Tannin

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  1. Tannin

    Goldstein's mechanics

    For all who are interested in gaining deeper insight into principles of mechanics (and eventually of most of the modern physics) I would like to recommend the book "Variational principles of mechanics" by C.Lanczos. In his introduction the author states that the book is not written for "superhuman beings" but for EVERYONE, and he keeps his promise. C. Lanczos was an assistant for A.Einstein and has many contributions in the field of mathematical physics. I think that understanding (and not merely following) Goldstein is much easier after Lanczos. His explanation of variational calculus is superb.
  2. Tannin

    Implicit Differentiation

    Let us assume that the equation [math]t(x,y)=0[/math] defines [math]y[/math] as an implicit function of [math]x[/math]. Why implicit? Because we cannot always easily express y as a function of x from the equation for t(x,y). For example,[math]c^3+y^3+3axy=0[/math], where c,a are constants. In order, to get y as a function of x we must solve cubic equation (difficult). Now without trying to find the dependence of y on x explicitly, let's just call the solution [math]y=y(x)[/math]. Now, constant 0 (of t(x,y)=0) may be thought of as a function of x, depending on it both directly and through [math]y(x)[/math]. Taking derivative we have: [math]\frac{d t}{d x}=\frac{\partial t}{\partial x}+\frac{\partial t}{\partial y} \frac{d y}{d x} [/math] Rearranging terms we have [math] \frac{d y}{d x}=-\frac{\frac{\partial t}{\partial x}}{\frac{\partial t}{\partial y}}[/math] Thus, we have found the derivative [math]\frac{d y}{d x}[/math] without finding y(x). So negative sign appears because of the constant in the RHS of the definition of the implicit function t(x,y)=const. Hope this answered your question. Also see Wikepedia: http://en.wikipedia.org/wiki/Implicit_function
  3. Let's begin with the Divergence Theorem [math]\nabla \cdot \vec{F}=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}=2z[/math] Now the easiest way to calculate the volume of revolution is to take as volume element a thin slab with circular cross section and thickness [math]dz[/math]. The area of the cross section is [math]\pi(r^2-z^2)[/math] and the resulting integral is [math]\int^5_3 \pi(25-z^2) 2 z dz=128 \pi [/math] Now for the surface integral: For the planar part the direction of the normal is [math]-\vec{k}[/math]. Thus [math]\int \int (xz \vec{i}+yz \vec{j} +\vec{k})\cdot (-\vec{k}) d\sigma=-\int \int d\sigma=-16 \pi [/math] For the spherical part we have to transform both the coordinates and unit vectors of [math]\vec{F}[/math] to the spherical coordinates [math]\vec{F}=r\sin{\theta}\cos{\phi}r\cos{\theta} (\vec{F}_r \sin{\theta}\cos{\phi} +\vec{F}_\theta \cos{\theta}\cos{\phi}-\vec{F}_\phi \sin{\phi} [/math] [math]+ r\sin{\theta}\sin{\phi}r\cos{\theta} (\vec{F}_r \sin{\theta}\sin{\phi} +\vec{F}_\theta \cos{\theta}\sin{\phi}+\vec{F}_\phi \cos{\phi}[/math] [math]+(\vec{F}_r \cos{\theta}-\vec{F}_\theta \sin{\theta})[/math] Now the surface element is the regular [math]r^2\sin{\theta}drd\theta d\phi[/math] directed along [math]\vec{F}_r[/math] After taking the dot product between the surface element and [math]\vec{F}[/math] only terms with [math]\vec{F}_r[/math] will survive. We have the following double integral (r=const) [math]\int_0^{2\pi}d\phi \int_0^{\arccos{3/5}} d \theta r^2 \sin{\theta} (r^2 \sin{^2 \theta} \cos{\theta} \cos{^2\phi}+r^2 \sin{^2 \theta} \cos{\theta} \sin{^2\phi}+\cos{\theta})=[/math] [math]=\int_0^{2\pi}d\phi \int_0^{\arccos{3/5}} d \theta \left(r^4 \sin{^3 \theta}\cos{\theta} + r^2 \sin{\theta}\cos{\theta} \right)=144 \pi[/math] Summing the contributions of the planar and spherical parts, we have [math]-16 \pi + 144 \pi=128 \pi[/math] The divergence theorem works!!! Regarding your solution : the direction of the normal to the spherical part when expressed through i,j,k vectors is indeed [math]\vec{r}(\theta,\phi)[/math] (6th line from above in your answer). There is no need to calculate the cross-product of [math]\vec{r}_\theta \times \vec{r}_\phi[/math] - this will only bring us again to the same radius vector. In order to continue you have to multiply the normal vector by the surface element of the spherical part. If you have a questions, I'd be happy to answer. Good luck!
  4. We shall prove in general that the equation [math]\frac{d^n y}{dx^n}-y=0[/math] has [math]y=e^x[/math] as it's solution. The proof is based on the mathematical induction. As usual it is done in two stages: 1. Show for n=1 [math]\frac{dy}{dx}=y[/math] As CPL.Luke has pointed out, the solution is [math]e^x[/math] 2. Assuming that for arbitrary natural [math]n[/math] the solution is [math]e^x[/math], i.e. [math] \frac{d^n y}{dx^n}=y[/math] we shall show that this is true for [math]n+1[/math] [math]\frac{d^{n+1} y}{dx^{n+1}}=\frac{d}{dx} \left( \frac{d^n y}{dx^n} \right) =\frac{dy}{dx}=y [/math] We are substituting the assumption instead of the parentheses. The proof is done.
  5. Tannin

    Exponential problem

    1. Graphical solution method: calculate the values of LHS (left hand side) and RHS at several x points, join the obtained points by smooth curves and you'll get rough estimates. One can see that there are 3 roots for [math]x^2=2^x[/math] 2. The next stage may be an expansion in Taylor series. Once you know the approximate location of the roots, you can expand both RHS and LHS in Taylor series around the estimated roots and solve the resulting algebraic equation For example expanding RHS of [math]x^2=2^x[/math] into Taylor series around zero and truncating after the quadratic term gives [math]2^x=e^{x\ln{2}}=1+x \ln{2} +\frac{x^2 (\ln{2})^2)}{2}+\ldots =x^2[/math] This quadratic equation has 2 roots: -0.78 and 1.69 As we can see, the quality of the solution deteriorates rapidly with the distance from the point of expansion. While the "left" (negative) root is not bad comparing to the root calculated by Ragib [math]\approx -\frac{23}{30} \approx -0.7667 [/math], the right one deviates significantly. In order to improve it, we should choose the point of expansion closer to the rough estimate of the root (obtained e.g. by graphical solution). Question to Ragib: How did you calculate the roots?
  6. Tannin

    Goldstein's mechanics

    In my opinion, the only problem of Goldstein's book is that the mathematical formalism used by the author differs very much from the formalism used in the modern research literature. Today's papers speak in the language of "manifolds","symplectic geometry", "differential forms" etc. And this is much more than some new notation - it's completely new unifying framework for thinking. So for all who is not afraid of mathematics, I would recommend to study this modern approach instead of tracking XIX century pathways. The definitive book introducing this new approach is Arnold V.I. "Mathematical foundations of classical mechanics",1989 (2ed) The book is rather slim and the exercises are very useful.
  7. Tannin

    Phospor that glows under infrared exposure ?

    I admit that I don't know the answer, but I can propose a mechanism: multiphoton absorption of "small" IR quants and subsequent fluorescence (or phosporescence) of the one "big" quant in UV-Vis range. For sufficiently high intensities, multiphoton absorption is not frequency selective. Also because we are not in the gas phase, but in the solid state, the upper level is actually a band. It would be very interesting to learn from the expert which compound they actually use and it this is the mechanism.
  8. Tannin

    Time dilation - a global hoax?

    For those who are interested what is SR explanation of the original kkris1 question. In the posts #5 and #39 I´ve shown the necessary diagrams and formulas. But kkris1 was (justly) not satisfied and in his post #53 he asks why the order of hitting the ceiling by vertical and slanted rays for stationary and moving observer is reversed (diagram in #39). I think this question is related to the fascinating point in SR about the order of events. Let´s begin with the simpler example. Borrowing the diagram from Landau, Lifshitz - Theory of fields - par.1. The frame X´Y´Z´ is moving right along the X-axis with some constant velocity V. From the point A two signals (light flashes) are sent in opposite directions. Because the velocity of light is equal in every inertial frame, they will reach points B and C (equidistant from A) simultaneously in the moving frame X´Y´Z´. But the same two events are not simultaneous in the stationary frame XYZ. Indeed point B moves toward the flash, while the point C runs away from the flash. Thus in the stationary frame XYZ signal will arrive to the point B before he reaches the point C - two events are not simultaneous now. Exactly, the same thing happens in kkris1 thought experiment. I bring here again the diagram for easier reference In the moving frame slanted ray hits the ceiling after the vertical ray. Transition to the stationary frame brings about reversal of the order of the events and now ¨originally vertical in the moving frame¨"ray hits the ceiling after the ¨originally slanted ray¨. While choosing proper frame velocity we can make both events to happen simultaneously. Using formula [math]\cos{\theta}=\frac{\cos{\theta^\prime}+\frac{V}{c}}{1+\frac{V}{c}\cos{\theta^\prime}}[/math] and substituting [math]\theta=90^\circ[/math] and [math]\theta^\prime=67.5^\prime[/math], we have V=0.38c. The angle 67.5 comes from the requirement that both rays will make equal angle with the vertical direction while preserving 45 degrees between them (read post #39). In this case both rays will hit the ceiling simultaneously and this will happen after [math]\frac{h}{c \sin{67.5}}[/math] seconds. The proper name offered by SR for these two events (a ray hits the ceiling) are ¨spacelike¨ (Landau, Lifshitz - par. 2) Because I am not an expert in SR, I invite the experts to check my calculations and point out the errors. Also, I would like to remind that there is no such thing as time dilation for the light rays - their world lines are always straight.
  9. Tannin

    Time dilation - a global hoax?

    Instead of trying to understand the ¨thought experiment¨ proposed by kkris1, this thread is going to overturn the SR and all the ¨conventional science¨ Making verbal statements (however appealing they may seem) is not enough for physics. Fortunately, we have our language of diagrams and formulas. I´ll try to explain what I´ve meant in my first post in this thread and I´ll be happy to hear your comments and corrections. Here is the diagram: Moving observer shoots simultaneously two rays: one is vertical and makes an angle [math]\theta=90^\circ[/math] with the floor (broken line) and the other slanted by the angle [math]\alpha=-45^\circ[/math] with regard to the ¨vertical¨ ray. The vertical ray hits the ceiling, while the slanted ray obviously reaches the point [math]\left(-\frac{h}{\sqrt{2}}, \frac{h}{\sqrt{2}}\right) [/math](assuming shooter to be the coordinate origin and h the ceiling height). Now, moving with the speed of [math]V=\frac{c}{\sqrt{2}}[/math] in the positive x-direction will rotate both velocity vectors by [math]45^\circ[/math] - light aberration (Landau, Lifshitz - formula 5.6). [math]\sin{\theta}=\frac{\sqrt{1-\frac{V²}{c²}}}{1+\frac{V}{c}\cos{\theta^\prime}} \sin{\theta^\prime}[/math] Definitely, the paths travelled by the rays cannot change - velocity of light is constant in every inertial frame and thus there is no time dilation for the light rays. So all we have - rotation of the velocity vectors with the angle [math]\alpha[/math] being conserved and equal to [math]45^\circ[/math], while the aberration angle [math]\theta´[/math] depends on the velocity of the moving observer. I don´t see any reason why the two-segment paths of rays (before and after reflection) shouldn´t behave the same way - rotation only (red lines in the diagram). Moreover, if we are taking the snapshot after [math]\frac{2h\sqrt{2}}{c}[/math] the second ray will hit the ground. The distances 0C and 0C´ will be equal in both frames of reference and the distance C0C´ will be equal to [math]\frac{2h\sqrt{2}}{c} \times \frac{c}{\sqrt{2}} =2h [/math], the distance travelled by the moving observer in this interval of time. So, I guess the origin of the kkris1 paradox was the assumption of time dilation for the light rays - ¨shorter distance in the longer time¨. In the above picture, everything is fixed ¨equal distances in equal time¨.
  10. Tannin

    vectors and kinematics

    Question 2: Draw two lines at the 90 degrees angle to each other. Now move only one line - the velocity of the point of the intersection, clearly, equals to the velocity of moving the line. If we move both lines, we get the point of intersection velocity as the hypotenuse of the right-angled triangle with the legs equal to the velocity of each line - use Pythagorean theorem. If initially the lines intersect at the angle alpha you should use the law of cosines (see Wikipedia - Triangle). Question 1: It is unclear from your problem, whether gravitation acts on the particles. If gravitation force does act on both particles remember the x-component of the velocity is conserved, and the y-component changes by -gt (where g is the free fall acceleration and t is the time passed). Create the scalar product of two vectors and make it equal zero (perpendicular) - from this condition you will be able to find the time by solving quadratic equation. Good luck!
  11. Tannin

    Time dilation - a global hoax?

    Hi, Kris! Thanks for the wonderful question - I have enjoyed to think about it. The whole discussion about the time dilation and velocity direction change (light aberration) refers to the case of the uniform motion in the constant direction (inertial frames). Indeed, the frames of the stationary and the moving observer are inertial: moving observer travels with the constant velocity along the x-axis. But, the frame connected with the light ray is not inertial: light ray changes its direction after the reflection in the mirror. Thus your intuition regarding time dilation and length contraction is not applicable here. Now let`s analyze light ray travelling one-way: from floor to ceiling. One moving observer shoots simultaneously two light rays: one vertically, the other - by the angle of 45 degrees. Clearly, for the time that the vertical ray hits the ceiling at the height h meters - point (0,h), the second ray wouldn`t reach the ceiling - it would reach only the point (-0.7h, 0.7h). Now for the stationary observer, the roles of the rays are changed: originally vertical ray would seem at 45 degrees (and reach the point (0.7,0.7) - not hitting the ceiling), and the originally slanted ray would seem vertical and hitting the ceiling. What has to be conserved is the ration between the paths of two rays (1:1) and the transition between the stationary and moving observers only rotates the velocity direction. We are reaching the conclusion, that the decision which ray hits the ceiling and which goes halfway, is up to the observer (stationary or moving). I hope that I´m not writing nonsense - these paradoxes are sometimes tricky to catch. Further reference is available in Landau, Lifshitz - Course of Theoretical Physics - Theory of Fields - par. 5
  12. Tannin

    Back to Hydrogen Experimentation

    Just to give the more accurate wording to the forementioned gas concentration required for explosion: The technical terms are: HEL - Higher Explosion Limit LEL - Lower Explosion Limit i.e. an explosion may occur only when the substance concentration is between these limits. Usually you can find these data in MSDS - Material Safety Data Sheet. They are freely available on the web and are "must" before you are going to perform your experiment. For example: http://www.llnl.gov/es_and_h/hsm/doc_18.04/doc18-04.html#2.0 According to this source: "Hydrogen has an unusually large flammability range. It can form ignitable mixtures between 4 and 75 percent by volume in air. The range for explosive mixtures is also very broad. Given confinement and good mixing, hydrogen can be detonated over the range of 18 to 59 percent by volume in air. The energy content per weight of a stoichiometric mixture of hydrogen and air is about the same as that for the explosive TNT, although much less energy is converted to pressure waves during detonation." Good luck and be careful!
  13. Tannin

    Mechanoluminescence?

    So far I've located only the definition: Mechanoluminescence - The emission of light as a response to a mechanical stimulus. http://perg.phys.ksu.edu/vqm/resources/vqm_glos.html This looks interesting and seems to have a lot of applications. I'll try to read and understand one of the research papers.
  14. Fourier's law for conductive heat tranfer states: q=-k grad T where q is the heat flux, k is thermal conductivity and grad T is the temperature difference (gradient). So, in the framework of the approximation provided by this law, the heat flux is equal for both glasses and thus they will reach the room temperature simultaneously.
  15. Tannin

    Charges of anions...

    Generally we are looking only on the last non-complete shell: phosphorus has electronic configuration of 1s2 2s2 2p6 3s2 3p3, but the complete shells play no part in the chemical bonding, so we are interested only in the p-electrons from 3p shell: we have 3 and we can take 3 more. Potassium is 1s2 2s2 2p6 3s1, but again we are looking only on the last non-complete shell and thus it has only one electron to donate. Another way to say the same thing is "completing octet - stable 8 electrons configuration". So everything you need to know is the number of column where the element is: P is in the 5th column, K is in the first. And then take the atoms in such a proportion that the sum of the electrons in their outer shells would equal 8. But, unfortunately, there is no such thing like a rule without the exceptions of the rule. The notable one is the "transition elements" (like Fe, Cr etc.). Here we are interested in the second from the last shell - e.g. 3d. And also another stable configurations (and not just octet) are possible. The reason is that the energies of 4s and 3d orbitals become very close, and thus the order of their filling is not so easy to figure. Also, the charge on the cations/anions may be defined in the different ways: the electrons are shared between the atoms and the charge (electron) transfer is never complete. So, it's rather a crude approximation to assume that K is 3+ cation. But this is our way - moving from one approximation to another and more answers are available in the fascinating field of quantum chemistry.