I have read the thread regarding 'the difference between the operators $\delta$ and $d$', but it does not answer my question.

I am confused about the notation for change in Physics. In Mathematics, $\delta$ and $\Delta$ essentially refer to the same thing, i.e., change. This means that $\Delta x = x_1 - x_2 = \delta x$. The difference between $\delta$ and $d$ is also clear and distinct in differential calculus. We know that $\frac{dy}{dx}$ is always an operator and not a fraction, whereas $\frac{\delta y}{\delta x}$ is an infinitesimal change.

In Physics, however, the distinction is not as clear. Can anyone offer a clearer picture?

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    $\begingroup$ Don't forget $\partial$ either! :-) $\endgroup$
    – b_jonas
    May 24, 2013 at 12:35
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    $\begingroup$ "We know that $\frac{dy}{dx}$ is always an operator and not a fraction, whereas $\frac{\delta y}{\delta x}$ is an infinitesimal change." The operator would be $d/dx$, not $dy/dx$. Also, it is actually valid to consider $dy/dx$ as the quotient of two infinitesimal numbers. That's how physicists, mathematicians, and engineers throught about it for hundreds of years after the invention of calculus, and Abraham Robinson proved ca. 1960 that it didn't lead to logical inconsistency. There is even a freshman calc book using this approach: math.wisc.edu/~keisler/calc.html $\endgroup$
    – user4552
    May 24, 2013 at 12:52
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    $\begingroup$ You should have mentioned your related question on math.SE. $\endgroup$ May 24, 2013 at 13:06
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    $\begingroup$ @b_jonas Don't forget $\eth$ either! :) $\endgroup$
    – Mike
    May 24, 2013 at 13:43
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    $\begingroup$ Related: physics.stackexchange.com/q/36150/2451 $\endgroup$
    – Qmechanic
    May 24, 2013 at 13:43

2 Answers 2


The symbol $\Delta$ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small.

The symbols $d,\delta$ refer to infinitesimal variations or numerators and denominators of derivatives.

The difference between $d$ and $\delta$ is that $dX$ is only used if $X$ without the $d$ is an actual quantity that may be measured (i.e. as a function of time) without any ambiguity about the "additive shift" (i.e. about the question which level is declared to be $X=0$). On the other hand, we sometimes talk about small contributions to laws that can't be extracted from a well-defined quantity that depends on time.

An example, the first law of thermodynamics. $$dU = \delta Q - \delta W$$ The left hand side has $dU$, the change of the total energy $U$ of the system that is actually a well-defined function of time. The law says that it is equal to the infinitesimal heat $\delta Q$ supplied to the system during the change minus the infinitesimal work $\delta W$ done by the system. All three terms are equally infinitesimal but there is nothing such as "overall heat" $Q$ or "overall work" $W$ that could be traced – we only determine the changes (flows, doing work) of these things.

Also, one must understand the symbol $\partial$ for partial derivatives – derivatives of functions of many variables for which the remaining variables are kept fixed, e.g. $\partial f(x,y)/\partial x$ and similarly $y$ in the denominator.

Independently of that, $\delta$ is sometimes used in the functional calculus for functionals – functions that depend on whole functions (i.e. infinitely many variables). In this context, $\delta$ generalizes $d$ and has a different meaning, closer to $d$, than $\delta$ in the example of $\delta W$ and $\delta Q$ above. Just like we have $dy=f'(x)dx$ for ordinary derivatives in the case of one variable, we may have $\delta S = \int_a^b dt\,C(t)\delta x(t)$ where the integral is there because $S$ depends on uncountably many variables $x(t)$, one variable for each value of $t$.

In physics, one must be ready that $d,\delta,\Delta$ may be used for many other things. For example, there is a $\delta$-function (a distribution that is only non-vanishing for $x=0$) and its infinite-dimensional, functional generalization is called $\Delta[f(x)]$. That's a functional that is only nonzero for $f(x)=0$ for every $x$ and the integral $\int {\mathcal D}f(x) \,\Delta[f(x)]=1$. Note that for functional integrals (over the infinite-dimensional spaces of functions), the integration measure is denoted ${\mathcal D}$ and not $d$.

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    $\begingroup$ I think that we could in principle consider the total heat supplied to/released from the system, as well as the overall work done, taken since some moment. The internal energy in this way is not much better, because we usually do forget about some parts of it (oscillational modes, rest mass, binding energies, etc). I am used to think that we dont write $dQ$ or $dW$ because this would imply that we deal with total differentials, that is, $Q$ and $W$ are 'functions of state'. They are not, however, and we write $\delta Q$ and $\delta W$ in order to stress this fact. $\endgroup$ May 24, 2013 at 14:57
  • $\begingroup$ Right, $Q$ and $W$ are not functions of state. That's equivalent to saying that the "total heat supplied to/released from the system" and "overall work done" since some moment depend on which moment we choose. $\endgroup$ May 25, 2013 at 11:01
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    $\begingroup$ Well, their sum also depends on this moment, but it nevertheless is a function of state -- it is $U$ plus some constant, which depends on the moment. $\endgroup$ May 25, 2013 at 11:08
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    $\begingroup$ Or difference, depending on your definitions. $\endgroup$ May 25, 2013 at 11:14

In many books, the difference between $d$ and $\delta$ is that, in the first case, we have the differential of a function and, in the second case, we have the variation of a functional.


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