Free Pearson Edexcel International A Level Physics lesson

Distances, stellar populations and evolution

Use geometry and luminosity before interpreting a star’s life.

What you’ll learn

  • Match the defined angle with its baseline and quote uncertainty as the angle gets small.
  • Explain both the physical distance calculation and where its luminosity calibration came from.
  • Use both HR coordinates to infer size and connect changes to stellar structure and fuel.

Use Earth’s orbit as a geometric baseline

A nearby star appears to shift against a distant background as Earth changes position in its orbit. The annual parallax angle p is conventionally half the largest six-month angular displacement. The triangle uses the Earth–Sun distance of one astronomical unit as its baseline, rather than the full diameter with that half-angle.

For a small angle in radians, d ≈ b/p with b = 1 AU. A parsec is the distance at which one AU subtends one arcsecond, giving d in parsecs = 1/p in arcseconds. One arcsecond is 1/3600 degree and 1 parsec is approximately 3.086 × 10¹⁶ m. Distinguish a parsec from a light-year, which is a distance travelled by light in one year.

Small parallax angles become difficult to measure, giving a useful nearby rung of a distance ladder rather than an unlimited method. The fractional distance uncertainty is approximately the fractional parallax uncertainty for small relative errors. Proper motion and instrumental effects must be separated from the annually repeating displacement; taking one photograph does not provide a parallax baseline.

d ≈ 1 AU/p(rad) d(pc) = 1/p(arcsec) 1 pc ≈ 3.086 × 10¹⁶ m

Symbols and units

  • p is the annual parallax half-angle, not gas pressure. Use radians in d ≈ b/p, or arcseconds in the numerical relation d(pc) = 1/p(arcsec).
  • b = 1 astronomical unit (AU) is the Earth–Sun baseline; d is distance in the same length unit as b when using radians.
  • 1 arcsecond is 1/3600 degree; 1 parsec (pc) ≈ 3.086 × 10¹⁶ m. A parsec is a length, not a time.
  • The maximum six-month apparent displacement is 2p in the stated geometry, pairing the full shift with the full orbital-diameter baseline.

How the equation explains the idea

  1. Use the half-angle that belongs with one AU: the example’s full shift 0.20 arcsec gives p = 0.20/2 = 0.10 arcsec.
  2. The parallax triangle gives tan p = b/d. For a small angle in radians, tan p ≈ p, so d ≈ b/p; the parsec definition packages this geometry as d(pc) = 1/p(arcsec).
  3. Insert p = 0.10 arcsec: d = 1/0.10 = 10 pc = 10 × 3.086 × 10¹⁶ = 3.086 × 10¹⁷ m, approximately 3.09 × 10¹⁷ m.
  4. Because d is inversely proportional to p, a small fractional parallax error gives approximately the same fractional distance error in magnitude. The approximation becomes unreliable when relative angle uncertainty is large.
Smaller annual parallax implies a greater distance

Conditions. Computed from d(pc) = 1/p(arcsec), normalized to p₀ = 0.100 arcsec = 4.848 × 10⁻⁷ rad and d₀ = 10.0 pc ≈ 3.086 × 10¹⁷ m. The parallax p is half the maximum six-month shift and uses baseline 1 AU. Only 0.2 ≤ p/p₀ ≤ 2 is plotted; p = 0 has no finite distance.

Parallax half-angle ratio p/p₀; Distance ratio d/d₀.

Parallaxes 0.20, 0.10 and 0.05 arcsec correspond to 5, 10 and 20 pc. The curve gets much steeper at small angles: the same absolute angular error gives a greater distance error. Using the full six-month shift instead of its half would halve the inferred distance.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Use a measured annual shift

  1. The maximum six-month displacement is 0.20 arcsec.
  2. Parallax p = 0.10 arcsec, so d = 1/0.10 = 10 pc.
  3. Distance is approximately 3.09 × 10¹⁷ m.

Halve the full shift before using the one-AU parallax relation.

Keep this idea. Match the defined angle with its baseline and quote uncertainty as the angle gets small.

Calibrate luminosity to extend the distance ladder

A standard candle has a luminosity that can be estimated independently of the received intensity. Combining that luminosity with I = L/(4πd²) gives d = √(L/(4πI)). It need not mean that every member of a class has exactly the same uncorrected luminosity; a measurable relationship may standardise the class.

For example, a calibrated period–luminosity relationship can estimate a Cepheid variable’s luminosity from its pulsation period. Geometrically measured nearby objects help calibrate more distant indicators, creating a linked distance ladder. A calibration uncertainty on an early rung can affect many later distances rather than averaging away independently.

Measure received intensity in a defined band or apply a justified full-spectrum correction, and account for dust extinction. Incorrectly adopting a larger luminosity makes the inferred distance larger. Because distance involves a square root, its approximate fractional uncertainty is half the combined fractional uncertainties in L and I, under a simple worst-case treatment; systematic calibration bias must still be discussed separately.

d = √(L/(4πI)) approximate δd/d = ½(δL/L + δI/I)

Symbols and units

  • L is independently calibrated luminosity in W; I is received intensity in W m⁻² and d is distance in m.
  • δL, δI and δd denote small positive uncertainty magnitudes in the corresponding quantities. Each δquantity/quantity ratio is dimensionless.
  • The displayed addition of fractional uncertainties is an approximate worst-case rule; independent random standard uncertainties are normally combined in quadrature instead.
  • 1 pc ≈ 3.086 × 10¹⁶ m. Luminosity and intensity must cover matching spectral ranges with extinction appropriately accounted for.

How the equation explains the idea

  1. Start with I = L/(4πd²). Multiply by d² and divide by I to isolate d² = L/(4πI); take the positive square root because distance is positive.
  2. Insert the example values: d² = (1.00 × 10²⁹)/(4π × 2.00 × 10⁻¹¹) = 3.979 × 10³⁸ m².
  3. Thus d = 1.995 × 10¹⁹ m. Dividing by 3.086 × 10¹⁶ m per pc gives approximately 646 pc. Without an independent L, the same I could describe many luminosity–distance combinations.
  4. Since d ∝ L¹ᐟ²I⁻¹ᐟ², the small-error worst-case fractional bound is approximately ½(δL/L + δI/I). A shared luminosity calibration bias remains systematic and does not vanish by adding more sources.

Worked example

A calibrated candle

  1. L = 1.00 × 10²⁹ W and I = 2.00 × 10⁻¹¹ W m⁻².
  2. d² = 1.00 × 10²⁹/(4π × 2.00 × 10⁻¹¹).
  3. d = 1.995 × 10¹⁹ m, about 646 pc.

Distance follows from an independently estimated luminosity.

Keep this idea. Explain both the physical distance calculation and where its luminosity calibration came from.

Read the HR diagram as a population and evolution map

A conventional Hertzsprung–Russell diagram plots luminosity upward and surface temperature increasing toward the left. The main sequence runs roughly from hot luminous upper-left stars to cool faint lower-right stars. Red giants occupy a cool but luminous upper region, while white dwarfs occupy a hot but faint lower-left region. These positions imply different radii through the Stefan–Boltzmann law.

Gravity contracts a cloud to form a protostar, increasing its internal temperature. Sustained core hydrogen fusion supports a main-sequence phase through an approximate balance of inward gravity and outward pressure. When core hydrogen becomes depleted, contraction and shell burning change the structure; outer layers can expand and cool while luminosity rises, moving the star toward a giant region.

Lower-mass stars can shed outer layers and leave a white dwarf that cools without sustained core fusion. Sufficiently massive stars pass through further burning stages and may undergo core collapse and a supernova, leaving a neutron star or black hole depending on their remnant conditions. A star does not simply travel along the entire main sequence as it ages: initial mass strongly influences its path and lifetime.

At fixed L: R ∝ 1/T² HR axes: luminosity upward; temperature hotter to the left

Symbols and units

  • L is stellar luminosity in W, R is radius in m and T is absolute surface temperature in K; σ is the Stefan–Boltzmann constant in W m⁻² K⁻⁴.
  • Subscripts cool and hot identify the two equal-luminosity stars; radius and temperature ratios are dimensionless.
  • On a conventional HR diagram, luminosity increases upward and temperature increases toward the left. These axes describe stellar properties, not age and time.

How the equation explains the idea

  1. Rearrange L = 4πR²σT⁴ to R² = L/(4πσT⁴), then take a positive square root: R = √(L/(4πσ))/T².
  2. At fixed L, the prefactor is constant, so Rcool/Rhot = (Thot/Tcool)². If Tcool = Thot/2, the radius ratio is 2² = 4.
  3. Emitting area scales as R², so the cooler equal-luminosity star has 4² = 16 times the area. Its power per area is 1/16 as large, leaving total luminosity unchanged.
  4. This establishes a size comparison, not a unique evolutionary history. Interpret the HR location together with mass, fuel and structure; the diagram is not one universal chronological track.

Worked example

Compare two equal-luminosity stars

  1. One star has half the surface temperature of the other.
  2. For equal L, Rcool/Rhot = (Thot/Tcool)² = 4.
  3. The cooler star needs sixteen times the emitting area and four times the radius.

Cool and luminous implies large radius, not necessarily low energy output.

Keep this idea. Use both HR coordinates to infer size and connect changes to stellar structure and fuel.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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