This should be easy Units of log graph?

[IN88]

I have a question that sounds easy but as of yet no one has been able to give me a satisfactory explanation. I made a plot of mobility (cm^2/Vs) against temperature (K), i then wanted to find an expression that showed the relationship between the two. What i did was took the natural logarithm of both mobility and temperature and plotted that. This gave me a straight line (R^2 = 0.9997). So i then rearranged this straight line equation to see how mobility related to temperature. I have attached a file with the plots, data and my working out. I came up with the equation: mobility = Const/(Temp^0.63). This equation does give me the values of mobility (in cm^2/Vs) when i put in temperatures (in Kelvin) that i was looking for. However, when i was thinking about the dimensions of the equation it did not seem to make physical sense. When you log or take the exponent of something, the something has to be dimensionless. I have asked some people about this, someone said that most of the time this sort of thing is just ignored and someone else told me to take the log of a ratio, to have T/T0 where T0 is some sort of reference temperature. However in my system there is no physics reason to have a reference temperature and i am happy with the equation that i have come up with as it gives me the values i was looking for. Its just the dimensions that i don't understand. If you log something with units, then take the exponent of it, does the thing you get back with have the same units as when you started? how do you go about logging things with units?

 

[Rap]

Strictly speaking, the logarithm of a dimensioned quantity is not consistent, you cannot do it. You have to divide by a standard unit if you are doing dimensional analysis, and if a and b have the same dimensions, you cannot say log(a/b)=log(a)-log(b). Usually you can just pick the standard unit to be unity, one unit of whatever system of units you are working in, so if you really had to justify your units when making, say, a log plot, you could say the horizontal axis is log(T/To) where To=1 degree, and the vertical axis is log(mu/muo) where muo equals 1 cm^2/volt/sec. Usually no one bothers with making this explicit.

 

[IN88]

Ok, thanks for the quick reply. So just to be clear its ok for me to write the equation:

mu = Const/(T^0.63)

and that makes sense? then if someone asks about units i just say that mu is actually mu/mu0, and T is actually T/T0 and that is the way it had to be in order for it to be dimensionless and so i could log it in the first place?

 

[netheril96]

Ok, thanks for the quick reply. So just to be clear its ok for me to write the equation:

mu = Const/(T^0.63)

and that makes sense? then if someone asks about units i just say that mu is actually mu/mu0, and T is actually T/T0 and that is the way it had to be in order for it to be dimensionless and so i could log it in the first place?

Your constant has a dimension.

 

[Rap]

Ok, thanks for the quick reply. So just to be clear its ok for me to write the equation:

mu = Const/(T^0.63)

and that makes sense? then if someone asks about units i just say that mu is actually mu/mu0, and T is actually T/T0 and that is the way it had to be in order for it to be dimensionless and so i could log it in the first place?

Well, you forgot the electric field. I would say

\mu=KE/T^a

where E is electric field, T is temperature, K is a (dimensioned) constant and mu is mobility. Its perfectly fine to write it this way, but taking logarithms of each side is, strictly speaking, not dimensionally correct. Basically, you can just ignore this detail, but to show you understand the dimensional problem, don't write anything like log(200 K), write log(200) instead. You can go along and write log(T) keeping in mind you are really writing log(T/To) where To=1 K. Nobody will fault you for writing it this way unless you are making a dimensional argument, which you are not. When making a graph, maybe it would be better to plot your data on log-log axes, that way the labels will still be T and mu, but the power law relationship will be clear.

 

[nasu]

The mobility is independent of the field, in the first order at least. This is the point in introducing it: the drift velocity is proportional to the field and the proportionality constant is called mobility.

 

[Rap]

The mobility is independent of the field, in the first order at least. This is the point in introducing it: the drift velocity is proportional to the field and the proportionality constant is called mobility.

ooops - right. I misremembered the equation. Anyway, the dimensional stuff is the same.