Graduate Calculating Lie Derivative for Case (ii)

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SUMMARY

The discussion focuses on calculating the Lie derivative for a specific case in differential geometry, particularly using the Fecko textbook. The process involves applying the pullback operator ##\phi^*## to a function ##\psi## and understanding how it transforms under the flow of a vector field. The Lie derivative is defined as $$\lim_{t \longrightarrow 0} \frac{ \phi^*_t f - f}{t}$$, with the procedure requiring the identification of integral curves and displacement of the curve by an infinitesimal amount. For case (i), the example provided shows that if ##f(x) = \exp(-x^2)##, then ##\mathcal L_V \exp(-x^2) = -2x \exp(-x^2)##.

PREREQUISITES
  • Understanding of differential geometry concepts
  • Familiarity with the Lie derivative and pullback operator ##\phi^*##
  • Knowledge of integral curves and vector fields
  • Basic calculus, particularly limits and derivatives
NEXT STEPS
  • Study the application of the pullback operator ##\phi^*## in differential geometry
  • Learn about integral curves of vector fields in detail
  • Explore the calculation of Lie derivatives in various contexts
  • Practice exercises from the Fecko textbook on differential geometry
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Students and researchers in differential geometry, particularly those learning about Lie derivatives and vector fields, as well as educators looking for exercises to enhance understanding of these concepts.

Abhishek11235
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I am relatively new to differential geometry. I am studying it from Fecko Textbook on differential geometry. As soon as he introduces the concept of lie derivative,he asks to do exercise 4.2.2 in picture. The question is,how do I apply ##\phi^*## to given function ##\psi## . I know that ##\phi^*## transport tensor fields against direction of flow. But how it does,I don't know.

Can anyone help me?

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Abhishek11235 said:
But how it does,I don't know.
What does your book say?
 
Consider case (i) first. An integral curve of V is Φ(t, x) = t + x because dΦ/dt = 1 and Φ(0, x) = x. What happens to a point x? Its shifted to t + x. What happens to a function ψ(x)?
 
kent davidge said:
Consider case (i) first. An integral curve of V is Φ(t, x) = t + x because dΦ/dt = 1 and Φ(0, x) = x. What happens to a point x? Its shifted to t + x. What happens to a function ψ(x)?
It becomes ##\phi= e^{-(x+t)^2}##. Is this the answer? Can you give procedure to calculate lie derivative like that for case 2?
 
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Abhishek11235 said:
It becomes ##\phi= e^{-(x+t)^2}##. Is this the answer?
yes, and the problem asks you to draw the graph of ##f##. You draw the curve as a function of ##t## for a fixed ##x##, which is the initial point. (Don't name the function ##\phi## to avoid confusion with the curve, which the problem already calls ##\phi##.)
Abhishek11235 said:
Can you give procedure to calculate lie derivative like that for case 2?
The Lie Derivative is something else. It's when you take the limit $$\lim_{t \longrightarrow 0} \frac{ \phi^*_t f - f}{t}.$$ What you need to do:

1 - Find the integral curves for the given vector field.
2 - Displace the curve by an infinitesimal amount and take the limit as I gave above.

For case (i) this is ##\phi(\epsilon, x) \equiv x' = x + \epsilon V##, but ##V = 1##, then ##x' = x + \epsilon##. So ##f(x') \approx f(x) + \epsilon df/dx## and ##\mathcal L_V f = df / dx##. In our case, ##f (x) = \exp (-x^2)##. So ##\mathcal L_V \exp (-x^2) = -2x \exp (-x^2)##.

Can you try yourself case (ii)?
 
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