# Symbols of derivatives

What is the exact use of the symbols $\partial$, $\delta$ and $\mathrm{d}$ in derivatives in physics? How are they different and when are they used? It would be nice to get that settled once and for all.

$$\frac{\partial y}{\partial x}, \frac{\delta y}{\delta x}, \frac{\mathrm{d} y}{\mathrm{d} x}$$

• For what I know, $\mathrm{d}$ is used as a small infinitisemal change (and I guess the straight-up letter $\mathrm{d}$ is usual notation instead of italic $d$, simply to tell the difference from a variable).
• Of course we also have the big delta $\Delta$ to describe a finite (non-negligible) difference.
• And I have some vague idea that $\partial$ is used for partial derivatives in case of e.g. three-dimensional variables.
• Same goes for $\delta$, which I would have sworn was the same as $\partial$ until reading this answer on Math.SE: https://math.stackexchange.com/q/317338/

Then to make the confusion total I noticed an equation like $\delta Q=\mathrm{d}U+\delta W$ and read in a physics text book that:

The fact that the amount of heat [added between two states] is dependent on the path is indicated by the symbol $\delta$...

So it seems $\delta$ means something more? The text book continues and says that:

a function [like the change in internal energy] is called a state function and its change is indicated by the symbol $\mathrm{d}$...

Here I am unsure of exactly why a $\mathrm d$ refers to a state function.

So to sum it up: down to the bone of it, what is $\delta$, $\partial$ and $\mathrm{d}$ exactly, when we are talking derivatives in physics.

Addition

Especially when reading a mathematical process on a physical equation like this procedure:

$$\delta Q=\mathrm{d}U+p\mathrm{d}V \Rightarrow\\ Q=\Delta U+\int_1^2 p \mathrm{d}V$$

It appears that $\delta$ and $\mathrm{d}$ are the same thing. An integral operation handles it the same way apparently?

• – Qmechanic Dec 17 '14 at 16:21
• @HDE226868 I don't believe so. This answer math.stackexchange.com/q/317338 gives the mathematical answer. But the conclusion is e.g. that $\delta$ is never used in math. I would like to know exactly what the symbols mean when used in physical equations. – Steeven Dec 17 '14 at 16:21
• @Steeven are you sure $\delta$ is never used in math? See functional derivative. It's not used in usual differential calculus though indeed. – Ruslan Dec 17 '14 at 17:03
• @Ruslan: No. But this 34.2k rep Math.SE user is: math.stackexchange.com/a/317345/13230 – Steeven Dec 17 '14 at 18:06
• @Steeven not quite. "never used in mathematics in particular context" is quite different than "never used in mathematics at all". – Ruslan Dec 17 '14 at 18:45

## 2 Answers

Typically:

• $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$
• $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ at constant $y$). This is sometimes denoted by $$f_{,x},\,f_x,\,\partial_xf$$
• $\delta$ is for small changes of a variable, for example minimizing the action $$\delta S=0$$ For larger differences, one uses $\Delta$, e.g.: $$\Delta y=y_2-y_1$$

NB: These definitions are not necessarily uniform across all subfields of physics, so take care to note the authors intent. Some counter-examples (out of many more):

• $D$ can denote the directional derivative of a multivariate function $f$ in the direction of $\mathbf{v}$: $$D_\mathbf{v}f(\mathbf{x}) = \nabla_\mathbf{v}f(\mathbf{x}) = \mathbf{v} \cdot \frac{\partial f(\mathbf{x})}{\partial\mathbf{x}}$$
• More generally $D_tT$ can be used to denote the covariant derivative of a tensor field $T$ along a curve $\gamma(t)$: $$D_tT=\nabla_{\dot\gamma(t)}T$$
• $\delta$ can also represent the functional derivative: $$\delta F(\rho,\phi)=\int\frac{\delta F}{\delta\rho}(x)\delta\rho(x)\,dx$$
• The symbol $\mathrm{d}$ may denote the exterior derivative, which acts on differential forms; on a $p$-form, $$\mathrm{d} \omega_p = \frac{1}{p!} \partial_{[a} \omega_{a_1 \dots a_p]} \mathrm{d}x^a \wedge \mathrm{d}x^{a_1} \wedge \dots \wedge \mathrm{d}x^{a_p}$$ which maps it to a $(p+1)$-form, though combinatorial factors may vary based on convention.
• The $\delta$ symbol can also denote the inexact differential, which is found in your thermodynamics relation$${\rm d}U=\delta Q-\delta W$$ This relation shows that the change of energy $\Delta U$ is path-independent (only dependent on end points of integration) while the changes in heat and work $\Delta Q=\int\delta Q$ and $\Delta W=\int\delta W$ are path-dependent because they are not state functions.
• For the hapless future student: note the word "typically" at the start of this answer ;) – DanielSank Dec 17 '14 at 20:26

First, I want to say that different people use different notation and I welcome any comments. I also feel as if I am about to enter a minefield.

Here the answer is made up with examples of use of $d$, $\partial$ and $\delta$.

I would say for $d$ that

$dV \over dx$

would be the total derivative in one dimension for $V(x)$ where the potential $V$ is a function of only one variable, $x$.

If $V$ is a function of two or more varaibles, say $x$ and $y$, then we have $V(x,y)$ and when it is differentiated with respect to $x$ and $y$ we get

$\partial V \over \partial x$ and $\partial V \over \partial y$

if we differentiate again we can get

$\partial^2 V \over \partial x^2$, $\partial^2 V \over \partial x\partial y$ and $\partial^2 V \over \partial y^2$ and so forth.

Finally, for $\delta$, I would say that $\delta$ represents something small, but not infinitessimal. So for example if $y=x^2$ and we increase $x$ by a small ammount to $x + \delta x$ the value of $y$ becomes $y + \delta y$ and we can write

$$y + \delta y = (x + \delta x)^2 = x^2 + 2x\delta x + \delta x^2$$

now because $y = x^2$ we can simplify this to give

$$\delta y = 2x\delta x + \delta x^2$$

and then divide both sides by $\delta x$ to get

$${\delta y \over \delta x} = 2x + \delta x \approx 2x$$

Now if we make the $\delta x$ vanishingly small (or infinitessimally small) we write it as $dx$ and our equation above becomes

$${d y \over d x} = 2x + d x = 2x$$

or

$${d y \over d x} = 2x$$

because $dx$ is so small it is effectively zero.

Finally, some other uses. In themodynamics we sometimes have $dU$ or $TdS$ where $d$ is meant to be 'a vanishingly small bit of'. The distinction between $\delta$ and $d$ you describe in the question is not one I was familiar with - it makes sense though as the author is wanting to draw the distinction between path dependent and path independent quantities - clearly in that example the both $d$ and $\delta$ are infinitessimal. In experimental physics, $\delta$ may be used to represent experimental error (or uncertainty) in a value e.g. $\lambda \pm \delta \lambda$ - this fits with $\delta$ being small, but not vanishingly small.

## protected by Qmechanic♦Dec 17 '14 at 17:09

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?