Proving the Relation in "Methods of Theoretical Physics

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SUMMARY

The discussion focuses on proving the relation regarding displacement being perpendicular to a surface as stated in "Methods of Theoretical Physics" by Morse and Feshbach. The key equation presented is $$\frac{dx}{\partial \psi /\partial x} = \frac{dy}{\partial \psi /\partial y} = \frac{dz}{\partial \psi /\partial z}$$. The solution approach involves demonstrating that the dot product of the displacement vector (dx, dy, dz) and the gradient of the scalar field (∇ψ) equals zero, confirming the perpendicularity condition.

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  • Understanding of vector calculus, particularly gradients and dot products.
  • Familiarity with the concepts of displacement in physics.
  • Knowledge of scalar fields and their properties.
  • Basic proficiency in mathematical notation and manipulation.
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  • Explore the implications of scalar fields in physics.
  • Learn about the geometric interpretation of dot products in relation to angles between vectors.
  • Review examples of displacement vectors in various physical contexts.
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This discussion is beneficial for physics students, educators, and anyone interested in theoretical physics, particularly those studying vector calculus and its applications in physical scenarios.

arbri
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Homework Statement


In "Methods of theoretical physics" from the authors Morse and Feshbach is writen:
" The displacement (dx,dy,dz) is perpendicular to the surface if the component displacemetn satisfy the equation:



How to prove this relation?



Homework Equations


the relevant equation is the equation of total change



The Attempt at a Solution

 
Last edited by a moderator:
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hi arbri! welcome to pf! :smile:
arbri said:
In "Methods of theoretical physics" from the authors Morse and Feshbach is writen:
" The displacement (dx,dy,dz) is perpendicular to the surface if the component displacemetn satisfy the equation:

$$\frac{dx}{\partial \psi /\partial x} = \frac{dy}{\partial \psi /\partial y} = \frac{dz}{\partial \psi /\partial z}$$

How to prove this relation?

that's just (dx,dy,dx)·∇ψ = 0 :wink:
 

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