Homogeneous Universe - Revision history

Cosmo All at 01:56, 13 December 2013

2013-12-13T01:56:55Z

Cosmo All at 17:21, 4 May 2013

2013-05-04T17:21:51Z

Oleg at 17:21, 4 May 2013

2013-05-04T17:21:16Z

Igor at 15:21, 11 October 2012

2012-10-11T15:21:56Z

Igor: /* Homogeneous and isotropic Universe, Hubble’s law */

2012-10-11T15:19:44Z

‎Homogeneous and isotropic Universe, Hubble’s law

Igor: /* Problem 2: global isotropy */

2012-10-11T15:19:11Z

‎Problem 2: global isotropy

Igor at 10:33, 10 August 2012

2012-08-10T10:33:48Z

Igor: /* Problem 5: Galilean invariance of the Hubble law */

2012-08-10T10:32:08Z

‎Problem 5: Galilean invariance of the Hubble law

Igor: /* Homogeneous and isotropic Universe, Hubble’s law */

2012-08-10T10:03:39Z

‎Homogeneous and isotropic Universe, Hubble’s law

Igor at 11:45, 1 August 2012

2012-08-01T11:45:51Z

@@ Line 66: / Line 66: @@
 Therefore the distribution law for velocities in the new frame has the same form with the same value of Hubble's parameter as in the original one.</p>
    </div>
-</div>
+</div></div>
-<div id="equ6"></div><div style="border: 1px solid #AAA; padding:5px;">
+<div id="equ6"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 6: Hubble law from homogeniety and isotropy ===
 Show that the Hubble's law represents the only form of expansion compatible with homogeneity and isotropy of the Universe.

@@ Line 1: / Line 1: @@
 [[Category:Dynamics of the Expanding Universe|1]]
-!!!
 <!-- = Homogeneous and isotropic Universe, Hubble’s law = -->
 __TOC__

@@ Line 1: / Line 1: @@
 [[Category:Dynamics of the Expanding Universe|1]]
+!!!
 <!-- = Homogeneous and isotropic Universe, Hubble’s law = -->
 __TOC__

@@ Line 23: / Line 23: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">Due to isotropy the distribution must coincide in any two points, because they can be transformed to each other by rotation by $180^\circ$ around the middle of the interval connecting them. The opposite is not true.[[#equ1 | see problem]].</p>
+     <p style="text-align: left;">Due to isotropy the distribution must coincide in any two points, because they can be transformed to each other by rotation by $180^\circ$ around the middle of the interval connecting them. The opposite is not true ,see [[#equ1 |the problem on homogeniety vs isotropy]].</p>
    </div>
 </div></div>
@@ Line 73: / Line 73: @@
 === Problem 6: Hubble law from homogeniety and isotropy ===
 Show that the Hubble's law represents the only form of expansion compatible with homogeneity and isotropy of the Universe.
-<div class="NavFrame collapsed">
+<!-- <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
      <p style="text-align: left;"></p>
    </div>
-</div></div>
+</div> --></div>

@@ Line 1: / Line 1: @@
 [[Category:Dynamics of the Expanding Universe|1]]
-= Homogeneous and isotropic Universe, Hubble’s law =
+<!-- = Homogeneous and isotropic Universe, Hubble’s law = -->
 __TOC__

@@ Line 19: / Line 19: @@
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 2: global isotropy ===
-Show that if some spatial distribution is everywhere isotropic then it is also homogeneous. Is the opposite true?.
+Show that if some spatial distribution is everywhere isotropic then it is also homogeneous. Is the opposite true?
 <div class="NavFrame collapsed">
    <div class="NavHead">solution</div>
@@ Line 30: / Line 30: @@
 <div id="equ3"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 3: examples ===
 What three-dimensional geometrical objects are both homogeneous and isotropic?

@@ Line 88: / Line 88: @@
      <p style="text-align: left;">The continuity equation takes the form:
 \[\frac{\partial \rho}{dt}
-	+\mbox{div}\left(\vec v\rho\right) = 0.\]
+	+\mbox{div}\left(\boldsymbol v\rho\right) = 0.\]
-It follows from the homogeneity of the Universe that $\rho$ can depend on time but it is independent on coordinates, i.e. $\rho  = \rho (t)$. Then in accordance with the Hubble's law  $\vec v = H\vec r$ one obtains
+It follows from the homogeneity of the Universe that $\rho$ can depend on time but it is independent on coordinates, i.e. $\rho  = \rho (t)$. Then in accordance with the Hubble's law  $\boldsymbol v = H\boldsymbol r$ one obtains
-\[\text{div}(\rho\vec v)
+\[\text{div}(\rho\boldsymbol v)
-	= \rho\text{div}(\vec v)
+	= \rho\text{div}(\boldsymbol v)
-	=\rho\text{div}(H\vec r) = 3\rho H.\]
+	=\rho\text{div}(H\boldsymbol r) = 3\rho H.\]
-Therefore, if the distribution $\rho$ was coordinate independent in some reference frame, then the expansion law $\vec v = H\vec r$ preserves the homogeneity of $\rho$ in all subsequent moments of time. Thus the initial homogeneity is conserved forever.</p>
+Therefore, if the distribution $\rho$ was coordinate independent in some reference frame, then the expansion law $\boldsymbol v = H\boldsymbol r$ preserves the homogeneity of $\rho$ in all subsequent moments of time. Thus the initial homogeneity is conserved forever.</p>
    </div>
 </div></div>

@@ Line 59: / Line 59: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;">Suppose the Hubble's law holds in the reference frame $K$, in which matter is at rest at the origin. Consider another frame $K'$, with origin at some point $A$, moving with velocity $\vec{v}_{A}$ relative to $K$. The quantities measured in the latter frame are denoted by primes. Then in $K'$ one has $\vec r' = \vec r - {\vec r_A}$ and
+     <p style="text-align: left;">Suppose the Hubble's law holds in the reference frame $K$, in which matter is at rest at the origin. Consider another frame $K'$, with origin at some point $A$, moving with velocity $\boldsymbol{v}_{A}$ relative to $K$. The quantities measured in the latter frame are denoted by primes. Then in $K'$ one has $\boldsymbol r' = \boldsymbol r - {\boldsymbol r_A}$ and
-\[\vec v' = \vec v - {\vec v_A}
+\[\boldsymbol v' = \boldsymbol v - {\boldsymbol v_A}
-	= H\vec r' = H\vec r - H{\vec r_A}
+	= H\boldsymbol r' = H\boldsymbol r - H{\boldsymbol r_A}
-	= H\left( {\vec r - {{\vec r}_A}} \right) = H\vec r'.\]
+	= H\left( {\boldsymbol r - {{\boldsymbol r}_A}} \right) = H\boldsymbol r'.\]
 Therefore the distribution law for velocities in the new frame has the same form with the same value of Hubble's parameter as in the original one.</p>
    </div>
-</div></div>
+</div>
 <div id="equ6"></div><div style="border: 1px solid #AAA; padding:5px;">
 === Problem 6: Hubble law from homogeniety and isotropy ===
 Show that the Hubble's law represents the only form of expansion compatible with homogeneity and isotropy of the Universe.

@@ Line 42: / Line 42: @@
 <div id="equ4"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 4: on Big Bang ===
+=== Problem 4: the Big Bang "explosion"===
 Why the notion of ''Big Bang'' regarding the early evolution of the Universe should not be treated too literally?
 <div class="NavFrame collapsed">
@@ Line 54: / Line 54: @@
 <div id="equ5"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 5: Hubble law Galilean invariance ===
+=== Problem 5: Galilean invariance of the Hubble law ===
 Show that the Hubble's law is invariant with respect to Galilean transformations.
 <div class="NavFrame collapsed">
@@ Line 80: / Line 80: @@
 <div id="equ7"></div><div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 7: homogeniety conservation ===
+=== Problem 7: preservation of homogeniety ===
 Show that if expansion of the Universe obeys the Hubble's law then the initial homogeneity is conserved for all its subsequent evolution.
 <div class="NavFrame collapsed">
@@ Line 135: / Line 135: @@
 <div id="equ11"></div><div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 10: age of the Universe ===
+=== Problem 10: the age of the Universe ===
 Estimate the age of the Universe basing on the observed value of the Hubble's constant (the Hubble time $t_H$).
 <div class="NavFrame collapsed">
@@ Line 146: / Line 146: @@
 <div id="equ13"></div><div style="border: 1px solid #AAA; padding:5px;">
-=== Problem 11: Olbers' paradox resolution ===
+=== Problem 11: Olbers paradox resolved ===
 Show that the model of the expanding Universe allows one to eliminate the Olbers' paradox.
 <div class="NavFrame collapsed">