Action Between Frames

At a glance

• The naive steady-motion scores are 𝑄1 =1, 𝑄2 =12, and 𝑄4 =14.
• The corrected steady-motion score is 𝑆𝑁 =1 for every 𝑁.
• For 𝑥(𝑡) =𝑡2, the frame actions are 𝑆1 =1, 𝑆2 =54, and 𝑆4 =2116.
• The continuous actions are 𝑆[𝑡] =1 and 𝑆[𝑡2] =43.
• Finer frames change the approximation, not the action assigned to the underlying continuous history.

Key idea

A discrete contribution must include both a rate and the duration over which that rate acts. The sum

is a Riemann sum for the continuous action ∫10(𝑥′(𝑡))2 𝑑𝑡. It becomes a property of the whole history as the slices become arbitrarily fine.

Worked solution, part by part

1. Diagnose the naive score

For the steady history 𝑥(𝑡) =𝑡, each of the 𝑁 equal intervals has

Therefore

So the table entries are

The puck's continuous motion is identical in all three recordings, but 𝑄𝑁 changes merely because the camera samples it differently. Indeed, 𝑄𝑁 →0 as 𝑁 →∞. That makes 𝑄𝑁 unsuitable as the action of the motion.

2. Use the normalized frame action

For 𝑥(𝑡) =𝑡, every interval has Δ𝑥𝑘 =Δ𝑡 =1/𝑁, so every interval speed is 1. Thus

for every positive integer 𝑁.

For 𝑥(𝑡) =𝑡2:

The values 1,54,2116 increase toward 43.

3. Pass to the continuous action

For 𝑥(𝑡) =𝑡, we have 𝑥′(𝑡) =1, so

This agrees exactly with every frame sum for the steady history.

For 𝑥(𝑡) =𝑡2, we have 𝑥′(𝑡) =2𝑡, so

This is the limit approached by the accelerating-history frame sums.

A complete interpretation is: Finer measurements change the numerical Riemann-sum approximation, while the limiting integral is the action attached to the continuous history itself.

Check and completeness

For steady motion, the corrected action remains exactly 1 at every listed resolution and in the integral. For accelerating motion,

so the finite values move in the expected direction toward the integral. All comparisons hold the endpoints and one-second duration fixed; only the frame partition or the history function changes.

Teaching note: connection to the first lab

For a free particle of mass 𝑚, the physical action is

Multiplying every history's normalized action by the same positive constant 𝑚/2 changes neither its ranking nor where it is stationary. In the first lab, every interval lasted exactly one second, so dividing by Δ𝑡 and multiplying by Δ𝑡 did not visibly change the squared-step formula.