What Do Partial Derivatives Tell Us in Thermodynamics and Beyond?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
Mandelbroth
Messages
610
Reaction score
23
In a thermodynamics question, I was recently perplexed slightly by some partial derivative questions, both on notation and on physical meaning.

I believe my questions are best posed as examples. Suppose we have an equation, [itex](\frac{\partial x(t)}{\partial t}) = \frac{1}{y}[/itex], where y is a function, not necessarily of t.

When we solve for y, does this become [itex]y = (\frac{\partial x(t)}{\partial t})^{-1}[/itex]?

In general, does [itex](\frac{\partial x(t)}{\partial t})^{-1} \neq \frac{\partial t}{\partial x(t)}[/itex]? I was under the impression that they are, in general, not equal.

Suppose we had a function that is dependent on both position in space (x,y,z) and time (t). I shall denote it f(x,y,z,t). If we take [itex]g = \frac{\partial f}{\partial t}[/itex], the others are said to be held constant. Can g be seen as the change in f with respect to time at a specific point, or is it simply talking about the average change in f over an infinitesimal change in time? I thought that the former of the two (change at a point) was true, but now I'm not sure.
 
Physics news on Phys.org
Mandelbroth said:
When we solve for y, does this become [itex]y = (\frac{\partial x(t)}{\partial t})^{-1}[/itex]?
Yes, you are allowed to invert both sides of an equation.

Mandelbroth said:
In general, does [itex](\frac{\partial x(t)}{\partial t})^{-1} \neq \frac{\partial t}{\partial x(t)}[/itex]? I was under the impression that they are, in general, not equal.
In general, if the same variables are held constant, you can do this. I'll show you what I mean; take the total derivatives of your functions, and assume that they are all functions of each other for generality:
[tex] \begin{align}<br /> dx &= \left(\frac{\partial x}{\partial t}\right)_y dt + \left(\frac{\partial x}{\partial y}\right)_t dy\\<br /> dy &= \left(\frac{\partial y}{\partial t}\right)_x dt + \left(\frac{\partial y}{\partial x}\right)_t dx<br /> \end{align}[/tex]
where the, for example, [itex]\left(\frac{\partial y}{\partial x}\right)_t[/itex] means [itex]t[/itex] means is held constant. Then you can eliminate [itex]dy[/itex] in the first relation:
[tex] \begin{align}<br /> dx &=\left(\frac{\partial x}{\partial t}\right)_y dt + \left(\frac{\partial x}{\partial y}\right)_t \left[\left(\frac{\partial y}{\partial t}\right)_x dt +\left(\frac{\partial y}{\partial x}\right)_t dx\right]\\<br /> &=\left[\left(\frac{\partial x}{\partial t}\right)_y + \left(\frac{\partial x}{\partial y}\right)_t\left(\frac{\partial y}{\partial t}\right)_x\right]dt +\left(\frac{\partial x}{\partial y}\right)_t \left(\frac{\partial y}{\partial x}\right)_t dx<br /> \end{align}[/tex]
Upon comparing the coefficient of [itex]dx[/itex], you can see that,
[tex] 1=\left(\frac{\partial x}{\partial y}\right)_t \left(\frac{\partial y}{\partial x}\right)_t [/tex]
so in other words:
[tex] \frac{1}{\left(\frac{\partial y}{\partial x}\right)_t} = \left(\frac{\partial x}{\partial y}\right)_t[/tex]
So, like I said earlier, this is true only if the same variable is held constant in each derivative.
Mandelbroth said:
Can g be seen as the change in f with respect to time at a specific point, or is it simply talking about the average change in f over an infinitesimal change in time?
This is the change in [itex]f(x,y,z,t)[/itex] with respect to time, at a specific spatial coordinate [itex](x,y,z)[/itex] (the points you are holding constant). In order to find a change in average [itex]f[/itex], you would need to use the average value of the function, like:
[tex] \frac{1}{V}\frac{\partial}{\partial t}\int_V f(x,y,z,t)d^3 x[/tex]
 
Last edited: