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Examples in Physics

Vectors and Analytical Geometry

Vectorial equation of a line

Problem

Derive the vectorial equation of a straight line which passes through two generic points, A and B.

Choose two concrete points A and B, and use that vectorial equation to obtain the parametric equation of the line.

Generate a 3D plot of this line.

Solution

The vectorial equation of a line is the equation which is satisfied by the position vector of any point on this line. We want to express this equation in terms of the position vectors of the points A and B. The parametric equation can then be constructed from the vectorial equation by taking the coefficients of the unit vectors.

1. To construct the vectorial equation of the line, let and be vectors pointing to A and B. The vector difference - is parallel to the line we want to represent.

(1)

Now let be the position vector of any point of this line; the vector is also parallel to the line so their cross product is equal to 0 (you can use &x, or the operator from the palette of Common Symbols)

(2)

This is the vectorial equation of the line which passes through A and B. The position vector of any point belonging to this line satisfies this equation.

2. To construct the parametric representation of this line, turn the generic points A and B into concrete points; that is, give values to the components of and . For example:

(3)

(4)

Regardless of the values of and , the position vector of a point is

(5)

Thus the vectorial equation for these particular points A and B is

(6)

Since the unit vectors [] are independent, their Coefficients in the vectorial equation constitute the parametric equations; you can get those coefficients by taking the scalar product

(7)

We now have a system of three equations in terms of three unknowns x, y, and z, but it still represents the equation of a line, so two of the variables can be expressed in terms of the third one, which acts as the parameter of the parametric equations. Thus, by solving the system you get the following parametric equations

Warning, solve may be ignoring assumptions on the input variables.
	(8)

The interpretation is: given a value for z, the values of x and y are those you see above.

3. To plot this parametric representation (see plots[spacecurve]), first select the curve's parameter:

(9)

Now construct the appropriate input for the spacecurve command, say for z from -4 to 4

(10)

Some options for improving the visualization are

(11)

Vectorial equation of a plane

Problem

Derive the vectorial equation of a plane which passes through three generic points A, B, and C.

Choose three concrete points A, B, and C on this plane and plot it.

Solution

The vectorial equation of a plane is the equation which is satisfied by the position vector of any point on the plane.

1. To construct this equation, let , , and be the vectors pointing to A, B, and C, and let be the position vector of any point on this plane. The differences and are vectors parallel to the plane we want to represent, and also the difference between and any of , or is a vector parallel to the plane. So the equation of the plane can be obtained by first taking the cross product of any difference involving , and to construct a vector perpendicular to the plane (you can use &x, or the operator from the palette of Common Symbols)

(12)

then taking the scalar product of with any of the differences parallel to the plane involving and equating it to zero. So for instance, one way of writing this vectorial equation of the plane is

(13)

2. To plot this plane, turn the generic points A, B, and C into concrete points; that is, give values to the components of , and . For example:

(14)

(15)

(16)

For we always have

(17)

The vectorial equation for these particular points A, B, and C is thus:

(18)

To verify that the surface represented by this equation contains the points A, B, and C, you can substitute the values of the coordinates of these points and check if the equation is satisfied. For example, these are the coordinates of the first point:

(19)

The following shows that, for these points , the equation is satisfied:



	(20)

That this surface is a plane is clear from the fact that Eq is linear in all of x, y, and z. One way of plotting this plane is with the command implicitplot3d.

(21)

Vectorial equation of a plane tangent to a sphere of radius a

Problem

Derive the vectorial equation of a plane tangent to a sphere of radius a.

Solution

The vectorial equation of this plane is the equation satisfied by the position vector of any point on it.

Let represent the position vector of any point on the plane, be a vector pointing to the center of the sphere, and be a vector pointing to the point B where the plane is tangent to the sphere. So the difference is a vector on the plane, and the difference is a vector from the center of the sphere to the point of contact between the sphere and the tangent plane (that is, a vector perpendicular to the plane). Hence, these two vectors are perpendicular, and so their scalar product is equal to zero.

(22)

This is already the vectorial equation of the tangent plane, only it is not yet expressed in terms of the radius of the sphere. Now since is a vector from the center of the sphere to the point of contact with the plane, the norm of this vector is the radius a of the sphere.

(23)

Expand both expressions to use them together.

(24)

(25)

(26)

So simplify one with respect to the other one, eliminating (see simplify/siderels)

>	$simplify(`+`(Physics:-Vectors:-`.`(r_, B_), `-`(Physics:-Vectors:-`.`(r_, A_)), `-`(``(`^`(Physics:-Vectors:-Norm(B_), 2))), Physics:-Vectors:-`.`(B_, A_)) = 0, {`+`(``(`^`(Physics:-Vectors:-Norm(B_...$

(27)

After collecting terms in the above, the requested vectorial equation can be more compactly rewritten as .

(28)

(29)

As an exercise, consider choosing three concrete values for the positions of A, B, and the radius a of the sphere, then insert these values into the derived equation and plot the sphere and tangent plane together (see plots[display] to merge the plots).

Volume element of a sphere

Problem

Determine the infinitesimal volume element of a sphere, expressed in spherical coordinates.

Solution

Let be the vectorial equation, parameterized by u, v, and w, of a generic 3-D geometric object; in this case, we are dealing with a sphere of generic radius r. The volume element is derived from the equation as .

We want this volume element to be expressed in spherical coordinates (); we can always choose these coordinates themselves as parameters u, v, w. Therefore, we are interested in the explicit form of (note the use of %diff, the inert form diff, and for the cross product you can use &x, or the operator from the palette of Common Symbols)

(30)

The first step is to write the vectorial equation for a sphere of radius r; that is the equation satisfied by the position vector of any point of the sphere. In spherical coordinates, with the origin of the reference system at the center of the sphere, this vectorial equation has its simplest form where points to any point of the sphere, r is the radial coordinate (constant over the sphere) and is the radial unit vector. So

(31)

From which the value of

(32)

can be computed directly, using the value command

(33)

Alternatively, one could compute this result one step at a time by making it explicit that depends on and For this purpose, change the basis in the vectorial equation to the Cartesian basis (), where all the unit vectors are constant and so the partial derivatives can be performed directly.

(34)

So, the answer introduced in (30) becomes:

(35)

(36)

Hence the volume element is

Parametrization of Curves, Surfaces and Volumes

Consider the following representing a curve in space

(37)

>	$C := {y = Physics:-Vectors:-`^`(x, 2), z = 0};$

$Typesetting:-mprintslash([C := {y = `*`(`^`(x, 2)), z = 0}], [{y = `*`(`^`(x, 2)), z = 0}])$

(38)

The parametric equations for this curve are

(39)

The right-hand sides of equations above are the components of the position vector in cartesian coordinates, from where a vectorial form of these equations is

(40)

The curve can also be passed in vector form

(41)

(42)

The equations of circle of radius on the plane can be written as

>	$C := {z = 0, Physics:-Vectors:-`+`(Physics:-Vectors:-`^`(x, 2), Physics:-Vectors:-`^`(y, 2)) = Physics:-Vectors:-`^`(a, 2)};$

$Typesetting:-mprintslash([C := {z = 0, `+`(`*`(`^`(x, 2)), `*`(`^`(y, 2))) = 1764}], [{z = 0, `+`(`*`(`^`(x, 2)), `*`(`^`(y, 2))) = 1764}])$

(43)

(44)

In vector notation,

(45)

Changing the basis of this vector, for example to cylindrical, we get

(46)

The same result can be obtained by specifying the basis

(47)

To parametrize surfaces and volumes you can use ParametrizeSurface and ParametrizeVolume, which work in the same way as ParametrizeCurve, only with two and three parameters respectively.

(48)

(49)

(50)

(51)

(52)

(53)

Integral Vector Calculus

Integral vector calculus is about computing path, surface and volume vector integrals. Those are integrals where the integrand is a scalar or vector function, and the computation is done from any description (algebraic, parametric, vectorial) of the region of integration - a path, surface of volume.

There are three kinds of line or path integrals:

Physics:-Vectors:-int(F, r_ = A_ .. B_)[path = C];

where and are points in space, the limits of integration, that belong to the curve over which the integral is performed. In the first integral, is a scalar function and the result of the integration is thus a vector. In the second and third integrals the integrand is a vector function, so that the dot product is a scalar and so is the integral, while in the third one is a vector and so is its integration over the region .

Likewise, there are three kinds of surface integrals

and two types of volume integrals

The line element in path integrals is expressed in terms of a parameter as

Using indexed notation for derivatives , the surface element in surface integrals is expressed in terms of parameters as

and the volume element in volume integrals as

The integrals in the three cases are computed by first expressing the integrand and the integration element in terms of the parameters using the parametric equations derived with ParametrizeCurve, ParametrizeSurface and ParametrizeVolume, then performing the vector product operations, then the integration.

Vectorial Path integrals

(54)

Consider the following scalar function , path and integration limits and

(55)

>	$C := {y = Physics:-Vectors:-`^`(x, 2), z = 0};$

$Typesetting:-mprintslash([C := {y = `*`(`^`(x, 2)), z = 0}], [{y = `*`(`^`(x, 2)), z = 0}])$

(56)

(57)

(58)

The line or path integral, shown here in inert form on the left-hand side and active, computed to the end on the right-hand side, is

(59)

The output on the right-hand side is a vector. Within int, to perform this integration the curve $C = {y = Physics:-Vectors:-`^`(x, 2), z = 0};$ is first parametrized using ParametrizeCurve

(60)

The output above is a sequence, first the parametric equations as an ordered list (order ) then the parameter, in this case . For the formulation of the integral to make sense, the limits of integration and must belong to this curve, i.e. satisfy the parametric equations for some value of the parameter, in this case and

(61)

(62)

To see the integral after being parametrized and before performing the integration you can use the option inert

(63)

Since there is a relation one-to-one between the integration limits and and the parameter's range, instead of indicating and you can also indicate the range of itself, getting the same result (7)

>	$(Int = Physics:-Vectors:-int)(F, r_, path = C, parameter = {t = 0 .. 1});$

(64)

When the parameter's range is passed, you can also use a shortcut notation, passing the second argument as an equation r_ = C, making more explicit that r_ is the vector representation of the region of integration , so the line element is the differential of the parametric equations of C. In this case indicating path = C is redundant and can be omitted.

>	$(Int = Physics:-Vectors:-int)(F, r_ = C, parameter = {t = 0 .. 1});$

(65)

The integration path and the limits of integration can be expressed in vector notation as well

(66)

(67)

(68)

(69)

Vectorial Surface and Volume integrals

The case of surface and volume integrals is analogous to that of line (path) integrals, but for two things: instead of one, there are two or three parameters, and instead of indicating integration limits, it is required that you indicate the parameter's ranges.

The following represents the surface of a sphere of radius centered at the origin

Error, (in Physics:-Assume) {} is a set without a property; it is not a valid assumption

(70)

(71)

From the definition of the vectorial surface element as

is a vector perpendicular to the surface of the sphere with magnitude equal to the radius . Hence, the vectorial surface integral of over the upper half of the sphere should be a vector perpendicular to the plane in the direction, with modulus equal to half the surface of a sphere specified by . Also, checking the form of the parametrization returned by ParametrizeSurface is useful to understand what the ranges for the parameters and need to be in order to represent the desired region.

(72)

From symmetry considerations, the integral of over the lower half of the sphere should have the same magnitude but opposite direction (), from where the integral over the whole sphere, so for , should be equal to 0

(73)

From this example we see that to get the area computing the integral of over the whole surface it is necessary to take as integrand the modulus of , that is its scalar product with a unit vector parallel to it. By definition of surface element, , so that the unit vector is given by

(74)

$Typesetting:-mfrac(Typesetting:-mrow(Typesetting:-mrow(Typesetting:-mi($

(75)

where is the vectorial form of the parametric equations of the surface. It is easy to see this vector is the radial unit vector . For that purpose you can use the option of ParametrizeSurface to get the vectorial form of the parametric equations for

(76)

Introduce this value of into the expression for and change basis to spherical

(77)

(78)

Putting all together, the area of a sphere of radius is given by this closed surface integral

(79)

In the case of the volume of a sphere, an algebraic representation of the region is

(80)

from where the volume of a sphere of radius is equal to the integral of the corresponding with parameters , and

(81)

Using ParametrizeVolume one can also see that the vectorial representation of this region is just the position vector written in spherical coordinates and basis

(82)

Vectors in Spherical Coordinates using Tensor Notation (advanced)

The following is a topic that appears frequently in formulations: given a 3D vector in spherical (or any curvilinear) coordinates, how do you represent and relate, in simple terms, the vector and the corresponding vectorial operations Gradient, Divergence, Curl and Laplacian using tensor notation?

I. The line element in spherical coordinates and the scale-factors

Start by setting the spacetime to be 3-dimensional, Euclidean, and use Cartesian coordinates







	(83)

In vector calculus,the line element is at the root of everything, and in Cartesian coordinates it has the simple form

(84)

To compute the line element in spherical coordinates, the starting point is the transformation

(85)

(86)

Since in are just symbols with no relationship to , start transforming these differentials using the chain rule, computing the Jacobian of the transformation . In this Jacobian J, the first line is

So in matrix notation,

Typesetting:-mrow(Typesetting:-mfenced(Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mtable(Typesetting:-mtr(Typesetting:-mtd(Typesetting:-mi(

(87)

To complete the computation of in spherical coordinates we can now use ChangeBasis, provided that we next substitute in the result, expressing the abstract objects in terms of .

In two steps:

(88)

The line element

(89)

This result is important: it gives us the so-called scale factors, the key that connects 3D vectors with the related covariant and contravariant tensors in curvilinear coordinates. The scale factors are computed from by taking the scalar product with each of the unit vectors , then taking the coefficients of the differentials (just substitute them by the number 1)

(90)

The scale factors are relevant because the components of the 3D vector and the corresponding tensor are not the same in curvilinear coordinates. For instance, representing the differential of the coordinates as the tensor , we see that the corresponding vector, the line element in spherical coordinates , is not constructed by directly equating its components to the components of , so

The vector is constructed multiplying these contravariant components by the scaling factors

This rule applies in general. The vectorial components of a 3D vector in an orthogonal system (curvilinear or not) are always expressed in terms of the contravariant components the same way we did in the line above with the line element, using the scale-factors , so that

where on the right-hand side we see the contravariant components and the scale-factors . Because the system is orthogonal, each vector component satisfies

The scale-factors do not constitute a tensor, so on the right-hand side we do not sum over j. Also, from

it follows that,

where on the right-hand side we now have the covariant tensor components .

This relationship between the components of a 3D vector and the contravariant and covariant components of a tensor representing the vector is key for translating vector-component to corresponding tensor-component formulas.

II. Transformation of contravariant and covariant tensors (can skip)

Define two representations for the same tensor: will represent A in Cartesian coordinates, while will represent A in spherical coordinates.

$Typesetting:-mprintslash([{A__c[j], A__s[j], Physics:-Dgamma[b], Physics:-Psigma[b], Physics:-d_[b], Physics:-g_[b, c], Physics:-LeviCivita[b, c, d], S[b], X[b]}], [{A__c[j], A__s[j], Physics:-Dgamma[...$

(91)

Transformation rule for a contravariant tensor

We know, by definition, that the transformation rule for the components of a contravariant tensor is , the same as the rule for the differential of the coordinates. Thus, the transformation rule from to computed using TransformCoordinates, should return the same relation as . However, the application of the command requires one to pay attention because, just like in , we want to isolate the Cartesian (not the spherical) components. This is like performing a reversed transformation. So we will use

where on the left-hand side we isolate the three components of A in Cartesian coordinates, and on the right-hand side we transform the spherical components from spherical (4^th argument) to Cartesian (3^rd argument), which according to the 5^th bullet of TransformCoordinates will result in a transformation expressed in terms of the old coordinates (here the spherical ). Expand things to compare with

(92)

We see that the transformation rule for a contravariant vector is, indeed, like the transformation for the differential of the coordinates.

Transformation rule for a covariant tensor

For the transformation rule for the components of a covariant tensor , we know by definition that it is , and so the same transformation rule for the gradient , where and so on. We can directly experiment with this by changing variables in the differential operators

(93)

This result, and the equivalent ones replacing x by y or z in the input above, can be computed in one go, in matrix and simplified form, using the Jacobian of the transformation computed after . Now we need to take the transpose of the inverse of J (because we are now transforming the components of the gradient )

rtable(1 .. 3, [Physics:-d_[x], Physics:-d_[y], Physics:-d_[z]], subtype = Vector[column]) = rtable(1 .. 3, [`+`(`*`(sin(theta), `*`(cos(phi), `*`(Physics:-d_[r]))), `/`(`*`(cos(phi), `*`(cos(theta), ...

(94)

The corresponding transformation equations relating the tensors and in Cartesian and spherical coordinates is computed with TransformCoordinates as in , just lower the indices on the left and right hand sides, i.e., remove the tilde ~

(95)

We see that the transformation rule for a covariant vector is, indeed, just like the transformation rule for the gradient.

Side Note: once the computation of these transformation rules is understood, we can also perform the inverse of as follows

(96)

III. Deriving the transformation rule for the Gradient using TransformCoordinates (can skip)

Turn the CompactDisplay notation for derivatives ON, so that the differentiation variable is displayed as an index:

The gradient of a function f, in Cartesian and spherical coordinates respectively, is given by

(97)

(98)

What we want now is to depart from in Cartesian coordinates and obtain in spherical coordinates using the transformation rule for a covariant tensor we previously computed with TransformCoordinates in . (An equivalent derivation which is simpler and has less steps is done in Sec. IV.)

Start by changing the vector basis in the gradient

(99)

We can see that in this result the coefficients of are the three lines in the right-hand side of after replacing the covariant components by the derivatives of f with respect to the j^th coordinate, shown here using indexed notation due to CompactDisplay being ON

(100)

(101)

So since is the inverse of , replace A by in (the formula computed using TransformCoordinates), then insert the result in to relate the gradient in Cartesian and spherical coordinates. We expect to arrive at the formula for the gradient in spherical coordinates .

$Typesetting:-mrow(Typesetting:-mfenced(Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mtable(Typesetting:-mtr(Typesetting:-mtd(Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mrow(Typesetting:-...$

(102)

(103)

Simplifying, we arrive at

(104)

(105)

IV. Deriving the transformation rule for the Divergence, Curl, Gradient and Laplacian, using Covariant derivatives

The Divergence

Introducing the vector A in spherical coordinates, its Divergence is given by

(106)



	(107)

(108)

We want to see how this result, , can be obtained departing from a tensorial representation of the object, this time the covariant derivative , and then using TransformCoordinates. For that purpose, first transform the coordinates and the metric introducing nonzero Christoffel symbols





	(109)

Side Note: despite having nonzero Christoffel symbols the space still has no curvature as all the components of the Riemann tensor are equal to zero

(110)

Now consider the divergence of the contravariant tensor, computed in tensor notation

(111)

(112)

Side Note: the covariant derivative expressed using the D_ operator can be rewritten in terms of the non-covariant d_ and Christoffel symbols as follows

(113)

Summing over the repeated indices in , we have

(114)

How is this related to the expression of the in ? The answer is in the relationship established at the end of Sec I between the components of the tensor and the components of the vector , namely that the vector components are obtained by multiplying the contravariant tensor components with the scale-factors . So in the above we need to substitute the contravariant with the vector components divided by the scale-factors

(115)

(116)

(117)

Comparing with , we see these two expressions are the same:

(118)

The Curl

The Curl of the the vector in spherical coordinates is given by

(119)

One could think that the expression for the Curl in tensor notation is the same as in a non-curvilinear system

But in a curvilinear system is not a tensor, so instead we need to use the non-Galilean form , where is the determinant of the metric. Moreover, since the expression has one free covariant index (the first one), to compare with the vectorial formula this index also needs to be rewritten as a vector component, as discussed at the end of Sec. I, by using

The formula for the vectorial Curl is thus expressed using tensor notation as

(120)

(121)

after which we replace the contravariant tensor components with the vector components using . Proceeding the same way we did with the Divergence, expand this expression. We could use TensorArray, but Library:-TensorComponents places a comma between components which makes things more readable in this case

(122)

Now replace the components of the tensor with the components of the 3D vector using

(123)

(124)

We see that these are exactly the components of the Curl

(125)

The Gradient

Once the problem is fully understood, it is easy to redo the computations of Sec.III for the Gradient, this time using tensor notation and the covariant derivative. In tensor notation, the components of the Gradient are given by the components of the right-hand side

(126)

where on the left-hand side we have the vectorial Nabla differential operator and on the right-hand side, since is a scalar, the covariant derivative becomes the standard derivative .

(127)

The above is the expected result

(128)

The Laplacian

Likewise we can compute the Laplacian directly as

(129)

In this case there are no free indices nor tensor components to be rewritten as vector components, so there is no need for scale-factors. Summing over the repeated indices,

(130)

Evaluating the Laplacian on the left-hand side,

$Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mrow(Typesetting:-mrow(Typesetting:-mrow(Typesetting:-msup(Typesetting:-mrow(Typesetting:-mi($

(131)

On the right-hand side we see the dAlembertian, in curvilinear coordinates; rewrite it using standard diff derivatives and expand both sides of the equation for comparison

(132)

This is an identity, the left and right hand sides are equal:

(133)

Mechanics

Static: reactions of planes and tensions on cables

Problem

A bar AB of weight w and length L has one extreme on a horizontal plane and the other on a vertical place, and is kept in that position by two cables AD and BC. The bar forms an angle with the horizontal plane and its projection BC over this plane forms an angle with the vertical plane. The cable BC is on the same vertical plane as the bar. Determine the reactions of the planes at A and B as well as the tensions on the cables.

Solution

There are two equations that contain information about the state of equilibrium of a system. The first one, assuming that the sum of the forces acting on the body are equal to zero, states that the center of mass of the body is not accelerated. The second one, assuming that the sum of the moments of the forces acting on the body (that is, the total torque) is zero, states that the rotation of the body around its center of masses unchanging (if it is not rotating, it stays that way). These two equations involve the reactions of the planes and the tensions on the cables, so from them we can obtain the solution to the overall problem. There is no friction so it is also clear that the reactions and are perpendicular to the planes, as shown in the figure, and the tensions and on the cables have direction AD and BC, respectively.

The steps to solve this problem are:

Determine each force acting on the bar and its application point

Equate the sum of the forces to zero.

Equate the sum of the moments to zero.

Solve these two vectorial equations for , , and , representing the reactions of the planes at the points of contact A and B and the tensions of the cables attached to the bar at A and B, respectively.

The forces acting on the bar are its weight and the reactions and tensions , , and . So the two equilibrium equations are:

(134)

(135)

where, in the input above, to enter the cross products you can use &x or the operator from the palette of Common Symbols. Set the origin and orientation of the reference system to project these vectors; any choice will do, but a good one will simplify the algebraic manipulations. We choose the origin at the point B, the vertical z axis in the direction of the reaction such that the y axis in the direction of , and the x axis in the remaining direction, anti-parallel to . With these selections, the vectors entering and are projected as follows:

(136)

where for simplicity we use abs to represent Norm, so that represents the norm of to be determined,

(137)

(138)

and where is to be determined. This reaction is applied to the bar at A, represented by ; its component along the x axis is obtained by projecting the segment BA onto the horizontal plane , resulting in BC, and then into the intersection of the two planes. So:

(139)

For the other vectors we have

(140)

(141)

where and are to be determined,

(142)

(143)

The two equilibrium equations now appear as

(144)

(145)

These two vectorial equations represent a system of six equations which can be obtained by equating each of the coefficients of , , and in each of the equations to zero; that is, taking the components of the vectorial equations along each axis:

(146)

(147)

So the system of equations to be solved is

(148)

The unknowns are

(149)

and the solution is

(150)

Work

Problem

A particle is submitted to a force whose Cartesian components are given by. Calculate the work done by this force when moving the particle along a straight line from the origin to a point ().

Solution

The work to be calculated is equal to the line integral of the force along the path (line) indicated:

where represents the integration domain.The steps to solve this problem are:

Determine the vectorial (parameter t) equation of the line over which we are going to integrate.

Express x, y, and z entering the components of in terms of the components of , after which the scalar product becomes dt and the work can be expressed as .

Compute the integral.

The work to be calculated is given by

(151)

The force acting on the particle is

(152)

The vectorial equation of a line where t is a parameter, is represented by the vector position of any of its points. This line passes through the origin and a generic point

(153)

In such a case, can be written directly as the product of the parameter t and the vector representing .

(154)

So t = 0 at the origin, and t = 1 when . To construct the scalar product , first express in terms of the vectorial equation of this line; that is, x, y, and z entering its components shall be taken from the components of .

(155)

(156)

Now compute .

(157)

So the line integral for the work W is now

(158)

When the particle is at we have t = 1, so the value of W to move the particle to is

(159)

(160)

Lagrangian for a pendulum

Problem

Determine the Lagrangian of a plane pendulum with a mass m in its extremity and a suspension point which:

a) moves uniformly over a vertical circumference with a constant frequency

b) oscillates horizontally on the plane of the pendulum according to .

Solution

a) A figure for this part of the problem is

The Lagrangian is defined as

(161)

where T and U are the kinetic and potential energy of the system, which is in this case constituted by a single point of mass m. The potential energy U is the gravitational energy

(162)

where g is the gravitational constant and the y axis is chosen to be along the vertical, pointing downwards, resulting in the gravitational force . The kinetic energy is:

$Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mn($

(163)

To compute this velocity, the position vector of the suspension point of the pendulum

(164)

must be determined. Choosing the x axis along the horizontal and the origin of the reference system at the center of the circle (see figure above), the x and y coordinates are given by:

(165)

(166)

(167)

(168)

This expression contains products of trigonometric functions, so a simplification consists of combining these products.

(169)

For the gravitational energy, expressed in terms of the parametric equations of the point of mass m, we have

(170)

So the requested Lagrangian is

(171)

(172)

Taking into account that the Lagrangian of a system is defined up to a total derivative with respect to t, we can eliminate the two terms that can be rewritten as total derivatives; these two terms are and so

(173)

_________________________________________________________________

b) The steps are the same as in part a:

(174)

(175)

$Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mn($

(176)

(177)

Now the only difference regarding part a) is in the expression of the y coordinate, which for this part b) is:

(178)

So the parametric equations in this case are

(179)

(180)

(181)

For the gravitational energy, expressed in terms of the parametric equations of the point of mass m, we have

(182)

So the requested Lagrangian is

(183)

(184)

(185)

The terms in L that can be expressed as total derivatives can be discarded are those involving and , so the Lagrangian is

(186)

Inertia tensor for a triatomic molecule

Problem

Determine the Inertia tensor corresponding to a triatomic molecule that has the form of an isosceles triangle with two masses in the extremes of the base and a mass in the third vertex. The distance between the two masses is equal to a, and the height of the triangle is equal to h.

Solution

The general formula for the Inertia tensor is given by (note the use of the abbreviation kd_ for KroneckerDelta)

(187)

where N is the number of particles, is the mass of each particle, is its position in a reference system with the origin at the "center of mass", is the component in the i^th direction of the position vector associated to the k^th particle in the reference system, and is the Kronecker delta, part of Physics. Set this definition of the InertiaTensor to be an indexing function for the InertiaTensor matrix.

(188)

For example, for a component in the diagonal, we have

(189)

and outside of the diagonal we have

(190)

At this point we can proceed to setting the particularities of this problem. The number of particles is 3.

(191)

Hence the matrix is

Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mtable(Typesetting:-mtr(Typesetting:-mtd(Typesetting:-mrow(Typesetting:-mi(

(192)

Two of the masses are equal

(193)

Now choose any system of reference (not at the center of mass) where we are going to project the position vectors of each atom as well as the center of mass . The vectors entering the definition of the Inertia tensor are related to and by

(194)

For , we choose a system of reference with the origin at the middle of the segment connecting the two atoms of mass equal to . Using Cartesian coordinates, we take the x axis along this segment and the z axis passing through the third atom of mass . So in this referential, the positions of the three atoms are

(195)

(196)

(197)

Indicate the real objects of this problem so that simplification steps further below can take that into account


Error, (in Physics:-Setup) invalid input: Physics:-Setup expects value for keyword parameter realobjects to be of type {"not given", Physics:-name, function, list({Physics:-name, function}), set({Physics:-name, function})}, but received {42, h, m[1], m[2]}

Compute the position of the "center of mass." By definition, it is

Typesetting:-mprintslash([R_[CM] := `/`(`*`(%sum(`*`(m[k], `*`(R_[k])), k = 1 .. 3)), `*`(%sum(m[k], k = 1 .. 3)))], [`/`(`*`(%sum(`*`(m[k], `*`(R_[k])), k = 1 .. 3)), `*`(%sum(m[k], k = 1 .. 3)))])

(198)

Evaluating these sums, we have the value of :

(199)

So these are the positions of the three particles viewed from the center of mass and expressed in terms of the known quantities , a and h:

(200)

The answer to the problem posed, that is the inertia tensor for this triatomic molecule, is now obtained by evaluating the abstract expression for the IT_Matrix at these values of the position vectors

$Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mtable(Typesetting:-mtr(Typesetting:-mtd(Typesetting:-mfrac(Typesetting:-mrow(Typesetting:-mn($

(201)

The equations of motion in curvilinear coordinates, tensor notation and Coriolis force

The formulation of the equations of motion of a particle in Cartesian coordinates using vector notation is simple. However, depending on the problem, it may be advantageous to use curvilinear coordinates and also tensor notation, for example when describing the motion of a particle as seen from a non-inertial system of reference (e.g. a rotating planet, like earth). When the particle's movement is observed from one such rotating referential, we also see "acceleration" that is not due to any force but to the fact that the referential itself is accelerated. The material below discusses and formulates these topics, and derives the expression for the Coriolis and centripetal force in cylindrical coordinates as seen from a rotating system of reference.

Cartesian coordinates

Vector notation

Generally speaking, the equations of motion of a particle are easy to formulate: the position vector is a function of time, the velocity is its first derivative, and the acceleration is its second derivative. For instance, in Cartesian coordinates

(202)

(203)

(204)

Newton's 2^nd law, that in an inertial system of reference when there is force there is acceleration and vice versa, is then given by

(205)

where represents the total force acting on the particle. This vectorial equation represents three second order differential equations which, given initial conditions, can be integrated to arrive at a closed form expression for .

Tensor notation

In Cartesian coordinates, the tensorial form of the equations is also straightforward. In a flat spacetime - Galilean system of reference - the three space coordinates form a 3D tensor, and so do its first and second derivatives. Set the spacetime to be 3-dimensional and Euclidean, and use lowercaselatin indices for the corresponding tensors







	(206)

The position, velocity, and acceleration vectors are expressed in tensor notation as in , and

(207)

(208)

(209)

Setting a tensor to represent the three Cartesian components of the force


$Typesetting:-mprintslash([{Physics:-Dgamma[b], F[j], Physics:-Psigma[b], Physics:-d_[b], Physics:-g_[b, c], Physics:-LeviCivita[b, c, d], X[b]}], [{Physics:-Dgamma[b], F[j], Physics:-Psigma[b], Physic...$	(210)

Newton's 2^nd law , now expressed in tensorial notation, is given by

(211)

The three differential equations represented by this tensorial form of are as expected

$Typesetting:-mprintslash([{F__x(t) = `*`(m, `*`(diff(x(t), `$`(t, 2)))), F__y(t) = `*`(m, `*`(diff(y(t), `$`(t, 2)))), F__z(t) = `*`(m, `*`(diff(z(t), `$`(t, 2))))}], [{F__x(t) = `*`(m, `*`(diff(diff(...$

(212)

Things are straightforward in Cartesian coordinates because the components of the line element are exactly the components of the tensor

(213)

and so the form factors (see related MaplePrimes post) are all equal to 1.

In the case of no external forces , and the equations of motion, whose solution is the trajectory, can be formulated as the path of minimal length between two points, a geodesic. In this case, as the spacetime is flat the geometry is Euclidean, as can be seen in the 3x3 identity metric matrix ; the previously mentioned two points lie on a plane, and the geodesic is a straight line. The differential equations of this geodesic are thus the equations of motion with , and can be computed using Geodesics

(214)

Curvilinear coordinates

Vector notation

In the case of curvilinear coordinates, for example with cylindrical or spherical variables, the form of these equations is obtained by performing a change of coordinates.

(215)

This change of coordinates leaves the z axis and its corresponding unit vector unchanged. By changing the basis and coordinates used to represent the position vector , the equation becomes

(216)

where, since in the coordinates depend on t, in not just and but also the unit vector depend on t. The velocity is computed as usual, by differentiating

(217)

The second term is due to the dependency of on the coordinate as well as the chain rule `and`(Physics:-Vectors:-diff(_rho(t), t) = Physics:-`*`(Physics:-Vectors:-diff(_rho, phi), Physics:-Vectors:-diff(phi(t), t)), Physics:-`*`(Physics:-Vectors:-diff(_rho, phi), Physics:-Vectors:-diff(ph... . The dependency of curvilinear unit vectors on the coordinates is automatically taken into account when differentiating due to the Setup setting geometricdifferentiation = true.

For the acceleration,

(218)

where the term involving comes from differentiating in while taking into account the dependency of on the coordinate This result can also be obtained by directly changing the variables in the acceleration in

(219)

Newton's 2^nd law becomes

(220)

In the absence of external forces, by equating the vector components to 0, that is the coefficients of the unit vectors of the acceleration , we get the system of differential equations for whose solution is the trajectory of the particle in cylindrical coordinates

(221)

$Typesetting:-mprintslash([{diff(phi(t), `$`(t, 2)) = `+`(`-`(`/`(`*`(2, `*`(diff(rho(t), t), `*`(diff(phi(t), t)))), `*`(rho(t))))), diff(rho(t), `$`(t, 2)) = `*`(rho(t), `*`(`^`(diff(phi(t), t), 2)))...$

(222)

In this formulation with , although (no acceleration in the direction) is naturally expected, the same cannot be said for the other two equations for and , as discussed below under Coriolis and Centripetal forces. The key observation at this point is that the right-hand sides of both unexpected equations involve , rotation around the z axis.

Tensor notation

Both and return the same equations when using tensor notation. When doing so, one can transform the position (168), velocity (169), and acceleration (170) tensors, but since these tensors are expressed as functions of a parameter (the time), it is simpler to only transform the underlying metric via TransformCoordinates. This has the advantage that all the geometrical subtleties of curvilinear coordinates, like scale factors and dependency of unit vectors on curvilinear coordinates, are automatically and succinctly encoded in the metric:





	(223)

The computation of geodesics assumes that the coordinates depend on a parameter. That parameter is passed as the first argument to Geodesics

(224)

These equations of motion are the same as the equations in which were computed using standard vector notation while differentiating and taking into account the dependency of curvilinear unit vectors on the curvilinear coordinates in and . One of the interesting features of computing with tensors is that, as previously mentioned, the geometrical subtleties of curvilinear coordinates are automatically encoded in the metric .

To understand how the geodesic equations are computed in one go in , one can perform the calculation in steps:

Make a function of directly in the metric

Compute - not the final form of the equations - but the intermediate form which expresses the geodesic equation using tensor notation, in terms of Christoffel symbols

Compute the components of this tensorial equation for the geodesic (using TensorArray)

For step 1, we have

Typesetting:-mrow(Typesetting:-msub(Typesetting:-mi(

(225)

Set this metric where





	(226)

For step 2, the geodesic equations in tensor notation with the coordinates depending on the time t are computed by passing the optional argument tensornotation

(227)

Step 3: compute the components of this tensorial equation

(228)

This is the expected result, the same as .

Having the tensorial equation is also useful for formulating the equations of motion in the presence of force. For this purpose, redefine the contravariant tensor to represent the force in the cylindrical basis


$Typesetting:-mprintslash([{Physics:-D_[b], Physics:-Dgamma[b], F[`~j`], Physics:-Psigma[b], Physics:-Ricci[b, c], Physics:-Riemann[b, c, d, e], Physics:-Weyl[b, c, d, e], Physics:-d_[b], Physics:-g_[b...$	(229)

Newton's 2^nd law

(230)

now using tensorial notation, becomes

(231)

Error, (in Physics:-TensorArray) free indices on both sides of the equation are different: found {~a} on the left-hand side and {~b} on the right-hand side

where we recall (see related MaplePrimes post) that, to obtain the vector components entering from these tensor components , we need to multiply the latter by the scale factors , and so the component of in the direction of is equal to .

Coriolis force and centripetal force

After changing variables the position vector in is expressed as

A distinction needs to be made here as to whether or not the unit vector depends on the time , with the former being the general case. When is a constant, the value of the coordinate (the angle between and the x axis) does not change; there is no rotation around the z axis. On the other hand, when , the orientation of , and thus the coordinate , changes with time, meaning that either the force acting on the particle has a component in the direction which produces rotation around the z axis or the system of reference itself is rotating around the z axis.

Likewise, the expression can also represent the position vector measured in the original Galilean (inertial) system of reference, where a force is producing rotation around the z axis, as well as the position of the particle measured in a rotating non-inertial system of reference. It follows that the transformation can also be interpreted in two different ways: representing the position of the particle in the original inertial system of reference, or representing a transformation from an inertial system of reference to a rotating non-inertial system of reference.

This equivalence between the trajectory of a particle subject to an external force as observed in an inertial system of reference, and the same trajectory with no external force observed from a non-inertial accelerated system of reference, is actually at the root of the formulation of general relativity, and is also well known in classical mechanics. The (apparent) forces observed in the rotating non-inertial system of reference, due only to its acceleration, are called Coriolis and centripetal forces.

The equations

that appeared in when in the inertial system of reference , are related to the Coriolis and centripetal forces in the non-inertial referential. To see this relation, following the presentation in [1], introduce a vector representing the rotation of the non-inertial referential around the z axis. When the non-inertial system rotates clockwise in the inertial system of reference, in the non-inertial system increases in value in the anti-clockwise direction.

(232)

According to [1], (39.7), the acceleration in the inertial system is expressed in terms of the quantities in the non-inertial rotating system by the sum of the following three vectorial terms. First, the components of the acceleration measured in the non-inertial system, which are given by the second derivatives of the coordinates multiplied by the scale factors, which in this case are given by (see this post in MaplePrimes)

(233)

Second, the Coriolis force divided by the mass, which is by definition given by

(234)

Third, the centripetal force divided by the mass, which is defined by

(235)

Adding these three terms,

(236)

where the right-hand side of is the same result computed in

(237)

So that

(238)

From this result, in the absence of external forces, and the acceleration measured in the rotating system is given by the sum of the Coriolis and centripetal accelerations

(239)

In conclusion: in the absence of external forces, the acceleration of a particle observed in a rotating (non-inertial) system of reference is not zero.

Expressing this equation in terms of we get