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 "<"

what does teh big bang have to do with evolution?

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.

anything can be moved with a strong enough magnetic field, all matter has protons, and electrons. which are effected by EM forces.

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>

With the radium paint, ZnS existed in there anyway. Radium salts on their own will not glow unless you have so much of it that it will kill you. The hands were painted with the RaCl2/ZnS mixture which led to the glow as the emission of radiation from the radium 'excited' the ZnS. Many of the watch hands quit glowing not because the Ra had decayed, but because the ZnS failed to function.

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

To make things harder' date=' I don't think so !

 

A water balloon, and a plain air balloon, both identical and inflated to the exact same size, will be affected by the same air drag resistance, and will have different weights.

For very sure, launch both from your Empire State building, and you will find that a long time after that pedestrian dries out, the air inflated one is still flying. Even if you do it in your bedroom with zero crosswind, they will not fall at the same speed.

It's the law of selective gravity.

Miguel[/quote']Also, the two balloons have extremely different densities, and hence, the buoyant force on one of them (as a fraction of its weight) is several times the other's. With the golf and bowling balls, the density of both are pretty high and in the same ballpark (no pun intended). The buoyant forces on them are negligible.

Hi.

What interesting thing to do with it ? It's a light green powder in a glass vial; the box is marked as "radium" and intended to mix with a clear glue also supplied in the kit.

Am still alive (I think) after that stuff has been in my house for 30 years.

Miguel

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).

I am sorry to say that I do not understand the metric tensor one bit.

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.

 

Isn't it the Kronecker Delta in Euclidean 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]

 

And wouldn't that be a collection of row vectors?

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.

A mole of air at STP, (for the sake of argument say it's all Nitrogen) will displace ~22.4 L (assuming it's close to ideal gas behavior) and have a mass of 28g. A mole of He will have a mass of 4g. This mass will be somewhat larger as you increase the pressure above atmosphere to fill the balloon. So you can lift ~ 1g/L, including the mass of the balloon[/u'];

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

They both accelerate until terminal velocity is reached then stay at that speed.

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").

one important note is that the dot prodcut can be generalized to any dimension, but the vector prodcut is perculiar to the 3 d world.

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/

This is from a 6 part physics homework problem' date=' I got part a-e, but part f (likely to be my grade ) is what I'm having some trouble with:

A mass [i']m[/i] slides down a frictionless incline and sticks to a spring (spring constant k ), initiating simple harmonic motion.

 

f) What is the angular frequency of the SHM?

 

I got [math](k/m)^{1/2}[/math] as the solution (someone please tell me how to get the Latex radical )

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.