# Does the logarithm of a non-dimensionless quantity make any sense?

A train consists of an engine and $$n$$ trucks. It is travelling along a straight horizontal section of track. The mass of the engine and of each truck is $$M$$. The resistance to motion of the engine and of each truck is $$R$$, which is constant. The maximum power at which the engine can work is $$P$$.

The train starts from rest with the engine working at maximum power. Obtain an expression for the time $$t$$ taken to reach a given speed $$v$$.

I wrote $$a(t)=\frac{P}{v(t)M(n+1)}-\frac{R}{M} \tag{1}$$

Putting $$(1)$$ into standard differential form: $$[M(n+1)v]dv+[(Rv(n+1)-P)]dt=0 \tag{2}$$ Since $$(2)$$ is non-exact, let $$M(n+1)v=f(v) \tag{3}$$ and $$(Rv(n+1)-P)=g(v) \tag{4}$$ Since $$\frac{1}{g}\left(\frac{\partial f}{\partial t}-\frac{\partial g}{\partial v}\right)=\frac{R(n+1)}{P-Rv(n+1)}=h(v) \tag{5}$$ i.e. a function of $$v$$ only.

The integrating factor to $$(2)$$ is then given by: $$I(v)=e^{\int h(v)dv}=e^{-\ln(P-Rv(n+1))}=\frac{1}{P-Rv(n+1)} \tag{6}$$ The final solution then looks something like this $$M(n+1)\left[\frac{-Rv(n+1)-P\ln(P-Rv(n+1))+P\ln(P)}{R^2(n+1)^2}\right]-t=0 \tag{7}$$

But doesn't the argument of $$\ln()$$ have to be some dimensionless quantities for it to make sense? (I got $$\ln(P)$$ and $$P$$ is not dimensionless in this case.)

Can someone please explain where my conceptual errors lie?

In equation (7) you have the expression $$−P\ln(P−Rv(n+1))+P\ln(P)$$

But since this a difference between two logarithms you can rewrite the expression (remember $$\ln a - \ln b = \ln \frac ab$$) as $$P\ln\left(\frac{P}{P−Rv(n+1)}\right)$$

Now you have the logarithm of a dimension-less quantity, as it should be.

The reason behind this problem is that you've not yet simplified the final expression.

For example, let's suppose you get a term $$\ln (f(v))$$ in your final indefinite integral, where $$f(v)$$ has dimensions and isn't dimensionless. This is, as you noted, weird as a logarithm's arguments should always be dimensionless. But, now if you apply the limits, you get

$$\ln(f(v))\biggr|_{v_1}^{v_2} = \boxed{\ln\left(\frac{f(v_1)}{f(v_2)}\right)}$$

Now, as you see, the boxed expression is perfectly valid. The argument in the logarithm is, as expected, dimensionless. So, there will never be a case where you'd encounter an expression like $$\ln(\text{quantity with dimension})$$ if you apply the limits and then analyze the expression.

$$I(v)=e^{\int h(v)dv}=e^{-\ln(P-Rv(n+1))}=\frac{1}{P-Rv(n+1)} \tag{6}$$

If $$v$$ is has non-trivial dimensions, then $$\int \frac{1}{v} dv = \ln\left|\frac{v}{D}\right|$$, where $$D$$ is equivalent to $$e^{-C}$$ in the dimensionless case: $$\int \frac{1}{x} dx = \ln\left|x\right| + C = \ln\left|x\right| - \ln e^{-C}$$.

If $$v$$ is in $$\left.\mathrm{m}\middle/\mathrm{s}\right.$$, for example, we could write:

$$\int \frac{1}{v} dv = \ln\left|\frac{v}{1\ \left.\mathrm{m}\middle/\mathrm{s}\right.}\right| + C$$

Yes, you can take the logarithm of a dimension. It's basically the same thing as taking the square of a dimension: it's mathematically valid and makes sense, but, obviously, it's not generally equivalent to the dimension itself. For example, both $$\mathrm{K}^2$$ and $$\ln{\left(\mathrm{K}\right)}$$ make sense, but neither is equivalent to $$\mathrm{K} .$$

If you get stuck, you can remember that $$\ln{\left(ab\right)} ~=~ \ln{\left(a\right)} + \ln{\left(b\right)} \,,$$ so you can rewrite any log of a scalar-dimension-having quantity as $$\ln{\left(x\right)} ~=~ \underbrace{\ln{\left(\frac{x}{\operatorname{dim}{\left(x\right)}}\right)}}_{\begin{array}{c}\text{dimensionless}\\[-25px]\text{factor}\end{array}} ~+~ \underbrace{\ln{\left(\operatorname{dim}{\left(x\right)}\right)}}_{\begin{array}{c}\text{isolated}\\[-25px]\text{units}\end{array}} \,.$$

The resulting math works like always, where the general rule's that both sides of an equation must be equal for the equation to hold. So if you end up with $$1 + \ln{\left(\mathrm{K}\right)} = 2 + \ln{\left(\mathrm{K}\right)} - 1 \,,$$ that's perfectly legal, as the $$ \ln{\left(\mathrm{K}\right)} "$$ cancels out on both sides, satisfying the equality. Of course, if the units don't cancel, then there's a dimensional error.

Note that the heuristic against adding terms with differing dimensions no longer holds. That heuristic only works when valid modes of constructing sums with terms of differing dimensions are avoided, which isn't the case here. This may lead to confusion in non-technical settings. For example, probably don't want to present a PowerPoint in a business meeting that refers to $$ 100 + \ln{\left(\mathrm{USD}\right)} ";$$ the math can be useful if you're doing calculations, but some folks find it confusing.

• This is non-standard, do you have a reference where it is worked out more rigorously? – fqq May 25 at 16:47
• Anyway, requiring to have the right "+ln(dim)" terms on both sides seems basically a way to require (the more standard) adimensionality – fqq May 25 at 16:48
• No, it's not "intro-level". The standard answer would be that log of dimensional quantities don't make sense. It might well be the case that you can define an appropriate logarithm function on some algebraic structure representing dimensional quantities, but it is not the usual logarithm and you have to prove its properties etc (e.g. the identity you write is valid for the natural logarithm, but fails already with complex numbers) – fqq May 25 at 16:56
• Otherwise, I agree that what you say works, but it's just formal/sloppy notation for keeping the adimensional quantities coherent. What's log(metre)? – fqq May 25 at 16:57
• I hadn't thought of things in terms of adding logarithms, but could see that it would make sense that if one regarded what would often be written as "5dBm" as instead "5+dBm", then converting signal it to dBW would require adding (dBW-3-dbM), thus yielding "2+dBW". – supercat May 25 at 18:16