Why do we use Taylor expansion expressing potential energy

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Discussion Overview

The discussion revolves around the use of Taylor expansion to express potential energy, particularly in the context of spring potential energy as described by Hooke's Law. Participants explore the theoretical underpinnings and practical implications of using Taylor series in this context.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions why Taylor expansion is preferred for expressing spring potential energy, noting that they often use the relationship F = -Kx and U(x) = -Gradient U(x).
  • Another participant references a PDF on Hooke's Law to provide additional context.
  • Several participants explain that Hooke's Law applies to ideal springs and is valid only for small stretches (small x). They note that Taylor expansion allows for the expression of more general potential energies beyond ideal springs.
  • It is discussed that any infinitely differentiable function, including potential energy, can be expanded in a Taylor series, which is particularly useful when expanding about the equilibrium position (x = 0).
  • Participants elaborate that when expanding potential energy around its minimum, the first derivative is zero, leaving the quadratic term as the dominant term, which corresponds to the "spring" potential energy.

Areas of Agreement / Disagreement

Participants generally agree on the utility of Taylor expansion in expressing potential energy, particularly in the context of small displacements in springs. However, the initial question regarding the necessity of Taylor expansion versus direct application of Hooke's Law remains open for further exploration.

Contextual Notes

The discussion does not resolve whether Taylor expansion is strictly necessary in all cases or if Hooke's Law can suffice in certain scenarios, indicating a potential area for further inquiry.

Ahmed Abdalla
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My textbook doesn’t go into it, can someone tell me why Taylor expansion is used to express spring potential energy? A lot of the questions I do I think I can just use F=-Kx and relate it to U(x) being F=-Gradiant U(x) but I see most answers using the Taylor expansion instead to get 1/2 kx^2. Usually it has to do with x being a small value but I never understood why the Taylor expansion is used
 
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The Hooke's law force, F = -Kx, defines an ideal spring. In a real spring, Hooke's law is valid only for small stretches (small x) of the spring. The Taylor expansion of the potential energy about the equilibrium position (in this case, x = 0), is used to express more general potential energies, not necessaily of an ideal spring. Since any function (inifinitely differentiable) can be expanded in a Taylor series, even the potential energy can be expanded in a Taylor series. If you are expanding the potential energy about its minimum (at x = 0), then
U(x) = U(0) + x (dU/dx)0 + (x2 / 2!) (d2U/dx2)0 + ...(higher order terms)
where the subscript 0 on the derivatives means that they are evaluated at x = 0. If x = 0 is a minimum, then the first derivative in the above expansion is zero, and that leaves the quadratic term as the dominant term in the Taylor expansion of the potential energy, and is identified as the "spring" potential energy.
 
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Chandra Prayaga said:
The Hooke's law force, F = -Kx, defines an ideal spring. In a real spring, Hooke's law is valid only for small stretches (small x) of the spring. The Taylor expansion of the potential energy about the equilibrium position (in this case, x = 0), is used to express more general potential energies, not necessaily of an ideal spring. Since any function (inifinitely differentiable) can be expanded in a Taylor series, even the potential energy can be expanded in a Taylor series. If you are expanding the potential energy about its minimum (at x = 0), then
U(x) = U(0) + x (dU/dx)0 + (x2 / 2!) (d2U/dx2)0 + ...(higher order terms)
where the subscript 0 on the derivatives means that they are evaluated at x = 0. If x = 0 is a minimum, then the first derivative in the above expansion is zero, and that leaves the quadratic term as the dominant term in the Taylor expansion of the potential energy, and is identified as the "spring" potential energy.
Thank you that explained it in great detail!
 

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