I know that the units of a Dirac Delta function are inverse of it's argument, for example the units of $\delta(x)$ if $x$ is measured in meters is $\frac{1}{meters}$.

But, my question is what are the units inside the Dirac Delta?

For example if you have $\delta (x - 1)$ is the exact expression $x$ inside the dirac delta unitless? If not, that means that the $1$ would actually have to have units of meters. That seems odd to me, which is why I think that inside the units are unitless and as a whole you give the dirac delta inverse units of the argument but I'm not sure. If someone could clear this up for me I'd really appreciate it, thanks!


3 Answers 3


Let's clear up one thing first: you can never add (or subtract) two expressions with different units. For example, in $\delta(x - 1)$, the 1 is unitless, which means $x$ has to be unitless as well.

Then there's the separate issue of the units of the expression inside the Dirac delta. The expression inside the delta is its argument, and as you know, the argument does not have to be unitless. So there's your answer. You can write $\delta(x - 1\text{ m})$, for example, and since $x - 1\text{ m}$ has units of length, the delta function itself will have units of inverse length.

The reason the arguments of many functions have to be unitless is that those functions can be expressed as a power series,

$$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$$

where the $a_i$ are just numbers.1 For example,

$$\begin{align} \exp(x) &= 1 + x + \frac{1}{2}x^2 + \cdots \\ \sin(x) &= x - \frac{x^3}{6} + \cdots \end{align}$$

If $x$ had units of, say, length, then you would be adding a number to a length to an area (length squared) to a volume etc., and as I said, that can't happen.

But the delta function is not one of these functions that can be expressed as a power series. In fact, it's not really a function at all. It's a distribution, defined implicitly by the integral

$$\int_a^b f(x) \delta(x - x_0)\mathrm{d}x = \begin{cases}f(x_0), & a \le x_0 \le b \\ 0,& \text{otherwise}\end{cases}$$

In order for this integral to come out with the same units as $f(x_0)$, the rest of the expression being integrated - that is, the combination $\delta(x - x_0)\mathrm{d}x$ - has to be unitless. And since $\mathrm{d}x$ has the same units as $x$, $\delta(x - x_0)$ has to have the unit that cancels that unit out.

1One might argue that you can define a function $f$ as a power series where the coefficients $a_n$ do have units of the appropriate type, and in that case the argument $x$ would have units. But you can always generalize such a function to a different function of a unitless variable: just write $a_n = f_0 b_n/x_0^n$ where $b_n$ is unitless and $x_0$ has the same dimension as $x$, and then express the function as $$f(x) = \sum a_n x^n = f_0\sum b_n (x/x_0)^n = f_0 g(x/x_0)$$ where $g(y) = \sum b_n y^n$. The function $g$ expresses the same functional relationship as $f$, but in terms of a unitless variable, and is thus more generally useful.


Hint: Use the identity

$$ \delta(ax)~=~\frac{1}{|a|}\delta(x) $$

(where $a\neq 0$ is a non-zero real constant) to transport units in and out of the Dirac delta function $\delta(x)$.


You are correct, the delta function itself has the inverse dimension of its argument. The question is now: where does the argument come from? What determines its units? In principle, one can make up arbitrary expressions with different units for each term. But can this be interpreted physically? No. In many applications in physics, the delta function serves to restrict some expression to a point in space(time), i.e.


where $\mathbf{r}$ is a general position vector and $\mathbf{r'}$ represents the expression to which you want to restrict it. This can be used for example to model the density of a point charge in electrodynamics. Both quantities in the argument represent a position, i.e. they have the dimension of a length and therefore can be measured in meters. It would not make sense to assign a position to for example a velocity.

Therefore if you write down an expression like $\delta(x-1)$, you have to make sure that x is dimensionless (provided that the $1$ is taken to be dimensionless as well).


You can also think of it like this:

The dirac delta function only gives a non-zero value if its argument vanishes. In your case, this is given by the condition $x-1=0$ or $x=1$. This is an equation. An equation can only hold if the terms on both sides of the equality sign are equal. This also means that the units have to match. You cannot equate for example seconds to meters, or meters to unitless (dimensionless) quantity. Therefore all terms in the argument of the delta function must have the same dimension/unit.

  • $\begingroup$ Is it possible in your expression $\delta(r-r')$ that both $r$ and $r'$ contain dimensions? Or are they strictly numbers/vectors with no dimensions? $\endgroup$ Oct 10, 2013 at 4:39
  • $\begingroup$ It is not only possible, but necessary, as I have pointed out above. $\endgroup$ Oct 10, 2013 at 5:51

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