Question about the definition of a partial derivative

[lerus]

I just started to study thermodynamics and very often I see formulas like this:
$$ \left( \frac {\partial V} {\partial T} \right)_P $$
explanation of this formula is something similar to:
partial derivative of $V$ with respect to $T$ while $P$ is constant.
But as far as I remember partial derivative is always calculated when all variables except one ($T$ in our case) are constant.

Is it correct that:
$$ \left( \frac {\partial V} {\partial T} \right)_P $$
the same as
$$ \left( \frac {\partial V} {\partial T} \right) $$

Thank you

(∂V∂T)P

 

[robphy]

Is it correct that:
$$ \left( \frac {\partial V} {\partial T} \right)_P $$
the same as
$$ \left( \frac {\partial V} {\partial T} \right) $$

Yes, BUT only if $V(T,P)$.

Since the physical quantity $V$ can be seen as
different functions of different pairs of independent variables,
it's best in thermodynamics to be explicit about what is being held constant.

 

[lerus]

Thank you for your reply
For instance, if we have function $V \left( T, P, X \right)$

is $\left( \frac {\partial V} {\partial T} \right)_P \equiv \left( \frac {\partial V} {\partial T} \right)$

Thank you.

 

[robphy]

If you don't specify what $V$ depends on and what is held fixed,
I can't compute any partial derivatives.

 

[lerus]

Thank you for reply.
For instance, we have function $V \left( T, P, X \right)$
When we calculate $\left( \frac {\partial V} {\partial T} \right)$
we change $T$ and keep $P$ and $X$ constant
When we calculate $\left( \frac {\partial V} {\partial T} \right)_P$ we do the same
Then why do we need $_P$ ?

Thank you

 

[robphy]

From Schroeder's Thermal Physics p. 31

Problem 1.45. As an illustration of why it matters which variables you hold fixed
when taking partial derivatives, consider the following mathematical example.
Let $w = xy$ and $x = yz$.
(a) Write $w$ purely in terms of $x$ and $z$, and then purely in terms of $y$ and $z$.

(b) Compute the partial derivatives
$$\left(\frac{\partial w}{\partial x}\right)_y \qquad\mbox{and}\qquad \left(\frac{\partial w}{\partial x}\right)_z$$
and show that they are not equal.
(Hint: To compute $\left(\frac{\partial w}{\partial x}\right)_y$ use a formula for $w$ in terms of $x$ and $y$, not $z$ .
Similarly, compute $\left(\frac{\partial w}{\partial x}\right)_z$ from a formula for $w$ in terms of only $x$ and $z$.)

(c) Compute the other four partial derivatives
(two each with respect to $y$ and $z$),
and show that it matters which variable is held fixed.

 

[lerus]

Thanks a lot for example,
I think I understand it better now
If variables $x, y, z$ were independent then $\left( \frac {\partial w} {\partial x} \right)_y \equiv \left( \frac {\partial w} {\partial x} \right) _z$
but if $x, y, z$ are not independent, then
$\left( \frac {\partial w} {\partial x}\right)_y \neq \left( \frac {\partial w} {\partial x}\right)_z$

Thank you

 

[PeroK]

Thanks a lot for example,
I think I understand it better now
If variables $x, y, z$ were independent then $\left( \frac {\partial w} {\partial x} \right)_y \equiv \left( \frac {\partial w} {\partial x} \right) _z$
but if $x, y, z$ are not independent, then
$\left( \frac {\partial w} {\partial x}\right)_y \neq \left( \frac {\partial w} {\partial x}\right)_z$

Thank you

Well, the root of the problem is general sloppiness in using the same symbol for different functions. The more mathematically precise solution, in this example, would be to define:
$$w(x, y) = xy, \ \text{and} \ \bar w(y, z) = y^2z$$And then it's clear what function is being differentiated.

 

[PeroK]

PS and then it's obvious and trivial that not all functions have the same partial derivatives!

 

[robphy]

Possibly interesting reading:
https://bridge.math.oregonstate.edu/papers/bridge.pdf
Bridging the Gap between Mathematics and the Physical Sciences
Tevian Dray and Corinne Manogue

2 An Example

Here’s our favorite example: Suppose $T(x, y) = k(x^2 + y^2 )$.
What is $T(r, θ)$?
We often ask this question of mathematicians and other scientists.
Some mathematicians say “$k(r^2 + θ^2 )$”. Many mathematicians refuse to answer, claiming that the question is ambiguous. Everyone else, including some mathematicians, says “$kr^2$”. One colleague, who holds a split appointment in mathematics and physics, simply laughed, then asked which hat he should wear when answering the question. What’s going on here?
...

And yes, a physicist really will write $T(x, y) = k(x^2 + y^2)$ for, say, the temperature on a
rectangular metal slab, and $T(r, θ) = kr^2$ for the same temperature in polar coordinates,
even though the mathematician would argue that the symbol $T$ is being used for two different
functions. This is not sloppy mathematics on the part of the physicist; it’s a different language.
$T$ is the temperature, a physical quantity which is a function of position;
the letters which follow merely indicate which coordinate system one is using to label the position.
This can be rigorously translated into the differential geometer’s notion of a scalar field,
or phrased more informally as:

Science is about physical quantities, not about functions.

So not only do other scientists speak a different language, they use the same vocabulary!
...

 

[PeroK]

Possibly interesting reading:
https://bridge.math.oregonstate.edu/papers/bridge.pdf
Bridging the Gap between Mathematics and the Physical Sciences
Tevian Dray and Corinne Manogue

I feel that needs that to be made clear for the student's benefit. Ultimately, you may know when you mean by $T$ in every context, but it's a source of confusion if the student is left to guess what you mean.

In this case, that $T$ is a physical quantity with different functional identities should be made clear from the outset. You can't do physics without calculus and calculus is carried out on functions. You need both physical quantities and functions.

Another good example is the "total" derivative. Physics texts often make a big play about the difference between a total derivative and a partial derivative. The total derivative, however, is simply the usual derivative of a plain old single-variable function.

The danger is that you end up with a sort of mystique about these things. Whereas, if the student can ultimately reduce things to the basics of calculus, then he/she is able to disentangle things for themselves and avoid misconceptions.

 

[PeroK]

Thanks a lot for example,
I think I understand it better now
If variables $x, y, z$ were independent then $\left( \frac {\partial w} {\partial x} \right)_y \equiv \left( \frac {\partial w} {\partial x} \right) _z$
but if $x, y, z$ are not independent, then
$\left( \frac {\partial w} {\partial x}\right)_y \neq \left( \frac {\partial w} {\partial x}\right)_z$

Thank you

If you have the time and inclination, you may be interested in my insight on demystifying the chain rule(!):

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

 

[vanhees71]

I feel that needs that to be made clear for the student's benefit. Ultimately, you may know when you mean by $T$ in every context, but it's a source of confusion if the student is left to guess what you mean.

I'd also say the mathematicians are here too soft. It's indeed a very bad habit of physicists to use the same symbol for different functions, distinguishing the functions only by the naming of the arguments. This can lead to serious confusion. What drives my nuts, e.g., is to use the same symbol for a function and its Fourier transform.

In thermodynamics, it's on the other hand a somewhat different story, because the various quantities have a fixed meaning, no matter as functions of which other variables you use them. Then it's, of course, important to list all (!) the independent variables to be held fixed when differenting with respect to the only other independent variable. So there is some justification for this quite confusing notation, although I think it contributes a lot to the difficulties everybody has with thermodynamics.

In this case, that $T$ is a physical quantity with different functional identities should be made clear from the outset. You can't do physics without calculus and calculus is carried out on functions. You need both physical quantities and functions.

Another good example is the "total" derivative. Physics texts often make a big play about the difference between a total derivative and a partial derivative. The total derivative, however, is simply the usual derivative of a plain old single-variable function.

The danger is that you end up with a sort of mystique about these things. Whereas, if the student can ultimately reduce things to the basics of calculus, then he/she is able to disentangle things for themselves and avoid misconceptions.

I couldn't agree more ;-).

 

[lerus]

If you have the time and inclination, you may be interested in my insight on demystifying the chain rule(!):

https://www.physicsforums.com/insights/demystifying-chain-rule-calculus/

Thank you, very interesting article. It is possible to argue with some of the statements made there, but still very interesting.
Thank you.