Day 30 - Calculus of Variations

Day 30 - Calculus of Variations#

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Announcements#

  • Midterm 2 is posted

  • Look for feedback from DC on projects

Seminars this Week#

MONDAY, March 31, 2025#

  • Condensed Matter Seminar 4:10 pm,1400 BPS, Justin Wilson, Louisiana State University, Title: Measurement and Feedback Driven Adaptive Dynamics in the Classical and Quantum Kicked Top

TUESDAY, April 1, 2025#

  • Theory Seminar, 11:00am., FRIB 1200, Kazuyuki Ogata, Kyushu University, Title: “Knock It Out of the Nucleus -Structure of Nuclei Revealed by Knockout Reactions”

  • High Energy Physics Seminar, 1:00 pm, 1400 BPS, Manel Errando, Washington University in St. Louis, Title: Extracting Meson Distribution Amplitudes from Nonlocal Euclidean Correlations at Next-to-Next-to-Leading Order

Seminars this Week#

WEDNESDAY, April 2, 2025#

  • Astronomy Seminar, 1:30 pm, 1400 BPS, Andy Tzandikas, Univ. of Washington, Title: Searching for the Rarest Stellar Occultations

  • PER Seminar, 3:00 pm., BPS 1400, Abigail Daane, Professor of Physics, South Seattle College, Title: The obstacles, stumbles, and growth in examining the “decolonization” of physics education

THURSDAY, April 3, 2025#

  • Colloquium, 3:30 pm, 1415 BPS, Alex Sushkov, Boston University, Title: Nuclear magnetic resonance at the quantum sensitivity limit

Clicker Question 30-1#

The generic segment, \(ds\), of a curve in 2D Cartesian coordinates is given by

\[ds = \sqrt{(dx)^2 + (dy)^2}\]

The integral of \(ds\) from \(s_1\) to \(s_2\) gives the length of the curve, \(l\). What is the correct expression for \(l\)?

  1. \(l = \int_{s_1}^{s_2} ds\)

  2. \(l = \int_{s_1}^{s_2} \sqrt{(dx)^2 + (dy)^2}\)

  3. \(l = \int_{s_1}^{s_2} \sqrt{1 + (dy/dx)^2} \, dx\)

  4. \(l = \int_{s_1}^{s_2} \sqrt{(dx/dy)^2 + 1} \, dy\)

  5. More than one of the above

Clicker Question 30-2#

I can explain why:

\[\int_{s_1}^{s_2} f((Y(x), Y'(x), x) \, dx > \int_{s_1}^{s_2} f((y(x), y'(x), x) \, dx\]

where \(Y(x) = y(x) + \alpha \eta(x)\), the true path plus an error term.

  1. Yes, I can explain why

  2. I think I can explain why

  3. I’m having trouble seeing why

  4. I don’t think I can explain why

Clicker Question 30-3#

For the function \(Y(x) = y(x) + \alpha \eta(x)\), where \(y(x)\) is the true path, \(\eta(x)\) is a small error term, and \(\alpha\) is a small parameter, what is the derivative of \(Y(x)\) with respect to \(\alpha\)?

\[\frac{dY}{d\alpha} = ?\]

  1. \(y(x)\)

  2. \(\eta(x)\)

  3. \(\eta'(x)\)

  4. \(\alpha \eta(x)\)

  5. \(y'(x) + \alpha \eta'(x)\)

Clicker Question 30-4#

For the function \(Y'(x) = y'(x) + \alpha \eta'(x)\), what is the derivative of \(Y'(x)\) with respect to \(\alpha\)?

\[\frac{dY'}{d\alpha} = ?\]

  1. \(y'(x)\)

  2. \(\eta'(x)\)

  3. \(\eta''(x)\)

  4. \(\alpha \eta'(x)\)

  5. \(y''(x) + \alpha \eta''(x)\)

Clicker Question 30-5#

The “surface term” that we computed for \(\int_{s_1}^{s_2} \eta'(x) \frac{df}{dy'} dx\) is:

\[\left[\eta(x)\dfrac{df}{dy'}\right]_{x_1}^{x_2}=0\]

I can explain why this surface term is equal to zero:

  1. Yes, I can explain why

  2. I think I can explain why

  3. I’m having trouble seeing why

  4. I don’t think I can explain why

  5. I don’t know what a surface term is

Clicker Question 30-6#

We completed this derivation with the following mathematical statement:

\[\int_{s_1}^{s_2} \eta(x) \left[\dfrac{\partial f}{\partial y} - \dfrac{d}{dx}\left(\dfrac{\partial f}{\partial y'}\right)\right] = 0\]

where \(\eta(x)\) is an arbitrary function. What does this imply about the term in square brackets?

  1. The term in square brackets must be a pure function of \(x\).

  2. The term in square brackets must be a pure function of \(y\).

  3. The term in square brackets must be a pure function of \(y'\).

  4. The term in square brackets must be zero.

  5. The term in square brackets must be a non-zero constant.