Is the Magnetic Vector Potential Dependent on a Constant Vector?

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SUMMARY

The discussion centers on the dependency of the Magnetic Vector Potential on a constant vector, specifically addressing problem 2.b) from the University of Washington's physics homework. The conclusion drawn is that for the vector potential A to remain unchanged (A = A'), the term Lambda must be a constant vector. If Lambda were not constant, the gradient would introduce additional terms to A, contradicting the observed behavior. The analysis suggests that the difference D = A - A' can be expressed as a gradient, reinforcing the necessity of Lambda being constant.

PREREQUISITES
  • Understanding of vector calculus, particularly gradients.
  • Familiarity with the concept of Magnetic Vector Potential in electromagnetism.
  • Knowledge of Cartesian coordinates and their application in physics.
  • Basic principles of vector fields and their properties.
NEXT STEPS
  • Study the properties of gradients in vector calculus.
  • Explore the role of Magnetic Vector Potential in classical electromagnetism.
  • Review the implications of constant vectors in physical equations.
  • Analyze examples of vector potential transformations in different coordinate systems.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and researchers interested in vector calculus applications in physical theories.

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Homework Statement



It's problem 2.b) on this page:

http://www.phys.washington.edu/users/schick/322A/322-08ps3.pdf

The Attempt at a Solution



So, what it looks like to me, since the only terms in the actual vector potential are all multiplied, Lambda would have to be a constant, so that when you take the gradient of it, it becomes 0 and doesn't change A'. So basically A=A' and Lambda = some constant vector.

If Lambda wasn't constant, then taking the gradient would give some additional terms added to A, whereas no addition takes place.
 
Physics news on Phys.org
Write A' in Cartesian coords.
Take the difference D=A-A'.
It should be easy to show that D is a gradient.
 

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