Why do we use Taylor expansion expressing potential energy

In summary, the Taylor expansion of the potential energy is used to express more general potential energies, not just for an ideal spring. This allows for a more accurate representation of the potential energy, taking into account small stretches of the spring. The expansion involves evaluating the derivatives at the equilibrium position and if the minimum is at x=0, the quadratic term becomes the dominant term and represents the "spring" potential energy.
  • #1
Ahmed Abdalla
10
2
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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  • #3
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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Likes Ahmed Abdalla
  • #4
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!
 

1. Why do we use Taylor expansion expressing potential energy?

The Taylor expansion is a mathematical technique used to approximate a function using a polynomial series. In the case of potential energy, it allows us to express a complex function in terms of simpler, polynomial functions. This makes it easier to analyze and understand the behavior of the potential energy in different situations.

2. How does Taylor expansion help in understanding potential energy?

Taylor expansion helps in understanding potential energy by breaking down a complex function into simpler components. This allows us to examine the behavior of potential energy at a specific point or over a small range, providing insight into the relationship between potential energy and other variables such as distance or position.

3. Can Taylor expansion be used for any type of potential energy function?

Yes, Taylor expansion can be used for any type of potential energy function as long as it is differentiable. This means that the function must have a well-defined derivative at all points within the range of interest.

4. What are the benefits of using Taylor expansion for potential energy?

One of the main benefits of using Taylor expansion for potential energy is that it allows us to approximate the function with a high degree of accuracy, even for complex functions. This makes it a powerful tool for analyzing and predicting the behavior of potential energy in various systems.

5. Are there any limitations to using Taylor expansion for potential energy?

One limitation of using Taylor expansion for potential energy is that it is only accurate within a small range around the point of expansion. This means that it may not accurately predict the behavior of potential energy for large variations in distance or position. Additionally, the accuracy of the approximation depends on the number of terms used in the expansion, so a higher degree polynomial may be needed for more accurate results.

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