In general one can convert from one unit to another (of the same dimension) by using (($U$,$A$) and ($V$,$B$) being compatible unit pairs in the following)

$xU=(x\ast f_{U\rightarrow A}+o_{U\rightarrow A})A=yA$

Oftentimes the offset $o_{U\rightarrow A}$ will have a value of zero, which reduces the conversion to a simple multiplication with a factor $f_{U\rightarrow A}$. For some units like temperatures, however, we need that offset as well.

There are a few compound units in use, which are combinations of a temperature and some other unit like joule per kelvin or degree Celsius day.

My question is how would I convert those compound units unambiguously?

To be more specific: Assume we want to convert from a (compound) unit $UV$ to a unit $AB$ and are given the following conversion formulas:

  • $xU=(x\ast f_{U\rightarrow A}+o_{U\rightarrow A})A$
  • $xV=(x\ast f_{V\rightarrow B})B$

The conversion formula from $x UV = y AB$ seems to depend on the order of conversions:

If converting first from $U$ to $A$ and then from $V$ to $B$ we end up with the following calculation

$xUV\\ =(x\ast f_{U\rightarrow A}+o_{U\rightarrow A})AV \\ =((x\ast f_{U\rightarrow A}+o_{U\rightarrow A})A\ast f_{V\rightarrow B})B \\ =(x\ast f_{U\rightarrow A}\ast f_{V\rightarrow B}+o_{U\rightarrow A}\ast f_{V\rightarrow B})AB$

Doing it the other way around, we end up with

$xUV\\ =(x\ast f_{V\rightarrow B})UB\\ =(x\ast f_{V\rightarrow B}\ast A\ast f_{U\rightarrow A}+o_{U\rightarrow A})B\\ =(x\ast f_{V\rightarrow B}\ast f_{U\rightarrow A}+o_{U\rightarrow A})AB$

So depending on the order of conversions we end up in results, which differ by

$\begin{array}{l}(x\ast f_{U\rightarrow A}\ast f_{V\rightarrow B}+o_{U\rightarrow A}\ast f_{V\rightarrow B})-(x\ast f_{V\rightarrow B}\ast f_{U\rightarrow A}+o_{U\rightarrow A})\\=o_{U\rightarrow A}\ast f_{V\rightarrow B}-o_{U\rightarrow A}\\=o_{U\rightarrow A}\ast(f_{V\rightarrow B}-1)\end{array}$

Can somebody explain to me, where I'm going wrong here or how this ambiguity is resolved in practice?


2 Answers 2


Note that those compound units are used to express rates of change: change in energy per unit of temperature, or change in temperature per unit of time (day). That is, $\Delta E/\Delta T$ and $\Delta T / \Delta t$. The offset term cancels out when taking the differences.

  • $\begingroup$ The above mentioned degree Celsius day does not follow that scheme. I'm not actually sure where this unit is applied, however it is part of the ontology I have to work with. Also there are others like Kelvin meter. $\endgroup$
    – Sirko
    Commented Apr 20, 2017 at 16:06
  • $\begingroup$ I have to admit that I am not familiar with the notation, and I'm not sure I'm following it correctly, and I may be missing the point of the question. But why do you compound these units? Aren't the conversions independent of each other? Don't you simply do each conversion on its own value and then multiply the results? If so, there's no problem with ordering. $\endgroup$
    – garyp
    Commented Apr 20, 2017 at 20:34
  • $\begingroup$ Regarding the reason: I have some formulae with already given units for the operands. Although I have not encountered a case so far, those input units might be compound units including scaled units. I was just considering the general case, where such compound units might need to be converted and then stumbled upon the described issue. At which point is the notation unclear? It's done just by me (no other reference), but imagined it to be straight forward. If there is a more common notation for this, I'm happy to adjust the question accordingly. $\endgroup$
    – Sirko
    Commented Apr 21, 2017 at 13:15
  • $\begingroup$ The notation is probably quite clear; I just haven't taken the time to completely understand it. But my question is: for the two example cases that you provided, why is not the answer $(xU)(xV)$? $\endgroup$
    – garyp
    Commented Apr 21, 2017 at 17:59
  • 1
    $\begingroup$ Conclusion: I would say that as far as physics is concerned, there is no ambiguity because either we are dealing with an absolute temperature and it has to be in Kelvin, or we are dealing with a difference of temperature, and the offset does not matter. $\endgroup$
    – user175150
    Commented Feb 4, 2019 at 8:58

Most of the time, the temperature quantity is a difference between two temperatures, not an absolute value. So, the offset is zero: \begin{align} \Delta T(F) &= T_2(F) - T_1(F) \\&= \left(\frac{9}{5}T_2(C) + 32\right) - \left(\frac{9}{5}T_1(C) + 32\right) \\&= \frac{9}{5}\left(T_2(C) - T_1(C)\right) \\&=\frac{9}{5}\Delta T(C) \end{align} This is true for quantities like entropy changes, heat capacities, and coefficients of thermal expansion.

An exception to this would be the universal gas law $PV = nk_BT.$ In this case, you need to use an absolute temperature scale where $T=0$ actually means absolute zero. So, your choices are Kelvin or Rankine, and the offset is zero, since zero Kelvin and zero Rankine mean the same thing.

  • $\begingroup$ Does this "most of the time" include weather reports and cookbooks? Even in the scientific literature I would strongly contest that first sentence - degrees centigrade are alive and well in plenty of scientific disciplines. $\endgroup$ Commented Apr 11, 2017 at 23:06
  • $\begingroup$ @EmilioPisanty For the examples that the questioner was concerned with, yes, temperature differences are more commonly used than the actual values--especially when using non-absolute values like Celsius. I gave the counter example to show that this is not always the case. But, in most cases where a temperature is used in a formula, either the quantity is a difference, or it needs to use an absolute scale like Kelvin. $\endgroup$
    – Mark H
    Commented Apr 11, 2017 at 23:28
  • $\begingroup$ @EmilioPisanty How does your last phrase contradict what is written in this answer? Mark points out that degrees centigrade are only used for temperature difference. Do you question that? If we have an absolute temperature instead, then it has to be in Kelvin for the formula to be correct in the first place. $\endgroup$
    – user175150
    Commented Feb 4, 2019 at 9:01

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