physics mechanics - lecture notes

8/14/2019 physics mechanics - Lecture Notes

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Physics 1. Mechanics

Math0. Useful mathematical formulae

Here you will find a (limited) number of useful mathematical formulae which you must know in

order to understand physics.

Let f = f(x) and g = g(x) and f = df/dx, . . . . Then

(f + g) = f + g, (f g) = fg + f g

Let f = f(y), and y = y(x), then

df

dx=

df

dy

dy

dx

Following two previous rules f

g

=

gf f g

g2

Differential:

df =

df

dx

dx

Taylor expansion

f(x) = f(x0) + (x x0)

df

dx

|x=x0

+ 12

(x x0)2

d2f

dx2

|x=x0

+ . . .

or for the infinite series

f(x) = f(x0) +i=1

1

n!(x x0)

n

dnf

dxn

|x=x0

Useful approximations for |x| 1

sinx x, cosx 1 x2/2, ex 1 + x. ln(1 + x) x, (1 + x) 1 + x

Let f = f(x, y). Partial derivative f/x is the derivative with respect to x while y is assumed

constant.

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Physics 1. Mechanics Math0

Total differential

df =

f

x

dx +

f

y

dy

First term of the Taylor expansion

f(x, y) = f(x0, y0) +

fx

|x=x0,y=y0

(x x0) +

fy

|x=x0,y=y0

(y y0)

Features of mixed derivatives for normal functions fx,y:

2f

xy=

2f

yx

Indefinite integral

fdx = f + C

Definite integral x2

x1

fdx = f(x2) f(x1)

Substitution. Let f = f(y) and y = y(x) then

f(y)dy =

f(y(x))

dy

dxdx

Integration by parts f gdx = f g

gfdx

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Lecture 0. Partial derivatives

1 Functions

Let x be some variable (like time), and for each x we measure another variable y which depends on x (like

a body temperature). The dependence of y on x is described by a function y = y(x). It is essential that x

be continuous, and that for each x from some interval x1 < x < x2 there exist a corresponding y.

2 Derivative

Let the argument x change from x1 to x2, x x2 x1. The function change is y = y(x2) y(x1). The

ratio y/x shows how fast the function changes on average when the argument changes on the interval

(x1, x2). Alternatively, let us choose a point x and shift from this point x. The above ratio then will be

written asy

x=

y(x + x) y(x)

x

as depends on x and x as well. If we now require x 0 (infinitesimal change) then the ratio (if it exists

at all) will show the local rate of change of the function y in the point x,

y dy

dx= lim

x0

y

x, (1)

and is called a derivative. In this notation dy/dx is not a ratio but a new function. However, it is obtained

by calculating the ratio of two infinitesimal values.

Starting with the obvious expression

y =

y

x

x

we can extrapolate this onto infinitesimal changes and obtain a differential:

dy = ydx =

dy

dx

dx, (2)

which is the change of y when x changes by infinitesimal dx 0 (but dx = 0 !).

Example: Let y(x) = xn, then (dy/dx) = nxn1, dy = nxn1dx.

Example: The function y = |x| does not have a derivative at x = 0.

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Physics 1. Mechanics Lecture 0

3 Partial derivatives

Let now y be a function of more than one variable xi, i = 1, 2, . . . , n (here n is the number of independent

variables). We define a partial derivative (y/xl) ofy with respect to the variable xl as a regular derivative,

taken as if y were function of xl only while other variables remain constant. In this case dy is produced by

the change of each xi independently, so that one has:

dy =

y

x1

dx1 +

y

x2

dx2 + . . . =

i

y

xi

dxi. (3)

Example: Let y = x21 + x1x2, then (y/x1) = 2x1 + x2, (y/x2) = x1, dy = (2x1 + x2)dx1 + x1dx2.

Q: Partially differentiate and write down the full differential for the following functions:

y = x21 + 2x1x22 + x

52, (4)

y = exp(x21 x22) [sin x1 + 2 cos(2x1x2)]. (5)

Each partial derivative is itself a function.

4 Some applications

4.1 Differentiation along a curve

Let y = y(x1, x2, . . .), and each argument is, in turn, a function of a single argument t: xi = xi(t). Eventually,

y is a function of t too, and one may ask what is the derivative (dy/dt). One way is to substitute all xi(t)

into the expression for y and, after an explicit form of y(t) is obtained, differentiate it as usual. This is not

always convenient. Instead, let us write the full differential dy in two forms:

dy =

dy

dt

dt, (6)

dy =i

y

xi

dxi (7)

However,

dxi =

dxidt

dt (8)

and therefore dy

dt

dt =

i

y

xi

dxi (9)

= [i

y

xi

dxidt

]dt (10)

dy

dt

=i

y

xi

dxidt

(11)

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4.2 Implicit function

Sometimes it is impossible to give an explicit form of a function y = y(x1, x2, . . .). Instead, and implicit

function is given in the form of an equation F(y, x1, x2, . . .) = 0, which is difficult or impossible to solve

with respect to y. In order to find derivatives of y we use again the full differential:

dF = 0 =

F

y

dy +

i

F

xi

dxi (12)

dy = [i

F

xi

dxi]/

F

y

(13)

y

xi=

F

xi

/

F

y

(14)

5 Taylor expansion

We use the full differential to approximate a function variation:

y =

y

x

x

dy

dx

x (15)

or, in a more appropriate form,

y(x) = y(x0) +

dy

dx

|x=x0(x x0) + . . . (16)

Here |x=x0 means that the derivative is evaluated in the point x = x0, and . . . means that the expression is

not exact and (hopefully) smaller terms should be added.For a function of many variables one has, respectively

y(xi) = y(xi,0) +i

y

xi

|xi=xi,0(xi xi,0) (17)

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Lecture 1. Coordinates

1 Coordinates. General principles

We start with the description of the motion of a point mass (point or particle ). First thing to do is

to learn how to describe the position of the point. In order to do that one needs coordinates. In the

simplest case of a motion along a line (not necessarily straight line) one needs to choose an origin O

and a method of assigning a coordinate, that is, a rule to establish a correspondence of each point of

the curve to a real number (positive of negative). In order to be useful the coordinate should change

continuously and, if possible, different points have to have different coordinates and each point has

to have only one coordinate.

O

Figure 1: Coordinate on a curve: one-dimensional space.

1.1 Examples

1.1.1 Straight line

Any point can be taken as the origin O. Let us choose the positive direction to the right (see Fig. 2),

and assign to each point a number corresponding to the distance from O with the sign + if the point

is to the right ofO, and if it is to the left. Let us denote this coordinate as x, then < x < .

Q: Is this choice unique ? Give an example of another choice. Try to find a coordinate x which

would change in the range 1 x 1 and yet cover the whole line.

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O x

Figure 2: Coordinates on a straight line

O

Figure 3: Coordinates on a circle.

1.1.2 Circle

A simple choice of the coordinate would be the angle measured counterclockwise from some radius

(see Fig. 3). In order to make this coordinate unique we have to require, for example, 0 < 2.

However, in this case the coordinate is not continuous at = 0. On the other hand, in many

applications it is more desirable to have the coordinate continuous. To do that we may say that the

coordinates and + 2n (where n is an arbitrary integer) describe the same point.

1.1.3 Closed curve with no intersections

Let us consider a simple closed curve which does not intersect itself. A simple coordinate choice is

the angle measured counterclockwise from some straight line intersecting the curve. In some cases

this coordinate may be ambiguous, and a better coordinate would be the curve length measured from

one chosen point in a chosen direction. In this case the coordinate is always positive.

2 Proceeding further

In the above examples we needed only one coordinate to describe a point position. In this case

it is said that the space is one-dimensional. If two-coordinates are necessary, the space is called

two-dimensional and ifn coordinates are needed, the space is called n-dimensional ofnD. The space

we live in is three-dimensional, that is, in general three coordinates are needed to describe a point

position. In what follows we shall denote a point as P and its coordinates as (xi), where i = 1, . . . , n.

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O

Figure 4: Coordinates on a closed curve with not self-intersections.

x1

x2

Figure 5: General 2D coordinates.

2.1 Examples of 2D coordinates

In the 1D case we had a single curve. In the 2D case we build two family of parallel curves (see

Fig. 5), one family for each coordinate. Parallel means that the curves from the same family do

not intersect. For each curve from one family the second coordinate is constant.

2.1.1 Cartesian coordinates

The two families are straight lines intersecting and 90. The coordinates (x, y) are chosen along each

line as in the 1D case of a straight line (distance with sign). The curves of x = const are straight

lines parallel to y-axis. The curves of y = const are straight lines parallel to x-axis.

2.1.2 Polar coordinates

One family is the radii coming out of the origin, the other is the circles with the origin as a center.

One coordinate, r, is a distance from the origin (nonnegative), the second is the angle as in the

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x

yP(x, y)

Figure 6: 2D Cartesian coordinates.

r

P(r, )

Figure 7: Polar coordinates.

case of a circle (see above). The curves of r = const are circles. The curves of = const are radii.

2.2 3D examples

For 3D case xi = const gives a surface.

2.2.1 3D Cartesian coordinates

Similar to the 2D case. x = x0 = const corresponds to a plane (2D space !) parallel to the y z

plane and crossing the x axis at x = x0.

2.2.2 Cylindrical coordinates

The coordinates (r,,z) are the polar coordinates (r, ) with the addition of the height z. r = const

are cylindrical surfaces around z-axis. z = const are planes parallel to x y plane.

Q: What are = const surfaces ?

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P(x , y , z)

y

x

z

Figure 8: 3D Cartesian coordinates.

P(r,,z)

r

z

Figure 9: Cylindrical coordinates.

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P(R, , )

R

Figure 10: Spherical coordinates.

2.2.3 Spherical coordinates

The coordinates (R, ,) are the distance from the origin, the angle measured downward from z-axis,

and the polar angle measured in the projection on the plane x y counterclockwise from x-axis.

Q: What are R = const, = const, and = const surfaces, respectively ? What are the

dimensions of the (sub)space r = const and = const ? What is this space ?

2.3 Relation between different coordinate systems

The same point can be assigned coordinates with the help of different coordinate systems, and it

is necessary to know the relation between those coordinates. In the following we assume that thecoordinate origin is the same for all chosen systems.

2.3.1 2D Cartesian - polar

The two systems are shown in the figure. It is accepted to choose x-axis as the base of the polar

coordinates. The following relation holds:

x = r cos ,

y = r sin .

(1)

2.3.2 3D

The three coordinate systems are shown in the same figure. One has

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x = r cos = R sin cos ,

y = r sin = R sin sin ,

z = z = R cos .

(2)

2.4 Distance element

Let P(xi) and P(xi + dxi), i = 1, 2, 3, be two (infinitesimally) close points. We shall denote the

distance between these points as ds and call it distance element. If the coordinates are cartesian, the

distance element is given by

ds2 = dx2 + dy2 + dz2, (3)

and such space is called Euclidean. The expression for the distance element will be different in

different coordinate systems. As an example we derive the distance element for spherical coordinates.

Let P(r,,) and P(r + dr, + d, + d) be to close points. If we knew dx,dy,dz in the cartesian

coordinates the distance element would be obtained from (3). However, in this case (r,,) are

independent variables, while (x ,y,z) are functions of these variables. Thus, we have to derive the

full differentials, according to the rule (??) and using the relations (2):

dx =

x

R

dR +

x

d +

x

d

=sin cos dR + R cos cos dR sin sin d,

(4)

and similarly for dy and dz. After substitution into (3) one gets

ds2 = dR2 + R2d2 + R2 sin2 d2. (5)

Q: Derive ds2

in cylindrical coordinates.

3 Curvilinear orthogonal coordinates

In all cases above the distance element had the form

ds2 =

i=1,2,3

h2i dx2

i (6)

Indeed, let us take the cylindrical coordinates with x1 = r, x2 = , and x3 = z. One has

ds2 = dr2 + r2d2 + dz2 (7)

that is,

h1 hr = 1, h2 = h = r, h3 hz = 1 (8)

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For spherical coordinates x1 = R, x2 = , x3 = one has

ds2 = dR2 + R2d2 + R2 sin 2d2 (9)

and

h1 hR = 1, h2 h = R, h3 h = R sin (10)

The coordinates where the distance element takes the form (6) are called orthogonal coordinates.

Cartesian coordinates with h1 = h2 = h3 = 1 are the special case of orthogonal coordinates. If atleast one of hi = 1 then the coordinates are called curvilinear coordinates.

Let us assume that we have previously introduced Cartesian coordinates (x1, x2, x3) = (x ,y,z)

but it would be more convenient to use some other orthogonal coordinates ( X1, X2, X3). We have to

derive the distance element with the use of these new coordinates. In order to do that we start with

ds2 = dx2 + dy2 + dz2 (11)

Unless we are given the relation between the new and old coordinates we cannot do anything further.

Thus, we assume that the functions

x = x(X1, X2, X3), y = y(X1, X2, X3), z = z(X1, X2, X3) (12)

are given. If so, we have

dx =x

X1dX1 +

x

X2dX2 +

x

X3dX3, (13)

dy =y

X1dX1 +

y

X2dX2 +

y

X3dX3, (14)

dx = zX1

dX1 +z

X2dX2 +

zX3

dX3, (15)

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Advanced 1. Curvilinear coordinates

1 Advanced material

Let us assume that we are interested in describing the same n-dimensional space with two different

coordinate systems, xi and x

i. For simplicity we assume that xi are cartesian, that is, ds2 =

idx2i .

Of course, there should be a relation between the two systems, that is, xi = xi(x

j), i = 1, . . . , n,j = 1, . . . , n. We are interested in the distance element which is ds2 =

i dx

2

i in the cartesian

coordinates. Now dxi is a full differential:

dxi =j

xixj

dxj , (1)

so that

ds2 = i

dx2i

=

i

j

xixj

dxj

k

xix k

dxk

=j,k

i

xixj

xix k

dxjdx

k

=jk

gjkdx

jdx

k

(2)

Now we can forget the initial cartesian coordinates and conclude that the most general form for the

distance element is

ds2 =i,j

gijdxidxj, (3)

where gij = gji. The construction gij is called a metric tensor. The metric tensor is symmetric

gij = gji, which means that it has 6 independent components in 3-dimensional space.

Q: How many independent components has gij in n-dimensional space ?

Q: Derive metric tensor for x = x y, y = y.

Q: Derive metric tensor for elliptical coordinates , , where 2 = x2/a2 + y2/b2, tan = y/x.

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Physics 1. Mechanics Advanced 1

In cartesian coordinates gij = ij, where

ij =

1 if i = j

0 if i = j(4)

In cylindrical coordinates (1 = r, 2 = , 3 = z) one has

g11 = 1, g22 = r2, g33 = 1,

g12 = g13 = g23 = 0.(5)

In spherical coordinates (1 = R, 2 = , 3 = )

g11 = 1, g22 = R2, g33 = R

2 sin2 ,

g12 = g13 = g23 = 0.

(6)

Coordinates, for which gij = 0 ifi = j are called orthogonal.

The space where it is possible to choose global (that is, which cover the whole space) cartesian

coordinates, such that gij = ij everywhere, is called Euclidean.

General remarks. The space ofspecial relativity is pseudo-Euclidean, since it is possible to choose

coordinates (1 = x, 2 = y, 3 = z, 4 = ct - time) so that g11 = g22 = g33 = 1, g44 = 1:

ds2 = dx2 + dy2 + dz2 c2dt2.

This space is called Minkowsky space. General relativity states that the space is locally Minkowskian.

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Lecture 2. Vectors

1 A simplest vector: Displacement

Let us assume that a particle starts moving from the point P1(x1, y1) and proceeds subsequently

to the point P2(x2, y2) and further to P(x3, y3) (we shall work in the two-dimensional space for

simplicity of graphical representation, see Fig. 1). As a result of the first move the coordinates

x

y

P1(x1, y1)

P2(x2, y2)

P3(x3, y3)

P1P2

P2P

3

P1P

2

Figure 1: Displacement vector.

change by (x2 x1, y2 y1). The same can be described by saying that the particle moved bythe distance

(x2 x1)2 + y2 y1)2 in certain direction. The two parameters (two coordinates or

distance and direction) completely determine the particle displacement from the initial point. We

shall say that a displacement vectorP1P2 is defined either as the two components (x2 x1, y2 y1)

or the distance+direction (see Fig. 2).The numbers x2 x1 and y2 y1 are called the vector x and y components, respectively. In

order to assign any physical meaning to this displacement vector we have to say what can be done

with it. First, let us notice thatP2P3 = (x3 x2, y3 y2) and P1P3 = (x3 x1, y3 y1) are also

displacement vectors, andP1P3 =

P1P2 +

P2P3 if we define the vector summation as the summation

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P1

P2

x1 x2

y1

y2

x

y

Figure 2: Displacement vector and coordinates.

of the corresponding components: x3

x1 = (x2

x1)+(x3

x2), and similarly for y. This means that

moving by some distance in some direction and afterwards by another distance to another direction

we as a result move by some third distance to some third direction, and the rule to find this resulting

move is known (and independent of the order of the two movements).

Let us denote the distance between P1 and P2 as |P1P2| and call that the magnitude of the vector.If we want to move from the point P1 in the same direction but by a distance |P1P2|, where is somereal number, we shall say that the movement is in the same direction, if > 0, and in the opposite

direction, if < 0. It is easy to see that the same can said as follows: P1P2 = ((x2x1), (y2y1)),

that is, multiplication of a vector by a number is multiplication of all its components by this number.

These two operations, vector summation and multiplication by a number, make the construction

something with direction meaningful. In fact, a construction may be called a vector only if these

two operations are properly defined.

2 Vectors more generally

In analogy with the displacement vector we shall call an object A a vector if a) it has several

components (3 in the three-dimensional space), b) the summation is defined, and c) multiplication by

a real number is defined. We shall write A = (Ax, Ay, Az) or A = (A1, A2, A3) or A = (Ai, i = 1, 2, 3).

Then, for each A and B, the vector sum C = A + B means that Ci = Ai + Bi, i = 1, 2, 3. For

each A and real , D = A means Di = Ai (we shall omit i = 1, 2, 3 where this is obvious). The

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x

y

P1

P2

P3

x1 x2

y1

y2

y3

x3

Figure 3: Multiplication by a number

important features of the summation and multiplication are the following:

A + B = B +A, (1)

(A+ B) = A+ B, (2)

( + )A = A+ A. (3)

The graphical representation of vectors would be the same as for the case of the displacement

vector. Fig. 4 shows the relation between the vector components (Ax, Ay) (REMEMBER!: it is the

same as (A1, A2) !), the vector magnitude |A| and the angleAx between the vector and x-axis for

the two-dimensional case:Ax = |A| cos(Ax), Ay = |A| sin(Ax). (4)

It is also clear that |A| =A2x + A2y.Geometrically vector summation is shown in Fig. 5.

3 Scalar product

Given two vectors A = (Ai) and B = (Bi) (see Fig. 6) one can construct a scalar

A B =i

AiBi. (5)

This construction is called scalar product or dot-product. Scalar is an object which does not depend on

the choice of coordinates. Let us check whether the construction (5) is indeed coordinate independent.

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x

y

Ax

Ay

A

Figure 4: Vector and components.

For simplicity we consider a two-dimensional case, for which A1

= |A| cos(Ax), A2 = |A| sin(Ax),B1 = |B| cos(Bx), and B2 = |B| sin(Bx). Substituting into (5), we have

A B = |A| |B| cos(AxBx), (6)or, noticing that AxBx = AB is the angle between A and B, one gets

A B = |A| |B| cos(AB). (7)In the right hand side of the last expression the vector magnitudes and the angle between the two

vectors are completely independent of the coordinate choice (coordinates or vector components do

not even appear there) which means that the scalar product is also coordinate independent, that is,

invariant. This is true not only in the two-dimensional space but in any dimensions.

Properties of the scalar product. From the very definition it is clear that

A B = B A, (8)(A) B = (A B), (9)A (B + C) = A B + A C. (10)

Vector magnitude (length). From the definition A A =i AiAi = i A2i = |A|2, so that|A| =

A A. (11)

This property allows to define a unit vector e such that e e = 1: for any nonzero vector A/|A|

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A

B

C

C= A + B

A

B

C

Figure 5: Vector summation.

will have a unit length. Of special use are the unit vectors along the axes of the cartesian coordinatesystem ei, i = 1, 2, 3 (or i = x,y,z - the same !).

With the help of these unit vectors we can represent any vector as follows (see Fig. 8):

A =i

Aiei. (12)

Now we have several equivalent representations of a vector:

A = (Ai, i = 1, 2, 3) =

iAiei. (13)

It is important to understand that all these representations mean exactly the same. Use the one

which is most convenient.

Unit vectors and vector decomposition. We saw above that with the use ofex, ey, and ez, a vector

A can be decomposed as

A = Axex + Ayey + Azez.

The choice of the unit vectors ei in the cartesian coordinates is global, that is, their direction does

not depend on the point. However, we can choose three unit vectors in each point of the space

independently and decompose a vector in the same way. The three unit vectors do not even have tobe orthogonal, that is, the condition ei ej = 0, ifi = j, is not necessary (it is very convenient thoughand we shall use it in what follows). The only necessary condition is that they are independent, that

is, ifa1e1+a2ej +a3e3 = 0, then a1 = a2 = a3 = 0. Thus, in general, we can choose three unit vectors

ei which are not position dependent (that is, their directions are different in different points) and

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A

B

(AB)Figure 6: Two vectors: notation.

decompose a vector as in (13). In practice, such unit vectors are usually related to some coordinateframe, that is, the direction is chosen along the tangential to the curves building the frame.

Example - spherical coordinates. In spherical coordinates a curve = const, = const, is a

radius, which means that we choose the unit vector e1 eR parallel to the radius-vector r:

eR =r

|r|= (sin cos , sin sin , cos )

= sin cos ex + sin sin ey + cos ez.

(14)

The vector e2 e is tangential to R = const and = const, and the vector e3 e is tangentialto R = const and = const. They can be found from geometrical considerations but it is instructive

to derive these two from the orthogonality conditions. We shall start with e which can be written as

e = a1ex + a2ey (it is always in a plane parallel to x y plane since = const. The two conditionsare

e eR = a1 sin cos + a2 sin sin = 0, (15)e e = a21 + a22 = 1. (16)

The solution is a1 = sin , a2 = cos , or a1 = sin , a2 = cos . In the special case eR = ex( = /2, = 0) one has e = ey, which shows that we have to choose the first set:

e = sin ex + cos ey.

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x

y

z

exey

ez

Figure 7: Unit vectors ei.

The third unit vector can be expressed ase = b

1ex + b

2ey + b

3ez and there are three conditions:

e eR = b1 sin cos + b2 sin sin + b3 cos = 0, (17)e e = b1 sin + b2 cos = 0, (18)e e = 1. (19)

The second condition gives b2 = b1 sin / cos . Substituting into the first equation, one gets

b1 sin / cos + b3 cos = 0,

and after substituting into the third condition one would get

b3 = sin , b1 = cos cos , b2 = cos sin .

Q: Why the sign of b3 is chosen as above and not b3 = sin ?

Now

e = cos cos ex + cos sin ey sin ez.

With the use of these unit vectors the radius-vector can be written as r = ReR.

Exercise. Find er, e, and ez for cylindrical coordinates.

Scalar product - scientific notation. We can use (12) and (8)-(10) to write the scalar product as

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x

y

A

ex

ey

A = Axex + Ayey

Figure 8: Vector, components, and unit vectors.

follows:

A B = (i

Aiei) (j

Bjej)

=i,j

AiBj(ei ej)

=i,j

AiBjij

=i

AiBi.

(20)

Here we used the properties of the unit vectors ei: ei ej = 1 is i = j and zero otherwise, and havedefined the famous Kronecker delta-symbol:

ij =

1, if i = j,0, if i = j. (21)

Properties of ij. The delta-symbol is fully symmetric: ij = ji . The delta-symbol is used to

filter indices:

j ijAj = Ai. (22)

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Useful formulae:

i

ii = 3, (23)j

ijjk = ik. (24)

Scalar product - angle between two vectors. From the relation A B = |A| |B| cos(AB) one

immediately finds:cos(AB) = A B

A AB B . (25)

Thus, if the components of the two vectors A and B are known it is straightforward to find the

angle.

Cosine theorem. Let there is a triangle built on two vectors running from the same point, A and

B. The third vector is then C= B A. The length ofC can be found using the scalar product:

C2 = (B A)2 = A2 +B2 2A B.

3.1 Decomposition for two vectors

Let there are two vectors, A and B, and we want to represent the vector B as a sum B = B +B,

where = B A, and = B A (see figure).

AB

B

B

Figure 9: Decomposition for two vectors.

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4 Vector product

Besides the above described scalar product, another object can be constructed which can be shown

to be a vector. This object is called a vector product C = A B and is derived according to thefollowing rule

C1 = A2B3 A3B2,C2 = A3B1

A1B3,

C3 = A1B2 A2B1.(26)

From the definition we can see that such vector product can be constructed only in three-dimensional

space. From (26) one has

e1 e2 = e2 e1 = e3,e2 e3 = e3 e2 = e1,e3 e1 = e1 e3e2,

(27)

and using A =i Aiei and similar representation for B it is possible to write

C= AB = det

e1 e2 e3

A1 A2 A3

B1 B2 B3

. (28)

Q: Prove (27) and (28).

Properties of the vector product. The following properties of the vector product follow from the

definition:

AB = B A,(A)B = (AB),A (B + C) = AB + AC.

(29)

Magnitude and direction. Let for simplicity A and B be in x y plane, that is, A3 = 0, B3 = 0.We shall also write

A1 = |A| cos(Ax), A2 = |A| sin(

Ax),

B1 = |B| cos(Bx), B2 = |B| sin(Bx). (30)Substituting (30) into (28) one finds (see Fig. 10):

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B

A

AB

90

Figure 10: Direction of the vector product.

C= AB

= |A||B| sin(AB)e3. (31)Q: Prove (31)

Q: Show that the length ofA B is equal to the area of the parallelogram built on the twovectors.

Q: Show that C A and C B, and A, B, and C obey the right hand rule.Vector product -scientific notation. One most important construction is defined usually: the

Levy-Chivita symbol

ijk =

1, if ijk = 123, or 231, or 3121, if ijk = 132, or 321, or 2130, if at least two from ijk are equal.

(32)

With this symbol the vector product C= AB can be written as follows:

Ci =j,k

ijkAjBk. (33)

Q: Prove (33)

Another, probably even better definition is 123 = 1 and it is completely antisymmetric: ijk =

jik = ikj . In other words, each transposition of two indices multiplies by 1.Q: How many independent components has ijk ?

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4.1 EXERCISES.

The below exercises with the indices and symbols are very important for better understanding.

1. Calculate

i ii and

ij ij.

2. Calculate

ij(AiBj AjBi)ij.3. Calculate

ijk ijk .

4. Calculate

jk ijkjk .

5. Calculate ijk ijkijk .6. Calculate m mijmkl. HINT: trym

mijmkl = K(ikjl iljk)

and find K.

7. Calculate

mn imnjmn. HINT: trymn

imnjmn = Xij

and find X.

5 More complex operations with vectors

Let A, B, and C are three arbitrary vectors. We know how to build a scalar from two vectors:

A B. This scalar can be multiplied by the third vector to give another vector

(A

B)C

which is parallel to C.

On the other hand, we can build a vector A B. This vector can be multiplied by the thirdvector either to give a scalar:

C (AB)

or to give a new vector

C (AB)

Q: Show that|C

(A

B)|

is a volume of the parallelepiped built on the three vectors.

Q: Show that

C (AB) = (CA) B

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Q: Derive the formula:

A (B C) = B(A C) C(A B). (34)

ATTENTION: this expressions are very important and useful and we will need them in future.

6 Displacement vector and curvilinear coordinates

Displacement vector may be used for derivation of the scale factors hi in curvilinear coordinates.

Indeed, dr2 = ds2, thus calculating

dr2 =

(

r

X1)dX1 + (

r

X2)dX2 + (

r

X3)dX3

2

= (r

X1)2dX2

1+ (

r

X2)2dX2

2+ (

r

X3)2dX2

3

+ 2(r

X1) ( r

X2)dX1dX2 + 2(

r

X1) ( r

X3)dX1dX3 + 2(

r

X2) ( r

X3)dX2dX3

we have

h21

= (r

X1)2, h2

2= (

r

X2)2, h2

3= (

r

X3)2,

(r

X1) ( r

X2) = (

r

X1) ( r

X3) = (

r

X3) ( r

X3) = 0

This becomes even more obvious if we write

r = xx + yy + zz,

r

X1=

x

X1x +

y

X1y +

z

X1z,

r

X2=

x

X2x +

y

X2y +

z

X2z,

r

X3=

x

X3x +

y

X3y +

z

X3z,

(r

X1)2 = (

x

X1)2 + (

y

X1)2 + (

z

X1)2,

. . .

and compare the obtained expressions with Lecture 1.

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differentials dxj as follows:

dxi =j

xixj

dxj , (4)

which is just the rule for the full differential.

Then a vector Ai, i = 1, . . . , n, is an object which transforms like the differentials:

Ai =j

xixj

Aj . (5)

A vector in n-dimensional space has n components.

It is possible to define a more general construction which is called tensor. For example, a 2-rank

tensor is a two-index object Tij (n2 components) which transforms as follows:

T

ij =kl

xixkxj

xl

Tkl. (6)

In previous lecture we introduced the metric tensor gij. Let us check whether it is indeed a tensor.

The distance element ds2 =

ij gijdxidxj is invariant under coordinate transformations, that is, ds2

is independent of the coordinate choice. Therefore, writing the same distance element in two different

coordinate frames, one has ij

gijdxidxj =km

gkmdx

kdx

m,

and usingdxi =

k

xixk

dxk, dxj =

m

xjxm

dxm,

we obtain km

ij

xixk

xjxm

gij

dxkdx

m.

Comparing this with

km g

kmdx

kdx

m we find that

gkm

= ij

xixk xj

xm g

ij, (7)

which does not look exactly as (6). Nevertheless, gij is a tensor. In fact, in curvilinear coordinates

exist tensors (and vectors too) of two kinds (they are not completely independent but related through

gij). We shall call them up and down (these are not real names which we will not use now). For

example, up-vector transforms as (5), while down-vector transforms as follows:

Ai =

j

xjxi

Aj . (8)

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Lecture 3. Motion

1 General concepts

Any change of the position with time is called motion. This definition implies that we have a good

method to measure time everywhere and always. For the time being let us assume that indeed such

method exists so that we can observe xi(t), that is, coordinates of a particle are functions of time.If the coordinates do not change we say that the particle is in the rest and consider that as a special

case of motion.

By providing xi(t) we describe the motion completely, since we know where is the particle at

each moment of the time. However, sometimes we want to know some particular things about

the motion, for example, how quickly the particle position is changing. For the description of this

quickly we define the velocity vi (dxi/dt). It is important to understand that, in general, the

velocity is defined using coordinates and can differ from our usual perception of the speed. Thus,

in cartesian coordinates one has the following components of the velocity: (dx/dt), (dy/dt), and

(dz/dt). In spherical coordinates the components are (dr/dt), (d/dt), and (d/dt), that is, even

do not have the same dimensions. If necessary, we shall emphasize this by calling this velocity a

coordinate velocity, and the other one (see below) a physical velocity. In general relativity, however,

it is not always possible to define globally the physical velocity as we understand it in our everyday

life (measured in km/h, for example), and one has to use the velocity introduced above.

In a similar way one introduces acceleration ai (d2xi/dt

2) which shows how quickly the velocity

is changing. One might go further and introduce higher derivatives. However, this is useful only

in very few cases, because of the structure of classical mechanics, where external influence (force)

determines acceleration.Once we know xi(t), a simple procedure (derivation) gives us subsequently the velocity and

acceleration. The inverse procedure is a little bit more complicated but still remains only a technical

one. Namely, if we know vi(t) and the initial conditions xi(t = t0) = xi0, one immediately gets:

xi(t) = xi0 +

tt0

vi(t)dt. (1)

ATTENTION: for not to make a mistake one has to make clear the distinction between the moment

of time t in which we would like to have xi(t), and the time variable t

which is the integration

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variable only. Similarly, if we know the acceleration ai(t) and the initial conditions vi(t = t0) = vi0,

we have:

vi(t) = vi0 +

tt0

ai(t)dt. (2)

Calculation of the distance is a little more complicated since we have to know how to write the

distance element ds. We shall delay the discussion of this until the next session where our ordinary

space is considered (in which it is always possible to choose cartesian coordinates). It has to be

mentioned, however, that both velocity and acceleration are vectors.

All points which are passed by the particle in space constitute a trajectory (or path or orbit).

Examples of trajectories are shown in Fig. 1

x

y

Figure 1: Trajectories in two-dimensional (left) and one-dimensional (right) cases.

2 Alternative approach

An alternative approach is usually used in the case where cartesian coordinates are possible (but not

necessarily chosen). In our three-dimensional space we can define a displacement vector (see previous

lecture). We can go even further and determine a particle position by the displacement vector from

the origin of the coordinates to the point where the particle is now. Following general principles

this vector is P0P, where P = (x,y,z) (in cartesian coordinates) and P0 = (0, 0, 0) (origin !). Thus,P0P = (x,y,z). This vector is denoted r and is called a position vector or radius-vector. Using this

vector, we can define velocity and acceleration in a more concise way:

v =dr

dt

, a =dv

dt

=d2r

dt2

. (3)

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Now (1)-(2) will take the following shape:

r(t) = r0 +

tt0

v(t)dt, (4)

v(t) = v0 +tt0

a(t

)dt

, (5)

It is now even sufficiently easy to calculate the distance. Indeed, from ds2 =

i dx2

i (in cartesian

coordinates), one has

ds

dt=

i

dxidt

21/2

= (i

v2i )1/2 = (v v)1/2 = |v|, (6)

so that

s =tt0

|v|(t)dt. (7)

In practice, in order to calculate the distance one always has to substitute

|v| =

dx

dt

2+

dy

dt

2+

dz

dt

21/2. (8)

In spherical coordinates one has to use

|v| =

drdt

2+ r2

ddt

2+ r2 sin2

ddt

21/2. (9)

Q: Write |v| in cylindrical coordinates.

2.1 Usage of unit vectors

The velocity vector can be written as

v = xex + yey + zez.

Here we use the short notation A dA/dt for any A. Instead of using the cartesian unit vectors

one could use another set of those, similarly as we did in Lecture 2. We shall show how it is done

on the example of two-dimensional polar coordinates. We define er = r/r as a unit vector along the

radius-vector (tangential to the curve = const, and e as a unit vector tangential to r = const

(there is no z - two-dimensional case). It is easy to find

er = cos ex + sin ey, e = sin ex + cos ey. (10)

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From these expressions we find

ex = cos er sin e, ey = sin ex + cos ey. (11)

Now the velocity

v =dr

dt= xex + yey, (12)

where

x = r cos , y = sin , (13)

and

x = r cos r sin ,

y = r sin + r cos .(14)

Substituting (11) and (14) into (12) we have

v = rer + re. (15)

Q: Express v in terms of eR, e, and e (spherical coordinates).

3 Motion - examples

3.1 Constant acceleration

Let a = const. This means that neither the magnitude nor the direction of the acceleration do not

change. From (4)-(5) one has

v = v0 + a(t t0), (16)

r = r0 + v0(t t0) +1

2a(t t0)

2. (17)

Q: Prove these expressions.

The most famous example of such motion is the motion near the earth surface, where the free-fall

acceleration g = const. Let us choose the coordinate system as in Fig. 2. Then g = (0,g) and one

has for the components and coordinates

vx = v0 cos , (18)

vy = v0 sin g(t t0), (19)

x = x0 + v0 cos (t t0), (20)

y = y0 + v0 sin (t t0) 1

2g(t t0)

2. (21)

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x

y

g

v0

Figure 2: Motion near the earth surface.

Trajectory Trajectory (path) is the curve along which the particle moves. The equation for the

trajectory can be obtained by excluding time from the equation for x and further substituting into

the equation for y:

y = y0 + (x x0)tan g(x x0)

2

2v20

cos2 . (22)

Q: Derive (22).

Distance In order to calculate the distance along the path (trajectory) we have to integrate

s =

tt0

|v(t)|dt

=

tt0

v20

cos2 + (v0 sin g(t t0))2dt.

(23)

3.2 Circular motion

This motion is in a plane. It is natural to use polar coordinates. In this coordinates r = const,while the angle = (t) is time dependent. The derivative = d/dt is (naturally) called angular

velocity (if the radius r is changing, dr/dt is called radial velocity). The velocity magnitude is found

from |v| =

(dr/dt)2 + r2(d/dt)2 = r. In most cases, however, this motion is described in terms

of r and v. In order to do that let us write down the relation between the cartesian and cylindrical

coordinates of the particle:

x = r cos , y = r sin , z = z, (24)

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from which (with the use of simple differentiation) one gets

vx = r sin , vy = r cos , vz = 0. (25)

Now we introduce the angular velocity vector = ez and notice that (25) can be written as

v = r. (26)

The physical sense of this vector is that it shows the rate of the angle change and also the direction

of the rotation axis.

Further differentiating (26) with respect to time one has

a =d

dt r + (dr/dt)

= d

dt r + v

=d

dt r + ( r)

(27)

Let us now write r = r + r, where r = zez , r = xex + yey . Then application of the

formulae of the vector algebra gives

a =d

dt r 2r. (28)

Q: Prove (28).

In the special case when the angular velocity does not change = const the centripetal acceler-

ation is a = 2r.

Let us now consider a more general case when = (t)ez. In the polar coordinates r = rer,

and ez er = e, so that (28) will give

a = e 2rer. (29)

The first term in the fight hand side of (29) is called tangential acceleration, the second term is calledcentripetal acceleration.

4 General

A particle motion is given by r(t) and the velocity is v = dr/dt. From the very definition the velocity

is tangent to the trajectory. What about the acceleration ? The acceleration is defined as a = dv/dt,

and, in general, it is neither parallel nor perpendicular to the velocity. However, we can decompose

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it similarly to what has been done in the case of the circular motion. For this to do let us write

v = vv, (30)

where v = |v| is the velocity magnitude, and v = v/v is the unit vector in the direction of the velocity.

Of course, so far we have not done anything new by writing the identity (30). Now, however, the

acceleration can be written as follows:

a =d(vv)

dt=

dv

dtv + v

dv

dt. (31)

We see that the first term in (31) is parallel to v v. The second term is perpendicular to the

velocity. Indeed, since v v = 1 one has

d

dt(v v) = 0 = 2v

dv

dt .

which means that dv/dt v. Thus, we have decomposed the acceleration into the tangential and

normal components:

at =dv

dtv, (32)

an = vdv

dt. (33)

The first one shows the change in the velocity magnitude, the second one gives the change of thedirection. The centripetal acceleration is the special case of the normal acceleration when a particle

moves along a circle. In the circular motion the magnitudes of the velocity and centripetal acceleration

are related through the circle radius: |a| = v2/R. We can define similarly the curvature radius of the

trajectory in the general case as R = v2/|an|, so that

R =1

v

dv

dt.

Let us now pay attention that the vector dv is a unit vector, while the vector dv/dt is not, in

general. We can build a new unit vector

N =dv/dt

|dv/dt|, (34)

which is called the vector of the normal. In order to complete these two vectors we have to add a

third one, which can be done as b = v N.

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Advanced 3. Motion in curvilinear coordinates

1 Advanced material

1.1 General coordinates

In general coordinates vi = dxi/dt and ds2 = ij dxidxj. Therefore,

ds

dt= (ij

gijdxidt

dxjdt

)1/2

= (

gijvivj)1/2

(1)

Thus, the velocity magnitude is |v| =

ij gijvivj. However, is velocity a vector ? From the previous

lecture we know what should be the transformation law when we change coordinates. Indeed,

v

i =dxidt =

j

xixj

dxjdt =

j

xixj v

j,

provided that the time t does not change when we change the coordinates. This means that the

velocity defined as vi = dxi/dt is a vector only if t is invariant under coordinate transformation. In

nonrelativistic classical mechanics t is an absolute time, which is the same for any coordinate choice,

so that vi is a vector.

If xi = xi(t) this defines a one-dimensional curve in the space. Is the choice of time the only

method to define a curve ? Obviously, any substitution t = t() is also good, if there is one-to-one

correspondence. Since s(t) is a monotonic function, s can be used for the curve parametrization withequal success. Let us say that s is the new time and see what happens if we define new velocity

as dxi/ds. The only effect of our choice would be in the magnitude of the velocity:

v v =ij

gijdxids

dxjds

= 1.

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1.2 Differentiation along the curve

In section 1.1 we encountered the derivative of the kind v(dA/dt) (where A is some vector). Such

derivative is called the derivative along the curve. Using the definition of the curve length ds/dt = v

we can rewrite such derivative in terms of derivation with respect to the path:

vd

dt= v

d

ds. (2)

With this definition the curvature radius takes the form

R1 =

dv

ds

. (3)

Since we live in a three-dimensional space, any curve is three-dimensional, in general, and one

curvature radius is not sufficient to describe the curve behavior. Indeed, a second curvature is

defined related to the twisting

T =

db

ds

(4)

(see section 1.1 for the definition of b and N).

The first and second curvatures are defined as C1 = 1/R, C2 = T, and the total curvature is

C2 = C21

+ C22

.

Exercise: Using N = b v and differentiating with respect to s show that

C=

dN

ds

.

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Lecture 4. Relativity

1 Reference frames

A reference frame is an observer with a coordinate system. This implies that a) some coordinates

are chosen to describe the position of any particle, b) there is some physical body relative to which

the motion is considered (it is possible to perform physically meaningful measurements), c) there isa method of time measurements. The last means that a clock should exist in each point of the space,

and all these clocks should be synchronized. They are synchronized by sending some signal from

some reference clock to all others, telling them what is the correct time in the reference point. Such

procedure requires that the clocks in all points of the space be identical. In Newtonian mechanics the

signal speed is assumed to be infinite, so that the synchronization is straightforward and easy. Let us

now assume that there are two frames, S and S, so that S moves with the velocity V relative to S.

We also assume that both observers put themselves in the coordinate origin of each of these systems,

and both choose cartesian coordinates building the axes so that x is parallel to x, y is parallel to

y, and z is parallel to z. If the signal velocity is infinite it is possible to synchronize the clocks in

S and S as well so that the two observers measure the same time: t = t. On the other hand, it is

clear (see Fig. 1) that for each point P, r = r + R, where r is the radius-vector ofP measured by

the observer S, r is the radius-vector ofP measured by the observer S, and R is the radius-vector

ofS (coordinate origin) measured by the observer S. Thus, one has the Galileo transformation in

the following form:

t = t, r = r + R. (1)

Differentiating (1) with respect to time t and taking into account that t = t, we immediately find

v = v + V, (2)

where v = dr/dt is the velocity ofP measured by S, v = dr/dt is the velocity ofP measured by S,

and V = dR/dt is the velocity ofS measured by S, that is, the relative velocity of the two frames.

The two frames, S and S, are said to be related by the Galileo transformation if V = const. In

Newtonian mechanics the following Galileo relativity principle holds: all physics laws have the same

shape in two systems related by the Galileo transformation. In particular, this means that the second

Newton law looks identical, F = ma and F

= m

a

, in both frames. Indeed, mass m is a scalar

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x

y

x

y

R

r

r

V

Figure 1: Galilean transformations.

and the same in both frames. The acceleration can be obtained by differentiating (2) with respect

to time,which gives a = a (if V = const). From this we derive the law of the force transformation

F = F.

In what follows we assume that the frame S is inertial. This, in particular, means that the force

in the second Newton law is caused by other physical bodies. In the absence of such bodies there is

no any force and a = 0. The frame S is also inertial if V = const. What happens if the frame S is

non-inertial, that is V is time-dependent ? Direct differentiation of (2) gives

a = a + A, (3)

where A is the acceleration of the frame S measured by S. Substituting in the Newton law, we have

Fext = ma = ma + mA,

where the index ext emphasizes the nature of the force: it is produced by other physical bodies.

If we want to have the Newton law i in the accelerated frame S similar to its usual form, that

is, F = ma, we have to define F = Fext mA. The second term in this expression is the inertia

force which is not caused by any physical body but is due to the the acceleration of the frame.

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x

y

x

y

R

r

r

a F = ma

Figure 2: Inertial force in the accelerated system. d

This expression can be also considered as the transformation rule for the force.

1.1 Examples of inertial forces

Hanging body. Let a body hang on a rope from a ceiling (see Fig. 3) and let the ceiling move

ma

mg

Figure 3: Hanging body in non-inertial frame.

with constant acceleration a. What would be the angle between the rope with the vertical ? The

simplest (and most effective) way is to consider the situation in the non-inertial accelerated frame,

where the position of the body is determined by the balance of the rope tension T (directed along

the rope), vertical gravity force mg and the inertia force ma. This means that the vector sum

g + (a) should be directed along the rope. It is clear now that tan = |a|/|g|.

Sliding down accelerated slope. What should be the acceleration of the slope to prevent sliding

of the body (see Fig. 4)? In the accelerated frame the vector sum of the gravity force mg and inertia

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instead we have to say that r = rr, where the index r means rotated from S into S. In the

following we shall not emphasize this difference which should, nevertheless, be kept in mind.

Our objective is to establish the relation between the velocities and accelerations measured in

both frames. Let us start with the simplest case where the particle P does not move at all in the

rotating frame, v

= 0. In this case, from the point of view ofS, the particle P simple experiencescircular motion with the constant angular velocity. We know that the observer Swould measure the

velocity v = r, and the acceleration a = v, The nature of this acceleration is simply the

rotation of the velocity vector. The radius-vector r and the velocity vector v rotate by the angle

dt during the infinitesimal time dt. Since the rotation of the both vectors is the same, the relation

between the velocity change dv and v is the same as between dr and r. The latter can be found

from what we already know: dr = vdt and v = r, so that dr = rdt. We conclude that for

the velocity vector one would have dv = vdt. Moreover, for any vector A rotating with S we

will have in the standing frame S: dA

=

Adt. In a more general way, if a vector

Ais changing

in the rotating frame thendA

dt=

dA

dt+ A (4)

where (dA/dt) is the rate of the vector change as measured by the standing observed and (dA/dt)

is the rate of the vector change as measured by the rotating observer and rotated into the standing

frame.

Let now the particle P move in the frame S with the velocity v in the moment t. The observer

S will measure in the same moment the velocity v = v + r. During dt this velocity changes

by dv

in the frame S

. The observer S will measure the velocity change dv which is due to a) thechange dv, b) the change due to the rotation of the velocity vector vdt = (v + r),

and c) due to the change of the rotational term d( r)m = dr = dr = vdt =. The

last contribution simply says that the particle is on another circler in the moment t + dt, so that its

rotation velocity is not the same as before. Combining all these contributions together, dividing by

dt and taking into account that dv/dt = a, dv/dt = a, we finally obtain

a = a + 2 v + ( r). (5)

The last term is already known to us: it is the centripetal acceleration and can be written as 2r.

The second term is called Coriolis acceleration. If now we wish to write the second Newton law in

the usual form F = ma, we have to write

F = Fext 2m v + m2r, (6)

where the first term is the force cause by other bodies, the second term is called the Coriolis force,

and the last one is the centrifugal force. The last two are inertia forces and are caused by the frame

rotation.

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The below figures show once again the change of the velocity in the rotating and nonrotating

frame.

r

v

x

y t

r

v

x

y t + dt

Figure 6: Velocity change in the rotating frame. The left figure shows the position r (blue) andvelocity v (red) of the particle in the moment t. The right panel shows what happens in the momentt + dt. The particle moved into the new position r + vdt and has the new velocity v + v.

r

v

x

y

x

y t

r

vx

y

x

y t + dt

Figure 7: Velocity change in the nonrotating frame. In order to find the velocity in the nonrotatingframe we have to rotate the velocity in the rotating frame by the angle and add r. The leftfigure shows the velocity in the moment t: the velocity v (red) is rotated by the angle , the positionr (blue) is rotated by the same angle (and gives r) and r (green) is added. The total velocity vis shown in magenta. The right figure shows the velocity in the moment t + dt: v + dv and r + dr

are rotated by the angle + d, and (r + dr) is added.

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Advanced 4. Rotations, general noninertial frames

1 Advanced material: Rotations

Two cartesian coordinate systems S(x,y,z) and S(x, y, z) with the common origin can be trans-

formed one into another by a rotation. Let us start with the rotation around z axis by the angle z.

Then the transformation reads

x = x cos z + y sin z,

y = x sin z + y cos z,

z = z.

(1)

It is convenient (and not only convenient !) to represent this transformation in a matrix form. Let

us define row

r =

x y z

,

and column

r =

x

y

z

,

and similarly for S.

The transformation (1) can be now written as follows:

x

y

z

=

cos z sin z 0

sin z cos z 00 0 1

x

yz

(2)

or in the following symbolic form

r = Rz(z)r, (3)

Rz(z) =

cos z sin z 0

sin z cos z 0

0 0 1

. (4)

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The construction Rz(z) is called an operator, and (4) is the matrix form of the rotation operator.

In a similar way we can find the operators (matrices) for rotations around x and y axes:

Rx(x) =

1 0 0

0 cos x sin x

0 sin x cos x

(5)

Ry(y) =

cos y 0 sin y

0 1 0

sin y 0 cos y

(6)

If we want to rotate, say, first around z by z and after that around x by x, the result will be

r = Rx(x)r

= Rx(x)Rz(z)r. (7)

In the component form (3) is written as follows:

xi =j

Rijxj. (8)

For the rows one would have

r = rR, (9)

(remember matrix multiplication !) or

xi =j

xjRji. (10)

Thus,

Rji = Rij R = RT, (11)

where RT is the transposed matrix: RTij = Rji.

For any rotation

i xixi =

k x

kx

k, which gives

k

xkxk =ijk

RkixiRkjxj =ijk

RTikRkjxixj =ij

ijxixj. (12)

Thus, k

RTikRkj = ij, (13)

or

RTR = I, (14)

where I is the unit matrix (all diagonal elements equal to one, all nondiagonal equal to zero).

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Definition. If for a matrix R exists a matrix R such that RR = I, the matrix R is called the

inverse matrix and denoted R1. Thus, for rotations we have R1 = RT.

Since the matrix product is noncommutative, that is, for two matrices A and B, in general,

AB = BA, the result of two successive rotations depends on their order. However, if we consider

infinitesimal rotations d, the situation is different. Using sin(d) , cos(d) 1, weobtain

Ri(d) = I + dJi, (15)

where

Jx =

0 0 0

0 0 1

0 1 0

(16)

Jy =

0 0

10 0 0

1 0 0

(17)

Jz =

0 1 0

0 1 0

0 0 0

(18)

Q: Prove that Ri(di)Rj(dj) = Rj(dj)Ri(di).

Now we can define the infinitesimal rotation angle vector as d = (dx, dy, dz) and a vector

of matrices R= (Rx,Ry,Rz) (and similarly for J), so that the general infinitesimal rotation will be

r = r + (J d)r, (19)

where J d is an operator which acts on r.

Q: Prove that (19) can be written as dr = d r.

2 Advanced material: noninertial frames

The Galilean relativity (Newtonian mechanics) state that there is a class of special frames (inertial

frames) in which the motion laws have the same shape. What happens in noninertial frames and

why they are worse than inertial ones ? Actually, if we want that the physical laws be universal we

should find a way to describe these laws in any frame. We already know how to include inertia forces

in a uniformly accelerated frame and in a uniformly rotating frame. Now let us try to do the same

in an arbitrary frame. In order to do that we shall write down the second Newton law in an inertial

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frame S(xi) (where xi are cartesian coordinates) in the form

md2r

dt2= F, (20)

whereF

is an external (true) force, caused by other bodies. Let the observer in a noninertialframe S also chooses cartesian coordinates xi, and we know the relation r(r

, t). The dependence on

t shows that this is not a simple coordinate transformation but the relation between the coordinates

in two frames with some relative motion. The only transformation between the two frames, which

conserves the scale, is the time dependent translation (this is not precise, we shall look into it more

closely later) and rotation:

r = X + Rr, (21)

where X(t) is the radius-vector of the coordinate origin of S according to the observer S. Now one

hasv = V +

dR

dtr + Rv, (22)

where V is the relative velocity of the two frames as measured by S. Differentiating once again with

respect to time we get

a = A +d2R

dt2r + 2

dR

dtv + Ra, (23)

where A = dV/dt. Left multiplying by R1 and taking into account (20) we have finally

ma

=R

1F

ext mR

1Am

R1

d2R

dt2r

2mR

1dR

dtv

. (24)

The first term in the right hand side of (24) is the true force measured in S, the second term is

the generalization of the inertia force due to the relative motion of the coordinate origin of the two

frames, the third term is the generalized centrifugal force, and the last term is the generalized Coriolis

force. We see that there no other inertia forces.

Exercise: Derive the usual centrifugal and Coriolis force for A = 0 and

R =

cos(t) sin(t) 0

sin(t) cos(t) 00 0 1

(25)

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Lecture 5. Particle dynamics I

1 Newtons laws, momentum, kinetic energy

This section is to remind the basics of the particle dynamics. We shall proceed in an inertial frame,

so that there are no inertia forces. The second Newton law says that the equation is proportional to

the forceF = ma. (1)

Momentum p is defined as follows:

p = mv, (2)

which immediately gives the second Newton law in a more general form:

F =dp

dt. (3)

Now the momentum change

p = p2 p1 =t2t1

F(t)dt. (4)

Kinetic energy is a scalar:

K =mv2

2=

mv v2

. (5)

The rate of change of the kinetic energy

dK

dt= mv dv

dt= mv a = (ma) v = F v. (6)

During the time dt the energy change is

dK = F vdt = F dr = dW. (7)

The scalar quantity dW is called the infinitesimal work. If the particle moves from r1 to r2 the

kinetic energy change K is equal to the work which is done along the trajectory:

K = W = r2

r1

F

dr. (8)

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The last integral is taken along the trajectory. It, and therefore, the work depends, in general, on

the path (and not only on the initial and final points r1 and r2). The work produced in unit time

P = dW/dt is called power. For our case it is

P = F

v. (9)

Third Newton law states that if two particles interact then F12 = F21. Here F12 is theforce which particle 1 exerts on particle 2. This has to be completed by the superposition principle:

if there are a number of particles acting on particle i the total force which particle i experiences is

the vectorial sum of the forces from all other particles:

Fi =j

Fji. (10)

Two additional important concepts are the concept of a physical system which may include a number

of particles, and the concept of the resulting force acting on the whole system, which is the vectorial

sum of all the forces acting on all parts of the system. If the system consists of a number of particles,

the resulting force would be

F =i

(Fi,ext +j

Fji), (11)

where Fi,ext is the external force (that is, not by any particle from the same system) on particle i,

and Fji is the force exerted by particle j on particle i (both i and j belong to the physical system

under consideration). Using the third Newton law Fij = Fji we obtain

F =i

Fi,ext, (12)

that is, the internal forces do not contribute to the resulting forced acting on the system as a whole.

2 Work as a path integral

Work is calculated differently from the integrals we have seen before. In the expression

W =

2

1

F dr

the integral is done along the curve. The force can be a function of r and t, in general, F = F(r, t)

and in this case the work can be calculated only if we know r(t) as follows

W =

t2t1

F(r(t), t) v(t)dt, (13)

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so that the integrand becomes a function of only one variable t.

In many cases, however, F = F(r) and is independent oft. If this is the case, the path (trajectory)

should be expressed in a parametric form.

2.1 Example: friction

We assume the a body is moving in a horizontal plane so that the friction force magnitude is constant

while it is always directed against the velocity F = fv/v. Now F vdt = fvdt = f ds, whereds is the length of the path. Thus, W = f s, where s is the length of the whole path made by theparticle.

2.2 Example: constant force

For a constant force F one has

W =

2

1

F dr = F

2

1

dr = F (r2 r1)

and does not depend on the path but only on the initial and final positions. This is an example of a

conservative force for which work does not depend on the path. In other words, a conservative force

does not produce work along any closed path.

3 Momentum conservation, center of mass

The total momentum of the system is defined as the vectorial sum of the momenta of all its parts:

P =i

pi =i

mvi. (14)

The change of the total momentum with time is given by dP/dt:

P

dt

=d

dti

pi

=i

dpidt

=i

Fi = Fext,

(15)

where Fext is the sum of the forces caused by the bodies which do not belong to the system (external

forces). If this resulting external force Fext = 0 the total system momentum does not change

dP/dt = 0, that is, the momentum is conserved.

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Let us now define the center-of-mass (CM) as the point with the radius-vector

Rcm =

i miri

i mi

=

i miriM

, (16)

where M is the total mass of the system. There does not have to be any particle in this point - itis completely imaginary. Differentiating (16) with respect to time we find that CM is moving with

the velocity

Vcm =

miviM

=P

M. (17)

Thus, the physical sense of the center-of-mass is that this is a point which may be assigned the total

momentum and the total mass of the system. From (15) we find the acceleration of the center-of-mass

as follows

acm = Fext/M. (18)

In the absence of external forces the center-of-mass moves with constant velocity. If it was in the

rest in the beginning it would remain in the rest if only internal forces are present.

While the total momentum can be attributed to the motion of the center-of-mass, the kinetic

energy cannot. Indeed, the total kinetic energy

K =i

Ki =i

mi2

v2i

= i

mi2

[(vi Vcm) + Vcm]2

=i

mi2

(vi Vcm)2 +i

mi(vi Vcm) Vcm +i

mi2

V2cm.

(19)

The second term in the last expression of (19) vanishes (SHOW THAT !), and the first term can be

written in terms of the velocity relative to the center-of-mass ui = vi Vcm, so that we get finally:

K =M

2V2cm +

i

mi2

u2i . (20)

Thus, the total kinetic energy of the system is the sum of the kinetic energy of the center-of-massand the kinetic energy of the relative motion of the system parts. The last contribution is called

internal energy.

4 Center-of-mass: continuous systems

Most of the systems we deal with do not consist of point-like particles, but are continuous bodies

of finite size. The generalization onto such systems is straightforward, although not always simple

for calculations. We divide the whole body onto small (infinitesimal) parts with the mass dm,

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radius-vector r, and velocity v, so that the sums are substituted by integrals:

M =

dm, (21)

P = vdm, (22)Rcm =

rdm/M. (23)

The integrals should cover all material, they are written in a symbolical form. In practice, one has

always to express the integrals in terms of coordinates.

4.1 Center-of-mass: examples

A rod 1. A zero thickness rod 0 < x < l of the mass m (see Fig. 1).

O xdx

Figure 1: Center of mass for a rod.

In the integral (23) dm = (m/l)dx ( = dm/dx = m/l is the linear density), and we have

Xcm =1

m

l0

xmdx

l=

l

2, (24)

as could be expected.

2. Same but the linear density is not constant = ax. Then

Xcm =

l0

xdx

l

0dx

=2l

3. (25)

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A triangle A uniform density triangle has the vertices (0, 0), (l, 0), and (l, h), and the mass m.

The surface density dm/dA = m/A, where A = lh/2 is the triangle area. We divide the triangleonto infinitesimal dxdy (see Fig. 2) so that for each of which dm = dxdy. We have (for any general

Ox

y

dx

dy

Figure 2: Center of mass for triangle.

surface density (x, y)

Xcm =

xdxdy dxdy , (26)Ycm =

ydxdy dxdy

, (27)

where the integral limits should be chosen according to the conditions. In our case

(. . .)dxdy

l0

hx/l0

dy(. . .)

dx, (28)

so that we have

Xcm =

l0

(hx/l0

xdy)dxl0

(hx/l0

dy)dx= 2l/3 (29)

Ycm =

l0

(hx/l0

ydy)dxl0

(hx/l0

dy)dx= h/3. (30)

A half-disk Given a uniform thin half-disk with the radius a (see Fig. 3). It is obvious that

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Ycm = 0. For Xcm we use (26):

Xcm =

a0

(a2x2a2x2 xdy)dx

a

0(

a2x2

a2x2 dy)dx

=4a

3. (31)

5 Reactive motion

From the general form of the second Newton law F = dp/dt and p = mv it follows that the velocity

can change if the mass changes. A rocket is based on this principle. Let us consider one-dimensional

motion of a rocket which ejects backwards gas with the velocity u relative to the rocket. Let v be

the velocity of the rocket in the moment t and v + dv is its velocity in the moment t + dt. Let also m

be its mass in the moment t and m + dm is its mass in the moment t + dt (dm < 0 since the rocket

ejects gas). The mass of the ejected gas is

dm (so that the total mass (m + dm) + (

dm) = m

does not change), while the velocity of the gas is v u (here v > 0 and u > 0). The momentumconservation (no external force) gives

mv = (m + dm)(v + dv) + (dm)(v u) dv = udm/m. (32)

Differentiating the last equation with respect to time one gets the rocket acceleration

a = u/m, (33)

where = dm/dt is the rate of the gas ejection.On the other hand, integrating (32) one gets

vf vi = u lnmimf

, (34)

where i and f mean initial and final, respectively.

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Ox

y

dx

y =

a2 x2

y =

a2 x2

Figure 3: Center of mass for halfdisk.

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Math1. Complex numbers

In this lecture we bring (without proof) the basics of the usage of complex numbers, which is

necessary for a physicist.

A complex number is a construction (it is widely accepted use z for general complex numbers)

z = x + iy, where x and y are real numbers, and i is a special number such that i2 = 1. For any

two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 we can find z1 + z2 and z1 z2 as follows:

z1 + z + 1 = (x1 + x2) + i(y1 + y2), (1)

z1z2 = (x1x2 y1y2) + i(x1y2 + x2y1). (2)

The absolute value of the complex number z = x + iy is

|z| =

x2 + y2.

The number x is called the real part and the number y is called the imaginary part:

x = Re z, y = Im z.

Thus, z = Re z +i Im z, and |z| =

(Re z)2 + (Im z)2. A complex conjugate (cc) number z is defined

as z = Re z Im z. It is easy to see that |z|2 = zz. Now the division of the complex numbers if

defined as follows:z1z2

=z1z

2

z2z2=

z1z

2

|z2|2

Exercise: Express z1/z2 in terms of x1, y1, x2, y2, i.

We can also write

Re z = (z + z)/2, Im z = (z z)/2i.

Any complex number can be represented as a point in the complex plane, where x axis is for Re z

and y axis is for Im z. The same point in the x y plane can be represented by its polar coordinates

r and , so that x = r cos , y = r sin . With this one can write

z = x + iy = |z|(cos + i sin ). (3)

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Physics 1. Mechanics Math1

This is completed with the (most important) Euler formula

cos + i sin = exp(i) = ei. (4)

With this one can write z = |z| exp(i), z = |z| exp(i), 1/z = (1/|z|)exp(i), z1z2 =

|z1||z2| exp[i(1 + 2)], zn = |z|2 exp(in), and so on.

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Lecture 6. Particle dynamics II

In this lecture we shall study several important cases of particle dynamics in electric and magnetic

fields.

1 Field

We say that an electric field E exists in the space if we can put an electric charge q (charged particle)

in any point and it will experience the force F = qE. The concept of the field is one of the central

concepts in physics. Since we shall deal with fields a lot in future it is worth to devote some tome to

this concept now.

We start with the example of the Earth and satellites around it. It is known that each two masses

are attracted to each other. For the time being the precise shape of the force is not important. Let

us say that the two bodies have masses M and m, M > m, and are in the positions R and r,

respectively. Then the force acting on, say m, is proportional to m and M and depends on R and r.

The same force but with the opposite direction acts on M. Although the magnitude of the force is

the same, the magnitudes of the accelerations are inversely proportional to masses, so that the larger

mass is less affected by this interaction. Let us assume for simplicity that M m, so that the effect

of the interaction on M can be neglected. Then we can regard the large mass M as a source of the

force acting on the small mass m. Thus, we can take this small mass (which is called test particle

or test mass), put it in an arbitrary point of the space and measure the force acting on it. Since we

fixed the position and the mass of the large body, the only meaningful dependence of the measured

force would be on the mass of the test particle and its position. If we put different test particles in

the same point, each would feel the force proportional to its mass, that is, the force can be writtenas Fm = mE, where E does not depend at all on the test particle mass m. If we take the same

test particle in two different points, say 1 and 2, it will experience forces F1 = mE1, and F2 = mE2,

which are, in general, different. That means that the vector E depends, in general, on the position,

E = E(r). If the large body moves somehow, this vector can depend on time too, E = E(r, t). Let

us now forget about the body which causes E and look at everything from the point of view of the

test particle. The test particle does know anything about what causes the forces, the only think it

knows is that in each point of space r and every moment t it would feel the force F = mE(r, t). The

test panicle (and we too) concludes that the vector E(r, t) exists in space even without presence of

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this test particle. Indeed, any test particle could be placed in any position at any time, and the

relation F = mE(r, t) does not change because of our bringing the test particle into consideration.

Thus, E(r, t) is real, depends only on other bodies (source) and is independent of test measurements.

We say that there is a field E(r, t). This field exists in space in time and produced by other bodies

which are not of interest to us and the motion of which is usually assumed to be known. The mass mwhich, being multiplied by the field, gives the force acting on the test particle, is called, in general, a

charge. When certain field is considered, the characteristic word should be added to make clear what

is studied. Thus, the field produced by the gravitational forces is called gravitational field and the

mass is the gravitational charge. In this lecture we will study the motion in the electric and magnetic

fields acting on the electric charge.

Fields in general Fields are not necessarily related to forces. In general, we say that there is

a field in space if some quantity can be measured at different positions and times, but the very

existence of which is not determined by such measurement. In the above consideration what was

measure was the force vector, thus the field was a vector field. A well-known example of another field

is temperature, which can be measured at any position and any time. Temperature is a scalar field

which exists even if we do not measure it. A test particle would be in this case a thermometer.

2 Motion in a constant electric field

When we consider a motion in a constant electric field we assume that the field vector E does not

depend on r and t, at least in that part of the space where our test particles moves. The electric

force is F = qE, where q is the electric charge of the particle (we omit the word test for brevity).

This case is not especially interesting since the second Newton law states

F = ma ma = qE a = qE/m = const,

that is, the particle moves with constant acceleration, the case we studied earlier.

3 Motion in a uniform oscillating field

A field is called uniform is it does not depend on r. It can still depend on time. In the section we

shall try the field of the form

E = E0 sin(t), (1)

where E0 = const is a constant vector (both the magnitude and direction does not change), and

is the frequency (or angular frequency). It is convenient to define a unit vector e in the direction of

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E0, e = E0/|E0|, so that E0 = E0e. The equation of motion reads

a =dv

dt=

d2r

dt2=

qE0m

e sin(t). (2)

It is convenient (this procedure will be widely used in future) to split the position vector r = xe + r,where r e. Then the equation of motion (2) splits into

d2rdt2

= 0, (3)

d2x

dt2=

qE0m

sin(t). (4)

Eq. (3) describes the motion with zero acceleration (constant velocity), that is, we immediately know

the answer:

r = r,0 + v,0(t t0). (5)

Here v,0 e. Thus, the particle moves with the constant velocity v,0 in the direction perpendicular

to the electric field direction (!!! it is only a part of the motion !!!).

Frame transformation. Here we introduce a very important method of studying particle dy-

namics. We should remember that it is always possible to switch to another inertial frame, and

the second Newton law will not change the form. If we now switch to the frame moving with the

velocity v,0 with respect to our original frame, we will see the particle not moving in the direction

perpendicular to the electric field. In fact, any motion with constant velocity can be excluded from

the consideration by switching to the corresponding moving inertial frame. (WARNING: we have

to know, in general, how the fields transform with the frame transformation. For the time being,

we shall leave this issue aside, assuming that our electric field remains the same.) Thus, in our case

we may say that the particle does not move in the perpendicular direction in the properly chosen

reference frame.

With all above, the only equation of motion to be solved is the equation (4) for single coordinate

in the direction of the electric field. This equation is easily integrated to give

vx = vx0 +qE0m

(1 cos(t)) , (6)

x = x0 +

vx0t +

qE0m

t

qE0m2

sin(t), (7)

where we added the initial conditions x = x0, vx = vx0 at t = 0. We see that the motion in x direction

is the motion with the constant velocity vd = vx0 + qE0/m and periodic oscillations. Again we can

switch to the reference frame moving in x direction with the velocity vd. In this frame the particle

only oscillates with the frequency .

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4 Constant magnetic field

The electric and magnetic field are the two parts of the electromagnetic field, and act on electric

charges. However, while the electric field acts on any charges, the magnetic field acts only on moving

charges. The magnetic force looks as follows (SI units)

F = qv B, (8)

where B is the magnetic field and v is the particle velocity.

[In the Gaussian (CGS) system of units magnetic and electric fields are measured in the same

units and the force is written as F = qvB/c, where c is the speed of light. In our course we shall

use SI units throughout.]

The force (8) is perpendicular both to the velocity and magnetic field direction. As an obvious

consequence, the magnetic force cannot change the kinetic energy of the particle, that is, the magneticforce does not produce work:

physics mechanics - lecture notes

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