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I was given the following instruction to follow when we find the units of the $\delta(x)$ function:

We can deduce these by inspecting the expression for its area: $∫_{−∞}^{∞}()=1$

Clearly the right-hand side of this is dimensionless

On the other hand, if $$ is a position, then $$ has the dimension of length

So the delta function must cancel out those length dimensions in order to leave a dimensionless right-hand side, i.e. $()$ must have dimensions of $(ℎ)^{−1}$

So we see a very important thing here: the physical dimensions of the delta function are the inverse of the physical dimensions of its argument.

A delta function of time, $()$, has dimensions $()^{−1}$; a delta function of momentum, $()$, has dimensions $()^{−1}$ and so forth

I thought that the right hand has the units (for example if $x$ is position) $m^2$. as we are finding the area under the graph. So in general the right hand would have the units of $(units)^{2}$

PhysicsQuestion
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    Unity can have dimension, it can be a unit charge (C), unit current (A), or unit something and delta function has to have a dimension of "density" for instance C/L, A/L. Sometimes considering the delta function is dimensionless is fine but the dimension needs to be fixed by its coefficient, namely the coefficient of the delta function needs to be physical quantity with dimension. – Yolbarsop Oct 11 '20 at 10:26
  • Yes, and this is what happens most of the time: the Dirac delta is a density without unit. The units can be hidden in other parts. – LL 3.14 Oct 12 '20 at 09:05

2 Answers2

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If $a(t)$ is acceleration and $t$ is time, will $\int a(t) \, dt$ have dimension $\text{length}^2$ since we are finding the area under the graph? No, it will not; it will have dimension $\text{length}/\text{time}$.

The interpretation of integral as area should not be taken too far.

md2perpe
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  • With acceleration, it's still true that you can multiply the units on the axes to get the units on the integral. (Or is your point that the integral can have units other than length squared, even though it's "an area"?) – Misha Lavrov Oct 11 '20 at 17:52
  • @MishaLavrov. Yes, my point is that the integral can have units other than length squared, and the "area" other dimension than length squared. – md2perpe Oct 11 '20 at 18:01
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Your are correct that the physical dimension of the delta function is the inverse of its arguments dimension.

The dimension of an integral is the product of the dimensions of the integrand and the integration variable.

Spencer
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