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I have a problem in analysis of laboratory data, when I have to take the logarithm of a physical quantity. I know that logarithm is a dimensionless quantity. I have to logarithmize the physical quantity $F=I/V^2$, where $I$ is the electric current and $V$ is voltage. We know that both current and voltage have uncertainty $\mathrm dI$, $\mathrm dV$. In order to find the uncertainty of $\ln(I/V^2)$, I use the formula with partial derivatives and i find that: $$ \mathrm Δ(ln(F))= \sqrt{\frac{\mathrm (dI)^2}{I^2} + 4\frac{\mathrm (dV)^2}{V^2}} $$ If I choose to write voltage and current in volts and amperes, I take a different result in the value of $$ln(F)=ln(\frac{\mathrm I}{V^2})$$than writing them in kilovolts and nanoamperes, for example, but the error has the same value as you can see! So I have a different relative error depending on unit of measurement!

Isn't that weird? What's the problem?

ang_88
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    Closely related: Why is it “bad taste” to have a dimensional quantity in the argument of a logarithm or exponential function?, and links therein. – Emilio Pisanty Dec 07 '17 at 13:26
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    Your formula for the uncertainties should be $\frac{dF}{F}=\frac{dI}{I}+2\frac{dV}{V}$. – Raskolnikov Dec 07 '17 at 13:37
  • https://physics.appstate.edu/undergraduate-programs/laboratory/resources/error-propagation Here on General function of multivariables it uses this formula – ang_88 Dec 07 '17 at 13:50
  • @Angelos Is your link basically using a Gaussian error model? – jjack Dec 07 '17 at 14:09
  • I made a big mistake in the initial text, i am not talking about uncertainty of F but the uncertainty of ln(F) – ang_88 Dec 07 '17 at 22:15
  • As long as the units of $dI$ match the units of $I$, and the units of $dV$ match the units of $V$, and you correct the formula to add in quadrature, you should get the right answer regardless of what units you use. If your answer is changing with units, then you're failing to match units in current or voltage. – Sean E. Lake Dec 07 '17 at 22:19
  • Uncertainty doesn't change with units, but the value of ln(F) depends on the units, isn't that correct? – ang_88 Dec 07 '17 at 22:23
  • @Angelos Yes, the value of $\ln(F)$ depends on units, but $$\operatorname{d}\ln(F) = \lim_{\Delta F\rightarrow 0} \left[\ln(F+\Delta F) - \ln(F)\right] = \lim_{\Delta F\rightarrow 0} \frac{\Delta F}{F}$$ doesn't. – Sean E. Lake Dec 07 '17 at 22:41
  • So, if i change the unit of measurement there will be a different relative error? – ang_88 Dec 07 '17 at 22:49
  • @Angelos No. The units cancel, as long as you use the same units for $F$ and $\mathrm{d},F$. – Sean E. Lake Dec 07 '17 at 22:53
  • You just said that the value of $ln(F)$ depends on units but $dln(F)$ doesn't. So, for every unit that i may choose, $dln(F)$ remains the same, but i have a different value for $ln(F)$. Relative error is the ratio of $dln(F)$ to $ln(F)$, which changes as $ln(F)$ changes and $dln(F)$ remains constant. What am i thinking wrong? – ang_88 Dec 07 '17 at 23:04
  • @Angelos Relative error is not $\frac{\operatorname{d}\ln(F)}{\ln(F)}$. Relative error is $\operatorname{d}\ln(F) = \frac{\operatorname{d}F}{F}$. – Sean E. Lake Dec 08 '17 at 00:06

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Your formula for the error is not correct, because it's not dimensionally consistent. Moreover, I could not find support for the formula you state in the link provided in the comments.

If $F=I/V^2$, $$dF=\left|\frac{\partial F}{\partial I}\right| dI+\left|\frac{\partial F}{\partial V}\right| dV=\frac{1}{V^2} dI+2\frac{I}{V^3} dV$$ The absolute values are due to the fact that errors always add. Therefore dividing by $F$: $$\frac{dF}{F}=\frac{dI}{I}+2\frac{dV}{V}$$

Usually, for complicated statistical reasons, it is more appropriate to add errors in quadrature, so that one should actually do: $$dF=\sqrt{\left(\frac{\partial F}{\partial I}dI\right)^2+\left(\frac{\partial F}{\partial V} dV\right)^2}=\sqrt{\left(\frac{1}{V^2} dI\right)^2+\left(2\frac{I}{V^3}dV\right)^2}$$ Dividing by $F$ one gets the result: $$\frac{dF}{F}=\sqrt{\frac{dI^2}{I^2}+4\frac{dV^2}{V^2}}$$ This is the method they use in the link you provided.

Both of the above formulas may be used depending on context. Note that both are dimensionally consistent (i.e. you only add terms with the same dimensions). I don't know if in your analysis you need the absolute error $dF$ or the relative error $dF/F$, that's for you to decide. But both can be obtained from the above. As to the matter of taking logarithms of dimensional quantities, it shouldn't matter as long as you use consistent units. See also the question linked in the comments.

John Donne
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  • If you're using the functional approach, there's no need to take derivatives, as in my answer. – JamalS Dec 07 '17 at 15:13
  • @JamalS There's many ways of doing this! I suggested this way because it's close to the approach of the website linked by the OP in the comments. Different ways are welcome :) – John Donne Dec 07 '17 at 15:18
  • There are disadvantages with the differential approach, as elucidated in the reference in my answer. – JamalS Dec 07 '17 at 15:19
  • In my opinion the OP was having problems with the formula he was using, not with the approach. I might be wrong – John Donne Dec 07 '17 at 15:22
  • I really made an important mistake in the initial text, I am really sorry, it didn't make any sense what i was trying to explain – ang_88 Dec 07 '17 at 22:17
  • It might be worth copying the links from the comments into your answer - comments have a tendency to disappear, this answer and the links deserves to stay. – Floris Dec 08 '17 at 17:19
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I would suggest using the functional approach to error propagation in this case. In general, in the majority of cases, it is what I would favour. For your case, the uncertainty in the $\log$ is,

$$\sqrt{\left[ \log \left( \frac{I+\delta I}{V^2}\right)-\log\left( \frac{I}{V^2}\right)\right]^2 + \left[ \log \left( \frac{I+\delta I}{V^2}\right)-\log\left( \frac{I}{(V+\delta V)^2}\right)\right]^2}$$

which simplifies nicely because of logarithms to,

$$ =\sqrt{[\log(I+\delta I) - \log(I)]^2 + 4[\log(V+\delta V) - \log(V)]^2}.$$

Notice there's no problem with dimensions in the uncertainty: we are taking logarithms of dimensionful quantities, but subtracting two logarithms is the same as taking the ratio of the argument, thus producing a dimensionless parameter. Explicitly,

$$=\sqrt{\log^2 \underbrace{\frac{I+\delta I}{I}}_{\mathrm{dimensionless}} + 4\log^2 \underbrace{\frac{V+\delta V}{V}}_{\mathrm{dimensionless}} }$$

For everything you'd need to know about errors for undergraduate experiments (assuming nothing fancy like Markov Chain Monte Carlo), see:

Measurements and their uncertainties: a practical guide to modern error analysis

JamalS
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