WHAT DOES MASS REALLY DO?

Hiya, Sean Lally, Physics guy here. Today on "How do we know that?" we're talking about what mass is all about. Believe it or not, this is trickier than it seems. First, what kind of mass are we talking about?

You probably know that objects experience a gravitational force (Fgrav) between them, and that the amount of attraction depends on their masses (m1 and m2) and the distance (d) between them. This is easy to represent symbolically (and is frequently referred to as Newton’s Law of Universal Gravitation (though he never wrote is as such):

Fgrav = G m1 m2 / d^2

Having more than 2 masses makes things a bit more complicated, but the idea and approach is similar. The other term, G, is a physical constant (6.67 x 10^-11 Nm^2/kg^2) related to just how strong the attraction is. You might be able to guess that, given the smallness of this constant, gravitational forces are only significant when the masses (or at least one of them) are great – things like stars and planets. Take two cupcakes out in space, away from strong gravitational forces, and they don’t really attract each other a whole lot. And what a waste of perfectly good cupcakes!

Now this mass, the mass that responds to gravitational forces, we refer to as gravitational mass.

But let’s talk more generally – if there is an unbalanced force, any force, acting on a mass, the mass will accelerate. In other words, if YOU push on a mass with a force greater than any resistive forces (friction, etc.), the mass will accelerate. The acceleration will be greater with greater forces, but it will be smaller with greater masses. There is a convenient way to represent this, too, with symbols. Enter Newton’s 2nd Law:

F = m a

Now this mass, the one that responds to forces in general, is called the inertial mass.

Maybe I’m not making a big enough deal out of this – these are two completely different ways to view the concept of mass: one idea of mass is gravitational (how the mass responds to gravitational forces) and the other is inertial (how the mass responds to any force at all).

Now here’s the punchline: the two masses, while conceptually very different, are numerically the same!* This idea is known as the Equivalence Principle, and it is the cornerstone of Einstein’s theory of general relativity. Let’s explore this.

We’ll use the classic physics example of the sealed elevator, so that you cannot see what’s happening outside of your box. And let’s imagine that you are conveniently carrying a tennis ball. You hold the tennis ball out at arms reach and release it - you notice (to your relief) that it falls and accelerates toward the floor. Quick, give 2 possible interpretations of what you see.

So, the ball could simply be accelerating normally – down, toward the ground. But, couldn’t the elevator be accelerating upward toward the ball? Is there a way to tell the difference?

The equivalence principle says NO.

On the other hand, if you want to change the rules a little, would it help if you could look out a (newly inserted) window? What would be evidence as to whether your box was accelerating or the ball was?

Let’s look at a more peculiar example. Get back in your box, but this time you have a flashlight. You direct the beam of light straight to the opposite wall. The spot lands exactly where you expect (directly across from the flashlight). What are the possible interpretation(s) of this?

Now imagine that your box is accelerating upwards. Draw the path that the light would take from your flashlight to the opposite wall.

Behold the genius of Einstein! The path is not straight, but rather, curved. And in the correct interpretation of the equivalence principle:

There is no difference between the light curving due to the accelerated box, and the light curving down as though it is affected by gravity.

In other words, light is affected by gravity – or better still, mass causes space to curve. Let’s look at this idea in more detail. You are no doubt familiar with the idea that the shortest distance between two points is a straight line. This is true in classic (Euclidean) geometry – the kind of geometry that you learn in high school. Another view of this is called Fermat’s Principle – light will take a path that causes the time of travel to be shortest.

But the case we’ve been discussing, the flashlight in the accelerated box, is clearly different. The path of light is NOT straight. If it is indeed affected by gravity – as “caused” by mass – then the space itself is not Euclidean. Rather, space is curved. You can even re-interpret the tennis ball problem – there is no force acting on it; it is simply following the shortest space-time path (a geodesic). Of course, for you to notice these effects, the gravitational force has to be substantial – due to stars, planets, galaxies, etc.

You might not be surprised to learn that there is a whole truckload of mathematics that could be attached to these ideas. We’ll leave that for you when you get to graduate school. It’s interesting to note that these were a great challenge to Einstein himself!

How about that – crazy stuff, eh? That’s all for today – see ya soon, see ya on the Moon!


ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009

WHY IS THE SKY DARK AT NIGHT?

Hiya, Sean Lally, Physics guy here. Today on "How do we know that?" we're talking about the night sky and what it looks like to us. So, here’s a classic question for you – why is the sky dark at night?

So, what do you think? Take a minute to think about it.

I’ll bet that you’re thinking about day and night being caused by the rotation of the Earth, toward and away from the Sun. And that is true.

However…. That’s not the end of the story.

Imagine going outside on a good night and counting each and every star you can see. How many stars would you count? Quick, what’s your guess?

Now, are you seeing all the stars that exist? No. How many stars do exist?

Did you guess millions? Billions? Billions of billions? You’re getting warmer.

The actual number is certainly not known, but there are on the order of 10^12 galaxies, and each may have on the order of 10^12 stars. That gives an ballpark estimate of 10^24 stars. That’s really, really, really big. No wait, that’s an understatement.

It is this many:

1,000,000,000,000,000,000,000,000 stars.

If you started counting from birth, one star per second for you entire life, could you count that high? How long would it take you? A month? A year? Your lifetime? Multiple lifetimes???

Oh by the way, how would one come up with this number? What’s a decent way to estimate the number of stars in the universe? Imagine that you have some technology – say, a nice telescope. Maybe even the Hubble Telescope!

So, back to the point of the lesson. With all of these stars, why is the sky dark at night. Think about the question again.

Or how about this question? Imagine staring at a forest full of trees. Given a dense forest, would you expect to light at the other end? Why or why not?

Back to the universe. 10^24 is a HUGE number of stars – one would think that with all of that light hitting us, the sky should be brilliant at night. And yet it is not. Hmmmmm.

Let’s think about this. What are some possible reasons why we do not see a brilliant sky at night?

• Do we have an unobstructed view of all stars, or are some blocked?
• Is there some interstellar matter that absorbs starlight?
• Are all stars of the same brightness, regardless of their distance from us?
• Is the universe populated with a mostly constant number density of stars?
• Is the universe infinitely large or infinitely old?
• Does the inverse-square law of light hold for distant stars, or does light obey some other law?
• Does the motion of distant stars affect the brightness that we see?

Can you think of others? What might be your answers to these questions? As it turns out, there are reasonable scientific answers to each of these.

This problem, often called “Olber’s Paradox” (though Henry Olber, who described it in 1823, was certainly not first person to do so), does not have a simple answer. Indeed, one really might expect to have a bright night sky. Why?

Imagine the Earth at the center of series of thin spheres of stars. Each sphere has a certain number of stars “on” it. As you get farther from Earth, the light from each individual star decreases (inverse square relationship: 1/4πr^2), but the number of stars per sphere actually increases (square relationship for surface area of sphere: 4πr^2). These two effects cancel each other out, so each sphere has effectively the same total brightness.

If the universe actually were infinite in size, we likely would have a bright night sky. So, it seems as though the universe is not infinite in size (though there are substantial mathematical objections to this argument). The better answer has to do with the age of the universe – it is NOT infinite. Moreover, stars have only existed for some of that time. And given an expanding universe, it is easy to see that starlight from many, many stars has simply not reached us yet. Enough time has not yet existed!

Well, that’s all for now. See ya soon, see ya on the Moon!

Further Reading

Barbara Ryden – Cosmology
(I have borrowed heavily from Ryden’s chapter 2 for this lesson.)
Edward Robert Harrison - Darkness at Night: A Riddle of the Universe
http://www.esa.int/esaSC/SEM75BS1VED_index_0.html

All text and images copyright 2009 Sean Lally, except where noted.

THE UNIVERSE IS DOING WHAT?

Hiya - Sean Lally, physics guy here. Today on "How do we know that?" we're talking about the universe at large. And I do mean 'large.'

We've just seen how speeds can be determined based on so-called "Doppler Shifts." Let's revisit what that is exactly. The Doppler Shift (z) is defined by the difference between the expected and observed wavelengths, divided by the expected wavelength. Better yet:

z = (λobs - λexp) / λexp

We can also represent z by this expression:

z = v / c

(This is a non-relativistic expression - that may not mean much to you now. Suffice it to say that galactic motion, while fast, does not rise to the level of being relativistic - which means super fast, near the speed of light.)

Also FYI - a negative z indicates a 'blue shift', while a positive z indicates a 'red shift.' Does that make sense to you? Think about it for a moment - a negative z means that observed wavelength is lower than the expected one. You could also say that the observed frequency is higher than the expected one - this is classic blue shift. Got it?

So, let's imagine that you have some data - a bunch of z-values for galaxies. And let's also imagine that they were nearly all greater than zero*. Quick, what could this mean?

* It's worth noting that a few galaxies have z's that are less than zero (blue shifted). These are in the local neighborhood of galaxies, the local group. We needn't worry about these now - they are, by far, in the minority of galaxies. Also - strictly speaking, we shouldn't exactly refer to the galaxy being red or blue-shifted. Typically, one looks at spectral lines from the galaxy. And importantly, we should mention that this pioneering work was first done by Vesto Slipher in the early 1900s.

A-ha - if nearly all galaxies are moving away from us, it strongly suggests that the universe is expanding.

Here's a tricky question - we know what things look like from OUR perspective. What would it look like if you were observing from somewhere else in the universe? For example, would you expect to see the Milky Way galaxy red-shifted if you were on a planet orbiting a distant star outside this galaxy?

Hmmmm.... so, everywhere you observe in the universe, it seems to be expanding away from everywhere else. What in the world can this mean? Is there a "priveledged position" in the universe? Or better yet - is there a "center" to the universe? Is it expanding from "somewhere?"

This idea is so profound and important that we call it the Cosmological Principle (or, occasionally, the Copernican Principle):

On large scales (say, 300 light-years or so), there is nothing special about any location in the universe.

Or if you prefer, the universe at large is homogenous and isotropic.

What does that mean, exactly? In short, the universe has no center or preferred direction at all (isotropic). Furthermore, it looks pretty much the same no matter where you look (homogeneous).

Here's one other very peculiar thing to consider. In 1929, Edwin Hubble had some data at his fingertips - both redshifts of galaxies (from which could be calculated speeds) and distances to these galaxies. What would you do with this data, if you had it? What would a scientifically curious person do with this kind of data? One could graph it, yes? But how, and what would any relationship mean?

Think about this - you have numerical data for redshifts of galaxies and their distances from us. How would you create a graph to show any relationship between them? Think about and discuss it with your friend.

Well, let’s cut to the chase - Hubble graphed redshift (basically, the velocity of the galaxy) versus its distance from us. What do you think he found?

Here's the basic graph he arrived at:

(See image above.)

What in the world can this mean?

Note also that this is a linear graph. Linear graphs have slopes (or steepness) that usually have some significance - what is the significance of this slope?

A widely used value for this slope, also called the Hubble Constant, is:

Ho = 70 +/- 7 km/s/Mpc

That's 70 kilometers per second per megaparsec. In other words - the further away the galaxy, the faster it is receding. Oh my.....

That's all for now, my friends - see ya soon, see ya on the Moon!

All text and images copyright 2009 Sean Lally, except where noted.

HOW FAST IS THAT MOVING?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at cosmic speeds – aka, how fast is that thing moving?

So, how DO you measure the speed of something? Well, let’s first agree on what speed is, shall we? Speed is the rate at which something covers a particular distance. For our purposes, we will not be splitting hairs of speed versus velocity. Velocity is very similar to speed, but it specifies a direction as well.

Back to speed (represented by a v):

Speed = distance / time

v = d / t

The units are usually something convenient like miles per hour, meters per second, kilometers per hour, furlongs per fortnight….

Let’s say you want to know how fast a car is moving? How might you measure this? Talk to your partner and come up with at least two possible ways to determine the car’s speed.

You probably came up with at least one method that involved a car moving a predetermined distance and you using a stopwatch. Imagine that the car is too far away to get a good look at it. Cars seem big when you’re close to them, but the further you are from the, the smaller they appear until finally they disappear from your sight. Really, if you’re a kilometer from the car, you won’t see it at all.

You might think that using a telescope or binoculars would help out here, and certainly they would up to a point. Of course, it would be really hard to measure the distance that the car is moving. After all, the further away you are, the shorter the car’s traveled distance seems to be (and, if you’re trying the technique learned earlier, the harder to measure the angle). You may have noticed this if you’ve ever watched an airplane pass overhear, or have looked at the ground below as you flew in an airplane. It really doesn’t seem as though you’re moving great distances in short times – in other words, it is hard to get a good estimate of how fast you are moving.

So, is there another way to measure speed?

Glad you asked. Why, yes there is, and it is a fantastic way. Let us consider one of “Physics Greatest Hits”, the Doppler Effect. Let’s experiment a little.

At this point, you will need one or both of the following:

Foam ball (or wiffle ball, cut open and packed with foam) with battery-powered buzzer inside

Tuning fork on a string – it’s really important that the tuning fork be securely fastened to the string; the fork should have a hole in the handle. The amount of string doesn’t matter too much – a meter or so. You will also need a tuning fork mallet to strike it.

You’ll also need a friend nearby. (But then, don’t you always?)

Let’s try out the foam ball. Turn on the buzzer. How would you describe the tone? Now, you are going to throw the ball to your friend. Listen carefully to the sound, and ask your friend to do the same. Throw and listen!

What do you notice? Have your friend throw it back to you and pay careful attention yet again.

You can also try this with the tuning fork. It should have a piece of string tightly attached to the handle. Seriously, a flying tuning fork can be a pretty dangerous thing! Strike the tuning fork and listen to the tone. Strike it again and swing it in a circle above your head. What do you notice? What does your friend notice? Does this matter whether or not the tuning fork is moving toward or away from your friend? Try it again until you notice any trend.

So, something is happening. What is it? Discuss with your friend. Ask other classmates to make sure that everyone agrees.

Change things up a bit. Does the speed of the ball or tuning fork make a difference? Have a third person stand somewhere to pay close attention. Write down all of your observations.

Let’s look at the details of what is really happening.

Think about this. Imagine that your friend is sitting on a chair singing a lovely note: “Laaaaaaaaaaaaaaaaaaaaaaaa…….” If you could actually SEE the note (as a wave), it might resemble this:

(See image 1 above.)

What’s going on here? Well, your friend (the happy face above) is sending out waves with a constant frequency (related to the pitch they are singing). Each wave is expanding like a sphere around your friend, just like good waves do. Let’s imagine that they are sending out 440 waves per second – that’s what musicians call the note ‘A’. Note that wherever you stand, at the points A, B, C or D, you will receive the same number of waves per second (440). In other words, you both agree on the frequency of the note. Only 3 waves are depicted above, the principle is the same.

Now let’s make it trickier – hooray! Imagine that your friend is moving to the right. Maybe they’re on a skateboard zipping by you. Things may get a little weird. Let’s see…

(See image 2 above.)

Things are a bit different now. Your friend is STILL sending out 440 waves per second, but she is moving to the right – that means that each new wave will start at a new location, to the right of where the previous one was released. Therefore, what you receive will depend on where you stand. Note that if you stand at point C, you will receive MORE than 440 waves per second (a higher pitch/frequency). At point B, less than 440 waves per second (a lower pitch/frequency). What your friend sings doesn’t change, but your reception of it does. And it’s not an illusion – you actually DO receive more or fewer waves per second, depending on where you stand. This phenomenon is called the Doppler Effect. You have noticed it for sound, but it also applies to light (and all electromagnetic radiation).

First, let’s talk about sound. You probably noticed that the pitch of sound changed as the ball (or tuning fork) moved. When it moved toward you, the pitch went up. When it moved away from you, the pitch went down.

The tuning fork may have been a bit stranger, as it was moving faster and in a circular path. You should have noticed that the pitch warbled up and down, depending on where it was in its path. Here’s a question - if you swing the fork in a circle, directly above your head, should you notice any effect? Why or why not? Try it and see?

And indeed, this is exactly the case with light or any electromagnetic radiation. If the emitter is moving away from you (relatively speaking), you will receive a lower frequency than you would otherwise. In the physics and astronomy biz, we call this a RED SHIFT.

Similarly, if the emitter is moving toward from you (relatively speaking), you will receive a higher frequency than you would otherwise. We call this a BLUE SHIFT.

By the way, “relatively speaking” refers to the motion between the emitter and the receiver. If you (the receiver) are moving toward the emitter, it is essentially the same as the emitter moving toward you. Of course, if you both move toward (or away from) each other, the effect is even greater.

This does NOT mean that the objects will appear red or blue – rather, a frequency is shifted lower (red shift) if the object is moving away from you. The term ‘red shift’ is used since red is at the low end of the visible spectrum. ‘Blue shift’ is used since blue is near the upper end of the visible spectrum. Incidentally, the term ‘violet shift’ is sometimes used in astronomy.

So perhaps you see where I’m going with this…. Could there be a mathematical way to represent speed as a function of wavelength change?

Oh my, yes.

f’ = f (v +- vd) / (v -+ vs)

Crazy formula, eh? Let’s figure out what these terms mean.

f’ Frequency measured by receiver
f Frequency actually emitted by emitter
v speed of sound (approximately 340 m/s in air)
vd speed of detector
vs speed of source/emitter

How about this +- or -+ business? This is a matter of personal preference for sign convention. I remember the word ‘TA TA,' standing for “toward-away toward-away”. If the detector is moving toward the source, use a plus sign. If the detector is moving away from the source, use a minus sign. In the denominator, it is reversed. If the source is moving toward the detector, use a minus sign. If the source is moving away from the detector, use a plus sign. Got it?

How about the expression for light?

f = fo sqrt [(1 – β)/(1 + β)]

where β = v/c, and sqrt means “square root”. Sign convention is a bit tricky here, too. The above formula is correct for the detector moving away from the source of light. If the detector is moving toward the source, you must reverse both signs in front of β.

For astronomy, an easier formula is often used:

v = ( | Δλ | / λo ) c

| Δλ | the absolute value (positive) of the change in wavelength
λo the emitted wavelength
c the speed of light

By the way, if you only know the frequency and not wavelength, it is easy to determine the wavelength using the simple relationship:

c = f λ

That is, speed of light = frequency x wavelength.

So to get back to our initial question – how can we know how fast something is moving, especially in the case when conventional means are just not practical? We compare a particular wavelength from the object’s spectrum to what we expect from a laboratory standard. Pop the numbers into the formula above and there you go – instant speed!

Further Reading

A good mathematical analysis of the Doppler Effect can be found in most college-level physics texts. Here is one recommendation:

Fundaments of Physics, 8th edition – Halliday, Resnick & Walker. See chapter 17.

ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009 (except where noted)

WHAT’S THAT STAR MADE OF?

Hi folks, Sean Lally here.

So here’s a question for you to ponder – how on Earth (literally) can we tell what a star is made of? We clearly can’t travel to a star – nearly all are way too far away to be reachable in our lifetimes with normal space travel, and more importantly – they’re really hot! (More on this later.)

Quick problem for you to consider: Traveling at a “typical” spaceship speed, say 25,000 miles per hour (that was an average speed for rockets heading to the Moon), how long would it take you to reach the nearest star, our very own Sun (about 93,000,000 miles away)?

Hmmmm….

Well, if keep things simple and just use the equation (v = d / t), what do you come up with?

How about traveling to the next closest star system, Alpha Centauri (4.37 light-years away)? How far would it take you to get there? It will be helpful to note that a light-year, the distance that light would travel in one year, is 5,878,630,000,000 miles.

Think we could make it there quickly?

So, all that aside – it’s not too practical to take a space-drive out to a star, even the nearest one, grab a cup of star-stuff and bring it back to Earth for analysis. So then, how do know what it’s made of?

For this exercise, you’ll need access to a spectrum tube power supplies and tubes containing several gases. You’ll also need spectroscopes or diffraction gratings.

Do you remember our earlier exercise about prisms? Well, prisms are wonderful are breaking up light into its constitutive parts, but it’s not the only game in town. Another tool that breaks light up is called a diffraction grating – that’s the central part of the modern spectroscope, though prisms were used for quite a long time (and still work quite well). Diffraction gratings are cheap and simple to produce in large quantities, making them somewhat preferable to prisms. And they perform the same task, break up light according to wavelength, though the process is different for diffraction gratings.

Let’s put aside the details of how this happens. A diffraction grating is made of many tiny slits through which light can travel. However, if the slits are similar in size to the wavelengths, light can break up and “interfere” with itself – crests of light waves adding to other crests, troughs adding to other troughs, and some crests and troughs canceling each other out. In short, light hits a diffraction grating, and the various waves add or cancel: a pattern emerges and the spread of the pattern is related to the wavelength of the light. (For more details of the process, see consult the references below.) Of course, the wavelengths have to be there in the first place to actually emerge from the grating. But like sunlight through a prism, the colors are spread out in a pattern related to their wavelengths.

Of course, the material casting off light may have many different wavelengths – you can only see them after they have passed through a grating or prism. But once they do, you can see a pattern that you could assign to that particular light-emitting stuff.

Let’s see how this works in practice.

Ask your teacher to help you set up the spectrum tube power supply – be careful, as these can generate 2000 volts or more!

Have a look at the spectra of several gases. I recommend Hydrogen, Helium, Oxygen, Carbon Dioxide among others. To do this, you will need to position yourself around a meter from the spectrum tube, and look through the diffraction grating (or spectroscope, if you are using one of those). What do you notice? Are there differences? Describe?

What in the world is going on here? At high temperatures, things glow and each element has its own unique “emission spectrum,” not unlike a fingerprint. So in principle, if you had a library of emission spectral images, you could tell what the particular gas is made of. It is, of course, more complicated with compounds (chemical collection of multiple elements), but the principle is the same. If you had a large enough set of fingerprints, you could conceivably compare your pattern to the set and then maybe tell what the stuff actually is. Behold the beauty and power of spectroscopy!

Additional resources

http://www.teachersource.com/

Good source for diffraction gratings, spectroscopes and other many other materials for teaching color.

http://chemistry.bd.psu.edu/jircitano/periodic4.html
http://www.800mainstreet.com/spect/emission-flame-exp.html
http://ioannis.virtualcomposer2000.com/spectroscope/amici.html#colorphotos

These include libraries of spectral images.


http://www.starspectroscope.com/
A small diffraction grating optic that you can attach to a telescope eyepiece to examine some stellar spectra of your own.

ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009 (except where noted)

WHAT COLOR IS THAT?

(http://www.webexhibits.org/causesofcolor/3.html)

Let’s go back to our friend, the Sun. How would you answer this question – what color is the Sun?

Think about it for a moment… ask a friend.

What do you think?

Well, it doesn’t really have ONE color. Let’s explore this idea. You have probably heard of prisms and you’ve probably seen rainbows. What do these things have in common? (That’s a tricky little question, by the way.)

The answer is a little strange. First, some background. Light, as you may know, travels at a constant speed, 3 x 10^8 m/s (186,000 miles/second).

c = 3 x 10^8 m/s

Note that the letter c is used to denote “constant”, though some prefer to think of it as c for celerity (aka speed).

Now, that’s not just fast – it’s FAST! As in….

F A S T ! ! !

That’s nearly 7 times around the Earth’s equator in one second, or up to the Moon and back in around 2.5 seconds. FAST. In fact, that’s as fast as anything could ever be – to make things weirder, the only things that can go that fast are things with NO mass – in other words, light. Plus, it’s not just one type of light that travels that fast – it’s ALL types of electromagnetic radiation, visible and otherwise. So, that means radio waves, microwaves, infrared light, ROYGBV, ultraviolet, x-rays… All travel at the same speed – IN A VACUUM. That’s the important part.

I’ll use visible light as an example, but this applies to all electromagnetic waves. Light travels at a constant speed in a vacuum, yes, but vacuums are hard to come by. If light enters something that is NOT a vacuum its speed changes (lowers) – sometimes a lot, sometimes a little, but always. Mind you, even dropping by 50% (as light’s speed does in many types of glass) keeps the speed still extremely fast (1.5 x 10^8 m/s). Basically, half of really fast is still really fast!

So, what does this have to do with the color of the Sun? Light from the Sun comes in many different wavelengths – each particular wavelength corresponding to a color (in the case of visible light) or if you prefer, an energy (in the case of all of the types of electromagnetic radiation). The Sun emits virtually every possible wavelength of electromagnetic radiation in some amount.

How do we know this?

For this part of the experiment, you will need a prism and the Sun – make sure to call ahead to see if the Sun is available.

What do you expect to see when your prism is brought into the open sunlight? Discuss with your friend before trying it.

So, what do you see? Is it what was expected? What can this mean?

If you have a regular piece of thick glass (not a prism), try it again. Does it have the same effect? Why?

As noted above, light travels at a constant speed in a vacuum. But when NOT in a vacuum, the story changes a little. Light travels at a speed that depends entirely on the medium – its so-called optical density (usually embodied in a term called “index of refraction”) – relative to light’s wavelength. In other words, different wavelengths of light travel at different speeds in a prism (or whatever it’s going through) and, therefore, emerge at (slightly) different times. Another way to think about this: different wavelengths are slowed down in matter by amounts that depend on the relative sizes of the wavelengths and atomic size of the medium.

A natural consequence of this slowing is the bending of the light as it exits the prism. Since violet is slowed more than red, red emerges first. But why the bending? Here is a common analogy. Think of light as you would a car on the road – if you start to go off the road to the right, maybe going from pavement to dirt, your right tires will experience different friction than the left tires. As a result, the tires on the left will have a different speed than those on the right – naturally, the car will turn somewhat. The story is similar for light. Different resistance means different speed and different amounts of bending.

So, back to our friend, the Sun. What color is it? If you said yellow or orange, you’re somewhat right. The truth is, it emits light of many colors – virtually every color you can imagine – AND nearly all other types of electromagnetic radiation. That’s right – from the Sun we get radio waves, microwaves, infrared light, ultraviolet light, x-rays and gamma rays. WOW! But why the yellow/orange color?

Why do you think it appears to be this color?

Well, it’s a combination of two things: Yellow/orange is the color given off with the greatest intensity, but also (and very importantly): this color is close to the most sensitive range of human sight. If we could see the Sun with microwave or x-ray sensitive eyes, we would see quite a different object in the sky.

Have a look for yourself. Find images at the Solar Data Analysis Center.

http://umbra.nascom.nasa.gov/images/

Now, the Sun doesn’t emit all types of electromagnetic radiation in equal amounts. In fact, it is close to being what physics types call a blackbody emitter.

(See image above.)

This curve (which is a pretty good representation of the Sun’s output) looks a little strange, doesn’t it? In this case, the peak is in the orange/yellow range. But look! There are plenty of other types of electromagnetic radiation emitted from the Sun, just in smaller amounts. But you know this already, don’t you. For example, why do you wear sunscreen?

That’s all for now. See you again, chromatic friends!

ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009 (except where noted)

THE SUN IS DOING WHAT?

(http://galileo.rice.edu/sci/observations/sunspot_drawings.html)

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at the Sun. You’ve heard of it, yes? It keeps us warm, it makes plants grow, it gave us the Beach Boys…

But did you know that it also rotates? It’s true, I tell you! It takes about 25-35 days to complete one rotation. Now I know what you’re thinking – big deal, the Earth rotates once per day. Well, the Sun has a diameter about 100 times greater than that of the Earth. At the Earth’s equator, the average speed is close to 1000 miles per hour. At the Sun’s equator, it’s nearly 3 times that speed.

So, how do we know that?

In Galileo’s heyday, the early 1600s (that’s around 400 years ago), it was ridiculous to imagine that the Sun was moving. It was dangerous enough to imagine that the Earth was moving. Galileo was tried and convicted for writing and teaching that very notion. Galileo may have not been the first person to use a telescope, but he’s the first person to write detailed observations through his telescope about the night sky. In a few short weeks, Galileo discovered that there were many more stars in the sky than previously thought, that the Moon was heavily cratered, that Venus went through phases (strongly suggesting that it rotated around the Sun), that Jupiter had satellites of its own, that Saturn had “ears” (later shown to be rings) and that the Sun had little spots on it!

Yikes! The Sun has spots? These were noted by projecting the image of the Sun onto a piece of paper. They had also been noted by Christoph Scheiner, Thomas Harriot, and Johannes and David Fabricius.

Can one see these spots easily? Well, yes and no. If you try to look at the Sun with the naked eye, then no – in fact, it’s a bad idea, in general. If you look through a designated solar telescope, a pretty costly device, then yes, it’s usually pretty easy. (Alternately, a scope fitted with an appropriate solar filter is also a good way to view the sun – and much cheaper!)

We will not assume that you have easy access to a solar telescope, or telescope fitted with a solar filter, so let’s first have a look at Galileo’s images.

http://galileo.rice.edu/sci/observations/sunspot_drawings.html

Now comes the impressive part. Galileo created the drawings in a series over the course of several days. Imagine them as a “flip book.” Or better yet, imagine that someone had created a film of the images.

Actually, you don’t have to imagine this at all. Look at the video clips (same page).

http://galileo.rice.edu/sci/observations/sunspot_drawings.html

What are the possible explanations of what was seen? Are there alternate explanations (other than the Sun rotating)?

So, why do you think so many of Galileo’s contemporaries were skeptical? What possible harm could come from imagining that the Sun rotated?

So, it seems as though the Sun rotates. Why might this be?

For that matter, why does the Earth rotate?

That’s it for today – have fun, see you in the Sun!

http://galileo.rice.edu/index.html
The Galileo Project, a great source for all things Galileo

http://galileo.rice.edu/sci/observations/sunspots.html

Galileo – Discoveries and Opinions of Galileo


Personal Solar Telescopes

http://www.coronadofilters.com/
http://www.solartelescope.net/
http://www.cloudynights.com/item.php?item_id=104
(Sunspotter – a modern classic)

ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009 (except where noted)

HOW FAR AWAY IS THAT?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at distances in the universe. So, how DO we know how far things are away from us? How far away is the Moon? How about the Sun? Stars? How can we know the distance to anything we can’t easily visit?

Think about this for a minute. If you want to know how far away your pencil is from you right now, how would you determine that?

How about if you want to know how far your science teacher is from you?

These aren’t super tricky. You’d probably use some type of measuring stick.

How about if you want to know how far you are from home? That might require a different type of “measuring stick” or different method altogether. Think about that one and come up with an idea or two.

I suppose you could drive there and just watch the mileage on your car’s odometer (trip counter). Or maybe you could just map it online and see what the program claims the mileage to be. Or perhaps you could use a GPS.

Here’s a tougher one: Imagine that you want to know how far across a river is, but (and this is the catch) you can’t actually cross it to measure directly. Hmmmmmm – how would you do that? We’ll go one step trickier – let’s say that you’re limited by the technology of 2500 years ago. That means, no GPS, no computers, no modern technology whatsoever.

And while you’re thinking about that one, imagine how exactly one would measure the distance to the Moon without going there. That means that, no, we can’t drive a spaceship there with a long measuring tape behind us. We also can’t drive our spaceship there and check the odometer when we’re done!

OK, both of these problems require the same type of thinking. Take a minute to discuss with your partner how one might be able to attack the river problem.

Have an idea? Here’s one low-tech way to “solve” the problem. (Hey, why did I put solve in quotation marks?)

(See image 1 above.)


So, you’re standing where the diamond is and you notice a tree across the river, indicated by the star. (It doesn’t have to be a tree, but you get the idea.)

Now let’s say that the only equipment you have is a protractor and meter stick. How might you find the distance across the river to the tree? Think about it. Discuss with your partner.

Here’s one approach that will work as long as you’re careful. It involves the use of a simple instrument that you construct, similar to a surveyor’s tool called a transit.

(See image 2 above.)


In this example, you mark a spot close to the shore directly across from the tree. Then you walk a measured distance away from this spot, ending at the diamond spot on the diagram. Here is where the protractor comes to use; also, it may be easier to put it on a notebook so that you can mark off the angle as you measure it. Get on the ground as low as you can. Aim the protractor such that the center (90 degrees) is aimed directly across the river. Try your best to determine the angle (θ) that the tree makes with respect to the shoreline – see the picture above. You may want to use a straw as a “sighting tube”. If so, make sure that the straw crosses the “origin” of your protractor and line it up with the tree. Make sure that you record the angle that the straw makes with respect to the horizontal base of the protractor (which is parallel to the shoreline). You may find it best to ask your teacher for help.

For most students, using a river won’t be very realistic – instead, use this method to calculate the distance across a road (but do very careful – watch for traffic!).

Yes, it’s a little tricky. That’s why I used the word solve in quotes above – this method, like any measuring method, is only as accurate as the tools and the person taking the measurements. In principle, though, it can be quite accurate.

Now, construct a scale diagram on paper. You measured a distance along the shoreline above. Decide upon a realistic scale for the size of your paper. For example, 1 m (of outside distance)= 1 cm (on paper).

Use your protractor to construct the angle that you measured above, in the same position on the diagram that it was outside. Draw additional lines to make a triangle, as shown above.

Now, measure the side of the triangle that represents the distance across the river. Convert it back to meters, using your scale. For example, if your scale was 1 m = 1 cm, and the line is 20 cm – the distance would be 20 m. Get it?

Alternate method – trigonometry!

An alternate method that does not require a scale diagram is to use the mathematics of trigonometry. Everything that you did above is repeated, but a scale diagram is not needed. Since you know one side of a right triangle and an angle adjacent to it, you may use the trigonometric ratios – specifically, the tangent function.

In case you do not know:

In a right triangle, the three sides can be defined as a, b and c. However, it’s sometimes more useful to call them opposite, adjacent and hypotenuse. The hypotenuse is the longest side of the right triangle – the side directly across from the right angle. But what side is opposite and what side is adjacent? This depends on the angle that you’re thinking about – a so-called reference angle.

(See image 3 above.)

Note that opposite means “opposite the reference angle”, and adjacent means “adjacent to the reference angle.”

In a right triangle, several ratios can be defined:

Sine (sin), cosine (cos) and tangent (tan) are the most common. For our purpose, tangent will be most useful. Here are the definitions of the ratios:

sin (θ) = opposite / hypotenuse

cos (θ) = adjacent / hypotenuse

tan (θ) = opposite / adjacent

(This is easy to remember with the pneumonic SOH CAH TOA: Sin equals Opposite over Hypotenuse, Cos equals Adjacent over Hypotenuse, Tan equals Opposite over Adjacent.)

Literally, we read this as (for example):

“Sine of theta (θ) is equal to the opposite side divided by the hypotenuse.” What this means is that no matter how big or small the triangle, the ratio of the sides associated with this angle will always have the same value. For example, the sine of 30 degrees is 0.5 – this means that no matter what the actual size of a right triangle, if it has a 30-degree angle in it, the ratio of side opposite this angle to the hypotenuse of this triangle will always be 0.5. Pretty neat, eh? Sin(θ), cos(θ) and tan(θ) are simply ratios of sides associated with particular angles (θ).

For tangent, our trig ratio of choice, “tangent of theta (θ) is equal to the opposite side divided by the adjacent.”

But how do we apply this to our river problem above?

You have the angle, measured with your protractor. Determine the tangent of this angle by using a graphing calculator (make sure it is in “degrees mode”). Your teacher may help you with this. You know the adjacent side of the triangle and now you can calculate the opposite side:

tan (θ) = opposite / adjacent

Using algebra, we can find that:

opposite = (adjacent) x [tan (θ)]

Try it!

How well did you do? Well, if you were measuring the distance across a river, it may be tricky to get the actual distance. However, if you used a road you could carefully measure the actual distance across the road – use a meter stick, trundle wheel, string or gullible friend to find the actual distance across the road.

Now find the percent that your value is different from the actual value:

[ (Your value) – (actual value) ] / (actual value)

Multiply this by 100 to make it a percent.

Now what is a good percent? That’s hard to say. Different experiments and methods have different amounts of acceptable error. There are no real absolutes here – usually, the scientific community decides on what is acceptable. But here is a rough guide for this particular experiment:

If your value is under 25% (which means that you measured at least 75% of the actual value), that is not bad for a quick experiment of this sort. Higher than that? Try it again.

What are sources of your error? Think about this and write some down. It will be very helpful to talk this over with your partner or teacher.

Astronomical Distances

But what does this have to do with astronomical measurements? The same principle applies, believe it or not. We can measure star angles from the Earth, using instruments that are similar to protractors – sextants, quadrants, etc. This was a classic technique used for centuries. It is usually easier, however, to measure how far stars are from other stars in the sky – this can be viewed as an angle if you think about one line from you to a star, and then another line from you to a different star. Imagine the angle that exists between these two lines.

But there is a little problem.

We still need a measurable distance like the shoreline above, and moving a short distance on the Earth doesn’t give us a significant distance (especially when the stars are pretty far away). So, we change the game a little – we wait for the Earth to move to another part of its orbit, having gone 2 AU. (An Astronomical Unit, or AU, represents the size of the semi-major axis of Earth’s orbit. It is also close to the average distance between Earth and Sun). The 2 AU becomes our known baseline. We take angular measurements at the two points in the orbit and come up with a parallax angle with respect to background stars. The technology is a little different, but the approach is very similar. This technique (called astrometry) works very well for many stars, particularly the nearby ones. Further stars have much smaller parallax angles, making this technique less useful; for the stars farther away, we have other methods for measuring the distances.

By the way, this technique was used to measure the distance to the Moon nearly 2500 years ago by Eratosthenes, among others. Measuring cosmic distances is tricky business, but it’s certainly not new.

That’s all for now – see ya soon, see ya on the Moon!

Further Reading

Kitty Ferguson – Measuring the Universe
Dava Sobel – Longitude
Barbara Ryden - Cosmology


ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009

SO, HOW BIG IS THE EARTH?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at the Earth yet again.

So, now we know that the Earth is a sphere. How big is that sphere?

Quick, how would you determine that? Think about it – ask a friend.

Here are a couple of possibilities:

Get in an amphibious vehicle (!) and drive/sail around the equator, arriving back where you started. Clearly, you should set your odometer (trip counter) to zero and note the reading upon your return to the starting point.

How about this? Get a gigantic piece of string and wrap it around the equator until it reaches the starting point. Alternately, you could wrap the string around adjoining lines of longitude, passing over the north and south poles. This would give you a slightly different value, but the principle is the same.

What’s wrong with these methods? Anything?

Neither of these methods has been utilized, but the size of the Earth has been known with great accuracy for nearly 2500 years! How did someone know this?

As you might expect, it has something to do with mathematics. Let’s go back to our friends, the ancient Greeks – namely, Eratosthenes of Cyrene (276 – 295 BCE). It was known by the locals that Syene (Aswan, in Egypt) was located at or near the Tropic of Cancer – that’s a circle of latitude around the Earth close to 23.5-degrees North. This location is important because the Sun is directly overhead at noon on the day of the Summer Solstice, and no object will cast a shadow then. (The Earth is inclined at an angle of 23.5-degrees, and yes, this has a lot to do with why there will be no shadown here on the Summer Solstice.) With this in mind, Eratosthenes devised a clever experiment. But first, how would you use this information?

How about if you knew of another location roughly along the same line of longitude a certain known distance from Syene? Any clues?

This other location, Alexandria, was north of Syene by some 5000 stades (close to 800 kilometers). At noon on the day of the Summer Solstice, Eratosthenes set up a large stick – it certainly did cast a shadow at noon (when the shadow was shortest), and he determined the size of this shadow and the angle from the top of the stick to the far side of the base. See diagram above, which is clearly NOT TO SCALE.

Think of this – the distance from Syene to Alexandria was known and the shadow and angle were easy to measure very accurately. How would you use this information to determine the size around the Earth? It’s a geometry problem. Think about it for a moment and discuss with your friend.

This part is not so obvious – the two rays of sunlight drawn above are very nearly parallel to each other. If the two sticks were sufficiently long enough, they would intersect at the center of the Earth – creating an angle between them. But this angle is the SAME as the angle between the top of the stick and the ray of sunlight. Again, it looks that way in this diagram, but the proof is a classic one in geometry. If you don’t know the geometry here, you may need to suspend disbelief.

Once you see this, the problem is a simple proportion. How many degrees are in a complete trip around a circle? 360, right? Consider this:

The angle of shadow (measured by Eratosthenes to be 7.2-degrees) is to the angle completely around a circle as the distance between Syene and Alexandria is to the complete distance around the Earth (circumference).

Wow, that’s a mouthful. Symbolically it is so much simpler:

7.2-degrees / 360-degrees = 800-km/ Circumference

I have taken the liberty of writing this as a proportion, though Eratosthenes thought of this just a little differently. Can you solve this for the circumference of the Earth?

So, is this a good estimate? Well, our current value for the approximate circumference of the Earth is 40,030 km (around 25,000 miles). Would you consider this technique to be pretty accurate?

So, what are sources of error in this experiment? And what assumptions did Eratosthenes make. There is a lot to think about here – talk this over with your friend.



Further Reading

Kitty Ferguson – Measuring the Universe

Nicholas Nicastro - Circumference



ALL TEXT AND IMAGES COPYRIGHT SEAN LALLY 2009

How do we know that the Earth is spherical?

Hiya, Sean Lally Physics Guy here. Today on “How do we know that?” we are looking at the spherical Earth. So, how DO we know that the Earth is spherical (or close to it)?

So, you live on Earth – at least I presume that you do. That seems pretty reasonable. You’ve known that the Earth is a sphere (or close to it) for as long as you can remember. However, this isn’t really intuitive, is it? Look around you – it’s probably pretty flat where you are. Drive through Ohio or Iowa – yeesh, that’s some flat Earth! How do we know that the Earth actually isn’t flat?

Think about this for a moment – talk with a friend. If you had to prove to someone that the Earth was spherical, someone not from Earth perhaps (?!?), how could you convince him or her? What would be sufficient evidence?

Write down ways that would convince you that the Earth is spherical.

Now look at your list again, but back up 2500 years. Which of these would likely have been plausible then?

Find a partner and explain your choice(s) to him or her. Listen to your partner’s choices. Are these convincing arguments? Is your partner convinced by your reasoning?

Let’s see how your list compares to some of the classic reasoning by the leading thinker of 2350 years ago – Aristotle.

Aristotle, like most ancient Greek philosophers, believed that there were only 4 elements – earth, air, fire and water. Knowing now that there are well over 100 chemical elements, this seems a little kooky. However, the discovery of the chemical elements is a relatively recent thing – chemistry, as we know it, is less than 200 years old. It’s not even as old as the United States! It seemed reasonable to the ancients that all things could be categorized as one of the “big 4”, or some combination of these “elements.” Actually, if you think about if for a moment, you could probably convince yourself that this is pretty smart – or at least not totally ridiculous. Mind you, Aristotle certainly didn’t create the idea of a spherical Earth – his teacher Plato had the notion of spheres within spheres, and others (particularly the Pythagoreans) before both of these gentlemen had spherical theories.

Ancient Greeks (the so-called natural philosophers – predecessors of today’s scientists) believed that these elements would (if given the chance) arrange themselves according to their weights. In an absolute sense, Earth is heavier than water, which is heavier than air, which is heavier than fire. So naturally, things made of earth (which include most things within your reach right now) would tend to collect below the others. Since things can move from all directions, it was natural to imagine that things should collect at a center – that is, arrange themselves in a spherical orientation. (Actually, as we understand gravitation today, that’s not too far removed from how we believe stars and planets are formed.)

Aristotle went even further – he imagined the ideal scenario in which there was a sphere of earth, surrounded by a sphere of water and then spheres of air and fire. Think about – it’s not totally crazy. In fact, it’s somewhat intuitive. After all, if you put something heavy into a swimming pool, it usually sinks, right? (Of course, you can probably come up with objections to Aristotle’s logic, but that’s ok – we don’t pay much attention to Aristotle’s science anymore.)

Are there other reasons to convince someone that the Earth is spherical? Yes, and these are easier to understand and accept.

Imagine that you live near water – like so many early civilizations. Imagine watching boats go away from the shore. What would you see as the boat got further away from you?

Imagine climbing a mountain and staring off at the far away horizon. What do you think you would notice?

Finally, consider the phenomenon of eclipses. Maybe you have not yet seen a lunar or solar eclipse, but when you finally do – wow, they are awesome things to behold. Try to find some pictures of solar eclipses – that is when the Moon gets in the way of the Sun. This happens every year and a half or so, and is usually only visible from a very small portion of the Earth, so it’s likely that you, your family or friends may have never seen one before. Draw what you would imagine this would resemble.

Ask your family or teachers if they have ever seen a solar eclipse. Find some images online. The first thing you notice is the Moon is a dark circle across the Sun (with the Sun’s corona extending around the Moon). This isn’t too surprising – you see the Moon all the time and it certainly looks round, at least during the full Moon phases.

But now look at images of lunar eclipses – that is when the Earth comes between the Sun and the Moon. The shadow of the Earth creeps across the face of the Moon. You may have never seen one directly, but draw what you might expect to see.

Another experiment you could try would be to look at the shadow that a ball casts on a wall when placed in the path of a bright lamp.

Now find some images of lunar eclipses online. Try to find images that show the Earth’s shadow creeping across the surface of the Moon, rather than the time that the shadow completely covers the Moon (though this is interesting, too). What do you notice about the shadow?

What would a lunar eclipse resemble if the Earth were flat? How about if the Earth were a flat disc (as Samuel Rowbotham argued in the 1800s)?

Is this convincing evidence that the Earth is spherical? Can you imagine other arguments that might have worked 2500 years ago?

In any case, Aristotle’s writings convinced nearly everyone (if they could read, of course) and the Earth as a sphere became part of the accepted knowledge base that virtually all people had. Often, Christopher Columbus is given credit for having proven that the Earth is spherical – this is simply not the case. That idea is a myth of more recent origin.

One more thing – I’ve been using the word spherical pretty loosely. The Earth is, in fact, not spherical – it is bulged a bit (oblate) at the equator. The equatorial diameter is about 40 km (27 miles) longer than the polar diameter. Consider how that compares to a billiard ball.

According to the Billiard Congress of America,

16.16. Balls and Ball Rack
All balls must be composed of cast phenolic resin plastic and measure 2 ¼ (+.005) inches [5.715 cm (+ .127 mm)] in diameter.

40 km may seem like a large difference in diameter. However, by percent of diameter, the Earth is closer to a perfect sphere than the average billiard ball.

Wow! Anyway, that’s all for now – see ya soon, see ya on the Moon!

Further Reading

The Beginnings of Western Science
David C. Lindberg

This is about the best text on early science that you will ever find.

Flat Earth?
The wikipedia entry on the Flat Earth Society is certainly worth your time, as is a book by Christine Garwood, titled Flat Earth.

More on the Earth, spherical or flat:

Aristotle - On the Heavens
R. Dicks – Early Greek Astronomy to Aristotle
David Lindberg - Beginnings of Western Science
http://www.bca-pool.com/cgi/site/framegate.cgi?url=http://www.bca-pool.com/play/tournaments/rules/rls_gen.shtml&cat=p

ALL TEXT COPYRIGHT SEAN LALLY 2009