timo

[math]e^{a+b} = e^a \cdot e^b[/math] for a start. Furthermore, [math] e^{i\alpha} = \cos \alpha + i \sin \alpha [/math].

By "introduction course" I mean a first-semester course so my advice still stands.

If I understood you correctly then it's going to be an introduction course, so - You will probably not need to know much about physics in advance. - Instead, if you really want to prepare then practice a bit basic math: Integration and differentiation, rearranging equations, solving systems of equations, exponential and logarithmic function.

What exactly do you want checked? - If you want to know whether you made a calculation mistake somewhere: No, I see none. - If you want to know whether what you wrote makes sense: Strictly speaking you did not even say what the letters stand for, what can be assumed as given, what you want to arrive at, etc. Of course it is relatively obvious in your case. But "obvious" is a nasty bugger that stabs you in the back when you expect it the least. E.g.: Are you aware that you assumed r being independent of time and of the implications that has? - If you want to know if you wrote it down in a correct manner: It is obvious what you mean, so basically "yes". Purists would possibly prefer rewriting statements like [math]v_a(t)= \int G \cdot \frac{m_b}{r^2} \,dt = G \cdot \frac{m_b}{r^2} \cdot t + v_0[/math] as [math]v_a(t)= v_0 + \int_0^t G \cdot \frac{m_b}{r^2} \,dt' = G \cdot \frac{m_b}{r^2} \cdot t + v_0[/math].

Investigate the argument under the square root seperately, first. It only depends on c and must be >= 0. For example you could plot the argument in gnuplot: f(x) = 1. - 6.*x + 9.*x*x - 4.*x*x*x set xrange[0:2] set yrange[-5:5] set grid plot f(x) t "argument under the square root" Sidenote: Mathematica, Maple and friends are great programs with nice capabilities. For some strange reason, gnuplot is better-suited for the task at hand in the majority of cases I encounter.

I did not check your equations but a priori the fact that you can write the solutions in a complicated manner does not contradict the statement that there is only the trivial solution. But your equations are not a solution, only a way to find it (and the approach you presented in your previous post seems quite ok to me). Perhaps you should set s=1 (see first step in my first post to see why this is sufficient) and then try to find constraints on c (by looking at the possible c for which both square roots exist). I would not be too surprised if that straightforwardly leads to c=1 which then directly leads to a=b=1 (from plugging into your equations). And of course there can be something wrong with my sketch; I did not actually perform the steps too carefully.

Hm, perhaps I should have explicitly stated what I mean by "solution": A solution is a set of numbers (a,b,c,s) for which the surface area A and the volume V of a cuboid with side-lengths a, b and c equal those of a cube with side-length s. So if for each solution a=b=c=s then there is no solution for which the cuboid is not a cube. So there is no non-cube cuboid that has the same surface area and volume as any cube.

I think there should only be the trivial solution. Sketch for the proof I'd try: - If (a,b,c,s) is a solution, then for any x>=0 (xa,xb,xc,xs) is a solution. Hence, it suffices to look for solutions with s=1. This fixes V=1, then. - Let c be a fixed value between zero and one, you can show that the V=1 cuboid has its minimal surface when a=b - Under the constraint that a=b, you can show that the V=1 cuboid has its minimal surface when a=b=c=1; i.e. when it is the cube. => A cuboid with V=1 will have at least the same surface as a V=1 cube. It will have the same surface when it is a cube. => There is only the trivial solution.

It's a variation of Godwin's law, also known as the Zuse Uncertainty principle:

One wavelength is probably [math]2\pi[/math].

I am still not entirely sure what you want to do but in any case you need to start with at least some definition for the scalar product for showing show it equals something. I was thinking that [math]\vec{a}\cdot \vec{b} = a_1b_1+a_2b_2[/math] can be considered the standard definition [for school, at least] and that you want to show the more geometric interpretation [math]\vec{a}\cdot \vec{b} = |\vec{a}| |\vec{b}| cos \theta[/math].

Try formally breaking up the vector [math] \vec a[/math] in a part parallel to [math]\vec b [/math] and a part perpendicular to [math]\vec b [/math]. Then use the properties of the scalar product (of course at some point you probably need the statement that the scalar product between perpendicular vectors is zero) and the definition of the cosine. I think that should work. You can in theory go via the parallelogram (however you came up with that idea). You'd need to break up the non-base into a part perpendicular to the base (-> the height) and a rest there. So it's kind of equivalent.

Sure. I am not exactly sure what you want to do and why? But for a start, explicitly calculating the correct integral in the interval containing x=0 and the value that the numerical integration would give is easy.

The only subinterval that contributes to the error is the one where you pass x=0.

Apart from the special titles it is an insignificant gimmick. They are roughly ordered from small to big. Apart from the special titles for experts, moderators and administrators, it depends on the number of posts. As soon as they reach moderator or administrator status, I think. Experts have a switch to chose between "resident expert" and "<field> expert" (e.g. "Physics Expert"), afaik.

J = A*T^2 * e^(-W/kT) [math] \Rightarrow W = - kT \log \frac{ J}{AT^2}[/math], "log" meaning the natural logarithm (often written as "ln").

Slightly more abstract: It converts the number into a symbol (an ordered collection of digits) that can be displayed on the screen and understood by the reader - though not necessarily in the way I sketched. This is not exclusive to computers: Actually, when you write "10" on a piece of paper you are writing down a symbol for a number. Sidenote: The symbol "[math]\frac{100}{10}[/math]" would be a different symbol for the same number.

It's rather easy to write a code constructing a string from a natural number, e.g.: std::string result=""; for (; value!= 0; value /= 10) { int last_digit = value - 10*(value/10); switch ( last_digit ) { case 0: result="0"+result; break; case 1: result="1"+result; break; ... case 9: result="9"+result; break; default: std::cout<<"You screwed up the digit "<<last_digit<<std::endl; abort(); } } if (result == "") result="0";

The simplest method probably is shooting a ray of electrons through a (perpendicular) magnetic field, visualizing the paths somehow (e.g. in a bubble chamber), measuring the resulting radii and calculating the mass from them. For simplicity, anything said below refers to single free particles. EDIT: Oh, and it also is just an attempt to explain the different meanings/definitions of mass in relativity, not the official physicists definition template or something like that. In relativity you have four coordinates, i.e. four directions at each point. Often, and for this post, you chose one of them to be "time coordinate" and give it the label 0 or t and the other three to be "space" with labels 1,2,3 or x,y,z. The (4-)momentum [think of the momentum as the property determining the motion] of the particle can be described by four numbers, [math]p^0, p^1, p^2, p^3[/math]. Assuming the most common choice for the coordinate system then apart from factors of c [which only determine the unit a quantity is measured in] these four values are interpreted as follows: [math]p^0[/math] is the (relativistic) energy E of the free particle, [math]\vec p = \left( \begin{array}{c} p^1 \\ p^2 \\ p^3 \end{array} \right)[/math] is the non-relativistic momentum or 3-momentum. The values for [math]p^0 \dots p^3[/math] depend on the choice of the coordinate system. However, there are properties of the momentum of that are not dependent on how you did chose your coordinate system [*]. One such property is the (pseudo-)magnitude [math]m_i[/math] of the momentum. In the coordinates I used above it is [math]m_i^2 = \left( p^0 \right) - \vec p^2 [/math]. Ok, now for some different choices of the term mass: - [math] m_i [/math] is the "invariant mass" (or "inertial mass" or "proper mass"). It is independent of a chosen coordinate system and only a property of the particle type. This then allows statements like "an electron has a mass of 511 keV". -[math] m_\gamma := p^0 [/math] (the energy) is the "relativistic mass". Some people seem to like this definition of mass for its property that it equals the energy [math]m_\gamma = p^0 = E[/math] (with canonic factors: [math] E=mc^2[/math]).[**] This then allows statements like "according to Einsteins famous E=mc², everything ultimately consists of pure energy." - In the special case that [math]\vec p = \vec 0 [/math], you can easily verify that [math] m_\gamma = m_i[/math]. So if you want "the best of both worlds" you can say that you define mass to be the energy of a particle for which [math] \vec p = 0[/math]. Under the constraints I put in the introduction and the additional constraint that [math]m_i > 0[/math], [math]\vec p=\vec 0[/math] means that the particle must be at rest. So you get the "rest mass" [math]m_0[/math]. - There are other possible meanings of mass that are related to other fields. So what is the "rest mass"? Perhaps think of it as a definition of mass that has the nice property of the invariant mass (i.e. it is a property of the particle, not the way you look at it), does not have the redundancy of "relativistic mass" and is perhaps easier to define for someone with little to no background in relativity. Note that when the momentum of a free particle is negligible compared to its mass, then the three definitions of mass give the same result. Footnotes: [*]: Starting from this insight you can even go as far as to demand that no physics at all should depend on the choice of a coordinate system. This leads to a very elegant formulations within relativity (note that "very elegant" does not need to mean "very useful" ) [**]: I dislike it for pretty much for the same reason.

Speaking for Germany (and probably a large part of Europe, too): It's referred to as "9/11". The numbers are even pronounced in English, i.e. "nine-eleven" and not "neun-elf". That is true at least for [a large] part of the media (actual usage of the US-term might vary, but I cannot really judge that). I would imagine it to be considered a name, not a date. Native english-speaking countries might handle it differently, though.

By breaking, i.e. transferring the (linear) kinetic energy of the car to rotational (kinetic) energy of the flywheel. Q: What is the difference between a nuclear engineer and a car engineer? A: The nuclear engineer tries to make sure that nothing can go wrong. The guy building cars knows that something will go wrong, anyways. As far as I know (from being told by my professor for experimental physics several years ago), the main reason not to use that kind of energy storage is not the problem of getting the system to work but that in case of a car accident you do not want to have heavy, fast-rotating iron wheels flying around. I don't think I'd want springs jumping around, either.

Yes. ?

Why history?

yes and thanks

It seems that the code tag is deactivated in for comments on user pictures. There probably is no reason why it should be deactivated there. Could you [where supposedly "you"="Capn", as always] activate it?