# What are the dimensions for the Heaviside Step Function?

The Heaviside step function is defined as

\begin{array}{ll} H(x) = 0 && \text{if }x<0 \\ H(x) = 1/2 && \text{if } x=0 \\ H(x) = 1 && \text{if } x>0 \end{array}

However, the function can also be defined as $$\int H(x) \phi'(x)dx = -\phi(0) \,. \tag{1}$$ It's clear that $$-\int \delta(x) \phi(x)dx = -\phi(0), \tag{2}$$ and a property of distributions gives $$\int \theta(x) \phi'(x)dx = -\int \theta'(x) \phi(x)dx \, .\tag{3}$$ Finally, the delta function always has the same dimensions as the inverse of its argument.

Given these facts, what are the dimensions for the Heaviside Step Function?

The Heaviside step function is given (and sometimes defined I think) to be the integral of the Dirac delta. Does that mean that its dimensions are that of the Dirac Delta's multiplied by the dimensions of the variable of integration? That would make it dimensionless. What am I missing here?

The step function is dimensionless. We can immediately see this using its basic definition. We can also see it using the distribution definition, $$\int H(x) \phi'(x) dx = -\phi(0).$$ The dimensions of the left-hand side are $$[H] \frac{[\phi]}{[x]} [x] = [H] [\phi]$$ where $[x]$ is the dimension of the variable of integration. The right hand side has dimension $[\phi]$, so $[H]$ is trivial, and $H$ has no dimensions.
• Also easily seen from the fact that $H(x>0)=1$ and therefore $[H(x>0)]=1$. – Emilio Pisanty Aug 15 '16 at 0:12
• I think maybe the $1$ could carry units though--like $1 kpr * 1 hr = 1k$ ?? :) – user45664 Aug 15 '16 at 0:29