Ohh, that makes sense. Because that's what velocity is ><'.

Does the force * time thing have anything to do with e=mc^2 ? And rocket fuel is using up m for e..? Or is e=mc^2 referring to something else?


Impulse is a word I rarely hear, so it sounds really odd. Thanks for the clarification!

No, E=mc^2 is mostly used for nuclear reactions, not chemical ones. You CAN theoretically use it empirically to determine the energy of a chemical bond by breaking that bond and measuring the change in the mass of the system -- I worked the math for this in an assignment in high school but I don't know if it actually gets used this way in practice.

Rather, rocket thrust works based on the principle of conservation of momentum. When you throw something off the back of the rocket, the equal-and-opposite-reaction pushes the rocket forward.

A simple example that you should be able to work given the knowledge you already have:

A drone weighing 250 grams when empty is armed with a pellet gun. Pellets weigh 1 gram each, and the gun can fire them at 200 meters per second. If you load the drone with a single pellet, how fast will it be going due to the recoil after it shoots?

(I haven't actually WORKED the answer.)

To make this more like a rocket, load the drone up with 10 shots and fire one per second. This takes more work -- you haven't studied the math that would make this easier, so you'd have to work out the math one shot at a time -- but you can see that it starts off at 259 grams going one way and 1 gram going the other way for the first shot, but 250 grams going one way and 1 gram going the other way for the last one -- the same impulse results in a different change in velocity.

Ohh, okay. I remember there being problems about using E=mc^2 for coal in some previous unit at school, but in real life it makes sense that it isn't useful.

(200 gm/s) / 251 g = 0.7968 m/s about.

200/251 + 200/252 + 200/253 + 200/254 + 200/255 + 200/256 + 200/257 + 200/258 + 200/259 + 200/260 = 7.8288 m/s about.

I'm glad it was only ten x'D.
Thanks for the drones problem! I didn't notice that the pellet added mass to the drone until later :x

Oh, no, no, E=mc^2 IS useful, just not for the kinds of chemistry you can do in a basic lab. Nuclear chemistry is a totally different beast.

Pellet doesn't add to the mass of the one-shot example, because it's already fired the shot by the time you calculate it. 10-shot example you should have gone from 250 to 259 instead of 251 to 260.

I'll check your math later, I've gotta run right now.

Okay, so your math is right if you fix the mistake I already mentioned.

As for a real-world example of E=mc^2... Photons don't have mass, because they're pure energy. They travel at the speed of light, never any faster or any slower. But we know that photons can transfer kinetic energy to other objects -- which means that photons must have momentum.

Except momentum is mass times velocity, and photons have no mass!

That's where E=mc^2 comes in. We can say that a photon's effective mass, for momentum purposes, is equal to its energy, divided by the speed of light.

woops. So it's actually

(200 gm/s) / 250 g

And

200/250 + 200/251 + 200/252 + 200/253 + 200/254 + 200/255 + 200/256 + 200/257 + 200/258 + 200/259

Oh wow, that's interesting--but also confusing. A photon doesn't have a mass but it does have an "effective" mass? Also, if E=mc^2 holds, then without a mass there's no energy..?

Now we're getting into relativity. Fun stuff!

E=mc^2 is actually a simplification of the full formula, which is E^2 = p^2*c^2 + (m*c^2)^2, where p is momentum and m is rest mass, by setting momentum equal to zero (that is, considering an object at rest). If you instead set mass to zero, like for a photon, then you get E = p*c, which is to say, the momentum of a photon times its speed (obviously, the speed of light) equals its total energy. Mass-energy equivalence means that a particle's mass while in motion is greater than its mass when at rest.

When it comes down to it, energy is energy, and energy is the thing that actually does stuff. So it doesn't matter if the energy comes from rest mass or momentum; you can combine the two into into a single measurement and treat them as interchangeable.

It should be noted that a photon's relativistic mass is very very small. It's possible to set up an experiment to observe the effect of photon momentum on Earth, but it requires sensitive tools, because friction (including drag from the air) will easily overwhelm it. However, a spacecraft was launched in 2010 that uses a solar sail for propulsion -- the only thing making it move is the light from the sun pushing on its sail like wind in a kite. And it's still accelerating today!

To give you an idea of just how small the forces involved are: A solar sail 800m x 800m square can catch enough sunlight to exhibit a force of only around 5 newtons.

How is this relativity..?
Granted, I don't know what relativity is, and everything I've seen that explained it shot out of my head like gibberish x'D

The full equation looks a lot like the Pythagorean formula :o. Why is it simplified though? Is it just so that I, as a student, can understand the formula (since we learned energy before momentum)?

...What about p = mv though? If the mass is zero, the p turns zero too


Oh wow, solar sails are like the stuff of fantasies xD. Makes me wonder though, if E makes up that equation, where things like thermal energy and the rest come in--though I haven't yet learned about any of those :x

This is relativity because energy-mass equivalence is something that only really matters at relativistic scales.

At normal scales, you can just use p=mv and just ignore the contribution that motion adds to the rest mass. I mean, look at that full equation -- the mass term has c^4, while the momentum term only has c^2. And c is a BIG number. You'd need a momentum in the neighborhood of 10^17 (or a mass on the order of 10^-17 kg) in your equation for those terms to be on the same order of magnitude.

p = mv can be true for massive objects with no velocity. That's why the simplification is used -- it lets you reason about the energy equivalence of the mass itself at scales where the relativistic contribution of its velocity is negligible.

Macroscopic measurements of different types of energy come from the collective behavior of huge numbers of particles. At subatomic scales -- a proton has a mass of less than 10^-30 kg -- many things like "heat" don't have meaning. Temperature, for example, is a measure of the average kinetic energy of the particles in the thing being observed. There are field forces (like gravity and magnetism) that still come into play, so potential energy is a thing that still comes into the picture, but that doesn't participate in energy-mass equivalence as far as I can tell in my research. (I actually had to go look this up because I didn't know.)

Solar sails DO totally seem like sci-fi stuff, don't they?

Beyond the idea of actually moving from sunlight, how simple they look is really amazing! Reminds me of reading books about space-ships XD. But sunlight pushing stuff is really cool :o

Thanks for all the answers!

What is scattering of light? In class we were talking about how "Rayleigh scattering" was gas molecules absorbing and re-emitting light in all directions, "with the proportion of light scattered related to its wavelength, yielding blue sky and red sunsets." And then: 'Because the Sun’s output begins to drop off at violet wavelengths, most of what reaches us due to scattering is blue and green, yielding familiar “sky blue.”'

So, lights are made of photons at different energies, and when they hit stuff like gas molecules they get "absorbed" then "re-emitted" (whatever those mean) and then the "proportion" (amount?) of light "scattered" (seen?) is based on the "wavelength" (of the light before it hit the gas?) and sunlight has photon particles with energies that make blue-light wavelengths which is why we see blue?


I end up nodding off a lot in physics class since it's later in the day. Hence the many holes in my learning :/

In your defense, Rayleigh scattering is usually taught very poorly at the high school level.

So before I start getting into the technicalities you're asking about... I'm just going to make one big statement: The sky is blue because the Sun is blue.

Now I'm going to roll it back to the beginning and build back up to that statement.

Though first, one quick definition, since I'm going to use it in my description and you were asking about it: A proportion is just a fraction -- a description of some part of a whole that has a certain property. If two things are proportional, then there's a relationship between them such that they go up and down together at related rates.

Part 1: Wavelengths of Light

As you know, what we perceive as "white" light is made up of a combination of colors. No individual photon of light is "white." (For that matter, no individual photon of light is magenta, either. Magenta doesn't exist except in your head.)

The color of light is determined by its wavelength, and the wavelength of light is directly determined by its energy. (For ANY wave, if you know two of wavelength, velocity, and energy, you can determine the third one. We know that light's velocity is c, so that means its wavelength and its energy have a directly inverse relationship with each other. When one goes up, the other goes down.) So whenever you see "color" or "wavelength" or "energy" you know it's all talking about the same thing.

The Sun is a fairly good approximation of a blackbody radiator. That means it emits photons in a continuous range of energies (this is called an "emission spectrum"), with the most of them at an energy level determined by the temperature of the blackbody (this is known as its "spectral peak") and then you get less of them the farther away from that energy level you go. The Sun burns at around 6000K, so it emits blue-green light the brightest, a little less violet light and yellow light, a little less near-ultraviolet light and red light, a little less far-ultraviolet light and infrared light, and so on. There's no gaps; every color in between is represented.

(Incandescent light bulbs are ALSO blackbody radiators, but since they're much cooler than the Sun, their spectral peak is in the infrared region of the spectrum, so most of the light coming from an incandescent bulb is heat, and most of the visible light is on the red end of the spectrum. Fluorescent and LED lights aren't blackbody radiators, so their emission spectra aren't smooth and continuous.)

Upshot: When your book talks about sunlight dropping off in the violet region, it's referring to the shape of this emission spectrum.

Part 2: Reflection

Throw a ball at a big flat surface. It bounces off at a predictable angle, right? Light does the same thing: shoot a beam of light at a big flat perfect surface (like a mirror) and it'll bounce off predictably.

Throw a ball at an uneven surface. Still bounces, but it isn't perfectly predictable; if you throw a bunch of balls, each one might bounce a different direction. You can estimate where they're likely to end up, but you can't predict any individual throw. Again, light does the same thing: shoot a beam of light at a sheet of paper, and while you know it's going to bounce, it's not going to bounce perfectly. It scatters, because each of the photons gets bounced in a little bit different direction.

Throw a ball at another ball. You can't even be sure if it's going to bounce, this time, because that's an awfully small target, and it's pretty likely that you're going to miss. You throw a bunch of balls, and most of them are going to go right on by without even so much as deflecting. And even if you hit, you can't be sure which way it's going to bounce because both balls are round, and you can't even be sure how hard it's going to bounce because the other ball is going to get knocked away too.

Now we're getting closer to the idea of Rayleigh scattering, but we're still not there yet.

Part 3: Crazy Physical Analogies That Actually Work!

Forgive me for that silly outburst of clickbait.

Blue light has a wavelength of 475 nanometers. A nitrogen molecule, like what the vast majority of our atmosphere is made up of, has a scattering diameter of a little under 1 femtometer -- around 600 million times smaller! That's like trying to hit a speck of dust with a beach ball, except the speck of dust weighs a trillion kilograms. (I actually calculated that number; I didn't just pick a big number arbitrarily -- I used E=mc^2 here!)

Now, fortunately for sanity, Newton's third law does in fact still apply at this scale. The principle of equal and opposite reaction still applies. (The first and second laws get fuzzy when you start approaching the speed of light. We already discussed the changes to the second law -- F = m*a is really an approximation of E^2 = m^2*c^4 + p^2*c^2 -- and the first law requires that you can construct a coordinate system with the object in question at relative rest, and you can't do this for photons because they're ALWAYS traveling at c no matter how you look at them.)

So given the third law, you can treat this situation like shooting said (indestructible, perfectly rigid) beach ball with said speck of dust traveling at the speed of light.

Surprisingly, the math actually does work out that way.

There's some fiddly electromagnetic fudge factors you have to take into account in order to define how big the balls are, but once you've established the effective scattering cross-section, Rayleigh scattering turns out to be a fairly clean formula that describes what you get when you average out all of the probabilities over a volume of space.

If you shoot the beach ball dead center, it's just going to go zipping directly in the opposite direction. If you hit it with a glancing shot on the side, both objects are going to ricochet in different directions.

Or, to drop the metaphor and go back to the actual question at hand: When a photon hits a nitrogen molecule, it'll bounce off in a direction based on how it hits and the relative sizes, and you can average out how far it's likely to deflect from its original straight-line path. And since different colors of light have different wavelengths and therefore different sizes, they have different statistics.

The Rayleigh scattering formula collects those statistics into a very convenient number: Given a beam of a given color of light shining through a unit volume of space, how intense is the light coming from the sides instead of passing straight through?

Part 4: Why Blue?

Now that we understand what Rayleigh scattering is, the next question is... Why blue?

Well... that's where those electromagnetic shenanigans I mentioned come into play. It turns out that the relevant cross-sectional radius has an inverse relationship with the wavelength, so that means the cross-sectional area has an inverse square relationship with the wavelength -- that is, counterintuitively, longer wavelengths mean a SMALLER target. (Actually, it's that higher frequencies make it less likely to have a collision, and higher frequencies mean lower wavelengths.) And then since the light can scatter in any direction, the amount of light coming in any one direction follows the inverse square law (brief description: a ray passing through the center of a sphere must pass through a point on the surface of that sphere, and surface area is 4pi*r^2), so Rayleigh scattering ends up being inversely proportional to the 4th power of the wavelength.

The important part of that last paragraph is that shorter wavelength -> more scattering.

For a ray of light that WOULDN'T otherwise pass in a straight line from the Sun to your eye, for you to be able to see it at all it would have to be scattered.

Remember that violet has the shortest wavelength of visible light, and red has the longest. So that means the red light from the sun is more likely to keep going instead of bouncing off of the air in order to come back and hit your eye.

Why isn't the sky purple, then? Because referring back to part 1, the Sun emits more blue light than it emits violet light, so even though a greater PROPORTION of purple light is getting scattered back to your eye, there's just not as much of it in the first place, so you see more blue.

Part 5: But You Said...!

Yep. I said the Sun is blue.

Sure doesn't look blue, does it? Looks yellow!

The light that IS coming directly from the Sun to your eye is ALSO getting scattered. The blue light is getting deflected off to the side more instead of reaching you. That leaves the light that gets scattered less -- the yellows and the oranges and the reds.

Why isn't it red, then, if red gets scattered least?

Same reason the sky isn't purple: the Sun emits more yellow light than red.

From outer space, the Sun looks white, and space looks black. There's no air in the way, so all of the colors of light mix together in our eye and we perceive white that might be ever so faintly tinted blue.

Part 6: Absorbing? Re-Emitting?

I didn't actually talk about this stuff above!

This is one of the reasons -- though by no means the ONLY reason -- I think Rayleigh scattering is taught poorly. The whole nitty-gritty details about absorption and re-emission is diving into too much detail on the WRONG PART of the phenomenon. It suggests that it's relevant to the macro-scale things we observe, when in fact you can understand how Rayleigh scattering works without ever needing to talk about it.

But to actually discuss it...

Photons don't actually BOUNCE off of things. Despite our fun little relativity discussion, they don't have mass; they're pure energy. So when a photon hits something, it transfers its energy into whatever it hits. But atoms can't hold arbitrary amounts of energy; they want to be at the lowest stable energy state they can, so if nothing happens to make the higher-energy state stable they quickly shed that extra energy in the form of another photon. And since the energy of the photon determines its wavelength, the photon that comes out is exactly the same color as the one that went in. (Obviously there are exceptions -- those exceptions are where chemistry happens.)

Photons also aren't just particles. They're also waves, and no macro-scale physical model can fully imitate ALL of the weirdness that implies. It's the wave nature of a photon that causes the counterintuitive behavior where longer wavelengths mean shorter cross-sections for collisions.

But you don't NEED to worry about that for a high-school level description.

Oh wow sections of information! Thanks for taking the time to answer with all that!

So the sky is blue because the sun is mostly blue and the blue wavelengths hit nitrogen more (and nitrogen makes up most of the air) because it's a higher frequency and shorter wavelength (I was confused by "Actually, it's that higher frequencies make it less likely to have a collision, and higher frequencies mean lower wavelengths." Isn't blue light a higher frequency than red light..?).

But why is it that sometimes the sky is pink or orange? Surely both the sunlight and the nitrogen in the sky stay the same? At night is it dark blue just because there's less light?

Does the sun having more energy cause it to have lots of blue wavelengths? Similar to how the blue bit of the fire is hotter--is that even related?

Why does Newton's first law require a coordinate system to be constructed? And what does "relative rest" mean? Why does it matter that the photon doesn't rest? Or does it mean that for some reason the first law requires there to be a moving photon to be compared to a not moving photon--which doesn't happen? And since photons are always moving at c...waves of light are basically how the photons are moving?

I think I actually meant to say "more likely" there and I messed up during one of my edits to the text.

Another way to think about it is that shorter wavelengths means there's less of a gap between the crests of the waves, so there's a greater probability that one of those crests will intersect with the target.

Nope, there's more sky between you and the sun as the sun approaches the horizon -- think about two concentric circles and a line passing between a point on the inner circle and a distant point. That means the blue light is getting scattered even further, leaving the redder shades of light to pass through to where you can see them.

Yup, got it in one.

Bingo, got it again! This goes back to the discussion of blackbody radiation. Objects that emit light because of how hot they are follow a very predictable pattern.

The first law is inertia, that is, an object won't change velocities unless a force acts upon it. A consequence of this fact is that -- in the absence of external forces -- you can pick any object you want, call it the center of your universe, define its velocity as zero, and judge the velocity of everything else relative to it.

The classic example is riding in a car. From your perspective inside the car, the seat, the steering wheel, the radio knobs, and the doors aren't moving -- they have a velocity of zero. You look out the window, and everything is moving backwards at 90mph. From the perspective of the cop on the side of the road, the street has a velocity of zero, and you're breaking the law. Newton's first law says that both of these descriptions are equally valid.

The second guess is closer.

If you pass a slowpoke going 70, then you, the slowpoke, and the cop agree that there's a 20mph difference in your velocities. (The cop still thinks you're breaking the law.)

The cop has a radar gun. After all, that's how he knows you're going 90mph faster than he is, and how he knows the slowpoke is going 70mph faster than he is. If he took off in his cruiser and got up to 100mph so he could catch up with you, the radar gun would say that you were going -10mph, and the slowpoke was going -30mph.

So far, so good. But here's where things get weird...

No matter who's measuring it, EVERYONE -- you, the slowpoke, the cop -- agrees that the radar beam coming out of that gun is going at the speed of light, c.

Not c - 70. Not c - 90. Just c. All the time. Always.

Even the alien spacecraft zipping by the planet at 0.3c sees the beam going c, not 0.7c. (The cop isn't going to try to write the alien a ticket. It gets off with a warning.)

So because photons always travel at c no matter how you look at them, you can't measure the velocity of things relative to the photon. Oh, sure, you can try, and you can get a number, and that number might even seem reasonable -- if the alien were in a drag race with a laser beam, the judge at the finish line would say that the laser beam was going 0.7c faster. But unlike you and the slowpoke, where everyone agreed you were going 20mph faster, the alien thinks the laser is going 1.0c faster than it.

The space police don't write speeding tickets; the interstellar speed limit is c, and you can't break it. On the other hand, drag racing a laser beam is reckless, so the cop comes up to the race at 0.5c. He thinks he's going roughly 0.2c faster than the alien (not EXACTLY 0.2c, but that's a lecture for another lesson), and he thinks the laser is going 1.0c faster than him, so the cop thinks that the laser is outrunning the alien by 1.2c. The cop doesn't have to write a speeding ticket for the laser, though, for the same reason it's not illegal to pass someone going 140mph relative to them on the highway.

So the judge, the alien, and the space cop disagree on how much faster the laser is than the alien.

That's why Newton's first law breaks down at relativistic speeds: when you're going that fast, then things look like they're going varying speeds relative to each other even if there AREN'T forces acting on them.

Oh wow, light is strange! I wonder why it happens that regardless of relativity, everyone sees it as 1.0c faster than themselves?
Uh, but if lightbeamA were racing lightbeamB would lightbeamA see lightbeamB as going 1.0c faster. And why Newton's first law doesn't work is because even without another force acting on it, the light travels 1.0c faster?

Err, so the Earth rotates around every day and around the time the skies turn pink/orange at the place, the place is further away from the sun. And because it is further away the sunlight has to go through more air stuffs to reach the person's eyes at that place. And because there is more sky, the blue's already been scattered a lot further away, now what's left to be scattered to reach the person is the red/orange light? And so those reach the person's eyes? But if that's so then why is it that during nighttime the sky is more bluish than reddish?

EDIT: Attempted to make it more coherent.

That's Einsteinian relativity instead of Newtonian relativity. The shortest description I can give you is that the speed of light is really better described as the speed of time, that is, a measurement of the relationship between time and space. If it sounds confusing that's because it is and I don't know if I can make it intuitive.

Answering this fully gets pretty deep into relativity and I'm not sure I can satisfactorily answer it without going overboard -- and then THAT risks being too confusing.

At the core of it, though, is that you can't consider things from a photon's perspective. The math breaks down; trying to set up that frame of reference ends up quite literally dividing by zero.

You might have heard about time dilation -- the faster you go, the slower time appears to pass for everyone else. When you hit the speed of light, the universe stops experiencing time at all from your perspective; if you COULD look through a photon's eyes, you'd see yourself crossing the entire universe without any time having passed at all. You couldn't see another photon moving because there would be no time for the photon to move in: your velocity formula of (change in position) / (change in time) ends up with a zero on the bottom.

It's not so much that Newton's first law doesn't work so much as it is that it doesn't apply. The law has a big "if" in it. At normal scales you can simplify that condition to "no net force" but the more formal statement of the law is "in an inertial reference frame" and light doesn't have one of those, for the above reason.

You got it.

Because the Earth itself is in the way. Only the light that gets scattered the most can reach you, because anything taking a more direct path is busy shining on the daytime side of the planet.

So..the dark blue I see at night is actually bits of blue light arriving from the other side of the planet :o?

Also, just a sort of random question out of curiosity: when a light in a room turns on, does the light actually travel/move to fill up the space or does everything just magically light up at once? Or, if I have a laser on top of one hill hitting a mirror over 186,282 miles away angled to hit some surface beside me, will there be a delay in my seeing that surface? Actually, this is probably a badly thought out question and stuff will take the time of light speed--and it just looks like a dark room fills up with light at once because light speed is so fast, and rooms so small.


I have a math review problem (midterms are coming up) that asks me to find a factor of (j-3)^2 - (k-5)^2 but I'm not sure how to do it. I tried to expand it, but just got j^2 - 6j - 16 - k^2 + 10k and it doesn't seem useful. Should I have expanded it at all?

Yes, exactly. If the atmosphere wasn't there, it would be completely black. Pictures taken from the moon support this.

Neither! Light doesn't fill space. When you turn the light on, the bulb begins to emit photons. Those photons bounce off of the things in the room and you see the ones that end up ricocheting into your eyes.

But yes, objects closer to the light source will be hit with photons before objects farther away from it.

This experiment has actually been carried out for real!

When they visited the moon, astronauts left behind a retroreflector on the surface. (Retroreflectors are the things like on street signs and bicycles that are textured so that no matter what angle incoming light hits the surface, at least some of it is sure to bounce towards your eyes.) If you have a powerful enough laser (it has to be powerful because the atmosphere is going to make it spread out and lose energy), and you shine it at that retroreflector, you won't see it light up for about 2.5 seconds. (Actually, you won't SEE it light up at all because the reflection is so weak by the time it gets back here, but it's measurable by instruments.)

This technique has given us a VERY precise figure on EXACTLY how far the moon is from the earth -- and how fast it's spiraling away from the planet (about 3.8cm/year).

No such thing as a bad question. It shows you're exploring the problem space and seeing what else there is to be found.

Though regarding the dark room, light speed isn't actually the only factor involved. The bulb itself takes time to warm up and start emitting light in the first place. So if it feels instant between hitting the switch and seeing things, that's just a sign of how bad humans are at perceiving very small intervals of time.

Nope, you shouldn't have expanded it. It's a formula you should have memorized already: a^2 - b^2 = (a + b)(a - b)

So Coda is imparting his worldly knowledge on everyone here?

I can help with Physics and Anthropology… if I remember to check the thread.

Ohh I completely forgot about a^2 - b^2 = (a+b)(a-b). Thank-you!
So (j-3)^2 - (k-5)^2 --> (j - 3 + k - 5)(j - 3 - k + 5) --> (j + k - 8)(j - k + 2) and these two are the factors!

Wow, it's amazing it's actually been done :o. I based that question off of some telegraph experiment I learned about last year, which I don't remember too clearly.
But, uh, the moon is spiraling away from the Earth..? Granted it's not a lot every year but if Earth survived long enough it's just going to spiral away?

Light bulbs emit photons and those go into my eyes so..the fact everyone in a room sees the whole room light up for an instant when the switch is flipped on and off is because light moves and bounces quickly?

You're welcome to contribute!

Yes, yes, yes, and yes.

The other reason that everyone see the bright flash of light when you turn to bulb on is because the light travels faster than your eyes can adjust to it. So the brightness is caused by the delay of your pupils adjusting from the light meeting your eyes. Also why most people blink when they walk outside or see a bright light, to protect your pupils while they adjust down to let in less light.

Is the light that I see not the same light that another person sees?

The pupils take time to adjust...do you mean that when the light first turns on and off, people don't see the room and just see the bright flash and that's because the eye can't make out the details because it needs to adjust? Why do pupils need to be protected :o?
I mean, it makes sense pupils need to be protected, but why is light harmful? Is it like the sunspots thing where when I look at a super bright object I see a blob in my eyes for a while? What is that?

Nope. It comes from the same source, but the photons are distinct.

Yes. A less-bright pulse of light, on the other hand, can do a fairly good job of giving you an impression of the room, but your body's reaction will make the darkness afterward feel even darker.

This is why you should try to avert your eyes from oncoming headlights when driving at night, by the way -- the bright light in your face diminishes your low-light vision.

That's retina burn. Mild cases just result in the nerves in your eyes getting overloaded for a while and sending off bad signals until they recover -- that's what the spots are. More serious cases, like looking into the sun for more than a brief moment, can lead to permanent blindness. This risk of damage is why we have reflexes to make us blink and look away, because we as a species depend so strongly on our sense of vision.

EDIT: Forgot to explain why light is harmful. Long story short, light is energy. Put too much energy into a living cell and it cooks. Your eye has a lens in it that concentrates light onto the retina, and your retina is made of very very sensitive cells, so it takes less light to cause damage there than to -- for example -- your skin. (On the other hand, the lens filters out the ultraviolet light that can cause DNA damage, so unless you've had cataract surgery or something that removes the lenses you don't have to worry about sun-induced retinal cancer like you have to worry about skin cancer.)

Ah, so that's why everyone can see in a room XD. Actually, when I think about it, once light hits someone's eyes, doesn't it go bouncing off (or probably re-emitted would be better) as the color of the eyes?

Ultraviolet light can cause DNA damage? I remember reading a long time ago about some person who was making pictures of how he was seeing UV light after surgery (I searched it up again: http://www.komar.org/faq/colorado-ca...et-color-glow/). So...any time he's seeing this UV light he's at risk for cancer from DNA damage? And in fact anyone with cataract surgery risks that?

And thanks Tiva and Coda about the eyes!

Do you mess around with Cameras? It is the same idea as your eyes. You have a lens that collects light to process an image. Too much light and it over saturates the image like a photo taken in high light. Too little and the image isn't sharp and colors less pronounced. When you turn on the light your eyes take a few seconds to process and close the aperture that lets in light aka your pupil.

I've only slightly messed around with one camera ^^;;. Buut even though I don't know what an aperture is, I can understand that at first, the eyes are adjusted to darker lights by having the pupil let in more light... Ohh! Then suddenly the light turns on and now there's too much light to see, for which the pupil needs to take time to adjust for because it's a body part, and then the light goes away before the pupil adjusts. And that's why people only see a flash and not the objects in the room!

Does that mean a less bright pulse of light gives an impression of the room because it requires less time for the eyes to adjust? Or that the eyes don't really need to adjust much but it does anyway and that's why it's harder to adjust back to less light when the pulse of light goes away?

You're basically right on all counts!

What is a frequency in graphing? And can tangent, cotangent, secant, or cosecant graphs have frequency? All I know is, if I need to answer what a frequency is on my test, I'm going to write the reciprocal of the period. But I don't actually know what the frequency is.


Also, why do arccos, arcsin, and arctan exist as functions? It's weird that arctan is the only one that extends forever along the x-coordinate. Whereas stuff like arccos(2) or arcsin(2) turns out undefined.

The frequency is how frequently (hence the name) a periodic function (such as sin) goes through its full cycle per unit time. That's why it's 1 over the period -- one unit of time, divided by how many units of time must elapse for a single cycle, is clearly equal to how many cycles are in a single unit of time.

arcsin, arccos, and arctan exist because they're inverse functions. If you know that sin x = some value, but you don't know what x is, then arcsin allows you to find that. And a property of an inverse function is that, over its domain and range, f(f'(x)) = f'(f(x)) = x. For example, if f(x) = 2x, then f'(x) = x/2, because f(f'(x)) = f'(f(x)) = 2(x/2) = (2x)/2 = x.

As for why arctan is the only one that extends forever over the x axis... well, like I said, these are inverse functions. Plot x = sin(y), x = cos(y), and x = tan(y) on a graph, but limit them to a single cycle of the functions, and you'll see that the graph of x = sin(y) is exactly the graph of y = arcsin(x), the graph of x = cos(y) is exactly the graph of y = arccos(x), and the graph of x = tan(y) is exactly the graph of y = arctan(x).

So only arctan goes on forever in both horizontal directions because only tan goes on forever in both vertical directions -- sin and cos only have values between -1 and +1.

These functions are ridiculously useful in real-world practice. I use arctan a lot.

Think about a right triangle with one point at the origin, the right angle on the x axis, and the third point at (x, y). Call the angle at the origin θ. Since it's a right triangle, you know SOH-CAH-TOA. And since tangent = opposite / adjacent, then you know that tan θ = y / x.

Now get rid of the triangle.

You've got an arbitrary point (x, y) in 2D space now. Draw a ray from the origin through that point. What angle does that ray make with the x axis?

tan θ = y / x
arctan tan θ = arctan (y / x)
θ = arctan(y / x)

Now, technically, this is ambiguous, because tan's range is only (-pi/2,pi/2], not (-pi,pi], so you have to look at the signs of x and y to determine which quadrant θ is actually in, but that's easy to do.

So in other words, arctan lets you figure out what angle you have to turn in order to face a given point. You can see why this would be SUPER useful in video games.

So..arcsin and arccos need to be functions because functions are more useful than relations (I think that's what they're called?) since those give just one output. Or maybe when a function gets inversed(?) people like them to remain functions and so limit them..? That doesn't really seem it though.
(if I just plot x = sin(y) or x = cos(y) I get a relation on desmos)
On the other hand, arctan can go on forever in the x direction and still give one output/be a function which is why they get to go on forever?

The bit about the video games was really interesting :o. Although, though I have a sense that angles are useful in video games I don't really know why figuring out an angle based on a coordinate would be useful.


I found this in my review sheet but I have no clue what it means:
Regular division: p = qd + r, r < d (dividend = quotient x divisor + remainder)
p(x) = q(x)d(x) + r(x) deg(r) < deg(d)

Taking d(x) = x - x_0,
p(x_0) = 0 <-> r(x_0) = 0 <-> p(x) = (x - x_0)q(x), i.e., (x - x_0)|p(x)

(It's the "Taking" part I don't understand. I'm pretty sure deg(r) < deg(d) is something about how remainders don't have the same "degree" as divisors(?), though I don't really know what a "degree" is. I used "<->" to replace the arrows in my review sheet. I'm not sure what "|" is supposed to be, though I vaguely remember reading somewhere it means "divides")


EDIT: Below those equations on dividing, the Remainder Theorem, Factor Theorem, Fundamental Theorem of Arithmetic, Fundamental Theorem of Algebra, and Rational Root Theorem are listed out...and they all refer to p(x) = (x - x_0)q(x). I guess it means to say that somehow, x - x_0 got rid of the remainder, somehow.?

Actually, that IS it. For the same reason, square root is constrained to only the POSITIVE square root of the number, even though its inverse operation, x^2, exists for all x.

I'm not sure what you mean by a relation in this context. Usually when I think of relations I'm thinking of sets.

Yes. Like I said, it's because tangent goes to infinity in both directions between -pi/2 and +pi/2.

You locate objects in space by their coordinates. A simple example would be a 2D game that counts pixels as the coordinate system. If you have a bird enemy that can dive-bomb the hero, then it might be flying along at y=600 while the hero is running on the ground at y=0, and when it gets close enough to see, it needs to figure out what angle to dive at in order to attack.


I had to look this up because I couldn't figure out the context: This is specifically talking about polynomial division.

The degree of a polynomial is the highest exponent in it -- the degree of "x + 1" is 1, the degree of "x^2 - x + 1" is 2, the degree of "x^3" is 3, etc.

The degree of the remainder in a polynomial division MUST be less than the degree of the divisor. Think about it by analogy with regular division -- regular division is like repeated subtraction, and if the remainder is larger than the divisor, then your quotient isn't big enough and there's still room to subtract more divisors from the dividend.

The double-headed arrow means "if and only if", and it means that if the left side is true then the right side is also true, and if the right side is true then the left side is also true. For example, "2x = x <-> x = 0" says "if 2x = x, then x = 0, and if x = 0, then 2x = x". On the other hand, "x = 2 <-> x^2 = 4" is invalid, because while "if x = 2 then x^2 = 4" is true, "if x^2 = 4 then x = 2" is false because x could also be -2. (You would just write "x = 2 -> x^2 = 4"; the single arrow is read "implies".)

I'm not super familiar with the property as it's written, but some searching around shows me that this is a property of a polynomial zero. From there, I understand a lot more about it.

"Taking" there could also be interpreted as "suppose" -- that is, suppose you have d(x) = x - x_0. The following must either all be true, or all be false:

1. p(x_0) = 0 -- that is, the original polynomial crosses the x axis at x_0, or equivalently, x_0 is a root of p(x)
2. r(x_0) = 0 -- that is, the remainder of p(x) / (x - x_0) is 0 at x_0
3. There exists some q(x) such that (x - x_0) * q(x) = p(x) -- equivalently, (x - x_0) is a factor of p(x).

This is a formal way of saying "(x - x_0) is a factor of p(x), if and only if p(x_0) = 0".

For example: We can show that the function (x^2 - x) = 0 at x = 1. This property lets us immediately know that (x^2 - x) is divisible by (x - 1). It doesn't tell us what the result of the division is, but we know that a result exists. But it equivalently means that if we know a degree-1 factor of a polynomial, we know that it must touch the x axis at that point.

It means that all of those theorems only hold if p(x) is evenly divisible by (x - x_0). If it's NOT evenly divisible (that is, if r(x) isn't zero) then the theorems don't apply.

Thanks a lot!
I think I meant to say "not a function" when I used the word "relation." Basically, I'm confused why this happens:

http://i.imgur.com/IW50pIc.png?1

I think the picture makes it self-evident why it's undefined outside of that narrow domain of x values -- it simply doesn't exist there, because the original function never takes on those values. And it's constrained to that range of y values in order to make it a proper function, because it can get really hard to work with expressions that can take multiple values simultaneously.

To review:

The domain of a function is the set of all x values that the function is defined for.
The range (sometimes "codomain") of a function is the set of all y values that the function can take on.

Since an inverse function is found by switching x and y in the formula, it also follows that the domain of a function constrains the range of its inverse, and the range of a function is the domain of its inverse.

Ohh okay, I think I finally get that arcsin and arccos are functions. Thank-you! I was struggling with the idea that they had to be functions at all.

Yep, they're functions. If you ever NEED to get a different value for arcsin or arccos for some reason, you can just add k * 2pi, where k is any integer (including negative), but I'm not aware of any circumstances where that's relevant. (You can add k * pi for arctan, and I already explained a meaningful example of when you might need to do that.)

I know that if I ever need to answer the question: "How do airbags protect you" for a physics test, I'll just say: "Because while change in momentum in the same, the time it takes for the force to reach you is lengthened, thus reducing the force you receive once it actually hits you (Δp = FnetΔt)"

But what is Δt actually..? Since momentum is conserved, is the change in momentum just to say that if momentum changed, the thing with the momentum most have caused a force to act over a certain amount of time? And so when two balls collide and the momentum has not changed there is no net force happening, but for each particular ball, maybe, the momentum has changed and so it experienced a net force?
Then is "impulse" just a thing which is for specific objects and not for entire systems?
And also, when ping pong balls do collide, since they bounce away with new velocities (and thus new momentums, and I guess accelerating backwards too) I have to assume that time has occurred during their hitting each other? Actually, I guess I should assume that when two things first touch there's no force whatsoever and then it takes time for a net force and changing directions to actually happen o~o?


EDIT: Unrelated
Is work done so long as there is a net force and is net work done so long as there is a change in velocity?

That's not quite a precise answer to the question.

The change in momentum is the same, but because the impulse is spread over a greater time, the amount of force acting on your body at any given moment in time is less.

Impulse isn't specific to a single object or to a whole system. It's specific to an interaction. No force is being exerted before the interaction begins, and no force is being exerted after the interaction ends, and no force is being exerted on other objects in the system that aren't participating in the interaction, but the objects and the system continue to exist both before and after the interaction being considered.

Yes, in physical reality, Δt is never zero -- there's always a span of time involved in a force being applied. If Δt COULD be zero, then a force would have to be infinite in order to cause any change in momentum.

For the ping-pong ball example, think about watching such an event in slow motion. You would see a period of time after they touch where the balls are deforming around the point of contact, squishing together before bouncing back and returning to their original shape. The more rigid the objects colliding are, the less deformation occurs, and therefore the less time they spend interacting, and therefore the greater the force must be.

That's why hard objects are brittle. They don't readily deform in collisions, so if the force of the impact exceeds the force holding the object together, it breaks.

EDIT:
No. Work is done as long as there is a net change in position. If the net change in position is zero, you could have exerted all the force you want for as long as you want and you wouldn't have done any work.

There's an argument to be said that work is done if there is a net force, but that's summing up the force vectors over the entire period of time being considered -- in order for the object to end up back where it started, every force applied to it would have needed an opposite force applied to it, and it doesn't matter if those are applied at the same time (meaning no change in velocity) or at different times (meaning velocity changes signs).

... Are you familiar with the difference between scalars and vectors?

Oh! That makes a lot more sense now. So impulse is an interaction during which force acts during a period of time, and therefore having a car crash is having the car suddenly stop/slow while the body's still moving forward and say there are no air bags coming out, the impulse is the interaction between the body and the car--which doesn't take much time to finish so the force can be easily greater than the force holding the body's bones/skin together(?) and that makes the person get injured. On the other hand, if there are airbags, that interaction lasts longer, and so the force is less, and hopefully less than the force holding the person's body together and so they're less likely to get injured?

Period of time? Meaning work should be one interaction(?) but net force could be the result of many vectors(?)

Scalars are just quantities while a vector contains a quantity and a direction is what I know. Uhh, and vectors can be drawn together, and net force being zero could mean vectors of force going in pretty much any direction so long as they end up where they started when put together?

Saying "force holding it together" was a handwave on my part. It doesn't actually work that way but it's close enough to give you the general idea. An object's strength (and there are several different kinds of strength to measure various different kinds of interactions) is measured in force per unit area (that is, pressure), and it measures how much pressure can be applied to the object before it undergoes a permanent deformation (such as bending or breaking) that it can't bounce back from. So air bags actually provide an ADDITIONAL benefit beyond just spreading out the impulse over time -- it ALSO spreads out the force over a larger surface area compared to hitting your steering column chest-first, meaning there's less pressure being applied to any one part of your body.

(Crumple zones in the structure of a car also protect the occupants by increasing the impulse time, since the cab of the car ends up decelerating more slowly than the point of contact. They don't do anything about increasing the surface area of the contact against your body like an air bag does, although they do additionally reduce the force by using up some of the kinetic energy by deforming -- after all, it takes energy to make the metal bend.)

I wouldn't describe work as one interaction. But if you're considering the end result of a series of actions and figuring out the work done by that, then you can add up the force vectors over time and get the same answer as if the integrated force was the only thing that happened. It's analogous to our earlier discussion of average velocity -- you're choosing to disregard the in-between details to come up with an overall description of what happened in the long run.

And yes, you're basically right about net force. To get into a technicality of the math, the operation isn't actually ADDING the vectors, but INTEGRATING them over time, because you could balance out a force of 2N over 1s with a force of 1N over 2s. As long as you're working with nice round-number quantities like that, you can get the right result by inspection (perhaps breaking up the 1N/2s impulse into two 1N/1s impulses so that you've got consistent time units) but generalizing it beyond that gets into calculus.