What Do Partial Derivatives Tell Us in Thermodynamics and Beyond?

Mandelbroth
Messages
610
Reaction score
24
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, (\frac{\partial x(t)}{\partial t}) = \frac{1}{y}, where y is a function, not necessarily of t.

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

In general, does (\frac{\partial x(t)}{\partial t})^{-1} \neq \frac{\partial t}{\partial x(t)}? 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 g = \frac{\partial f}{\partial t}, 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 y = (\frac{\partial x(t)}{\partial t})^{-1}?
Yes, you are allowed to invert both sides of an equation.

Mandelbroth said:
In general, does (\frac{\partial x(t)}{\partial t})^{-1} \neq \frac{\partial t}{\partial x(t)}? 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:
<br /> \begin{align}<br /> dx &amp;= \left(\frac{\partial x}{\partial t}\right)_y dt + \left(\frac{\partial x}{\partial y}\right)_t dy\\<br /> dy &amp;= \left(\frac{\partial y}{\partial t}\right)_x dt + \left(\frac{\partial y}{\partial x}\right)_t dx<br /> \end{align}<br />
where the, for example, \left(\frac{\partial y}{\partial x}\right)_t means t means is held constant. Then you can eliminate dy in the first relation:
<br /> \begin{align}<br /> dx &amp;=\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 /> &amp;=\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}<br />
Upon comparing the coefficient of dx, you can see that,
<br /> 1=\left(\frac{\partial x}{\partial y}\right)_t \left(\frac{\partial y}{\partial x}\right)_t <br />
so in other words:
<br /> \frac{1}{\left(\frac{\partial y}{\partial x}\right)_t} = \left(\frac{\partial x}{\partial y}\right)_t<br />
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 f(x,y,z,t) with respect to time, at a specific spatial coordinate (x,y,z) (the points you are holding constant). In order to find a change in average f, you would need to use the average value of the function, like:
<br /> \frac{1}{V}\frac{\partial}{\partial t}\int_V f(x,y,z,t)d^3 x<br />
 
Last edited:

Similar threads

Replies
6
Views
2K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
3
Views
3K
Replies
1
Views
2K
Replies
3
Views
3K
Replies
9
Views
2K
Back
Top