Can the argument of a function contain physical units?

Question

Consider the function of temperature as a function of position, $T(x,y,z)$. Is the argument of the function dimensionless? I mean, when we want to find the value of the function at the position $x=1m$, $ y=1m$, $z=1m$ should we do the substitution $$T(1m,1m,1m)$$ or the substitution $$T(1,1,1)$$Does it make sense to consider the function $T(x,y,z)$ as a mapping from $\mathbb{R^3}$ to $\mathbb{R}$, considering also that the output value must have units of temperature?

Answer 1

Yes, it makes sense. The input is a position, which you will denote with a value with a length unit.

As a simple example, imagine you have a 1D region whose temperature varies linearly with position, we would write

$$T(x)=\alpha x+T_0$$ where $\alpha$ would have dimensions of temperature per length.

Of course, if you want to express your input in terms of some dimensionless variable $\xi=x/x_0$ then that is fine too. Then you would have

$$T(\xi)=\alpha'\xi+T_0$$ where $\alpha'=\alpha x_0$ now just has units of temperature.

So, in conclusion, either is fine as long as what actually defines your function is consistent with dimensions/units.

Answer 2

The only important thing is that physical equation must contain terms whose physical dimensions are homogeneous. It is a consequence of such a request that the arguments of any mathematical special functions (elementary or not) which can be defined through a power series must be dimensionless. But the same doesn't hold if the dependence on the variables is through a function composition so that every actual argument of a special function is the ratio of the an independent variable with dimension and suitable constant with the same physical dimension. Formally it is equivalent to say that any compact formula like $ f(x,y,z)$ should be intended as $f(x,y,z)=\Phi(x/\alpha,y/\beta,z/\gamma)$, with $\alpha,\beta,\gamma$ being constants homogeneous to $x,y,z$ respectively.

Notice that in the case of a logarithm, the formula may need an implicit term to be dimensionally correct. For instance, if $x$ is a quantity with dimensions, it is perfectly possible to have a quantity defined as $$ A = \log \int \mathrm{exp}(-x/\lambda ) \mathrm{d} x, $$ provided only differences of $A$ have a physical meaning.

Answer 3

It depends on the form of the function. Another answer gives the example of a temperature that varies linearly with position (perhaps a rod whose ends are connected to different-temperature baths),

$$ T(x)= \alpha x + T_0 $$

in which the dimension of the position $x$ depends on the units chosen for $\alpha$. A more famous example comes from the kinematics of motion with constant acceleration:

$$ x(t) = x_0 + v_0 t + \frac 12 a t^2 $$

In this case, a derivation of the relationship from calculus gives the coefficients natural meanings and natural units. An important part of the first-year physics experience is the problem where this kinematic equation is the right relationship to use, but the initial velocity is given in miles per hour, the acceleration given in meters per second squared, and the relevant time given in minutes. The beginning student puts "the numbers" into "the formula" and gets "the answer," which is wrong. This kinematic equation is a relationship among dimensionful quantities.

(The experience is traumatic for some aspiring physicists; you can usually provoke an involuntary shudder by whispering "let $g$ equal 32 ft/s$^2$" in their ear.)

However, there is a caveat about dimensionful quantities which arises when you start solving problems with differential equations. For example, the intensity of a radiation field as you move a distance $\ell$ into an absorber is given by

$$ I(\ell) = I_0 e^{ -n\sigma\ell} $$

in which $I_0$ is an initial intensity, $n$ is a number density, and $\sigma$ is a cross-sectional area. But you will never find a relationship like $e^x$ where $x$ is dimensionful, because exponentiation is a transcendental function, not an algebraic one. The mathematical exponential obeys a power-series relationship,

$$ e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots, $$

and the only way to for this series to be dimensionally consistent is for $x$ to be a dimensionless number.

The same restriction holds for the arguments of all transcendental functions, including trig functions and logarithms. This is why physicists tend to prefer the radian (defined as a ratio of two lengths) over the degree for measuring angles --- especially if the subject of discussion is a phase, rather than a geometric angle.