Chapter 8: Accelerating Reference Frames and The … · 8-1 Chapter 8: Accelerating Reference...

8-1

Chapter 8: Accelerating Reference Frames and The

Curvature of Spacetime 8.1 Flat and Curved Spacetime

A familiar example of the visualization of curved or warped spacetime is the rubber sheet analogy. A flat rubber sheet with a rectangular grid painted on it is stretched, as shown in figure 8.1(a). By Newton’s first law, a free particle, a small rolling ball m moves in a straight line as shown. A bowling ball is then placed on the rubber sheet distorting or warping the rubber sheet, as shown in figure 8.1(b). When the small ball m is rolled on the sheet it no longer moves in a straight line path but it now curves around the bowling ball M, as shown. Thus, gravity is no longer to be thought of as a force in the Newtonian tradition but it is rather a consequence of the warping or curvature of spacetime caused by mass. The amount of warping is a function of the mass.

Figure 8.1 Flat and curved spacetime

In the case of special relativity, where one frame of reference was moving at a constant

velocity with respect to another frame, we found that the spacetime diagram became skewed instead of the usual orthogonal diagram we are use to. We also saw that in the spacetime diagram of figure 2.1(e) we saw that the world line of an accelerated particle was curved through spacetime. So what does acceleration do to a spacetime diagram?

First let us plot the displacement of a falling object (an accelerated object) on the Earth, the Moon, and the planet Jupiter, as shown in figure 8.2. Each plot is a curve showing that the falling object is accelerated. That is, the slope of the distance versus time graph represents the velocity. As can be seen in the diagram, the slope is constantly changing with time implying that the velocity is constantly changing with time, implying that the motion is accelerated.

The acceleration g on the Moon is quite small (gMoon = 1.62 m/s2) and is shown in the diagram as the pink curve, which is almost a straight line. The acceleration g on the Earth (gEarth = 9.80 m/s2) is shown in the diagram as the blue curve, which is a curved line. The acceleration g on the planet Jupiter (gJupiter = 23.2 m/s2) is shown in the diagram as the tan colored curve, which has a much larger curvature. Hence the acceleration of gravity is greatest on Jupiter, and smallest on the Earth’s moon, with the Earth itself somewhat in between. The mass of Jupiter is 1.89 1027 kg, the mass of Earth is 5.97 1024 kg, and the mass of the Moon is 7.34 1023 kg. Notice that the acceleration is greatest where the mass of the parent planet or moon is greatest. Hence the mass of the gravitating body has a direct relation to the size of the acceleration of any body close to the gravitating body. (Recall from College Physics that the acceleration due to gravity on a planet or a moon was given by

2GMgr

= (8.1)

Hence the greater the mass M of the planet, the greater the acceleration due to gravity g.)

Chapter 8: Accelerating Reference Frames and The Curvature of Spacetime

8-2

Displacement versus Time

0

500

1000

1500

2000

2500

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00

Time - seconds

Dis

pla

cem

en

t -

mete

rs

g Earthg Moong Jupiter

Figure 8.2 A plot of the displacement versus time for an accelerating body on Jupiter, the

Earth, and the Earth’s Moon.

Since the plot of the distance x versus the time t shows the world line of an object in spacetime, the greater the acceleration implies the greater the curvature of that world line. If we look at this from the point of view of mass, the greater the mass, the greater will be the curvature of the world line in spacetime, hence the greater the curvature of spacetime itself. That is, mass causes spacetime to be curved. Also, if we attach a reference frame to the falling body, then we have an accelerated reference frame. And again, the greater the acceleration, the greater is the curvature of the accelerated reference frame in spacetime. We will see therefore, that accelerated reference frames will be curved frames in spacetime.

If mass warps spacetime how do we measure the warping? The warping results in a curvature of spacetime. How do we measure the curvature of spacetime? Let us start by determining the concept of curvature, then we will determine the curvature of a surface and eventually gravitate to the curvature of spacetime.

8.2 The Concept of Curvature The problem between the Euclidity of space and the non-Euclidity of spacetime can be

likened to early man’s observation of earth. To him the earth was flat, and nobody could convince him otherwise. All you had to do was look out at the surface of the earth and except for a few mountains; it was obvious to anyone that the earth must be flat. Today we would say that in a small-scale region, or “local” region, the earth appears flat, but in the large (or global view) it is a sphere.

In the same way our concept of spacetime, is limited to “local” regions and it is difficult for us to see the non-Euclidean “global” region of spacetime. Is spacetime flat or is it curved? How do we determine the curvature of spacetime? We have already said that matter warps spacetime. But how do we measure this warping of spacetime? For that matter, how do we measure the curvature or non-curvature of space itself? Let us start at the beginning and see what we mean by curvature.

Let us start by locating a point A along a curve, figure 8.3(a). O Is the origin of our coordinate system and the point A is located by the vector r. Let us draw 3 unit vectors at the point A.

t is a unit vector that is tangent to the surface at A and points in the direction that the object is moving at this particular instant of time. It is called the unit tangent vector because it is tangent to the curve at the point A. Even though it is tangent to the curve it is not necessarily perpendicular to the vector r. If the curve were a circle, then t would be perpendicular to r, but for the general case it is not.


8-3

r

O

bn

t

x

y A’

(b)

r

O

b

n

t

x

y A

(a) Figure 8.3 Location of a point on a curve.

n is the principle normal vector and is normal or perpendicular to the tangent vector at

the point A and points away from the curve. In general, n is not parallel to r. If the curve were a circle, then n would be parallel to r, but for the general case it is not. The vectors t and n form a plane, which is in the xy-plane.

b is the unit binormal vector and is perpendicular to the unit vectors t and n and therefore perpendicular to the xy-plane. Later we will also let the curve bend out of the xy-plane and look at the curvature then.

As you move along the curve, these unit vectors also change their direction in space as they follow the path of the curve.

If the initial curve were like the one in figure 8.3(b), then the curvature would obviously be different. In this case we would consider a point A’ along the curve. O is still the origin of our coordinate system and the point A’ is located by the vector r. We again draw 3 unit vectors at the point A’.

t is the unit vector that is tangent to the surface at A’ and points in the direction that the object is moving at this particular instant of time. It is still called the unit tangent vector because it is tangent to the curve at the point A’.

n is the principle normal vector and is normal or perpendicular to the tangent vector at the point A’ and points upward away from the curve. The vectors t and n again form a plane, which is in the xy-plane.

Notice that in Figure 8.3(a) n points upward away from the surface of the curve, while n points downward in Figure 8.3(b) away from the surface of the curve. Hence, n will vary in its direction depending on the curve considered, but it will always be perpendicular and outward from the curved surface.

b is the unit binormal vector in figure 8.3(b) and is again perpendicular to the unit vectors t and n and therefore perpendicular to the xy-plane.

Hence the three unit vectors will always be there but their direction will depend upon the curvature of the curve.

Let us now consider the motion of an object through space as the object moves from the original point A to a second point B as in figure 8.4. The tangent vector can be written as

dd

=rts

(8.2)

To prove this, the vector r specifies the position A while the vector (r + dr) locates the point B, figure 8.4. The direction of dr specifies the direction of the moving object at the particular instant of time, while the magnitude of the vector is dr. As you can see in the diagram, dr is the chord between A and B, while ds is the portion of the arc of the curve between these two points. In the limiting process as the angle θ between the two points A and B approaches 0, dr = ds, therefore dr/ds is a unit vector pointing in the direction of the tangent at a particular point, which we already stated in equation 8.2. Hence, the vector t is indeed a unit vector.


8-4

Figure 8.4 Motion of a body along the curve.

As the object moves along the curve, the unit tangent vector will also change. This can be seen in figure 8.5. The rate of change of t along the curve is a measure of the curvature of

Figure 8.5 The rate of change of the unit tangent vector. the curve itself. The rate of change of the unit tangent vector as the body moves from A to B is given by ∆t/∆s. This is called the average curvature of the arc, which is just the rate of change of t along the arc, which is the rate of change of the direction along the arc of the curve, figure 8.5(b). The actual curvature, K, at a point along the arc is defined as the limiting process of this rate of change or

2 1

0 0K = ' lim lim

s s

ds s ds∆ → ∆ →

− ∆= = =

∆ ∆t t t tt (8.3)

The quantity dt/ds is sometimes called the curvature vector. In the limit, when ∆s approaches zero, the angle θ also approaches zero, and the vector t2 − t1 becomes perpendicular to the tangent vector t1, as can be seen in figure 8.5(c). But if it is perpendicular to t, it is in the same direction as the unit normal vector n, as shown in figure 8.3(a) or (b). t’ is a unitary vector.

We can also show that t’ is perpendicular to t mathematically by using the definition of the dot product. That is

t t = 1 Upon differentiation

( ) 0d d dds ds ds

• = • + • =t tt t t t

Therefore

0dds

• =tt

which also shows that

' 0dds

• = • =tt t t (8.4)

O x

y AB

ds

θ 2rr1

rdr12r - =t1

t2

t1

t2 t2 t1-θ t1t2

t2 t1-θ

(a) (c)(b)

r

O x

y ABrd

ds

r + rdθ


8-5

Recall that when the scalar product of two vectors is equal to zero, the two vectors are perpendicular to each other. Hence, the vector t’ is perpendicular to t.

dds

⊥tt (8.5)

The faster t turns, the larger the components of t’. The length of the vector t’ is a measure of the rate of change of t along the curve and is a measure of the curvature of the curve itself. Also note that

2

2' d d d dds ds ds ds

= = =

t r rt (8.6)

Note that since dt/ds is perpendicular to the unit vector t, it is also parallel to the unitary

vector n. Hence we can also write the normal vector n in terms of the vector t’ as

''

=tnt

but t’|= K

Therefore, '

K=

tn

Or t’ = K n (8.7)

where K is called the curvature of the curve. Just as we used figure 8.5 to show how the unit tangent vector changed with the curvature of

the curve, we can use figure 8.6 to show how the unit normal vector changes with the curvature of the curve.

Figure 8.6 The rate of change of the unit normal vector.

As the object moves along the curve, the unit normal vector will also change. This can be seen in figure 8.6. The rate of change of n along the curve is also a measure of the curvature of the curve itself. The rate of change of the unit normal vector as the body moves from A to B is given by ∆n/∆s. This is also called the average curvature of the arc, which is just the rate of change of n along the arc, which is the rate of change of the normal to the direction along the arc of the curve, figure 8.6(b). The actual curvature, K, at a point along the arc is defined as the limiting process of this or

2 1

0 0K = ' lim lim

s s

ds s ds∆ → ∆ →

− ∆= = =

∆ ∆n n n nn (8.8)

(a) (c)(b)O x

y A

Bds

θ 2rr1

t1

t2

n1

n2

θn1

n2

n2 1- n

θ

n1 n2

n2 1- n


8-6

The quantity dn/ds is also sometimes called the curvature vector. In the limit, when ∆s approaches zero, the angle θ also approaches zero, and the vector n2 − n1 becomes perpendicular to the normal vector n1, as can be seen in figure 8.6(c). But if it is perpendicular to n, it is in the same direction as the unit tangent vector t, as shown in figure 8.3(a) or (b). n’ is a unitary vector.

We can also show that n’ is perpendicular to n mathematically by using the definition of the dot product. That is

n n = 1 Upon differentiation

( ) 0d d dds ds ds

• = • + • =n nn n n n

Therefore

0dds

• =nn

which also shows that

' 0dds

• = • =nn n n (8.9)

Recall that when the scalar product of two vectors is equal to zero, the two vectors are perpendicular to each other. Hence, the vector n’ is perpendicular to n.

dds

⊥nn (8.10)

The faster n turns, the larger the components of n’. The length of the vector n’ is a measure of the rate of change of n along the curve and is a measure of the curvature of the curve itself. Also note that

2

2' d d d dds ds ds ds

= = =

n r rn (8.11)

Note that since dn/ds is perpendicular to the unit vector n, it is also parallel to the tangent

vector t. Hence we can also write the tangent vector t in terms of the vector n’ as

''

=ntn

but |n’|= K

Therefore, '

K=

nt

Or n’ = K t (8.12)

where as before, K is called the curvature of the curve.

Using the tangent vector t and the normal vector n, we can write the binormal vector b as the vector product

b = t n (8.13)

The curvature here is all taking place in a plane and we will show examples of this in the next section. Later we will also look at the case where the curve can come out of the plane and we then have to look at another curvature; the curvature of the surface that is coming out of the plane.


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8.3 The Curvature of a Circle and an Ellipse As an example of the concept of curvature, let us consider the curvature of the circle shown in figure 8.7. Let us find the curvature of the circle between points A and B. We draw the radius R of the circle out to point A and then another to point B. We then draw the tangent vector t1 at the point A and the tangent vector t2 at the point B. When the radius R rotates from A to B it sweeps out the angle θ as

(a) (b)

O x

yA

Bθ

t1

t2R

R ∆s

t1

t2 t2 t1-θ

∆s

Figure 8.7 Curvature of a Circle.

shown. Since the tangent vectors t are perpendicular to the radius R, as R moves through the angle θ, the tangent vector also turns through the same angle θ. Hence the angle between the tangent vectors t1 and t2 is also the angle θ. This is shown in figure 8.7(b). (Figure 8.7(b) is slightly enlarged in order to see this.)

In general, the relation between the arc length and the angle turned through is given by

s = r θ (8.14)

and as can be seen in figure 8.7(b) the arc length s is just equal to the change in the tangent vector ∆t, hence equation 8.14 becomes

∆t = r θ (8.15) Also the radius r, from figure 8.7(b), is equal to the magnitude of the tangent vector t, which is just equal to 1 because the tangent vector t is a unit vector. Equation 8.15 becomes

∆t = 1 θ (8.16)

The curvature of the circle is given by equation 8.3 as

K = s

∆∆

t

Substituting equation 8.16 into equation 8.3 gives

K = s s

θ∆=

∆ ∆t (8.17)

But from equation 8.16 and figure 8.7(a) we see that


8-8

∆s = r θ = R θ

or θ = ∆s/R (8.18)

Replacing equation 8.18 into equation 8.17 gives for the curvature of the circle

/ 1K = s Rs s s R

θ∆ ∆= = =

∆ ∆ ∆t (8.19)

Hence the curvature of a circle is given by

1K = R

(8.20)

Therefore, the curvature of a circle at any point equals the reciprocal of the radius of the circle, and hence is the same for all points on the circle. Thus for a particular circle, the radius of curvature is a constant.

Example 8.1

Find the curvature of a circle of radius (a) R = 1 cm, (b) R = 2 cm, (c) R = 5 cm, (d) R = 1/2 cm, (e) R = 1/10 cm, (f) R = .

Solution

The curvature of the circle is found from equation 8.20. a. For R = 1 cm, the curvature is K = 1 = 1 = 1 cm−1 R 1 cm b. For R = 2 cm, the curvature is K = 1 = 1 = 0.5 cm−1 R 2 cm c. For R = 5 cm, the curvature is K = 1 = 1 = 0.2 cm−1 R 5 cm d. For R = 1/2 cm, the curvature is K = 1 = 1 = 2 cm−1 R 1/2 cm e. For R = 1/10 cm, the curvature is K = 1 = 1 = 10 cm−1 R 1/10 cm f. For R = cm, the curvature is K = 1 = 1 = 0 R cm

To go to this Interactive Example click on this sentence.

Notice that because of the reciprocal nature of the curvature, when the radius of curvature of the circle is small, the curvature is large; when the radius of curvature of the circle is large, the curvature is small, as would be expected. Also notice that K = 0 for R = . For this case the arc of the


8-9

circle is essentially a straight line and hence a straight line has zero curvature. Thus if K = 0, the surface is flat. When K is a large number we have a large curvature of the surface.

For an arbitrary curve, K is not a constant but varies from point to point. As an example, consider the ellipse shown in figure 8.8. Notice that the curvature is not a constant but varies as you go around the ellipse. To determine the curvature at any arbitrary point on the ellipse, place a circle that is tangential to the ellipse at that point. The radius of curvature of that circle is then measured, and the curvature K of the ellipse at that point is just the reciprocal of the radius of curvature of that tangential circle, equation 8.23. For another point on the ellipse draw another tangential circle for that point. Again, the radius of curvature of this second circle is then measured,

R 1

R 2

Figure 8.8 Determining the radius of curvature at any point on an ellipse.

and the curvature K of the ellipse at this point is again just the reciprocal of the radius of curvature of that tangential circle. In this way the curvature at any point on the ellipse can be determined. The drawn circle is sometimes called the osculating circle.

As an example, the radius of the osculating circle R1 in figure 8.8 is 0.90 cm giving a curvature of K1 = 1/R1 = 1/0.90 cm = 1.11 cm−1 at that point, while the radius of the osculating circle R2 is 1.70 cm giving a curvature of K2 = 1/R2 = 1/1.70 cm = 0.588 cm−1 at that point.

Using this technique of the osculating circle, the curvature at any point of any arbitrary curve can be determined.

8.4 Intrinsic Curvature

Our description of curvature so far is called external or extrinsic curvature. That is, we needed another external dimension to measure the curvature. For example for two-dimensional curvature we needed the third dimension to measure the curvature. If we look at a sphere we see a ball, but we also see the space exterior to the sphere. How do we measure the curvature of three-dimensional space? Or how would we measure the curvature of four-dimensional spacetime? The techniques that we just showed, although good, will not do the job for us.

In order to measure curvature in higher dimensions we need to abandon the extrinsic view and come up with an intrinsic or internal measuring technique. Gauss redefined the concept of curvature in terms of measurements made within the surface itself so that measurements are internal. What can we measure on a surface that will remain a constant? If we take a piece of paper, make two points A and B and draw a line between them, then if we bend the paper into any shape or curvature without tearing, shrinking, stretching or creasing the paper, the arc length of any curve on the paper will remain unchanged. In general, lengths, angles, areas would also be unaffected.


8-10

Gauss showed that the length of any curve drawn on the surface will remain unchanged. “Fortunately, Gauss had shown, in what was his crowning achievement in surface theory, his Theorema Egregium (or “extraordinary theorem”), that the curvature of a surface can actually be redefined (and computed) in terms of measurements made within the surface, and so is an intrinsic quantity, the surrounding space is not needed.”1

11 12

21 22ij

g gg

g g

=

The measurement of a length in a Cartesian coordinate system is given by

ds2 = dx2 + dy2 + dz2 (8.21)

while a measurement of a length in a Cylindrical coordinate system is given by

ds2 = dρ2 + ρ2dφ2 + dz2 (8.22)

In general the measurement of a length in a generalized coordinate system is given by

ds2 = gij dxi dxj (8.23)

where equation 8.23 is called the Fundamental Quadratic Form and gij is called the metric tensor and is given by

(8.24)

We will go into more detail with this in the next section.

8.5 Curvilinear Coordinates in Two Dimensions – The Differential Quadratic Form We have seen that when we deal with inertial frames of reference moving through spacetime, the subject matter of special relativity, we must deal with skewed coordinate systems, figure 8.9. We have shown that because of the skewed nature of the coordinate system, we had to deal with the contravariant and covariant components of vectors. To deal with the problem of general relativity, we have seen that mass warps spacetime, that is, spacetime becomes curved in the region of the mass, hence the straight line nature of the skewed coordinate system of special relativity must change to a

O

2

1

Figure 8.9 Skewed coordinate Figure 8.10 Curvilinear coordinate system system of special relativity. of general relativity.

1 Farber, Richard L. “Differential Geometry and Relativity Theory, An Introduction” Marcel Decker, Inc.


8-11

curvilinear skewed coordinate system, as shown in figure 8.10. This type of coordinate system is called a Gaussian coordinate system. As we showed in section 3.1 the coordinates x1 and x2 are called contravariant coordinates, because the components of a vector are found by dropping a line from the tip of the vector r, parallel to the x2 axis, to the xl axis. The component found this way was called the contravariant component of the vector r and was designated as x1. The second component was found by dropping the line from the tip of the vector r, parallel to the x1 axis, to the x2 axis. The component found this way was also called the contravariant component of the vector r and was designated as x2. The unitary base vectors a1 and a2 were called covariant vectors.

The necessity for using a Gaussian coordinate system is because the straight-line mesh system can only be traced out on a plane and could never be drawn on a curved surface. As an example, on a sphere the coordinate lines are great circles for the longitude lines and parallel circles for the latitude lines, and the coordinates of a point on a sphere are then given in terms of its latitude and longitude.

Let us now consider a point P in our Gaussian coordinate system and a neighboring point Q, as shown in figure 8.11. The vector displacement dr from the point P to the point Q is given by

1 2

1 2d dx dxx x

∂ ∂= +

∂ ∂r rr (8.25)

But what is 1/ x∂ ∂r ? This partial derivative tells you how the vector quantity dr varies as you move along the x1 axis. At the position P it points along the direction of the x1 axis, and hence we can define it as the unitary vector

1 1x∂

=∂

ra (8.26)

Figure 8.11 Infinitesimal displacement vector dr in curvilinear coordinates.

In the same way we ask what is 2/ x∂ ∂r ? Now this partial derivative tells you how the vector quantity dr varies as you move along the x2 axis. At the position P it points along the direction of the x2 axis, and hence we can define it as the unitary vector

2 2x∂

=∂

ra (8.27)

The unitary vectors a1 and a2 are tangent to the x1 and x2 axes respectively. The unitary vector a1 will vary along the x1 axis, and the unitary vector a2 will vary along the x2 axis, but they will have a constant value at a particular point (x1, x2). Because they are the result of a derivative, the unitary vectors a1 and a2 are probably not unit vectors. However, these unitary vectors a1 and a2 form a base system, and the infinitesimal vector dr, equation 8.25 can now be written as

dr = a1 dx1 + a2 dx2 (8.28)

The distance between point P and point Q can be found from

2 1 2 1 2

1 2 1 2ds d d dx dx dx dx = • = + • + r r a a a a (8.29) 2 1 1 1 2 2 1 2 2

1 1 1 2 2 1 2 2ds dx dx dx dx dx dx dx dx= • + • + • + •a a a a a a a a

But from the definition of the scalar product

2

1

Qdr

P

a 2

1a


8-12

a1 a2 = a2 a1 and, of course,

dx1dx2 = dx2dx1 therefore

2 1 1 1 2 2 21 1 1 2 2 22ds dx dx dx dx dx dx= • + • + •a a a a a a (8.30)

Using equations 8.26 and 8.27, we now make the following definitions

11 1 1 1 1gx x

∂ ∂= • = •

∂ ∂r ra a (8.31)

12 21 1 2 2 1g g= = • = •a a a a (8.32)

22 2 2 2 2gx x∂ ∂

= • = •∂ ∂

r ra a (8.33)

Replacing equations 8.31 through 8.33 into equation 8.30 gives

ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2 (8.34)

Equation 8.34 is called the first differential quadratic form2

11 12

21 22ij

g gg

g g

=

. The full significance of this form was first recognized by Gauss, who pointed out that all the metrical properties of any surface can be expressed in terms of its coefficients, gij. Equation 8.34 was derived for a two dimensional surface. We can generalize equation 8.34 for any number of dimensions by writing it in the form

ds2 = Σ gij dxidxj (8.35)

Where the Greek letter Σ means that the sum will be made over the indices i and j. Einstein simplified the notation by leaving out the summation sign but saying whenever the dummy index is used twice in the expression, it will be summed over, hence

ds2 = gij dxidxj (8.36)

The quantities gij, called the metrical coefficients, are sometimes written in a matrix format as

(8.37)

Equation 8.36 gives the length of the quantity ds as an invariant quantity, because it comes from the product of the covariant metrical coefficients gij and the contravariant coordinates dxidxj.

The matrix equation 8.37 is sometimes called a tensor equation. A tensor is, in general, an n-dimensional matrix. Equation 8.37 is a tensor of order two, which means that essentially it has two dimensions associated with it. That is, the rows extend in the x-direction, while the columns extend in the y-direction, hence two dimensions, and for this case, it is a tensor of order two. If we were working in three dimensions, x, y, and z, then the tensor would have three columns and three rows, but it would still be a tensor of order two. A tensor of order three can be visualized as a three dimensional matrix. That is, besides the two dimensions of rows and columns shown in equation 8.37, it would also have rows and columns going into the page, which is the third dimension. Hence, besides the rows in the x-direction, there would also be rows in the z-direction, which is into and out of the page. Hence the matrix would have three dimensions and would be a tensor of order three. Even though you can have tensors of order four and higher, it is not as easy to visualize them.

2This is sometimes referred to as the fundamental quadratic form.


8-13

Let us now consider some special cases. If dr is taken only in the direction of x1, then we have

11 11 ds g dx=

While if dr is taken only in the direction of x2, then we have

22 22 ds g dx=

The angle between the two unitary vectors a1 and a2 is found from the definition of the dot

product, as ( )1 2 1 2 1 2cos ,• =a a a a a a

Hence,

( ) 1 21 2

1 2

cos , •=

a aa aa a

(8.38)

But we already showed in equation 8.32 that a1 a2 = g12

Hence, equation 8.38 becomes

( ) 121 2

1 2

cos , g=a a

a a (8.39)

The magnitude |a1| is found from ( )1 1 1 1 1 1 1 1cos ,• = =a a a a a a a a

But from our definition in equation 8.31

a1 a1 = g11 (8.31) Hence

|a1|2 = g11 and

1 11g=a (8.40) The magnitude |a2| is found similarly as

( )2 2 2 2 2 2 2 2cos ,• = =a a a a a a a a

But from the definition in equation 8.33 a2 a2 = g22 (8.33)

Hence |a2|2 = g22

and 2 22g=a (8.41)

Replacing equations 8.40 and 8.41 into equation 8.39 yields the angle between the two unit vectors as

( ) 121 2

11 22

cos , gg g

=a a (8.42)

Notice from equation 8.42 that if g12 = 0 over the surface, then cos(a1,a2) = 0 and therefore the coordinate axes x1 and x2 are orthogonal.

Let us now consider some examples of different coordinate systems.


8-14

(a) An orthogonal Cartesian coordinate system.

Example 8.2

Find the first differential quadratic form for an orthogonal Cartesian coordinate system.

The infinitesimal displacement vector dr for a Cartesian coordinate system is shown in figure 8.12. The vectors a1 and a2 are unit vectors. The first differential quadratic form is found from equation 8.34 as

ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2

Since this is an orthogonal system x1 and x2 must be orthogonal, or perpendicular, to each other everywhere, which implies that a1 is perpendicular to a2. Hence, from the definition in equation 8.32, we have that

g12 = a1 a2 = a1 a2 cos900 = 0 (8.43) and equation 8.34 becomes

ds2 = g11 dx1dx1 + g22 dx2dx2 (8.44)

Let us now determine g11 for a Cartesian coordinate system. Recall from the definition in equation 8.31 that

11 1 1 1 1gx x

∂ ∂= • = •

∂ ∂r ra a (8.31

But what is ∂r/∂x1 for a Cartesian coordinate system? For a Cartesian coordinate system, x1 = x and x2 = y, and r = ix + jy. Therefore

( )1

x y x yx x x xx

∂ +∂ ∂ ∂ ∂= = = +

∂ ∂ ∂ ∂∂

i jr r i j (8.45)

Figure 8.12 Infinitesimal displacement vector dr in Cartesian coordinates.

But

1 and 0x yx x

∂ ∂= =

∂ ∂

Therefore

1x∂

=∂

r i (8.46)

Replacing equation 8.46 into 8.31 gives

11 1 1 1 1 1gx x

∂ ∂= • = • = • =

∂ ∂r ra a i i

Therefore, for an orthogonal coordinate system we have

g11 = 1 (8.47)

Solution

Qdr

Pa 2

1a

2

1

j

i


8-15

A similar calculation will yield g22 for a Cartesian coordinate system. Recall from equation 8.33 that

22 2 2 2 2gx x∂ ∂

= • = •∂ ∂

r ra a (8.33)

But what is ∂r/∂x2 for a Cartesian coordinate system? For a Cartesian coordinate system, x1 = x and x2 = y, and r = ix + jy. Therefore

( )2

x y x yy y y yx

∂ +∂ ∂ ∂ ∂= = = +

∂ ∂ ∂ ∂∂

i jr r i j (8.48)

But

0 and 1x yy y

∂ ∂= =

∂ ∂

Therefore

2x∂

=∂

r j (8.49)

Replacing equation 8.49 into 8.48 gives

22 2 2 2 2 1gx x∂ ∂

= • = • = • =∂ ∂

r ra a j j

Therefore, for an orthogonal coordinate system we also have

g22 = 1 (8.50)

Replacing equations 8.47 and 8.50 into equation 8.44 yields

ds2 = dx1dx1 + dx2dx2 (8.51) But since x1 = x and x2 = y, equation 8.51 becomes

ds2 = (dx)2 + (dy)2 (8.52)

Equation 8.52 is the first differential quadratic form for an orthogonal coordinate system, and we see that for this case it is nothing more than the Pythagorean theorem.

The matrix of the metrical coefficients gij, equation 8.37, for the orthogonal coordinate system is written in the form

11 12

21 22

1 001ij

g gg

g g

= =

(8.53)

Also note that the quantities gij are constants for the Cartesian coordinate system

(b) A skewed coordinate system.

Example 8.3

Find the first differential quadratic form for a skewed coordinate system.

Solution


8-16

The infinitesimal displacement vector dr for a skewed coordinate system is shown in figure 8.13. The vectors a1 and a2 are unit vectors. The first differential quadratic form is found from equation 8.34 as

ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2 From the definition in equation 8.31 we have

11 1 1 1 1gx x

∂ ∂= • = •

∂ ∂r ra a (8.31)

But in the skewed coordinate system already treated in chapter 4, we found that

r = x1a1 + x2a2 (4.2) Therefore

1 21 21 2

1 21 1 1 1 1x xx x

x x x x x∂ ∂∂ ∂ ∂

= + + +∂ ∂ ∂ ∂ ∂

a ar a a (8.54)

In a skewed coordinate system the unit vector a2, which points in the direction of the x2 axis, does not vary as you move along the x1-axes, hence the term

21 0

x∂

=∂a (8.55)

And similarly, the unit vector a1, which points in the direction of the x1 axis, does not vary as you move along the x1-axes, hence

11 0

x∂

=∂a (8.56)

Figure 8.13 Infinitesimal displacement

vector dr in a skewed coordinate system.

Also since x2 and x1 are independent axes and are not functionally related to each other, the derivative

2

1 0xx

∂=

∂ (8.57)

and finally 1

1 1xx

∂=

∂ (8.58)

Replacing equations 8.55 through 8.58 into equation 8.54 yields

11x∂

=∂

r a (8.59)

Hence,

11 1 11 1gx x

∂ ∂= • = •

∂ ∂r r a a

Since a1 is a unit vector we have for our skewed coordinate system

g11 = 1 (8.60)

Qdr

P

a 2

1a

2

1


8-17

A similar calculation will yield g22 for a skewed coordinate system. From the definition in

equation 8.33 we have

22 2 2 2 2gx x∂ ∂

= • = •∂ ∂

r ra a (8.33)

But as we already showed r = x1a1 + x2a2 (4.2)

Therefore 1 2

1 21 21 22 2 2 2 2

x xx xx x x x x

∂ ∂∂ ∂ ∂= + + +

∂ ∂ ∂ ∂ ∂a ar a a (8.61)

In a skewed coordinate system the unit vector a2, which points in the direction of the x2 axis, does not vary as you move along the x2-axes, hence the term

22 0

x∂

=∂a (8.62)

And similarly, the unit vector a1, which points in the direction of the x1 axis, does not vary as you move along the x2-axes, hence

12 0

x∂

=∂a (8.63)

Also since x2 and x1 are independent axes and are not functionally related to each other, the derivative

2

2 1xx

∂=

∂ (8.64)

and finally 1

2 0xx

∂=

∂ (8.65)

Replacing equations 8.62 through 8.65 into equation 8.61 yields

22x∂

=∂

r a (8.66)

Hence,

22 2 22 2gx x∂ ∂

= • = •∂ ∂

r r a a

Since a2 is a unit vector we have for our skewed coordinate system

g22 = 1 (8.67)

To determine the metrical coefficient g12 we rearrange equation 8.42, into the form

( )12 11 22 1 2cos ,g g g= a a (8.68)

The angle between the two unit vectors a1 and a2 is the angle α of the skewed coordinate system, and using equations 8.60 and 8.67 equation 8.68 becomes

12 cosg α= (8.69)


8-18

Substituting equations 8.60, 8.67, and 8.69 into equation 8.34 gives

ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2 ds2 = dx1dx1 + 2 cosα dx1dx2 + dx2dx2

ds2 = (dx1)2 + 2 cosα dx1dx2 + (dx2)2 (8.70) Equation 8.70 is the first differential quadratic form for a skewed coordinate system.

The matrix of the metrical coefficients gij, equation 8.37, for the skewed coordinate system is written in the form

11 12

21 22

1 coscos 1ij

g gg

g gα

α

= =

(8.71)

Note, that since the angle α is a measure of the skewedness of the coordinate system, it is equal to a constant for a particular system, and hence the quantities gij are also constants for the skewed coordinate system. Note that as a special case, if the angle of skewedness α = 900 then the coordinates would be perpendicular to each other, and since the cosα = cos900 = 0, then equation 8.71 would reduce to equation 8.53 which gives the value of gij for orthogonal coordinates. That is an orthogonal coordinate system is a special case of a skewed coordinate system when α = 900.

(c) A Polar Coordinate System.

Example 8.4

Find the first differential quadratic form for a polar coordinate system.

Figure 8.14 Infinitesimal displacement vector dr in a polar coordinate system.

The infinitesimal displacement vector dr for a polar coordinate system is shown in figure 8.14. The

Solution

Q

dr

P

41

24

21 = constant

22 = constant

= constant

23 = constant

r

drr +

21

11

31

a 2 a 1


8-19

variable x1 represents the radial distance r from the center of the circle and x2 represents the angular distance θ from the x-axis. Vectors a1 and a2 are the unitary vectors and point in the directions of increasing r and θ respectively.

The locus of all points a distance x11 from the origin is a circle around the origin of radius x11, and the locus of all points a distance x12 from the origin is a circle around the origin of radius x12, and so on for the circles of radii x13, x14, etc. Hence everywhere along x1, the distance r is a constant.

Since x2 represents the angular distance θ from the x-axis, the straight line x21 represents all points that make an angle of θ = x21 = 00, while the straight line x22 represents all points that make an angle θ = x22 = 22.50 with respect to the x-axis. Similarly, the straight-line x23 represents all points that make an angle of θ = x23 = 45.00, while the straight-line x24 represents all points that make an angle θ = x24 = 67.50 with respect to the x-axis, etc. Hence, everywhere along x2 the angle θ is a constant.

The point P is located by the position vector r, given by

r = r r0 = x1 r0 (8.72)

where r0 is a unit vector in the direction of increasing r. (The unit vector r0 is the same as the unitary vector a1). The point Q is located by the position vector r + dr. The infinitesimal displacement vector dr between the points P and Q for a polar coordinate system is shown in figure 5.14.

The first differential quadratic form is again found from equation 8.34 as

ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2 The metric coefficient g11 is found from equation 8.31 as

11 1 1 1 1gx x

∂ ∂= • = •

∂ ∂r ra a (8.31)

where ( ) ( )0 0

01

r rr

r r r rx∂ ∂∂∂ ∂

= = = +∂ ∂ ∂ ∂∂

r rr r r (8.73)

But 0 0r

∂=

∂r (8.74)

because r0 is a unit vector in the direction of increasing r and hence does not change as r varies. Also

1rr

∂=

∂ (8.75)

Replacing equations 8.74 and 8.75 into equation 8.73 gives

01x∂

=∂

r r (8.73)

Replacing equation 8.73 into equation 8.31 yields

11 0 01 1 1gx x

∂ ∂= • = • =

∂ ∂r r r r

Hence g11 = 1 (8.76)


8-20

The metric coefficient g22 is found from equation 8.33 as

22 2 2 2 2gx x∂ ∂

= • = •∂ ∂

r ra a (8.33)

where ( ) ( )0 0

02

r rr

x θ θ θ θ∂ ∂∂∂ ∂

= = = +∂ ∂ ∂ ∂∂

r rr r r (8.77)

But

0θ

∂=

∂r (8.78)

because the distance r is independent of the angle θ. But what is 0 ?θ

∂=

∂r Consider the circle of radius

ro in figure 8.15(a). When the unit vector ro turns through an angle dθ, it also sweeps out the arc

(a) (b) Figure 8.15

length ds. Now if the angle dθ is small, as it is since it is a differential angle, then the chord dr0 is approximately equal to the arc length ds. That is, since the distance moved along a circle of radius r0 is

s = r0 θ its differential is

ds = r0 dθ (8.79)

And as we just pointed out, the arc ds is approximately equal to the chord dr0 as seen in figure 8.15(a). The direction of the vector dr0 is perpendicular to the vector r0 itself. Hence we can define a new unit vector θ0 which is perpendicular to r0, and points in the direction of increasing θ, as seen in figure 8.15(b). We can now rewrite equation 8.79 as

dr0 =|r0|dθ θ0 Since r0 is a unit vector, its magnitude is |r0| = 1. Hence

dr0 = dθ θ0 which we can rewrite as

0

θ∂

=∂r

θ0 (8.80)

Replacing equation 8.78 and 8.80 into equation 8.77 yields

d ro

d θ

r 0

r 0

ds P r

r 0 θ 0 θ


8-21

( )0 002

rr r r

x θ θ θ θ∂∂ ∂∂ ∂

= = + = =∂ ∂ ∂ ∂∂

r rr r r θ0 (8.81)

Replacing equation 8.81 into equation 8.33 yields

22 2 2gx x∂ ∂

= •∂ ∂

r r = rθ0 rθ0 = r2(θ0 θ0) = r2

Hence g22 = r2 (8.82)

To determine the metrical coefficient g12 we use equation 8.68.

( )12 11 22 1 2cos ,g g g= a a (8.68) However the two unitary vectors a1 and a2 for this case are perpendicular to each other. Hence

( ) ( )01 2cos , cos 90 0= =a a

and g12 = g21 = 0 (8.83)

Substituting equations 8.76, 8.82, and 8.83 into equation 8.9 gives

ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2 ds2 = g11 drdr + 2 g12 drdθ + g22 dθdθ

ds2 = dr2 + r2 (dθ)2 (8.84)

Equation 8.84 is the first differential quadratic form for a polar coordinate system. The matrix of the metrical coefficients gij, equation 8.24, for the polar coordinate system is

written in the form 11 12

221 22

010 ij

g gg

g g r

= =

(8.85)

Notice that since g22 is not a constant, but varies as r2, the mesh system will vary with the coordinate r. The greater the value of r, the greater the area enclosed within the grid lines x1, x2.

Also note from equation 8.33 that

22 2 2 2 2gx x∂ ∂

= • = •∂ ∂

r ra a

Hence

2 2x∂

=∂

ra (8.86)

But equation 8.86 is equal to equation 8.81. That is,

2 rx∂

=∂

rθ0 (8.81)

Hence a2 = r θ0 (8.87)

Notice that the unitary vector a2 is in the same direction as the unit vector θ0. However, notice that the unitary vector a2 is not a unit vector. It is equal to r times the unit vector θ0.

Also note from equation 8.31 that


8-22

11 1 1 1 1gx x

∂ ∂= • = •

∂ ∂r ra a (8.31)

Hence

1 1x∂

=∂

ra (8.88)

But equation 8.88 is equal to equation 8.73.

01x∂

=∂

r r (8.73)

Hence a1= r0 (8.89)

Notice that in this case, the unitary vector a1 is equal to the unit vector r0. That is, the unitary vector a1 is a unit vector. Hence, for the case of polar coordinates, the unitary vector a1 is a unit vector, while the unitary vector a2 is not a unit vector.

Notice that the mesh system of each coordinate system that we considered up to now was a

constant mesh system. That is, for the orthogonal coordinate system shown in figure 8.12, the coordinates x1 and x2 generated a series of squares of equal areas. For the skewed coordinate system in figure 8.13 the coordinates x1 and x2 generated a series of quadrangles each of equal area.. Of course the skewed areas of figure 8.13 are not equal to the square areas of figure 8.12, but the area mesh for the particular coordinate system was a constant. Now notice the mesh system of the polar coordinates shown in figure 8.16. The mesh system is no longer a constant and the area varies with the coordinate r. The larger the value of r, the greater the area enclosed within the grid lines x1, x2. This occurs because the metric coefficient g22 is no longer a constant but varies as r2.

Figure 8.16 Area of the mesh system of polar coordinates.

(d) Skewed Coordinates of Special Relativity.

Example 8.5

Write down the differential quadratic form for the spacetime diagram of special relativity, and list the metrical coefficients gij.


8-23

The differential quadratic form for spacetime of special relativity is the invariant interval,

ds2 = dx2 − c2 (dt)2 (8.90) Equation 8.90 is the first differential quadratic form for spacetime of special relativity.

The matrix of the metrical coefficients gij, equation 8.37, for special relativity is written in the form

11 122

21 22

010 ij

g gg

g g c

= = − (8.91)

Notice that g22 is a constant, equal to the negative of the square of the speed of light, -c2.

For the four dimensions of flat spacetime, the invariant interval would be

ds2 = dx2 + dy2 + dz2 − c2 (dt)2 (8.92)

and the matrix of the metrical coefficients gij, equation 8.37, for the four dimensional flat spacetime of special relativity is written in the form

= = −

11 12 13 14

21 22 23 24

31 32 33 342

41 42 43 44

1 0 0 00 1 0 00 0 1 00 0 0

ij

g g g gg g g g

gg g g gg g g g c

(8.93)

In the standard notation of general relativity, equation 8.92 becomes

ds2 = gij dxidxj (8.94)

Equation 8.94 gives the length of the quantity ds as an invariant quantity, because it comes from the product of the covariant metrical coefficients gij and the contravariant coordinates dxidxj.

We still must find the invariant interval for curved spacetime and relate it to the gravitational mass that is causing the curvature of spacetime. 8.6 The Differential Quadratic Form for a Rotating Disk Let us consider a rigid disk that is rotating at a constant angular velocity ω, as shown in figure 8.17(a). Figure 8.17(a) represents the motion as observed from the moving frame of reference while Figure 8.17(b) represents the motion as observed from the stationary frame of reference. The symmetry of the problem suggests that we use a cylindrical coordinate system to describe the motion of the disk. The coordinates that we would normally use are

x = r cos θ y = r sin θ

z = z

Solution


8-24

But since the disk is rotating at the constant angular velocity ω, the angle is changing with time. So we will use the coordinate θ + ωt as the angle. Hence the coordinates that we will use are

x = r cos(θ + ωt) (8.95) y = r sin(θ + ωt) (8.96) z = z (8.97)

So we are using the three coordinates r, θ, z, and t to describe the motion. To get the fundamental quadratic form, we will need the quantities dx2, dy2, and dz2. To determine the quantity dx we differentiate equation 8.95 as

dx = r[−sin(θ + ωt)(dθ + ωdt)] + cos(θ + ωt) dr or

dx = −rsin(θ + ωt)(dθ + ωdt) + cos(θ + ωt) dr (8.98)

and dx2 comes from squaring equation 8.98 as

dx2 = r2sin2(θ + ωt)(dθ + ωdt)2 + cos2(θ + ωt) dr2 − 2rsin(θ + ωt)(dθ + ωdt) cos(θ + ωt) dr (8.99)

To determine the quantity dy we differentiate equation 8.96 as dy = rcos(θ + ωt)(dθ + ωdt)] + sin(θ + ωt) dr (8.100) and dy2 comes from squaring equation 8.100 as

Figure 8.17 The rotating disk.

dy2 = r2cos2(θ + ωt)(dθ + ωdt)2 + sin2(θ + ωt) dr2 + 2rcos(θ + ωt)(dθ + ωdt) sin(θ + ωt) dr (8.101)

And of course dz = dz (8.102)

and dz2 = dz2 (8.103)

Combining equations 8.99, 8.101, and 8.103 gives dx2 + dy2 + dz2 = r2sin2(θ + ωt)(dθ + ωdt)2 + cos2(θ + ωt) dr2 − 2rsin(θ + ωt)(dθ + ωdt) cos(θ + ωt) dr

+ r2cos2(θ + ωt)(dθ + ωdt)2 + sin2(θ + ωt) dr2 + 2rcos(θ + ωt)(dθ + ωdt) sin(θ + ωt) dr + dz2

dx2 + dy2 + dz2 = r2sin2(θ + ωt)(dθ + ωdt)2 + r2cos2(θ + ωt)(dθ + ωdt)2

+ cos2(θ + ωt) dr2 + sin2(θ + ωt) dr2 + dz2

dx2 + dy2 + dz2 = r2 [sin2(θ + ωt) + cos2(θ + ωt)] (dθ + ωdt)2 + [sin2(θ + ωt) + cos2(θ + ωt)] dr2 + dz2

But sin2(θ + ωt) + cos2(θ + ωt) = 1

Therefore dx2 + dy2 + dz2 = r2 (dθ + ωdt)2 + dr2 + dz2

X

Y

Z

Pθ

r

ω

(b)

z

y

xr

ω

θrcosθ

r sinθ P

(a)


8-25

or dx2 + dy2 + dz2 = r2 (dθ2 + 2ωdθdt + ω2dt2) + dr2 + dz2 dx2 + dy2 + dz2 = r2 dθ2 + 2ω r2dθdt + ω2 r2dt2 + dr2 + dz2 (8.104)

The fundamental quadratic form for this rotating disk becomes

ds2 = dx2 + dy2 + dz2 – c2dt2 ds2 = r2 dθ2 + 2ω r2dθdt + ω2 r2dt2 + dr2 + dz2 – c2dt2

ds2 = dr2 + r2 dθ2 + dz2 + 2ω r2dθdt + (ω2 r2 – c2)dt2 ds2 = dr2 + r2 dθ2 + dz2 + 2ω r2dθdt + (ω2 r2

2 22 2 2 2 2 2 2 2

2 = + + + 2 1 rds dr r d dz r d dt c dtcωθ ω θ

− −

– 1) c2dt2 c2

(8.105)

Equation 8.105 gives the fundamental quadratic form for the rotating disk. The matrix of the metrical coefficients gij, equation 8.37, for the four dimensional spacetime is written in the form

2 211 12 13 14

21 22 23 24

31 32 33 34 2 22

41 42 43 44 2

1 0 0 00 0 20 0 1 0

0 0 0 1

ij

g g g g r rg g g g

gg g g g

r cg g g gc

ω

ω

= =

− −

(8.106)

As a special case, let us consider a person at rest on the disk and rotating with the disk. For this rotating observer, who is at rest on the rotating disk, the coordinates dθ = 0, dr = 0, and dz = 0. For this observer, the fundamental quadratic form, equation 8.105, becomes

2 2

2 2 22 = 1 rds c dt

cω

− −

(8.107)

For an observer at rest at the very center of the disk in the stationary frame of reference, Figure 8.17(b), his fundamental quadratic form would be given by

ds2 = dX2 + dY2 + dZ2 – c2dT2 (8.108)

This observer is totally at rest because r = 0 at the center of the disk and v = ωr, so if r = 0 then v = 0. (In fact, figure 8.17(b) shows the disk more like a Merry Go-Round, with a stationary ground at the center of the disk.) After one complete revolution, his spacial coordinates would not have changed because his coordinate system is at rest. Hence dX = dY = dZ = 0, and the time interval he would read on his own clock which is at rest, would give him the value dT. Hence,

ds2 = – c2dT2 (8.109)

Since the interval ds is an invariant, we can equate equations 8.107 and 8.109 as

ds2 = ds2 2 2

2 2 2 221 r c dt c dT

cω

− − = −


8-26

22

2 2

2

1

dTdtrcω

=

−

(8.110)

Taking the square root of both sides of the equation we get

2 2

2

1

dTdtrcω

=

−

(8.111)

But v = ωr, and hence v2 = ω2r2. Substituting this into equation 8.111 gives

2

2

1

dTdtvc

=

−

(8.112)

The time dT is the time interval as observed by the stationary observer at the center of the rotating disk, which we call dt0, the proper time for the clock. The rotating observer reads his clock time to be dt. With this notation, equation 8.112 would be written as

( )=

−

0

2 2

1 /

dtdtv c

(8.113)

But this is exactly the Lorentz time dilation equation we developed in chapter 1. So we have again shown that an accelerated body (the disk is undergoing a centripetal acceleration as it is rotating) has the same time dilation as an inertial body.

Everything we have said about time dilation for a rotating disk, also has an equivalent length contraction, as in special relativity. However for an object on the rotating disk the distance from the center of the disk to the object represents a length that is moving perpendicular to the motion and hence does not experience the length contraction of special relativity. However a length that is perpendicular to the radius is a length in motion and will experience the Lorentz-Fitzgerald contraction.

8.7 The Differential Quadratic Form for Spherical Coordinates Let us consider a sphere as shown in figure 8.18. We can represent a point on the sphere by the coordinates x, y, z or by the spherical coordinates r, θ, φ. The symmetry of the problem suggests that we use the spherical coordinate system to describe the system. The coordinates that we use can be seen from figure 8.18 as

x = r sinφ cosθ (8.114) y = r sinφ sinθ (8.115) z = r cosφ (8.116)

So we are using the three coordinates r, θ, φ, and t to describe the system. To get the fundamental quadratic form, we will need the quantities dx2, dy2, and dz2. To determine the quantity dx we differentiate equation 8.114 as

dx = r[sinφ(−sinθ dθ) + cosθcosφ dφ] + sinφcosθ dr (8.117)


8-27

Figure 8.18 Spherical Coordinates.

or

dx = rsinφ(-sinθ dθ) + rcosθcosφ dφ + sinφcosθ dr dx = −rsinφsinθ dθ + rcosθcosφ dφ + sinφcosθ dr

and finally dx = sinφ[cosθ dr − rsinθ dθ] + rcosθcosφ dφ (8.118)

Its square, dx2, comes from squaring equation 8.118 as

dx2 = sin2φ[cosθ dr − rsinθ dθ]2 + [rcosθcosφ dφ]2

+ 2 sinφ[cosθ dr − rsinθ dθ][rcosθcosφ dφ] (8.119)

dx2 = sin2φ[cos2θ dr2 + r2sin2θ dθ2 − 2 cosθ dr rsinθ dθ] + r2cos2θcos2φ dφ2 + 2 sinφ[cosθ dr − rsinθ dθ][rcosθcosφ dφ]

dx2 = sin2φ[cos2θ dr2 + r2sin2θ dθ2 − 2 cosθ dr rsinθ dθ] + r2cos2θcos2φ dφ2

+ [2 sinφcosθ dr − 2 sinφrsinθ dθ][rcosθcosφdφ]

dx2 = sin2φ[cos2θ dr2 + r2sin2θ dθ2 − 2 cosθ dr rsinθ dθ] + r2cos2θcos2φ dφ2 + [2 sinφcosθ dr rcosθcosφdφ − 2 sinφrsinθ dθ rcosθcosφdφ]

dx2 = sin2φ cos2θ dr2 + sin2φ r2sin2θ dθ2 − 2 sin2φcosθ dr rsinθ dθ + r2cos2θcos2φ dφ2

+ 2 sinφcosθ dr rcosθcosφdφ − 2 sinφ rsinθ dθ rcosθcosφdφ

The quantity dx2 is

dx2 = sin2φcos2θ dr2 + r2 sin2φsin2θ dθ2 + r2cos2θcos2φ dφ2 − 2r sin2φcosθsinθ dr dθ + 2r sinφcosφcos2θdr dφ − 2r2 sinφcosφsinθcosθ dθ dφ (8.120)

To determine the quantity dy we differentiate equation 8.115 as

y = r [sinφsinθ] (8.115) dy = r[sinφcosθ dθ + sinθcosφ dφ] + [sinφsinθ]dr (8.121)

and dy2 comes from squaring equation 8.121 as


8-28

dy2 = r2[sinφcosθ dθ + sinθcosφ dφ]2 + [sinφsinθ]2 dr2 + 2r[sinφcosθ dθ + sinθcosφ dφ][sinφsinθ]dr

dy2 = r2 [sin2φcos2θ dθ2 + sin2θcos2φ dφ2 + 2sinφcosθ dθ sinθcosφ dφ] + sin2φsin2θ dr2 + 2r[sinφcosθ dθ sinφsinθ dr + sinθcosφ dφ sinφsinθ dr]

Finally, dy2 = r2 sin2φcos2θ dθ2 + r2 sin2θcos2φ dφ2 + 2 r2 sinφcosθ dθ sinθcosφ dφ

+ sin2φsin2θ dr2 + 2rsinφcosθ dθ sinφsinθ dr + 2rsinθcosφ dφ sinφsinθ dr (8.122) To determine the quantity dz we differentiate equation 8.116 as

z = r cosφ (8.116)

dz = r (−sinφ dφ) + cosφ dr dz = − rsinφ dφ + cosφ dr (8.123)

and dz2 comes from squaring equation 8.123 as

dz2 = r2sin2φ dφ2 + cos2φ dr2 − 2rsinφ dφ cosφ dr or

dz2 = r2sin2φ dφ2 + cos2φ dr2 − 2rsinφcosφ dr dφ (8.124) Combining equations 8.120, 8.122, and 8.124 gives dx2 + dy2 + dz2 = sin2φcos2θ dr2 + r2 sin2φsin2θ dθ2 + r2cos2θcos2φ dφ2

− 2r sin2φ cosθsinθ dr dθ + 2r sinφcosφcos2θ dr dφ − 2r2 sinφcosφsinθcosθ dθ dφ + r2 sin2φcos2θdθ2 + r2 sin2θcos2φ dφ2 + 2r2 sinφcosθsinθcosφ dθ dφ + sin2φsin2θ dr2

+ 2rsinφcosθ dθ sinφsinθ dr + 2rsin2θcosφ dφ sinφ dr + r2sin2φ dφ2 + cos2φ dr2 − 2rsinφcosφ dr dφ (8.125)

But the terms

2rsinφcosφcos2θ dr dφ + 2rsin2θcosφ dφ sinφ dr = 2r sinφcosφ[sin2θ + cos2θ] dr dφ and

sin2θ + cos2θ = 1 Therefore

2rsinφcosφcos2θ dr dφ + 2rsin2θcosφ dφ sinφ dr = 2r sinφcosφ dr dφ Therefore equation 8.125 becomes dx2 + dy2 + dz2 = sin2φcos2θ dr2 + r2 sin2φsin2θ dθ2 + r2cos2θcos2φ dφ2 − 2r sin2φcosθsinθ dr dθ + 2r sinφcosφ dr dφ − 2r2 sinφcosφsinθcosθ dθ dφ + r2 sin2φcos2θ dθ2 + r2 sin2θcos2φ dφ2 + 2r2 sinφcosθsinθcosφ dθ dφ + sin2φsin2θ dr2

+ 2rsinφcosθ dθ sinφsinθ dr + r2sin2φ dφ2 + cos2φ dr2 − 2rsinφcosφ dr dφ (8.126) But

2r sinφcosφ dr dφ − 2r sinφcosφ dr dφ = 0 Therefore equation 8.126 beccomes dx2 + dy2 + dz2 = sin2φcos2θ dr2 + r2 sin2φsin2θ dθ2 + r2cos2θcos2φ dφ2 − 2r sin2φcosθsinθ dr dθ

+ r2 sin2φcos2θ dθ2 + r2 sin2θcos2φ dφ2 + sin2φsin2θ dr2 + 2rsinφcosθ dθ sinφsinθ dr + r2sin2φ dφ2 + cos2φ dr2 (8.127)

But sin2φcos2θ dr2 + sin2φsin2θ dr2 = sin2φ[sin2θ + cos2θ]dr2 = sin2φ dr2


8-29

Hence, equation 8.127 becomes dx2 + dy2 + dz2 = sin2φ dr2 + r2 sin2φsin2θ dθ2 + r2cos2θcos2φ dφ2 − 2r sin2φcosθsinθ dr dθ + r2 sin2φcos2θ dθ2 + r2 sin2θcos2φ dφ2 + 2rsinφcosθ dθ sinφsinθ dr

+ r2sin2φ dφ2 + cos2φ dr2 (8.128) But

r2 sin2φsin2θ dθ2 + r2 sin2φcos2θ dθ2 = r2 sin2φ[sin2θ + cos2θ]dθ2 = r2 sin2φ dθ2 Therefore, equation 8.128 becomes dx2 + dy2 + dz2 = sin2φ dr2 + r2 sin2φ dθ2 + r2cos2θcos2φ dφ2 − 2r sin2φcosθsinθ dr dθ

+ r2 sin2θcos2φ dφ2 + 2rsinφcosθ dθ sinφsinθ dr + r2sin2φ dφ2 + cos2φ dr2 (8.129) But

sin2φ dr2 + cos2φ dr2 = [sin2φ + cos2φ]dr2 = dr2 Hence, equation 8.129 becomes dx2 + dy2 + dz2 = dr2 + r2 sin2φ dθ2 + r2cos2θcos2φ dφ2 − 2r sin2φcosθsinθ dr dθ + r2 sin2θcos2φ dφ2 + 2rsinφcosθsinφsinθ dr dθ + r2sin2φ dφ2 dx2 + dy2 + dz2 = dr2 + r2 sin2φ dθ2 + r2cos2θcos2φ dφ2 + r2sin2φ dφ2 − 2r sin2φcosθsinθ dr dθ

+ r2 sin2θcos2φ dφ2 + 2rsinφcosθsinφsinθ dr dθ (8.130) But

2rsinφcosθsinφsinθ dr dθ = 2rsin2φcosθsinθ dr dθ and

− 2r sin2φcosθsinθ dr dθ − 2rsinφcosθsinφsinθ dr dθ = 0 Therefore,

dx2 + dy2 + dz2 = dr2 + r2 sin2φ dθ2 + r2cos2θcos2φ dφ2 + r2 sin2θcos2φ dφ2 + r2sin2φ dφ2 (8.131)

But r2cos2θcos2φ dφ2 + r2 sin2θcos2φ dφ2 = r2cos2φ dφ2[sin2θ + cos2θ]

= r2cos2φ dφ2

Therefore, dx2 + dy2 + dz2 = dr2 + r2 sin2φ dθ2 + r2cos2φ dφ2 + r2sin2φ dφ2 (18.132)

But r2cos2φ dφ2 + r2sin2φ dφ2 = r2[cos2φ + sin2φ]dφ2 = r2dφ2

Equation 18.132 becomes

dx2 + dy2 + dz2 = dr2 + r2dφ2 + r2 sin2φ dθ2 (8.133)

Equation 8.133 is called the fundamental quadratic form for spherical coordinates. Since the fundamental quadratic form for spacetime is

ds2 = dx2 + dy2 + dz2 – c2dt2

The fundamental quadratic form for spacetime in spherical coordinates is

ds2 = dr2 + r2dφ2 + r2 sin2φ dθ2 – c2dt2 (8.134) The matrix of the metrical coefficients gij, equation 8.37, for the four dimensional spacetime is written in the form


8-30

φ

= = −

11 12 13 142

21 22 23 242 2

31 32 33 342

41 42 43 44

1 0 0 00 0 00 0 sin 00 0 0

ij

g g g gg g g g rgg g g g rg g g g c

(8.135)

We should point out here that the angle φ in figure 8.18 is between r and the z-axis, while the

angle θ is between rsinφ and the x-axis. That is θ is in the x-y plane while φ is perpendicular to that plane. In some books these angles are reversed. That is, the angle θ is made the angle between r and the z-axis, while the angle φ is made to be the angle between rsinθ and the x-axis. In this case the fundamental quadratic form takes the form

dx2 + dy2 + dz2 = dr2 + r2dθ2 + r2 sin2θ dφ2 (8.136)

And the fundamental quadratic form for spacetime in spherical coordinates then becomes

ds2 = dr2 + r2dθ2 + r2 sin2θ dφ2 – c2dt2 (8.137)

In this form the matrix of the metrical coefficients gij, equation 8.37, for the four dimensional spacetime is then written in the form

= = −

11 12 13 142

21 22 23 242 2

31 32 33 342

41 42 43 44

1 0 0 00 0 00 0 sinθ 00 0 0

ij


(8.138)

Which is the correct form? Either one can be used depending on the way the angles are defined

in figure 8.18. In most standard textbooks in the lower level of mathematics and physics courses, the form of equation 8.133 is used. In some very advanced books in relativity the form of equation 8.136 is used. It is in this form that the standard form for Schwarzschild’s equation is written. We will see more about this in the next chapter.

Summary of Basic Concepts Warped spacetime Matter causes spacetime to be warped so that the world lines of particles in spacetime are curved. Gravity is no longer to be thought of as a force in the Newtonian tradition but it is rather a consequence of the warping or curvature of spacetime caused by matter. Hence, matter warps spacetime and spacetime tells matter how to move. Extrinsic Curvature

Is spacetime flat or is it curved? How do we determine the curvature of spacetime? But how do we measure this warping of spacetime? For that matter, how

do we measure the curvature or non-curvature of space itself? The original approach to a description of curvature is called external or extrinsic curvature. That is, we need another external dimension to measure the curvature. For example for two-dimensional curvature we needed the third dimension to measure the curvature. If we look at a sphere we see a ball, but we also see the space exterior to the sphere. How do we measure the curvature of three-dimensional space? Or how would we measure the curvature of four-dimensional spacetime? In order to measure curvature in higher dimensions we need to abandon


8-31

the extrinsic view and come up with an intrinsic or internal measuring technique. Curvature of a circle The curvature of a circle at any point equals the reciprocal of the radius of the circle, and hence is the same for all points on the circle. Thus for a particular circle, the radius of curvature is a constant. Notice that because of the reciprocal nature of the curvature, when the radius of curvature of the circle is small, the curvature is large; when the radius of curvature of the circle is large, the curvature is small. If the curvature is zero then the radius of the circle is infinite, hence for this case the arc of the circle is essentially a straight line and hence a straight line has zero curvature. Thus if K = 0, the surface is flat. When K is a large number we have a large curvature of the surface. Curvature of an ellipse

For an ellipse the curvature is not a constant but varies as you go around the ellipse. To determine the curvature at any arbitrary point on the ellipse, place a circle that is tangential to the ellipse at that point. The radius of curvature of that circle is then measured, and the curvature K of the ellipse at that point is just the reciprocal of the radius of curvature of that tangential circle. For another point on the ellipse draw another tangential circle and just do the same thing. The drawn circle is called an osculating circle. This technique can be used to determine the curvature of any surface. Intrinsic Curvature Gauss showed that the length of any curve drawn on the surface will remain unchanged, and that the curvature of a surface can actually be redefined and computed in terms of measurements made within the surface, and so is an intrinsic quantity, the surrounding space is not needed.

Gauss said that all the metrical properties of any surface can be described in

terms of the fundamental quadratic form, or the differential quadratic form:

ds2 = gij dxidxj

where gij are called the metrical coefficients. Curvilinear Coordinates in Two Dimensions – The Differential Quadratic Form We have seen that when we deal with inertial frames of reference moving through spacetime, the subject matter of special relativity, we must deal with skewed coordinate systems. To deal with the problem of general relativity, we have seen that mass warps spacetime, that is, spacetime becomes curved in the region of the mass, hence the straight line nature of the skewed coordinate system of special relativity must change to a curvilinear skewed coordinate system. This type of coordinate system is also called a Gaussian coordinate system. The Differential Quadratic Form for a Rotating Disk A rotating disk is an example of accelerated motion. In particular, the acceleration is called centripetal acceleration. The fundamental quadratic form for this rotating disk becomes

θ

ωω θ

− −

2 2 2 2 2

2 22 2 2

2

= + +

+ 2 1

ds dr r d dzrr d dt c dtc

The time dilation on the accelerated rotating disk is the same as in special relativity and there is a tangential length contraction but not a radial one. The fundamental quadratic form for spacetime in spherical coordinates is ds2 = dr2 + r2dφ2 + r2 sin2φ dθ2 – c2dt2 (8.134)

If the angles φ and θ are reversed the fundamental quadratic form then becomes



8-32

Summary of Important Equations

Tangent vector dd

=rts

(8.2)

Curvature, K, at a point along the arc

2 1

0 0K = ' lim lim

s s

ds s ds∆ → ∆ →

− ∆= = =

∆ ∆t t t tt (8.3)

dt/ds is the curvature vector.

2 1

0 0K = ' lim lim

s s

ds s ds∆ → ∆ →

− ∆= = =

∆ ∆n n n nn (8.8)

Curvature of a circle 1K = R

(8.20)

binormal vector b = t n (8.13) Fundamental Quadratic Form - measurement of a length in a generalized coordinate system ds2 = gij dxi dxj (8.23)

The metric tensor 11 12

21 22ij

g gg

g g

=

(8.24)

Unitary vectors 1 1x∂

=∂

ra (8.26)

2 2x∂

=∂

ra (8.27)

Infinitesimal vector dr = a1 dx1 + a2 dx2 (8.28) Metrical coefficients

11 1 1 1 1gx x

∂ ∂= • = •

∂ ∂r ra a (8.31)

12 21 1 2 2 1g g= = • = •a a a a (8.32)

22 2 2 2 2gx x∂ ∂

= • = •∂ ∂

r ra a (8.33)

First differential quadratic form ds2 = g11 dx1dx1 + 2 g12 dx1dx2 + g22 dx2dx2

(8.34) ds2 = Σ gij dxidxj (8.35) ds2 = gij dxidxj (8.36)

Metrical coefficients 11 12

21 22ij

g gg

g g

=

(8.37)

Angle between the two unit vectors

( ) 121 2

11 22

cos , gg g

=a a (8.42)

11 12

21 22

1 001ij

g gg

g g

= =

First differential quadratic form Orthogonal coordinate system

ds2 = (dx)2 + (dy)2 (8.52) gij, for the orthogonal coordinate system

(8.53)

Skewed coordinate system

ds2 = (dx1)2 + 2 cosα dx1dx2 + (dx2)2 (8.70) gij, for the skewed coordinate system

11 12

21 22

1 coscos 1ij

g gg

g gα

α

= =

(8.71)

Polar coordinate system

ds2 = dr2 + r2 (dθ)2 (8.84) gij, for the polar coordinate system

11 122

21 22

010 ij

g gg

g g r

= =

(8.85)

Spacetime of special relativity

ds2 = dx2 − c2 (dt)2 (8.90) gij, for special relativity

11 122

21 22

010 ij

g gg

g g c

= = − (8.91)

Four dimensions of flat spacetime, the invariant interval would be

ds2 = dx2 + dy2 + dz2 − c2 (dt)2 (8.92) the metrical coefficients gij,

= = −

11 12 13 14

21 22 23 24

31 32 33 342

41 42 43 44

1 0 0 00 1 0 00 0 1 00 0 0

ij

g g g gg g g g

gg g g gg g g g c

(8.93) In the standard notation of general relativity, equation 8.92 becomes

ds2 = gij dxidxj (8.94) The fundamental quadratic form for the rotating disk

ds2 = dx2 + dy2 + dz2 – c2dt2


8-33

θ

ωω θ

− −

2 2 2 2 2

2 22 2 2

2

= + +

+ 2 1

ds dr r d dzrr d dt c dtc

(8.105)

The matrix of the metrical coefficients gij, for the rotating disk is

2 211 12 13 14

21 22 23 24

31 32 33 34 2 22

41 42 43 44 2

1 0 0 00 0 20 0 1 0

0 0 0 1

ij

g g g g r rg g g g

gg g g g

r cg g g gc

ω

ω

= =

− −

The fundamental quadratic form for spherical coordinates dx2 + dy2 + dz2 = dr2 + r2dφ2

+ r2 sin2φ dθ2 (8.133) The fundamental quadratic form for spacetime in spherical coordinates ds2 = dr2 + r2dφ2 + r2 sin2φ dθ2 – c2dt2

(8.134)

The matrix of the metrical coefficients for four dimensional spacetime is

φ

= = −

11 12 13 142

21 22 23 242 2

31 32 33 342

41 42 43 44

1 0 0 00 0 00 0 sin 00 0 0

ij


(8.135) If the angles φ and θ are reversed the fundamental quadratic form then becomes


and the metrical coefficients gij, then becomes

= = −

11 12 13 142

21 22 23 242 2

31 32 33 342

41 42 43 44

1 0 0 00 0 00 0 sinθ 00 0 0

ij


(8.138)

Questions for Chapter 8

1. What does it mean to say that

spacetime is warped? 2. How can you measure how much

spacetime is warped? 3. Explain the concept of curvature. 4. What is the difference between extrinsic

curvature and intrinsic curvature? 5. Why must you use intrinsic curvature

to study spacetime?

6. What is the fundamental quadratic form?

7. A disk rotating at a constant angular velocity has the same equations as the Lorentz equations. What do you think would happen to these equations if there was an angular acceleration of the disk?

To go to another chapter, return to the table of contents by clicking on this sentence.

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