FRW metric derivation: constraints from isotropic and homoge

[binbagsss]

I don't understand the reasoning for any of the three constraints imposed.

why would $dtdx^i$ terms indicate a preferred direction? what if there was identical terms for each $x^i$ would there still be a specified or preferred direction? (or is it that in this case we could rename $t$ to make these terms redundant?)

I don't understand why if the $g_tt$ component were to depend on $x^i$ homogeneity would be broken? (also again, what if we had some function symmetrical in all $x^i$; would it still be broken?)Consistent with not understanding this I don't understnad why $h^{ij}$ can not have time-dependence.

The definitions I have are:

. Homogeneous is defined as "the same in all locations" while isotropic means "the same in all directions

ta

 

[PeterDonis]

why would $dtdx^i$ terms indicate a preferred direction?

Because the coefficient that multiplies them, $B_i$, defines a preferred direction: the direction of the vector $B_i$.

what if there was identical terms for each $x^i$

That just means that the vector $B_i$ points in a direction that has equal $x$, $y$, and $z$ components. That's not the same as not having a preferred direction.

I don't understand why if the $g_{tt}$ component were to depend on $x^i$ homogeneity would be broken?

Because homogeneity means things are the same at every spatial point at a given instant of time. If $g_{tt}$ depends on $x^i$, then the "time dilation factor" is not the same at every spatial point at a given instant of time.

I don't understnad why $h^{ij}$ can not have time-dependence.

That's not what the quote you gave says. It says that the time dependence of $h_{ij}$ can only be in an overall scale factor $a(t)$, so that $h_{ij} = a(t) \gamma_{ij}$, where $\gamma_{ij}$ does not depend on time.

 

[binbagsss]

That just means that the vector $B_i$ points in a direction that has equal $x$, $y$, and $z$ components. That's not the same as not having a preferred direction.
.

So say $2dxdt+2dydt+2dzdt$ ; $B_i=2$; what is the preferred direction?
I thought $B_i$ is a component of the tensor $g_{uv}$ and is therefore a scalar and not a vector?

 

[binbagsss]

That's not what the quote you gave says. It says that the time dependence of $h_{ij}$ can only be in an overall scale factor $a(t)$, so that $h_{ij} = a(t) \gamma_{ij}$, where $\gamma_{ij}$ does not depend on time.

so the definition for isotropic is again at a given time it must look the same in all directions?

 

[PeterDonis]

So say $2dxdt+2dydt+2dzdt$ ; Bi=2

No, $B_i$ is a vector with components $(2, 2, 2)$. Big difference.

I thought $B_i$ is a component of the tensor $g_{uv}$ and is therefore a scalar and not a vector?

First, components of a tensor are not scalars. They are numbers, but they are not invariant under coordinate transformations, so they're not scalars (a scalar is defined as a number that is invariant under coordinate transformations, like the invariant norm of a 4-vector).

Second, the notation your source is using is confusing if you are trying to keep components straight. It is splitting up the 4x4 symmetric matrix $g_{uv}$, with 10 components, as follows: the number $g_{00}$ (1 component), the 3-vector $B_i$ (3 components) and the 3x3 symmetric matrix $h_{ij}$ (6 components). The term $B_i dt dx^i$ is actually a summation: it's three terms added together (using the Einstein summation convention), so $B_i$ in their notation represents a 3-vector.

 

[binbagsss]

No, $B_i$ is a vector with components $(2, 2, 2)$. Big difference.

ok, but what is the preffered direction?

 

[PeterDonis]

what is the preffered direction?

The direction pointed at by a vector with components $(2, 2, 2)$.

 

[PAllen]

I find the whole discussion in the source quoted in the OP problematic. In any manifold at all, some finite (often large) spacetime region can be put into Gaussian Normal coordinates such that g00 is 1 and g0i is zero. In other words, in the terms of the OP snippet, A = 1, B = 0, can be achieved for some region of any spacetime at all, around any point. What isotropy/homogeneity require is that h is factorable into a function of time and a spatial submetric that does not depend on time and that has constant curvature (for some foliation).

See, for example MTW, exercise 27.2

[edit: for an interesting example, Lemaitre coordinates represent a Gaussian Normal system for the Schwarzschild metric, covering one exterior [in relation to Kruskal coordinates] and the portion of one interior accessible to a collapsed body. The Schwarzschild geometry is clearly not homogenous, let alone isotropic about every point.]