hobz

Thanks for the tip!

I can see the price is $27.70 which hardly scares anyone.

I don't think there is experimental evidence to support your statement. As far as I can tell, a slit experiment shows us that the more we know about a photon's position (going through the slit) the less we know about where it's going. That is, we get a wider distribution on the photographic plate.

Saying that each individual photon has an uncertainty associated with it, is no different than saying each throw of a die has an uncertainty associated.

Cheers mate!

I dont have this fresh in mind, but Gauss quadrature is merely the integral of a polynomial that is fitted to the given function points

Me neither, but doens't it strike you as amazingly simple and easy to use the G-Q?

[...] but not actually very interesting (since I don't have the details in my mind heheh), and I dont have the interest to refresh them either .

Sounds like it is really NOT worth the effort!

As a certain greek states (as well as people who know me): All I know is that I do not know anything!

No arrogance intended. The insight required to understand G-Q is sheer genius!

I think you are right; but I'm not sure we agree on the meaning of "practical".

Good one

Simon Singh has a good and accessible introduction to coding. Check out http://en.wikipedia.org/wiki/The_Code_Book

In this book, Singh describes an encryption method similar to what you propose (if I understand you correctly), and how to analyze such an encrypted message. Much of the work involved in "cracking" the code, could be performed by a computer such as comparing the frequency of letters in the encrypted message to that of the standard language in which the message was written.

There are four lectures by Richard Feynman.

Ah yes. These lectures were "compiled" so to speak into the book QED: The strange theory of matter and light, which I incidentally just finished reading. Thanks for posting their origin. Never seen the Feynman at work

This guy claims that time is quantized in what he calls "tempons".

Anyways, the -on in tempon has a particle ring to it, although it is most certainly not a particle.

I have used this way of approximating the integral of a function. It is pretty straight forward as it incorporates the sum of a product of the function to be integrated and a weight function.

The weight function is a Legendre polynomial which is a solution to Legendre's differential equation.

My problem is I do no understand how these solutions are connected in approximating the integral of a function?

Anyone can use Gaussian quadrature once the solutions to the L. polynomials have been calculated, but understanding how Gauss saw a connection from a differential equation to a (very nice way) of approximating an integral is not apparent.

I took an introduction course in numerical analysis, and asked my teacher about this, to which he replied: "It is not worth the time to examine how the connection is established".

Can anyone here shed some light on this matter?

Thanks!

Assuming equally symmetric riders of the same density the heavy guy wins. The drag coefficient is probably the same or very close and the volume of the riders is proportional to their weight.

 

The frontal area of the heavier guy is, however, proportionally less.

 

Advantage heavy guy!

Normally I would agree, but your statement incorporates that the area of the heavier guy is proportionally less with respect to his weight. However, the weight was "canceled out" as the math explains.

I found the following video lectures, from MIT's Open Course Ware - three courses explaining the basic physics - very helpful.

http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/CourseHome/index.htm

http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/CourseHome/index.htm

http://ocw.mit.edu/OcwWeb/Physics/8-03Fall-2004/CourseHome/index.htm

And look for "The Mechanical Universe" which is an older series of videos, that has very good animations explaining all kinds of physics.

I have yet to discover a video course on QM. Hoping OCW will add one to their repertoire.

That's been addressed. If drag and friction are ignored, there is no mass dependence. Friction and drag account for all of the difference.

The force calculation you did seems to dependent on mass?

The wave packet is the description of the photon, it does not physically confine it.

I see! So by assuming a very definite momentum for the photon, which I interpret as the ability of predicting where the photon will "go", we get a infinite sinusoid that tells us that the photon is anywhere in space!

And the inverse is also true. If the photon (in it's wave packet interpretation) is located in some region in space, it has a variety of momenta that makes it impossible to predict where it will "go"?

The same in other terms.

Knowing where the photon will "go" (or along which path it will travel if you will), makes it impossible to know where in space it is located.

Knowing where the photon is located, makes it impossible to know where it is "going"?

Can't really do that without specific drag coefficients, friction, etc. You could just guess, but that would just be another "I think," and might not realistically reflect what two different bikers would actually experience. My guess would be that mass is roughly proportional to volume, while drag is roughly proportional to surface area. Weight, then, would increase faster than drag, favoring the bigger rider. I would also guess that other friction that would be negligible compared to air resistance in this scenario, but I don't know without actually going out and testing it.

You (not necessarily you, although your name suggest you are used to tedious work) could construct two functions of mass, drag coeff. and friction coeff. and compare the two. You could also assume equal (or negligible) drag- and friction coefficients, since the question was concerned with the mass.

Heavier riders usually have an advantage when going down hill compared to lighter riders. This advantage is almost the opposite of what the lighter riders have up hill. That is, the lighter riders reach the top first, but the heavier rider can catch up with him after reaching the foot of the hill again. This is based on my experience with watching the Tour on TV, and is not science.

Perhaps we SHOULD go through the math, since there seems to be alot of "I think"'s?

A photon will also have an inherent uncertainty associated with it.

Is this uncertainty related to the wave packet? That is, the region in space "occupied" by the wave packet is related to the probability of the photon "actually" being there?

(I realized that the uncertainty could be interpreted as a property of the photon, and thus there is not "actual" location -- the photon is confined (by the wave packet) to a certain region of space by probability, but in a sense is in all locations probable at the same time.)

By "equivalent" I meant, are they identical. Mea culpa for not being precise enough.

I am not sure that it is a classical sinusoid. I speculate that one could obtain the probability of the photon being along a direction by squaring the wave packet?

In the Feynman lectures on physics, I encountered a sum of sinusoids having almost identical frequency. The resulting sinusoid had a very small location in space, so to speak. My question is; is this resulting wave packet identical to what is called a photon?

Consequently, the location in space is dependent on the number of sinusoids that are summed. If the number of sinusoids is finite, the wave packet will be periodic, and there by the same "photon" (assuming that the answer to my first question is "yes") is spread, like knots on a rope.

What makes up all these sinusoids? (I have encountered a similar discussion in the "Beyond The Mechanical Universe"-video series, but they fail to tell us of the origin of the sinusoids. (They simply say, "If waves of different wavelengths are added... especially if the wavelengths are close together... the result can be a kind of wave concentrated in a limited region of space.")

The wavelengths that are added are [math]\frac{h}{\lambda}[/math].

No, you can derive [math]\frac{d}{dx}\sin{x}=\cos{x}[/math] just from the definition of the derivative and the angle sum formulas.

 

i.e. using [math]\frac{df(x)}{dx} = \lim{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/math]

 

and

 

[math]\sin{a\pm b} = \sin{a}\cos{b}\pm\cos{a}\sin{b}[/math]

and

[math]\cos{a\pm b} = \cos{a}\cos{b} \mp \sin{a}\sin{b}[/math]

Ahh, cool stuff. Thanks! Will look into it.

The nth Taylor polynomial requires the nth derivative. The "Taylor expansion" or "Taylor series" requires all of them.

 

One answer to your question is that if we define sine and cosine by their Taylor series, we can get the nth derivative by differentiating the series term by term:

If sin(x) is defined by sin(x)= x- x^3/3!+ x^5/5!+ ..., then it is obvious that sin(0)= 0, sin'(x)= 1- 3x^2/3!+ 5x^4/5!+...= 1- x^2/2!+ x^4/4!= cos(x), sin"(x)= -2x/2!+ 4x^3/4!+ ...= -(x- x^3/3!+...)= -sin(x) etc. You could then show that y= sin(x) satisfies the differential equation y"= -y as well as the initial condition sin(0)= 0, sin'(0)= 1 and so, by the existance and uniqueness" theorem for initial value problems, is identical to the "traditional" sine function.

 

 

Given any infinitely differentiable function we can define its Taylor's series at, say, 0 (the "McLaurin series"). It does NOT follow that the series converges to that function. For example, the function

[math]f(x)= e^{-\frac{1}{x^2}}[/math] for [math]x\ne 0[/math], f(0)= 0 is infinitely differentiable at 0. All derivatives are equal to 0 at x= 0 so its Taylor's series about x= 0 is identically equal to 0. That converges, of course, for all x but is equal to f(x) only for x= 0.

This solution requires knowledge of how to construct the Taylor series in the first place. That is, knowing that sine and cosine are closely related in calculus. Historically one must have had a feeling or hunch that sin'(x) = cos(x).

So basically, linear implies "well understood" and "easily solvable" (though not by my personal interpretation of the term "easy". Perhaps easier solvable is more accurate).

I realized that my initial question was not very accurately put forward. "Stuff" is rather closely related to "undefined".

Can I (and how can I learn how to) transform e.g. classical mechanics into LA? Say a system of springs, masses and dampers connected in parallel and series?

Another example, the movement of a pendulum that is, where the output is location at time t.

What is the physical significance of the eigenvectors? E.g. what do eigenvectors (and values) say about a system? My limited insight inhibits me from defining "system" very accurately, but let's say that a configuration of springs and masses can be transformed into LA, what then do eigen- tell me?

What can eigen- tell me in general (or by an example)?

A note.

The Taylor expansion of e.g. sine, requires the n'th derivative to be known.

Thus before the sine can accurately be constructed (using Taylor), we must know beforehand that the first few derivatives are cos x, - sin x, - cos x, sin x, cos x. But how can we know this (exactly) when we not yet have defined any way of calculating these functions (using Taylor)?

It seems like there is some hidden fundamentals to linear algebra that supersedes the systems of linear equations. For example, in DSP, complex exponentials are considered eigenvectors to LTI systems. This is very fascinating, and I would like to know if this is based on systems of linear equations, or there is some way of transferring stuff into linear algebra.

Thanks for clarifying! I would seem out of the six different possibilities, it just happens to be the sine, cosine and tangent that are most often used?

I am revisiting some of the ol' trigonometry, and while I was reading about in various places, it occurred to me that the authors fairly quickly introduce definitions of the trigonometric function.

I am left with definitions like, [math]\sin \theta = \frac{opp.}{hyp.}[/math], which stems from the fact that in a right angled triangle, the ratio of the triangle ABC (with the hypotenuse AB = 1 and angle between AB and AC = [math]\theta[/math]) relates to the ratio of any scaled version of the triangle having the same angle. If we assume that there is a larger triangle ADE (hypotenuse = AD), then from Euclid (if I remember correctly) the relation [math]\frac{BC}{AB} = \frac{DE}{AD}[/math] defines [math]\sin \theta[/math] as BC with AB being 1.

Now what I don't understand is, why could [math]\sin \theta[/math] not have been chosen to be the inverse ratio? And likewise with cosine? What is so special about the ratios where we divide by the hypotenuse, contra the ratio where we divide by the "end"-height?

Thanks!

I know that the electric field is zero. But does that mean that the net sum of electric FORCE is zero? That is, if I put one test charge inside the sphere (if you like), would all the forces acting on that charge sum up to zero, or would there simply BE no forces?

Check out

If I was informed to be exposed to an electric field of high strength, how little metal would is require to protect my self?

I understand that the electrons on the outside of the cage, would align themselves in a manner that would be related to the strength of the field. Suppose my field was so strong, that it would require all free electrons to be at exactly the same position in space, because the field was so strong that it compressed the electrons into one point.

According to QM, this would not be possible. How can we "prevent" this from happening by imposing some law on the phenomenon? It would require infinite energy?

Thanks for posting, but it doesn't really answer any of my questions.