Tuesday, July 2, 2013

Tidal force Taylor series Newton's law of universal gravitation centrifugal force

http://marty-green.blogspot.ca/2012/01/quadrupole-field-tidal-forces-and.html

http://en.wikipedia.org/wiki/Tidal_force

Why I hate physics

Sunday, January 1, 2012

The Quadrupole Field: Tidal Forces and Particle Beams

People have an awful time eplaining the tides. Everyone knows the moon attracts the oceans, making them bulge out on one side: but how do you explain the second bulge on the wrong side of the earth? The tides go in and out twice a day, not once. How do you explain it?

The mistake people make is to think that it’s the pull of the moon that causes the tides. It is and it isn’t. There is absolutley no tidal force in a steady gravitational field. The tidal force occurs because the moon’s gravity is not a constant over the volume of the earth. It get’s weaker as you get farther. It’s the variational component of the steady field which is responsible for the tides. You can draw a picture of it easily:

The big grey arrow represents the average magnetic field in the vicinity of the earth, and it has absolutely no influence on the tides. The four little black arrows, representing the deviation from the steady field, are entirely responsible for the bulges in the ocean. You can see in the picture that the distortion of the ocean is exactly in line with the four small arrows. These arrows represent what is called the “quadrupole component” of the field, and they cause the tides.

It is very significant to notice that the up/down arrows are the same strength as the left-right arrows…at least I’ve drawn them that way. In real life, they are exactly half strength…because there are two more off-axis arrows in the actual field, representing the component in and out of the page. The TOTAL strength of the four off-axis arrows is equal to the strenght of the on-axis arrows. This is not an accident, and it’s not something that applies only to gravitational fields. It’s a basic property of all electric and magnetic fields that is also known as Gauss’s Law: flux in is equal to flux out.

I bring this up because it applies in a huge way to the Stern Gerlach experiment. Everyone, and I mean everyone explains Stern-Gerlach by saying the field is stronger near the pointy magnet and weaker near the flat magnet. Yes it is, but what about the off-axis component? It’s just like the moon’s gravity! In fact, it’s even more so, because for the moon, we have three-dimensional geometry, so the off-axis field lines are weaker by a factor of 2; but in the two-dimensional geometry of the long pointed magnets, the off-axis component (which everyone ignores!) is exactly as strong as the on-axis component. Just as I’ve drawn in the picture.

And just as the steady-state component of the moon’s gravitational field has absolutely no effect on the tides, why should the steady-state component of the magnetic field have any effect on the silver atoms? Everything that happens in Stern-Gerlach has to be explainable on the basis of the quadrupole component of the field. Why should it be different from the tides?

But if this is true, the implications are huge. Everybody says that the beam splits in two. Even Feynman talks about splitting the beam, and then taking one component of the split beam and putting it through a second Stern-Gerlach magnet rotated through 90 degrees; he says then it will split in two again. He even goes so far as to acknowledge that the experiment has never been done quite this way, but he doesn’t doubt that it is theoretically possible.

And why shouldn’t it be possible? Everyone knows the beam splits in two. But does it??? In the original Stern-Gerlach experiment, there was no pencil-shaped beam: the beam came out of an oven through a slit, not a pinhole, and the orientation of the slit was perpendicular to the axis of the magnet. The silver atoms came out in a fan-shape, and spread into two lines on the detection plate…not two dots, but two lines. If I am correct, and I can’t see why I wouldn’t be, there is no way to split a pencil beam into two dots. Because the quadrupole field has just as much splitting force in the left-right direction as it does in the up-down direction.

That's why you can't expect the beam to split into two dots. If you haven't been following my blog regularly, you might have missed my post from last month when I calculated the spreading of an actcual beam through the quadrupole field. Instead of two dots, you get a donut that looks like this:

That's the pattern for a beam that goes in with it's spins polarized in the up direction. For an unpolarized (random) beam, you just get a simple donut pattern.

There is no physical way to construct a magnetic field that gets stronger from top to bottom without at the same time have it show just as much variation in the left-right direction. So it is impossible to split a pencil-shaped beam into two dots. Yes, you can split a fan-shaped beam into two lines, but that is not at all the same thing. In particular, all of Feynmann’s thought-experiments that he works through in Vol. 3 Chapter 5…are based on a non-physical misinterpretation of the apparatus. Feynmann got it wrong. Oh, it’s all right for the point he’s making about adding amplitudes and transformation of basis states…all that is immaculate. But the actual experiment does not exist.

Tidal force

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Figure 1: Comet Shoemaker-Levy 9 in 1994 after breaking up under the influence of Jupiter's tidal forces during a previous pass in 1992.

The tidal force is a secondary effect of the force of gravity and is responsible for the tides. It arises because the gravitational force exerted by one body on another is not constant across its diameter. The nearest side is attracted more strongly than the farthest side. Thus, the tidal force is differential. Consider the gravitational attraction of the moon on the oceans nearest the moon, the solid earth and the oceans farthest from the moon. There is a mutual attraction between the moon and the solid earth which can be considered to act on its centre of mass. However, the near oceans are more strongly attracted and, since they are fluid, they approach the moon slightly, causing a high tide. The far oceans are attracted less. The attraction on the far-side oceans could be expected to cause a low tide but since the solid earth is attracted ( accelerated ) more strongly towards the moon, there is a relative acceleration of those waters in the outwards direction. Viewing the Earth as a whole, we see that all its mass experiences a mutual attraction with that of the moon but the near oceans more so than the far oceans, leading to a separation of the two.
In a more general usage in celestial mechanics, the expression 'tidal force' can refer to a situation in which a body or material (for example, tidal water, or the Moon) is mainly under the gravitational influence of a second body (for example, the Earth), but is also perturbed by the gravitational effects of a third body (for example, by the Moon in the case of tidal water, or by the Sun in the case of the Moon). The perturbing force is sometimes in such cases called a tidal force^[1] (for example, the perturbing force on the Moon): it is the difference between the force exerted by the third body on the second and the force exerted by the third body on the first.^[2]

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Explanation[edit]

Figure 2: The Moon's gravity differential field at the surface of the Earth is known (along with another and weaker differential effect due to the Sun) as the Tide Generating Force. This is the primary mechanism driving tidal action, explaining two tidal equipotential bulges, and accounting for two high tides per day. In this figure, the Moon is either on the right side or on the left side of the Earth (at center). The outward direction of the arrows on the right and left indicates that where the Moon is overhead (or at the nadir) its perturbing force opposes that between the earth and ocean. The inward direction of the arrows at top and bottom indicates that where the Moon is 90 degrees away from overhead. Its effect is perpendicual and does not change the strength of the Earth's attraction on the ocean.

When a body (body 1) is acted on by the gravity of another body (body 2), the field can vary significantly on body 1 between the side of the body facing body 2 and the side facing away from body 2. Figure 2 shows the differential force of gravity on a spherical body (body 1) exerted by another body (body 2). These so-called tidal forces cause strains on both bodies and may distort them or even, in extreme cases, break one or the other apart.^[3] The Roche limit is the distance from a planet at which tidal effects would cause an object to disintegrate because the differential force of gravity from the planet overcomes the attraction of the parts of the object for one another.^[4] These strains would not occur if the gravitational field were uniform, because a uniform field only causes the entire body to accelerate together in the same direction and at the same rate.

Effects of tidal forces[edit]

Figure 3: Saturn's rings are inside the orbits of its largest moons. Tidal forces oppose the material in the rings from coalescing gravitationally to form moons.^[5]

In the case of an infinitesimally small elastic sphere, the effect of a tidal force is to distort the shape of the body without any change in volume. The sphere becomes an ellipsoid with two bulges, pointing towards and away from the other body. Larger objects distort into an ovoid, and are slightly compressed, which is what happens to the Earth's oceans under the action of the Moon. The Earth and Moon rotate about their common center of mass or barycenter, and their gravitational attraction provides the centripetal force necessary to maintain this motion. To an observer on the Earth, very close to this barycenter, the situation is one of the Earth as body 1 acted upon by the gravity of the Moon as body 2. All parts of the Earth are subject to the Moon's gravitational forces, causing the water in the oceans to redistribute, forming bulges on the sides near the Moon and far from the Moon.^[6]
When a body rotates while subject to tidal forces, internal friction results in the gradual dissipation of its rotational kinetic energy as heat. If the body is close enough to its primary, this can result in a rotation which is tidally locked to the orbital motion, as in the case of the Earth's moon. Tidal heating produces dramatic volcanic effects on Jupiter's moon Io. Stresses caused by tidal forces also cause a regular monthly pattern of moonquakes on Earth's Moon.
Tidal forces contribute to ocean currents, which moderate global temperatures by transporting heat energy toward the poles. It has been suggested that in addition to other factors, harmonic beat variations in tidal forcing may contribute to climate changes.^[7]
Tidal effects become particularly pronounced near small bodies of high mass, such as neutron stars or black holes, where they are responsible for the "spaghettification" of infalling matter. Tidal forces create the oceanic tide of Earth's oceans, where the attracting bodies are the Moon and, to a lesser extent, the Sun.
Tidal forces are also responsible for tidal locking and tidal acceleration.

Categories of gravitational forces[edit]

There are four different ways to categorize the gravitational forces acting on the earth while it rotates around the sun. The first category is there are the tides resulting from the combined pull of the Moon and the Sun. The pull of the Moon and the Sun affects the movement of the atmosphere and oceans, as well as stress in the Earth's crust. Second category considers the changing positions of other planets in relation to the Earth. The third category of gravitational forces that affect tidal forces involves planetary motions affecting the Sun's circulation and solar activity. Fourth and final category considers orbital affects of those motions. Changes in the speed of the Earth's orbit may affect small movements of the Sun, so basically small periodic influences on the Earth's climate and tidal forces.^[8]

Mathematical treatment[edit]

For a given (externally-generated) gravitational field, the tidal acceleration at a point with respect to a body is obtained by vectorially subtracting the gravitational acceleration at the center of the body (due to the given externally-generated field) from the gravitational acceleration (due to the same field) at the given point. Correspondingly, the term tidal force is used to describe the forces due to tidal acceleration. Note that for these purposes the only gravitational field considered is the external one; the gravitational field of the body (as shown in the graphic) is not relevant. (In other words the comparison is with the conditions at the given point as they would be if there were no externally-generated field acting unequally at the given point and at the center of the reference body. The externally-generated field is usually that produced by a perturbing third body, often the Sun or the Moon in the frequent example-cases of points on or above the Earth's surface in a geocentric reference frame.).

Figure 4: Graphic of tidal forces. The top picture shows the gravity field of a body to the right, the lower shows their residual once the field at the centre of the sphere is subtracted; this is the tidal force. See Figure 2 for a more detailed version

Tidal acceleration does not require rotation or orbiting bodies; for example, the body may be freefalling in a straight line under the influence of a gravitational field while still being influenced by (changing) tidal acceleration.
By Newton's law of universal gravitation and laws of motion, a body of mass m a distance R from the center of a sphere of mass M feels a force $\vec F_g$ equivalent to an acceleration $\vec a_g$ , where:

$\vec F_g = - \hat r ~ G ~ \frac{M m}{R^2}$ . . . , and. . . $\vec a_g = - \hat r ~ G ~ \frac{M}{R^2}$ . . . ,

where $\hat r$ is a unit vector pointing from the body M to the body m (here, acceleration from m towards M has negative sign).
Consider now the acceleration due to the sphere of mass M experienced by a particle in the vicinity of the body of mass m. With R as the distance from the center of M to the center of m, let ∆r be the (relatively small) distance of the particle from the center of the body of mass m. For simplicity, distances are first considered only in the direction pointing towards or away from the sphere of mass M. If the body of mass m is itself a sphere of radius ∆r, then the new particle considered may be located on its surface, at a distance (R ± ∆r) from the centre of the sphere of mass M, and ∆r may be taken as positive where the particle's distance from M is greater than R. Leaving aside whatever gravitational acceleration may be experienced by the particle towards m on account of m's own mass, we have the acceleration on the particle due to gravitational force towards M as:

$\vec a_g = - \hat r ~ G ~ \frac{M}{(R \pm \Delta r)^2}$

Pulling out the R² term from the denominator gives:

$\vec a_g = - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{(1 \pm \Delta r / R)^2}$

The Maclaurin series of 1/(1 + x)² is 1 – 2x + 3x² – ..., which gives a series expansion of:

$\vec a_g = - \hat r ~ G ~ \frac{M}{R^2} \pm \hat r ~ G ~ \frac{2 M }{R^2} ~ \frac{\Delta r}{R} - \cdots$

The first term is the gravitational acceleration due to M at the center of the reference body $m$ , i.e., at the point where $\Delta r$ is zero. This term does not affect the observed acceleration of particles on the surface of m because with respect to M, m (and everything on its surface) is in free fall. When the force on the far particle is subtracted from the force on the near particle, this first term cancels, as do all other even-order terms. The remaining (residual) terms represent the difference mentioned above and are tidal force (acceleration) terms. When ∆r is small compared to R, the terms after the first residual term are very small and can be neglected, giving the approximate tidal acceleration $\vec a_t$ (axial) for the distances ∆r considered, along the axis joining the centers of m and M:

$\vec a_t$ (axial) $~ \approx ~ \pm ~ \hat r ~ 2 \Delta r ~ G ~ \frac{M}{R^3}$

When calculated in this way for the case where ∆r is a distance along the axis joining the centers of m and M, $\vec a_t$ is directed outwards from to the center of m (where ∆r is zero).
Tidal accelerations can also be calculated away from the axis connecting the bodies m and M, requiring a vector calculation. In the plane perpendicular to that axis, the tidal acceleration is directed inwards (towards the center where ∆r is zero), and its magnitude is $| \vec a_t$ (axial) $| /2$ in linear approximation as in Figure 2.
The tidal accelerations at the surface of planets in the Solar System are generally very small. For example, the lunar tidal acceleration at the Earth's surface along the Moon-Earth axis is about 1.1 × 10⁻⁷ g, while the solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about 0.52 × 10⁻⁷ g, where g is the gravitational acceleration at the Earth's surface. Modern estimates put the size of the tide-raising force (acceleration) due to the Sun at about 45% of that due to the Moon.^[9] The solar tidal acceleration at the Earth's surface was first given by Newton in the 'Principia'^[10]

Relation with centrifugal force[edit]

If a secondary body orbits a primary body, the forces that could tear the second body apart if its strength and internal gravity are not enough, are the tidal force and the "centrifugal force" associated with any rotation of the secondary body about its axis.^[citation

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