# Physicists, not mathematicians, can multiply both sides with $dx$ - why?

The following question is asked without malicious intentions - it's not intended as a flamebait!

In my physics textbooks (Young & Freedman in particular) I have often seen derivations of equations that use multiplication of, say, $dx$ on both sides, and then integrating.

Taking courses both at the math and physics department I've noticed that real mathematicians often frown at this. Why is that?

• Why, mathematicians do it too. They just prefer to understand what exactly they're doing before doing it. Commented Jun 23, 2011 at 7:55
• Could you please give an example of a derivation in your textbook upon which you think "real mathematicians" frown? I can offer one explanation that is relevant: mathematicians often love to ensure that everything they do is completely rigorous. Although rigor alone is not mathematics, it is necessary. For example, a good number of students learn about the change of variables rule in integration without knowing the formal proof of this rule. And certainly many people do not stop and look up the hypotheses of the relevant theorem every time they apply the rule ... Commented Jun 23, 2011 at 7:59
• ... However, I am fairly certain that many mathematicians would frown upon a student applying the change of variables rule all the time without really knowing that there is a theorem (with proof) behind the rule. Unfortunately, this is how one is taught about the rule in schools these days. Commented Jun 23, 2011 at 8:01
• Hm, I have never seen mathematicians frown upon multiplying both sides of an equation by $dx$. However, I have seen a few frown upon "canceling" the $dx$ term in expressions like $$\frac{dy}{dx}\,dx.$$ See math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio. Commented Jun 23, 2011 at 8:02
• I second Amitesh Datta, without a concrete example it is hard to guess what the physicists have done. There are several different contexts and explanations possible. Commented Jun 23, 2011 at 9:42

Suppose you have the equation $$f(x) = g(x).$$ You could "multiply both sides by $dx$" $$f(x) \, dx = g(x) \, dx$$ and then integrate over some interval $$\int_a^b f(x) \, dx = \int_a^b g(x) \, dx.$$ However, unless you have learned about differential forms, the second equation above is meaningless. This can easily be avoided by simply integrating both sides of $f(x) = g(x)$ with respect to $x$ just as you learned in calculus.

Now consider the differential equation $$f(y) \frac{dy}{dx} = g(x).$$ Here, many people will "multiply by $dx$" to get $$f(y) \, dy = g(x) \, dx$$ and then integrate both sides. Again, without the machinery of differential forms, this statement is meaningless. Here we can avoid this by integrating the original equation with respect to $x$: $$\int f(y) \frac{dy}{dx} \, dx = \int g(x) \, dx$$ or to use different notation, $$\int f(y(x))y'(x)\, dx = \int g(x) \, dx.$$ Then as long as certain hypotheses are satisfied, the change-of-variables (substitution) theorem says that this is just $$\int f(y) \, dy = \int g(x) \, dx$$ so we are back to where the first method lead us.

The point of this is that manipulations involving "multiplying by $dx$" are usually shorthand for some more "rigorous" method. However, if you are aware of (and comfortable with) the mathematics going on behind the scenes, then there is no loss of rigor. "Real mathematicians" don't frown upon using these kinds of manipulations themselves, they just frown upon students using them without knowing why they work.

• Great answer so far. What exactly are those "certain hypotheses?" Commented Jun 23, 2011 at 23:43
• If you are referring to the change of variables part, there are a few different COV theorems depending on the integral (Riemann, Lebesgue, etc.) and the types of functions you are dealing with. A weak version for the Riemann integral is: If $\phi\in C^1$ and strictly increasing on $[a,b]$, and if $f$ is integrable on $[\phi(a),\phi(b)]$, then $\int_{\phi(a)}^{\phi(b)} f(t) \, dt = \int_a^b f(\phi(x))\phi'(x) \, dx$. Commented Jun 24, 2011 at 0:22
• What about the methods of "homogeneous equations" and "exact equations"? In these methods they often multiply by $dx$ to get to the "diferential form" of the DE. How can we understand this? Commented Sep 25, 2017 at 1:40
• I feel like the word "meaningless" here is harmful. They're not meaningless. It's just that their meaning is subtle and requires explication. Commented Jun 13 at 11:20

Because it is considered as non rigorous. Calculating with infinitesimals is something physicists and mathematicians have done since at least the time of Fermat, even long before that. But it always felt awkward to work with infinitesimals as they seemed to lead to contradictory statements if not manipulated carefully.

In the 19th century, under the impulse of Cauchy, calculus started to be formalized and the calculus of infinitesimals was replaced by the more precise and rigorous $\epsilon-\delta$ methodology. Physicists didn't take much notice however since calculating with differentials still worked and was more practical.

In the 1960's however, Abraham Robinson developed a rigorous approach to manipulating infinitesimals now called non-standard analysis. At the time it was controversial, nowadays, I think it is more accepted. That doesn't mean physicists really care that much, but in a way, Robinson vindicated their approach. Although you still need some work to make things rigorous.

• I don't quite see how that's relevant — differential forms have nothing to do with infinitesimals (in modern approach, at least) Commented Jun 23, 2011 at 8:33
• I don't see how differential forms are involved with the OP? Because he is speaking about integration? Then I guess the theory of differential forms is another approach towards treating expressions with $dx$. Commented Jun 23, 2011 at 8:37
• Even the title of OP mentions differential form ($dx$)!.. Well, frankly speaking, it's not very clear what OP is asking about. But multiplication both sides (of PDE, say) by $dx$ and integrating (almost verbatim from OP) makes perfect sense in the context of differential forms... Commented Jun 23, 2011 at 8:43
• OK, he uses $dx$, but doesn't mention differential forms. But I agree that the question makes sense in the context of differential forms. I just thought it also makes sense from the point of view of the theory of infinitesimals, which is how $dx$ was first conceived by Leibniz. Commented Jun 23, 2011 at 8:48

I think the earlier answers to this question missed the mark.

In physics, you never know exactly what you're studying. There's always uncertainty in your knowledge of the initial conditions of a system, and even if you could plug exact numbers into your model, the model itself is just an approximation.

Increasing the precision of numerical integration will yield a more accurate approximation up to a point, but beyond that point (somewhere around the atomic level, say), more precision makes the model worse. Fortunately, the result often doesn't drift too far from the right answer even in the infinitesimal limit, which means calculus is useful. But it's not correct. We know how every continuum model breaks down except for the spacetime continuum, and there are plenty of reasons to think that it breaks down too.

To put it another way, the rigorous methods of calculus are derived rigorously from assumptions that are incorrect (as pertains to the real world), so they are not actually rigorous (in the context of doing physics).

In physics, an integral is a large but finite sum, and $$dx$$ is a small but nonzero value. Since it's nonzero, there's no reason you can't multiply and divide by it. The justification for ignoring terms of order $$dx^2$$ is not that they disappear in a limit, but that they're close enough to zero that the difference won't be significant amidst all of the other noise in the problem. The Dirac delta function is a function: it's a Gaussian with a width small enough that you can't see it through your slightly blurry spectacles. And so on. These approximate methods can't be proven correct, but that's okay, because in science you can never know that anything you're doing is correct.

• This is a very simplistic view of what physicists are studying. (Putting aside the hilarious claim that they don't quite even know what.) In fact: in recent years I almost feel that I learned more interesting maths from physicists than from many mathematicians. Commented Aug 2, 2023 at 20:00
• "To put it another way, the rigorous methods of calculus are derived rigorously from assumptions that are incorrect ". This is incorrect and this answer is very poor. The reason we can treat derivatives like fractions is justified by the chain rule and by differential forms. Commented Aug 2, 2023 at 20:00
• Professional physicists also do rigorous mathematics. That's not what the question was about, or what this answer is about. I feel like there's an impedance mismatch between the question and this site—and my answer, apparently, given the reaction to it so far. Commented Aug 2, 2023 at 20:16
• I gather that your answer is not about rigorous mathematics, hence not about professional physics. I would totally agree with that. Commented Aug 2, 2023 at 20:20
• @KurtG. If you want the mathematician's version of the "fuzzy spectacles", I think it's the intuitionistic theory of the continuum. But I still think that formalizing what professional physicists do (and they do do that other stuff, whatever your last comment said) is somewhat beside the point. Physics is about the real world; even when you're doing physics by fixing a formal system and deriving machine-checkable results in it, that's not what it's about. (And that's why this was the wrong site for this question.) Commented Aug 2, 2023 at 21:20