SME
University Physics III: Waves, Optics & Modern · Problem type
The cosmic background’s temperature with redshift
Use this for “The cosmic microwave background is 2.725 K today. How hot was it at z = 1,100, and where does today’s spectrum peak?”
Chart
3,000
K:
λₘₐₓ
=
b
T
=
2,898
μm·
K
3,000
K
=
0.9659
μm
Tinted: the visible band, 380 to 750 nm
This is an example. Type your own values to start a new problem.
T (K)
=
T₀ (K)
×
(1
+
z
)
λtoday
Peak wavelength today
worked out (not typed) · mm
λₜₕₑₙ
Peak wavelength then
worked out (not typed) · μm
Formulas
T
=
T₀
×
(1
+
z
)
3,000.225 = 2.725 × (1 + 1,100)
λtoday
=
2.898
T₀
1.063
=
2.898
2.725
λₜₕₑₙ
=
2,898
T
0.9659
=
2,898
3,000.225
Assumptions
•
The background is blackbody light whose wavelengths all stretch by 1 + z as space expands, so it stays a blackbody and its T falls as 1 ÷ (1 + z).
•
Wien’s law λₘₐₓ = b ÷ T with b = 2.898 × 10⁻³ m·K, which is 2.898 mm·K or 2898 μm·K.
•
Today T₀ = 2.725 K; near z = 1,100 the universe had cooled to about 3,000 K and light first travelled freely.
Step-by-step
WE KNOW
T₀ = 2.725 K, z = 1,100
FIND
temperature then (T), peak wavelength today (λtoday), peak wavelength then (λₜₕₑₙ)
Step 1 · Find temperature then (T)
Use T = T₀ × (1 + z)
Every
wavelength
has
stretched
1
+
z
times
since
then,
and
T
goes
as
1
λ
,
so
the
light
was
1
+
z
times
hotter.
T
=
2.725
×
(1
+
1,100)
T
=
2.725
×
1,101
T
=
3,000.225
K
Step 2 · Find peak wavelength today (λtoday)
Use λtoday = 2.898 ÷ T₀
Wien’s law with b = 2.898 mm·K: the peak today, in the microwaves.
λtoday
=
2.898
2.725
λtoday = 1.063 mm
Step 3 · Find peak wavelength then (λₜₕₑₙ)
Use λₜₕₑₙ = 2,898 ÷ T
Wien’s law again with b = 2898 μm·K: the hotter the light, the shorter its peak.
λₜₕₑₙ
=
2,898
3,000.225
λₜₕₑₙ = 0.9659 μm
Check
Put the numbers back in. Both sides should match.
3,000.225 = 2.725 × (1 + 1,100) ✓
1.063
=
2.898
2.725
✓
0.9659
=
2,898
3,000.225
✓
Topic check
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