about it on your behalf.

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Well, this is the most common thing for which mathematicians make fun of physicists. Because we don't bother to cancel out derivatives, and we "NEVER" check if we can imply some rule on our equations. The thing is, that almost all functions, which can appear in nature or real life systems are, in most times, continuous and differentiable. There are, for sure, some special cases. But for most simple tasks, eg. mechanic, this is quite valid.

So in the case of $v$. In order to define velocity, the object has to change its position in some amount of time. And furthermore, we don't have infinite speed in real life. This implys, that $dx/dt$ has allways that some non infinite value. From this it follows, that $v$ can be rewritten as function of either $t$ or $x$.

I am not sure if there is a special case or not, but for physicists it is not important, because in 99.9% this will be true. If there are special cases, they could be "obviously strange". You should have in mind, that at least in theory, we always check our calculations with experiment, so we have an experimental proof instead of a mathematical one (generally).

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    $\begingroup$ I think your argument is flawed... just because $dx/dt$ is never infinite doesn't mean you can always write $v$ as a well defined function of $x$. A simple counterexample: throw a ball straight up in the air and then catch it. You would need to define separate functions $v(x)$ to fully describe that, as there are different velocities at the same point in space at different times. $\endgroup$ – BioPhysicist Aug 13 at 12:59
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    $\begingroup$ @BioPhysicist and also v(x) would not be defined for the points in space higher than the ball reached when thrown up. $\endgroup$ – Peteris Aug 13 at 16:16
  • $\begingroup$ @BioPhysicist I believe that can be resolved by this Math SE answer. In your example, the relation between $v$ and $x$ looks like a rotated absolute value function. That's not a function, but if you pick a point on the curve, it locally looks like a function (except at the apex). That lets $dv/dx$ be defined; the difficulty is now locating the points, since $x$ doesn't work. We can evaluate $dv/dx$ given $t$, and if we already have a point we can consider small displacements $dx$. Like this answer says, in physics, this should usually work out. $\endgroup$ – HTNW Aug 13 at 17:28
  • $\begingroup$ @BioPhysicist generaly speaking, I should than describe $v$ as $\frac{d\textbf{r}}{dt}$. But scince we are talking about the ways mathematic is used for calculations in physics, the system is normaly described in as few dimensions as possible. I can allways say, that $x$ points directl upwards... If we want to be more fundamental we should describe system with generalised coordinates $q_1, q_2, .. q_n$... $\endgroup$ – Vid Aug 13 at 17:31
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    $\begingroup$ “that almost all functions, which can appear in nature or real life systems are in most times continuous and differentiable” – definitely exaggeration. Lots of processes like collisions between hard objects, phase transitions and chemical reactions are discontinuous (at least unless you zoom in very closely, which may be a theoretical escape but is generally completely impractical). Just, physicists don't tend to worry about that: they restrict themselves to some subdomain or submanifold where the behaviour is continuous, and leave the general case for engineers/chemists etc. to worry about. $\endgroup$ – leftaroundabout Aug 13 at 22:59

It is true, that in nature there is only one true independent variable, time. All others are "pseudo-independent". They are variables humans bless as independent in order to answer what-if scenarios and to establish mathematical models of systems byways of separation of variables. The common term for these "pseudo-independent" quantities is generalized coordinates.

Looking at a complex mechanical system, like a human launching a ball while riding on a skateboard. First, we decide what the degrees of freedom are and assign generalized coordinates to them. These are simple measurable quantities of distance, angle or something else geometrical forming a generalized coordinate vector $$\boldsymbol{q} = \pmatrix{x_1 \\ \theta_2 \\ \vdots \\ q_j \\ \vdots} \tag{1}$$ In this example there are $n$ degrees of freedom. All the positions of important points on our mechanisms can be found from these $n$ quantities. If there are $k$ kinematic hardpoints (such as joints, geometric centers, etc) then the $i=1 \ldots k$ cartesian position vector is some function of the generalized coordinates and time $$ \boldsymbol{r}_i = \boldsymbol{\mathrm{pos}}_i(t,\, \boldsymbol{q}) \tag{2}$$

Here comes the chain rule part. With the assumption that (2) is differentiable with respect to the generalized coordinates, and that contact conditions do not change due to separation, or loss of traction, the velocity vectors of each of the hardpoints is found by the chain rule

$$ \boldsymbol{v}_i = \boldsymbol{\mathrm{vel}}_i(t,\,\boldsymbol{q},\,\boldsymbol{\dot{q}}) = \frac{\partial \boldsymbol{r}_i}{\partial t} + \frac{\partial \boldsymbol{r}_i }{\partial x_1} \dot{x}_1 + \frac{\partial \boldsymbol{r}_i }{\partial \theta_2} \dot{\theta}_2 + \ldots + \frac{\partial \boldsymbol{r}_i }{\partial q_j} \dot{q}_j + \ldots \tag{3} $$ where $q_j$ is the j-th element of $\boldsymbol{q}$, and $\dot{q}_j$ its speed (being linear or angular).

The above is not a division of infinitesimals, but the multiplication of a partial derivative $\tfrac{\partial \boldsymbol{r}_i }{\partial q_j}$ with the particular coordinate degree of freedom speed $\dot{q}_j$.

Maybe you are more comfortable with this more rigorous notation using partial derivatives that what you have seen so far. The term partial derivative means, take the derivative by varying only one quantity and holding all others constant. This is what allows us to use pseudo-independent quantities $q_j$ for the evaluation of the true derivative with time (the one actual independent quantity).

The same logic is applied to higher derivatives as well

$$ \boldsymbol{a}_i = \boldsymbol{\rm acc}_i(t,\boldsymbol{q},\boldsymbol{\dot q}) = \frac{\partial \boldsymbol{v}_i}{\partial t} + \ldots + \frac{ \partial \boldsymbol{v}_i}{\partial q_j}\, \dot{q}_j + \ldots + \frac{ \partial \boldsymbol{v}_i}{\partial \dot{q}_j} \,\ddot{q}_j \tag{4} $$

The last part might be a bit confusing, but when you express it in terms of actual degrees of freedom it might be clear. Consider the degree of freedom $\theta_2$ and its time derivatives $\omega_2$ and $\alpha_2$. Then the terms $\frac{ \partial \boldsymbol{v}_i}{\partial \theta_2} \omega_2 $ and $\frac{ \partial \boldsymbol{v}_i}{\partial \omega_2} \alpha_2 $ are more clear I hope, as $\boldsymbol{v}_i$ depends on both the position $\theta_2$ and the speed $\omega_2$.

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  • $\begingroup$ I sort of get the equations but I don't get how it connects to the question I had asked, it would be nice if you could leave a statement in the end about the big picture overview of what the equations mean $\endgroup$ – Buraian Aug 13 at 21:41
  • $\begingroup$ You feel reservations in changing $v$ from a function of time to a function of displacement. I share this reservation, but I also show how it is useful mathematically by braking apart the complex problem of time evolution into multiple solvable problems of time and configuration evolution and putting all back together with the chain rule. I feel like the paragraph between (2) and (3) attempts to directly address your concerns. $\endgroup$ – John Alexiou Aug 15 at 0:57
  • $\begingroup$ Have to -1 despite appreciating the bulk of the explanation. The concern's the strong statements about time being a "true"-independent variable, suggesting that there's some fundamental quality distinguishing time. $\endgroup$ – Nat Aug 15 at 3:04
  • $\begingroup$ @Nat - I guess it is "proper" time I am talking about and it irreversibility that makes it independent. $\endgroup$ – John Alexiou Aug 15 at 17:35

I like this question and there already some good answers. I'm not going to repeat those, but I wanted to add a couple of points focused on the second part of your question regarding "swapping" differentials.

The first is that the presence of a differential quantity is an abstraction that is usually only useful as an intermediate step in calculating something else. By that, I mean you never measure something like $\rho\ dV$ directly. You can only hope to measure:

  1. The integral of that quantity $\int \rho\ dV$ over some volume (equivalently, you back off of the abstraction and measure $\rho \Delta V$ for some finite volume $\Delta V$) --OR--
  2. The "ratio of differentials" (being deliberately loose for the moment), which in the limit is a derivative. So an expression like $f(t) dt = g(x) dx$ gets "divided through" to be $f(t) = g(x) (dx/dt) = g(x)v(t)$. We believe we know how to measure changes in quantities and gradients.

That's relevant to the second part of your question about "swapping" differentials because when that's done legitimately, it typically works because you're ultimately going to put that expression under an integral sign, and the notation conveniently reflects (some might prefer to say the notation is easily abused when applying) the integration-by-substitution rule $$ \int_a^b f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(u) du$$ which you could rewrite in Leibnitz notation for $u = g(x)$ and get the appearance that you are swapping or canceling differentials.

Since the integration-by-substitution rule is basically the chain rule in reverse, however, all of this begs your initial question of why the chain rule is valid in physics. For that, I refer back to the other already-good answers.

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that means we can write velocity as a function of distance

That's not quite the intended meaning. Rather, the intended meaning is:

If you could write velocity as a function of distance in the domain of interest, then the equation would hold.

It's up to you to deduce whether that assumption can be met satisfactorily in the problem, but usually it's quite obvious that it can.

One way to see this is that you can artificially restrict the domain to the portion of space and time that is of interest and disregard the rest of the domain, and then argue that this assumption would hold there.
(Notice I basically just rephrased the continuity notion of a limit here.)

The only way for this to be false in your particular example is to have multiple velocities at a given point in time (or no velocity at all), which generally wouldn't make sense in the (continuous) everyday world we're familiar with.

And if the discussion is about some unusual boundary condition where you can't take a limit on all sides and show the problem is continuous, then you wouldn't read such a claim about that situation without some kind of other (implicit or explicit) indication as to why it's true.

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