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Does $^\circ\mathrm{C}^2$ have any physical consistency? I know that $\mathrm{K}^2$ is a valid unit even though I don't know in which context it can be used. But I feel strange about $^\circ\mathrm{C}^2$ because of the offset. My intuition about unit behaviour is that it is not a valid unit at all. I am looking for an scientific argument to answer if the unit $^\circ\mathrm{C}^2$ would be valid or not? Thanks.

Update

I am developing units and quantity classes for a project. I implemented units as a vector of dimensions and a scale factor before a reference unit. I also implemented an offset in order to convert my $^\circ\mathrm{C}$ into $\mathrm{K}$. Now I am implementing multiplication and division. Then the question has risen.

No, I never met $^\circ\mathrm{C}^2$ before, and I believe it will never happen. I suspect it is because units are based on some kind of vector-space where the origin is a key concept which ill-defined units such as $^\circ\mathrm{C}$ break.

Update 2

Reading this post helped a bit.

I am not sure I understood what NowIGetToLearnWhatAHeadIs have said. Anyway, distance have an origin when you measure it, because you make the origin of your rule to coincide with some arbitrary origin (a point of space that will endorse the responsibility of origin).

Measurements of area by the means of a rule will have consistency not mater if your rule is Imperial or International as long as you use units to track your computations.

But this will not be correct if you decide to start with the first digit graduation of your rule instead of the origin of your rule. Eg.: I always use $1\,\mathrm{cm}$ or $1\,\mathrm{in}$ to set my origin as I just read the rule as if I was using it the right way. Then, the area will be meaningless until you adjust for all your measurements of the origin shift you falsely introduced.

This is how I am now perceiving $^\circ\mathrm{C}$ units and I really doubt it is meaningful to square Celsius or Fahrenheit if they are taken as "absolute" measurement. If they are difference temperatures it looks right because you withdraw the offset.

Now asking the SE Physics community: Are my last new statements correct? Thank you.

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    $\begingroup$ Can you give us some context? What brings up the question? Have you seen degrees C squared somewhere? $\endgroup$
    – Sean
    Mar 26, 2015 at 13:13
  • $\begingroup$ I am seeing closing vote because it is unclear what I am asking. The question is simple, is it valid to square $^\circ\mathrm{C}$. Answer are, Yes or No and Why! $\endgroup$
    – jlandercy
    Mar 26, 2015 at 13:28
  • $\begingroup$ Well note that $T_2(C)-T_1(C) = T_2(K) - T_1(K)$, so in that case (i.e., a difference between two temperatures) you can have that $C^2=K^2$. $\endgroup$
    – Kyle Kanos
    Mar 26, 2015 at 13:32
  • $\begingroup$ I know that temperature difference will vanish the offset and then, only then, $^°\mathrm{C}$ will means the same that $\mathrm{K}$. But what if you are not dealing with temperature difference. $\endgroup$
    – jlandercy
    Mar 26, 2015 at 13:35

3 Answers 3

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If you are using an absolute temperature, you should use Kelvin. For instance, when using the Stefan-Boltzmann Law, $$P=A\epsilon\sigma T^4$$ it wouldn't make sense to have units of $^\circ C^4$; only units of $K^4$ physically make sense here.

However, if you are using a temperature difference, then both Celsius and Kelvin are equally valid because a temperature difference in Celsius is equal to one in Kelvin. So when using the heat equation, $$\Delta Q=c_p\rho\Delta T$$ you can safely use Celsius as easily as Kelvin because it is a temperature difference. This means that in situations where you are called to use the difference in temperature to some higher power, like $\Delta T^2$, you could just as easily use Celsius as Kelvin without worrying about problems. I can't think of any instances where that is required, but the point is that you can do it.

TL;DR If you're using absolute temperature squared, you can't use Celsius; it doesn't make sense. If it's temperature difference you need, then in all cases feel free to switch between the two at will.

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  • $\begingroup$ Yes this is how I understand temperature units. So you are confirming that the offset is a problem and implies the units must be handled with care. $\endgroup$
    – jlandercy
    Mar 31, 2015 at 12:58
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    $\begingroup$ @jlandercy absolutely. I thought about trying to explain why using Celsius for absolute temperature can wind up not making sense, but that would be like explaining why using a frog in the place of a calculator doesn't make sense $\endgroup$
    – Jim
    Mar 31, 2015 at 13:25
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The unit $^\circ\mathrm{C}^2$ does make sense. It represents a difference in temperature squared. You are worried that it is invalid because the origin is shifted. This is not a problem though, because we use $\mathrm{m}^2$ all the time, and there is no natural origin at all for positions.

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  • $\begingroup$ But we do use the terminology, "Traffic tied up on I-90 for 5 miles, east and west of the accident at Mile 38". $\endgroup$
    – DJohnM
    Mar 26, 2015 at 14:32
  • $\begingroup$ I agree.I do not believe that dimensional analysis has to do with the physical meaning of the units. $\endgroup$
    – TheQuantumMan
    Mar 26, 2015 at 18:26
  • $\begingroup$ I have updated my question because I am not sure I have understood what you answered. $\endgroup$
    – jlandercy
    Mar 30, 2015 at 16:44
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When I wrote a units package, I put degFinterval and degCinterval as units, specifically units that are intended to be squared and multiplied and divided. I put "interval" in the name to make it absolutely clear to users that they should only be multiplying temperature differences (which don't have a fixed origin) rather than raw temperatures (which do). (In the code, degCinterval turns out to be an exact synonym of Kelvin, of course.)

This is basically consistent with what you said, and with NowIGetToLearnWhatAHeadIs's answer. :-D

Sometimes people use the terminology "degrees celsius" is a specific temperature, "celsius degrees" is a temperature interval or difference. Then only the latter should be squared. I'm not sure that this terminology is really universal though :-D

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  • $\begingroup$ I have read your code, thanks for sharing. Anyway I need something more dynamic that is able to keeps tracks of user modifications during runtime and overhead are not a concern in my application. $\endgroup$
    – jlandercy
    Apr 2, 2015 at 9:37
  • $\begingroup$ In S.S. Stevens well cited 1946 paper "On the Theory of Scales of Measurement" he coined the term "Interval" specifically to describe measurements that CANNOT be multiplied sensibly. Because of Stevens, your choice of unit name degFInterval etc might not be as "absolutely clear" as you believe unfortunately. $\endgroup$
    – Christopher Bruns
    Jun 30, 2021 at 15:43

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