CLASSICAL-MECHANICS · Interactive Practice

Defining Reality: The Second and the Meter

IKey Formulas

Formula Name Description
1 s=9,192,631,770ΔνCs1\ \text{s} = \dfrac{9{,}192{,}631{,}770}{\Delta\nu_{Cs}} Definition of the second Fixed number of cesium periods
ΔνCs=9,192,631,770 Hz\Delta\nu_{Cs} = 9{,}192{,}631{,}770\ \text{Hz} Cesium frequency Exact by definition
c≡299,792,458 m/sc \equiv 299{,}792{,}458\ \text{m/s} Speed of light Defined constant (not measured)
1 m=c⋅1299,792,458 s1\ \text{m} = c \cdot \dfrac{1}{299{,}792{,}458}\ \text{s} Definition of the meter Distance light travels in that time
λ=cf\lambda = \dfrac{c}{f} Wavelength–frequency Light wavelength from its frequency

IIInteractive Visualizations

Visualization 1 — Clock precision as real-world drift

How much does each timekeeper drift, and how long until it is wrong by a full second?

💡 A cesium clock is a million times steadier than the spinning Earth — six orders of magnitude — which is why the world stopped timing by the planet and started timing by an atom.

Visualization 2 — The meter as a light-travel time

Let light travel for a slice of time and it sweeps out a precise distance: d=c td = c\,t.

💡 We measure time far more precisely than length, so fixing cc turns every length into a time measurement — the reason cc is defined, not measured.

Visualization 3 — Wavelength from frequency, λ=c/f\lambda = c/f

Raise the frequency and the wave compresses: λ\lambda and ff are inversely locked through cc.

💡 Feed 606606 nm back into f=c/λf = c/\lambda and you recover 4.949×10144.949\times10^{14} Hz — the wavelength and frequency versions of the meter are the same statement.

IIIQuiz Questions

Question 1

The modern definition of the second is based on cesium-133. How many periods of the cesium radiation make up exactly one second?

✅ Correct! One second = 9,192,631,770 cesium periods.

❌ Not quite. That value belongs to a different definition. The second uses the cesium frequency of 9,192,631,770 Hz.

Show solution

Solution:

By definition since 1967: 1 s=9,192,631,770ΔνCs,ΔνCs=9,192,631,770 Hz1\ \text{s} = \frac{9{,}192{,}631{,}770}{\Delta\nu_{Cs}}, \qquad \Delta\nu_{Cs} = 9{,}192{,}631{,}770\ \text{Hz}

One second equals 9,192,631,770 periods of the cesium hyperfine transition radiation. This number is exact by definition — not rounded.

The distractors: 86,400 is the number of seconds in a day (old astronomical definition), 299,792,458 is the speed of light, and 1,650,763.73 is the krypton-86 wavelength count used for the 1960 meter.

Question 2

Since 1983, the speed of light c=299,792,458c = 299{,}792{,}458 m/s is a quantity that scientists measure in the laboratory and refine over time.

True or False?

✅ Correct! Since 1983, cc is defined exactly, not measured.

❌ Not quite. The 1983 pivot fixed cc by definition. It is a defined constant, and the meter is derived from it.

Show solution

Solution: False.

In 1983 scientists fixed the speed of light by international agreement: c≡299,792,458 m/s(exact, by definition)c \equiv 299{,}792{,}458\ \text{m/s} \quad (\text{exact, by definition})

The speed of light is no longer a measured quantity — it is a defined constant. The meter is then derived from it as the distance light travels in 1299,792,458\tfrac{1}{299{,}792{,}458} of a second. Any experiment that appears to measure cc is really testing the quality of our length and time standards.

Question 3

Using λ=c/f\lambda = c/f with c=299,792,458c = 299{,}792{,}458 m/s, estimate the wavelength of the krypton-86 orange line whose frequency is approximately f=4.949×1014f = 4.949 \times 10^{14} Hz.

Which value is closest?

✅ Correct! λ = c/f ≈ 606 nm, the orange krypton line.

❌ Not quite. Divide c by f: 299,792,458 ÷ (4.949×10¹⁴) ≈ 6.06×10⁻⁷ m = 606 nm.

Show solution

Solution:

λ=cf=299,792,458 m/s4.949×1014 Hz\lambda = \frac{c}{f} = \frac{299{,}792{,}458\ \text{m/s}}{4.949\times10^{14}\ \text{Hz}}

λ≈6.058×10−7 m=605.8 nm≈606 nm\lambda \approx 6.058\times10^{-7}\ \text{m} = 605.8\ \text{nm} \approx 606\ \text{nm}

This is the orange krypton-86 line used to define the meter in 1960. The 1960 standard set the meter at 1,650,763.73 of these wavelengths, which reverses back to the same frequency — confirming the definitions are consistent.

Question 4

When you chain the definitions together, the meter can be written as 1 m=30.6633149⋅cΔνCs1\ \text{m} = 30.6633149 \cdot \dfrac{c}{\Delta\nu_{Cs}}.

What does this equation reveal about the relationship between the meter and the second?

✅ Correct! Both units trace back to ΔνCs — the meter contains the second.

❌ Not quite. The equation shows both the meter and second depend on the cesium frequency ΔνCs — they are linked, not independent, and no physical artifact is involved.

Show solution

Solution:

Starting from the meter and substituting the second's definition:

1 m=c299,792,458⋅1 s=c299,792,458⋅9,192,631,770ΔνCs1\ \text{m} = \frac{c}{299{,}792{,}458}\cdot 1\ \text{s} = \frac{c}{299{,}792{,}458}\cdot \frac{9{,}192{,}631{,}770}{\Delta\nu_{Cs}}

Combining the constants: 9,192,631,770299,792,458=30.6633149\frac{9{,}192{,}631{,}770}{299{,}792{,}458} = 30.6633149

1 m=30.6633149⋅cΔνCs\boxed{1\ \text{m} = 30.6633149 \cdot \frac{c}{\Delta\nu_{Cs}}}

Both the meter and the second trace back to a single cesium frequency ΔνCs\Delta\nu_{Cs}. They are not independent — the meter's definition literally contains the second's. This reflects the modern philosophy of defining units by fixing the fundamental constants of nature.

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