Tải bản đầy đủ - 0 (trang)
Masses, Distances and Times in the Universe

Masses, Distances and Times in the Universe

Tải bản đầy đủ - 0trang

Masses, Distances and Times in the Universe



Lu =











Rm



n



L ( Rm )

4π R 2 dR = 4π Rm2 L ( Rm )n[ R]∞Rm → ∞

( R / Rm )



51



(3.1)



But, before making reasonable estimates of the mass for a typical star

(like our Sun), a typical galaxy (like our Milky Way) and the observable

universe, let us remind ourselves what is the mass of the Earth and the

mass of the Sun, using simple newtonian arguments. The mass of the

Earth can be obtained using the quantity of the gravitational acceleration

on its surface g ≈ 980 cm/sec2, the quantity of Newton’s universal

gravitational constant, G ≈ 6.67 × l0-8 cm3 g-1 sec-2, and the Earth’s radius,

RT ≈ 6378 km, which is fairly well known since the time of Erathostenes.

The values of g and G can be obtained without much difficulty from

experiments performed on Earth’s surface. We have



F = mg = G



mMT

RT2



g

, i.e. MT =   RT2 = 5.97 × 1027 g

G



(3.2)



On the other hand the mass of the Sun can be obtained from the fact

that, to keep the Earth in its elliptic but very nearly circular orbit around

the Sun, it is necessary that the effective centrifugal force be exactly

compensated by the gravitational attraction,



MT (2π RST / TT )2

M M

(2π / TT )2 3

= G T 2 S , i.e. M S =

RST

RST

G

RST



(3.3)



33



= 1.99 × 10 g

Where RTS is the Earth-Sun distance, and use has been made of the

fact that νT = 2πRST / TT, i.e., the Earth’s average velocity, obtained

dividing the length of the orbit, 2πRST, by the period of time necessary to

complete one revolution, TT = 365 × 24 × 60 × 60 sec.

In what follows we will show how very simple physical arguments3

can give us excellent estimates of the mass of stars and galaxies. We will

use the dimensionless quantities



α ≡ e2 / ℏc ≃ 1/137



(3.4)



52



The Intelligible Universe



and



α g ≡ Gm 2p / ℏc ≃ 6.3 × 10 −39



(3.5)



which measure, respectively, the strength of the electrostatic interaction,

and the much smaller gravitational interaction, in terms of physical

constants (see table in Sec. 2.2). We will also make use of the virial’s

theorem,4 which states that for a system in equilibrium in a central field,

Ekin =



1

Epot

2



(3.6)



where Ekin = kinetic energy, Epot = potential energy. A star is made up of a

plasma of electron and ions (basically protons in the hydrogen burning

phase of the life of a star, which is the longest one) and we can equate

the kinetic energy of particles in the surface of this plasma which we

assume to have spherical shape to a representative electrostatic ionization

energy, so that



( Ekin )S ≈ k BT ≈



e2

α rC



(3.7)



The gravitational potential energy of a proton with mass mp, on the

other hand, is given by

m M

(3.8)

( Epot )S ≈ G Rp S

S

Then, making use of the virial theorem, Eq. (3.2), we can write,

dividing by ħc both sides of the equation,

2

1  e2  1  Gm p  N S



 





α  ℏc  2  ℏcc  RS /rC



(3.9)



where NS is the number of protons within the sphere of radius Rs, and,

therefore

NS ≃



4π 3 4π 3

RS

rC , i.e. ( RS /rC ) ≃ N S1 / 3

3

3



(3.10)



Substituting (RS/rS) = NS1/3 in Eq. (3.9), and taking into account the

definitions in Eq. (3.4) and (3.5), we get



Masses, Distances and Times in the Universe



 1

NS ≃  2

 αg





53



3/2













≃ 5.65 × 1057



(3.11)



which is the number of protons (baryons) in the plasma sphere which

makes up the star. The star’s mass, hence, will be given by

M S = N S m p ≃ 9.4 × 1033 g



(3.12)



which is in fairly good agreement with the solar mass evaluated above,

Eq. (3.3). Note that the expression for the mass of a typical star as given

by Eqs. (3.11) and (3.12) involves only general physical constants, like

the combination αg = Gmp2/ ℏ c and nip, a truly remarkable result. Figure 3.1

depicts the equilibrium configuration of a star.

The mass of a protogalaxy, i.e. a compact galactic mass presumably

in the very first moment after formation, can be estimated in a similar

way. In this case gravitational pressure can be assumed to be

compensated by radiation pressure from the heated up plasma, and one can

conceive a near equilibrium situation when the gravitational free fall

time, tg, for particles in the surface of the quasi-spherical matter

distribution, is comparable5 to the characteristic dissipation time, td, of

radiant energy coming away from the hot plasma. If tg « td, the

protogalactic plasma would collapse towards the center of mass before

reaching quasi-equilibrium conditions. If, on the other hand, tg » td,

radiation pressure would push the protogalactic matter away from the

center.

The gravitational free fall time can be given in terms of the

protogallactic mass MG, (which must be the same as the gallactic mass),

and the protogallactic radius RPG, (which may be much smaller than the

future galactic radius), taking into account the newtonian expression for

uniformly accelerated motion,

2

RPG ≃ 12 (GMG / RPG

)tG2



(3.13)



54



The Intelligible Universe



Fig. 3.1. Equilibrium configuration of a star leading to an estimate of its mass.



M S ≈ (2 / α g )3 / 2 m p ≈ M sun

from which tg can be eliminated as

3

t g ≃ (2 RPG

/ GMG )1/ 2



(3.14)



The characteristic dissipation time can be estimated dividing the total

radiation energy, Er, enclosed by the protogallactic sphere, by the rate at

which radiation energy is given up at its surface, i.e.,

Er

1 RPG

td ≃



(3.15)

2

4

c 4π RPG (σ T ) 3 c

where as given by Stefan–Boltzmann’s law, the energy density per unit

volume in terms of the absolute temperature T, is equal to (σT4), and,

therefore

 4π 3 

(3.16)

Er = 

RPG  (σ T 4 )

 3



Equating tg and td, and taking into account that MG = NG mp, where

NG in the number of protons in the whole protogalaxy, one gets

3

2

2 RPG

18 RPG  m p c 

1 RPG



,

i

.

e

.

N







G

GN G m p 9 c 2

 Gm2p   ℏ 





 ℏc 







Now we can make use of the fact that



(3.17)



Masses, Distances and Times in the Universe



 m p c  m p  me c  m p 1





 ℏ  = m  ℏ  = m αr



e 

e

c







55



(3.18)



and taking into account that



(RPG / rc ) ≈ N G1 / 3



(3.19)



a relationship fully analogous to that in Eq. (3.10), we can finally get

3/2



N G ≃ (18m p /me )



 1



 αα g















3/ 2



≃ 3.3 × 1067



(3.20)



the number of protons (baryons) in the whole protogalaxy. The galaxy

mass is, therefore



MG = N G m p ≃ 5.5 × 10 43 g



(3.21)



This is of the order of 1010 typical star masses, since, as shown in

Eq. (3.12), Ms ≈ 9.4 × 1033 g. Again, the expression of MG as given by

Eqs. (3.20) and (3.21) involves only numerical factors of order unity and

combinations of general physical constants, such as (mp/mc) ≈ 1836, the

mass ratio of proton and electron, α = e2/ ℏ c ≈ 1/137 (dimensionless),

αg = Gmp2/ ℏ c ≈ 6.3 × 10-39 (dimensionless), and mp, the proton mass. It

should be mentioned that Eq. (3.21) agrees fairly well with estimates of

the total mass of our own galaxy, the Milky Way or Via Lactea, which

contains about 1011stars, and therefore has a mass



M VL = 1011 M SUN ≈ 2 × 1044 g



(3.22)



The mass of the observable universe, MU cannot be calculated, at least

for the time being, in a way similar to those used to estimate Ms and MG,

because, as we know, the whole universe is expanding at fantastic speed,

and we are in no way near an equilibrium situation. However one can

make a rough estimate of its total mass if one takes seriously current

observational estimates of its present average density, ρ0, its present

radius, R0 ≈ τ0/C, and its spatial curvature k < 0, corresponding to an open

universe with less mass than that necessary for a hypothetical future

phase of contraction after the present expansion.

Using, for instance,



56



The Intelligible Universe



ρ0 ≃ 2 × 10 −31 g/cm 3



(3.23)



R0 ≃ 1.6 × 10 28 cm (τ 0 ≃ 1.75 × 1010 years)



(3.24)



one can substitute ρ 0, and R0 in the expression of MU to get



MU ≃ ρ0V0 ≃ ρ0



4π 3

R0 ≃ 3.8 × 1055

3



(3.25)



which is of the order of 1012 galactic masses, equivalent roughly to the

actually observable number of galaxies. It may be noted that, strictly

speaking, the expression used for the volume

V0 ≃



4π 3

R0

3



(3.26)



corresponds to the euclidean volume (flat space |k| = 0), somewhat

different from the more general relationship7



V0 ≃ 









4π r 2 dr  3

 R0

0 (1+ |k | r 2 )1 / 2



1



(3.27)



which would be the correct general relativistic expression for a noneuclidean open space. However, the definite integral within square

brackets, which is elementary and can be found in tables, becomes (4π/3)

if |k |2 = 0, and does not differ by more than 15% from this value for |k | =

+1, or |k | = −1, corresponding, respectively, to a closed and an open

universe. Errors in the estimates of ρ 0 and R0 are larger than 15%. The

results thus obtained for number of baryons and masses of succesively

larger objects in the cosmos are summarized in Table 3.1.

Table 3.1

Object or System

Star

Galaxy

Observable Universe



Calculated number of

barions

Ns ≈ 5.6 × 1057

NG ≈ 3.3 × 1067



Mass (gr)

Ms ≈ 9.4 × 1033

MG ≈ 5.5 × 1043

MU ≈ 3.8 × 1055



Number of units in

larger object

(MG/Ms) ≈ 1010

(MU/MG) ≈ 1012

(MU/MU) ≈ 1



Masses, Distances and Times in the Universe



57



3.2. Distances

As mentioned in the historical introduction, the oldest estimates of

astronomical distances go back to the time of Erathostenes and

Aristarchus of Samos. While Erathostenes, using a very simple method,

made a very good estimate of the Earth’s radius, within of 5% the correct

value, Aristarchus went from the Earth to the moon and from there to the

sun, in two steps, so to speak. The latter’s arguments were correct, but

some of the observational data he used lacked the necessary accuracy,

and, consequently, his numerical estimates for sizes and distances to

moon and sun were seriously flawed. We can reproduce the basic

reasoning of Erathostenes and Aristarchus and use modern observational

data to obtain successively the Earth’s diameter, the distance to the moon

and the distance to the sun.



Fig. 3.2. Erathostenes method to measure the Earth’s circumference. The distance from

S (Syene, Southern Egypt) to A (Alexandria) was SA ≈ 5000 stades ≈ 800 km, and

the angle between the vertical and the direction of sunrays at A, on the particular day

and time at which no shadow was present at S, could be measured as a = 7.2°. Therefore

2πRT = (α/360) AS ≈ 40,000 km.



It may be noted that the greek geometer had the good luck of relaying

on the distance between two cities, Syene, directly under the ecliptic in

Southern Egypt, and Alexandria, in the Mediterranean cost, connected by

the Nile river, the only great river running South to North almost along a

meridian line in the Northern Hemisphere. Thanks to this happy

coincidence he was able to measure with very good accuracy the distance

between both cities, 5000 stades. The angle made by sunrays with the



The Intelligible Universe



58



vertical at Alexandria (in the same year’s day and time in which no

shadow could be seen at noon at Syene) was 7.2º, i.e. one fiftieth of a full

circle, and consequently, the Earth’s circumference was estimated as

250,000 stades or 40,000 km.

Figure 3.3 shows another happy coincidence, perspicaciously noted

by Aristarchus, which paved the way for an estimate of the size of the

diameter of the moon’s orbit. As he noted, it takes, in a lunar eclipse,

almost the same time for the moon to begin to reappear after full



Fig. 3.3. Lunar eclipse. As noted by Aristarchus the path of the moon within the cone of

full shadow produced by the Earth is just about two lunar diameters. This fact pointed the

way to an estimate of the Earth-moon distance in terms of the lunar diameter.



occultation as it took her to disappear after making contact with the

shadow cone produced by the Earth. This implies that the length of the

path travelled during the eclipse was about two lunar diameters (more

exactly n = 1.8 times the lunar diameter). On the other hand the maximum

time for a lunar eclipse is tEL = 1 hour, 42 min. i.e. 102 min, relatively

small in comparison with the time needed for the noon to complete a full

orbit around the Earth, which is TL = 27.3 days. Therefore, assuming that

the orbital velocity is constant, we have

t EL

nDL

=

=, i.e.

TL 2π DTL



dTL  27.3 × 24 × 60  1.8 

=



 ≃ 110.45

102

DL 

 2π 



(3.28)



Masses, Distances and Times in the Universe



59



where DL is the lunar diameter and dTL the Earth–moon distance. On the

other hand the apparent angular diameter of the moon as seen from the

Earth’s surface is directly measurable, and is given by



θL =

dTL



DL

= 9.205 × 10 −3 rad

1

− DT

2

i.e.



dTL

1 1 DT

=

+

DL θ L 2 DL



(3.29)



where DT is the Earth’s diameter. Combining Eqs. (3.28) and (3.29) one

gets

DT

= 0.276

DL



(3.30)



which means that the moon’s linear diameter in fairly large, between

one forth and one third of the Earth’s linear diameter. Also, introducing

Eq. (3.30) into Eq. (3.28), one gets

dTL

= 30.46

DL



(3.31)



which, taking into account that DT = (40,000/π) km leads to dTL = 384 ×

106 m as the Earth–moon distance.

Finally, to estimate the Earth-sun distance in terms of the Earth-moon

distance, Aristarchus considered the triangle made up by Earth, moon

and sun when exactly half moon is viewed from the Earth. The angle a

subtended from the center of the Earth to the centers of moon and sun is

very approximately the same, although slightly smaller, than the

corresponding angle subtended from the point nearest the moon on the

earths surface (see Fig. 3.4). While it is only eight arc minutes (8’)

smaller than a right angle, due to the very large distance from sun to

Earth, this difference was not beyond the observational capabilities even

at Aristarchus time but he did measure it with a large error.



The Intelligible Universe



60



Fig. 3.4. Triangle made up by Earth, Moon and Sun when exactly half moon is viewed

from the Earth, which allows an estimate of the Earth–sun distance, dTS, in terms of the

Earth–moon distance, dTL, and the angle α.



From Fig. 3.4 it can be seen that

cos α =



dTL

dTS



(3.32)



and, taking into account that the observable value of a is 89 degrees and

52 minutes, very close to a right angle, one gets



dTS =



dTL

= 149 × 109 m

cos α



(3.33)



a very large distance, indeed.

Also, taking into account that the angular dimension of the sun is

close to that of the moon, it may be expected that the linear diameter of

the sun be about (dTS/dTL) times the angular diameter of the moon. Using

previous results and the fact that



θs =



DS

D

≃ S ≃ 12.88 × 10 −3 rad

1

dTS − 2 DT dTS



(3.34)



one can readily check that

DS

= 109.1

DT



(3.35)



i.e. that the size of the sun is more than one hundred times that of the

Earth, or RS = RSUN = 6.96 × 105 km.



Masses, Distances and Times in the Universe



61



These were the first steps of man into its cosmic neighbourhood but,

as we will see immediately, these large distances were puny in

comparison with the far vaster distances between stars and between

galaxies.

The radius of a typical star can be calculated from Eq. (3.10) as

Rs = N S1 / 3rc ≃ 8.9 × 105 Km



(3.36)



where Ns ≈ 5.65 × 1057 and rc ≡ (1/α) ( ℏ /mec) = 0.5 × 10-8 cm have been

used.

This is very close to RSUN as given by Eq. (3.36). The size of a typical

star remains fairly constant during most of its long lifetime. On the other

hand the same argument applied to a protogalaxy would give, using now

NG ≈ 3.3 × 1067, something like

RG = N G1/ 3rc ≃ 1.6 × 10 9 Km



(3.37)



which is far smaller than the radius of an actual galaxy of comparable

mass. Fortunately we can use observational data on the rotation speeds of

outermost stars in an actual spiral galaxy (Andromeda) to get a realistic

value for the present radius of a typical spiral galaxy, RG.

Notwithstanding the dark matter (or missing mass) question, it can be

argued that the centrifugal acceleration of the outermost star is

compensated by the acceleration of gravity towards the galactic center,

and, therefore



M S vS2

M M

= G S2 G , i.e.

RG

R G

RG =



GMG

vS2



= 9.1 × 10 21 cm = 9.1 × 1016 Km (3.38)



where G = 6.67 × l0-8 cm-1 g-1 sec2, MG = 5.5 × l043 g and vs = 200 ×

l05 cm/sec have been used. This is equivalent to

RG c

365 × 24 × 60



2



= 9.17 × 10 3 light year



(3.39)



Tài liệu bạn tìm kiếm đã sẵn sàng tải về

Masses, Distances and Times in the Universe

Tải bản đầy đủ ngay(0 tr)

×