Comparing different scale factor functions in the same graph

[hedgehug]

The FLRW metric accounts for a case with the constant scale factor $a(t)=\text{const}=1$, which is the equivalent to Minkowski metric. I was told that empty spacetime with any coordinate parametrization is equivalent to Minkowski spacetime, and that includes constant $a$ FLRW for spatially flat universe as well maximally negatively curved expanding models with the linear $a(t)\propto t$.

I was also told that $dt$ in the Minkowski metric is not the same $dt$ as in the FLRW metric, but we have Milne line that is Minkowski in disguise in the same graph with all the other scale factor functions, and they are all functions of the same coordinate time on the horizontal axis.

If the time coordinate $t$ in $a(t)=\text{const}$ (Minkowski) is not the same t as in $a(t)\neq \text{const}$ (FLRW), then it's also different for a scale factor $b(t)\neq a(t)$ that is also changing in time. It would imply that time coordinate in the FLRW metric depends on the scale factor function. I think it's ridiculous, since we compare them all in the same graph with the same coordinate time on the horizontal axis. Do you think otherwise?

By admitting that I was wrong about the different time coordinate of the scale factor line for linearly expanding, negatively curved, empty universe with the Milne metric, I'm also telling you that if you continue to claim that $t$ in $a(t)=\text{const}$ is different from $t$ of all the other scale factor functions, then you'll be also saying that coordinate time depends on the scale factor function.

 

[PAllen]

I was also told that $dt$ in the Minkowski metric is not the same $dt$ as in the FLRW metric, but we have Milne line that is Minkowski in disguise in the same graph with all the other scale factor functions, and they are all functions of the same coordinate time on the horizontal axis.

A better wording is simply that the time coordinate in Milne coordinates is different than the time coordinate in standard Minkowski coordinates. This is simply because the foliations are different. In Milne coordinates, you are slicing Minkowski spacetime into hyperbolic spatial slices, while in standard coordinates, you are using flat spatial slices.

The second part of your comment refers to some diagram from another thread. Please repost it here so we don't have to find your other thread.

If the time coordinate $t$ in $a(t)=\text{const}$ (Minkowski) is not the same t as in $a(t)\neq \text{const}$ (FLRW), then it's also different for a scale factor $b(t)\neq a(t)$ that is also changing in time. It would imply that time coordinate in the FLRW metric depends on the scale factor function. I think it's ridiculous, since we compare them all in the same graph with the same coordinate time on the horizontal axis. Do you think otherwise?

I don't think it is meaningful to say the time coordinate is the same between different manifolds. A graph such as you describe is simply picking a parameter based on t to classify a family of spacetimes. There is no implication time coordinates are the 'same'.

The only case you can really compare time coordinates is with different coordinates for the same spacetime. The only case of this within the FLRW family of metrics is between a(t)=constant with flat spatial slices, and a(t)=t with hyperbolic spatial slices. Here they are clearly different because the coordinates are different. All other cases are simply non-comparable.

 

[hedgehug]

All other cases are simply non-comparable.

Comparison of non-comparable cases:

 

[PAllen]

Non-comparable cases:
View attachment 369080

Thanks.

 

[hedgehug]

But you know we're comparing them, right? That's what this graph is for - to compare them.

 

[PeterDonis]

I was also told that $dt$ in the Minkowski metric is not the same $dt$ as in the FLRW metric, but we have Milne line that is Minkowski in disguise

More precisely, Milne is one "wedge" of MInkowski in different coordinates. So the Milne $dt$ is not the same as the Minkowski $dt$. You were told that also in the previous thread.

the time coordinate $t$ in $a(t)=\text{const}$ (Minkowski)

No, that's not Minkowski coordinates, that's Milne coordinates. You can't put the Minkowski metric in Minkowski coordinates in the FLRW form, so you can't even define an $a(t)$ for the Minkowski metric in Minkowski coordinates.

if you continue to claim that t in a(t)=const is different from t of all the other scale factor functions

Nobody has claimed that. The way $t$ is defined in FLRW coordinates is the same for all spacetime geometries that can be described by those coordinates: it's the proper time of comoving observers.

coordinate time depends on the scale factor function.

No, both coordinate time and the scale factor $a(t)$ as a function of coordinate time, given that you have chosen coordinates in the FLRW form on a spacetime that can be described by thosed coordinates (so that there is even a scale factor $a(t)$ to begin with), depend on the spacetime geometry you are describing in FLRW coordinates.

That's why the curves all look different in the graph you are obsessing about.

Comparison of non-comparable cases

They're not "comparable" in the sense that they are all different spacetime geometries. But given that you can write all of their metrics in the same form, you can formally compare the scale factor functions $a(t)$ as a function of the time coordinate $t$ between them. That's all the graph does. The graph makes no claim that there is any physical comparison between different spacetime geometries.

Why would you want to make such a "formal" comparison? In order to compare the different models with observations. In other words, each of those curves is a prediction from a different model about how a graph of scale factor vs. coordinate time will look, if we take our observations and use them to derive an observed scale factor $a(t)$ as a function of coordinate time. That allows us to see how well each model matches observations. The $\Lambda C D M$ model is our current "best fit" model because its predictions best match observations. (And of course how its curve looks on that graph is just one of many, many predictions the model makes.)

 

[hedgehug]

@PeterDonis you're so full of s*h*i*t, that I have only one more request - just ban me already.