Timeline for How do we know that we only need three second derivatives to represent all the second derivatives of a thermodynamic system?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 27 at 9:38 | vote | accept | Raffaella | ||
| Mar 27 at 9:36 | comment | added | Confuse-ray30 | @Raffaella The math behind this is regarding the space of parameters as a manifold, the thermodynamical potential as a scalar field and the derivatives (locally) are just the canonical basis vectors of your tangent space. Every other derivative can be obtained (clearly, because it is a basis of the vector space) as linear combinations of these (in this case 3) basis vectors. | |
| Mar 27 at 9:33 | comment | added | Confuse-ray30 | @Raffaella Define whatever is here as a new function, say $g$ and $h$ and repeat procedure. | |
| Mar 27 at 9:33 | comment | added | Raffaella | And why we need exactly three second derivatives to express other derivative? Is there any math behind it? Anyway, thank you for your answer!! | |
| Mar 27 at 9:27 | comment | added | Raffaella | What if second derivative? Would it be the same? | |
| Mar 27 at 9:18 | history | edited | Confuse-ray30 | CC BY-SA 4.0 |
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| Mar 27 at 8:47 | comment | added | Raffaella | That's exactly what I want to ask: why can any other derivative along some direction in parameter space be expressed as a linear combination of these | |
| Mar 27 at 8:19 | history | answered | Confuse-ray30 | CC BY-SA 4.0 |