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In my introductory physics book there is this equation:

$$U=\frac{PV}{(\gamma -1)}$$

that is transformed to

$$dU=\frac{Pdv+Vdp}{(\gamma -1)}$$

It seems like a derivative operation, but where are the differentials at the denominators?

What kind of operation is this? And under which conditions can be applied? Thanks.

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  • $\begingroup$ Yes, of course, if $\gamma$ is a constant. $\endgroup$ – Michael Rozenberg Dec 22 '17 at 7:21
  • $\begingroup$ You can think of it as a derivative with respect to an unnamed variable or as a formal multiplication of $dU/dt = (P \, dV/dt + V \, dP/dt)/(\gamma-1)$ with $dt$. $\endgroup$ – md2perpe Dec 22 '17 at 7:36
  • $\begingroup$ My teacher showed us that the derivative of x(t) in dt is dx(t)/dt and then he learn us how to calculate it. But in this case there arent differential in the denominator. What theorems can i use to calculate this type of differential equation ? $\endgroup$ – Poiera Dec 22 '17 at 7:36
  • $\begingroup$ There is no reason to turn the uppercase into lowercase. $\endgroup$ – Yves Daoust Dec 22 '17 at 20:36
  • $\begingroup$ The lower/upper case is just extremely bad and confusing practice. You have differential $d(pV)=Vdp+pdV$, the product rule. The denominator is a constant, which is apparent from the physics context (the ratio of specific heats for a given gas). $\endgroup$ – orion Dec 22 '17 at 20:49
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It seems as though $U$ is regarding potential (internal) energy. This is always regarded as a state function, which means it can be written as an exact differential (or equivalently that the function is path independent, or conservative). More precisely, for $U = U(P, V)$,

$$dU = \frac{\partial U}{\partial P}dP + \frac{\partial U}{\partial V} dV$$

More generally, if $F(x_1,..., x_n)$ is a state function, then:

$$dF = \frac{\partial F}{\partial x_1}dx_1 +...+ \frac{\partial F}{\partial x_n} dx_n$$

In your case, we have $U = \frac{PV}{\gamma - 1}$, and if we treat $\gamma$ as a constant,

$$\frac{\partial U}{\partial P} = \frac{V}{\gamma - 1} \\ \frac{\partial U}{\partial V} = \frac{P}{\gamma - 1}$$

So that

$$dU = \frac{PdV + VdP}{\gamma - 1}$$ As desired.

Edit:

Basically, you can parametrize $P, V$ as you like, say with respect to some arbitrary variable $t$. The reason this works here is that $U$ is path independent, which means if you start at $P_0, V_0$ and move to $P_1, V_1$, the energy will change from $U_0$ to $U_1$ regardless of which path you moved along in the $PV$-plane.

Parametrizing $P,V$ as $P = P(t), V = V(t)$, and taking a derivative of your equation with respect to $t$ you get

$$\frac{dU}{dt} = \frac{V \frac{dP}{dt} + P \frac{dV}{dt}}{\gamma - 1}$$ By exploiting the chain rule.

Multiplying through by $dt$ and using the fact that a function parametrized as $F = F(t)$, has small change: $$dF = \frac{dF}{dt}dt$$ Plugging this in with $F = U, V, P$, you again get the result you were looking for.

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  • $\begingroup$ Nice. If i want to get back to the original equation using integral, do i have to use the indefinite one ? Why is it possible to integrate even there are different functions in the equation ? Is it a special type of differential integration ? $\endgroup$ – Poiera Dec 22 '17 at 7:43
  • $\begingroup$ @Poiera: Actually, the integral comes naturally because $d(PV) = PdV + VdP$ (this is just the chain rule written differently) so that the equation can also be written as $dU = d(PV)/(\gamma - 1)$, so that integrating will just give you the original equation. $\endgroup$ – infinitylord Dec 22 '17 at 7:51
  • $\begingroup$ In this case P and V are functions of which variables ? $\endgroup$ – Poiera Dec 22 '17 at 11:49
  • $\begingroup$ @Poiera: I've edited to answer that question $\endgroup$ – infinitylord Dec 22 '17 at 20:36
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We starts with $$U=\frac{PV}{(\gamma -1)}$$

I'll just differentiate this w.r.t a dummy variable $x$, assuming $\gamma$ is a constant.

$$\frac{dU}{dx}=\frac{P\frac{dV}{dx}+V\frac{dP}{dx}}{(\gamma -1)}$$

Multiplying $dx$ throughout, $$dU=\frac{PdV+VdP}{(\gamma -1)}$$ as shown.

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If $\gamma$ is a constantand $U,P,V$ are function for the same variable, say $x$, then you get:

$$U=\frac{PV}{(\gamma -1)}\\\dfrac{dU}{dx}=\dfrac{d}{dx}\frac{PV}{(\gamma -1)}=\frac{P\dfrac{dV}{dx}+V\dfrac{dP}{dx}}{(\gamma -1)}$$hence $$dU=\frac{PdV+VdP}{(\gamma -1)}$$

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  • $\begingroup$ Is it necessary that the functions are function for the same variable ? $\endgroup$ – Poiera Dec 22 '17 at 7:39
  • $\begingroup$ @Poiera no, you should be familiar with the chain rule no? So if, for example one of them is a function of $t$ and the other is for $x$(let's say that $U$ is a function of $x$) then you get $U(x)=\frac{P(x)V(t)}{\gamma-1}$, now if $t$ is a function of $x$, then you can do it using the chain rule. And if you familiar with partial derivative you can say that if $t$ and $x$ has no relation then $dV=0$ so we can still say this is true. One note, $\gamma$ has to be a constant when looking at the variable we take derivative of, it can be a function that has no relation to, in my example, $x$ $\endgroup$ – ℋolo Dec 22 '17 at 7:47
  • $\begingroup$ The way @infinitylord showed how to do it is much more elegant than this way, it also explain why it works even if it is not the same variable very clearly. $\endgroup$ – ℋolo Dec 22 '17 at 7:55

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