l**n
发帖数: 67 | 1 Tha past has passed and i want to know the future.
So the first question i'll ask is what will be
the fate of our universe. Mankind is of no hope.
They will die off sooner or later, by i still want
to know the destiny of our universe, even if every-
thing is doomed to disappear.
Let's ask the question in a explicit way.
Will the universe expand forever or recollapse
eventually?
In cosmology, the answer is highly depends on the
value of cosmological constant. Let's denote it by
\Omega_\La | c**********g
发帖数: 222 | 2 >>>> this is wrong. what offset the gravity of the sun is the gas pressure
( acturally, it is the gradient of the gas pressure). The pressure is positive.
when we mean that the pressure is positive, we mean that P1,P2 and P3 the trace
of the energy-monentum tensor (\rho+P1+P2+P3) is positive. Here, \rho is the energy
density
>>> Also, in NS, gravity is balanced by the gradient of the neutron denenerate
pressure. Again, this pressure itself is positive. | q**i
发帖数: 174 | 3 I like it. Keep it going, please.
a quick question: if a blackhole keeps sucking things in without
anything escaping it (forget about Hawking's blackhole
radiation for the moment), wouldn't it get bigger and bigger (
having larger and larger event horizon)? Wouldn't one day
the blackhole will be there universe?
【在 l**n 的大作中提到】
: Tha past has passed and i want to know the future.
: So the first question i'll ask is what will be
: the fate of our universe. Mankind is of no hope.
: They will die off sooner or later, by i still want
: to know the destiny of our universe, even if every-
: thing is doomed to disappear.
: Let's ask the question in a explicit way.
: Will the universe expand forever or recollapse
: eventually?
: In cosmology, the answer is highly depends on the
| l**n
发帖数: 67 | 4 well, i must confess that i am not so clear here. What i mean negative
is just negative with respect to gravity. I didn't know this negative here
has certain meaning in GR or whatever theory it is. But i still think
that the fussion process plays an important role in balancing the gravity
in Sun besides the pressure of the gas. Maybe a less controversial example
is the Jupiter where the gas pressure alone balances the gravity. But in a
star, fussion does matter.
【在 c**********g 的大作中提到】
: >>>> this is wrong. what offset the gravity of the sun is the gas pressure
: ( acturally, it is the gradient of the gas pressure). The pressure is positive.
: when we mean that the pressure is positive, we mean that P1,P2 and P3 the trace
: of the energy-monentum tensor (\rho+P1+P2+P3) is positive. Here, \rho is the energy
: density
: >>> Also, in NS, gravity is balanced by the gradient of the neutron denenerate
: pressure. Again, this pressure itself is positive.
| c**********g
发帖数: 222 | 5
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The point is that in fact pressure is a tensor, so the sign has no much definite
meaning. Also, in relativity, it can be covariant or contravovariant (the sign
is opposite). SO, we need to pay attention to the meaning of the sign.
In the case of the cosmological constant, the negative pressure will produce
a repulsive gravity! This is the reason that it will accelarate the expansion of
the universe.
I still do
【在 l**n 的大作中提到】
: well, i must confess that i am not so clear here. What i mean negative
: is just negative with respect to gravity. I didn't know this negative here
: has certain meaning in GR or whatever theory it is. But i still think
: that the fussion process plays an important role in balancing the gravity
: in Sun besides the pressure of the gas. Maybe a less controversial example
: is the Jupiter where the gas pressure alone balances the gravity. But in a
: star, fussion does matter.
| i*******n
发帖数: 166 | 6
They are in fact different concepts.
\Lambda is a constant in field equation of gravity.
And vaccuum energy is a concept in QED.
Surperisingly, \Lambda has the same properties in the equation
as the vaccum energy. That's the reason that people relate it
to vaccuum energy. They may be found to be unified in the quantumn
gravity theory.
Here negative means the abnormal eqation of state of this
kind of materials. pressure = - density (expansion of this
kind of "gas" don't do wo
【在 l**n 的大作中提到】
: well, i must confess that i am not so clear here. What i mean negative
: is just negative with respect to gravity. I didn't know this negative here
: has certain meaning in GR or whatever theory it is. But i still think
: that the fussion process plays an important role in balancing the gravity
: in Sun besides the pressure of the gas. Maybe a less controversial example
: is the Jupiter where the gas pressure alone balances the gravity. But in a
: star, fussion does matter.
| l**n
发帖数: 67 | 7 I agree i was wrong here. But i still don't quite understand that
why this cosmological constant with negative pressure is needed
to ensure a static universe. If the total angular momentum(fishy?)
of the universe is not zero, is it possible to get a static universe
without \Lambda? The earth doesn't necessarily fall into the sun
as long as it has angular momentum, right? | c**********g
发帖数: 222 | 8
The SIGN of the cosmological constant Einstein introduced in order to offest
the expansion of the universe is OPPOSITE to the SIGN of the current
favorite cosmological constant.
In the cosmological dynamical equation: H^2+k/a^2=\rho. here, a is
the scale fator of the universe, H is hubble constant. k=-1,0,1.
\rho=\rho_matter+\rho_lambda is the cosmological density. static universe
requires H=0. when k=0, \rho=0. a negative cosmologicl constant (but pressure
is positive) can satisfy this
【在 l**n 的大作中提到】
: I agree i was wrong here. But i still don't quite understand that
: why this cosmological constant with negative pressure is needed
: to ensure a static universe. If the total angular momentum(fishy?)
: of the universe is not zero, is it possible to get a static universe
: without \Lambda? The earth doesn't necessarily fall into the sun
: as long as it has angular momentum, right?
| f*******d
发帖数: 339 | 9 I don't think you are right here. The sign of the cosmological constant introduced
by Einstein is actually the same as has been discovered.
In Einstein's static model, k=+1 (a finite Universe), so
\rho=\rho_m + \rho_\Lambda >0.
On the other hand, there is also an equation which is
\ddot{a}/a + H^2 +k/a^2 = -p
A static Universe requires H=0, \ddot{a}=0, so
k/a^2 = -p,
for k=+1, it requires p<0, which is provided by a POSITIVE cosmological constant,
as has been discovered.
【在 c**********g 的大作中提到】
:
: The SIGN of the cosmological constant Einstein introduced in order to offest
: the expansion of the universe is OPPOSITE to the SIGN of the current
: favorite cosmological constant.
: In the cosmological dynamical equation: H^2+k/a^2=\rho. here, a is
: the scale fator of the universe, H is hubble constant. k=-1,0,1.
: \rho=\rho_matter+\rho_lambda is the cosmological density. static universe
: requires H=0. when k=0, \rho=0. a negative cosmologicl constant (but pressure
: is positive) can satisfy this
| c**********g
发帖数: 222 | 10 in fact, thing is subtle here.
The two equations are:
(1) H^2+k/a^2=\rho
(2) \ddot{a}/a=-(\rho+3P).
static universe requires that a=constant and H=0. so, \rho=constant.
so \rho_matter=0 and \rho_lambda=k/a^2;
but then, (2) \ddot{a}/a=-2 \rho_lambda.
so the only static solution is \rho_matter=0 and \rho_lambda=0, namely,
an empty universe.
so, I am confused that how can Einstein build a static universe.
在 fisherdad (渔父) 的大作中提到: 】 | f*******d
发帖数: 339 | 11
You are confused because you made the wrong assumption \rho_m=0,
Einstein did not make this assumption because we knew it is
not zero. Instead, from the two equations you wrote down,
the solution is (after set H=0, k=+1)
(1) 1/a^2=\rho_\Lambda + \rho_m
(2) 0=\rho_m+\rho_\Lambda + 3 P_\lambda
set P_\lambda=-\rho_Lambda, one obtain
\rho_Lambda=-1/2 \rho_m, and
1/a^2=1/2 \rho_m.
【在 c**********g 的大作中提到】
: in fact, thing is subtle here.
: The two equations are:
: (1) H^2+k/a^2=\rho
: (2) \ddot{a}/a=-(\rho+3P).
: static universe requires that a=constant and H=0. so, \rho=constant.
: so \rho_matter=0 and \rho_lambda=k/a^2;
: but then, (2) \ddot{a}/a=-2 \rho_lambda.
: so the only static solution is \rho_matter=0 and \rho_lambda=0, namely,
: an empty universe.
: so, I am confused that how can Einstein build a static universe.
| c**********g
发帖数: 222 | 12
I always think that a is changing. Then because \rho_matter proportional
a^(-3). then.... I made this stupid mistake. Now since a is just a constant.
Everything is OK. Thanks!
【在 f*******d 的大作中提到】
:
: You are confused because you made the wrong assumption \rho_m=0,
: Einstein did not make this assumption because we knew it is
: not zero. Instead, from the two equations you wrote down,
: the solution is (after set H=0, k=+1)
: (1) 1/a^2=\rho_\Lambda + \rho_m
: (2) 0=\rho_m+\rho_\Lambda + 3 P_\lambda
: set P_\lambda=-\rho_Lambda, one obtain
: \rho_Lambda=-1/2 \rho_m, and
: 1/a^2=1/2 \rho_m.
| c**********g
发帖数: 222 | 13
I always think undersence that a is changing (though I am discussing a static universe). Then because \rho_matter proportional a^(-3). then.... I made this stupid mistake. Now since a is just a constant. Everything is OK. Thanks!
【在 f*******d 的大作中提到】
:
: You are confused because you made the wrong assumption \rho_m=0,
: Einstein did not make this assumption because we knew it is
: not zero. Instead, from the two equations you wrote down,
: the solution is (after set H=0, k=+1)
: (1) 1/a^2=\rho_\Lambda + \rho_m
: (2) 0=\rho_m+\rho_\Lambda + 3 P_\lambda
: set P_\lambda=-\rho_Lambda, one obtain
: \rho_Lambda=-1/2 \rho_m, and
: 1/a^2=1/2 \rho_m.
|
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