Units into Log and Exponential Functions

[terryphi]

Hello,

Something I've wondered about for some time is what happens to units once we pump them into a exponential or log function.

For example in neutron attenuation

I(x) = I_0 * exp(-Sigma * x)

I feel like this something I should know, but I just don't get it.

I suspect what's happening is that the units stay the same since basically all the function is doing is changing the magnitude of the scalar quantity associated with the unit, and not necessarily what the unit measures.

If that's so is there a word I can use to describe a function that does change a inputs unit (such as the aformentioned) and one that does not?

 

[ZapperZ]

Hello,

Something I've wondered about for some time is what happens to units once we pump them into a exponential or log function.

For example in neutron attenuation

I(x) = I_0 * exp(-Sigma * x)

I feel like this something I should know, but I just don't get it.

I suspect what's happening is that the units stay the same since basically all the function is doing is changing the magnitude of the scalar quantity associated with the unit, and not necessarily what the unit measures.

If that's so is there a word I can use to describe a function that does change a inputs unit (such as the aformentioned) and one that does not?

This is a bit puzzling to understand.

If I've read it correctly, then you might have a problem understanding that the argument for the exponential above must be dimensionless. This means that whatever units x has, sigma must have the inverse of that unit.

The same with logarithm.

Zz.

 

[terryphi]

Sorry,

an exponential function can only take a dimensionless argument?

 

[jtbell]

Yes.

 

[terryphi]

Sorry, I've never heard this, and I've used several equations which I'm relatively sure have had dimensional arguments.

Can you give some sort of explanation?

 

[S.Daedalus]

In a physical quantity, magnitude and dimension are inseparable -- whatever function you have of its magnitude also applies to its dimension. Trivial example: Square area A = x². If you know that [x] = m, you also know that [A] = m², i.e. that area is measured in m².

However, e^x is problematic -- what, exactly, is the exponentiation of a dimension supposed to mean? To better see the problem, expand into a powerseries: f(x) = e^x = 1 + x + x²/2! + x³/3! + ... If x now has dimension [dim], the first summand is dimensionless, the second of dimension [dim], the third of dimension [dim]², and so on. So, what dimension is whatever quantity is denoted by f(x) supposed to have? There's no unique assignment possible. Only if x is dimensionless does one get a unique dimension for f(x) -- which is none, as well.

 

[Nabeshin]

To better see the problem, expand into a powerseries: f(x) = e^x = 1 + x + x²/2! + x³/3! + ... If x now has dimension [dim], the first summand is dimensionless, the second of dimension [dim], the third of dimension [dim]², and so on. So, what dimension is whatever quantity is denoted by f(x) supposed to have? There's no unique assignment possible. Only if x is dimensionless does one get a unique dimension for f(x) -- which is none, as well.

Note that this extends to other non-polynomial functions too, like the sine/cosine functions, and anything that can be made out of them (cosh,sinh,spherical harmonics, bessel functions, etc.).

 

[ZapperZ]

Sorry, I've never heard this, and I've used several equations which I'm relatively sure have had dimensional arguments.

Can you give some sort of explanation?

What if you show us what equations you used, and I'm sure we can point out where you made your mistake of thinking they had a dimension.

Zz.

 

[sophiecentaur]

Sorry,

an exponential function can only take a dimensionless argument?

I imagine you would find it hard to evaluate e^{(5 bannanas)}.
That's why an exponential must be dimensionless.

 

[terryphi]

What if you show us what equations you used, and I'm sure we can point out where you made your mistake of thinking they had a dimension.

Zz.

Heh, I just realized you're right. Every function I can think of is dimensionless!

Thanks PF :D

 

[terryphi]

Ah, I've found one that doesn't fit.

The Gieger-Nuttall law.

http://scienceworld.wolfram.com/physics/Geiger-NuttallLaw.html

 

[jtbell]

The Geiger-Nuttall is an empirical numerical formula, obtained basically from curve-fitting to data. The values of the coefficients depend on the specific units used for time and energy.

Note that you can re-cast it into a form in which the argument of the logarithm is dimensionless, by writing copies of the equation for isotopes 1 and 2 and subtracting one from the other:

$$\ln {\frac{\lambda_2}{\lambda_1}} = -a_1 \left( \frac{Z_2}{\sqrt{E_2}}<br /> - \frac{Z_1}{\sqrt{E_1}} \right)$$

In this version the coefficient $a_2$ drops out. I don't have any appropriate textbooks to check here at home, but I suspect the derivation of the Geiger-Nuttall law by Gamow et al. produces something like what I wrote.

 

[DrDu]

I know of some examples where the question of dimensions is also not obvious namely physical chemistry where e.g. you can find definitions in textbooks like "the pH is the negative decadic logarithm of the concentration of hydronium ions".
However, in more careful texts you will see that what enters is always the quotient of the concentration (or activity to be more precise) and the concentration in some standard state, so that the argument of the logarithm is in fact dimensionless.

 

[sophiecentaur]

Surely a quantity may have dimensions but a number can have none. Is there anything more to be said?

 

[terryphi]

So is that a good measure of a book? If they have logs with units?

 

[sophiecentaur]

The book may have a lot of good and correct iknfo in it. It just didn't make it clear that the exponential was dimensionless.