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I saw it on numberphile. Something something cosine of theta. I have no idea, but that sounds like triganometry to me, and pi is all over triganometry.
The pi is coming from that cosine.The cosine of a variable angle alpha drifts between 1 and -1, and the zero point is where alpha equals pi/2. That transition from positive to negative can be seen as the change in direction of the big ball.
I still have no idea how to understand this, but writing that down helped a little.
https://www.youtube.com/watch?v=abv4Fz7oNr0
Here it is. Should be fairly easy to test in any game engine. Too bad I don't know how to use any of them.
Nah, try phi.Quote:
Originally Posted by CoccoBill
Here's another mind blowing thought experiment.
Imagine an infinite forest, where all the trees are planted in a square grid. Also imagine the trees have zero thickness, but we can see them.
Now, can you see through the forest?
I'll let you ponder that while I write up the mind blowing bit.
What do you mean they have zero thickness but we see them? You're talking shit.
The answer is yes, we can see through the forest.
Imagine a graph, and we're standing at the point where the x and y axes meet... we can call that (0,0). We look at the tree at (1,1), and we can see it, but it blocks tree (2,2) and tree (3,3) etc all the way to infinity. Note these coordinates divide to give us the same number... 1.
Now imagine looking at tree (1,2), and draw a straight line from (0,0) to (1,2) and conitnue. Next tree we hit is (2,4), then (3,6). You can start to see what's going on here.
If you look at tree (1,7), then the next blocked one is (2,14), then (3,21)...
Blocked lines of sight basically represent rational numbers. And rational numbers statistically make up 0% of all the numbers.
So from that, we can imply that we can see through the forest because there are infinitely more irrational numbers than rational numbers.
Now for the clincher... imagine there's a route through the forest that is the "optimal" route to go in order to be as far away as possible, on average, from the nearest tree.
That is the most irrational number, and that number is phi... the golden fucking ratio.
So if there's a contender for the most random number, it's surely the most irrational one.
Of course I'm talking shit, it's a thought experiment. They have zero thickness because as soon as you give them actual thickness, the thought experiment doesn't work and we can't see through the woods.Quote:
Originally Posted by oskar
But if you can't see them at all, then how can you know it's there? You have to both know where it is while not being able to actually see it. Just work with it.
Back to the pi question:
I don't see where the theta comes from. There are no angles in this problem; it's 1-dimensional.
I see the side note that (1-x)/(1+x) is approximately equal to cos theta, and he's calling that a Taylor Series of cos... but...
A) that doesn't look like the Taylor series expansion of cosine.
B) He never defines theta, so it's hard to tell what he's even saying.
A) Check out google images for cosine taylor expansion.
Taylor Series for cos at very small angles is just to set cos(theta) = 1.
At slightly larger angles, we'd say cos(theta) = 1 - (1/2)*(theta)^2
We keep adding higher even powers of theta as we add more terms. We never add a theta to the denominator.
So ... obv. he's a doctorate and I'm not, but I don't see how that's a Taylor Series expansion of cosine.
B)
What the hell does it mean to say f(x) = g(theta) unless you define theta. You don't get to arbitrarily choose any old inputs when you do a series expansion. The whole point is that you're using the same inputs in a function that, while it's less precise, is easier to work with. You gotta at least state how to transform x into theta when you tell me how g(theta) is identical to f(x).
Yeah you're talking another language there.
I did think this myself, but I think the velocity vector is the angle... when there's an acceleration, it's the same as a change of angle.Quote:
I don't see where the theta comes from. There are no angles in this problem...
I mean, I have no idea, but I get that triganometry will be everywhere here.
Well, I was thinking that he's writing out the equations for the velocities as a function of x. Then in so doing, he sees the (1-x)/(1+x) thing, and he's all, "Oh hey! That's just a form of cosine under a certain domain. So he replaces that expression with cos(theta), as a simplification.
That's not it, though.
I plotted (1-x)/(1+x) to see what it looks like, and it's nothing like cos(x), aside from the fact that it has a value of 1 when x = 0.
The first thing that made me skeptical was noticing that the (1+x) in the denominator makes it divide by 0 when x = -1, which is not cosiney at all.
It doesn't have a slope of 0 when x = 0, which pretty much makes it a bad approximation of cos, even when |x| << 1.
A vector encodes 2 pieces of information, its direction and magnitude. It cannot be defined with only 1 datum (except for the 0-vector, which has ambiguous direction).
I think he makes a mistake and admits so shortly after, it should be cos (n) x, where n=0 in the case of the x16 ball, n=1 where the ball is x1600 and so on.
I dunno if that's why you're seeing a discrepency.
I watched that numpherphile video thinking I could get involved with what you were talking about only to find out that you were talking about something completely different, fml.
Clue me in as to what you're talking about please (I'm drinking but may still provide slightly useful).
edit - If you're talking about the two balls problem (which if we were you'd think Ongs mum would be an expert) then 6:30 into the video I'm thinking that you may be able to make any number you want given some arbitrary setting of numbers which would mean you could make any number you want but as pi is special making pi seems more special than any other number.
Bold is probably absolute nonsense now I looked into it in more detail.
After thinking I've found what you're talking about I completely agree variables come out of nowhere and aren't defined, the bigger version of his worksheet is very hard to follow and I'm sure there are a couple more variables such as X instead of x which appear from nowhere too.
The only bit I can work out is that when theta is small you can say
cos(theta) = 1 - (theta^2)/2
and if you replace theta with n theta (assuming it's still always small enough)
cos(n * theta) = 1 - (n * theta^2)/2
WAIT ONE MINUTE!!!! (and this isn't the reason but it makes a link)
If you draw a right angled triangle with cos(y) = (1+y)/(1-y) you can say the hyp = 1+y and the adj = 1-y
If you then try and work out what the opposite side is you can find out that it is 2root(y)
The theta comes from eulars formula
e^(ix) = cos(x) + isin(x)
Which isn't actually an angle in terms of the problem itself it's just a byproduct of solving some part of the equation linking it back to argand diagrams. So the theta comes from this and is why it seems to appear out of nowhere.
Hence where the right angled triangle comes from and hence the relationship I mentioned in my previous post.
I got this from his full solution here http://www.numberphile.com/pi/Pi_Balls_Ed.pdf
Which is in absolutely no way obviously related (or linked) in the video itself.
edit - FUN FACT
When I was doing further maths at a-level the link e(ix) = cos(x) + isin(x) made it so I didn't have to remember lots things I was meant to learn for rote in normal a-level maths. A few times I got comments form the teacher saying I have no idea what you use to solve these problems but you seem to get them right every time.
I should mention that I absolutely hate 99% of numberphile videos for this exact reason they go from casual talking about a problem, some simple solution, ????, final answer and you're meant to take something away from that. It's bullshit.
I didn't take anything much from that rant except you hate numberphile.
I don't like all of the numberphiles, it depends who the professor in question is. Some of them are annoying. But I do like the format... Brady asking professors questions, more focus on professors. Sometimes Mathologer comes along and corrects them, but he's all about himself and his t-shirts. (I do like Mathologer too). Sure there are some where they end and I'm thinking "is that it" but I do enjoy his channel. I've just discovered his physics channel, so that's my entertainment sorted for a few days at least.
People rarely appreciate genius during it's time. I imagine people will look back 1000 years on at these forums and appreciate the real insight it gave into a genius breaking down his thought process.Quote:
Originally Posted by OngBonga
tl;dr I told MMM where theta comes from and he'll put the effort in to relate everything back nicely.
I see.Quote:
Originally Posted by Savy
Since we know that (1-x)/(1+x) will always lie between 0 and 1 for x <1, and we define x = m/M <= 1/16, it fits the domain.
Since cos(anything) will always return a value from -1 to 1, the range 0 to 1 is contained therein. Therefore cos(anything) can yield the same answer as the fraction, provided the correct input.
In this case, he's not really concerned with any physical representation of theta, just that it can encode the same result if chosen wisely to correspond to x, which he sets out in definition by saying cos(theta) = (1-x)/(1+x). It's not an approximation, it's a straight correlation he creates out of convenience.
Then the magic is that the result utilizes this transformation from x to theta without needing to really be back transformed. He shows that the complete kinematic equations for any time are simplified down to a sine and cosine function describing the 2 bodies' motions, given only the input of how many collisions they've had, and their relative masses.
He back transforms the answer just to pin down values of x = m/M, but that final result is kinda moot. We know that there are some values of x which have this property, the exact values are not really special.
The pi kinda comes from the conservation of energy statement.
(1/2)*SUM( m_i*(v_i)^2 ) = (1/2)*SUM( m_f*(v_f)^2 )
It's a sum of squares on either side of the equals sign. Any sum of squares is identical in from to a distance, squared.
The distance, squared in this case is the total energy, which is conserved, i.e. the same before and after any collision. In a sense (which I will explain multiple times), this defines a radius or diameter, which is equal for all states of masses and velocities.
If we look at 2 masses, then it's a sum of 2 squares on each side of the equation. This is like 2 right triangles with the same hypotenuse.
We can re-write it as follows to illustrate this (ignoring the factors of 1/2, which all cancel out):
( SQRT(m1_i)*v1_i )^2 + ( SQRT(m2_i)*v2_i )^2 = {hypotenuse}^2
AND
( SQRT(m1_f)*v1_f )^2 + ( SQRT(m2_f)*v2_f )^2 = {hypotenuse}^2
I'm using the square root of the mass to more clearly show that these are just a^2 + b^2 = c^2 equations.
Note that {hypotenuse}^2 is just an awkward way to write {Total Kinetic Energy}
Since these are equal to each other, the injection of setting them each equal to a third thing is trivial.
It just might help to see that these are kinda like 2 right triangles which share a hypotenuse.
Now, there is a fun thing about right triangles and circles. Draw a circle and then a diameter across that circle. Call the endpoints of that diameter A and B. Pick any point on the circle and call it point C. Construct the triangle ABC. It is always a right triangle with the angle at point C being the 90 degree angle.
I.e. the triangle ABC is always a right triangle whose hypotenuse is the diameter of the circle.
Perhaps you recall from grade school geometry that the center of a circle which circumscribes a right triangle is coincident with the midpoint of the hypotenuse.
Same thing.
The conservation of energy statement says that the initial conditions (the initial masses and velocities) imply a triangle, of sorts, by defining the length of the hypotenuse, which is the total kinetic energy. Since we're talking about idealized elastic collisions, we assert that the total kinetic energy after the collision is always the same as the total kinetic energy before the collision. Hence for any collision, the before and after conditions are described by the 2 sides of a triangle, which have a peculiar relation of sharing a hypotenuse.
***
note: I explain the same concept in multiple ways in this post. I am not covering how or why the pi comes out of the scenario in the way that it does. I am only showing why the choice of modeling this with a circle or with sines and cosines is sensible and not an arbitrary choice.
I kinda followed along with that trignonmetry. I dunno if I was imagining correctly, but I pictured a circle with two right triangles inside, sharing a hypotenuse. Triangle A has 16x the area that triangle B has, and with each collision, triangle B increases in size and triangle A decreases. After pi collisions, the areas of the trianlges are the same.
As we scale up, and the mass of the ball (area of the triangle relative to the other) increases hundred-fold, the number of collisions until we have equality, is pi increased ten-fold.
Fuck knows why, but it's cool.
The area of the triangles is a red herring. The side-lengths (which vary while the hypotenuse length does not) is all that matters to motivate the construction of a circle.Quote:
Originally Posted by OngBonga
This is enough to invoke pi into this discussion without any ambiguity as to why it arises.
Draw a line whose length is the total KE of the 2 masses. Since one of the masses starts at rest, this is simply a line of length (1/2)m1_i*v1_i^2.
Call the ends of the line A and B, so that AB = KE_tot.
Each of those points is always an endpoint for exactly 1 side of any triangle for which the line AB is a hypotenuse.
For reference sake, call the 3rd vertex of the triangle C.
The locus of all points C which fit this construction creates a circle with the hypotenuse AB as a diameter.
I.e. a circle whose center is at the midpoint of AB and whose radius is AB/2 describes all the combinations of KE in m1 and m2 which share the same total KE.
In our problem, call the KE of mass 1 the length of side CA, and the KE of mass 2 the length of side CB.
In the initial condition, our "triangle" has CA = AB, and CB = 0, so it's not really a triangle, as such.
This tells us that all of the KE is contained in just one mass, mass 1, and that mass 2 is at rest with no KE.
After the first collision, CA is shorter, and CB is longer, but still under the constraint that CA^2 + CB^2 is the same is was before the collision.
I.e. the total KE is conserved, which is to say, the hypotenuse formed after the collision is the same length AB.
This is true again after the 2nd collision and all collisions.
Using this picture, the "allowed" relations between the energies of the 2 masses sweep out a semi-circle.
(It's a semi-circle because we're only talking about cons'n of E and we need to invoke cons'n of momentum to avoid squaring out the negative numbers.
When we invoke cons'n of momentum, it becomes a full circle, with negative velocities indicating motion in the opposite direction.)
All of this leads to a picture of a circle with a diameter drawn across and the lengths CA and CB which satisfy cons'n of energy all put point C on that circle.
***
It's worth pointing out in all of this that we're ignoring the collisions between the smaller mass and the wall (infinite mass, if you like). This is justified, since the effect of the wall being perfectly rigid (infinite mass) is to never accelerate due to collisions with anything, and all of the KE in the small ball prior to the collision is present in the small ball after the collision. The ONLY outcome of the collisions with the small ball and the wall is to reverse the direction of the small ball, without changing its mass or speed.
If a black hole is a singularity, how can it have measurable spin? It would take literally 0 seconds to complete one rotation, no matter how "fast" it was rotating.
A black hole has a singularity at its center. A black hole is space-time volume surrounding a singularity where the curvature of spacetime is such that space and time are somehow flip-flopped. I.e. we are normally free to move in space, but move inexorably in one direction in time. Across a black hole's event horizon, that's flipped (or so the math predicts). Meaning we would be free to move about in time, but always inexorably moving in 1 direction - toward the singularity. (Y'know, math... sometimes you piss me off.)
I think of the singularity as the defining feature of the black hole, but not the black whole.
Good catch on this one:
What can it mean for a particle with no physical dimensions to rotate?
How would we know / what would we measure to determine whether or not it was rotating?
Now, let yourself move away from black holes and think about an electron. An electron has no measurable size, but we still talk about the spin of an electron. That's one reason I'm always quick to point out that QM spin is not other spins.
Aside from that, are you bothered by the center of a spinning rigid object?
I was talking about this with my friend earlier, and I decided that the singularity must have a very small volume, rather than zero... ie a sphere with radius of one Planck length. I guess the same must be true of an electron.Quote:
Now, let yourself move away from black holes and think about an electron.
Can it be tought of more as 4D spin? I can visualise the roatation of a hypercube, is this sort of what electrons do?Quote:
That's one reason I'm always quick to point out that QM spin is not other spins.
Good question. I'll assume a perfect sphere under no other influence. What is the exact centre doing? It must be rotating because the rest of the sphere is doing so, and in a measureable way.Quote:
Aside from that, are you bothered by the center of a spinning rigid object?
I'd think the exact center point is with a high probability empty, so it wouldn't be doing anything. If there happens to be an electron or a muon or whatever other weird crap, idk.Quote:
Originally Posted by OngBonga
The Planck length is a trick of math, and it doesn't describe or limit anything.
Remember, length is a subjective property.
If something is the "smallest length," then what happens under GR space contraction?
Back on topic:
Even if electrons were spheres (they're not) with radius one plank length (they're not), the outer surface might as well be infinitely far from the center as far as this discussion is concerned.
It's not 4D spin. It's a quantum property. It's not in any way analogous to anything actually spinning.
It's called that because in the early days of figuring particles out, people noticed that electrons have an intrinsic magnetic moment. They hypothesized that if the electron was a rotating sphere or taurus or anything of charge, that is a current loop, and current loops create magnetic fields. They were wrong. There are serious problems with that model, and plenty of evidence suggests it's not good. Still, the name stuck. Spin.
It's really no worse than calling quarks by color words. Color doesn't exist on that scale. Color is something that is produced by charged particles changing states. I.e. photons are emitted when charged particles change states. An electron could change states to produce a photon of any color, so the notion that a particle has a color is just another one of those things in QM where your expectations have to take a back seat.
Practically, you're probably right, and the thing at the center would have a complicated motion; it wouldn't be sitting there.Quote:
Originally Posted by CoccoBill
Still, I think ongie's talking about the center of a theoretically homogenous solid.
A homogenous solid is "infinitely cuttable without losing any of its properties."
It's a model frequently used in materials science and fluid dynamics, so most of engineering.
Even if it's not reality, the question is interesting.
So my 8 year old educated me this week. Turns out that the hot-ness of a planet is not entirely correlated with its proximity to the sun. She had to do a little project on Venus and it turns out that's the hottest planet.
Seems odd that Mercury, which millions of miles closer to a massive fireball burning at about a jillion degrees Fahrenheit would be less hot.
Initial research suggest that the reason for this is Venus's very thick atmosphere that retains heat. But if that's the case....wouldn't Venus have just incinerated itself? Or is that actually what's happening...it' just takes a long time? Is venus burning up?
If you're stood five meters from a fire wearing a bikini, or ten meters from the fire while wearing thermal underwear, a wooly jumper and a blanket, when do you suppose you'd be warmer?
I dunno about Venus burning up. It has a weaker magnetic field than we do... one reason that might be the case is because it doesn't have a liquid core like we do. So its core could be cooler than ours.
Venus isn't going anywhere though, and it's not gonna turn from a rocky planet into a gassy one. So no, I doubt very much it's burning up.
I have no idea, I'm just happy to see a post in this thread.
Another point to consider is that heat doesn't travel through space like it does through air. We feel warmth from the sun because of radiation, not because of convection.
I'm not 100% sure on this, but I think the light intensity at Venus is 4x weaker than Mercury (being twice the distance, inverse square law, 2 squared is 4). But when you consider heat moving through a fluid in a gravity field, well now most of the heat is going up, so you'll get much less than a quarter of the heat at twice the distance, unless you're above the heat source, in which case you'll get much more than a quarter.
Electromagnetic radiation travelling through space doesn't have this problem.
So it's not just a case of standing further away from a fire while wearing insulating clothing... alsd the fire itself is super-efficient at dissipating heat through the medium.
It does seema little crazy when you think in millions of miles, but distance in this sense is irrelevant, you have to see it as relative distances. Venus is approximately twice as far from the sun as Mercury is. So being five or ten meters from a fire is a resonable analogy.
Atmospheres in general are more transparent to high energy light than low energy light. The UV light from the sun penetrates the atmosphere and gets absorbed by some atom or molecule. That atom or molecule may emit a single photon of the same energy as was absorbed, or it may emit multiple photons, each of lower energy, but whose total energy is the same as the original UV photon. The final step is back to the beginning. The lower energy photons cannot penetrate the atmosphere to radiate back out into space, so their energy becomes a part of the thermal energy of the atmosphere. This is the greenhouse effect.
To summarize: UV light from the sun makes it through the clouds to the planet, but the IR light is reflected off. Then the UV light that made it through gets stepped down to IR light via absorption and emission. Now it's trapped inside the cloud layer, and will reflect back toward the surface. Even though 90% of the suns energy is reflected away into space (for Venus), the 10% that gets in is largely stuck there.
Note: if this was the whole story, Venus would be increasing in temperature without bound, which is clearly not the case. I suspect blackbody radiation of the upper atmosphere is what's missing. As the atmosphere heats, it emits more and more radiation. I.e. hot things glow brighter and brighter as they get hotter. My guess is that the atmospheric heating increases to the point where the blackbody radiation counters it, then it will hold that temperature.
If that process isn't enough to counter the incoming heat, then we could see evaporative cooling, like on Mercury.
Mercury has no atmosphere because it's so close to the sun. Any atmosphere it would accumulate is heated to the point where the average speed of the atmospheric molecules is greater than the tiny planet's escape velocity. The atmosphere effectively boils off.
Interestingly, Mercury is the planet with the greatest difference between it's max and min temperatures. Since it's orbital and axial rotations are fairly close (in a 3:2 relation), the day length is on par with the year length. Meaning that the planet has a really hot side and a really cold side. There are craters near the poles which are in permanent shadow, and extremely cold despite the proximity to the sun.
Is Venus burning up?
The Venusian atmosphere is full of CO2 and water, both greenhouse gasses, neither flammable. You get CO2 and water if you burn methane (natural gas). There's also a lot of sulfuric acid and other nasty stuff in the atmosphere. Even if the atmosphere were methane and oxygen, it's not clear that it burning would make it go away. It would depend on the amount of atmosphere, the mixing of the methane and oxygen, and the mass of the planet.
Or am I completely misunderstanding the part about Venus burning up?
I don't think so.Quote:
Originally Posted by MadMojoMonkey
This was pretty much my question.Quote:
Note: if this was the whole story, Venus would be increasing in temperature without bound
Would people overheat in space?
Depends on where they are in space and what protection they're wearing.Quote:
Originally Posted by OngBonga
Assuming proper space suit, and in orbit near Earth, then no. Astronauts and Cosmonauts and professional space people at large do it all the time and are fine. They are exposed to high levels of ionizing radiation just by being above the atmosphere. Even though, professionally, this is the highest expected exposure to harmful radiation of any job, it's still less radiation than a typical cigarette smoker exposes themselves to on a regular basis.
If no space suit, then the side exposed to the sunlight would be exposed to intense UV and gamma rays, both irradiating you and cooking you (?)*.
The side in shadow would be ridiculously cold.
*UV causes sunburn, but IR is generally considered "heat." I'm not sure if the atmosphere absorbs a lot of IR light. A brief investigation has me scratching my head because different diagrams are showing different absorption characteristics of the atmosphere. Regardless, the IR would be more above the atmosphere than below it. I'm just not sure how much more.
I'm assuming naked and very very far away from a sun.
There's nothing in space to dissipate heat, so I'm thinking maybe we burn up.
Blackbody radiation dissipates heat. This is how cooling fins work in space.Quote:
Originally Posted by OngBonga
This is how the ISS keeps its temp regulated. Some of those big panels on it are solar panels, but some of them are BB radiation surfaces.
None of which answers the real questions, here.
How much power is generated by a person's body heat?
How much power can be dissipated via BB radiation from a person's body?
The first is about 100 Watts for a person whom is active, but not working up a sweat. That can go up to about 150 W for a person working out or whatever.
The 2nd is a bit harder to get at. A typical human has about 2 square meters of skin, and is at a temperature of about 300 K.
P = A*{sigma}*T^4
where P is the power emitted by the blackbody (under a whole lot of simplifying assumptions) in Watts, A is the surface area of the blackbody in m^2, {sigma} is the Stefan-Bolzmann constant, 5.67(10)^(-8) W/(m^2*K^4), and T is the temperature in Kelvin.
Plugging that in,
P = {2 m^2} * 5.67(10)^(-8) W/(m^2*K^4) * {300 K}^4
P ~= 920 W
So you're initially emitting 920 W, but only generating as much as 150 W, and you're rapidly cooling.
Wow, I really didn't expect blackbody radiation to be significant, like it would take years to lose heat by this method. I expected body functions to generate more heat than was lost.
Thanks for the reply.
MMM I want to get into science.
Any suggestions on books to read? I'm thinking 1/2nd year undergrad stuff would be the right level or very interesting more general public books if you have any suggestions.
Well, my bias is clear, but I'd suggest getting a nice, thick, illustrated intro physics book.
I recommend physics as a starting point for all sciences, because I don't know of a science degree that doesn't require introductory physics.
It shouldn't be too hard to find used textbooks online.
Randall D. Knight - Physics for Scientists and Engineers is a good choice. It's available used for ~$20 after a brief google search, so practically free considering it's over 1,200 pages.
David J. Griffiths - Introduction to Quantum Mechanics is extremely well written, but does rely heavily on your background understanding of linear algebra and some other physics concepts.
Peter V. O'Neill - Advanced Engineering Mathematics is another one of my "always on the reading list" books. It's another huge compendium of a textbook, but it covers so much math that is put to regular use in any engineering field, like diff EQ and linear algebra.
Brian Bradie - A Friendly Introduction to Numerical Analysis will be extremely useful if you'll be using computers to gather and analyze data. The quirkiness of dealing with a discrete number system and not a continuous number line is interesting to unravel, and can sink your research if you aren't aware of the pitfalls.
***
Every book I've mentioned so far is in rotation for my bathroom reading materials. They are not books I used for a class and never opened again.
A good text on probability and statistics could be extremely useful, but I assume you're already solid on those topics.
Other than that, science covers a wide range of fields. There's every corner of engineering, chemistry, biology, geology, etc. Is there any field in specific which attracts you?
Physics, everything else isn't science.Quote:
Originally Posted by MadMojoMonkey
I love you and your enthusiasm for science. Thank you for the suggestions. I want to get more into particle physics. I'm having a slight crisis in life that I'm wasting myself that I go through every year or so.
Happy BirthdayQuote:
Originally Posted by Savy
Birthday isn't till August.Quote:
Originally Posted by BananaStand
https://www.youtube.com/watch?v=qNnFI0_kl8I
I occasionally have a crisis where I worry that I'm not wasting myself enough.
Well, I didn't mention that you need to be familiar with and competent at vector calculus for most of physics. Knight goes pretty easy on the calc., but you need to know vector math and what a derivative is and how to find the derivative of some simple functions. Same for integrals. That stuff is covered in O'Neill.Quote:
Originally Posted by Savy
You'll need to be able to think your way through a cross product in order to understand torque and how electric and magnetic fields interact. The right hand rule comes up often enough and its something students tend to struggle with.
If you're good with those foundations, then start with Knight and Griffiths. Feel free to jump right into Griffiths Intro to QM if you like, but keep the Knight text on the side just to reference. Otherwise, start with Knight and keep O'Neill as reference.
Aw, shucks! :h:Quote:
Originally Posted by Savy
A chick just took me to a place that sells nitrogen ice cream. What did I just eat?
Overpriced ice cream.Quote:
Originally Posted by BananaStand
All those books are like £50+. Which isn't much but it really puts me off. At teh same time stats and probability is probably (or maybe I'm misunderstanding) my weakest part as I never really did any.Quote:
Originally Posted by MadMojoMonkey
Did Trump already bulldoze all libraries there?
It was really fucking good though.Quote:
Originally Posted by Savy
Ice cream.Quote:
Originally Posted by BananaStand
Diatomic nitrogen, N2, is inert and composes more than 3/4 of the air we breath. Chemically, it's about as safe to work with as anything gets.
The Liquid Nitrogen (LN2) is used as a coolant and nothing more. It doesn't bond with anything in the ice cream, or change the food at all. It chills the cream and evaporates into normal air. (Note that it's air lacking all oxygen, and the N2 will sink to the floor, so asphyxiation warnings are due. Use it in a well-ventilated area, and you're fine.)
If done with skill, LN2 ice cream is a bit smoother and creamier than other ice creams, due to the rapid cooling making it less likely for ice chunks to form as it cools. Ice still forms, but not in macroscopic bits. You need to mix the cream rapidly while adding the LN2 to do this, and if mixing by hand, it probably wont be enough to get this effect.
We make it regularly in the physics department as a fun cap to a meeting. (Not faculty meetings, but anything involving new or prospective students, and any student group can make it whenever they like.)
LN2 is readily available in any large city. It's produced as a by-product of more desirable liquids, like liquid oxygen, which is used in welding. It is incredibly affordable, usually running less than $1 per liter (~$0.30 per L in St Louis). You need to provide your own container, and don't mess around with this. If you're not an experienced welder, don't even try to make one yourself. You'll want to buy a Dewar, which is a fancy name for a lab-quality thermos.
***
Our recipe:
1 qt half and half
2 Tbsp vanilla extract
1/2 cup sugar
Add any flavorings. For chocolate ice cream, add chocolate powder or syrup (~1/3 cup).
All quantities are rough guidelines, and personally, I think a bit of salt does a lot for the taste.
Mix ingredients and stir in a bowl. Slowly pour in the LN2 while stirring the liquid, and it will set into ice cream in a matter of seconds. If it starts to melt, just start stirring again and add a bit more LN2.
Using heavy cream instead of half and half will make a richer tasting ice cream, but some people don't want the added fat.
I didn't really love my prob-stats book, so I'm not as eager to recommend it as those other titles.Quote:
Originally Posted by Savy
I'm not sure what to tell you about the prices. Those books cost hundreds new, and the Knight and O'Neill texts are over 1,200 pages each.
I did find a copy of Knight for $20 when I googled it, so maybe just keep checking from time to time until you find a price you're comfortable with.
There's no real reason for you to get the most recent edition of any of those texts. Maybe older books will be cheaper? Very little tends to change from one edition to the next in math-based text books. The pictures and order of things is usually the biggest change.
***
Of course, libraries do loan out books, as was noted by Coccobill. College libraries may or may not be part of the public library system of the city, and may or may not be open to the general public. Check into those, if you're going this route.
Yeah it's 100% me wanting something for nothing. I don't disagree. I'm like ohh I'd like to half ass learning physics and it cost me nothing, I'm entitled bro.Quote:
Originally Posted by MadMojoMonkey
I should check out my uni library and see how much I can get away with now I'm a graduate.
(Late post. Sorry. This was open on my work computer when I came in today. I thought I sent it on Friday.)
There's always MIT OpenCourseware on YouTube.Quote:
Originally Posted by Savy
Phys I
https://www.youtube.com/watch?v=Uo28...gtgOt8LGH6tJbr
Phys II
https://www.youtube.com/watch?v=rtlJ...OILaOC2hk6Pc3j
Too late, dead to me.Quote:
Originally Posted by MadMojoMonkey
http://www.physics-astronomy.org/201...break.html?m=1
MMM, could you put your speculation hat on and explain what, if confirmed, would be the real world consequences of this? Obviously "slight" problems with current theories, but do you see any practical applications?
I thought they dismissed this years ago as measurement errors?
Even so, I can speculate somewhat. They say the neutrinos took 60 nanoseconds less "than light would have done". Yes, but is that light in a vacuum? Or light along the same route the neutrino took?
The neutrino is so small it doesn't interact with anything. It doesn't collide with anything, it doesn't lose energy. As far as the neutrino is concerned, dense space is essentially a vacuum. The photon, however, only moves at c in a vacuum... well what is a vacuum to a photon?
I suspect what we're seeing is the neutrino finding a straighter line than a photon can find.
Yeah. It would unhinge causality if anything can move faster than light. It's already true that the order of events is ambiguous, and observer dependent. This is easily demonstrated so long as you accept that light has a finite speed.Quote:
Originally Posted by CoccoBill
If we synchronize 2 clocks in the Einstein way, i.e. we put something exactly in the middle in-between the 2 clocks, it sends out a flash, and the 2 clocks start counting from the moment the flash reaches them. Now, these 2 clocks read the same time if you're at the point where the flash was emitted. At some pre-determined time, the clocks themselves emit a flash. The point in the middle sees these flashes simultaneously, but any observer closer to one clock than the other will see the flashes at different times. QED. The order of events is a matter of perspective.
However, using math, all observers could show that the flash emitted from one clock could not have caused the flash from the other.
If something can move faster than light, then we can no longer say with certainty if event A could not have caused event B.
Sometimes I want to start a "Ask a Wugy an Economics Question" thread (in honor of this superb and timeless thread), yet I think that every answer would be "let's see what the supply and demand model predicts".
Also I wouldn't try to answer questions before you, because I wouldn't give a fuck.
Hey mojo, can I ask you a favour? Can you take a look at this list, and put a vague value on it, assuming it's all second hand but in good, clean working condition...
Sorry, it's quite a list, but I have good reason to ask. I'll tell you how much I paid for it all after...
Pyrex test tubes (short) 125 x 16ml original boxing
Tiny watch glasses approximately 100
Still heads approximately 100
200 test tubes not in original box Pyrex
SCHOTT AG Duran approximately 40
Short test tubes approximately 250
Semi micro sinisters approximately 10
Gas pipe spares approximately 100
250ml tubes approximately 20
Still heads short approximately 12
500ml Pyrex tubes approximately 50
Test tube scissor holders approximately 100
Thin-walled boiled tubes approximately 70
Fluted flannel approximately 50
Till heads approximately 40
5 boxes of 20 thin walled boiling tubes 5 x 20
Still heads approximately 125. 16 mm x 0.8 mm wall
Drying tubes approximately 20
Pyrex test tubes short approximately 90
Pyrex quick fit test tubes 10 ml approximately 20
Glass sintered crucibles Pyrex approximately 50
Test tubes stoppers approximately 50
Acetone pump spares approximately 60
A mixture of test tubes approximately 100
Fluted flannel approximately 40
Micro test tubes long app 40
Thin-walled test tubes app 250
Pyrex circular chemical holders 100ml bowls with lips app 15
Pyrex 500ml chemical holders approximately 13
Box of 21 filled funnels
Box of fluted funnels approximately 20
Box of fluted funnels approximately 30
Box of thinned walled test tubes approximately 50
50ml dropping funnels with stopper approximately 100
Watch glass 3 sizes approximately 200
Pipe clay triangles approximately 200
Crucibles approximately 200
Small watch glasses approximately 300
Chemistry pegs approximately 250
Metal tongs approximately 200
Box of large pestle and mortars approximately 30
Filter funnels approximately 50
Box of desiccators approximately 30
Test tube racks approximately 50
Volume metric approximately 40
Ong's starting a meth lab? Now i see why he moved to the country.
Haha I have been tempted to google "how to make meth" but figured I should get a VPN first.
I've seen Breaking Bad. Seems like a lot of shitty work to me, and I can't be doing with that much heat. Weed? Sure, slap on the wirsts, don't do it again. Meth? Bye bye freedom, hello sore arse.
I'm going to list it all individually on ebay, it's basically my new business venture. But... I'm probably going to have my own small lab, for fun. I mean, why not? So I might make a tiny bit of meth, just to see if I can make it blue. But only enough that if I got caught, it's clearly not intended for sale. And I'd blame TV, and then sue Vince Gilligan.
Can you wait till they finish/cancel better call saul?
Cancel? Not a chnance. I'd have thought there's only room for one more series after this one. Gale is unimpressed with the meth he's been testing, I imagine that's a hint to the timeline, ie he'll be getting a quality sample soon.
Chemists are kinda particular about what qualifies as "used" when it comes to glassware. If it's been used for anything, even to hold water, they consider it dirty. Basically, if they weren't the one to open the box on a new item, they don't trust its history.
Good luck with the sales, but don't expect anything near top dollar on the glassware, and probably the crucibles and anything that would come into direct contact with an experimental chemical.
Fair enough. I didn't pay much for it... £500 for the entire lot, which is, what, $600? I dunno, but being conservative, I figured that even used it's worth a lot more than that. I estimated if it were all new, we're talking north of £5k, but I realise it isn't worth anywhere near that unless it's in better condition that I'm expecting. I pick it up on Wednesday, so there's still an element of unknown. I get it won't sell to proper chemists, but schools looking to save a few quid on test tubes? Why buy new stuff when they're "dirty" after their first use? Also, there's surely a large amateur chemist market. I'll clean everything first with diluted bleach then rinse thoroughly with water. That's sufficient for most applications, surely?
So long as I make my £500 back within two or three months, and I still have plenty of stuff left, I'll be happy.
A lot of the stuff like this selling on ebay is coming from China, so people are having to wait a month for their stuff to arrive.
Wut??Quote:
Originally Posted by OngBonga
Dickheads like me who want to make pops and fizzes. I don't give a fuck if a test tube is "100% sterile" or merely "clean by food hyegeine standards". If I can drink water from it, it's clean enough for me, so it then becomes a matter of money. If brand new it's £5, and "merely clean" it's £1, I'll take the latter.Quote:
Originally Posted by BananaStand
wut? speak english pleaseQuote:
Originally Posted by OngBonga
Have you ever owned a chemistry set?Quote:
Originally Posted by BananaStand
No. But if I had, it would have been when I was like 9 years old.Quote:
Originally Posted by OngBonga
You think there is a large adult market for "pops and fizzes" ?
Not so long ago, I bought a strip of magnesium ribbon for the sole purpose of setting it on fire.Quote:
Originally Posted by BananaStand
Ok "large" market probably wasn't the best choice of words, but cmon, don't be a pedant about language, otherwise you have no recourse when you use words like "irrefutable" when you mean "presumed".
I think you may be projecting your experience on to the rest of the world. You might be disappointed to find that there are probably actually very few adults who enjoy making fake lava and sticky slime in their living rooms.
If you're gonna resell shit on ebay, try iPhones, or dildos, or gold coins.