Calculating Lie Derivative for Case (ii)

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Discussion Overview

The discussion revolves around the calculation of the Lie derivative in the context of differential geometry, specifically focusing on an exercise from Fecko's textbook. Participants are exploring how to apply the pullback operator ##\phi^*## to a given function ##\psi## and the implications of integral curves of vector fields on this process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to apply the pullback operator ##\phi^*## to the function ##\psi## and seeks assistance.
  • Another participant prompts for clarification by asking what the textbook states regarding the application of ##\phi^*##.
  • A participant describes an integral curve of a vector field and discusses how a point is shifted and how a function ##\psi(x)## transforms under this shift.
  • There is a proposal that the function becomes ##\phi = e^{-(x+t)^2}##, and a request for a procedure to calculate the Lie derivative for a different case is made.
  • A later reply provides a definition of the Lie derivative and outlines steps to calculate it, including finding integral curves and displacing them by an infinitesimal amount.
  • Participants discuss the specific case (i) and its implications for the function ##f(x) = \exp(-x^2)##, leading to a calculation of the Lie derivative.
  • One participant encourages others to attempt the calculation for case (ii) themselves.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants express varying levels of understanding and propose different approaches to the problem. There is an ongoing exploration of the concepts without a definitive resolution.

Contextual Notes

Participants are working with specific examples and definitions from the textbook, which may limit the generalizability of their discussions. The calculations and transformations discussed depend on the definitions and assumptions made about the vector fields and functions involved.

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?

IMG_20190127_160545_484.jpeg
 

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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?
 
Last edited:
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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