Selected Questions- Answers From All Experts Astronomy Forum(Stellar Brightness- Magnitude Scale)

Question: Can you please explain how stellar brightness is discerned and especially the basis fo rthe stellar magnitude scale which still mystifies me?

Answer:

The easiest way to grasp stellar brightness is to use a simple analogy. Say I have  a 100 watt bulb and place it at a distance of 1 m, then what would its relative brightness be if placed twice as far away? By the inverse square law for light (the intensity of a light source decreases as the inverse square of its distance):

(d1/d2)² = B2/ B1

Where in this case, d1 = 1 m, B1 = 100 w and d2 = 2 d1 = 2m. Then, it’s obvious one needs to find B2.

B2 = B1 (d1/d2)²

But (d1/d2)²  = (1/2)² = ¼


Therefore: B2 = 100 w (¼) = 25 watts

In the astronomical context, one ought to also be able to use the same method in applying the inverse square law to stars. 

For example, say you are given this problem:  Beta Crucis has an apparent magnitude of +0.8 and this is from 261 light years. Find its absolute magnitude.

You'd first need to know  the basis of absolute magnitude (M abs) as that it is just  apparent magnitude (m) defined at a distance of 10 parsecs. If there are 3.26 light years per parsec this means, 10 pc (3.26 ly/pc) = 32.6 light years. Then, for Beta Crucis:

(d1/d2) = 261 ly/ 32.6 ly = 8

I.e. Beta Crucis must be eight times nearer for taking M (abs).

Then (by inverse square law for light):

(d2/d1)²   =  (8)² = 64

So:

Thus, B1= 64 B2. In other words, Beta's brightness at 32.6 light years is roughly 64 times more than its brightness at 261 light years . This converts to a magnitude difference of: 4.5. (See the diagram below for a stellar magnitude scale - with the Sun at the extreme left or brightest end, with an apparent magnitude of (-27):

The layout of the scale is arithmetical, like the algebraic number line, but when related to brightness one obtains ratios not differences. And these ratios are based on a logarithmic relation. For example, the apparent magnitude of the Sun makes it apparently the brightest object in the sky. But this is only because it is at 1 A.U. distance. IF “moved” to 32.6 light years (the standard candle distance to obtain absolute magnitude) then one obtains +4.8 magnitude.

So the new magnitude would be: (0.8 – 4.5) = -3.7.

That is, we locate +0.8 on the scale, then march off 4.5 units on it in direction of increased brightness.

Thus, its real magnitude -  in terms of its actual distance of 261 Ly-    makes it somewhat dimmer than it would be at the absolute magnitude distance.

The basic gist of the  magnitude scale is that every 5 increments (UNITS) in magnitude difference translates into 100 times difference in brightness, because each succeeding magnitude is different from the earlier one by 2.512 times, since:(2.515)⁵ = 100.

Thus a star of +1 magnitude is 100 times brighter than a star of +6 magnitude on the scale, since (6 – 1)= 5.

To make sure you understand the magnitude scale I attach the following problem:

Examine carefully the star magnitude positions on the given magnitude scale.

Which is brighter, Sirius or Castor, and by how many times? Which is brighter, Zosma or Sigma Draconis and by how many times? If a new star Alpha Stellaris is discovered and is found to be 15.6 times brighter than Sirius, estimate how many times brighter this new star is than Sigma Draconis.

"Math Drives The Universe?" Of course! (First of 3 Parts)

The ASTRONOMY magazine article entitled 'Speaking the Language of the Universe' (December, p. 24) is one every math phobe needs to read- especially people who also appreciate the wonders of the night sky.

But sadly, as the piece by Bob Berman observes, while "professional astronomers use math all the time" - see e.g. some of the (free)  journal articles here:

http://iopscience.iop.org/0004-637X/743/2

Hobby astronomers generally "don't have math in their bones".  As Berman adds:

"They hate the subject. Perhaps it reminds them of school. Square roots and standard deviations bog down articles for them, and science writers oblige by leaving out equations altogether."

Which is sad beyond measure, because indeed, as Berman goes on to show, math is the language of the universe. Sure it's easy to simply stand beneath the stars and simply gaze at galaxies, planets and the Moon through a telescope - Oooing and ahhing all the time. But while wonder and awe can be the keynote at that level, it can't lead to any understanding of what those objects actually are, what they're made of, how they move or how they might affect our planet. For that,  deeper analysis is needed, i.e. into the relationships between physical parameters that define them.

Just to know how distant an object is - say a nearby star- depends on math, and it is critical given that the distance can allow other properties to be determined. Consider the famous "parallax method" to obtain the distance to nearby stars. This method  can apply to all stars within a distance of maybe 50 parsecs (1 pc = 3.26 Ly) or those for which a measurable parallax angle p exists.  The geometry shown below is useful to this end:

The angle p is obtained by taking photographs of the same star six months apart (i.e. from opposite sides of Earth's orbit) and comparing the two positions. One can thereby obtain the distance, D from:

D = r/ tan (p)

The relationship is such that for p = 1 arcsec the distance of the star would be 1 parsec (e.g. par-allax sec-ond). an angle of 1 arcsec = 1" = 1/3600 degree. So we see it is an extremely tiny angle. similarly, if the angle p = 1/10" then D = 10 parsecs, so we perceive a reciprocal relationship such that D = 1/p", though we must ensure the units are consistent.

In many applications, such as the budding undergrad student meets in 1st year Astronomy, the parallax angle p is merged with the equation for the "distance modulus" -  which makes use of the absolute magnitude M (see previous blog on this) and apparent magnitude m. In this way, estimates of the star's energy output, and brightness can be made.

Then, if D is the distance, the usual expression for distance modulus is:

(m - M) = 5 log (D/10) = 5 log D - 5 log 10 = 5 log D - 5

But: D = 1/p
so:

(m - M) = 5 log (1/p) - 5

Or: (m - M) + 5 = 5 log p

A sample problem, such as an Astronomy 201-202 student would have to do for homework, is as follows:

Barnard's star has an absolute magnitude of +13.2 and an apparent magnitude m = +9.5. Find its distance in LIGHT YEARS.

The solution is based on using the parsec form of the distance modulus:

Then:

(m - M) = 5 log (1/p) - 5

(9.5 - 13.2) = 5 log(1/p) - 5

-3.7 = 5 log (1/p) - 5

5 log (1/p) = (5 - 3.7) = 1.3

log (1/p) = (1.3)/5 = 0.26

Taking anti-logs:

1/p = D = 1.81 pc

But 1 pc = 3.26 Ly, so D = (1.81 pc)(3.26 Ly/pc) = 5. 9 LY

A problem to try on your own:

The star Pollux in the constellation Gemini has a parallax angle p = 0."093. Find its distance in light years.

(For those who need a refresher on stellar magnitudes go to:  http://brane-space.blogspot.com/2011/07/tackling-simple-astronomy-problems-3.html

Next: Math examples that Astronomy majoring sophomores might confront.

A Spectacular New Comet Next Month? Possibly!

 Comet Ison heads sunward. Will it be spectacular? Don't make book on it yet.


Okay, there hasn't been a whole lot of publicity in the mainstream media concerning it, but that's with good reason. Astronomers were burned to a far-thee-well when Comet Halley arrived in 1986 and turned out to be a huge dud.  In Barbados, for example, after months of touting special Comet Halley tours, the Barbados Astronomical Society was burned when the visitors finally showed up in droves. Tourists coming to the Harry Bayley Observatory were disappointed and to quote some long time members, "Never again!" Predicted to arrive with a resounding brilliant  'bang', Halley actually appeared with a downer 'whimper'- and it bummed everyone out, astronomers and visitors alike.

Never again would a comet arrival be publicly hyped. Especially a comet heading sunward that might have most of its tail "burned off" as viewed from Earth, and so barely be seen with the naked eye.  So you can understand why astronomers aren't hyping Comet Ison (designated C/2012 SI) as it approaches for what may turn out to be a spectacular sight by the end of next month. How spectacular? That depends on whose POV you adopt, that of the optimists or pessimists. According to ASTRONOMY magazine ('Comet Ison Blazes Into Glory', November, p. 54):

"Optimists tout numbers that have  many amateur astronomers giddy with excitement. If they're right the comet could rival the full Moon when it passes closest to the Sun at perihelion November 28 and remain as bright as the planet Jupiter for several days around then."

Make no mistake that if the optimists are correct, the Comet WILL be spectacular - rivaling or exceeding Halley in its glorious 1910 arrival (and what many of us expected also in 1986).  The full Moon, with a magnitude of (-12.5) is the brightest object visible after the Sun, so if Comet Ison attains that level of brightness, no one will be able to ignore it. And that includes the "End of Time" folks who will undoubtedly be yelping again about "signs and wonders" and the approach of Armageddon.  But this is inevitable.

For astronomers, the comet will be a boon, and amateur astronomical societies around the world ought to be able to capitalize on it, to engender more wonder of the skies and in particular, solar system objects like comets, meteor showers, and also the planets.

But we haven't finished. What about the pessimists, which many of us became after Halley duped us in 1986. According to ASTRONOMY (ibid.):

"Pessimists peddle figures that would bring tears to the cheerleading squad of an undefeated football team during Homecoming. If they prove prophetic, Ison may not even reach naked eye visibility and will go down in the annals of astronomical history with Comet Kohoutek (C/1973 E1) which failed to live up to its billing as 'comet of the century' 40 years ago."

That last assessment is sobering to be sure. But as a realist, it seems to me the actual appearance is most likely to be somewhere between the two extremes: perhaps on a par with Comet West, see e.g.

Which was discovered in August, 1975, and reached a magnitude of (-3) from Feb. 25-27 and bright enough to be seen in daylight. The  magnitude of (-3) is some 9.5 magnitudes higher than for the full Moon brightness forecast by optimists,  which means it is about (2.5)^9.5  =   6, 030 times dimmer than Ison is predicted to be. See e,g, http://brane-space.blogspot.com/2011/07/solutions-to-simple-astronomy-problems.html

What got the optimists so revved up? Probably Ison's intensity from when it was first discovered, back on Sept. 12, 2012, beyond the orbit of planet Jupiter. It then beamed at magnitude +19, "an intensity almost unheard of for such a distant comet", according to the Astronomy piece. So, the optimists clearly have surmised that if it was that bright at such a distance, wow, it will be millions of times brighter by the time it's much closer to the Sun and within naked eye view of Earth.

The extraordinary brightness also made astronomers realize Ison's nucleus must be covered with ices sublimating directly into gas. by virtue of the solar heat, and if the rate of outgassing continued Earth viewers would be in for a real treat.

However, as the Astronomy article points out,  the projections for more rapid brightening didn't reach the expectations spurred by the earlier outgassing estimates. In particular, Ison appeared two magnitude dimmer (i.e. (2.5)²   =  6.25 times less bright)  than expected when it disappeared into the Sun's glare.  This has led a number of researchers to conjecture this dimming is a result of the fact less volatile gases are now in play. Thus, the extravagant brightening near the discovery date was due to really volatile gases like CO and CO2 - which preponderantly sublimate at lower temperatures. Now, high temp. gases are in play which don't translate into brightness intensity.  Once water ice is left in the nucleus, the comet will fall further behind the brightness curve- hence leading to the pessimists' prediction.

Will Ison soon appear bright in the early morning light? We have to wait and see. Brightness estimates put out by the IAU's Minor Planet Center, show Ison at 7th or 8th magnitude in early November. This will be enough to be seen in a good pair of binoculars. If it exceeds that magnitude threshold there's reason to be optimistic.

For those with binoculars (or even their naked eyes- if they're lucky) a good observational test comes up on November 1st. On that date Ison rises 4 hours before the Sun and reaches 30 degrees above the eastern horizon at morning twilight. For readers with a star atlas (like Norton's) it will be located  12 degrees southwest of Denebola, a (+2) magnitude star in the constellation Leo. Those observing at the time will also be able to see the planet Mars, some 7 degrees north-northwest of Ison, and at a magnitude +1.5.

I will observe on the date, and report back in my blog for November 1st with my magnitude estimate. Stay tuned!

TEACHING ALGEBRA II THE RIGHT WAY! USING APPLICATIONS!

According to the author of the HARPERS piece ‘Wrong Answer’, one gripe of Algebra II students is the ‘bait and switch’ tactics often used in their textbooks. The author cites a caption introducing Chapter 8 of the Algebra II Common Core text which displays a Basilisk lizard racing across the surface of a body of water – and there’s a hint (in the caption) of applying algebra to find the “surface tension” and application  – but nothing manifests. It’s a dead end!

The author, Nicholson Baker,  then exposes the scam:

“There’s nothing about surface tension or walking on water in Chapter 8, and indeed the caption would puzzle an expert on reptilian locomotion- since basilisk lizards don’t actually rely on surface tension to run on water. They’re like heavy striders”

He then goes on to explain the real physics, which attributes the lizards’ water skimming ability to “relying on the momentary inertia of the boluses of water beneath its long toed feet.”  In other words, certain discrete semi-separate masses of water (boluses) possess the property of inertia (resistance to displacement) when the reptile scampers over them – thereby allowing it to appear to skim the surface of the water if fast enough. But again, no algebra is shown to be applied.
 

Instead of teaching slavishly (and abstractly – i.e. via formal definitions), to a text, why aren’t various applications emphasized to really get students engaged? In this and two more blogs I will show a variety of ways to apply algebra to drum up of the interest of Algebra II students.  Each application requires only a passing or basic introduction to certain algebraic principles without going hog wild into abstraction, i.e. trying to get a kid to differentiate between continuous and non-continuous rational functions. In the case of an applications –based course, the question of whether functions employed are continuous will arise naturally – say in the course of discussions, or the applications.

 

 

1. Application to the Number Line:
 

One of the foremost concepts any student of algebra needs to master is the number line. There have been various ways to present it, but the most sterile is the teacher simply drawing a boring line on a blackboard and filling in numbers - positive and negative- at regular intervals. A better way is to introduce an immediate application, which demands the use of such numbers - say the stellar magnitude scale shown above.

 

 

If the students are told there is a way to tell the brightness of one star from another, they ought to immediately become curious. You then let them know the brightest stars are assigned negative numbers and the less bright positive ones. A class exercise would be to mark out an extended magnitude scale, say from (-26.5) at one end, to maybe +10 or +15  at the other. You can then use a star atlas to write the stellar magnitudes on the board - or on whatever medium is applicable - and let the kids fill them in.

 

 2. Kinematic applications:
 

Number line too boring?

 

Then one can really get into the 'nitty gritty' with the application to kinematics and dynamics. In this case, the motion of some object, say motor vehicle, is recorded on a distance - time or other (velocity v. time) graph, such as shown below and the results interpreted.

The top graph gives the displacement s in terms of the time as: s = 2t. The bottom graph gives the velocity based on this motion, which will be: v = s/t = 2t/t = 2, say in meters per second. Since there is no initial velocity, u = 0, then (v – u)/ t = 0 and the slope the v-t graph is zero. Meanwhile, the graph of the s-t graph (top) has constant slope. In either of these instances one can infer there is uniform motion and hence Newton’s 1^st law of motion applies.

 

 

 The basic equations of kinematics can then be derived or given, allowing all sort of interesting problems to be worked out. These equations include:

 

 

1) v = u + at

 

 

which relates the final velocity to u, the initial velocity and then to the acceleration a and time it is in motion, t.

 

2) s = ut +  ½ at²

 

This relates the distance traveled to the initial velocity u, the time t, and any acceleration a, If the initial velocity u = 0 then, s = ½ at²

3) v²  =  u²   + 2as
 

A different form relating final velocity to initial velocity u, and distance covered, as well as acceleration.  Remember, that when one moves on to calculus, the primary bugbear of all students is algebraic manipulation (say using partial fraction decomposition),  not recalling continuous and discontinuous functions and their properties.  A good exercise for students is to change the subjects for the different kinematic equations, which the teacher can propose for different circumstances.

 

For example, take equation (2). Now, apply it to a falling body, say a rock, dropped from a tree 20 m high. Then we know s = 20 m, and the initial velocity u = 0. So, how long it will it take to reach the ground? You need to solve for t, the time. (You can take g = 10 ms^-2  to keep it simple, or use 9.8 ms^-2 .)

 

 

In the case of equation (3) maybe a car initially moving at 30 m/s brakes to 10 m/s and does so in two seconds. How would you find the deceleration a? Would you need more than one kinematic equation to find it?

 

 

Application problems:
 

 

1) A car moves at 10 m/s for a period of 5 seconds then slows to 5 m/s for the next 10 seconds. Draw a graph showing the complete motion. What is the speed of the car after 2.5 seconds? What is its speed after 8 seconds? What is the total distance covered over 15 seconds? Over the first 5 seconds?

 

 

2) A go Kart starts off with an initial uniform velocity of 2 m/s for 5 s. It then accelerates to 3 m/s in 1 second and sustains that for 3 seconds. What is the distance it covers?

 

3) A car is traveling along a level road at 50 km/h, then it speeds up to 80 km/h over a period of 5 minutes. Find the acceleration in m/s/s and the distance covered in the time, in m if it maintains the final velocity for 1 minute. (Take 1000 m = 1 km, 1 h = 3600 s)

Tackling Simple Astronomy Problems (3)

One of the most fundamental astronomy areas, which also gave teachers and students doing CXC Astronomy fits, concerned the matter of relative and absolute brightness and stellar magnitudes. (Specifically, the stellar magnitude scale). Of course, it's always best to begin such a topic with prosaic examples.

I tended to use the simple, everyday example of comparing two light bulbs. Thus, if I have a 100 watt bulb and place it at 1 m, then what would its relative brightness be placed twice as far away? By the inverse square law for light (the intensity of a light source decreases as the inverse square of its distance):

(d1/d2)^2 = B2/ B1

Where in this case, d1 = 1 m, B1 = 100 w and d2 = 2 d1 = 2m. Then, it’s obvious one needs to find B2.

B2 = B1 (d1/d2)^2

But (d1/d2)2 = (1/2)^2 = ¼

Therefore: B2 = 100 w (¼) = 25 watts

What appeared to trip up many students was the inverse nature of the proportion (and for many others, the squaring of the distance term).

In the astronomical context, the student ought to be able to use the same method in applying the inverse square law to stars. But actually doing it proved to be the bugbear! For example:

Beta Crucis has an apparent magnitude of +0.8 and this is from 261 light years. Find its absolute magnitude.

The student is first given the basis of absolute magnitude as that apparent magnitude defined at a distance of 10 parsecs, and if there are 3.26 light years per parsec this means, 10 pc (3.26 ly/pc) = 32.6 light years. Then, for Beta Crucis:

(d1/d2) = 261 ly/ 32.6 ly = 8

Then:

(d1/d2)^2 = (1/8)^2 = 1/64

So:

B2/ B1 = 1/64

But in this case the object is being brought closer, so we want to find B1! Thus, B1= 64 B2. In other words, the star’s brightness at 32.6 light years is roughly 64 times more than its brightness at 261 light years which converts to a magnitude difference of: 4.5. (See Fig. 2 for a stellar magnitude scale - with the Sun at the extreme left or brightest end, with an apparent magnitude of (-27).

So the new magnitude would be: (0.8 – 4.5) = -3.7.

That is, we locate +0.8 on the scale, then march off 4.5 units on it in direction of increased brightness.

Ah, the magnitude scale and magnitude difference, introducing a logarithmic scale on top of the ratios! More headaches!

The layout of the scale is arithmetical, like the algebraic number line, but when related to brightness one obtains ratios not differences. And these ratios are based on a logarithmic relation. For example, the apparent magnitude of the Sun makes it apparently the brightest object in the sky. But this is only because it is at 1 A.U. distance. IF “moved” to 32.6 light years (the standard candle distance to obtain absolute magnitude) then one obtains +4.8 magnitude.

Thus, its real magnitude in terms of the absolute standard (absolute magnitude) makes it somewhat dim.

The basic gist of the scale is that every 5 increments (UNITS) in magnitude difference translates into 100 times difference in brightness, because each succeeding magnitude is different from the earlier one by 2.512 times, since:(2.515)^5 = 100. Thus a star of +1 magnitude is 100 times brighter than a star of +6 magnitude on the scale, since (6 – 1)= 5.

Problems:

(1) (a) Using the scale in Fig. 2 as a basis construct a simple stellar magnitude scale ranging from (-10) to (10).

i)Sirius has a magnitude of (-1.4), mark it on the scale and label it.
ii)Castor has a magnitude of (+1.6) mark its place on the scale and label.
iii)Sigma Draconis has a magnitude of (+4.7), mark and label it.
iv)Zosma has a magnitude of (+2.55), mark in and label it.

(b)Which is brighter, Sirius or Castor, and by how many times? Which is brighter, Zosma or Sigma Draconis and by how many times? A new star Alpha Stellaris is discovered which is found to be 15.6 times brighter than Sirius. Locate its approximate position on the scale and label it. Estimate how many times brighter this new star is than Sigma Draconis.

(2) One of the stars shown on the star map of Fig. 1 is Regulus in the constellation Leo. It has an apparent magnitude of (+1.35) and an absolute magnitude of (-0.6). Using this information only, estimate its actual distance from Earth.

(3) The Sun's absolute magnitude is +4.8. How much farther would it have to be from us to appear with that magnitude in the night sky? (The Sun's mean distance from us currently is 150 million kilometers for which the apparent brightness is (-27).

(4) Two stars in the same constellation are named Alpha and Gamma. Both have the same absolute magnitude. Alpha is at a distance of 32.6 light years and has an apparent magnitude of +0.8. Gamma is at a distance of 261 light years.

a)Using the inverse square law of light, estimate how many times Gamma is dimmer than Alpha.

b)Based on the stellar magnitude scale, estimate Gamma’s apparent magnitude.