0

In my thermodynamics class, I came across an equation:

$$\Delta S_2 - \Delta S_1 = C_v\ln\frac{T_2}{T_1}$$

And many equations involving changes of two changes. Now my question is do mathematicians have any neat notations to represent such changes?

In calculus, I do have learnt that the second derivative $\frac{d^2y}{dx^2}$ does represent the rate of change of rate of change (which is similar to the change of change?) for some continuous functions. But I can't stuff that notation here properly. I did try it out when went back to its derivation of the formula, and it goes something like:

$$\frac{dS}{dT} = \frac{C_v}{T}$$ $$\frac{d^2S}{dT^2} = C_v\ln T$$

Which makes sense for a continuous function. My next guess was to convert the second derivative to a discrete form was:

$$\Delta^2S = C_v\Delta\ln T$$

But I'm not convinced because I have just exterminated the $dT^2$ without any reason.

Akilan SS
  • 62
  • 2
    For the discrete case we discuss finite differences. Differences of differences are called second differences. Many of the calculations from calculus can be used analogously with the discrete case. For example if the second differences are constant this implies it's a quadradic or if the third differences are constant then it implies it's a cubic. https://en.wikipedia.org/wiki/Finite_difference – CyclotomicField Dec 08 '23 at 16:38
  • In your first equation. What $\Delta S_2$ and $\Delta S_1$ mean? – QuantumSuperfield Dec 08 '23 at 16:38
  • @QuantumSuperfield $\Delta S_2$ represents the change in entropy at some time T_2, and same for the other. – Akilan SS Dec 08 '23 at 17:07
  • @CyclotomicField After looking up the wiki, I'm impressed about myself that I got close to the actual notation. – Akilan SS Dec 08 '23 at 17:12
  • 2
    @AkilanSS To be honest, I do not remember have seen you first equation in any book. On the other hand, $\Delta S=S_2-S_1=C_v\ln\displaystyle\frac{T_2}{T_1}$ is well-known. – QuantumSuperfield Dec 08 '23 at 17:17
  • 2
    @AkilanSS If $\displaystyle\frac{dS}{dT}=\frac{C_v}{T}$, then $\displaystyle\frac{d^2S}{dT^2}=-\frac{C_v}{T^2}$. You need to fix that. – QuantumSuperfield Dec 08 '23 at 17:58

1 Answers1

1

You can certainly use $ \Delta ^ 2 $ to indicate a change of a change. Other people may not know what it means, but it should be easy to explain.

As for the derivation, you don't just eliminate the $ \mathrm d T ^ 2 $, you also change $ \ln T $ to $ \Delta \ln T $, so something more is going on.

Changing a differential to a discrete difference is a feature of linear approximation, but since we're willing to use logarithms, we don't need any approximation here, as long as $ C _ v $ is constant. We have $ \mathrm d S / \mathrm d T = C _ v / T $, so $ \mathrm d S = C _ v \, \mathrm d ( \ln T ) $, and integrating that gives us $ \Delta S = C _ v \, \Delta ( \ln T ) $ exactly. If the temperature $ T $ changes from $ T _ i $ to $ T _ f $, then $ \Delta ( \ln T ) = \ln T _ f - \ln T _ i = \ln ( T _ f / T _ i ) $, so we have $$ \Delta S = C _ v \ln \biggl ( \frac { T _ f } { T _ i } \biggr ) \text . $$ (This is the equation in the second comment by @QuantumSuperfield only with different subscripts because I don't want to clash with the meaning of your subscripts.) Then if this happens on two occasions, we have both $ \Delta S _ 1 = C _ v \ln ( T _ { 1 f } / T _ { 1 i } ) $ and $ \Delta S _ 2 = C _ v \ln ( T _ { 2 f } / T _ { 2 i } ) $; subtracting these gives $$ \Delta S _ 2 - \Delta S _ 1 = C _ v \ln \biggl ( \frac { T _ { 2 f } T _ { 1 i } } { T _ { 2 i } T _ { 1 f } } \biggr ) \text . $$ At no point did I make any approximation besides assuming that $ C _ v $ is truly constant throughout all of this (and assuming that both processes are reversible at constant volume so that $ \mathrm d S / \mathrm d T = C _ v / T $ applies in the first place).

So I'm struggling to make this exact equation match your equation. If $ T _ 1 $ and $ T _ 2 $ are the temperatures at which the two processes occur, and this is a sensible thing to say, then we're saying that $ T _ { 1 f } \approx T _ { 1 i } $ and $ T _ { 2 f } \approx T _ { 2 i } $, but this doesn't give anything like your equation. Instead, it allows us to write $ \Delta ( \ln T ) \approx \Delta T / T $, giving $$ \Delta S _ 2 - \Delta S _ 1 \approx C _ v \biggl ( \frac { \Delta T _ 2 } { T _ 2 } - \frac { \Delta T _ 1 } { T _ 1 } \biggr ) = C _ v \frac { T _ 1 \Delta T _ 2 - T _ 2 \Delta T _ 1 } { T _ 1 T _ 2 } \text . $$ That's not what you want at all.

Another simplification would be if the two processes occur right after each other, so that $ T _ { 2 i } = T _ { 1 f } $. But that still leaves us with three temperatures: an initial one, an intermediate one, and a final one. Perhaps we get back to the original temperature after the second process, so that $ T _ { 1 i } = T _ { 2 f } $ as well. Then if I write $ T _ 1 $ for $ T _ { 1 f } $ and $ T _ 2 $ for $ T _ { 2 f } $ (which is backwards from what I'd normally write but fits your equation better), I get $$ \Delta S _ 2 - \Delta S _ 1 = C _ v \ln \biggl ( \frac { { T _ 2 } ^ { \! 2 } } { { T _ 1 } ^ { \! 2 } } \biggr ) = 2 C _ v \ln \biggl ( \frac { T _ 2 } { T _ 1 } \biggr ) \text . $$ That's tantalizingly close, but different! If this is a situation that even could apply to your equation, then (unless $ T _ 1 = T _ 2 $ as well) it can't be right.

tl;dr: The answer to your question about notation is that $ \Delta ^ 2 $ is fine. But I have no idea how to fix the derivation. Possibly it was supposed to be $ \Delta S = C _ v \ln ( T _ 2 / T _ 1 ) $ all along.

Toby Bartels
  • 4,679
  • I'm glad my doubt regarding the notation was resolved. That aside, (maybe it's inappropriate for me to talk physics here in MSE) the equation $\Delta ^2 S = C_v ln \frac{T_2}{T_1}$ when talking about Entropy change during a chemical reaction. – Akilan SS Dec 09 '23 at 10:45