DQW

Thanks for the "feeler" matt. This is what I shall settle for until I have the need to revisit this in greater detail (ie: actually have to solve Lebesgue integrals) I had "almost" convinced myself to walk the following path : topological spaces -> sigma-algebras -> delta-rings, measures -> measure spaces, riemann, lebesgue measures -> riemann, lebesgue integrals. PS : the "=" following the sum of the widths of the discs should be a "<"

Good point. There's nothing biological in this thread (yet). It probably belongs in Physics or GD. Edit : Mokele's post wasn't up when I started on this...so ignore.

deleted

If you want a concise intro to lebesgue integrals, I'd recommend : Mathematics for Physicists (Dover) by Dennery & Krzywicki, Chapter III (Function space, orthogonal polynomials and fourier analysis), page 184 -188 ("Elementary Introduction to the Lebesgue Integral") In some searching around that I did a little while ago, when struck by the same problem, I found the following references : 1. Strook, D. W. A Concise Introduction to the Theory of Integration, 2nd ed. Boston, MA: Birkhäuser, 1994. 2. Henstock, R. The General Theory of Integration. Oxford, England: Clarendon Press, 1991. 3. Kestelman, H. Modern Theories of Integration, 2nd rev. ed. New York: Dover, 1960. I have no idea about the quality or usefulness of these last three references. I'd be grateful too, if matt would throw in his suggestions.

That's hardly an answer. You would expect electrons and protons to respond in opposite manners to an applied field.

deleted... ...I think I may have posted exactly what matt had left as an exercise for the OP.

Whoops ! Forgot to divide ! <looks around for rock to crawl behind>

Aha ! Didn't know that. But I'd heard something about spurious stuff - guess they just entirely skipped the RaCl2, eh ?

You could find out if you stuff really is radium - when radium paints were the in thing several years ago, the cheap substitute (often sold as radium to jack up profits) was zinc sulfide (I think). ZnS is photoluminescent, and its emission intensity will go up after exposure to the UV in sunlight. Take your bottle out to the sun (or hold it under a bright light), keep it there for an hour and take it back to a dark place. If it glows much brighter than before, it is more likely just ZnS. To make really sure there is really no Ra in it, check with a Geiger Counter (radium dust, if inhaled, will give you lung cancer).

Do you understand what a metric is ? Very roughly speaking, a metric is a function that describes the "distance" between points in some set. As a function g(x,y) a metric must satisfy certain properties, such as non-negativity, symmetry and something that states that the distance of any point from itself must be zero (and the converse). Now, this function may also be viewed as a tensor, in which case this tensor must also obey a bunch of corresponding properties (for instance, it must be symmetric and positive definite). In any space, the components of the metric tensor tell you how to calculate general infinitesimal displacements in that space. Yes, the components of the metric tensor for R^n are Kronecker Deltas. This simply follows from the generalization of Pythgoras. [math](dl)^2 = \sum_{i,j} g_{ij}~dx_i~ dx_j[/math] For R^n, you simply have [math] (dl)^2 = \sum_{i=1}^n (dx_i)^2 [/math] So, from above, this gives [math]g_{ij} = 1,~if~i=j~and~~ g_{ij}=0~otherwise [/math] This is nothing but the Kronecker Delta [imath]\delta _{ij} [/imath] I don't understand this question. To me it looks like the metric tensor for Eucliedean space will simply be the identity matrix [imath]\mathbf{1}_n [/imath] (but I might be mistaken). But so far, this is only something to give you an intuitive picture based on more familiar stuff. To really understand the metric tensor, you must know how it is rigorously defined. This is not something I'm entirely comfortable with (haven't looked into it in ages), so I'd leave it to the likes of matt, if you have more questions.

I have to say this is the most ridiculous question (requiring that you "throw in a graph") I've come across in a while. But in any case, something that may come close is the solubility of oxygen in water as a function of temperature.

How did you get that ? Are the balloons that heavy (around 23 to 24 g each) ?

Yes, that's correct. I didn't mean to imply that the only force was a decelerating force. Naturally, the accelerating force (mg downwards) was omitted because the acceleration due it it was equal for both balls. So, I was thinking entirely in the downward accelerating frame. For a sphere (in laminar flow), the terminal velocity goes like [imath]\sqrt{r} [/imath]. But for balls thrown off buildings and such, they will not likely reach terminal velocity. YT : I didn't factor in the effect of the dimples (which is important); the drag force on a sphere depends on the radius (and density) of the sphere. The time difference between the landings will depend on the height from which the balls are thrown. For a height of about 5m (or about 16 ft, roughly a second floor window), my rough estimate says that the time difference (on a windless day) will not be more than 40 milliseconds, if the golf ball were perfectly spherical. I can't say anything about the magnitude of the effect of the dimples, though I'm sure this is well documented and must be findable.

If you state the problem explicitly, it'll make more sense. The problem states (and this is my guess based on the OP) that the block does not slide down the incline and you are asked to find the minimum coefficient of friction that ensures this. As the incline gets steeper, the component of the weight down the incline increases (with the sine of the angle) and the normal reaction gets smaller and smaller (with the cosine). It will naturally require more stickiness (or a greater coefficient of friction) to keep the block from sliding down (since the frictional force is the product of the normal reaction and the coefficient). When the incline is nearly vertical, the weinght of the block is essentially mg and the normal reaction is tiny. The coefficient of friction must be truly enormous to keep the block from sliding down. When the incline is exactly vertical, the normal reaction is exactly zero. The force down the plane, however, is exactly mg. The frictional force upwards needs to balance this downward force for the block to remain stationary. In other words, you want to find a number, which when multiplied by zero, gives mg. Or : [math]0 \cdot \mu_s = mg [/math] But you know that, by definition [math]0 \cdot x = 0 [/math] for any finite number, x. Thus, [imath]\mu_s[/imath] can not be a finite number number ("and hence, must be infinite").

Just gave this a moment of thought now...and I can't see why not. Given any n-1 vectors in n-space, why can't I find a vector that is normal to them all (even, if necessary, by solving the determinant) ?

As Swansont pointed out the reason that old windows are bottom heavy is not because glass flows (that really has been debunked). In fact the time constant over which typical silicate glass flows is roughly of the order of millions of years (or longer). However, glass has liquid-like properties beyond its simply being an amorphous solid. Because glass has properties that make it solid-like as well as some that make it liquid-like, it falls under a different category of phases, called (you guessed it) "glasses". Their physics is quite complex because they are nonequilibrium creatures ! http://math.ucr.edu/home/baez/physics/General/Glass/glass.html http://hypertextbook.com/physics/matter/glass/

You're good ! [math] \omega = \sqrt{\frac{k}{m}} [/math]

Selena, when you speak of "the next wave of communications", does it necessarily mean FTL communication ? When I read the OP, my first thought was about communication using quantum encryption. This is theoretically feasible and currently is technologically undoable, but easily lends itself to Sci-Fi. Also, it is possibly the only defense to Quantum Computation, which will make breaking conventional ciphers a child's play. <note: I've lapsed in sci-fi talk here>

I suspect the OP believes the neutrino carries a charge. Neutrinos (and antineutrinos) are chargeless.

Yes, it is a cross product. Note also, that the only way to "multiply" vectors and get a vector, is by means of a cross product. By convention : [math]\vec{F} = q~(\vec{v} \times \vec{B} ) [/math]

The deceleration from drag on a spherical body (a ball) goes like 1/r2 or 1/r, depending on the nature of air-flow past the ball. In either case, the smaller ball experiences a greater deceleration due to drag. So, the golf ball slows down more and hence, lands second. Note : insane_alien, the drag force is greater on the bigger ball, but the deceleration is greater on the smaller ball.

I think I know how to project a vector onto another vector or onto any subspace of the parent space, but I'm not sure I understand what your question is asking of you. For instance, what does [(1,0,0,0),(3,4,0,0)] represent - a pair of vectors, or something else ?

I do not wish to detract from the approach that matt is taking with this subject, but (since were talking about uses) I thought I'll throw in an example (from physics, sorry matt) to help illustrate when one might use a dot or cross product. Consider an electric dipole of dipole moment [imath]\vec{p}[/imath] in an electric field [imath]\vec{E}[/imath]. There are two quantities that immediately spring to mind in this case, the interaction energy U (between the dipole and the field) and the torque [imath] \vec{ \tau }[/imath] (felt by the dipole). Both quantities have units which are the product of the units of the moment and the field. However, the torque is a vector quantity and the energy is a scalar. So, it should not come as a surprise that the torque is given by the vector (cross) product, while the energy is given by the scalar (dot) product. [math]\vec {\tau} = \vec{p} \times \vec{E} [/math] [math]U = - \vec{p} \cdot \vec{E} [/math] It is in this context that it makes some sense to refer to these operations as "multiplications". But, as cautioned earlier, they are not at all the same as the usual multiplication that is defined on a field like the reals (which are scalars). Why a dot product produces a scalar while a cross product produces a vector has been answered somewhere before, I believe - because that is how they are defined.

3. [imath]\LaTeX[/imath] for matrices : see example below [math] \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right) [/math] click on the matrix to see code.