Undergrad Can the Chain Rule be Applied to Simplify Divergence in Entropy Equation?

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SUMMARY

The discussion focuses on the application of the chain rule to simplify the divergence in the entropy equation for a Newtonian fluid, specifically using Fourier conduction law. The equation presented is \(\frac{1}{T} \nabla \cdot (-\kappa \nabla T) = - \nabla \cdot (\frac{\kappa \nabla T}{T}) - \frac{\kappa}{T^2}(\nabla T)^2\). Key insights include the use of the product rule \(\nabla\cdot (fV)=(\nabla f)\cdot V+f (\nabla\cdot V)\) with \(f=1/T\) and \(V=\nabla T\), as well as the chain rule for \(\nabla (1/T)=-\frac{\nabla T}{T^2}\). These mathematical tools are essential for correctly simplifying the divergence expression.

PREREQUISITES
  • Understanding of the entropy equation in thermodynamics
  • Familiarity with Fourier conduction law
  • Knowledge of vector calculus, specifically divergence and gradient operations
  • Proficiency in applying the chain rule and product rule in calculus
NEXT STEPS
  • Study the derivation of the entropy equation for Newtonian fluids
  • Learn about Fourier conduction law and its implications in thermodynamics
  • Explore vector calculus applications in fluid dynamics
  • Practice using the product rule and chain rule in various mathematical contexts
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Students and professionals in physics, engineering, and applied mathematics, particularly those focusing on thermodynamics and fluid dynamics.

BeeKay
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I am looking at the derivation for the Entropy equation for a Newtonian Fluid with Fourier Conduction law. At some point in the derivation I see

\frac{1}{T} \nabla \cdot (-\kappa \nabla T) = - \nabla \cdot (\frac{\kappa \nabla T}{T}) - \frac{\kappa}{T^2}(\nabla T)^2

K is a constant and T is a scalar field. It seems obvious that there is some way to use the chain rule on the middle term to get the left and right terms, but I frankly don't exactly understand the "rules" of how to use it with the divergence. I know you can't just factor out \frac{1}{T} from the middle term, but I'm not sure how to actually simplify that middle expression. Any help is appreciated.
 
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I think the relevant product rule is ##\nabla\cdot (fV)=(\nabla f)\cdot V+f (\nabla\cdot V)##. Try applying this with ##f=1/T## and ##V=\nabla T##.

You will also want to use the chain rule for ##\nabla (1/T)=-\frac{\nabla T}{T^2}##.
 
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