What is Euler's Criterion in plain simplified english

  • Thread starter Thread starter jenzao
  • Start date Start date
  • Tags Tags
    English
jenzao
Messages
48
Reaction score
0
What is "Euler's Criterion" in plain simplified english

This sentence in textbook reads: "If ∂M/∂y = ∂N/∂x the differential is exact (Euler’s Criterion)."

What does the equation mean, and also I am not familiar with the backwards number 6 symbol --what is that?
thanks
 
Physics news on Phys.org


This has no "plain English" definition :) You should grab some multivariate analysis textbook. Or try this: http://en.wikipedia.org/wiki/Partial_derivative"
 
Last edited by a moderator:


Given that f is a function of several variables, \frac{\partial f}{\partial x} is the partial derivative of f with respect to the variable x. It is the same as the normal derivative of f if we take all other variables to be constants.
Suppose we have a function f(x,y). The differential of f(x,y) is defined to be df(x,y) = \frac{\partial f}{\partial x}(x,y) dx + \frac{\partial f}{\partial y}(x,y) dy. If you have an expression M dx + N dy, it is called an exact differential if it is the differential of some f(x,y). The Euler criterion is one way of telling whether this is the case; it is easy to see that it should be the case.
 


"If you have an expression M dx + N dy, it is called an exact differential if it is the differential of some f(x,y). "

Could you please provide an example of this?
 


jenzao said:
"If you have an expression M dx + N dy, it is called an exact differential if it is the differential of some f(x,y). "

Could you please provide an example of this?
Suppose you have the vector field g(x,y) = (-x/r3/2) dx - (y/r3/2) dy, where r = \sqrt{x^2 + y^2}. Then g(x,y) is an exact differential, because g(x,y) = df(x,y) where f(x,y) = r-1/2. Ie., f(x,y) is the potential of a central force g(x,y). You may also see f = -\nabla g.
 


If you don't know about partial derivatives, why are you worrying about "Euler's Criterion" and "exact differentials"? The latter requires that you be well versed in partial derivatives.
 
Back
Top