timo

Judging from how you solved the problem with the 10% of the original population I´d say you are missing an important equation for logarithms: ln(x*y) = ln(x)+ln(y). As a little style-advice: Don´t write things like ln(e), write 1 directly (or skip it if it´s a factor). It makes your calculation much more readable. @MattC: I´m pretty sure it was a typo. But looking at your result you could have figured out that it couldn´t be correct. I´ll send you a PM. It doesn´t matter whether it is in preparation for an exam or homework. The point is that you don´t learn as much from seeing other people´s solutions as from solving it yourself. And I´m pretty convinced he can solve it himself - there´s not so much possibilities to take logarithms here, anyways.

I don´t think you should have posted the solution Matt. There´s a an error (probably a typo) in your calculation anyways, so perhaps you should consider removing it completely. It´s pretty naive to think people won´t look at solutions when they are given instead of really doing all of the calculation themselves. The question isn´t that hard, anyways and I think the confusion only stems from the free parameter you called a constant (which in fact for calculation terms might be the better way to think of it).

Either you didn´t post the whole question or k is simply a free parameter. Just solve for the time t, then.

Electrons are neither elements nor compounds. They are particles. I still don´t know if they are a material.

In this case there´s two things that come to my mind: 1) Meteors. They certainly added .... errr ... material to earth after it was created. The amount might be small (also depends on what age you think earth has) but there certainly were meteor impacts. 2) Plants. They need sunlight to grow. The sunlight comes from the sun. Here, you can either say "all the trees around me wouldn´t be there without the sun therefore not all material comes from earth alone" or you could say "but the raw materials for the plants were here before and the sun just helped converting them". There is not always a simple answer to a yes/no question. If you just need the answer for your personal knowledge, I don´t think it´s nessecary to have a definite "yes" or "no", anyways. Not everything is black or white.

I was in school when I was 18. For your questions: Q1: One could start pedantic discussions about what "natural material found from the earth" means or whether it still counts as a "yes" if solar energy (which comes from the sun) was used in the creation process. Perhaps others will do that. In the sense I think you meant the question I´d answer it with "yes" (where should it come from otherwise, anyways?). Q2, Q3: It boils down a bit to what you mean by "materials". In fact, I don´t know a scientific definition for the term. I don´t know about your education level but perhaps taking a look at the "fundamental concepts" section of the wikipedia link below and following the links at the terms provides a starting point for you to read up on what you are interested in. Namely the terms "Elements", "Compounds" and "Substance" might be interesting at first. Link: http://en.wikipedia.org/wiki/Chemistry

I stopped reading somewhere in paragraph two. I had a comment here about an error I thought you had in equation 2. But now when I look at it again, I figure out that I misread it. In fact, I don´t get eq. 2 at all. Neither side of the equation is an energy. Also, if it´s supposed to be energy squared, then I don´t see how you come up with the left-hand side. It´s the energy of what, actually?

Standard physics is 4D. There are theories ("attempts of theories" fits better, I think) having more dimensions than four but it´s not nessecarily eleven. Afaik, none of those approaches with additional dimensions have any experimental evidence, yet. Eleven dimensions is the minimum for String Theories with Supersymmetry, if I remember that correctly. Don´t know about the rest of your questions.

I can repeat it for a 3rd time if you want. I´m just not sure that this specific question is your problem. I´d rather expect that you´re confused by that x and y are not entries of a vector but only some numbers without much "meaning". So for a 3rd time (and again with different terms): If two vectors are non-parallel, then they are lineary independent. The DEFINITION of linear independence of vectors is that the only way a linear combination of them becomes zero is the trivial one: Each coefficient of the linear combination is zero. It´s not that I don´t want to help you - it´s just that I don´t really know where your problem lies. Perhaps if you posted step-by-step how far you got before running into a problem that would help. The first step is using 3 r1 = 2 r2 and rewriting it to the form A*a + B*b = 0.

I could, but since I already explained that in my first post I don´t think that this actually is your problem. Again: The only possibility for a linear combination of two non-parallel vectors a and b to equal the zero-vector is the linear combination 0*a + 0*b. This is what I think your problem in understanding is, actually. You are not given the equations 3 r1 = 0 and 2 r2 = 0 but the equation 3 r1 = 2 r2. This is equivalent to the equation 3 r1 - 2 r2 = 0 which was my starting point in above. Because they are the coefficients of the linear combination mentioned in above - "A(x,y)" and "B(x,y)" are just names I gave to them. For example, A(x,y) = 3(x + 4y) - 2(y - 2x -2).

I haven´t tried solving for the solution but atm I don´t see much problem with the question - except perhaps for one tricky thing you´ll possibly encounter many times again. I´d try solving the question the following way: You are given [math] 3 \vec r_1 - 2 \vec r_2 = \vec 0 [/math] Plug in the definitions of r1 and r2 and rewrite it to the following form: [math] A(x,y) \vec a + B(x,y) \vec b = \vec 0 [/math] where A(x,y) and B(x,y) are some terms dependent on x and y (the prefactors). Now, here comes the trick: Since a and b are not colinear, the only option way to have a linear combination of them resultng in the zero-vector is having both coefficients equal zero: [math] 0 \vec a + 0 \vec b = \vec 0 [/math]. Therefore, both A(x,y) and B(x,y) have to equal zero: A(x,y)=0 B(x,y)=0 That´s two equations for two unkowns x and y. Solve them. Like I said, I didn´t actually solve that problem so perhaps there´s some trapdoors hidden in it - but I doubt it.

Somehow, I´m not really convinced about that. None. Because I live in a city. A bike. But up to today I was too lazy bringing mine to Dresden or buying a new one.

... or the only one bored enough to reply ... I hate Sylvester. Assuming no air resistance and earth to be perfectly spherical, then "yes". You´d be falling down (and accelerate) until you arived at the center of earth. There, there´s no gravitational force pulling you anywhere (-> the gravitational force of a spherically symmetric body with uniform mass-density goes proportional to the radius and points inward) but you´d still have the velocity (-> kinetic energy) you gained when accelerating. Therefore, you´d pass through the center towards the other end of earth. On the way there you´d constantly decelerate ("fall up") because gravitation tried to pull you down to the center. When there´s no air resistance and if earth is perfectly spherical, conservation of energy guarantees that your trip ends exactly on the other side of earth - but you should hold on to some handle or so or you´d be pulled back again. If there is air resistance, you would lose some of your energy to the friction. Then, after you had passed the center of earth you wouldn´t have enough energy left to reach the other end of earth. You´d stop somewhere on the way being sucked back to the center. This process would repeat for some time untill you´d end up floating in the center of earth ( -> dampened oscillation). Having no (effective-) force (at the center of earth in this case) means that veloctiy remains unchanged, not that there is no more movement (-> the time-derivative of velocity is proportional for the force: dv/dt = a = F/m). If you already arive with some velocity (like when you fell down to the center of earth) you´ll simply keep the velocity (for the timspan of being in the area without force).

I´d hereby like to nominate you for the SFN "Creativity in Physics 2005"-award. You are possibly the first person in here I see, that manages to a) post in the Modern/Theoretical Physics forum b) and talk about electromagnetism c) and speaks the magic words "I have a theory that.." (slight deviations as in your case also count). but ..... d) doesn´claim that according to your theory, gravity is in fact electromagnetism. Now, when we completely ignore the probably very weak forces created and the anatomy of the bird your idea actually doesn´t sounds as stupid as ... erm ... as it sounds: As Klaynos pointed out, you have the Lorentz force F = qv x B with q being the charge the bird aquires when it rubs its feet, v its travelling velocity and B the magnetic field of earth. If the bird is flying directly north or directly south, there will be no force. But if it moves nonparallel to earth magnetic field, there will be a force as v x B doesn´t vanish anymore, then. So the bird simply had to always fly in such a direction in which the force it feels on its feet vanishes. It will either get to the warm south or end up as a penguin teasing icebears.

I don´t have much time right now, so I´ll stick to the questions that can be answered quickly: Above? Not at all. There´s a pretty wide range of methods to solve differential equations numerically. The easiest one is the following: 1) Start with a starting position x(0) and a starting velocity v(0). 2) chose a sufficiently small timestep dt. 3) Set t = 0. 4) x(t+dt) = x(t) + v(t)*dt, v(t+dt) = v(t) + g(x)*dt where g(x) might be the gravitational acelleration posted by Klaynos, for example (depeds on the system) 5) t = t+dt 6) goto 4 until you reached a t which is sufficiently large for your purposes. Or did you mean how the movement equation looks like as a differential equation? That´s d²x/dt² = g(x). Note that Klaynos and the follow-up poster(s) assumed that g is negative. That´s true if one assumes only a 1D problem with the positive direction beiong defined as "away from the attracting mass". But in general, the acelleration g is a vector anyways and for simple problems you know where it points - therefore knowing the magnitude (the positive value) is sufficient.

Every particle of the ensemble would be in the lowest possible energetic state. I don´t see anything particulary dangerous in that. A Bose-Einstein Condensate, for example, has almost all particles in the lowest attainable state - just not all.

By what definition? Temperature is not the same as energy, in case that was your thought.

Inobtainability of 0 K does not stem from a quantum mechanical condition. It´s a statistical argument. I can picture in my mind that my cup of coffee mutates into a giant gremlin that eats my neighbour's dog. But hopefully we can agree that this is not likely to happen ... because I don´t have any neighbours who have a dog. The world simply doesn´t always work as you imagine, even though that´s very sad. EDIT: I think I have a better analogon than the gremlin: You can theoretically dig a 10 meter hole in the earth. You can also dig 20 meter deep. Obviously, you can always dig deeper. Therefore, you can dig an ininitely deep hole in the earth. Now, if you think about the big picture (the earth being a sphere) you´ll notice that it somehow isn´t possible.

As far as I can see, that "non" completely answered your question since all except the first sentence in your post refered to "what if it is sensical". I don´t know what Einstein said or imagined. There are so many stories about him that I wouldn´t expect they´re all true. Anyways, you have to keep in mind that Relativity wasn´t as established and rounded out in Einstein´s times as it is today. You didn´t ask for an explanation in your thread so why are you asking for it here? Nevertheless, the short answer is: The movement direction of the photon (which would be the time-direction in its frame of rest) is self-perpendicular and has length zero. This causes problems constructing a proper base.

Your link leads me to http://www.microsoft.com.

So then I´ll be the first one who complains about "Modern/Theoretical Physics" - especially since noone seems to really know what it is supposed to be. Since you seem to like it, perhaps you can tell me what you think it is or even what it should be (perhaps we could agree on a better name for it). Saying "everything that doesn´t fit into Relativity or QM can go there" doesn´t seem like a reasonable statement now that we have the main forum specifically for threads that don´t fit into one of the sub-categories. Also from my point of view, most of the threads in the "Modern/Theoretical Physics" actually would belong to either QM or Relativity. Take for example your recent thread about Feynman Diagrams: I wouldn´t know any place where they appear other than Quantum Theories. Why didn´t you post it under "Quantum Mechanics"? Not that I really bother but maybe it helps me understanding what you or others think "Modern/Theoretical Physics" is. Apart from my dislike for "Modern..." there´s one point I´d see nessecary (and which is covered by the proposal Dave posted): I really think one of the subforums should explicitely mention Electrodynamics in its description.

Because if you placed an equal triangle next to the one you have you´d get a parallelogramm which has the area base*height.

You´re right. It´s surprisingly hard to find any info on the internet. I was browsing around a bit hoping to find a good website explaining the effect but sadly found nothing suitable. Ok, I´ll start by giving you the plain math so that we have a starting point. For simplification, let´s reduce spacetime to a two-dimensional one with only the coordinates t and r being the time measured by an infinitely distant observer and the distance to the singularity measured by that observer, respectively. Also note that I´ll use fractions of lightspeed as unit for velocity. There is a quantity called "invariant line element" ds² which plays a central role in relativity. When it´s integrated over a trajectory (a path taken through spacetime) of a particle, it gives the time experienced by that particle. This value is the same in any coordinate system. Massless particles take paths which have ds²=0, massive particles take paths with ds²=1. In the case of Special Relativity, the line element is calculated by ds² = dt² - dr² where dt and dr are ... well, let´s hope no mathematican sees this post ... small displacements in time and radius. At this point, you can see that massless particles move at lightspeed while massive ones nessecarily move slower: For massless particles you get 0 = dt² - dr² => dt=dr => v := dr/dt = 1 and with a similar reasoning v= dr/dt < 1 for massive particles. In the case of General Relativity however, ds²=dt²-dr² is not nessecarily the case. In fact, the actual rule for calculating this line element usually depends on the point in spacetime you´re at. Sidenote before I´ll proceed further: This is not only some weird effect of GR - the fact that the calculation rule (called "metric") depends on coordinate system AND point in spacetime is absolutely fundamental and the whole inner structure of spacetime is encoded in this calculation rule. In the case of a nonrotating Black Hole, one often uses the Schwarzschild Coordinate system. The Schwarzschild coordinates have the nice property, that a displacement (dt=1, dr=0) actually is what a distant obverver experiences as time and a displacement (dt=0, dr=1) actually is what this observer would call a displacement in space - this is not nessecarily true for an observer that is not far away from the singulariy (which would have been my explanation if you had already known the math). In these coordinates, the line element has the form [math] ds®^2 = \left(1-\frac{R}{r}\right) dt^2 - \frac{dr^2}{1- \frac{R}{r}} [/math] with R being the Schwarzschild Radius (Event Horizon) and r>R. This relation has an explicit dependence on the position as I already anticipated. We´re almost there and there´s two possibilities to see that any objects coming closer to the Event Horizon appear to slow down: a) Chose a fixed dr (one, for example). Hopefully, you´ll see that the dt needed to make ds²=0 or ds²=1 must increase as r decreases. Therefore, v=dr/dt decreases. b) explicitly solving for v=dr/dt. For the sake of simplicity, I´ll only do this for a massless particle (ds²=0): [math] \begin{array}{lrcl}& 0 &=& \left(1-\frac{R}{r}\right) dt^2 - \frac{dr^2}{1- \frac{R}{r}} \\ \Leftrightarrow & \frac{dr^2}{1- \frac{R}{r}} &=& \left(1-\frac{R}{r}\right) dt^2 \\ \Leftrightarrow & \frac{dr^2}{dt^2 \left(1- \frac{R}{r}\right)^2} &=& 1 \\ \Rightarrow & v = \frac{dr}{dt} &=& \left(1- \frac{R}{r}\right) \end{array}[/math] For decreasing r>1 (object getting closer to the Event Horizon), even a massless particle like a photon seems to become slower and slower and ultimately stop at the Event Horizon. This result might seems rather surprising as you´d expect massless particles to travel at a velocity of 1. The reason for this behaviour is that the chosen coordiantes are only physically meaningfull for positions very distant from the singularity at r=0.

Ok, first off: Forget redshift. Sure, there is redshift but that´s not the cause of the apparent slowdown of particles near the event horizont. Neither does the time the photon takes to reach an observer play any role. You´d have the same effect even with "instananeous" transformation of the position-information. Then, before I twist my brain around a suitable answer, I´d like to ask you the following question first: Do you already know the math involved and ask for how the rather surprising result can be understood or haven´t you seen the math yet? Sadly, neither your post nor the information in your profile give a hint on this.