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z

@vz rdrd'dz.

in dz given by @z

The volume of the in nitesimal part of space shown in gure (4.2) is given by rdrd'dz.

Problem d: Use the fact that the divergence is the ux per unit volume to show that in

cylinder coordinates:

v = 1 @r (rvr ) + 1 @v' + @vz :

@ (4.15)

r r @' @z

r

Problem e: Use this result to re-derive equation (4.8) without using Cartesian coordi-

nates as an intermediary.

4.5. IS LIFE POSSIBLE IN A 5-DIMENSIONAL WORLD? 37

In spherical coordinates a vector v can be expended in the components vr , v and v'

in the directions of increasing values of r, and ' respectively. In this coordinate system

r has a di erent meaning than in cylinder coordinates because in spherical coordinates

p

r = x2 + y2 + z 2.

Problem f: Show that in spherical coordinates

v = r12 @r r2vr + r sin @@ (sin v ) + r sin @v'

@ 1 1 (4.16)

@'

r

4.5 Is life possible in a 5-dimensional world?

In this section we will investigate whether the motion of the earth around the sun is stable

or not. This means that we ask ourselves the question that when the position of the

earth a perturbed, for example by the gravitational attraction of the other planets or by

a passing asteroid, whether the gravitational force brings the earth back to its original

position (stability) or whether the earth spirals away from the sun (or towards the sun).

It turns out that the stability properties depend on the spatial dimension! We know that

we live in a world of three spatial dimensions, but it is interesting to investigate if the orbit

of the earth would also be stable in a world with a di erent number of spatial dimensions.

In the Newtonian theory the gravitational eld g(r) satis es (see ref: 42]):

(r g) = G (4.17)

;4

where (r) is the mass density and G is the gravitational constant which has a value of

6:67 10;8 cm3 g;1 s;2 . The term G plays the role of a coupling constant, just as the

1/permittivity in (4.12). Note that the right hand side of the gravitational eld equation

(4.17) has an opposite sign as the right hand side of the electric eld equation (4.12). This

is due to the fact that two electric charges of equal sign repel each other, while two masses

of equal sign (mass being positive) attract each other. If the sign of the right hand side

of (4.17) would be positive, masses would repel each other and structures such as planets,

the solar system and stellar systems would not exist.

Problem a: We argued in section (4.3) that electric eld lines start at positive charges

and end at negative charges. By analogy we expect that gravitational eld lines end

at the (positive) masses that generate the eld. However, where do the gravitational

eld lines start?

Let us rst determine the gravitational eld of the sun in N dimensions. Outside the

sun the mass-density vanishes, this means that (r g) =0. We assume that the mass

density in the sun is spherically symmetric, the gravitational eld must be spherically

symmetric too and is thus of the form:

g(r) = f(r)r : (4.18)

In order to make further progress we must derive the divergence of a spherically symmetric

vector eld in N dimensions. Generalizing expression (4.16) to an arbitrary number of

dimensions is not trivial, but fortunately this is not needed. We will make use of the

qPN 2

property that in N dimensions: r = i=1 xi .

CHAPTER 4. THE DIVERGENCE OF A VECTOR FIELD

38

Problem b: Derive from this expression that

@r=@xj = xj =r : (4.19)

Use this result to derive that for a vector eld of the form (4.18):

(r g) =Nf(r) + r @f : (4.20)

@r

Outside the sun, where the mass-density vanishes and (r g) =0 we can use this result to

solve for the gravitational eld.

Problem c: Derive that

Ar

g(r) = rN;1 ^ (4.21)

;

and check this result for three spatial dimensions.

At this point the constant A is not determined, but this is not important for the coming

arguments. The minus sign is added for convenience, the gravitational eld points towards

the sun hence A > 0.

Associated with the gravitational eld is a gravitational force that attracts the earth

towards the sun. If the mass of the earth is denoted by m, this force is given by

Fgrav = rAm ^

N;1 r (4.22)

;

and is directed towards the sun. For simplicity we assume that the earth is in a circular

orbit. This means that the attractive gravitational force is balanced by the repulsive

centrifugal force which is given by

mv2 ^ :

Fcent = r r (4.23)

In equilibrium these forces balance: Fgrav + Fcent = 0.

Problem d: Derive the velocity v from this requirement.

We now assume that the distance to the sun is perturbed from its original distance r

to a new distance r + r, the perturbation in the position is therefore r = r ^. Because

r

of this perturbation, the gravitational force and the centrifugal force are perturbed too,

these quantities will be denoted by Fgrav and Fcent respectively, see gure (4.3).

Problem e: Show that the earth moves back to its original position when:

( Fgrav + Fcent ) r < 0 (stability) : (4.24)

Hint: consider the case where the radius is increased ( r > 0) and decreased ( r < 0)

separately.

4.5. IS LIFE POSSIBLE IN A 5-DIMENSIONAL WORLD? 39

v + Î´v

v

Fcent

Fgrav

Î´r

r

Fcent + Î´ Fcent

Fgrav + Î´Fgrav

Figure 4.3: De nition of variables for the perturbed orbit of the earth.

Hence the orbital motion is stable for perturbations when the gravitational eld satis es

the criterion (4.24). In order to compute the change in the centrifugal force we use that

angular momentum is conserved, i.e. mrv = m(r + r)(v + v). In what follows we will

consider small perturbations and will retain only terms of rst order in the perturbation.

This means that we will ignore higher order terms such as the product r v.

Problem f: Determine v and derive that

3mv2 r

Fcent = (4.25)

r2

;

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