phys216 practical astrophysics lecture 4 – photometry 1 module leader: dr matt darnley course...

PHYS216 Practical Astrophysics

Lecture 4 – Photometry 1

Module Leader:Dr Matt Darnley

Course Lecturer:Dr Chris Davis

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Luminosity and FluxLuminosity (L) - total Power emitted in all directions over all wavelengths (Joules/sec, or Watts)

Flux – Luminosity emitted per unit area of the source (f), or detected per unit area by the observer (F), over all wavelengths (W m-2)

Total luminosity, L, given by:

L = 4 π R★2 f

where f = surface flux and R★ = stellar radius.

At a distance D from the source, if the measured Flux (power received per unit detector area) = F then:

L = 4 π D2 F

therefore:

D2 F = R★2 f F / f = R★

2 / D2

This is the Inverse Square Law for radiationRemember! Surface Area of a Sphere = 4 π R2 ;

if R doubles, surface area quadruples…

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The Magnitude SystemMagnitude scale

– Hipparchos (2nd Century):

• the brightest stars - magnitude of 1 • the faintest stars - magnitude of 6

However, in terms of the amount of energy received, a sixth magnitude star is approx 100 times fainter, due to the eye's non-linear response to light.

Norman Pogson formalize the magnitude system in 1856:

• 6th magnitude star should be precisely 100 times fainter than a 1st mag star

• each magnitude corresponds to a change in brightness of 1001/5 = 2.512

The bottom line:Magnitude is proportional to the log10 of Flux.

Remember:The GREATER the magnitude, the FAINTER the object!

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The Magnitude SystemRelative magnitudes:

m1 - m2 = -2.5 log10(F1/F2)

i.e. difference in magnitude between two stars is given by the ratio of fluxes.

If m2 = 0, then..

Apparent magnitude, m:

m = -2.5 log10(F / F0)

F0 - flux from zeroth magnitude star, Vega.

This equation can also be re-witten:

m = -2.5 log10F + Z

where Z is the zero-point (described later).

Q. If star A is 100x brighter than star B, what’s the magnitude difference?

Q. How much brighter is Rigel than Bellatrix?

Bellatrixm = 1.6

Rigelm = 0.1

Saiphm = 2.1

Betelgeusem = 0.4

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The Magnitude SystemApparent magnitude, m (from previous slide):

m = -2.5 log10(F / F0) nb. F0 - flux of “zeroth mag” star Vega

m1 - m2 = -2.5 log10(F1/F2)

Absolute magnitude, M - apparent magnitude a star at a distance of 10 parsecs (pc)

Remember, F α 1/d2

so: m - M = -2.5 log10(102 / d2) = -2.5 log10(d-2) - 2.5 log10(102)

- where d is in parsecs

Distance Modulus: m - M = 5 log d - 5

For example:

• if the distance modulus, m - M = 0, d = 10 pc• if the distance modulus, m - M = 5, d = 100 pc • if the distance modulus, m - M = 10, d = 1000 pc

etc…

Q1. What is the Absolute magnitude, M, of

• Sirius, the brightest star in the sky: m = -1.5 mag, d=2.6 pc?

• The Sun, which is actually the brightest star in the sky: m = -26.7 mag, d = 5.10-6 pc

Q2. How much brighter would Sirius be if both stars were at a distance of 10 pc?

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Optical imaging through FiltersStars have different magnitudes at different wavelengths, i.e. when viewed through different filters/in different “wavebands”

Top-left: RATCam on the Liverpool TelescopeAbove: RATCam’s filter wheelBottom-left: CCD detector in RATCam

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Filter setsU 3600 ÅB 4300 ÅV 5500 ÅR 6500 ÅI 8200 ÅZ 9000 Å J 1.25 mmH 1.65 mmK 2.20 mmL 3.7 mmM 4.7 mmN 10.5 mmQ 20.9 mm

NB. 1 Angstrom (Å) = 10-10 m; 9000 Å = 0.9 10-6 m = 0.9 mm

• Wavelengths listed above correspond to the centre of the filter’s transmission. • Filter bandwidths typically 20% (i.e ~ 0.2 dl l) in the optical; 10% in the IR. • Infrared bands correspond to atmospheric “windows”.

Left –standard optical filter profilesBelow – IR filters plotted against atmospheric transmission.

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Johnson-Morgan-Cousins vs SloanThe most widely used Photometric System: Johnson-Morgan-Cousin UBVRI system

Modified by Bessel in the 1990s to better match the performance of CCDs.

The magnitude of an object through a given filter, is referred to as

mB, mV, mR …

or simply by B, V, R….

For example, Bellatrix has apparent magnitudes:

U = 0.54 mag B = 1.42 mag V = 1.64 mag K = 2.38 mag

Is Bellatrix a Red or Blue star?

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Johnson-Morgan-Cousins vs Sloan

However, in 1996…

The Sloan Digital Sky Survey (SDSS) introduced a new set of optical filters:

u’ g’ r’ i’ z’

Have:

• broader bandwidths than J-M-C • higher transmission • bandwidths don’t overlap in wavelength.

Ideal for measuring the red-shifts of galaxies – see right.

The SDSS map of Galaxies out to redshift z=0.15 between -1.5o < d < 1.5o. Each dot is a galaxy containing perhaps 100 billion stars…

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The Magnitude SystemThe Apparent magnitude, m - specific to the waveband through which it is observed.

For example:

Betelgeuse has U = 4.3 mag, B = 2.7, V = 0.42 mag, J = -3.0 mag, K = -4.4 mag

Vega has U = 0 mag, B = 0 mag, V = 0 mag, J = 0 mag, etc..

Debris disk around Vega (HST image)

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The Magnitude SystemApparent magnitude:

m = -2.5 log10(F / F0) where F0 is the flux of Vega.

If you know the Flux of Vega, F0 , in each filterIf you measure the Flux of a star on your CCD, through the same filters ….. You can work out the apparent magnitude of that star.

Q1. Calculate the Apparent U,B and V mags, mU, mV, mB of Rigel Q2. Calculate the Absolute Magnitudes, MU, MB, MV (assume a distance, d = 250 pc)

Rigel (the bright blue star in Orion):

• FU,Rigel = 3.47.10-9 W m-2 , FU,Vega = 2.09.10-9 W m-2

• FB,Rigel = 4.58.10-9 W m-2 , FB,Vega = 4.98.10-9 W m-2

• FV,Rigel = 4.30.10-9 W m-2 , FV,Vega = 4.80.10-9 W m-2

In ancient Egypt, Rigel’s name was…Seba-en-Sah, which means Foot Star or maybe Toe Star!

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Flux vs Flux DensityFlux – • Specific to the waveband, but also the photometric system, because the filters

have a different central wavelength and band-pass (width)

Flux Density – Flux per unit wavelength, Fl (in W m-2 nm-1 or W m-2 Angstrom-1) Flux per unit frequency, Fn (in W m-2 Hz-1).

Approximate Flux Density by dividing Flux by the“width” of the filter (nm, Angstrom, Hz).

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Flux vs Flux DensityFlux – • Specific to the waveband, but also the photometric system, because the filters

have a different central wavelength and band-pass (width)

Flux Density – Flux per unit wavelength, Fl (in W m-2 nm-1 or W m-2 Angstrom-1) Flux per unit frequency, Fn (in W m-2 Hz-1).

Approximate Flux Density by dividing Flux by the“width” of the filter (nm, Angstrom, Hz).

E.g. Rigel:

- FB = 4.58.10-9 W m-2

B-band filter: Dl = 72 nm, Dn = 1.17.1014 Hz

Therefore:

- F ,l B = 6.36.10-11 W m-2 nm-1

- F ,l B = 3.91.10-23 W m-2 Hz-1

www.astropixels.com

Rigel

Flux density in the centre of the B-band

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Spectral Energy DistributionFlux Density vs Wavelength gives Spectral Energy Distribution (spectrum).

The curve is called a Blackbody spectrum and is defined by the Planck function.

Spectrum peaks at a different wavelength depending on the temperature of the star.

The sun has a surface temperature of 5,800 K; it’s a yellow star.

Rigel (blue) : 11,000 K Betelgeuse (red) : 3,500 K

Fl

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An aside: Rigel vs. the Lightbulb (!?)How does flux from a BRIGHT STAR compare to flux from a 60 W LIGHT BULB?E.g. Rigel’s B-band flux - FB,Rigel = 4.6x10-9 W m-2.

1. 60W - power consumed by the bulb! Incandescent light bulb ~ 10% efficient• total flux radiated (across all wavelengths) ≈ 6W

2. Bulb burns at 3,000 K (i.e. like Betelgeuse), most of energy is radiated in the IR! • ~ 10% radiated in the optical (between 300 nm and 800 nm)• ~ 20% of this optical wavelength range covered by B-band filter.

The Luminosity of our bulb in the B-band is therefore:

LB,bulb = 6 x 0.1 x 0.2 = 0.12 W

3. Finally, if we assume our bulb is 1 km away:.

L = 4 π D2 F

… so …

FB,bulb = 0.12 / (4 π 10002 )

FB,bulb = 9.5x10-9 W m-2

B filter

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Colour and Colour IndexColour - defined in terms of the ratio of fluxes in different wavebands. - corresponds to a difference in magnitudes in two different bands

e.g. mB - mV = (B - V ), where (B - V ) is referred to as a the `colour index'.

[ Remember: m1 - m2 = -2.5 log10(F1/F2) ]

Any colour index can be constructed, e.g.

Betelgeuse:B – V = 2.7 – 0.42 = +2.28 (optical)J – K = -3.0 – (-4.4) = +1.4 (near-IR)

Rigel:B – V = 0.09 – 0.12 = -0.03J – K = 0.206 – 0.213 = -0.07

Note: colours can be negative, or even zero, like Vega!

Betelgeuse

Rigel

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Colour and Colour Index(B - V ) - most frequently used optical colour index - a measure of the effective temperature, Teff, of a star.

(Teff is the temperature of a blackbody that would emit the same amount of radiation – typically the temperature near the surface of a star)

For example:

For the Sun: mB = -26.14, mV = -26.78hence

(B - V)sun = 0.64 - Teff ≈ 5,700 K

For Vega: mB = mV = 0.0 (by definition)hence

(B - V)vega = 0.0 - Teff ≈ 9,900 K

Fl

For Betelgeuse: mB = 2.70, mV = 0.42hence

(B – V)bet = 2.28 - Teff ≈ 3,600 K

B - V = -2.5 log10(FB / FV)

Sun

Betelgeuse

[email protected]@0.55mm

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Empirical Formula for Colour IndexThe observed relationship between (B - V) and Teff for Main Sequence stars is given by:

So Colour gives you the Temperature:

• Teff corresponds to a single value of (B - V)

• Can’t use B or V (or any other magnitude) alone to measure Teff.

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Bolometric Luminosity (and Bolometric Corrections)

Teff is related to the total (or bollometric) luminosity,

L = 4pR*2 s Teff

4 - s is the Stephan-Boltzman constant

L is the intrinsic or absolute (not apparent!)brightness of the star: it represents the total outflow of radiation per second from the star (at all l).

Can often determine total (or bolometric) magnitude (apparent or absolute) via a simple bolometric correction:

mbol = mV - BC and hence Mbol = MV - BC

Bolometric correction, BC, is a function of (B - V) or Teff.

BC is ~zero for a star with a Teff = 5700 K, i.e. like our sun.

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Most of the sun’s energy is radiated in the optical.

The sun doesn’t radiate much light in the ultraviolet or infrared, so your eyes aren’t sensitive in these parts of the electromagnetic spectrum.

The Sun has MV = 4.82 and BC= 0.07, Hence: Mbol = 4.82 –

0.07 = 4.75

(absolute bolometric magnitude).

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What about using other colour indices, e.g. (U - B) to measure temperature?

The relationship between (U - B) and Teff is not monotonic; the spectral energy distribution deviates from a simple black body in the U-band.

The same value of U-B gives multiple values of B-V !

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Affects of interstellar dustLight absorbed and scattered by interstellar dust:

• absorbed light - re-emitted in the far-IR • scattered light - absorbed and re-emitted

Both cause extinction - objects appear fainter and “redder”

For small dust grains:

scattering cross-section - sscat l-4

absorption cross-section - sabs l-1

Overall extinction cross-section is:

sext l-1 l-2

Red light (longer wavelengths) is less extinguished than Blue light - hence the term reddening.

Tiny interstellar dust particles (image: A Davis, U Chicago)

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A blue sky (and red sunset)These processes also occur in the Earth’s atmosphere.

• Molecules in the air much smaller than the l of light, so • Scattering is more efficient at the shorter wavelengths

Blue sunlight is scattered many times; reaches our eyes with nearly equal intensity from every direction. We therefore see the sky as blue.

A sunset is red because the “reddening” is most extreme at this time of day.

Images from www.teachastronomy.com

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Extinction and ReddeningAV = absorption in the V band, in magnitudes – the visual extinction

Absorption in other bands is different because of dependency on wavelength, e.g.

AU = 1.53 AV - absorption in UV is more than in VAB = 1.32 AV - absorption in BLUE is (a bit) more than in VAK = 0.11 AV - absorption in IR is much less than in V

Extinction in the IR ~10-times less than at V…

B68 - optical B68 - IR

Colour “excess”Extinction changes the colour of a star!

Colour Excess, E(B - V) - also referred to as reddening - is the additional (B - V) colour caused by this wavelength-dependent extinction, so:

E(B - V ) = AB - AV = 1.32AV - AV = (1.32 - 1) AV

E(B - V ) = 0.32 AV

AV = 3.1 E(B - V )

The greater the extinction, Av , the greater the affect on the B-V colour!

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Nice blue star…

mB = 7.1 mag mV = 7.6 mag

(B-V) colour = 7.1 – 7.6 = -0.5 mag

A negative number; star is brighter in the blue (B) band!

Nice blue star with a cloud in front!

AV of cloud = 3.0 mag

E(B-V) = 0.32.AV = 0.96 mag(B-V) is now = -0.5 + 0.96 = 0.46 mag

A positive number; star is brighter in V and looks a bit reddish…

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Extinction and ReddeningFrom effective temperature, Teff , can calculate the intrinsic colour, (B - V)0 , compare this to the observed colour, and thus calculate the colour excess.

E(B - V ) = (B - V)observed - (B - V)0

From E(B - V ) we can calculate extinction, AV and correct the V-band magnitude.

For example:

G2V star has Teff = 5520 K; mB = 15.3 (observed); mV = 14.1 (observed).

Q. What is the extinction-corrected apparent V-band magnitude of this star?

Remember… AV = 3.1. E(B - V )

mV (extinction corrected) = mV (observed) - AV

Remember from afew slides back?

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Extinction and ReddeningFrom effective temperature, Teff , can calculate the intrinsic colour, (B - V)0 , compare this to the observed colour, and thus calculate the colour excess.



For example:


• Its intrisic (B-V) colour (based on Teff ) (B – V)0 = 0.68 mag• Its observed colour is (B - V)observed = 1.2 mag• Colour excess, E(B - V ) = 1.2 – 0.68 E(B – V) = 0.52 mag• Visual extinction, AV = 3.1. E(B - V ) AV = 1.61 mag• Extinction-corrected V-band mag of the star mV = 12.5 mag


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Extinction and ReddeningFrom effective temperature, Teff , can calculate the intrinsic colour, (B - V)0, compare this to the observed colour, and thus calculate the colour excess.



For example:


• Its intrisic (B-V) colour (based on Teff ) (B – V)0 = 0.68 mag• Its observed colour is (B - V)observed = 1.2 mag• Colour excess, E(B - V ) = 1.2 – 0.68 E(B – V) = 0.52 mag• Visual extinction, AV = 3.1. E(B - V ) AV = 1.61 mag• Extinction-corrected V-band mag of the star mV = 12.5 mag

Finally, when using the Distance Modulus equation it is important to account for extinction:

(mV,observed - AV ) - MV = 5 log d - 5

i.e. must correct apparent magnitude, m, for extinction before calculate absolute magnitude, M


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Atmospheric Absorption (and Airmass corrections)

As we saw earlier, extinction occurs in the earth’s atmosphere too!

The altitude of your target affects how much light gets through to the telescope

Distance traveled through the atmosphere, d:

d ≈ h/cos z = h sec z

‘sec z’ is known as the airmass.

- at an airmass of 1: z=0o - at an airmass of 2: z= 60o

Attenuation is proportional to sec z, and: mz = m0 - C sec z

C is a constant which depends on l (but changes from site to site, and time to time).

…To be discussed in more detail next time!

m0 mz

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End..

See you next week….

phys216 practical astrophysics lecture 4 – photometry 1 module leader: dr matt darnley course...

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