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PETE GREENWOOD Module 5 - Astrophysics Option Resolving power “A measure of how good your eyes (or any optical instrument) are at distinguishing as separate two distant objects with a small angular separation.” When light passes through an aperture it diffracts and produces a pattern on the opposite side to the source. Generally, the smaller the aperture the greater the diffraction. For a circular aperture the pattern is like this: snake.phys.lsu.edu/ ~gclayton/astrostuff.html The resolving power (Θ) is given by: x/y where x is the distance between the two objects and y is the distance between the objects and the eye (or optical instrument). Θ = x/y For Example: If two car headlights are 1.5m apart and we can distinguish the two lamps at a distance of 5km the resolving power is given by: 1.5 / 5000 = 3x10-4 rad Rayleigh Criterion for resolving two sources If the diffraction patterns of two light sources overlap we can say that they are “close” to each other (although this may only be in terms of angular separation). When the maximum of one diffraction pattern overlaps with the minimum of another we can say the two sources are just resolvable, this is when the angle of separation; Θ = λ/d. repairfaq.ece.drexel.edu/.../ module5.html Provided the angular separation of two sources is greater than λ/d the two sources can be resolved (we can see them as two separate sources). Where d is the diameter of the aperture and λ is the wavelength of the electro-magnetic radiation. 1 PETE GREENWOOD Telescopes For best resolution the diameter of the entrance aperture must be as large as possible, this is achieved by using optically smooth (defects are smaller that the wavelength of light) concave reflectors. Radio-Telescopes These “look” at and receive radio waves instead of light-waves. Since radio-waves have a much much much longer wavelength than light as large a collector as possible is needed to obtain the maximum resolving power (Θ=λ/d) for the telescope. For example: Radio telescopes at Joderal Bank have a diameter of 76m, if we estimate the wavelength of the radio waves to be 1m the resolving power is: 1/76 = 1.3x10-2 The above telescope could resolve a 10p coin (2.4cm diameter) at 16km since the angle subtended by the coin is: 2.4x10-2 / 16x104 = 1.5x10-6 rad This gives a resolving power of: 550x10-4 / 5 = 1.1x10-7 Radio astronomy is often favoured over traditional forms of astronomy since: Observations can be made during the day Objects can be viewed that are optically invisible Quasars can be observed The gathering power is huge (proportional to the diameter2) The process is not impeded by the dust and gas of our atmosphere Wire mesh can be used for the reflector (provided the holes are no bigger than 1/20th of the wavelength of the radio waves to be studied) this make the reflector lighter and less affected by wind and rain Quantum efficiency of detectors Quantum efficiency is the ratio of the number of responses in the detecting medium to the number of arriving photons (or quanta of radiation). Photographic film is 4% efficient The eye is 1% efficient Using photographic film to capture images of space means that faint sources must be tracked for a long time or large aperture instruments must be used. Tracking must end a sunrise! Charge Coupled Devices (CCD’s) as used in digital cameras have a quantum efficiency of up to 70%! A CCD is a silicon chip that is divided into picture elements (pixels). The following process summarises the mode of action for a CCD: 1. Photons from the image hit the CCD 2 PETE GREENWOOD 2. Electrons are released (the number released is proportional the to brightness/ intensity of the light) 3. The camera produces an image identical to the electron pattern 4. When the exposure is complete the charge is processed to produce an image Cosmology The evolution of the universe has been deduced by observing the light from stars. Stars consist of the same elements as exist on Earth, this provides evidence that all the stars and planets originated in the same place. In a laboratory, the emission lines in the spectrum of hydrogen have characteristic wavelengths resulting in characteristic colours. These same lines can be observed in the light from stars. Often the spectra from stars appear to have slightly different wavelengths; this is due to the Doppler Effect. The Doppler Effect In sound, relative motion of a source and observer causes an apparent change in the frequency of the sound detected by the observer. This change is due to The Doppler Effect and is a function of the relative velocity of the source and observer: Formula: Δf / f = v/c v velocity of approach Δf change in frequency f frequency of emission (calculated in lab) c speed of e/m waves Δf = (v/c) x f The equivalent for wavelength is: Δλ / λ = -v/c Δλ = (-v / c) x λ or From this we can see that the wavelength is apparently shortened when object and observer are getting closer together. Binary Systems In these systems two stars orbit about their common centre of mass. If it is possible to observe them in their plane of orbit the Doppler shift from each star will change during their orbit. These changes may be superimposed on a gross effect caused by the recession of the system as a whole. Hubble’s Law Hubble found a link between the velocity (v) of recession of distant galaxies and their distance (d) from us. By plotting a graph showing the velocity of recession against their distance from us he was able to show a dramatic relationship; The further away a galaxy is, the higher its velocity. Hubble was also able to show that the motion of the galaxies was not random, but in general the galaxies are all moving apart from each other. v is proportional to d 3 PETE GREENWOOD This simple proportional relationship leads to: v = H0 x d Where H0 is Hubble’s constant (65km.s-1.Mpc-1) v is usually in km.s-1 and d is in megaparsec (Mpc) There are two explanations for Hubble’s law where the light from a star shows red-shift: 1. The star may be moving away from us, like many stars in our galaxy and subsequently showing a red-shift (an example of the Doppler effect). Or 2. Space itself may be expanding. If this were the case, all the stars and galaxies would be moving with it and therefore they would all appear to be receding from one another. It is this second explanation that we use as it also supports the BIG BANG theory which we use to explain the existence of our universe. The Parsec (pc) "Parallax of one arc second." The astronomical unit (a.u.) is the distance between the Earth and the Sun (150 000 000 km) www.definity-systems.net/ ~apw/astro/units.html One parsec (pc) is the perpendicular distance from the Earth and Sun you must be for the angular separation of the Earth and Sun to be 1arc.second. 1 arc.second = 1 / 3600 of a degree (1° = 60min = 60sec therefore 1° = 1 x (60minx60sec) = 3600 arc.seconds) 1pc = 3.26ly 1 light year (ly) = the distance light travels in a year Parallax 4 PETE GREENWOOD Parallax is defined as ½ the angle through which a stars direction changes as the earth moves from 1 extremity of its orbit to another (measured in arc seconds). The age of the universe Using Hubble’s constant (H0) as 65km.s-1.Mpc-1 we can estimate the age of the universe since: H0 = speed of recession / distance H0 = v / d We can estimate the age of the universe to be 1 / H0. Since 1 Mpc = 3.1x1022 m, the age of the universe is approximately: 3.1x1022 m / 65000 m.s-1 = 4.8x1017 s To calculate the age of the universe in years we need to divide by: (60s x 60min x 24hrs x 365.25 days) = 31557600 Number of seconds in a year Which gives us: 4.8x1017 / 31557600 = 1.5x1010 Years = 15 Billion Years Optics - lens ray diagrams When drawing a lens ray diagram for a Bi-convex lens the following points must be remembered: A ray of light passing through the optical centre of the lens does not deviate from its path All rays of light parallel to the principle axis pass through the principle focus Focal Length Lens Power =f = metres = 1/f = Dioptres (m) (D) Object outside 2f A real image is created between f and 2f, it is inverted and smaller than the object. Used in cameras and in the eye whilst reading Object at 2f A real image is created at 2f, the image is the same size as the object and inverted. Used in photocopiers 5 PETE GREENWOOD Object between f and 2f The real image is outside 2f, inverted and magnified. Used in projectors, and as the objective lens in microscopes. Object at f The image is at infinity as parallel rays of light are formed. Used in spotlights and headlamps. Object between f and the lens A virtual image is formed on the same side of the lens as the object. The image is the same way up as the object and magnified. Used in magnifying glass’, eye lenses’ of optical instruments for distant viewing e.g. telescopes. http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/u14l5da.html Real images can be focused onto a screen whereas virtual images cannot; virtual images are the only images that are the right way up. The thicker the lens is in the middle compared with the edges, the shorter is its the focal length. Objects far away – telescopes When looking at objects at huge distances any two rays of light coming from that object will appear parallel. An image is subsequently produced in the focal plane of the objective lens; it is this image that is focused by the second “eye-piece” lens. 6 PETE GREENWOOD The eyepiece lens is positioned so that the image from the objective lens is at its focal length. Therefore the image from the eyepiece lens is at infinity, and the telescope is said to be in “normal adjustment” (this will not strain the eye through prolonged usage). webs.mn.catholic.edu.au/.../ hsc_astrophysics.htm The final image produced by the eyepiece lens is inverted, but this is not an issue as space has no up / down since the orientation of an object is only relative to gravity. Lens (and mirror) Formula http://www.physics.carleton.ca/~watson/1000_level/Waves_and_Optics/Gifs/Convex_Lens_Formula.gif The thin lens formula is: 1/u + 1/v = 1/f Any units can be used for this equation provided the same units are used throughout the equation. Rules for the lens equation: If the object is real the u is positive (if the object is virtual then u is negative) If the image is real than v is positive (if the image is virtual then v is negative) If the lens is convex f is positive (if the lens is concave then f is negative) Manipulation 1/u + 1/v = 1/f 7 PETE GREENWOOD Multiply both sides by v: Therefore: v/u + v/v = v/f v/u = v/f -1 Linear Magnification Linear magnification equals Or M = v/f -1 M = v/u Linear magnification is not very useful for application with telescopes since the distance from the object cannot easily be calculated. For these situations we use angular magnification. Angular Magnification Angular magnification is calculated using two angles, β and α and the equation: angular magnification = β / α α The angle subtended at the unaided eye by two rays of light form the top and bottom of an object β The angle subtended at the unaided eye by the image from the telescope © Peter Greenwood 2004 By creating triangles from; the angles α and β, the respective focal lengths and the image height we can say that: tan α = Image height / Focal length Objective tan β = Image height / Focal length Eye-piece Because α and β are very small we can say that: (When α and β are in radians) Therefore: tan α/ β is equal to α/ β α = Image height / Focal length Objective α = I / Fo β = Image height / Focal length Eye-piece β = I / Fe 8 PETE GREENWOOD Angular Magnification manipulation We can re-arrange the equation for angular magnification as follows: M=β/α M = (I / Fe) / (I / Fo) M = (I / Fe) x (Fo / I) M = Fo / Fe If: And we substitute in the values from above: Which means that: By cancelling the I’s we get: To get the largest angular magnification for telescopes it is best to use an objective lens with a large focal length and an Eyepiece lens with as small a focal length as possible. There are however limits to this as large lens suffer from both chromatic (distortions of the colours) and spherical aberration (distortion of the shape) due to the diffraction that goes on within the lens. Lens (and mirror) Equation (1 / u) + (1 / v) = (1 / f) u = distance from object to lens v = distance from image to lens f = focal length of the lens If the image is real then v is positive If the image is virtual then v is negative For a converging lens (or mirror) the focal length is positive For a diverging lens (or mirror) the focal length is negative Limitations of Refracting telescopes Spherical Aberration may occur, this is where imperfections in the glass due to impurities and the irregular forming of the lens due to forces acting on the glass during cooling (for example gravity) cause distortions of the image as the lens is not a perfect curve (not all light is transmitted through the lens). Due to the different wavelengths and frequencies of the visible spectrum the different colours of light are refracted by different amounts (blue more than red) this is known as chromatic aberration and can results with coloured fringes appearing around objects. The advantage of a refracting telescope is that it is a sealed unit and therefore there is no need for cleaning. Refracting telescopes can be made smaller and are cheaper than reflecting telescopes. 9 PETE GREENWOOD Reflecting telescopes Cassegrain telescope This design uses two mirrors and a convex eyepiece lens. The primary mirror is a (parabolic) concave shape that focus’ incoming rays of light onto the secondary mirror, a (parabolic) convex shape. The magnification of a reflecting telescope is given by: M = Fo / Fe where Fo is the equivalent focal length of the two mirrors combined and Fe is the focal length of the eye-piece lens. As shown in the diagram below the secondary mirror is placed within the focal length of the primary mirror. This secondary mirror reflects the image to the eyepiece where a convex lens focuses the image into the eye of the observer. http://zebu.uore gon.edu/~js/glo ssary/reflecting _telescope.html The Advantage s of a reflecting telescope are: No chr omatic aberration as no refraction occurs (except in the eye piece) No spherical aberration (if a parabolic mirror is used) Greater resolving power as larger apertures can be used Disadvantages of a reflecting telescope: Maintenance is needed due to the open nature of the telescope (dusting the mirror) Spectral Classes of Stars Stars have been classified into groups according to the dominant lines in their absorption and emission spectrum. The classes had the following letters ascribed to them by the ancient Greeks: O B A F G K M (Oh Be A Fine Girl Kiss ME!) 10 PETE GREENWOOD Hertzsprung-Russell Diagram imagine.gsfc.nasa.gov/. ../LC_main_p8.html Absolute magnitude is the magnitude of the star at 10pc (large negative numbers indicate the greatest magnitude) Most modern stars are plotted on the graph based upon their luminosity and surface temperature. Main Sequence stars are the most common; these are shown in the central curve of the graph which extends from highly luminous blue (top left) to the to very dim red (bottom right) stars. Stars leave this main sequence when the dominant fusion progress changes; when this occurs they will become one of the following stars: Red Giants and Supergiants are relatively cool and therefore emit a small amount of light per unit surface area. This means that be as bright as they are they must have a very large surface area. White Dwarfs are very hot dim stars; they are very dense due to a small diameter (suggested by their low luminosity). These are quite common but difficult to observe. Large stars can suffer catastrophic collapse that causes a sudden temperature rise which may result in a supernova explosion which can result in the expulsion of a large fraction of the stars mass. If a supernova explosion occurs the remaining mass of the star will collapse under its own gravity getting smaller and smaller until it is no more than a point of infinite density. 11 PETE GREENWOOD The origin of Absorption Spectra In general absorption spectra consist of a continuous emission spectra crossed by dark lines at certain points that correspond to emission spectral lines of certain elements that exist between the source of the continuous spectra and the observer. In a simple case, an atom of an element may undergo excitation by the absorption of a photon. On return to its ground state the photon is re-emitted in a random direction. The observed spectrum is therefore darkened at the point corresponding to the wavelength of the photon: instruct1.cit.cornell.edu/.../ astro101/lec13.htm If the surface of a star is very hot, high energy photons will leave its surface. This will cause absorption spectra to be observed at the short wave end of the continuum. In a class O star, the temperature is high enough to ionise significant numbers of helium atoms. Helium absorption spectra are seen from these stars. Hydrogen in the first excited state (n=2) can give rise to visible absorption lines (Balmer series) because n=2 becomes the “normal” ground state at high temperatures. In cooler stars, absorption spectra are more obvious at the long wavelength parts of the spectra. These are caused by absorption from molecules and transitions in molecular vibrational energy states. Finding Spectral Class by Dominant Absorption Lines Firstly use the H-R Diagram to find the absolute magnitude (M) and examine the absorption spectra to measure the apparent magnitude. Using: We can rearrange to give: m-M = 5log(d/10) (m-M)/5 = log(d/10) Therefore: 10[ (m-M)/5 ] = d/10 10 x 10[ (m-M)/5 ] = d 10[ {(m-M)/5}+1 ] 12 PETE GREENWOOD Magnitude and Intensity Magnitude is a measure of how bright a star is. There are, however, two different ways of indicating a stars magnitude; apparent magnitude and absolute magnitude. The scale we use to measure magnitude is based on that created by the ancient Greeks which ran from 1 to 6. On the ancient Greek scale 1 was the brightest star they could view and 6 was the dimmest. The problem with this scale is that with the creation of optical aids (such as the telescope) we can now see stars that are much further away than those the ancient Greeks could see. These stars now visible to us fall on both sides of this scale; some are dimmer than the dimmest stars the ancient Greeks could view, whilst others have absolute magnitudes much greater than those seen by the ancient Greeks. The Brightest stars visible to us are those with the largest negative magnitudes and the dimmest stars have the highest positive numbers. We can now see stars with a range of values from -25 through to +25. Apparent magnitude is how bright a star is when viewed from any location, this means that apparent magnitude is relative to the location of the star from the viewpoint. A star of apparent magnitude 2 is two and a half times brighter than a star of apparent magnitude 1, just as a star of apparent magnitude 3 is two and a half times brighter than a star of apparent magnitude 2. This logarithmic scale is used as the eye responds to light in a logarithmic fashion. Apparent magnitude and light intensity (I) are linked by the equation: m = -2.5 x log I + const. The above equation is used to compare the apparent magnitudes of 2 stars with their intensities (in J.S-1.m-2). The use of an unknown constant does not matter since the constant will cancel out when comparing two stars of known intensities or magnitudes: m1 = -2.5 x log I1 + const. m2 = -2.5 x log I2 + const. m2 - m1 = -2.5 x log I2+ const. - -2.5 x log I1+ const. m2 - m1 = -2.5 (log I2 - log I1) m2 - m1 = -2.5 log (I2 / I1) NB: this is the most useful form of the equation!! We can re-arrange the above equation as follows: Since: m2 - m1 = -2.5 log (I2 / I1) We can say: (m2 - m1) / 2.5 = log (I2 / I1) Therefore: 10x x [(m2 - m1) / 2.5] = (I1 / I2) 13 PETE GREENWOOD For Example: “Compare the apparent magnitudes of the Andromeda Galaxy (m=4.8) and the Crab Nebula (m=8.4)” (m2 - m1) / 2.5 = log (I2 / I1) (8.4 - 4.8) / 2.5 = log (I2 / I1) 10x (8.4 - 4.8) / 2.5 = 10x 1.44 therefore = 27.54 This means that the Andromeda Galaxy is 27.54 times as intense as the Crab Nebula. Absolute Magnitude is used as a method of comparing the magnitudes of distant stars using a common scale; it is the brightness of a star when it is at a distance of 10 parsecs from the sun. It is important to remember that although this 10 pc is measured from the sun and not the earth, the difference in the two measurements is negligible due to the huge distance the stars are from the earth. Absolute magnitude can be calculated by using the spectral class of the star and crossreferencing this with the main sequence band on the H-R diagram, this method is open to huge variation due to the fact that the main sequence is a band on the H-R diagram and not a line. Another method is to use the equation that relates the absolute and apparent magnitude (although this is only reliable for stars closer than 10Mpc (10 000 000pc): m - M = 5 x log (d/10) By re-arranging the equation on the previous page we can calculate the distance a star is from us (if we know the absolute and apparent magnitudes): d = 10 x 10(m - M) x 1/5 Alpha Centuri has apparent magnitude 0.1 and is at a distance of 1.32pc from the sun, find the absolute magnitude: m - M = 5 x log (d/10) 0.1 - M = 5 x log (1.32/10) M = 0.1 - 5 x log (1.32/10) M = 4.497 or 4.5 (2 d.p.) The above result is very similar to our sun. Compare the absolute magnitudes of the two stars Sirius (apparent magnitude -1.46 at a distance of 2.7pc from the sun) and Betelgeuse (apparent magnitude +0.50 at a distance of 94pc from the sun) M = m - 5 log (d/10) Sirius Betelgeuse M = -1.46 - 5 log (2.7/10) M = +0.50 - 5 log (94/10) = 1.38 = -4.37 This shows that Betelgeuse is the brightest star in terms of absolute magnitude (although not, in this case, in terms of apparent magnitude) 14 PETE GREENWOOD Deriving: m - M = 5 x log (d/10) If i is the intensity of light reaching the earth from a star from a star when it is d pc from Sun, and I is the intensity when the same star is 10pc from the Earth we can say that since: Intensity is proportional to 1/distance2 By using: i is proportional to 1 / dpc2 I is proportional to 1 / 10pc2 and We can see that: (I / i) = (d / 10)2 Now if we take logs of both sides: log (I / i) = log (d / 10)2 Which is the same as: (log I - log i) = 2 log d/10 Since: m = -2.5 x log i + const. and M = -2.5 x log I = const. M - m = -2.5 (log I - log i) From previous page: M - m = -2.5 (log I - log i) M - m = -2.5 x 2 x log (d/10) M - m = -5 x log (d/10) Therefore: m - M = 5 x log (d/10) Escape velocity and Black holes Escape velocity is the instantaneous launch velocity that would just allow an object to escape the gravitational influence of another object. It can be shown that: Ve = √ (2GM/R) (For an object of mass M, radius R. where G is the gravitational constant 6.67x10-11 N m2 kg-2) For massive objects of a small radius (high density) the velocity of escape (V e) becomes very large. For a black hole Ve ≥ c (velocity of light). 15 PETE GREENWOOD For a more massive black hole Ve ≥ c at a distance above the physical surface. The radius of the sphere formed by the points where the velocity of escape is equal to the velocity of light is known as the event horizon. The radius of this sphere is called the Swarzschild Radius. Within this, the velocity of escape is greater than the speed of light i.e. light cannot escape. At the Schwarzschild radius Ve = c therefore c = √ (2GM/R) this can be rearranged to give the equation in terms of the Swarzschild Radius: R = 2GM / C2 © Peter Greenwood 2004 Black Body Radiation and Wien’s Displacement Law A black body is a perfect absorber; it will absorb all wavelengths of radiation falling on it. At room temperature it will appear black since it does not reflect lights. A black body is capable of radiating or emitting all wavelengths of radiation provided its temperature is high enough. The graph here shows the energy distribution from a black body λ indicates the wavelength of most intense radiation emitted from a black body, it is not the radiation of longest wavelength. This is often referred to as λmax. As the temperature of different stars increases, the wavelength of the most intense radiation increases (shifts to the right of the scale). www.fofweb.com/Subscription/ Science/Helicon.a... At 4000K the whole visible spectra is emitted by black bodies. The emission is however weighted towards the red end of the spectra, at much higher temperatures (around 20000K) the spectra is shifted towards the blue end of the spectra. λmax x Tk = const. constant = 0.00289m.k. m.k. = metre-kelvin “Show that a temperature of around 4000K is needed for λmax to be in the visible spectrum.” T= 4000K λmax needs to be approx: 550x10-9m for the MIDDLE of the spectrum 640-740x10-9 for the RED end of the spectrum 580x10-9 for the YELLOW area of the spectrum 420-480x10-9 for the BLUE end of the spectrum 16 PETE GREENWOOD λmax x Tk = 0.00289m.k. λmax = 0.00289m.k / Tk. λmax = 0.00289m.k / 4000K λmax = 4128.6K Stefan’s Law Luminosity = Power from all the surface of a star = Energy per second per m2 x total surface area of a star = Joule seconds (J.s-1) or Watts (W) Stephan’s law states: Power (P) is proportional to Surface Area (A) Power (P) is proportional to Surface Temp4 (T4) P = σ x A x T4 The total power radiated by a star black body is proportional to the fourth power of is absolute temperature. Where σ is Stephan’s constant σ = 5.67x10-8 W.m-2.K-4 Quasars Quasars (quasi-stellar objects) are extremely luminous objects at immense distances (upward of 3 billion light years) from earth. They are extraordinary because they appear to be very small but they outshine whole galaxies. Quasars were discovered by analysing the location of a strong radio source. The spectrum examined exhibited a series of lines that no-body could identify. These spectral lines were found to be the result of incredible red-shift which indicated the quasar was moving away from Earth at 15% of the speed of light (45000km.s -1). Upon further examination quasars were confirmed to look like one singular star instead of a whole galaxy, indeed the size of the quasars was so small that the angular diameter of the objects could not be measured. Quasars are: Billions of light years away As bright as a thousand galaxies Much smaller than a galaxy This presents the following uncertainties or controversies: For quasars to be as bright as they are they must be consuming far more energy than could be contained in object of their size unless it is far denser than anything we have come across (except black-holes). There is some scepticism about the use of Red-shifts to determine the distance of quasars from the Earth. Some quasars seem to be receding at incredible speeds of up to 90% of the speed of light. Notes produced by Peter Greenwood: reproduced by his kind permission. 17