Why should arguments of certain functions be dimensionless?

[Nirmal Padwal]

Homework Statement

Solutions to physical problems often involve functions like
$$ cos u, sin u, e^u, ln u, · · · $$
where the argument $u$ is a scalar that depends on physical quantities such as time,
frequency, mass, etc. Explain why u must be dimensionless (that is, independent of
the system of units used for mass, length, and time).

Relevant Equations

.

This is my approach:

These quantities namely mass, length, and time, are all additive in nature. $2 m + 3m = 5 m$. If the argument of the functions mentioned in the problem statement is not dimensionless, then mass, length or time do not remain additive in the image space of the functions. $e^{2 m} + e^{3m} \neq e^{5m}$

Furthermore, if the arguments are dimensionless, the functions which were originally dimensionless have to be now themselves will have to be associated with physical quantities.

Is this reasoning valid?

 

[wrobel]

$\cos(u)=1-u^2/2+\ldots$ so you can not summate 1 and dimensional variable

Regarding log the situation is little bit more interesting. You can write $\log a-\log b$ if $a,b$ are of the same dimension

 

[Nirmal Padwal]

Thank you for your answer. It makes more sense than mine. So was my approach completely incorrect?

 

[haruspex]

Is this reasoning valid?

No. You could say exactly the same if the arguments are dimensionless.

The thing you are asked to prove is not strictly true. Consider e.g. $e^{\lambda t}$, where t is a time and λ is a rate, like fissions per unit time. $\lambda t$ is dimensionless, so ok. But we could rewrite it as $(e^{\lambda })^t$. This appears to violate the principle.
The resolution is that $e^{\lambda }$ has dimension, but rather a strange one. Where λ has dimension T^-1, we could write the dimension of $e^{\lambda }$ as $e^{T^{-1}}$. When this is subsequently raised to the power of t, provided the time units are the same, it all comes out in the wash.

 

[Nirmal Padwal]

Thank you for answering

 

[Orodruin]

The main result in dimensional analysis, the Buckingham pi-theorem, explains this. It is a quite powerful statement about the possible forms of physical relationships.

Edit: I also wrote an insight on this some time ago.

 

[haruspex]

The main result in dimensional analysis, the Buckingham pi-theorem, explains this. It is a quite powerful statement about the possible forms of physical relationships.

Edit: I also wrote an insight on this some time ago.

I don't see that it follows from that theorem. See post #4.

 

[haruspex]

Thank you for answering

I should also have discussed the trig functions.
If we are defining cos(x) as the limit of $1-\frac 1{2!}x^2+\frac 1{4!}x^4..$ then clearly x must be dimensionless since the different powers of x would have different dimension.
... unless, that is, it has some funny kind of dimension such that it becomes dimensionless when raised to an even power. See https://www.physicsforums.com/insights/can-angles-assigned-dimension/

 

[Orodruin]

I don't see that it follows from that theorem. See post #4.

It does, the Buckingham pi-theorem essentially states that any physical relationship must be possible to write as a level curve of a function of a number of dimensionless quantities $f(\pi_1,\ldots,\pi_k) = 0$. This means that if the variable I am interested in is in $\pi_1$ only (this can always be arranged), then
$$
\pi_1 = g(\pi_2, \ldots, \pi_k),
$$
where $g$ is a dimensionless function of dimensionless variables. In other words, if you are interested in the variable $a$ where $\pi_1 = a/b$ ($b$ being some combination of other variables that make $\pi_1$ dimensionless), then
$$
a = b g(\pi_2, \ldots, \pi_k).
$$
Here, since $g$ is a dimensionless function of dimensionless variables $\pi_i$, you have a normalising variable of the correct dimension ($b$) multiplied by a dimensionless function of dimensionless combinations.

Edit: In other words, any functional form (apart from a multiplicative factor) must be a function of dimensionless parameters.

 

[haruspex]

It does, the Buckingham pi-theorem essentially states that any physical relationship must be possible to write as a level curve of a function of a number of dimensionless quantities $f(\pi_1,\ldots,\pi_k) = 0$. This means that if the variable I am interested in is in $\pi_1$ only (this can always be arranged), then
$$
\pi_1 = g(\pi_2, \ldots, \pi_k),
$$
where $g$ is a dimensionless function of dimensionless variables. In other words, if you are interested in the variable $a$ where $\pi_1 = a/b$ ($b$ being some combination of other variables that make $\pi_1$ dimensionless), then
$$
a = b g(\pi_2, \ldots, \pi_k).
$$
Here, since $g$ is a dimensionless function of dimensionless variables $\pi_i$, you have a normalising variable of the correct dimension ($b$) multiplied by a dimensionless function of dimensionless combinations.

Edit: In other words, any functional form (apart from a multiplicative factor) must be a function of dimensionless parameters.

Ok, my mistake was failing to read this part properly: "Solutions to physical problems".
I am used to performing the sanity check that if e.g. I have a term e^expression then expression should be dimensionless. But I am aware that in principle it need not be, because the term as a whole need not represent a physical quantity (as in my example in post #4).