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I'm in doubt whether the equal sign in an expression like "$0 ^\circ \mathrm{C} = 273.15 \,\mathrm{K}$" is fair because, normally, if $A=B$, then, say, $2A=2B$, which is hardly applicable to "$0 ^\circ \mathrm{C} = 273.15 \,\mathrm{K}$".

So is it okay to use the "$=$" if we cannot actually perform multiplication or division by the same number on both sides? Which sign would suit better here if the "$=$" does not work? Thanks a lot in advance!

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    $\begingroup$ Thank you, guys! Some follow-up questions. • I see that the words like "matches", "corresponds to", "is" would work, but the "=" is really used in this context, right? • How do we call equations, which left-hand and right-hand sides cannot be multiplied or divided by the same number? Is there a specific term? $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 4:37
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    $\begingroup$ Just to note - you can't multiply degrees C in general. The equality is somewhat beside the point. That is, the multiplication operation is basically meaningless for degrees C. You can multiply change in degrees C (because this is basically equivalent to changes in Kelvin). $\endgroup$
    – johnnyb
    Commented Aug 17, 2022 at 13:23
  • $\begingroup$ In the same vein, $0 M_\text{bol} = 30128000000000 PW$ where $M_\text{bol}$ is absoute bolometrric magnitude, and PW is petawatts Something similar can also be done with 0 pH = (something concentration of H ions} $\endgroup$
    – James K
    Commented Aug 18, 2022 at 20:42

5 Answers 5

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I'm in doubt whether the equal sign in an expression like "$0 {\ ^\circ\mathrm{C}} = 273.15 {\ \mathrm{K}}$" is fair because, normally, if $A = B$, then, say, $2 A = 2 B$, which is hardly applicable to "$0 {\ ^\circ\mathrm{C}} = 273.15 {\ \mathrm{K}}$".

I think that the equals sign is totally appropriate, and the rule "if $A = B$, then $2 A = 2 B$" is perfectly applicable to this equation.

The only tricky part is that, unlike most unit symbols, the unit symbol $^\circ\mathrm{C}$ can't be treated as though it were a quantity that is multiplied by the number next to it. It must be treated as an operator which is written on the right-hand side of the quantity that it operates on.

As a result of this, the rule for multiplying a temperature in degrees Celsius by a scalar is strange. Specifically, the rule is that

$$n (t {\ ^\circ\mathrm{C}}) = (n t + (n - 1) 273.15) {\ ^\circ\mathrm{C}}.$$

But in any case, by applying this rule, we can see that

$$ \begin{align} 2 (0 {\ ^\circ\mathrm{C}}) &= (2 \cdot 0 + (2 - 1) 273.15) {\ ^\circ\mathrm{C}}\\ &= (2 \cdot 0 + 1 \cdot 273.15) {\ ^\circ\mathrm{C}}\\ &= (0 + 273.15) {\ ^\circ\mathrm{C}}\\ &= 273.15 {\ ^\circ\mathrm{C}}\\ &= (273.15 + 273.15) {\ \mathrm{K}}\\ &= (2 \cdot 273.15) {\ \mathrm{K}}\\ &= 2 (273.15 {\ \mathrm{K}}).\\ \end{align} $$

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    $\begingroup$ Hi Tanner Swett! That's awesome! Thank you a ton! $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 12:59
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    $\begingroup$ Your formula (which gives $100°C=1.37\times0°C$) makes sense if we consider Celsius as an alternative/proxy measure of heat energy. But Newton's Law of Coolng gives that it is temperature difference, rather than temperature per se, that is proportional to heat energy. $\endgroup$
    – ryang
    Commented Aug 17, 2022 at 13:25
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    $\begingroup$ @ryang: Yes, $100\sideset{^\circ}{}{\mathrm{C}} = \frac{373.15}{273.15} \times 0 \sideset{^\circ}{}{\mathrm{C}}$ is correct. $\endgroup$
    – Nat
    Commented Aug 19, 2022 at 7:30
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    $\begingroup$ @Nat I wasn’t referring to correctness/wrongness: I was implicitly merely pointing out that while Kelvin is naturally amenable to being treated as a proxy for heat energy (so, $2\times5K=10K$), Celsius was not designed, and is not typically understood, this way, and if we wish to force Celsius into this framework, then Tanner’s formula does make sense. OTOH, $2\times5°C=10 °C$ also absolutely makes sense, but, its interpretation is different. To be clear: if you find yourself having to double a °C value, then you are almost certainly doing something wrong (misinterpreting the problem). $\endgroup$
    – ryang
    Commented Aug 19, 2022 at 10:01
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    $\begingroup$ One solution is creating this strange multiplication rule, but another solution would be that multiplication is just not defined for these objects. Mathematics deals a lot with equivalent quantities. One might say that two object are "equal", even when no multiplication is defined on them. $\endgroup$
    – Rd Basha
    Commented Aug 19, 2022 at 17:19
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Let the rod's temperature be $x$ Kelvin and $y$ °Celsius, and length be $p$ inches and $q$ metres.

  1. Notice that while the units inch and metre can be singular/plural and be manipulated algebraically $$\text{inch}=\frac{\text{metre}}{39.27},\tag1$$ $\text‘y$°C’ is more a state than a quantity, the units Kelvin and °Celsius don't have singular/plural forms, and $$\text{°C}\ne\text{K}-273.$$

  2. These relationships hold: $$p=39.27q,\\y=x-273.$$

    Although $p$ and $q$ reside on different scales, from $(1),$ their respective units are such that $$p\text{ inches}=q\text{ metres}.$$

    $x$ and $y$ also live on different scales; unfortunately, $$y\text{°C} \ne x\text{K}.$$

  3. The last equality fails because its right side is proportional to kinetic energy, while its left side is not; the zero points of the Celsius and Kelvin scales do not even coincide.

    If we think of ‘hotness’ as a nonnegative quantity that starts from nought, then doubling $x$K is indicating that the rod has doubled in hotness; on the other hand, doubling $y$°C is signalling that its temperature difference from water's melting point has doubled.

  4. Hence, the conversion between $x$K and $y$°C is via correspondence rather than equality.

    enter image description here

    (image borrowed from this Answer)

  5. On the other hand, °Rankine is proportional to Kelvin, $$\text{K} = 1.8\text{°R},$$ and $500\text{°R}$ indeed equals $278\text{K}.$

  6. Steeven:

    Is the issue that the mathematical zero-level has to correspond to a physical zero-level, or is the issue simply that the two mathematical zero-levels are not coinciding?

    The latter: this issue is purely mathematical; after all we can construct a modified Celsius called Pelsius such that $1\text{°P}=3\text{°C}.$

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    $\begingroup$ Thank you, Ryang! $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 8:56
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    $\begingroup$ I expanded point 3: the point is that when interpreting the problem, we need to keep making sense, for example, clearly knowing the meaning of tripling $30°C$ as opposed to tripling $300K$ and knowing that $-300°C\times2$ makes no physical sense. This is not generally a problem, since what we usually want to triple is not temperature per se, but temperature difference or amount of heat energy. $\endgroup$
    – ryang
    Commented Aug 17, 2022 at 13:08
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    $\begingroup$ I meant to write that “$−200°𝐶×2$ makes no physical sense.” $\endgroup$
    – ryang
    Commented Aug 18, 2022 at 7:37
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EDIT: Apparently the SI convention is to use °C to refer to temperature difference from 273.15K, which would make the OP correct and my previously written answer (below) incorrect: you can't use an equals sign at all, unless both sides of the equality are $T- 273.15K$.

Degrees Celsius is an operator, not a variable. It goes on the right hand side of what it operates on, against mathematical convention for single symbol operators, because of tradition in common use, but that's what it is. It means "take this number and add it to the zero point of the Celsius scale". If we call the zero point $x$ then the operator C acts on the variable $y$ to make $y+x$. If you are precise in treating the degrees Celsius symbol as an operator that is positioned unconventionally, the equation and all algebraic manipulations hold. If you are imprecise and treat the degrees Celsius symbol as a function or a variable, the equation is neither true nor false, it merely inherits the nonsense in the expression, exactly as if you interpreted 2+2 as 2×(+2) and inferred 6×(+2)=12 from 2+2=4.

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    $\begingroup$ @Error403 No worries; my query was directed at g.s., because they too have mentioned "operator". $\endgroup$
    – ryang
    Commented Aug 19, 2022 at 11:49
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    $\begingroup$ @Error403 In this Answer, I defined precisely how % is a postfix operator. In mathematics, an operator takes input(s) and returns output(s). In the expression '7°C', in what sense is '°C' operating on '7'? Are '°C' and 's' "operators" just because it is meaningful to append them as units '°C' and 's' to the number '7'? And since °C is an "operator" (as all units are), therefore "0 °C=273.15K" is a proper equality? $\endgroup$
    – ryang
    Commented Aug 19, 2022 at 14:56
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    $\begingroup$ @ryang, thank a bunch again! $\endgroup$
    – Error 403
    Commented Aug 19, 2022 at 14:58
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    $\begingroup$ @Error403 In short, while I appreciate that calling '°C' an operator is seductive, I'm not sure that this is useful. Furthermore, I don't think Tanner's algorithm/formula (which basically repurposes the meaning of °C) has anything to do with °C being an operator: neither implies the other. In other words, the clause "As a result of this" in their Answer isn't logical, I think. To be clear, neither calling °C an operator, nor Tanner's algorithm/formula, is wrong; it's all about making sense of what you're doing. $\endgroup$
    – ryang
    Commented Aug 19, 2022 at 15:03
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    $\begingroup$ @ryang that's a fair assessment. A proper mathematician starts every statement by rigorously defining every symbol he's going to use. Physicists get to skip that for SI units, but should be more like proper mathematicians for secondary symbols like °C if they're going to do math with them. I'll modify my answer. $\endgroup$
    – g s
    Commented Aug 19, 2022 at 15:33
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This is comparing apples & oranges in general.

Example 0 : Water & Tumbler

A tumbler having 50% water is half-full & half-empty :
0.5 full = 0.5 empty. Multiply each side by 2 to get : 1.0 full = 1.0 empty !

What is wrong here is that we equated the water Part with the empty Part. Naturally, we will get contradictory answers on further manipulations.
This is comparing apples & oranges.

Example 1 : Age & Date

Some Person who was born in 2000 will be 22 years in 2022 AD , But we can not claim that this Person will 2x22=44 years old in 2x2022=4044 AD.

We can not assign a meaning to multiplying Age & Date , & then comparing with each other.
This is comparing apples & oranges.

We can say Age(2022)=22 , 2xAge(2022)=2x22=44 , Age(4044)=2044 or in general Age(X)=(X-2000) which is not Age(2X)=2(X-2000).

Example 2 : Celsius & Kelvin

Absolute 0 is a Point in the Temperature Curve. Multiply (& Dividing) is simply scaling this curve. With a Particular scaling, we Define it Kelvin. Let X=20 Kelvin be a Point on the Kelvin Curve. Multiply by 10 to get 200 Deci-Kelvin.
We have a meaning to use here: 20K=200 Deci-K.

Add (or Subtract) some Constant on the Temperature Curve & we have some other Curve (maybe shifting) and we can not multiply this & then compare with Original Temperature Curve.
This is comparing apples & oranges.

We can , of course, give a new name for this new Curve, eg Celsius.

We can say C=K(X) , where C is Celsius & K is a function converting some temperature in Kelvin to Equivalent temperature in Celsius.
This gives 2C=2K(X) , not 2C=K(2X)

Example 3 : Length & Area

With Side=3 , Area of Square = 9.
We can not claim that with Side=2x3 , Area=2x9=18.
This is comparing apples & oranges.

We can say A=f(S) where A is the Area & f converts length S into Area of Square.
This means, 2A=2f(S), which is true. We can not say 2A=f(2S) which is not true.

Summary :

0C=273.15K actually means that there is a function which converts K to C.
Check Page 3 Equation 3 of this Article which gives the Precise formula.

C=f(K)
0=f(273.13)

Multiply to 2 to get:
2C=2f(K)
0=2f(273.13)

We can not conclude this:
2C=f(2K)
0=f(2x273.13)

This is comparing apples & oranges in general.

When can we use "=" :
(1) When we are not going to Mathematically manipulate it further, Eg by squaring or using trigonometric functions.
[[ In case we want to Mathematically manipulate, we must ensure Mathematical Equality by using the conversion formula ]]
(2) When we are using Differences in temperature , we can multiply without contradictions. Difference in temperature is 2 Units. Double it to get Difference of 8 Units. This is valid when we are using linear scale.

What else can we use to avoid Issues :
(1) We could use words like "Equivalent" & "is same as"
(2) We could also use symbols like $\equiv$ & $\leftrightarrow$

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    $\begingroup$ Hi Prem! Yes, we cannot multiply on both sides for the reasons you have described, right. The question is that whether it is totally fine to use the equals sign in "0 °C = 273.15 K" when actually meaning functions but not the arguments? $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 5:53
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    $\begingroup$ Perhaps a symbol like the "↦" would work better? Sorry for a tautology, but the equal sign should denote equality. No? :) $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 6:01
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    $\begingroup$ I have updated this answer , to address your concerns. $\endgroup$
    – Prem
    Commented Aug 17, 2022 at 11:54
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    $\begingroup$ Great, Prem! Thank you a bunch for your input! $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 12:06
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The equal sign is fair, the multiplication is not.

$^\circ C$ does not have a meaningful meaning of multiplication, for any value. One requirement for multiplication to be meaningful, you basically need a zero that has a philosophical meaning of zero (i.e., that there is no quantity of something). The Kelvin scale satisfies this criteria (at $0K$ there is no physical movement), but the Celsius scale does not ($0^\circ C$ is an arbitrary point).

Now, you can multiply changes in $^\circ C$, because changes by definition have a philosophical zero ("no change" actually is in fact zero). But, interestingly, changes in $^\circ C$ is equivalent to changes in $K$.

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    $\begingroup$ Super! Thank you for your input, Johnnyb! $\endgroup$
    – Error 403
    Commented Aug 17, 2022 at 13:35
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    $\begingroup$ "But, interestingly, changes in °C is equivalent to changes in K." Isn't this by design: that °C and K are affinely related with gradient 1, so that they change at the same rate. °F, on the other hand,... $\endgroup$
    – ryang
    Commented Aug 17, 2022 at 13:37
  • $\begingroup$ @ryang - correct. $\endgroup$
    – johnnyb
    Commented Aug 17, 2022 at 16:41
  • $\begingroup$ Is the issue that the mathematical zero-level has to correspond to a physical zero-level, or is the issue simply that the two mathematical zero-levels are not coinciding? $\endgroup$
    – Steeven
    Commented Aug 28, 2022 at 19:20
  • $\begingroup$ @Steeven I expanded my Answer above in reply to your Question. $\endgroup$
    – ryang
    Commented Aug 29, 2022 at 2:47

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