Can I pull a time derivative outside of a curl?

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SUMMARY

The discussion centers on the mathematical operation involving the curl of the electric field E and the time derivative of the magnetic field B, specifically in the context of Maxwell's equations. Participants confirm that it is permissible to pull the time derivative outside of the curl operation due to the property of interchangeability of partial derivatives. This principle allows for the manipulation of the equation ∇ x E = -∂B/∂t without violating mathematical rules. The relevant equation is ∇ x (∇ x E) = ∇ x -∂B/∂t.

PREREQUISITES
  • Understanding of vector calculus, specifically curl operations.
  • Familiarity with Maxwell's equations and their implications in electromagnetism.
  • Knowledge of partial derivatives and their properties.
  • Basic proficiency in mathematical notation used in physics.
NEXT STEPS
  • Study the properties of curl and divergence in vector calculus.
  • Learn about the implications of Maxwell's equations in electromagnetic theory.
  • Explore the concept of interchangeability of partial derivatives in more complex scenarios.
  • Review examples of applying curl operations in physics problems.
USEFUL FOR

Students of physics, particularly those studying electromagnetism, mathematicians working with vector calculus, and educators teaching advanced calculus concepts.

L_landau
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Homework Statement


For the equation ∇ x E = -∂B/∂t I took the curl of both sides to get

∇ x (∇ x E) = ∇ x -∂B/∂t

I feel like it'd be very wrong to pull out the time derivative. Am I correct?
 
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L_landau said:

Homework Statement


For the equation ∇ x E = -∂B/∂t I took the curl of both sides to get

∇ x (∇ x E) = ∇ x -∂B/∂t

I feel like it'd be very wrong to pull out the time derivative. Am I correct?
No, you can pull out the time derivative.
 
L_landau said:

Homework Statement


For the equation ∇ x E = -∂B/∂t I took the curl of both sides to get

∇ x (∇ x E) = ∇ x -∂B/∂t

I feel like it'd be very wrong to pull out the time derivative. Am I correct?

Yes, this is just the property of interchangeability of partial derivatives: ##\partial / \partial t \; \partial / \partial \psi f(\psi,t) = \partial /\partial \psi \; \partial / \partial t f(\psi,t)##.
 

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