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Transcript
Stellar Parallax &
Electromagnetic Radiation
Stellar Parallax
• Given p in arcseconds (”), use
d=1/p to calculate the distance
which will be in units “parsecs”
• By definition, d=1pc if p=1”, so
convert d to A.U. by using
trigonometry
• To calculate p for star with d given
in lightyears, use d=1/p but
convert ly to pc.
• Remember: 1 degree = 3600”
• Note: p is half the angle the star
moves in half a year
Our Stellar Neighborhood
Scale Model
• If the Sun = a golf ball, then
–
–
–
–
–
Earth = a grain of sand
The Earth orbits the Sun at a distance of one meter
Proxima Centauri lies 270 kilometers (170 miles) away
Barnard’s Star lies 370 kilometers (230 miles) away
Less than 100 stars lie within 1000 kilometers (600 miles)
• The Universe is almost empty!
• Hipparcos satellite measured distances to nearly 1
million stars in the range of 330 ly
• almost all of the stars in our Galaxy are more distant
Luminosity and Brightness
• Luminosity L is the total power
(energy per unit time) radiated
by the star, actual brightness of
star, cf. 100 W lightbulb
• Apparent brightness B is how
bright it appears from Earth
– Determined by the amount of
light per unit area reaching Earth
– B  L / d2
• Just by looking, we cannot tell
if a star is close and dim or far
away and bright
Brightness: simplified
• 100 W light bulb will look
9 times dimmer from 3m
away than from 1m away.
• A 25W light bulb will look
four times dimmer than a
100W light bulb if at the
same distance!
• If they appear equally
bright, we can conclude that
the 100W lightbulb is twice
as far away!
Same with stars…
• Sirius (white) will look 9
times dimmer from 3
lightyears away than from 1
lightyear away.
• Vega (also white) is as
bright as Sirius, but appears
to be 9 times dimmer.
• Vega must be three times
farther away
• (Sirius 9 ly, Vega 27 ly)
Distance Determination Method
• Understand how bright an object is
(L)
• Observe how bright an object appears (B)
• Calculate how far the object is away:
B  L / d2
So
L/B  d2 or
d  √L/B
Homework: Luminosity and Distance
• Distance and brightness can be used to find
the luminosity:
L  d2 B
• So luminosity and brightness can be used to
find Distance of two stars 1 and 2:
d21 / d22 = L1 / L2 (since B1 = B2 )
i.e. d1 = (L1 / L2)1/2 d2
Homework: Example Question
• Two stars -- A and B, of
luminosities 0.5 and 2.5 times the
luminosity of the Sun, respectively -- are
observed to have the same apparent
brightness. Which one is more distant?
• Star A
• Star B
• Same distance
Homework: Example Question
• Two stars -- A and B, of luminosities 0.5 and 2.5 times the
luminosity of the Sun, respectively -- are observed to have the
same apparent brightness.
How much farther away is it than the other?
• LA/d2A = BA =BB = LB/d2B  dB = √LB/LA dA
•  Star B is √5=2.24 times as far as star A
“Light” – From gamma-rays to radio waves
• The vast majority of information we have about
astronomical objects comes from light they either
emit or reflect
• Here, “light” stands for all sorts of
electromagnetic radiation
• A type of wave, electromagnetic in origin
• Understanding the properties of light allows us to
use it to determine the
– temperature
– chemical composition
– (radial) velocity
of distant objects
Waves
• Light is a type of
wave
• Other common
examples: ocean
waves, sound
• A disturbance in a
medium (water, air,
etc.) that propagates
• Typically the medium
itself does not move
much
crest
Wave Characteristics
wavelength
2 x amplitude
trough
direction of wave motion
• Wave frequency: how often a crest washes over you
• Wave speed = wavelength ()  frequency (f)
Electromagnetic Waves
• Medium = electric and magnetic field
• Speed = 3 105 km/sec
Electromagnetic Spectrum
Energy:
low

medium

high
Electromagnetic Radiation:
Quick Facts
• There are different types of EM radiation, visible
light is just one of them
• EM waves can travel in vacuum, no medium needed
• The speed of EM radiation “c” is the same for all
types and very high ( light travels to the moon in 1
sec.)
• The higher the frequency, the smaller the
wavelength ( f = c)
• The higher the frequency, the higher the energy of
EM radiation (E= h f, where h is a constant)
Visible Light
• Color of light determined
by its wavelength
• White light is a mixture
of all colors
• Can separate individual
colors with a prism
Three Things Light Tells Us
• Temperature
– from black body spectrum
• Chemical composition
– from spectral lines
• Radial velocity
– from Doppler shift
Temperature Scales
Fahrenheit
Centigrade
Kelvin
459 ºF
273 ºC
0K
32 ºF
0 ºC
273 K
Human body
temperature
98.6 ºF
37 ºC
310 K
Water boils
212 ºF
100 ºC
373 K
Absolute zero
Ice melts
WUP: What is the Blackbody Curve and
how does it depend on temperature?
• jj: The blackbody curve describes the distribution of
reemitted radiation from a blackbody (an object that
absorbs and reemits all radiation falling upon it. The peak
of the frequency on a blackbody curve is directly
proportionate to to the temperature.
• K: Blackbody curve is an idealized body that absorbs all
electromagnetic radiation. It needs to remain at a constant
temperature but the spectrum is determined by the
temperature alone. As the temperature decreases so does
the intensity and its peak will move to a longer
wavelength.
Black Body Spectrum
• Objects emit radiation of all frequencies,
but with different intensities
Ipeak
Higher Temp.
Ipeak
Ipeak
Lower Temp.
fpeak<fpeak <fpeak
Cool, invisible galactic gas
(60 K, fpeak in low radio
frequencies)
Dim, young star
(600K, fpeak in infrared)
The Sun’s surface
(6000K, fpeak in visible)
Hot stars in Omega Centauri
(60,000K, fpeak in ultraviolet)
The higher the
temperature of an object,
the higher its Ipeak and fpeak
14
Wien’s Law
• The peak of the intensity curve will move
with temperature, this is Wien’s law:
Temperature * wavelength = constant
= 0.0029 K*m
So: the higher the temperature T, the smaller the
wavelength, i.e. the higher the energy of the
electromagnetic wave
Example
• Peak wavelength of the Sun is 500nm, so
T = (0.0029 K*m)/(5 x 10-7 m) = 5800 K
• Instructor temperature: roughly 100 °F =
37°C = 310 K, so
wavelength = (0.0029K*m)/310 K
= 9.35 * 10-6 m
= 9350 nm  infrared radiation
≈ 10 μm = 0.01 mm
Measuring Temperatures
• Find maximal intensity
 Temperature (Wien’s law)
Identify spectral lines
of ionized elements
 Temperature
Color of a radiating blackbody as a
function of temperature
• Think of heating an iron bar in the fire: red
glowing to white to bluish glowing