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# A 1.2-liter tank at a temperature of 27°C contains a mixture of 1.5 moles of oxygen, 1.2 moles of nitrogen,

###### Question:

A 1.2-liter tank at a temperature of 27°C contains a mixture of 1.5 moles of oxygen, 1.2 moles of nitrogen, and 1.6 moles of hydrogen. What is the total pressure of the mixture inside the tank?

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## Chemistry (4th Edition)

The correct formula for Nickel (II) Perchlorate is $Ni(ClO_{4})_{2}$.
Sincle the first element is Nickel, the formula for Nickel is Ni and then if you combine with Perchlorate, the anion becomes $Ni(ClO_{4})_{2}$.

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# Application of Geometric Sequence and Series

In this tutorial we discuss the related problems of application of geometric sequence and geometric series.

Example:

A line is divided into six parts forming a geometric sequence. If the shortest length is $3$cm and the longest is $96$cm, find the length of the whole line.

Solution:
Given that
${a_1} = 3$, ${a_n} = 96$, $n = 6$

Since   ${a_n} = {a_1}{r^{n – 1}}$

we have
$96 = 3 \times {r^5}$         $\Rightarrow r = 2$

For the length of the whole line, we have
${S_n} = \frac{{{a_1}\left( {{r^n} – 1} \right)}}{{r – 1}}$ (As $r > 1$)
${S_n} = \frac{{3\left( {{2^6} – 1} \right)}}{{2 – 1}} = 3\left( {64 – 1} \right) = 189$cm

Therefore, the whole length of the line is equal to $189$cm.

Example:

The value of an automobile depreciates at the rate of $15\%$ per year. What will the value of an automobile be after three years if it is purchased for $4500$ dollars?

Solution:
Let ${a_1} = 4500$ = purchased value of the automobile.

The value of automobile at the end of the first year
$= {a_1} – {a_1}\left( {\frac{{15}}{{100}}} \right) = {a_1}\left( {1 – 0.05} \right) = {a_1}\left( {0.85} \right)$

The value of the automobile at the end of the second year
$= {a_1}\left( {0.85} \right) – {a_1}\left( {0.85} \right)\left( {\frac{{15}}{{85}}} \right) = {a_1}\left( {0.85} \right)\left[ {1 – \frac{{15}}{{100}}} \right]$
$= {a_1}\left( {0.85} \right)\left( {1 – 0.85} \right) = {a_1}\left( {0.85} \right)\left( {0.85} \right) = {a_1}{\left( {0.85} \right)^2}$

Similarly, the value at the end of the third year
$= {a_1}{\left( {0.85} \right)^3}$

Hence, the value of the automobile at the end of $3$ years after being purchased for $4500$
$= 4500{\left( {0.85} \right)^3} = 4500\left( {0.6141} \right) = 2763.45$


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# American Institute of Mathematical Sciences

September  2003, 9(5): 1133-1148. doi: 10.3934/dcds.2003.9.1133

## Homoclinic bifurcations, fat attractors and invariant curves

 1 Departamento de Matemática, Facultad de Ciencias, La Hechicera, Mérida, 5101, Venezuela

Received  October 2000 Revised  November 2002 Published  June 2003

Here we show examples of homoclinic bifurcations which can be perturbed to produce invariant curves and attractors with high Hausdorff dimension.
Citation: Leonardo Mora. Homoclinic bifurcations, fat attractors and invariant curves. Discrete and Continuous Dynamical Systems, 2003, 9 (5) : 1133-1148. doi: 10.3934/dcds.2003.9.1133
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# Which data define an $SU(3)$ structure?

Let $(z_1,\overline{z}_1,\dots, z_n,\overline{z}_n)$ be the coordinates on $\mathbb{R}^{2n}$ given by the identification $\mathbb{R}^{2n} \simeq \mathbb{C}^n$. Define $$g =\left| dz_1 \right|^2+ \dots + \left| dz_n \right|^2 \\ \omega = \frac{i}{2} (dz_1 \wedge d \overline{z}_1+ \dots + dz_n \wedge d \overline{z}_n) \\ \gamma=dz_1 \wedge \dots \wedge dz_m$$ The subgroup of $GL(2n)$ that stabilizes $g$ and $\omega$ is $U(n)$ ("2-out-of-3-property"). The subgroup of $GL(2n)$ that additionally stabilizes $\gamma$ is $SU(n)$.

In one source an $SU(3)$ structure on a manifold $M$ is given by specifying a $2$-form $\omega$ and a $3$-form $\gamma$ so that in any point $p \in M$ there exists an element in the frame bundle $L \in GL(M)_p$ that pulls back $\omega$ and $\gamma$ to the forms above.

Question: A priori, this should not be an $SU(3)$-structure, because the stabilizer of the forms $\omega$ and $\gamma$ from above is bigger than $SU(3)$. Is this correct?

Is it maybe assumed that the manifold is a Riemannian manifold, and that $L \in GL(M)_p$ also pulls back the Riemannian metric to the $g$ defined above? I can see that this would then be an $SU(3)$-structure. Or is this a specialty of (complex) dimension 3, that here the stabilizer of $\omega$ and $\gamma$ is already equal to $SU(3)$?

The answer is: A carefully chosen $3$-form and $2$-form in fact do have stabiliser $SU(3)$.
Let $V$ be a real vector space of dimension 6 over $\mathbb{R}$ and let $e^1$, $\dots$, $e^6$ be a basis of $V^*$. Define $$\phi = e^1 \wedge e^3 \wedge e^5- e^1 \wedge e^4 \wedge e^6- e^2 \wedge e^3 \wedge e^6- e^2 \wedge e^4 \wedge e^5.$$ (This is $Re (dz^1 \wedge dz^2 \wedge dz^3)$ after choosing a suitable complex structure on $V$) The oriented stabiliser of $\phi$ is $SL(3,\mathbb{C})$, i.e. $$GL_+(V) \cap Stab_{GL(V)} \phi = SL(3,\mathbb{C}).$$ (A proof for this can be found in the appendix of R. Bryant: On the geometry of almost complex 6-manifolds) If $\sigma \in \Lambda^2 V^*$ is non-degenerate, it has stabiliser $Sp(6)$ the endomorphism which stabilise both forms are $Sp(6) \cap SL(3,\mathbb{C})=SU(3)$.


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On the Relative Motion of the Earth and the Luminiferous Ether

 On the Relative Motion of the Earth and the Luminiferous Ether  (1887)
 American Journal of Science, 1887, 34 (203): 333–345, Online

The discovery of the aberration of light was soon followed by an explanation according to the emission theory. The effect was attributed to a simple composition of the velocity of light with the velocity of the earth in its orbit. The difficulties in this apparently sufficient explanation were overlooked until after an explanation on the undulatory theory of light was proposed. This new explanation was at first almost as simple as the former. But it failed to account for the fact proved by experiment that the aberration was unchanged when observations were made with a telescope filled with water. For if the tangent of the angle of aberration is the ratio of the velocity of the earth to the velocity of light, then, since the latter velocity in water is three-fourths its velocity in a vacuum, the aberration observed with a water telescope should be four-thirds of its true value.[1]

On the undulatory theory, according to Fresnel, first, the ether is supposed to be at rest except in the interior of transparent media, in which secondly, it is supposed to move with a velocity less than the velocity of the medium in the ratio $\scriptstyle{\frac{n^2-1}{n^2}}$, where n is the index of refraction. These two hypotheses give a complete and satisfactory explanation of aberration. The second hypothesis, notwithstanding its seeming improbability, must be considered as fully proved, first, by the celebrated experiment of Fizeau,[2] and secondly, by the ample confirmation of our own work.[3] The experimental trial of the first hypothesis forms the subject of the present paper.

If the earth were a transparent body, it might perhaps be conceded, in view of the experiments just cited, that the inter-molecular ether was at rest in space, notwithstanding the motion of the earth in its orbit; but we have no right to extend the conclusion from these experiments to opaque bodies. But there can hardly be question that the ether can and does pass through metals. Lorentz cites the illustration of a metallic barometer tube. When the tube is inclined the ether in the space above the mercury is certainly forced out, for it is imcompressible.[4] But again we have no right to assume that it makes its escape with perfect freedom, and if there be any resistance, however slight, we certainly could not assume an opaque body such as the whole earth to offer free passage through its entire mass. But as Lorentz aptly remarks: "quoi qu’il en soit, on fera bien, à mon avis, de ne pas se laisser guider, dans une question aussi importante, par des considérations sur le degré de probabilité ou de simplicité de l’une ou de l’autre hypothèse, mais de s’addresser a l’expérience pour apprendre à connaitre l’état, de repos ou de mouvement, dans lequel se trouve l’ether à la surface terrestre."[5]

In April, 1881, a method was proposed and carried out for testing the question experimentally.[6]

In deducing the formula for the quantity to be measured, the effect of the motion of the earth through the ether on the path of the ray at right angles to this motion was overlooked.[7] The discussion of this oversight and of the entire experiment forms the subject of a very searching analysis by H. A. Lorentz,[8] who finds that this effect can by no means be disregarded. In consequence, the quantity to be measured had in fact but one-half the value supposed, and as it was already barely beyond the limits of errors of experiment, the conclusion drawn from the result of the experiment might well be questioned; since, however, the main portion of the theory remains unquestioned, it was decided to repeat the experiment with such modifications as would insure a theoretical result much too large to be masked by experimental errors. The theory of the method may be briefly stated as follows:

Let sa, fig. 1, be a ray of light which is partly reflected in ab, and partly transmitted in ac, being returned by the mirrors b and c, along ba and ca. ba is partly transmitted along ad,

and ca is partly reflected along ad. If then the paths ab and ac are equal, the two rays interfere along ad. Suppose now, the ether being at rest, that the whole apparatus moves in the direction sc, with the velocity of the earth in its orbit, the directions and distances traversed by the rays will be altered thus:— The ray sa is reflected along ab, fig. 2; the angle bab, being equal to the aberration =a, is returned along ba/, (aba/ =2a), and goes to the focus of the telescope, whose direction is unaltered. The transmitted ray goes along ac, is returned along ca/, and is reflected at a/, making ca/e equal 90—a, and therefore still coinciding with the first ray. It may be remarked that the rays ba/ and ca/, do not now meet exactly in the same point a/, though the difference is of the second order; this does not affect the validity of the reasoning. Let it now be required to find the difference in the two paths aba/, and aca/.

Let V= velocity of light.

v= velocity of the earth in its orbit,
D=distance ab or ac, fig. 1.
T=time light occupies to pass from a to c.
T =time light occupies to return from c to a/, (fig. 2.)

Then $\scriptstyle{T=\frac{D}{V-v}}$, $\scriptstyle{T_{/}=\frac{D}{V+v}}$. The whole time of going and coming is $\scriptstyle{T+T_{/}=2D \frac{V}{V^2-v^2}}$, and the distance traveled in this time is $\scriptstyle{2D \frac{V^2}{V^2-v^2}=2D \left(1+\frac{v^2}{V^2}\right)}$, neglecting terms of the fourth order. The length of the other path is evidently $\scriptstyle{2D \sqrt{(1+\frac{v^2}{V^2}}}$ or to the same degree of accuracy, $\scriptstyle{2D \left( 1+\frac{v^2}{2V^2}\right)}$. The difference is therefore $\scriptstyle{D\frac{v^2}{V^2}}$. If now the whole apparatus be turned through 90°, the difference will be in the opposite direction, hence the displacement of the interference fringes should be $\scriptstyle{2D \frac{v^2}{V^2}}$. Considering only the velocity of the earth in its orbit, this would be $\scriptstyle{2D \times 10^{-8}}$. If, as was the case in the first experiment, $\scriptstyle{D =2\times 10^{6}}$ waves of yellow light, the displacement to be expected would be 0.04 of the distance between the interference fringes.

In the first experiment one of the principal difficulties encountered was that of revolving the apparatus without producing distortion; and another was its extreme sensitiveness to vibration. This was so great that it was impossible to see the interference fringes except at brief intervals when working in the city, even at two o'clock in the morning. Finally, as before remarked, the quantity to be observed, namely, a displacement of something less than a twentieth of the distance between the interference fringes may have been too small to be detected when masked by experimental errors.

The first named difficulties were entirely overcome by mounting the apparatus on a massive stone floating on mercury; and the second by increasing, by repeated reflection, the path of the light to about ten times its former value.

The apparatus is represented in perspective in fig. 3, in plan in fig. 4, and in vertical section in fig. 6. The stone a (fig. 5) is about 1.5 meter square and 0.3 meter thick. It rests on an annular wooden float bb, 1.5 meter outside diameter, 0.7 meter inside diameter, and 0.25 meter thick. The float rests on mercury contained in the cast-iron trough cc, 1.6 centimeter thick, and of such dimensions as to leave a clearance of about one centimeter around the float. A pin d, guided by arms gggg, fits into a socket e attached to the float. The pin may be pushed into the socket or be withdrawn, by a lever pivoted at f. This pin keeps the float concentric with the trough, but does not bear any part of the weight of the stone. The annular iron trough rests on a bed of cement on a low brick pier built in the form of a hollow octagon.

At each corner of the stone were placed four mirrors dd ee fig. 1. Near the center of the stone was a plane-parallel glass b. These were so disposed that light from an argand burner a, passing through a lens, fell on b so as to be in part reflected to d,; the two pencils followed the paths indicated in the figure, bdedbf and bd/e/d/bf respectively, and were observed by the telescope f. Both f and a revolved with the stone. The mirrors were of speculum metal carefully worked to optically plane surfaces five centimeters in diameter, and the glasses b and e were plane-parallel and of the same thickness. 1.25 centimeter; their surfaces measured 5.0 by 7.5 centimeters. The second of these was placed in the path of one of the pencils to compensate for the passage of the other through the same thickness of glass. The whole of the optical portion of the apparatus was kept covered with a wooden cover to prevent air currents and rapid changes of temperature.

The adjustment was effected as follows: The mirrors having been adjusted by screws in the castings which held the

mirrors, against which they were pressed by springs, till light from both pencils could be seen in the telescope, the lengths of the two paths were measured by a light wooden rod reaching diagonally from mirror to mirror, the distance being read from a small steel scale to tenths of millimeters. The difference in the lengths of the two paths was then annulled by moving the mirror e/. This mirror had three adjustments; it had an adjustment in altitude and one in azimuth, like all the other mirrors, but finer; it also had an adjustment in the direction of the incident ray, sliding forward or backward, but keeping very accurately parallel to its former plane. The three adjustments of this mirror could be made with the wooden cover in position.

The paths being now approximately equal, the two images of the source of light or of some well-defined object placed in front of the condensing lens, were made to coincide, the telescope was now adjusted for distinct vision of the expected interference bands, and sodium light was substituted for white light, when the interference bands appeared. These were now made as clear as possible by adjusting the mirror e/, then white light was restored, the screw altering the length of path was very slowly moved (one turn of a screw of one hundred threads to the inch altering the path nearly 1000 wavelengths) till the colored interference fringes reappeared in white light. These were now given a convenient width and position, and the apparatus was ready for observation.

The observations were conducted as follows: Around the cast-iron trough were sixteen equidistant marks. The apparatus was revolved very slowly (one turn in six minutes) and after a few minutes the cross wire of the micrometer was set on the clearest of the interference fringes at the instant of passing one of the marks. The motion was so slow that this could be done readily and accurately. The reading of the screw-head on the micrometer was noted, and a very slight and gradual impulse was given to keep up the motion of the stone; on passing the second mark, the same process was repeated, and this was continued till the apparatus had completed six revolutions. It was found that by keeping the apparatus in slow uniform motion, the results were much more uniform and consistent than when the stone was brought to rest for every observation; for the effects of strains could be noted for at least half a minute after the stone came to rest, and during this time effects of change of temperature came into action.

The following tables give the means of the six readings ; the first, for observations made near noon, the second, those near six o'clock in the evening. The readings are divisions of the screw-heads. The width of the fringes varied from 40 to 60 divisions, the mean value being near 50, so that one division means 0.02 wave-length. The rotation in the observations at noon was contrary to, and in the evening observations, with, that of the hands of a watch.

 Noon Observations. 16. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. July 8 44.7 44.0 43.5 39.7 35.2 34.7 34.3 32.5 28.2 26.2 23.8 23.2 20.3 18.7 17.5 16.8 13.7 July 9 57.4 57.3 58.2 59.2 58.7 60.2 60.8 62.0 61.5 63.3 65.8 67.3 69.7 70.7 73.0 70.2 72.2 July 11 27.3 23.5 22.0 19.3 19.2 19.3 18.7 18.8 16.2 14.3 13.3 12.8 13.3 12.3 10.2 7.3 6.5 Mean 43.1 41.6 41.2 39.4 37.7 38.1 37.9 37.8 35.3 34.6 34.3 34.4 34.4 33.9 33.6 31.4 30.8 Mean in w. l. .862 .832 .824 .788 .754 .762 .758 .756 .706 .692 .686 .688 .688 .678 .672 .628 .616 .706 .692 .686 .688 .688 .678 .672 .628 .616 Final mean .784 .762 .755 .738 .721 .720 .715 .692 .661 P. M. Observations. July 8 61.2 63.3 63.3 68.2 67.7 69.3 70.3 69.8 69.0 71.3 71.3 70.5 71.2 71.2 70.5 72.5 75.7 July 9 26.0 26.0 28.2 29.2 31.5 32.0 31.3 31.7 33.0 35.8 36.5 37.3 38.8 41.0 42.7 43.7 44.0 July 12 66.8 66.5 66.0 64.3 62.2 61.0 61.3 59.7 58.2 55.7 53.7 54.7 55.0 58.2 58.5 57.0 56.0 Mean 51.3 51.9 52.5 53.9 53.8 54.1 54.3 53.7 53.4 54.3 53.8 54.2 55.0 56.8 57.2 57.7 58.6 Mean in w. l. 1.026 1.038 1.050 1.078 1.076 1.082 1.086 1.074 1.068 1.086 1.076 1.084 1.100 1.136 1.144 1.154 1.172 1.068 1.086 1.076 1.084 1.100 1.136 1.144 1.154 1.172 Final mean 1.047 1.062 1.063 1.081 1.088 1.109 1.115 1.114 1.120

The results of the observations are expressed graphically in fig. 6. The upper is the curve for the observations at noon, and the lower that for the evening observations. The dotted curves represent one-eighth of the theoretical displacements. It seems fair to conclude from the figure that if there is any displacement

due to the relative motion of the earth and the luminiferous ether, this cannot be much greater than 0.01 of the distance between the fringes.

Considering the motion of the earth in its orbit only, this displacement should be $2D\tfrac{v^2}{V^2}=2D \times 10^{-8}$. The distance D was about eleven meters, or 2×107 wave-lengths of yellow light; hence the displacement to be expected was 0•4 fringe. The actual displacement was certainly less than the twentieth part of this, and probably less than the fortieth part. But since the displacement is proportional to the square of the velocity, the relative velocity of the earth and the ether is probably less than one sixth the earth's orbital velocity, and certainly less than one-fourth.

In what precedes, only the orbital motion of the earth is considered. If this is combined with the motion of the solar system, concerning which but little is known with certainty, the result would have to be modified; and it is just possible that the resultant velocity at the time of the observations was small though the chances are much against it. The experiment will therefore be repeated at intervals of three months, and thus all uncertainty will be avoided.

It appears, from all that precedes, reasonably certain that if there be any relative motion between the earth and the luminiferous ether, it must be small; quite small enough entirely to refute Fresnel's explanation of aberration. Stokes has given a theory of aberration which assumes the ether at the earth's surface to be at rest with regard to the latter, and only requires in addition that the relative velocity have a potential; but Lorentz shows that these conditions are incompatible. Lorentz then proposes a modification which combines some ideas of Stokes and Fresnel, and assumes the existence of a potential, together with Fresnel's coefficient. If now it were legitimate to conclude from the present work that the ether is at rest with regard to the earth's surface, according to Lorentz there could not be a velocity potential, and his own theory also fails.

Supplement.

It is obvious from what has gone before that it would be hopeless to attempt to solve the question of the motion of the solar system by observations of optical phenomena at the surface of the earth. But it is not impossible that at even moderate distances above the level of the sea, at the top of an isolated mountain peak, for instance, the relative motion might be perceptible in an apparatus like that used in these experiments. Perhaps if the experiment should ever be tried in these circumstances, the cover should be of glass, or should be removed.

It may be worth while to notice another method for multiplying the square of the aberration sufficiently to bring it within the range of observation, which has presented itself during the preparation of this paper. This is founded on the fact that reflection from surfaces in motion varies from the ordinary laws of reflection.

Let ab (fig. 1) be a plane wave falling on the mirror mn at an incidence of 45°. If the mirror is at rest, the wave front after reflection will be ac.

Now suppose the mirror to move in a direction which makes an angle a with its normal, with a velocity $\omega$. Let V be the velocity of light in the ether supposed stationary, and let cd be the increase in the distance the light has to travel to reach d. In this time the mirror will have moved a distance $\tfrac{cd}{\sqrt{2} cos \alpha }$. We have $\tfrac{cd}{ad}=\tfrac{\omega\sqrt 2\cos\alpha}{V}$ which put = r, and $\tfrac{ac}{ad} = 1 - r$.

In order to find the new wave front, draw the arc fg with b as a center and ad as radius; the tangent to this arc from d will be the new wave front, and the normal to the tangent from b will be the new direction. This will differ from the direction ba by the angle $\theta$ which it is required to find. From the equality of the triangles adb and edb it follows that $\theta = 2\phi$, $ab=ac$,

$\tan {adb} = \tan {\left(45^o - \frac{\theta}{2} \right)} = \frac{1-\tan{\frac{\theta}{2}}}{1+\tan{\frac{\theta}{2}}} = \frac{ac}{ad}=1-r$,

or neglecting terms of the order $r^2$,

$\theta=r+\frac{r^2}{2}=\frac{\sqrt{2}\omega \cos \alpha}{V}+\frac{\omega^2}{V^2}\cos^{2}\alpha$

Now let the light fall on a parallel mirror facing the first, we should then have $\theta_/ = \tfrac{-\sqrt{2} \omega \cos \alpha}{V}+\tfrac{\omega^2}{V^2}\cos^2\alpha$ and the total deviation would be $\theta+\theta_/=2\rho^2\cos^2\alpha$ where $\rho$ is the angle of aberration, if only the orbital motion of the earth is considered. The maximum displacement obtained by revolving the whole apparatus through 90° would be $\Delta=2 {\rho}^2 = 2 = 0\cdot004''$. With fifty such couples the displacement would be 0•2". But astronomical observations in circumstances far less favorable than those in which these may be taken have been made to hundredths of a second; so that this new method bids fair to be at least as sensitive as the former.

The arrangement of apparatus might be as in fig. 2; s in the focus of the lens a, is a slit; bb cc are two glass mirrors optically plane and so silvered as to allow say one-twentieth of the-light to pass through, and reflecting say ninety per cent. The intensity of the light falling on the observing telescope df
would be about one-millionth of the original intensity, so that if sunlight or the electric arc were used it could still be readily seen. The mirrors bb/ and cc/, would differ from parallelism sufficiently to separate the successive images. Finally, the apparatus need not be mounted so as to revolve, as the earth's rotation would be sufficient.

If it were possible to measure with sufficient accuracy the velocity of light without returning the ray to its starting point, the problem of measuring the first power of the relative velocity of the earth with respect to the ether would be solved. This may not be as hopeless as might appear at first sight, since the difficulties are entirely mechanical and may possibly be surmounted in the course of time.

For example, suppose (fig. 3) m and m/, two mirrors revolving with equal velocity in opposite directions. It is evident that light from s will form a stationary image at s/ and similarly light from s/ will form a stationary image at s. If now the velocity of the mirrors be increased sufficiently, their phases still being exactly the same, both images will be deflected from s and s/, in inverse proportion to the velocities of light in the two directions; or, if the two deflections are made equal, and the difference of phase of the mirrors be simultaneously measured, this will evidently be proportional to the difference of velocity in the two directions. The only real difficulty lies in this measurement. The following is perhaps a possible solution : gg/ (fig. 4) are two gratings on which sunlight is concentrated. These are placed so that after falling on the revolving mirrors m and m/, the light forms images of the gratings at s and s/, two very sensitive selenium cells in circuit with a battery and a telephone. If everything be symmetrical, the sound in the telephone will be a maximum. If now one of the slits s be displaced through half the distance between the image of the grating bars, there will be silence. Suppose now that the two deflections having been made exactly equal, the slit is adjusted for silence. Then if the experiment be repeated when the earth's rotation has turned the whole apparatus through 180°, and the deflections are again made equal, there will no longer be silence, and the angular distance through which s must be moved to restore silence will measure the required difference in phase.

There remain three other methods, all astronomical, for attacking the problem of the motion of the solar system through space.

1. The telescopic observation of the proper motions of the stars. This has given us a highly probably determination of the direction of this motion, but only a guess as to its amount.

2. The spectroscopic observation of the motion of stars in the line of sight. This could furnish data for the relative motions only, though it seems likely that by the immense improvements in the photography of stellar spectra, the information thus obtained will be far more accurate than any other.

3. Finally there remains the determination of the velocity of light by observations of the eclipses of Jupiter's satellites. If the improved photometric methods practiced at the Harvard observatory make it possible to observe these with sufficient accuracy, the difference in the results found for the velocity of light when Jupiter is nearest to and farthest from the line of motion will give, not merely the motion of the solar system with reference to the stars, but with reference to the luminiferous ether itself.

Notes

1. It may be noticed that most writers admit the sufficiency of the explanation according to the emission theory of light; while in fact the difficulty is even greater than according to the undulatory theory. For on the emission theory the velocity of light must be greater in the water telescope, and therefore the angle of aberration should be less; hence, in order to reduce it to its true value, we must make the absurd hypothesis that the motion of the water in the telescope carries the ray of light in the opposite direction!
2. Comptes Rendus, xxxiii, 349, 1851; Pogg. Ann. Ergänzungsband, iii. 457, 1853; Ann. Chim. Phys., III, lvii, 385, 1859.
3. Influence of Motion of the Medium on the Velocity of Light. This Journal, III, xxxi, 377, 1886.
4. It may be objected that it may escape by the space between the mercury and the walls; but this could be prevented by amalgamating the walls.
5. Archives Néerlandaises, xxi, 2me livr.
6. The Relative Motion of the Earth and the Luminiferous Ether, by Albert A. Michelson, this Jour., III xxii, 120.
7. It may be mentioned here that the error was pointed out-to the author of the former paper by M. A. Potier, of Paris, in the winter of 1881.
8. De l'Influence du Mouvement de la Terre sur les Phen. Lum. Archives Néerlandaises, xxi, 2me livr., 1886.

 This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1931, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.


text:
# Composite Function

• July 27th 2009, 05:38 AM
cloud5
Composite Function
How to do these?
The $e^x2-1$ is suppose to be e power of x then x power of 2-1, 2-1 is located above the x.

Find the function f
1. $g:x\rightarrow \ln x$ and $gf:x\rightarrow 2 \ln (x+1)$

2. $g:x\rightarrow e^x$ and $gf:x\rightarrow e^x2-1$

1. $f:x\rightarrow(x+1)^2$

2. $f:x\rightarrow x^2-1$
• July 27th 2009, 06:55 AM
artvandalay11
Quote:

Originally Posted by cloud5
How to do these?
The $e^x2-1$ is suppose to be e power of x then x power of 2-1, 2-1 is located above the x.

Find the function f
1. $g:x\rightarrow \ln x$ and $gf:x\rightarrow 2 \ln (x+1)$

2. $g:x\rightarrow e^x$ and $gf:x\rightarrow e^x2-1$

1. $f:x\rightarrow(x+1)^2$

2. $f:x\rightarrow x^2-1$

1.
$
g(x)=ln (x)
g(f(x))=2ln(x+1)=ln(x+1)^2$
by the rules for logs

So f(x) must be $(x+1)^2$ as g(?)= ln ? so the $?=f(x)=(x+1)^2$

2. $g(x)=e^x$
$g(f(x))=e^{x^2-1}$ so again g(?)= $e^{x^2-1}$ and you should be able to look at that and see that it's $x^2-1$
• July 27th 2009, 06:58 AM
skeeter
Quote:

Originally Posted by cloud5
How to do these?
The $e^x2-1$ is suppose to be e power of x then x power of 2-1, 2-1 is located above the x.

Find the function f
1. $g:x\rightarrow \ln x$ and $gf:x\rightarrow 2 \ln (x+1)$

2. $g:x\rightarrow e^x$ and $gf:x\rightarrow e^x2-1$

1. $f:x\rightarrow(x+1)^2$

2. $f:x\rightarrow x^2-1$

1. $g(\textcolor{red}{x}) = \ln{\textcolor{red}{x}}$

$g[f(x)] = 2\ln(x+1)$

$g[\textcolor{red}{f(x)}] = \ln{\textcolor{red}{(x+1)^2}}$

$f(x) = (x+1)^2$

2. $g(\textcolor{red}{x}) = e^{\textcolor{red}{x}}$

$g[\textcolor{red}{f(x)}] = e^{\textcolor{red}{x^2-1}}$

$f(x) = x^2-1$


text:
# How do I show that any finite-dimensional (absolute) CW-complex $X$ is locally contractible?

I know that it holds even if $$X$$ has infinite dimension, but I am looking for a specific argument in the finite-dimensional case.

• Many such questions boil down to the following trick. Problem: Given an open subset of the $n$-skeleton. Can we construct an open subset of the $n+1$-skeleton such that the intersection with the $n$-skeleton is the given set? So we choose attaching pushouts and we leave out the midpoints of the $n+1$- cells. Then we have a retraction of the $n+1$-skeleton without the midpoints to the $n$-skeleton. Now we can take the preimage of the given set. Nov 10 '20 at 19:34

see Proposition A.4 p. 523. I do not think there is any significant simplification taking $$X$$ finite-dimensional.


text:
# Analysis Seminar: Orr Shalit (Technion) " On the matrix range of random matrices"

Date:
Wed, 04/12/201912:00-13:00
Location:
Ross 70

Title: On the matrix range of random matrices
Abstract: In operator theory, one often attaches to an operator (or to a family of operators) certain invariants by which the operator can be understood. The spectrum $\sigma(A)$ of an operator $A$ is an obvious example. Another classical example, which is somewhat less familiar, is the numerical range of an operator: it is the set of all complex numbers obtained from the operator by applying a vector state to it. Unlike the spectrum, the notion of numerical range easily extends to families of operators: the numerical range of a d-tuple $A = (A_1, \ldots, A_d)$ is the collection of all tuples of complex numbers of the form $(\phi(A_1), \ldots, \phi(A_d))$, where $\phi$ is a state. It is a subset of the d-dimensional space, which carries some information about $A$.
My talk will be about a matrix-valued analogue of the numerical range, called the matrix range of an operator tuple. I will explain what is the matrix range, what it is good for, and then I will report on recent work in which we prove that there is a certain "universal" matrix range, to which the matrix ranges of a sequence of large random matrices tends to, almost surely. In particular, our results imply that the numerical range of a sequence of certain N-by-N random matrices tends in the Hausdorff metric to the closed disc as N goes to infinity, almost surely. The key novel technical aspect of our work is the continuity of the matrix range of a continuous field of operators, and a certain quantitative matrix valued Hahn-Babach type separation theorem.
Based on joint work with Malte Gerhold.


text:
Errata are corrected as found and incorporated into newly printed copies of the book. This means that not all the errata listed below exist in your copy of the book. If you see any errors not listed below, please let me know.

• p10. The coefficient for usageurban should be beta_3
• p29. The log-likelihood equation should have the final subscript i on the eta and not on the ‘e’.
• p36. Replace
gdf <- group_by(wcgs, cut(linpred, breaks=unique(quantile(linpred, (1:100)/100))))

with

gdf <- group_by(wcgs, ntile(linpred,100))

The original fails because the largest and smallest values get sorted into an NA category. The ntile function chooses the breakpoints so that there are a roughly equal number of observations per bin. We choose 100 bins. This should be varied according to the size of the dataset.

• p40. For the same reasons as the comment on p36, replace
gdf <- group_by(wcgsm, cut(linpred, breaks=unique(quantile(linpred, (1:100)/101))))

with

gdf <- group_by(wcgsm, ntile(linpred,100))
• p40. Last line of R code should read hldf <- summarise(gdf, y=sum(chd=='yes'), ppred=mean(predprob), count=n())
• p41. The degrees of freedom for Hosmer-Lemeshow statistic should be J-2 (not J-1)
• p41. The change in the binning as explained for p40. results in a change to the H-L statistic:
> c(hlstat, nrow(hldf))
[1] 127.99 100.00
> 1-pchisq(127.99, 100-2)
[1] 0.02262
• p42. The change in the H-L statistic as described on the previous page now means the result in statistically significant. Nevertheless, we are not inclined to change the model because we have no clear idea how to improve it and the deficiency is not serious. This change in the H-L statistic due to a change in the binning demonstrates a problem in with its stability. We are not convinced of its value.
• p55. The R code to compute the Pearson chi-squared statistic is missing. It is sum(residuals(lmod,type="pearson")^2)
• p64. There is a single zero response value in beta regression model in this example. As explained in the help page for betar() this causes some difficulties. If this point is excluded, then the fit becomes similar to the quasibinomial as claimed in the text.
• p170. The robust package has been superceded by the robustbase package.
• p188. The scale of the Gamma in the compound Poisson-Gamma distribution is $\lambda^{-1}=\phi(p-1)\mu^{p-1}$ if defined in the same sense as on p175.
• p198. In the first equation needs some added parentheses in the denominator as in: $(2\pi)^{n/2}$.
• p198. The last equation should read $K^Ty \sim N(0,\sigma^2 K^T V K)$
• p204. The behavior of lme4 has changed since publication. Now when the MLE of a variance component approaches the boundary of the parameter space (usually zero), it prints a warning message. Since this occurs frequently in the boostrap sampling, many warning messages are produced. But these are warnings and not an errors - the results are OK.
• p215. The refit() function in the current (April 2016) lme4 no longer works as expected. Replace:
rmodr <- refit(rmod, ryield)

with

rmodr <- lmer(ryield ~ treat + (1|blend), penicillin)

The replacement achieves the same intended purpose but is slightly less efficient than the original. - p378 LHS of formula for the likelihood ratio test statistic is missing a minus


text:
Question

# The rate of a reaction gets doubled when the temperature changes from $$7^{0}C$$ to $$17^{0}C$$. By which factor will it change for the temperature change from $$17^{0}C$$ to $$27^{0}C$$?

A
1.81
B
1.71
C
1.91
D
1.76

Solution

## The correct option is C 1.91$$log_{ 10 }\displaystyle \frac { k_{ 2 } }{ k_{ 1 } } =\displaystyle \frac { 2.303E_a }{ R } \left( \displaystyle \frac { T_{ 2 }-T_{ 1 } }{ T_{ 1 }T_{ 2 } } \right) \\ \displaystyle \frac { R\ log\ 2\times 280\times 290 }{ 2.303\times 10 } = E_a\\ log \displaystyle \frac { k_{ 4 } }{ k_{ 3 } } =\displaystyle \frac { 2.303\times R\ log\ 2\times 280\times 290 }{ 2.303\times 10\times R } \left( \displaystyle \frac { 10 }{ 290 \times 300 } \right) \\ log_{ 10 }\left(\displaystyle \frac { k_{ 4 } }{ k_{ 3 } } \right) =\displaystyle \frac { 280 }{ 300 } log\ 2\quad\\ \displaystyle \frac { k_{ 4 } }{ k_{ 3 }}=1.91$$Chemistry

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text:
# Is ERNIE output skewed by statistical tests?

ERNIE is a hardware random number generator used to generate winning Premium Bond numbers in the UK. Wikipedia says: "ERNIE's output is independently tested each month by an independent actuary appointed by the government, and the draw is only valid if the output passes tests that indicate it is statistically random." But doesn't it mean that, if ERNIE output was truly and fully random (i.e. generated from a unitary distribution over a set of integers from 1 to N), rejecting some of the draws would actually distort the randomness?

-

No.

The final 'it' in the quoted section refers to the word 'output' not to the word 'draw'. Thus the "independent actuary" tests the output of the machine to check that it is working properly. If so, the draw is made. They do not test the actual draw - that would be pointless!

Note: I know nothing about ERNIE or the process itself. I am inferring this from the wikipedia page together with a basic assumption - which may well be unfounded - that the British government couldn't be that stupid. Combining this with the fact that if the sentence is grammatically correct - perhaps another unfounded assumption - then the 'it' refers, as I said, to the 'output' not to the 'draw', I conclude that the 'output' and the 'draw' are two separate things.

-

Yes. In the same way that flipping a fair coin (with equal probabilities of getting heads of tails) eight times in a row is likely to come up all heads 1/256 times, or all tails 1/256 times. The psychological perception of a sequence with a run of 8 heads or 8 tails is that it is so unlikely as to never occur at all; whereas we mathematicians see the likelihood of a run of 8 in 8 flips as occuring with 2/256 or just under 1% of the time.

The opposite error is true, and also occurs with some frequency in biomedical experiments and medical experiments. The standard for accepting a result in a clinical or medical trial is for $p<0.05$: that there is less than a 5% probability that the results occured by chance. Thus, one in twenty times, it is possible that a random occurence or set of occurences will be perceived or accepted as being statistically valid when it is not.

But it also depends on how much data (how many draws) are in the sample being gauged for randomness. The smaller the sample size, the more likely you are to discard a valid but unreasonable appearing "true" random sequence. So my answer is really a qualified "maybe".

Shouldn't the validity or "true randomness" of the method be the gauge, along with a check to see that the algorithm is properly implemented? The problem, of course, with software is that bugs can creep into the implementation at any point:

• the compiler could be messed up, generating incorrect code,

• the ALU (the arithmetic logic unit) in the central processing unit could be incorrectly implemented, e.g. the Pentium chip had the floating-point co-processor which incorrectly calculated some particular multiply operations depending upon the operands,

• the algorithm specified in the program requirements may be correct, but may be implemented incorrectly,

• the algorithm may be correct for 32-bit integer math calculations, but incorrect if the system uses 64-bit arithmetic, or vice versa,

I wonder if the software code and hardware is vetted as well there as it is in Las Vegas by the Nevada Gaming Commision which oversees gambling and the electronic machinery for slots and electronic poker, etc.

-
So you're actually saying "YES" and "NO" based on how the actuary performs the output test? – ABh Aug 28 '10 at 20:58

I would think not, but there is not really enough information to be sure.

A) Has the output ever been rejected? If not then no bias!

B) Suppose the US government gave out prizes by a 3 step process.

1. a set of 10,000 9 digit numbers is generated.
2. The set is tested for statistical randomness
3. If it passes, find which are valid social security numbers and split the prize money among those people (or give each \$X) OTHERWISE go back to step 1.

That would be fair if the tests were not biased against particular individual numbers. Tests such as those described in http://en.wikipedia.org/wiki/Diehard_tests would be fine.

It seems as if maybe that is the kind of thing going on (based on a quick perusal of the Wikipedia article on ERNIE)

I suppose that in my example every individual would have a fair chance of winning but (if it is the case that the social security number indicates age and/or birthplace) there might be non-randomness in that a case of almost all winners born in UTAH might be rejected for bunching and thus never happen.

-
"Has the output ever been rejected? If not then no bias!" This is not true -- the bias (in the underlying distribution) is still there, it's just not detectable from the data. – JBL Aug 28 '10 at 0:12
(Of course my previous comment is based on the assumption that rejecting output actually does alter the underlying distribution.) – JBL Aug 28 '10 at 0:13
The question isn't that clearly defined. My, not that well expressed points were A) If rejection only happens on a machine malfunction or a 1 in 10^10 coincidence, then no distortion of randomness has happened in our lifetime (and any that did would be less than epsilon) B) Suppose a machine on each run generates 1000 huge numbers one after another. The machine is actually perfect but we reject a set if the numbers come out in increasing order (or with too few or too many inversions). Then we would reject valid runs but would have a uniform distribution on 1000 integer sets. – Aaron Meyerowitz Aug 28 '10 at 5:22
It also seems to depend strongly on exactly what sort of test or audit is going to be performed by the actuary to assess this RNG (random number generator). Measure the smoothness of the distribution? Cross-correlation? How many data items are in the output to be assessed? Is there absolute certainty that the machine performs the same way and utilizes exactly the same algorithm and code steps when it is creating output for testing as opposed to creating output to be an actual draw? Is the code itself audited? Has the code been verified to be correct and random by specific criteria? – ABh Aug 28 '10 at 21:03
Aaron, I don't think we disagree much (if at all). I just intended to point out that if the test being applied alters, say, every one in 1000! draws but does so in a way that destroys uniformity over the set of interest (so, not as in your example), then the result is biased, even though in the lifetime of the universe this bias will never result in any outcomes that differ from what would have happened if the selection were truly uniform. – JBL Aug 28 '10 at 22:58


text:
# Why does the chlorine atom in hydrogen chloride acquire a slightly negative charge?

And in the $H - C l$ molecule, chlorine polarizes electron density towards itself to give a $\stackrel{+ \delta}{H} - \stackrel{\delta -}{C} l$ dipole. And in aqueous solution, the polarization goes to completion to give hydronium and chloride ions......
$\stackrel{+ \delta}{H} - \stackrel{\delta -}{C} l \stackrel{{H}_{2} O}{\rightarrow} {H}_{3} {O}^{+} + C {l}^{-}$
What we write as ${H}_{3} {O}^{+}$ is a conception only, and the actual acidium species is probably ${H}_{7} {O}_{3}^{+}$ or similar.


text:
# Numerical Integration (Simpsons 1/3 Rule) implementation in c programming.

## Simpson's 1/3 Rule

#### Simpson’s 1/3 rule is based on approximating the integrand by a second order polynomial .It is an extension of Trapezoidal rule where, the trapezoidal rule was based on approximating the integrand by a first order polynomial, and then integrating the polynomial over interval of integration.

It calculates the value of the area under any curve over a given interval by dividing the area into equal parts. It follows the method similar to integration by parts.  It is used to estimate the value of a definite integral.

In order to integrate any function f(x) in the interval (a,b), follow the steps given below:
Ø  Select a value for n, which is the number of parts the interval is divided into. Let the value of n be an even number.
Ø  Calculate the width, h = (b−a) / n.
Ø  Calculate the values of x0 to xn as x0 = a, x1 = x0+ h, x n-1 = xn-2 + h, xn = b.
Ø  Consider y = f(x). Now find the values of y (y0 to yn) for the corresponding x(x0 to xn) values.
Ø  Substitute all the above found values in the Simpson's Rule Formula to calculate the integral value.

The code for Simpson’s 1/3 rule implementing in C is given below:
#include<stdio.h>
#include<conio.h>
float f(float x)
{
return(1/(1+x));
}
void main()
{
int i,n;
float x0,xn,h,y[20],so,se,ans,x[20];
printf("\n Enter values of x0,xn,h: ");
scanf("%f%f%f",&x0,&xn,&h);
n=(xn-x0)/h;
if(n%2==1)
{
n=n+1;
}
h=(xn-x0)/n;
printf("\n Refined value of n and h are:%d %f\n",n,h);
printf("\n Y values: \n");
for(i=0; i<=n; i++)
{
x[i]=x0+i*h;
y[i]=f(x[i]);
printf("\n %f\n",y[i]);
}
so=0;
se=0;
for(i=1; i<n; i++)
{
if(i%2==1)
{
so=so+y[i];
}
else
{
se=se+y[i];
}

}
ans=h/3*(y[0]+y[n]+4*so+2*se);
printf("\n Final integration is %f",ans);

getch();
}

Input & Output:

Application:
In numerical analysis, Simpson's rules are used in Ship Stability and naval architecture, to calculate the areas and volumes of irregular figures. This rule is also used in computer engineering .It provides the exact result for a quadratic function or parabola.

Numerical Integration (Simpsons 1/3 Rule).Simpson's 1/3 Rule implementation in c programming.


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
The contraharmonic mean of HSD matrices. (English) Zbl 0641.15009

The contraharmonic mean of two positive semidefinite Hermitian matrices A and B is defined by the relation $C\left(A,B\right)=A+B-2\left(A:B\right),$ where $A:B=A{\left(A+B\right)}^{-1}B$ is the so-called parallel addition introduced by W. N. Anderson jun. and R. J. Duffin [J. Math. Anal. Appl. 26, 576-594 (1969; Zbl 0177.049)]. The dual of the contraharmonic mean of A and B is given by ${C}^{\text{'}}\left(A,B\right)=C{\left({A}^{-1},{B}^{-1}\right)}^{-1}·$ It is shown that

${C}^{\text{'}}\left(A,B\right)=A:B+2\left(A:B\right)C{\left(A,B\right)}^{-1}\left(A:B\right)=\left(A{\left(A:B\right)}^{-1}A\right):\left(B{\left(A:B\right)}^{-1}B\right)·$

With the aid of the contraharmonic mean and its dual the authors study fixed point problems, the monotonicity behaviour of C(A,B), an infinite family of means for positive semidefinite Hermitian matrices that generalize C(A,B), inverse mean problems, and connections between C(A,B) and least square problems.

Reviewer: A.R.Kräuter

##### MSC:
 15A45 Miscellaneous inequalities involving matrices 15A24 Matrix equations and identities 15A27 Commutativity of matrices


text:
# Vector calculus answer check

1. Aug 12, 2008

### terryfields

i think i get this now but just checking as i have the exam tomorrow and wont do this part if im getting it wrong.

(2) (a) find a vector perpendicular to a=i+2j-2k and b=-2i+3j+5k
just use the cross product? and get 15i+j+7k

(b) (i) find the equation of the plane through position vector (2,1,1) and perpendicular to the vector(3,-1,2)

is this Plane II=T(3,-1,2) + (2,1,1)

determin the perpendicular distance of plane II from the orogin

no idea here??

(ii) Find the equation of the plane III through three points (2,-2,1)=a (4,-1,6)=b and (3,-3,2)=c i use a-b and a-c and then do the cross product of these two answers to get (6i,-3j,-3k) am i right in thinking that it doesnt matter which 2 i use to get this cross product e.g i could use b-c b-a etc to find the normal then i have the answer planeIII=T(6,-3,-3)+(2,-2,1) (plus any point)

(iii) find the angles between the 2 planes, i'll wait for some feedback before i attempt this, thanks

2. Aug 12, 2008

### Defennder

Yes correct.

The equation of any plane is given by $$\vec{n} \cdot \vec{P_0 P} = 0$$ where n is the normal vector to the plane, OP is the position vector (x,y,z) and P0 is any reference point on the plane. The perpendicular distance from the plane can be found by assuming that the normal vector can be scaled by some constant so that it "touches" the origin.

You're right it does not matter, but make sure the two vectors you use are not parallel or anti-parallel (why?).

3. Aug 12, 2008

### terryfields

sorry i dont understand this second part what is the vector P0P? dont understand what i have to do with the normal vector here to find the equation of the plane?

4. Aug 12, 2008

### terryfields

and so ive got the 2 plane equations wrong right?

5. Aug 12, 2008

### Defennder

The P0 simply refers to any point on the plane taken as a reference. For example if it is given that the point (1,2,3) lies on the plane, you can let P0 be the reference point.

That is equation of a line, if I interpret your T to mean the parameter which varies. The vector equation of a plane has two parameters, not one. You're not required to express the equation of a plane in vector form, are you? If not just stick to the x,y,z notation alone.

6. Aug 12, 2008

### terryfields

so on this b(i) how do i go about finding (a) any point on the plane and (b) the normal vector as the only information i am given is a position vector (2,1,1,) and a perpendicular vector (3,-1,2)

7. Aug 12, 2008

### Defennder

The position vector given is a point on the plane. As for the normal vector, what is definition of "normal"?

8. Aug 12, 2008

### terryfields

but i have to do N.P0P doesnt that mean N dot product with the distance between the position vector and any point on the plane? if so isnt the distance between the position vector and itself going to be zero? as for the normal vector isnt that the opposite of the perpendicular? but i have no clue how to find it

9. Aug 12, 2008

### Defennder

Yes it will equate to zero. That is how you get an equation. The normal vector to the plane is also perpendicular to the plane.


text:
# Homework Help: Electricity field, finding charge

1. Jan 8, 2010

### cogman

1. The problem statement, all variables and given/known data
A charged particle of mass m is suspended on a massless-string in the presence of a uniform electric field. When the electric field is $$\vec{E} = \lambda \hat{i} + \mu \hat{j} N/C$$ the ball is in equilibrium at $$\angle \theta$$ ($$\lambda > 0$$ and $$\mu > 0$$)

Determine the charge q on the object in terms of m, g, $$\lambda, \mu, and \theta$$

2. Relevant equations
$$F_{g} = m * g$$
$$F_{e} = \vec{E} * q$$

3. The attempt at a solution
I'm not to confident in what I have (hence the reason I'm posting here) but I believe it looks something like this.

First I would set up my coordinate system such that the x axis is perpendicular to the string and the y axis is parallel. With that, I could say that the two main forces acting on the particle are that of gravity and the electric field. Thus

$$F_{g} = F_{e}$$
$$mg * (cos(\theta)\hat{i} + sin(\theta)\hat{j}) = E * q$$
$$mg * (cos(\theta)\hat{i} + sin(\theta)\hat{j}) = (cos(\theta)(\lambda + \mu)\hat{i} + sin(\theta)(\lambda + \mu)\hat{j})q$$

And that is about as far as I get before I start panicking. The major issue is, you can't divide vectors, so this is obviously has some serious flaw to it.

2. Jan 8, 2010

### diazona

The electric field
$$\vec{E} = \lambda \hat{i} + \mu \hat{j}$$
is a given quantity, so all parts of it are specified as part of the problem - including the unit vectors $\hat{i}$ and $\hat{j}$. The point is, since those are given, you can't just arbitrarily decide that, say, $\hat{j}$ points along the string. You have to use the coordinate system you're given. (Well, you could always rotate the whole problem into another coordinate system, but that would be completely unnecessary here.)

3. Jan 11, 2010

### cogman

Ok, taking your advice (sorry for the delay in posting) here is my second stab at it.

$$F_{e} + F_{g} + F_{T} = 0$$
$$F_{e} = q\lambda \hat{i} + q\mu \hat{j}$$
$$F_{g} = -mg \hat{j}$$

We Should be able to say that

$$F_{T}\hat{i} = -q\lambda \hat{i}$$
because gravity doesn't contribute anything in the $$\hat{i}$$ direction. For the $$\hat{j}$$ we would say
$$F_{T}\hat{j} = -q\mu \hat{j} + mg\hat{j}$$

With this in mind, we can bring up the formula that

$$|F_{T}|*sin(\theta) = |F_{g}\hat{j} + F_{e}\hat{j}|$$

For this, we need to find $$|F_{T}|$$, given our previous definition, that is simply

$$\sqrt{(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}}$$

thus

$$\sqrt{(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}} * sin(\theta) = \sqrt{(mg)^{2} + (q\lambda)^{2}}$$

From there, it is just algebra to solve for q. Squaring everything, I get

$$(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}) * sin^{2}(\theta) = (mg)^{2} + (q\lambda)^{2}$$

$$(q\lambda)^{2} + (q\mu)^{2}) * sin^{2}(\theta) - (q\lambda)^{2} = (mg)^{2} - (mg)^{2} * sin^{2}(\theta)$$

$$q^{2}(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2}) = (mg)^{2} - (mg)^{2} * sin^{2}(\theta)$$

$$q^{2} = \frac{((mg)^{2} - (mg)^{2} * sin^{2}(\theta))}{(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2})}$$

so for a final answer it would look something like

$$q = \sqrt{\frac{((mg)^{2} - (mg)^{2} * sin^{2}(\theta))}{(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2})}}$$

noting that there is no longer a - charge as the problem stated that mu and lambda are always positive.

Did I do the math right and make the correct assumptions?

4. Jan 11, 2010

### vitruvianman

see the papers of myron evans (alpha inst. for adv. lrn.)

5. Jan 11, 2010

### cogman

Thanks, I'll probably check it out later, however, Theoretical physics is probably not what my teacher wants to see in an introductory Electro-magnetic physics course.

I'm mostly just trying to confirm that I've done the math correctly (and if not, where did I go wrong)

6. Jan 12, 2010

### cogman

Turns out I'm an idiot. :)

long story short, I was trying to take the magnitude of a vector in the i and j direction by trying to square each component of that i and j vector. You can't do that. the final answer comes to be something like

$$|F_{T}|sin(\theta) = q\lambda$$

$$|F_{T}|cos(\theta) = mg - q\mu$$

$$q\lambda cot(\theta) = mg - q\mu$$

$$q = \frac{mg}{\lambda cot(\theta) + \mu}$$

7. Jan 14, 2010

lloll