December 30th, 2016, 08:38 PM  #11  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81  Quote:
$\displaystyle \int_{1}^{2}du = \int_{P_{1}}^{P_{2}}du = u(P_{2})u(P_{1})$ $\displaystyle \int_{P_{1}}^{P_{2}}du= \Delta u_{1}+\Delta u_{2}+.....+\Delta u_{n}$, n $\displaystyle \rightarrow \infty$, along arbitrary continuous path. $\displaystyle \int_{P_{1}}^{P_{2}}du = (u_{2}u_{1}) + (u_{3}u_{2}) + ......+ (u_{n}u_{n1}) = (u_{n}u_{1}) = u(P_{2})u(P_{1})$ Potential is not the solution context of the problem, it is an example of the solution.  
December 30th, 2016, 09:31 PM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,681 Thanks: 2175 Math Focus: Mainly analysis and algebra  Since you haven't been given a curve, you need to carry out the check to be sure that any curve will do. But once you know that any curve will do, you also know that you just need to evaluate the potential function at the endpoints. Actually calculating the integral over any particular path is a complete waste of effort. It's only worth doing to demonstrate that the integral is indeed independent of the path.

December 31st, 2016, 07:45 AM  #13  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81 
Why do you keep referring to it as a potential function, OP doesn't. The mathematics is "independence of path," which no calculus text introduces as a potential function. Google it. Potential is an application, just like entropy, where independence of path of $\displaystyle \int$ dQ/T establishes the entropy function S. Quote:
 
December 31st, 2016, 07:51 AM  #14 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,681 Thanks: 2175 Math Focus: Mainly analysis and algebra  
December 31st, 2016, 09:55 AM  #15  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81  Heat and Entropy, and path independence Quote:
$\displaystyle \int_{1}^{2}dQ$ depends on work interchange with the system. In general, $\displaystyle \oint$ dQ = W for a cycle, heat in equals work out. If W=0, $\displaystyle \oint$ dQ = 0 and $\displaystyle \int$ dQ is independent of path so that Q = Q(P) and $\displaystyle \int_{1}^{2}dQ = Q(P_{2})Q(P_{1})$. In general, $\displaystyle \int dQ/T$ is independent of path* and so defines a function S (entropy) of coordinates, so that between any two thermodynamic states, $\displaystyle \int_{1}^{2}dS = S(P_{2})S(P_{1})$. Never really grasped this until this thread, which put things together for me. Have to thank OP. *As shown by taking a system through a cycle while interchanging heat with heat reservoirs using Carnot cycles. Also demonstrated mathematically by Caratheodory. Last edited by zylo; December 31st, 2016 at 09:57 AM.  
December 31st, 2016, 12:22 PM  #16  
Senior Member Joined: Sep 2015 From: CA Posts: 1,201 Thanks: 613  Quote:
In the very next subsection, it's not even given a separate section number, the authors discuss conservation of energy and potential functions and how this function they didn't give a name to in the previous section is called a potential function.  
December 31st, 2016, 01:02 PM  #17 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81 
So we have for a system cycle, heat in = work out, $\displaystyle \oint dQ = \oint dW$. Suppose heat is zero so that $\displaystyle \oint dW=0$. Then $\displaystyle \int_{1}^{2}dW$ is independent of path and in this case $\displaystyle \int_{1}^{2}dW =W(P_{2})W(P_{1})$ and W is called the potential. For example, if you slide a block with friction, $\displaystyle \int_{1}^{2}dW = \int_{1}^{2} f\cdot dr$ is dependent on path and $\displaystyle \oint dW$ = heat generated $\displaystyle \neq$ 0. $\displaystyle \int_{1}^{2}dW = \int_{1}^{2}dQ+ \int_{1}^{2}dU$ (first law of thermo) but $\displaystyle \oint dW = \oint dQ$ because internal energy U is a state function and Q isn't. NOTE: $\displaystyle \oint$ dW = 0 iff $\displaystyle \int$ dW is independent of path because $\displaystyle \oint dW = \int_{1}^{2}dW+\int_{2}^{1}dW = \int_{1}^{2}dW\int_{1}^{2}dW = 0$ (two different paths make up the cycle). 
December 31st, 2016, 01:08 PM  #18  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,681 Thanks: 2175 Math Focus: Mainly analysis and algebra 
Apostol II introduces the concept of independence of path in section 10.10, gives and proves the result $$\int_{\mathbf a}^{\mathbf b} \nabla \phi \cdot d\mathbf \alpha = \phi(\mathbf b)  \phi(\mathbf a)$$ in section 10.11 as the second fundamental theorem of calculus for line integrals. Section 10.12 opens with the sentence Quote:
Last edited by v8archie; December 31st, 2016 at 01:18 PM.  
January 2nd, 2017, 09:22 AM  #19 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81  Independence of Path and Exact Differentials
To clean things up, the basic ideas are: Definition of line integrals: $\displaystyle \int_{C}H(x,y,z)ds=\sum H(x_{k},y_{k},z_{k})\Delta s_{k}$ $\displaystyle \int_{C}H(x,y,z)dx=\sum H(x_{k},y_{k},z_{k})\Delta x_{k}$ $\displaystyle \int_{C}H(x,y,z)dy=\sum H(x_{k},y_{k},z_{k})\Delta y_{k}$ $\displaystyle \int_{C}H(x,y,z)dz=\sum H(x_{k},y_{k},z_{k})\Delta z_{k} $ If x,y,z are independent variables, $\displaystyle \int dx=x, \int dy=y, \int dz=z$. If dH is a dependent variable given by $\displaystyle dH \equiv H_{1}(x,y,z)dx+H_{2}(x,y,z)dy+H_{3}(x,y,z)dz$, then $\displaystyle \int_{C} dH$ is a line integral dependent on path, unless $\displaystyle \int_{C}$ dH is independent of path $\displaystyle (\int_{C} dH = H)$ iff $\displaystyle dH=\frac{\partial H}{\partial x}dx+\frac{\partial H}{\partial y}dy+\frac{\partial H}{\partial z}dz$, which says dH is an exact differential and H=H(x,y,z) exists. 

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