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Consider:

$$ dU = T dS - PdV$$

Now, how would I differentiate both sides with temperature...?

Or more simply:

$$ dy = f'(x) dx$$

Can I differentiate this expression above? (I'm not talking about moving dx to the denominator on left)

More particularly:

$$ (dy)' = (f'(x) dx)'$$

does the above expression have any meaning?

tryst with freedom
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2 Answers2

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If you have the basic knowledge about differential form and exterior differentiation, then $\mathrm{d}\mathrm{d}U$ do have its mathematical meaning. $$ \mathrm{d}\mathrm{d}U=\mathrm{d}T\wedge\mathrm{d}S-\mathrm{d}P\wedge\mathrm{d}V=(\dfrac{\partial T}{\partial S}\mathrm{d}S+\dfrac{\partial T}{\partial V}\mathrm{d}V)\wedge\mathrm{d}S-(\dfrac{\partial P}{\partial S}\mathrm{d}S+\dfrac{\partial P}{\partial V}\mathrm{d}V)\wedge\mathrm{d}V\\=(\dfrac{\partial T}{\partial V}+\dfrac{\partial P}{\partial S})\mathrm{d}V\wedge\mathrm{d}S $$ But I do not know the physical meaning of this.

In thermal physics, maybe the Legendre transformation is more common. $$ \mathrm{d}(U-TS)=-P\mathrm{d}V-S\mathrm{d}T $$ move $\mathrm{d}T$ to the left and we can differentiate again.

user823011
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  • what exactly is this $ d(dU)$ mean? differential of differential? – tryst with freedom Oct 01 '20 at 10:55
  • First, do you know what a "differential" means? – user247327 Oct 01 '20 at 11:03
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    $\mathrm{d}U$ is 1-order differential form, therefore $\mathrm{d}\mathrm{d}U$ is 2-order, and the mathematical operation here is called exterior differentiation. Maybe my answer doesn't solve your question, after all, I have never seen the application of exterior differentiation in thermal physics. – user823011 Oct 01 '20 at 11:04
  • I did not see your comment yesterday. My knowledge of differentials come from this video playlist: https://www.youtube.com/watch?v=cFscZ9c0AIk&list=PL8erL0pXF3JYCn8Xukv0DqVIXtXJbOqdo – tryst with freedom Oct 02 '20 at 17:34
  • Your answer definitely has what I asked for but I want to know what is the meaning of expression rest is ok – tryst with freedom Oct 02 '20 at 17:35
  • I do not have much confidence in explaining exterior differentiation in just a few words, related knowledge maybe can be found in some mathematical analysis textbooks. In my knowledge, differential $\mathrm{d}(\cdot)$ always meanings a linear map,and the meaning here is the same. The equation above shows the relation between the small variations of different variables. Although learning differential form needs much kwowledge about analysis and algebra, the calculation rules are simple, as is shown. – user823011 Oct 04 '20 at 00:41
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I'm not exactly sure of your question, but you can indeed differentiate a differential. It is a slightly different notation, in that the typical notation for a second derivative doesn't allow for this, but if you adopt a different notation for the second derivative, it works perfectly fine. The typical notation for the second derivative of $y$ with respect to $x$ is $\frac{d^2y}{dx^2}$. Using a notation that allows for taking second differentials, this becomes $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$.

The process of taking a differential of a differential is straightforward. The differential of a differential (say $dx$) is simply $d^2x$. In other words, $dx$ has the meaning $d(x)$ and $d^2x$ has the meaning $d(d(x))$. The thing to keep in mind is that if you have something like $x\,dx$, this has to be differentiated with the product rule. So $d(x\,dx) = x\,d^2(x) + dx^2$.

So, if you are wanting to differentiate your equation above, you would get:

$$ dU = T\,dS - P\,dV \\ d(dU) = d(T\,dS - P\,dV) \\ d^2U = d(T\,dS) - d(P\,dV) \\ d^2U = T\,d^2S + dT\,dS - P\,d^2V - dP\,dV $$

If you wanted that in terms of a second derivative, you would have to get it in the form of the second derivative I noted above.

You can find out more about using higher order differentials algebraically from my paper, "Extending the Algebraic Manipulability of Differentials."

johnnyb
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  • What meaning does these derived quantites have? – tryst with freedom Oct 11 '20 at 04:31
  • If you got them into the form specified for the second derivative, it would be the second derivative of some quantity with respect to another. If you wanted the second derivative with respect to temperature, it might be easier to go back to the first differential, divide both sides by $dt$, and then take the derivative from there, but that would get quite messy, as every fraction would wind up using the quotient rule. – johnnyb Oct 11 '20 at 12:37