I am taking an introductory level class, Physics with Calculus, using Priscilla Laws' Workshop Physics. The activity guide has asked me to prove that:

$$ma\,\mathrm{d}x = mv\,\mathrm{d}v$$

My physics instructor informed me that the correct method to prove this is:

  1. Start with:


  2. Knowing that:

    $$a = \frac{\mathrm{d}v}{\mathrm{d}t}$$

  3. Therefore:

    $$ma\,\mathrm{d}x = m\frac{\mathrm{d}v}{\mathrm{d}t}\mathrm{d}x$$

  4. Rearrange using laws of multiplication:

    $$m\frac{\mathrm{d}v}{\mathrm{d}t}\mathrm{d}x = m\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}v$$

  5. Knowing that:

    $$v = \frac{\mathrm{d}x}{\mathrm{d}t}$$

  6. Therefore:

    $$m\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}v = mv\,\mathrm{d}v$$

  7. Therefore:

    $$ma\,\mathrm{d}x = mv\,\mathrm{d}v$$

But every single mathematician I have talked to agrees; this is magic math. It is not good practice to, quite frankly, abuse infinitesimals in this way. And yet, the book suggesting that we do this has been under the careful eye of physicists nationally for years -- Surely if doing the above were wrong, someone would have pointed it out by now, and insisted that it be removed. By this reasoning, clearly the above is not necessarily wrong per se, but it is incredibly clear that there is a lot more mathematical complication going on under the hood which makes the above possible. It is my goal here to understand why in the name of goodness the above works, and what rules had to be maintained in order to ensure its validity. This brings me to the first part of my question -- Why does the above work? What could go wrong using the methods used above? How can I avoid making mistakes when manipulating infinitesimals in that way?

Unfortunately, being an entry level calculus student, I am frankly not equipped to deal with such topics as partial derivatives, differential equations, or infinitesimals/hyperreals. Even so, I have been determined to understand, at least informally, the underlying math behind the above. And what little "knowledge" I have amassed thus far implies something which I find disturbing.

Before I explain what I, personally, see wrong with the above, please allow me to informally define what I understand differentials to be, as well as a more explicit notation through which to write them. I am sure that an official notation exists for this, but I do not know enough in mathematics to use such a notation.

When we have $y = f(x)$, and say $f'(x) = \frac{\mathrm{d}y}{\mathrm{d}x}$, what we are "really" saying is this:

The ratio between the infinitely small change in $y$ at some solution set $x$ resulting from a corresponding infinitely small change in $x$, and the infinitely small change in $x$ at some location $x$ from a corresponding infinitely small change in $x$.

The above sentence translates into mathematical notation, literally as:

$$\lim_{h\to 0} \biggl[\frac{f(y + h) - f(h)}{x + h - x}\biggr]$$

where $\mathrm{d}y = \lim_{h\to 0} [f(x + h) - f(x)]$, and $lim_{h\to 0} [dx = (x + h - x)]$.

Now, this is all fine and dandy, until we have a relationship such as:

$$y = xz = f(x, y)$$

Suddenly $\mathrm{d}y$ could be either: $\mathrm{d}y = (f(x + h, z) - f(x, z))$ or: $\mathrm{d}y = (f(x, z + h) - f(x, z))$

(In fact, there exists a third situation where $\mathrm{d}y = y + h - y$ if $y$ is what we are taking the derivative/integral with respect to.)

With these rather tedious ambiguities, it seems prudent to not only state the variable whose change we are observing, but to also state which variable we are changing to create this change. Furthermore, it would be nice if we could also keep track of what values all of our variables start with. Therefore, I shall define a new syntax to make this simpler:

if $y = f(x)$ then:

$$f'(A.x) = \frac{\mathrm{d}y_{A_x}}{\mathrm{d}x_{A_x}}$$

Where $A$ is the solution set vector (containing values of $x$, and $y$, and for instance $A.x$ is the $x$ value contained by $A$) where all of our variables start, and where $\mathrm{d}y_{A_x}$ means:

Starting with the values contained in $A$, and changing $x$ by some infinitesimal $h$, what is observed change in $y$?

and when $\mathrm{d}x_{A_x}$ means:

Starting with the values contained in $A$, and changing $x$ by some infinitesimal $h$, what is observed change in $x$?"

By these definitions, we see that our notion of $f'(x)$ hasn't changed -- we are merely being more specific about what's changing, and what changes we are observing.

Now, using this new notation, and trying to use the same proof that my physics instructor suggested, we run into some issues. Executing the first four steps:

  1. Start with:


    ($\mathrm{d}x_{A_x}$ was chosen (as opposed to, for instance, $\mathrm{d}x_{A_t}$) because it was originally the trailing $\mathrm{d}x$ of an integral taken with respect to $x$)

  2. Knowing that:

$$a = \frac{\mathrm{d}v_{A_t}}{\mathrm{d}t_{A_t}}$$

  1. Therefore:

$$ma\,\mathrm{d}x_{A_x} = m\frac{\mathrm{d}v_{A_t}}{\mathrm{d}t_{A_t}}\mathrm{d}x$$

  1. Rearrange using laws of multiplication:

$$m\frac{\mathrm{d}v_{A_t}}{\mathrm{d}t_{A_t}}\mathrm{d}x_{A_x} = m\frac{\mathrm{d}x_{A_x}}{\mathrm{d}t_{A_t}}\mathrm{d}v_{A_t}$$

We end up with the expression $m\frac{\mathrm{d}x_{A_x}}{\mathrm{d}t_{A_t}}\mathrm{d}v_{A_t}$. Now, in the method suggested by my physics instructor, we wold normally try to replace $\frac{\mathrm{d}x}{\mathrm{d}t}$ (or in this case, $\frac{\mathrm{d}x_{A_x}}{\mathrm{d}t_{A_t}}$) with $v$ -- but there's a problem:

$$v = \frac{\mathrm{d}x_{A_t}}{\mathrm{d}t_{A_t}}\text{ NOT }\frac{\mathrm{d}x_{A_x}}{\mathrm{d}t_{A_t}}$$

In order for us to replace $\frac{\mathrm{d}x_{A_x}}{\mathrm{d}t_{A_t}}$ with $v$, $\mathrm{d}x_{A_x}$ would have to equal $\mathrm{d}x_{A_t}$! That's an implication that I am not certain is true, and this is what bothers me about the method prescribed by my instructor. This brings me to the second half of my question: Am I correct in thinking that the method prescribed by my instructor is wrong for this reason? Are the concepts I am conveying here valid, or is there another way of viewing this? Is there an alternative method which I could use to prove the same thing, but does not fall prey to these problems? Does $\mathrm{d}x_{A_x} = \mathrm{d}x_{A_t}$? Will $\mathrm{d}x_{A_x}$ ALWAYS equal $\mathrm{d}x_{A_t}$ by definition, or do they only equal in this case, because of the nature of kinematic motion?

  • 1
    $\begingroup$ 1/ I think you should ask on math.SE if you want to know about the math. 2/ It's perfectly legitimate and is just an instance of the rule of substitution. 3/ "d" stands for the total differential, thus $dy=({\partial y}/{\partial A})dA+ ({\partial y}/{\partial x})dx$. $\endgroup$ – Raskolnikov Jan 19 '14 at 9:43
  • $\begingroup$ Raskolnikov: I would agree with you, but I was afraid that they might tell me to ask here since it involves concepts in physics as well. Is this post being here a nuisance to the physics.SE community? I may end up asking it on both sites if either one fails to provide clarity, and if that would not be deemed an unsavory action $\endgroup$ – Georges Oates Larsen Jan 19 '14 at 9:46
  • $\begingroup$ I don't think it will be considered a nuisance. Your main question is about the maths, not the physics. I don't see any physics here. $\endgroup$ – Raskolnikov Jan 19 '14 at 9:47
  • 1
    $\begingroup$ Raskolnikov: Alright, then I shall take your advice and ask on math.SE tomorrow! (It is very late right now) $\endgroup$ – Georges Oates Larsen Jan 19 '14 at 9:54
  • 1
    $\begingroup$ "The above sentence translates into mathematical notation, literally as: - - where $dy=\lim_{h\to 0} [f(x+h)-f(x)]$" Literally that mathematical notation means that $dy=0$ which probably is not what you mean. $\endgroup$ – JiK Jan 19 '14 at 11:26

You say that you don't like the proof that

$$ ma\,\mathrm{d}x = mv\,\mathrm{d}v $$

because it "abuses infinitesimals" and isn't rigorous. But an expression like $ma \, \mathrm{d}x$ is already impossible to define rigorously without using some pretty advanced math! What you should really be proving is something like

$$ \int ma\,\mathrm{d}x = \int mv\,\mathrm{d}v $$


$$ ma\,\frac{\mathrm{d}x}{\mathrm{d} t} = mv\,\frac{\mathrm{d}v}{\mathrm{d} t} $$

which can be proven by a simple, completely rigorous application of the chain rule.


I am sorry, I have not read your whole question.

One way to make sense of $dx$ etc. is to view them as differential forms. Then calculations like these have a precise meaning, you just do not want to divide by forms.

Let us consider $x$, $v$ and $a$ to be functions of $t$. Then $x'(t)=v(t)$, $v'(t)=a(t)$. Hence $$dx = v\,dt,\qquad dv=a\,dt,$$ and $$a\,dx=av\,dt=v\,dv.$$ The meaning of this equation under this interpretation is simply $$a(t)x'(t)=v(t)v'(t).$$

  • $\begingroup$ Don't you have some conceptual problems here? I mean, $x(t)$ is a function from $\mathbb{R}$ to $\mathbb{R}$, but $v(t)$ goes from $\mathbb{R}$ to the tangent space $T\mathbb{R}$, and how do you define the differential of a tangent field in an appropriate way? $\endgroup$ – Daniel Robert-Nicoud Jan 19 '14 at 11:29
  • $\begingroup$ Obviously, if you look at $v$ as a function $\mathbb{R}\to\mathbb{R}$ too, then no problems arise. $\endgroup$ – Daniel Robert-Nicoud Jan 19 '14 at 11:30
  • $\begingroup$ @DanielRobert-Nicoud, you are right, this is a very one-dimensional viewpoint. $\endgroup$ – Carsten S Jan 19 '14 at 11:36

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