Unit conversion and derivatives

[Guest]

This is about a simple unit conversion but I just seem to be getting lost.

I am told that Z is a "dimensionless coordinate with units of length" L i.e. L might 5m for example.
I'm then given a function to calculate a distance f(Z), which I think is also in units of L.
I need to calculate this function, but I also need the input and output to be in SI units.
I also need to take the 2nd derivative of f(Z), but wrt. Z measured in SI units.

Am I right in thinking the following (where " ' " denotes SI units):

Z' = L * Z

f'(Z') = L * f(Z)

and the second derivative I want is just
L * (d/dz)^2 (f(Z))

Or is it more complicated than that?

[Guest]

Who told you that? " a dimensionless coordinate with units of length" makes no sense. A "unit of length" is a "dimension" in this sense! If L is a constant then, yes, [tex]\frac{dLf(z)}{dz}= L\frac{df}{dz}[/tex] and [tex]\frac{d^2lf(z)}{dz}= L\frac{d^2f}{dz^2}[/tex]. That has nothing to do with the "units".

The question of what units the derivative and second derivative have depends on the units for f(z). If, for example, f has units of "length squared" and z has units of "length" then [tex]df/dz[/tex] has units of (length squared/length)= length and [tex]d^2f/dz^2[/tex] has no units. If L also has units of length then [tex]Ldf/dz[/tex] has units of length squared and [tex]L d^2f/dz^2[/tex] has units of length.

[Gedi2000]

Hi, ai was the one who asked the question earlier. Yes, the wording didn't make much sense to me either but I got there in the end, just wanted the thank the response.

My question wasn't really about dimensional analysis like length or area etc. it was about whether my length is going to be in metres or in L's. I think in this case the scaling by L was correct.

[Guest]

"Metres" are a unit of length. But what does "units of L" mean? Do you mean "the same units that L is measured in"? You say that L is measured in metres but in order to determine the units of [tex]\frac{dL}{dz}[/tex] or [tex]\frac{d^2L}{dz^2}[/tex] you need to know the units of z also.

Since units will "pass through" limits, the units of [tex]\frac{df}{dx}[/tex] will be the units of f divided by the units of x and [tex]\frac{d^2f}{dx^2}[/tex] will be the units of L divided by the units of x squared.

For instance if f is measured in "metres squared" and x is measured in metres then [tex]\frac{df}{dx}[/tex] has dimension of metres and [tex]\frac{d^2f}{dx^2}[/tex] is dimensionless. If f is measured in metres and t is measured in seconds, then [tex]\frac{df}{dt}[/tex] has dimensions of "metres per second" and [tex]\frac{d^2f}{dt^2}[/tex] has units of "metres per second per second" or "metres per second squared".