# 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.

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

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.