Is the Cosmological Principle Limited to Space Only?

[Chalnoth]

Formation and evolution of galaxies are rather local events compared to the scales we are dealing with;the others are my solution to paradox.

As I keep saying, the population of galaxies at z=1 is, for instance, very different from the population of galaxies at z=0.01. The reason for this is that there are more older galaxies in the nearby universe, and more younger galaxies in the early universe. Clusters are more numerous and larger in the nearby universe. Active galactic nuclei are more common a bit further away (that is, AGN's are typically characteristic of younger galaxies).

 

[AWA]

As I keep saying, the population of galaxies at z=1 is, for instance, very different from the population of galaxies at z=0.01. The reason for this is that there are more older galaxies in the nearby universe, and more younger galaxies in the early universe. Clusters are more numerous and larger in the nearby universe. Active galactic nuclei are more common a bit further away (that is, AGN's are typically characteristic of younger galaxies).

I thought we agreed before that the universe at z=0.01 is not yet homogenous so it is obviously different than the universe at z=1.
Besides the examples you are giving about galactic age shouldn't affect the universe density at large scale. The clusters part I would have to check it. I'm not sure that has been statistically shown to happen.

 

[Chalnoth]

I thought we agreed before that the universe at z=0.01 is not yet homogenous so it is obviously different than the universe at z=1.

You said it. I didn't agree to it.

What you have to bear in mind is that there is a fundamental difference between the typical length scale of homogeneity and actually statistically demonstrating it (given an appropriate threshold). To statistically demonstrate it, you need a region much larger than the scale of homogeneity (because you have to show that all such regions of said size are statistically identical, to within some pre-defined threshold).

It's not such a surprise to me that there's argument about this, because it's mathematically a difficult thing to demonstrate, and the threshold of homogeneity is arbitrary anyway.

Besides the examples you are giving about galactic age shouldn't affect the universe density at large scale. The clusters part I would have to check it. I'm not sure that has been statistically shown to happen.

Well, density is one of the most significant things to evolve with redshift. The average density of the matter in our universe at z=1 is eight times the density today. Baryon Acoustic Oscillation observations, which measure the typical separation between galaxies at different redshifts, are a good measurement of how this density changes with redshift.

 

[Calimero]

Sure, the distance-redshift relation is not linear, so?.

So it depends on input parameters.

Borrowing a little from cosmology textbook stuff, this is from Hobson's General Relativity:

"In general relativity the concept of a ‘moment of time’ is ambiguous and is replaced by the notion of a three-dimensional spacelike hypersurface. To define a ‘time’ parameter that is valid globally, we ‘slice up’ spacetime by introducing a series of non-intersecting spacelike hypersurfaces that are labelled by some parameter t. This parameter then defines a universal time in that ‘a particular time’ means a given spacelike hypersurface. We may construct the hypersurfaces t = constant in any number of ways. In a general spacetime there is no preferred ‘slicing’ and hence no preferred ‘time’coordinate t.
According to Weyl’s postulate, there is a unique worldline passing through each (non-singular) spacetime point. The set of worldlines is sometimes described as providing threading for the spacetime."

In our case one of the three spatial dimensions of the spacelike hypersurfaces represents visually (thanks to light's nature) the timelike worldline passing thru each spacetime point defined by a specific redshift.

Thanks. Very nice quotation.

I thought we agreed before that the universe at z=0.01 is not yet homogenous so it is obviously different than the universe at z=1..

There is the source of your confusion. Please look http://www.math.lsa.umich.edu/mmss/coursesONLINE/Astro/Ex2.2/" . You are comparing homogeneity of two different time slices, which, obviously, when compared are not the same, and talking about 'not yet' homogeneous universe. Universe is just as much homogeneous now as it ever was, just on different scale.

 

[AWA]

There is the source of your confusion. Please look http://www.math.lsa.umich.edu/mmss/coursesONLINE/Astro/Ex2.2/" . You are comparing homogeneity of two different time slices, which, obviously, when compared are not the same, and talking about 'not yet' homogeneous universe. Universe is just as much homogeneous now as it ever was, just on different scale.

Thanks, cool site.
A couple of remarks: first, that is a simulation school exercise and they stress it that as a simulation it is not expected to match the real universe.
Second, according to GR (and as pointed out in the quote from my last post) you can slice up spacetime arbitrarily (general covariance, remember?) : "We may construct the hypersurfaces t = constant in any number of ways. In a general spacetime there is no preferred ‘slicing’ and hence no preferred ‘time’coordinate t."
Given a statistically significant number of slices you should find homogeneity across the sufficiently long time-like worldline formed by the statistically large stack of spacelike slices if each of the different time spacelike slices is itself homogenous. If you don't agree with this, I should remind you that GR is to this day the best theory to understand the universe that we have.

 

[AWA]

The average density of the matter in our universe at z=1 is eight times the density today.

Please, back up that figure with some reliable reference. Specifying how do you exactly measure the universe density time spacelike slice at precisely z=1.

 

[Chalnoth]

Please, back up that figure with some reliable reference. Specifying how do you exactly measure the universe density time spacelike slice at precisely z=1.

Matter density scales as a^{-3}. But 1+z = 1/a, so matter density scales as (1+z)^3.

Basically, proposing that the matter density wasn't eight times its current value at z=1 requires proposing a universe that has radial-dependent density for an equal-time slicing. This is, in principle, a rather difficult thing to accurately determine, but suffice it to say our theories that use a homogeneous matter distribution work, while alternative theories proposed to explain certain unpleasant aspects of the homogeneous theories don't.

 

[AWA]

Matter density scales as a^{-3}. But 1+z = 1/a, so matter density scales as (1+z)^3.

Right, that is a model-dependent calculation, not an observation. That is what is apparently contradicted (and thus where I see the paradox) by the hypothetical future SDSS 3D galaxy map up to a high redshift, that we expect to be homogenous. But perhaps, I'm misunderstanding something and cosmologists don't expect to find that map statistically homogenous. If so, please explain.

Basically, proposing that the matter density wasn't eight times its current value at z=1 requires proposing a universe that has radial-dependent density for an equal-time slicing.

I see what you mean, yes, that seems to follow logically.
The universe is a strange place, and is full of apparent contradictions, wish we knew it better, but let's be humble (and honest), we are barely starting to grasp it.

This is, in principle, a rather difficult thing to accurately determine, but suffice it to say our theories that use a homogeneous matter distribution work, while alternative theories proposed to explain certain unpleasant aspects of the homogeneous theories don't.

Ultimately, it seems to come down to a practical matter.
BTW, I don't know what are those theories proposed to explain unpleasant aspects of homogenous theories, and what these unpleasnt aspects are. Would you elaborate a little on this?

 

[Chalnoth]

Right, that is a model-dependent calculation, not an observation. That is what is apparently contradicted (and thus where I see the paradox) by the hypothetical future SDSS 3D galaxy map up to a high redshift, that we expect to be homogenous. But perhaps, I'm misunderstanding something and cosmologists don't expect to find that map statistically homogenous. If so, please explain.

You have to look into it in a bit more detail. For example, if you combine nearby measurements of the Hubble constant and supernovae with WMAP data, you end up with a nearly-flat universe.

If you then proceed with the assumption of flatness and make use of BAO data, you get a connection between length scales at different redshifts. If this link between length scales at different redshifts doesn't line up, then that would be evidence that the assumption of homogeneity was wrong. Basically, this length scale being at the expected place with the assumption of flatness is a reasonably direct test of the relationship z+1 = 1/a.

There are all sorts of different ways you can do this sort of experiment, but the basic idea here is that if you make a series of assumptions, and multiple independent experiments measure the same set of parameters based upon those assumptions, you gain confidence that those assumptions are, in fact, true, at least in an approximate sense. One of those foundational assumptions is homogeneity.

Ultimately, it seems to come down to a practical matter,it seems.
BTW, I don't know what are those theories proposed to explain unpleasant aspects of homogenous theories, and what these unpleasnt aspects are. Would you elaborate a little on this?

In this case, some have attempted to explain away the accelerated expansion by proposing a universe that has radial-dependent density. It turns out that such proposals are ruled out by observation.

 

[AWA]

You have to look into it in a bit more detail. For example, if you combine nearby measurements of the Hubble constant and supernovae with WMAP data, you end up with a nearly-flat universe.

If you then proceed with the assumption of flatness and make use of BAO data, you get a connection between length scales at different redshifts. If this link between length scales at different redshifts doesn't line up, then that would be evidence that the assumption of homogeneity was wrong. Basically, this length scale being at the expected place with the assumption of flatness is a reasonably direct test of the relationship z+1 = 1/a.

There are all sorts of different ways you can do this sort of experiment, but the basic idea here is that if you make a series of assumptions, and multiple independent experiments measure the same set of parameters based upon those assumptions, you gain confidence that those assumptions are, in fact, true, at least in an approximate sense. One of those foundational assumptions is homogeneity.

This is all understood and fine, I'm just taking that assumption to its last logical consequences if we take relativity seriously, and if we agree that if you observe long distances spaces you are also observing the past, one cannot be homogenous if the other isn't too, and viceversa. As they say, you can't have one without the other.

This leads to some contradiction with standard cosmology, so when in doubt, of course we choose standard cosmology, right?

 

[AWA]

You have to look into it in a bit more detail.

If you then proceed with the assumption of flatness and make use of BAO data, you get a connection between length scales at different redshifts. If this link between length scales at different redshifts doesn't line up, then that would be evidence that the assumption of homogeneity was wrong. Basically, this length scale being at the expected place with the assumption of flatness is a reasonably direct test of the relationship z+1 = 1/a.

Searching with the word BAO in arxiv, the first random paper I read casts shadows over BAO signal measures.: http://arxiv.org/PS_cache/arxiv/pdf/1009/1009.1232v1.pdf

 

[Chalnoth]

This is all understood and fine, I'm just taking that assumption to its last logical consequences if we take relativity seriously, and if we agree that if you observe long distances spaces you are also observing the past, one cannot be homogenous if the other isn't too, and viceversa. As they say, you can't have one without the other.

This leads to some contradiction with standard cosmology, so when in doubt, of course we choose standard cosmology, right?

This is getting tiring. We do not observe homogeneity in the radial direction. We don't expect to, because the radial direction is also looking backwards in time.

What we observe instead is a universe that looks like the nearby universe is a later version of the far away universe. In other words, it's as if looking outward in space is looking through a succession of homogeneous equal-time slices. This is the standard cosmology. This is what we observe. None of our observations contradict this, and it is fully self-consistent.

As for the BAO paper, if you look at their data, the apparent 3\sigma deviation is represented in figure 6, where you can clearly see that the discrepancy comes down to the signal being rather noisier than their simulations estimate, which would be indicative of not properly taking something into account in the simulations.

A perhaps better paper for this particular issue is this one:
http://arxiv.org/abs/0705.3323

...because this shows the combination of WMAP, supernova, and BAO data. The relevant plot is fig. 13, where you see that the contours all converge on the same point in parameter space when these data are taken together.

 

[AWA]

This is getting tiring. We do not observe homogeneity in the radial direction.

Oh, but you don't have to answer if you get tired. There are more people in this forum.

We do not observe homogeneity in the radial direction.We don't expect to

Are you sure? we expect to find spatial homogeneity. why on Earth would you want to leave out one spatial dimension just beats me. I mean that's pretty bizarre, how do you keep one spatial direction inhomogenous and the others homogenous, and still keep isotropy?

What we observe instead is a universe that looks like the nearby universe is a later version of the far away universe. In other words, it's as if looking outward in space is looking through a succession of homogeneous equal-time slices.

See posts 49 and 55.

 

[Chalnoth]

Are you sure? we expect to find spatial homogeneity. why on Earth would you want to leave out one spatial dimension just beats me. I mean that's pretty bizarre, how do you keep one spatial direction inhomogenous and the others homogenous, and still keep isotropy?

1. Many cosmologists, early on, expected homogeneity in both time and space. This was disproven when Hubble measured the expansion of the universe.
2. Since when we look far away, we are looking back in time, we do not expect to see homogeneity in that direction, because an expanding universe changes with time.

See posts 49 and 55.

So, you're still confused about the simultaneity thing? The expansion of the universe itself creates a notion of "universal time". If you use coordinates that move with the expansion, then observers that are stationary with respect to the coordinate system each see the universe as being isotropic from their point of view.

This is, ultimately, what we mean by spatial homogeneity: if I go anywhere else in the visible universe, and adjust my velocity to move along with the local matter there, the universe will look isotropic to me. A homogeneous universe is defined as one in which you can do this: you can move anywhere within the universe, set your velocity to some value, and see an isotropic universe. You can then define the time coordinate globally in such a way that at the same time, separated observers see the same properties of the universe (such as the CMB temperature). In these coordinates, the properties of the universe are the same everywhere in space, but change with time.

You can change to a different set of coordinates, of course, and things won't necessarily look constant in space any longer. You'll still get the right answers for any observable you calculate, but you won't see the homogeneity.

 

[AWA]

Since when we look far away, we are looking back in time, we do not expect to see homogeneity in that direction, because an expanding universe changes with time.

This indeed is getting repetitive,once again this makes no sense in GR. You have some source where this is explicitly stated? that homogeneity is forbidden in one spatial direction?

This is, ultimately, what we mean by spatial homogeneity: if I go anywhere else in the visible universe, and adjust my velocity to move along with the local matter there, the universe will look isotropic to me. A homogeneous universe is defined as one in which you can do this: you can move anywhere within the universe, set your velocity to some value, and see an isotropic universe. You can then define the time coordinate globally in such a way that at the same time, separated observers see the same properties of the universe (such as the CMB temperature). In these coordinates, the properties of the universe are the same everywhere in space, but change with time.

You can change to a different set of coordinates, of course, and things won't necessarily look constant in space any longer. You'll still get the right answers for any observable you calculate, but you won't see the homogeneity.

You won't? spatial homogeneity is not a physical observable? it is just a convenient perspective only watchable with some privileged coordinates?

 

[Chalnoth]

You won't? spatial homogeneity is not a physical observable? it is just a convenient perspective only watchable with some privileged coordinates?

Yes, spatial homogeneity is only something that is watchable in some privileged coordinates. The only sort of homogeneity that would be visible in any coordinates is space-time homogeneity. We don't get to do that for homogeneity that is only in space, unfortunately.

This doesn't mean that spatial homogeneity is meaningless, however. Yes, it only appears in some special choice of coordinates. However, it isn't something that you can do in any sort of universe you might conceive. Remember the definition I laid down previously: if, at any point in space, one can construct a hypothetical observer that will see an isotropic universe, then we can call that universe homogeneous in space.

I could easily construct a universe that doesn't have this property. For instance, if we imagine a universe that is very dense in the direction of both poles of the Earth, but has very little matter in the directions outward from the Earth's equator, that would be a very anisotropic universe. The north/south direction would be picked out as a special direction. But what's more, there is no choice of observer located on Earth that could see that distribution as being isotropic.

In the end, this model of a homogeneous universe isn't a direct observable (because we can't move far enough away to check isotropy from different spatial locations), but it does have observable consequences. Namely, it states that the expansion of the universe should follow the Friedmann equations. When we measure the expansion of our universe using many different sorts of observations, and continually come up with the same answer every time, we gain confidence that the Friedmann equations are valid, at least approximately, which means we gain confidence that our universe is genuinely homogeneous in space (for a specific choice of coordinates).

 

[AWA]

Yes, spatial homogeneity is only something that is watchable in some privileged coordinates. The only sort of homogeneity that would be visible in any coordinates is space-time homogeneity. We don't get to do that for homogeneity that is only in space, unfortunately.

This doesn't mean that spatial homogeneity is meaningless, however. Yes, it only appears in some special choice of coordinates. However, it isn't something that you can do in any sort of universe you might conceive. Remember the definition I laid down previously: if, at any point in space, one can construct a hypothetical observer that will see an isotropic universe, then we can call that universe homogeneous in space.

I could easily construct a universe that doesn't have this property. For instance, if we imagine a universe that is very dense in the direction of both poles of the Earth, but has very little matter in the directions outward from the Earth's equator, that would be a very anisotropic universe. The north/south direction would be picked out as a special direction. But what's more, there is no choice of observer located on Earth that could see that distribution as being isotropic.

In the end, this model of a homogeneous universe isn't a direct observable (because we can't move far enough away to check isotropy from different spatial locations), but it does have observable consequences. Namely, it states that the expansion of the universe should follow the Friedmann equations. When we measure the expansion of our universe using many different sorts of observations, and continually come up with the same answer every time, we gain confidence that the Friedmann equations are valid, at least approximately, which means we gain confidence that our universe is genuinely homogeneous in space (for a specific choice of coordinates).

Ok, I see now clearly the source of our disagreement and of my "false paradox". Actually there is no paradox at all.

I guess the moral of the story is that one must not take GR to seriously because that is considered naive at best and against standard cosmology at worse.
But all books on GR stress general invariance (covariance), all of them say we can choose coordinates arbitrarily, which also means of course we can privilege some coordinates for the sake of convenience, but that convenience in no way means the results be get with that coordinates are physically real unless they can be reproduced with other choices of coordinates and metrics. I guess I also took too seriously the interchangeability of spacetime dimensions that relativity teaches us.

Now you tell me that spatial homogeneity, even though it is a property as physical as it can be, only appears with a determinate choice of coordinates that produce a certain privileged slicing of spacelike hypersurfaces, and that this homogeneity disapears if we try to make it coordinate invariant when we change the coordinates, appearing instead a sort of statistical homogeneity wrt both space and time (spacetime) and inhomogeneity or radial density dependence in the purely spatial hypersurface, and both of this things are forbidden by standard cosmology and astronomical observations and I have to take your word on this, no matter what GR says because you know more than me and standard cosmology says so and I'm a responsible citizen.

I declare the paradox solved unless someone else finds this a bit odd too or has some new input. Thanks a lot.

 

[Chalnoth]

General Relativity itself respects general covariance. But the specific distribution in our universe does not. In fact, it's pretty easy to prove that normal matter/radiation cannot respect general covariance, because the only covariant stress-energy tensor is one that behaves like vacuum energy.

Therefore, the very existence of matter ensures that the universe will look different in different coordinate systems. The general covariance of General Relativity ensures that you get the same results for the behavior of said matter no matter what coordinate system you use. And the math will be made easier if we use coordinates that follow any symmetries that exist in the matter distribution.

For instance, if you are doing physics on the surface of the Earth and not moving very far, it is convenient to approximate the Earth as a perfectly-flat surface. This set of coordinates will start to be wrong if we move too far along the Earth's surface or too far above it, but it is a convenient choice as long as we don't do these things.

If you are instead, for example, attempting to put a satellite into low-Earth orbit, it becomes convenient to use spherical coordinates centered at the center of the Earth, and ignore the effect of bodies further away. This set of coordinates will start to be wrong if you get too close to the Moon, or far enough from the Earth that the Sun's gravity becomes more important.

If you are instead interested in describing the motions of the planets, it becomes convenient to use spherical coordinates centered on the Sun.

And so on and so forth. You can use General Relativity in each case. In each case, exploiting the symmetries of the physical matter distribution makes the math easier.

This is what we are doing when we use FRW coordinates: we are exploiting a particular symmetry of the average matter distribution of our universe, namely spatial homogeneity. Allowing our coordinate choice to follow this symmetry makes the math easier.

 

[AWA]

General Relativity itself respects general covariance. But the specific distribution in our universe does not. In fact, it's pretty easy to prove that normal matter/radiation cannot respect general covariance, because the only covariant stress-energy tensor is one that behaves like vacuum energy.

I wish some expert relativist would confirm this, maybe some guy from the relativity forum, as I consider it not exact but that might be due to my poor knowledge of GR. I'll try to think about it some more.

Therefore, the very existence of matter ensures that the universe will look different in different coordinate systems.

We must have some different understanding of isotropy as different coordinates systems can mean rotating the observer point of view and this should be invariant if there is isotropy.

This is what we are doing when we use FRW coordinates: we are exploiting a particular symmetry of the average matter distribution of our universe, namely spatial homogeneity. Allowing our coordinate choice to follow this symmetry makes the math easier

Only remember this coordinate-dependent spatial homogeneity hasn't been completely confirmed by empirical observations. Close but not yet.

 

[Chalnoth]

I wish some expert relativist would confirm this, maybe some guy from the relativity forum, as I consider it not exact but that might be due to my poor knowledge of GR. I'll try to think about it some more.

Well, it's really trivial to see that this is true in special relativity. The Lorentz transformations in special relativity are the set of transformations that leave the following matrix unchanged:
\begin{array}{rrrr} 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -1 \end{array}

(This can also be identified as the metric of Minkowski space-time.)

Since the stress-energy tensor transforms between coordinate systems in the same way as the metric, to get a stress-energy tensor that also doesn't change when you perform a Lorentz transform, you need that stress-energy tensor to be proportional to the metric. That is:

\begin{array}{rrrr} \rho &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -\rho &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -\rho &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -\rho \end{array}

In other words, you need pressure that is equal to the negative of the energy density, a condition which no known matter field satisfies, but which vacuum energy does (some scalar fields get close, but the relationship isn't exact).

We must have some different understanding of isotropy as different coordinates systems can mean rotating the observer point of view and this should be invariant if there is isotropy.

Perhaps I wasn't entirely clear. The point is that the existence of matter ensures that at least some coordinate transformations lead to changes a different-looking universe. Obviously there can still be other symmetries in the universe such that certain particular types of coordinate change may leave everything looking the same. As you mention, isotropy means that rotating your coordinate change has no effect. And homogeneity means that performing a spatial translation on your coordinate system has no effect (for a particular equal-time slicing of the universe).

Only remember this coordinate-dependent spatial homogeneity hasn't been completely confirmed by empirical observations. Close but not yet.

Isotropy has, however, and that is also coordinate-dependent. One need only have a different velocity and the isotropy no longer appears.

 

[AWA]

Well, it's really trivial to see that this is true in special relativity. The Lorentz transformations in special relativity are the set of transformations that leave the following matrix unchanged:
\begin{array}{rrrr} 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -1 \end{array}

(This can also be identified as the metric of Minkowski space-time.)

Since the stress-energy tensor transforms between coordinate systems in the same way as the metric, to get a stress-energy tensor that also doesn't change when you perform a Lorentz transform, you need that stress-energy tensor to be proportional to the metric. That is:

\begin{array}{rrrr} \rho &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -\rho &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -\rho &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -\rho \end{array}

In other words, you need pressure that is equal to the negative of the energy density, a condition which no known matter field satisfies, but which vacuum energy does (some scalar fields get close, but the relationship isn't exact).

Good stuff, thanks. I agree with this. How would this change when applied to GR?

Isotropy has, however, and that is also coordinate-dependent. One need only have a different velocity and the isotropy no longer appears.

So according to you, not only spatial homogeneity is coordinate dependent, but also isotropy.
But I understood that according to SR no physics experiment should allow us to distinguish between different uniform velocities ("Special principle of relativity: If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K."), that is why we don't notice the Earth's rotational or translational speed. If what you say is true one would notice its relative velocity as special since at different velocities one could perform experiments like these: http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/experiments.html#Tests_of_isotropy_of_space and according to the results distinguish different velocities.

is not someone out there that sees it this way too?, please speak up.

 

[Chalnoth]

Good stuff, thanks. I agree with this. How would this change when applied to GR?

In GR, a covariant tensor is one that is proportional to the metric:

T_{\mu\nu} = \rho g_{\mu\nu}

So according to you, not only spatial homogeneity is coordinate dependent, but also isotropy.

Yes, absolutely. If you're moving with respect to the CMB, for instance, you will see the CMB ahead of you blue-shifted, and the CMB behind you red-shifted. And this is exactly what we do see: we measure our velocity with respect to the CMB as being about 630km/sec. The anisotropy induced by this velocity is at roughly the 0.1% level in terms of temperature difference in either direction. When we remove the effect of a 630km/sec velocity from the CMB, we end up with a universe that is isotropic to about one part in 100,000.

But I understood that according to SR no physics experiment should allow us to distinguish between different uniform velocities ("Special principle of relativity: If a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K."), that is why we don't notice the Earth's rotational or translational speed. If what you say is true one would notice its relative velocity as special since at different velocities one could perform experiments like these: http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/experiments.html#Tests_of_isotropy_of_space and according to the results distinguish different velocities.

The correct statement is that you can't distinguish between uniform velocities without comparing against other things. That is, you can't build a closed experiment to measure your velocity. But you certainly can measure your velocity with respect to the Sun, the Earth, the Milky Way, the cosmic microwave background, or anything else you choose to measure your velocity against.

 

[AWA]

The correct statement is that you can't distinguish between uniform velocities without comparing against other things. That is, you can't build a closed experiment to measure your velocity.But you certainly can measure your velocity with respect to the Sun, the Earth, the Milky Way, the cosmic microwave background, or anything else you choose to measure your velocity against.

Please don't start with the dirty strawman game, who could argue with this, I never suggested you couldn't measure your velocity and you know it. I said the special principle of relativity states that physical laws should be the same in every inertial frame of reference, or exactly what you state in the quote " That is, you can't build a closed experiment to measure your velocity." And I offered you a public reference with closed experiments that would give you different results with or without spatial isotropy, that according to you would then give different results with different uniform velocities so you could in principle distinguish them.

 

[JDoolin]

Well, it's really trivial to see that this is true in special relativity. The Lorentz transformations in special relativity are the set of transformations that leave the following matrix unchanged:
\begin{array}{rrrr} 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -1 \end{array}

I think only the Identity would leave that matrix unchanged. Perhaps you mean to say it would leave its determinant unchanged.

 

[Chalnoth]

Please don't start with the dirty strawman game, who could argue with this, I never suggested you couldn't measure your velocity and you know it. I said the special principle of relativity states that physical laws should be the same in every inertial frame of reference, or exactly what you state in the quote " That is, you can't build a closed experiment to measure your velocity." And I offered you a public reference with closed experiments that would give you different results with or without spatial isotropy, that according to you would then give different results with different uniform velocities so you could in principle distinguish them.

These experiments don't say anything about the isotropy of the matter in our universe. By "spatial isotropy" they mean instead that the laws of physics are the same no matter the orientation of your experimental apparatus. These are rather different concepts.

 

[Chalnoth]

I think only the Identity would leave that matrix unchanged. Perhaps you mean to say it would leave its determinant unchanged.

The way that you transform a matrix between coordinate systems is by sandwiching it between the matrix that transforms a vector, like so:

A&#039; = UAU^T

There is definitely a set of matrices U which, upon transforming the matrix representation of the Minkowski metric, leave the Minkowski metric unchanged. These matrices are a representation of the Poincaré group.

If you still have a difficult time seeing how this can be, consider the matrix product (reduced to two dimensions to make the math easier):

\[\left(\begin{array}{cc}<br /> \cosh \phi &amp; -\sinh \phi \\<br /> -\sinh \phi &amp; \cosh \phi \end{array}\right)\left(\begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1 \end{array}\right)\left(\begin{array}{cc}<br /> \cosh \phi &amp; -\sinh \phi \\<br /> -\sinh \phi &amp; \cosh \phi\end{array}\right)\]

 

[AWA]

These experiments don't say anything about the isotropy of the matter in our universe.

Yeah, right. Do you mean that matter here is special, that the physics here is different than in other points of the universe, that our instruments are special and follow special laws of physics? That is not a very popular opinion in modern cosmology. You yourself have said many times that isotropy is ubiquitous in our universe, that otherwise it would go against the Copernican principe. ( the no special place principle)

By "spatial isotropy" they mean instead that the laws of physics are the same no matter the orientation of your experimental apparatus.

Exactly. That is what they mean.That is why I presented them. Once again I ask you: Do you mean then that the matter here on Earth is different than in the rest of the universe? that our instruments have something special that wouldn't work outside the earth? Double strawman alarm!

These are rather different concepts.

You seem to have this muddled. If there is isotropy it is the same here and everywhere, think about it.

 

[JDoolin]

The way that you transform a matrix between coordinate systems is by sandwiching it between the matrix that transforms a vector, like so:

A&#039; = UAU^T

There is definitely a set of matrices U which, upon transforming the matrix representation of the Minkowski metric, leave the Minkowski metric unchanged. These matrices are a representation of the Poincaré group.

If you still have a difficult time seeing how this can be, consider the matrix product (reduced to two dimensions to make the math easier):

\[\left(\begin{array}{cc}<br /> \cosh \phi &amp; -\sinh \phi \\<br /> -\sinh \phi &amp; \cosh \phi \end{array}\right)\left(\begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1 \end{array}\right)\left(\begin{array}{cc}<br /> \cosh \phi &amp; -\sinh \phi \\<br /> -\sinh \phi &amp; \cosh \phi\end{array}\right)\]

Hmmm. That's interesting, but I'm not sure it's physically meaningful. What you are doing there is boosting, performing a transformation (making the time negative) then boosting again in the same direction. I guess since you made the time negative, maybe it boosts it back the way it was?

Yes-- you're right, almost. If you stretch it out along the t=x axis, then flip it vertically, then stretch it out on the t=x axis again, you do indeed get the same thing you started with, except it would be mirror-imaged along the t=0 axis. So you don't get the Minkowski Metric Unchanged--You get the Minkowski metric running backwards in time.

I think you could get something similar with rotation; turn it, flip it, turn it the same way again. You'd still end up with a mirror image, of course.


but the Lorentz transform itself only involves,

<br /> \begin{pmatrix}<br /> \cosh \phi &amp; -\sinh \phi \\<br /> -\sinh \phi &amp; \cosh \phi<br /> \end{pmatrix}<br />​

and the inverse Lorentz Transform would be

<br /> \begin{pmatrix}<br /> \cosh( -\phi) &amp; -\sinh (\phi) \\<br /> -\sinh (-\phi) &amp; \cosh (-\phi)<br /> \end{pmatrix}<br />​

It works just like a rotation matrix. You wouldn't say you had applied a "rotation" if you rotated it and then rotated it back. You just rotate it, and leave it that way.

 

[Chalnoth]

Hmmm. That's interesting, but I'm not sure it's physically meaningful. What you are doing there is boosting, performing a transformation (making the time negative) then boosting again in the same direction. I guess since you made the time negative, maybe it boosts it back the way it was?

When you do the same operation with tensors, this is how it works:

\eta&#039;_{\mu\nu} = \Lambda_\mu^\alpha \Lambda_\nu^\beta \eta_{\alpha\beta}

The reason, then, why you multiply the transformation matrix twice when transforming a matrix is because you have two indices to transform.

Another way of looking at it is to consider how the matrix is used. The metric sets up a dot product between vectors. You get the space-time distance between two events like so:

s^2 = d^T A d

...where A is set to the Minkowski metric as before, and d is the space-time four-vector that is the displacement between the events. A tiny bit of math verifies that:

s^2 = d_t^2 - d_x^2 - d_y^2 - d_z^2

...which is a valid space-time distance in special relativity. Now, if we want to transform to a different coordinate system, we perform a Lorentz transformation on the displacement like so:

d&#039;= \Lambda d

..where \Lambda is a transformation matrix representing the Lorentz transformation we want to perform. One thing that we know is that Lorentz transformations leave space-time distances unchanged. This means:

s^2 = d^T A d = d&#039;^T A d&#039;

Some quick math reveals:

d&#039;^T A d&#039; = d^T \Lambda^T A \Lambda d

So what this means is that a valid Lorentz transformation is one where:
A = \Lambda^T A \Lambda
...because this will leave the space-time distances unchanged.

 

[JDoolin]

When you do the same operation with tensors, this is how it works:

\eta&#039;_{\mu\nu} = \Lambda_\mu^\alpha \Lambda_\nu^\beta \eta_{\alpha\beta}

Okay, what does that mean? \Lambda and \eta are what we were earlier calling U and A? Can you write it with the summation notation, because, regardless of how "genius" Einstein's idea of not writing the summations it makes it virtually impossible for me to read. I can see now that this summation notation does represent \Lambda \cdot \eta \cdot \Lambda^T or vice versa, because I don't know whether the subscript or the superscript represents the row or column index.

The reason, then, why you multiply the transformation matrix twice when transforming a matrix is because you have two indices to transform.

<br /> \begin{matrix}<br /> D=A \cdot B\cdot C\\<br /> \Leftrightarrow \\<br /> d_{mn}=\sum_{j=1}^{4} a_{mj}<br /> \left (\sum_{i=1}^{4}<br /> b_{ji}c_{in} \right )<br /> \\ <br /> =\sum_{j=1}^{4} <br /> \sum_{i=1}^{4}a_{mj}b_{ji}c_{in}\\ <br /> =a_{mj}b_{ji}c_{in} (Einstein Notation) <br /> \end{matrix}​

Changing your notation changes nothing, except that now, there's no way to determine what you are saying, because you are not telling me the thirty-two quantities:

a_{11},a_{12},a_{13},a_{14},a_{21},...,a_{44},b_{11},b{12},b_{13},b_{14},b_{21},...,b_{44}​

The transformation:

\[\left(\begin{array}{cc}\cosh \phi &amp; -\sinh \phi \\-\sinh \phi &amp; \cosh \phi \end{array}\right)\left(\begin{array}{cc}1 &amp; 0 \\0 &amp; -1 \end{array}\right)\left(\begin{array}{cc}\cosh \phi &amp; -\sinh \phi \\-\sinh \phi &amp; \cosh \phi\end{array}\right)\]​

is NOT a lorentz Transformation. It is a reflection across an axis.

Another way of looking at it is to consider how the matrix is used. The metric sets up a dot product between vectors. You get the space-time distance between two events like so:

s^2 = d^T A d

...where A is set to the Minkowski metric as before, and d is the space-time four-vector that is the displacement between the events. A tiny bit of math verifies that:

s^2 = d_t^2 - d_x^2 - d_y^2 - d_z^2

...which is a valid space-time distance in special relativity. Now, if we want to transform to a different coordinate system, we perform a Lorentz transformation on the displacement like so:

d&#039;= \Lambda d

..where \Lambda is a transformation matrix representing the Lorentz transformation we want to perform. One thing that we know is that Lorentz transformations leave space-time distances unchanged. This means:

s^2 = d^T A d = d&#039;^T A d&#039;

Some quick math reveals:

d&#039;^T A d&#039; = d^T \Lambda^T A \Lambda d

So what this means is that a valid Lorentz transformation is one where:
A = \Lambda^T A \Lambda
...because this will leave the space-time distances unchanged.

Yes. Reflecting an image around any axis will preserve the distances. It doesn't mean that it is physically meaningful. Sure, maybe, reflection is part of the Poincaire group, and maybe by some technical definition, reflection is a member of the Lorentz Transformations. But it is NOT a Lorentz Boost.

 

[Chalnoth]

Okay, what does that mean? \Lambda and \eta are what we were earlier calling U and A?

Yes, this is the more proper notation. The second-rank tensor Lambda is the tensor which transforms a vector from one coordinate system to another like so:

v&#039;_\mu = \Lambda_\mu^\alpha v_\alpha

The second rank tensor \eta is the Minkowski metric.

As for which is the row and which is the column, well, that is just up to whichever configuration makes the linear algebra match the sums properly.

Can you write it with the summation notation, because, regardless of how "genius" Einstein's idea of not writing the summations it makes it virtually impossible for me to read.

I don't understand what's so hard. Just assume any indices that appear twice in a single expression are summed over. It's not really "genius", it's just convenient for making more complicated formulas readable.

The transformation:

\[\left(\begin{array}{cc}\cosh \phi &amp; -\sinh \phi \\-\sinh \phi &amp; \cosh \phi \end{array}\right)\left(\begin{array}{cc}1 &amp; 0 \\0 &amp; -1 \end{array}\right)\left(\begin{array}{cc}\cosh \phi &amp; -\sinh \phi \\-\sinh \phi &amp; \cosh \phi\end{array}\right)\]​

is NOT a lorentz Transformation. It is a reflection across an axis.

It's a Lorentz boost with \phi = \ln [\gamma(1 + \beta)], with \gamma being the usual special relativity gamma, and \beta being the velocity of the boost as a fraction of the speed of light.

You can read more on this notation here:
http://en.wikipedia.org/wiki/Lorentz_transformation

 

[JDoolin]

Yes, this is the more proper notation. The second-rank tensor Lambda is the tensor which transforms a vector from one coordinate system to another like so:

v&#039;_\mu = \Lambda_\mu^\alpha v_\alpha

The second rank tensor \eta is the Minkowski metric.

Earlier, you defined that tensor \eta as {{-1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}}

I cannot see how that tensor in any way resembles the Minkowski Metric. The Minkowski Metric is simply a Cartesian Coordinate system with time. The tensor you defined simply describes a reflection along the t=0 hyperplane. You'll have to try to make the connection between these two completely unrelated concepts for me.

As for which is the row and which is the column, well, that is just up to whichever configuration makes the linear algebra match the sums properly.

Why not have a notation that is unambiguous? If you have to transfer it to matrices and then figure out how to multiply the linear algebra sums to figure out whether the subscript and superscript represent rows or vectors, why not just leave it in matrix form in the first place?

I don't understand what's so hard. Just assume any indices that appear twice in a single expression are summed over. It's not really "genius", it's just convenient for making more complicated formulas readable.

I don't think it was "genius" either. At least we have that in common. And it would be okay if you didn't arbitrarily start putting in superscripts and subscripts randomly. There should be a clear order for row, colum, page, book, edition, etc.

It's a Lorentz boost with \phi = \ln [\gamma(1 + \beta)], with \gamma being the usual special relativity gamma, and \beta being the velocity of the boost as a fraction of the speed of light.

No, what you did is NOT a Lorentz Transformation. What you did was a Lorentz Transformation followed by a reflection followed by another Lorentz Transformation. You know this evaluates to the reflection at the end. (what you erroneously call the tensor representing the Minkowski Metric)

So the whole process evaluates to a reflection in time. NOT a Lorentz Transformation.

You can read more on this notation here:
http://en.wikipedia.org/wiki/Lorentz_transformation

There appears to be little overlap between our knowledge. Whereas I can understand what is going on in all the pictures and animations on that page, your knowledge seems to be constrained to this part:

http://en.wikipedia.org/wiki/Lorentz_transformation#Spacetime_interval

...which, quite frankly, looks like word-salad to me.

 

[Chalnoth]

Earlier, you defined that tensor \eta as {{-1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}}

I cannot see how that tensor in any way resembles the Minkowski Metric. The Minkowski Metric is simply a Cartesian Coordinate system with time. The tensor you defined simply describes a reflection along the t=0 hyperplane. You'll have to try to make the connection between these two completely unrelated concepts for me.

This isn't about reflection. The metric defines the dot product between vectors:

\vec{v} \cdot \vec{w} = v^\alpha v^\beta \eta_{\alpha\beta}

If you remember your special relativity, you may remember that, for instance, the space-time distance between points is:

s^2 = t^2 - x^2 - y^2 - z^2

Or that the mass of a particle is:

m^2 = E^2 - p_x^2 - p_y^2 - p_z^2

It shouldn't be difficult to verify that the metric I gave previously gives the proper dot product for four-vectors in Special Relativity.

Why not have a notation that is unambiguous? If you have to transfer it to matrices and then figure out how to multiply the linear algebra sums to figure out whether the subscript and superscript represent rows or vectors, why not just leave it in matrix form in the first place?

Because in General Relativity, we have to work with third-rank and fourth-rank tensors, not just first and second-rank ones. And the notation is perfectly unambiguous, by the way. You just have to use a little bit of thought to translate between the different ways of doing things, when it's possible at all to translate to standard linear algebra.

I don't think it was "genius" either. At least we have that in common. And it would be okay if you didn't arbitrarily start putting in superscripts and subscripts randomly. There should be a clear order for row, colum, page, book, edition, etc.

The superscripts and subscripts aren't arbitrary at all. In fact, a vector with an upper index is different from a vector with a lower index. Specifically,

v_\beta = v^\alpha \eta_{\alpha\beta}

So with \eta being the metric for Minkowski space, this means that the difference between the vector with the lower index and the one with the upper index is that the spatial components take on negative values. This doesn't mean, by the way, that the vector is mirrored, just that the vector with the upper index and the one with the lower index use a different sign notation (and in General Relativity, the two vectors can be very different, since the metric can be a function of time and space, and have off-diagonal components).

You may have noticed that before, I only combined an upper index with a lower one? This is specifically because when you're using this notation, that's the only kind of operation you can perform. If you want to sum over a pair of indices that are both lower or both upper, you first have to raise or lower one of them with the metric. This is why the metric appears in the dot product:

v^2 = v^\alpha v_\alpha = v^\alpha v^\beta \eta_{\alpha\beta}

No, what you did is NOT a Lorentz Transformation. What you did was a Lorentz Transformation followed by a reflection followed by another Lorentz Transformation. You know this evaluates to the reflection at the end. (what you erroneously call the tensor representing the Minkowski Metric)

Once again, the \Lambda matrix represents a Lorentz transformation, so that:

v&#039;^\alpha = \Lambda^\alpha_\beta v^\beta

We know that Lorentz transformations do not change the dot product. This means, for instance, the quantity:

m^2 = E^2 - p_x^2 - p_y^2 - p_z^2

...will evaluate to the same mass no matter which reference frame you perform the operations in.

This means that:

m^2 = p^\alpha p^\beta \eta_{\alpha\beta} = p&#039;^\alpha p&#039;^\beta \eta_{\alpha\beta}

This evaluates to:

p^\alpha p^\beta \eta_{\alpha\beta} = p^\alpha \Lambda_\alpha^\mu p^\beta \Lambda_\beta^\nu \eta_{\mu\nu}

Which implies:

\eta_{\alpha\beta} = \Lambda_\alpha^\mu \Lambda_\beta^\nu \eta_{\mu\nu}

...since the previous equation must hold for all choices of the momentum 4-vector p^\alpha. This last equation is equivalent to the matrix operations I gave previously, and it defines the possible ways to transform between different coordinate systems in Special Relativity, which includes rotations, translations, and boosts.

 

[JDoolin]

If the only thing you know about trigonometry is that r^2=x^2+y^2+z^2 then you don't really know much anything about trigonometry. And if the only thing you know about Special Relativity is that s^2=(ct)^2-x^2-y^2-z^2 you don't really know much about Special Relativity.

Though I have done it before, I can actually find no epistemological value in taking a single event (t,x,y,z) and dot product it with itself and a metric to achieve t^2-x^2-y^2-z^2. I don't think one gains any insight into the Special Theory of Relativity by doing such a thing.

Here is my own way of deriving the metric:

http://www.wiu.edu/users/jdd109/swf/Plot6to9.swf

This might be better suited to a paper than a video, but I think you can see there are ways to get to it by reason, rather than assertion. Perhaps after one understands why the metric is such as it is, then the mathematical shortcut has some value.

But I think that you are mis-applying matrices where they really don't belong, and refusing to use them where the really do belong.

 

[Chalnoth]

If the only thing you know about trigonometry is that r^2=x^2+y^2+z^2 then you don't really know much anything about trigonometry. And if the only thing you know about Special Relativity is that s^2=(ct)^2-x^2-y^2-z^2 you don't really know much about Special Relativity.

Though I have done it before, I can actually find no epistemological value in taking a single event (t,x,y,z) and dot product it with itself and a metric to achieve t^2-x^2-y^2-z^2. I don't think one gains any insight into the Special Theory of Relativity by doing such a thing.

This particular dot product gives you the proper time between events squared, such that if particle traverses at displacement (x, y, z) in time t at constant velocity, its clock will move by an amount given by the square root of the above dot product. The dot product of a momentum 4-vector with itself gives you the mass squared. Other 4-vectors will give you other coordinate-independent invariant quantities.

And by the way, all of special relativity is encoded in the metric. Once you have the metric, you can derive everything else. It's not necessarily easy, as some of the derivations will be a little non-obvious, and it's certainly more abstract than more visual demonstrations, but it's all there.

 

[AWA]

The Lorentz transformations in special relativity are the set of transformations that leave the following matrix unchanged:
\begin{array}{rrrr} 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -1 \end{array}

(This can also be identified as the metric of Minkowski space-time.)

Since the stress-energy tensor transforms between coordinate systems in the same way as the metric, to get a stress-energy tensor that also doesn't change when you perform a Lorentz transform, you need that stress-energy tensor to be proportional to the metric. That is:

\begin{array}{rrrr} \rho &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -p &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -p &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -p \end{array}

In other words, you need pressure that is equal to the negative of the energy density, a condition which no known matter field satisfies, but which vacuum energy does (some scalar fields get close, but the relationship isn't exact).

So , the vacuum tensor obviously satisfy it since all its components are zero and the electromagnetic tensor too because it has no trace so it has a Poincare group, right?
But massive fields wouldn't because they would have to have pressure components with opposite sign to the energy density component. I would like to understand better why is negative pressure considered unphysical for matter.

 

[Chalnoth]

So , the vacuum tensor obviously satisfy it since all its components are zero and the electromagnetic tensor too because it has no trace so it has a Poincare group, right?

The stress-energy tensor for vacuum energy is most definitely not one with all components equal to zero. But it is a stress-energy tensor that is proportional to the metric, and so transforms like the metric. The traceless electromagnetic tensor isn't proportional to the metric and so doesn't transform like the metric, and thus looks very different in different coordinate systems, just like a matter stress-energy tensor.

But massive fields wouldn't because they would have to have pressure components with opposite sign to the energy density component. I would like to understand better why is negative pressure considered unphysical for matter.

Well, matter just doesn't have negative pressure. If it were, then you could place a higher density of matter in a box than exists outside that box, and the pressure would pull inward on the sides of the box. But this isn't what happens: instead the pressure pushes outward.

 

[AWA]

The stress-energy tensor for vacuum energy is most definitely not one with all components equal to zero. But it is a stress-energy tensor that is proportional to the metric, and so transforms like the metric. The traceless electromagnetic tensor isn't proportional to the metric and so doesn't transform like the metric, and thus looks very different in different coordinate systems, just like a matter stress-energy tensor.


Well, matter just doesn't have negative pressure. If it were, then you could place a higher density of matter in a box than exists outside that box, and the pressure would pull inward on the sides of the box. But this isn't what happens: instead the pressure pushes outward.

I probably didnot express it correctly, I was considering that when in the vacuum solutions of GR we make Tab=0, so I figured that all the components are zero in this case, on the other hand, I realize that vacuum is supposed to have some energy, very high according to QFT, so I'm a little confused on this.
As for the electromagnetic tensor, I thought the electromagnetic tensor could be formulated in a covariant form: http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

 

[Chalnoth]

I probably didnot express it correctly, I was considering that when in the vacuum solutions of GR we make Tab=0, so I figured that all the components are zero in this case, on the other hand, I realize that vacuum is supposed to have some energy, very high according to QFT, so I'm a little confused on this.
As for the electromagnetic tensor, I thought the electromagnetic tensor could be formulated in a covariant form: http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

Yes, covariant. Not invariant.

The stress-energy tensor for vacuum energy is invariant under Lorentz transformations. Covariant means you're describing the same physics, just with different numbers. Covariant things can look very different in different coordinate systems. Invariant means that not even the numbers change, and so things look the same no matter what your position or velocity.

In the electromagnetic case, for instance, what a stationary observer sees as only being an magnetic field will be seen by most moving observers as a combination of electric and magnetic fields. However, both observers will describe the exact same paths for electrons moving in said field (once you correct for the relative coordinate system differences). The identical electron paths stem from the covariance of the theory. But the behavior still looks different to different observers.

 

[George Jones]

Maxwell's equations, written in terms of the electromagnetic field tensor, are covariant; Einstein's equation of GR is covariant. This means that the equations have the same structural form in all coordinate systems.

The metric tensor is invariant under Lorentz transformations, but, even in SR, is not invariant under general coordinate transformations.

 

[Chalnoth]

The metric tensor is invariant under Lorentz transformations, but, even in SR, is not invariant under general coordinate transformations.

Yes, sorry I didn't make this clear. For example, the metric (and therefore the stress-energy tensor for vacuum energy) looks rather different in Cartesian coordinates vs. spherical coordinates. Changed my post to clarify this point.

 

[AWA]

Yes, sorry I didn't make this clear. For example, the metric (and therefore the stress-energy tensor for vacuum energy) looks rather different in Cartesian coordinates vs. spherical coordinates. Changed my post to clarify this point.

Well, this is actually a deviation or distraction from the main discussion.
It is quite obvious that General covariance affects the GR field equations, not individual tensors, therefore the specific covariance or invariance (or lack of) of the matter stress-energy tensor is totally irrelevant. This tensor gives us the quantities of the components of the source of the gravitational field, nothing to do directly with the specific distribution of the universe.
Actually general covariance is not concerned directly with the specific distribution of matter of the universe, either, which is more of an empirical issue.
We are the ones that introduce from outside in the GR field equations assumptions concerning this distribution, this assumptions(spatial isotropy and homogeneity) are in part due to philosophical-historical reasons and in part due to empirical observations that are also constrained by a certain interpretation of these observations (redshift, statistical treatment of galaxy surveys like SDSS, 2MASS, etc).

 

[Chalnoth]

Well, this is actually a deviation or distraction from the main discussion.

No, it's not really. The point is that under any transformations in the Poincaré group, which includes Lorentz transformations, rotations, and translations, the stress-energy tensor for vacuum energy doesn't change. This is a very non-trivial statement, even in General Relativity, because you can reduce any space-time point to Minkowski space in some local region around that point.

This is relevant to the discussion because previously you expressed concern that the FRW metric was establishing a universal time coordinate. The demonstration that only the stress-energy tensor for vacuum energy can be invariant under Lorentz transformations indicates that there is no way to formulate a metric for the universe that includes any matter but which also doesn't include some sort of universal time coordinate.

Now, the particular choice for this universal time coordinate will be somewhat arbitrary, but if the system at hand has any symmetries to exploit, then some choices of the time coordinate will be much more convenient than others. In this situation, there is a specific sort of translational symmetry, where at every location in space, there is a particular time at which the space looks the same as some specific time at every other location in space. In other words, the universe is homogeneous in space for a specific choice of time coordinate. Exploiting this symmetry leads to much simpler equations.

 

[AWA]

This is relevant to the discussion because previously you expressed concern that the FRW metric was establishing a universal time coordinate. The demonstration that only the stress-energy tensor for vacuum energy can be invariant under Lorentz transformations indicates that there is no way to formulate a metric for the universe that includes any matter but which also doesn't include some sort of universal time coordinate.

I can't quite follow the logic from the invariant vacuum tensor to the necesity of including some sort of universal time coordinate in the presence of matter. Precisely GR is about the possibility to formulate any metric. Obviously this metric will include a time coordinate that you can consider "universal" for a number of practical reasons, but that doesn't make it really "universal", unless you want to go back to the concept of absolute time from Newton, but wasn't that what Einstein tried to change?

Now, the particular choice for this universal time coordinate will be somewhat arbitrary, but if the system at hand has any symmetries to exploit, then some choices of the time coordinate will be much more convenient than others.

If the system has them, sure.

In this situation, there is a specific sort of translational symmetry, where at every location in space, there is a particular time at which the space looks the same as some specific time at every other location in space. In other words, the universe is homogeneous in space for a specific choice of time coordinate. Exploiting this symmetry leads to much simpler equations.

That is right, if the assumptions are well founded, and I'm not saying there are no good reasons to believe it. As I said there is empirical observations that seem to indicate it, and philosophical reasons to expect it. But, we must also be prepared for surprises.

 

[Chalnoth]

That is right, if the assumptions are well founded, and I'm not saying there are no good reasons to believe it. As I said there is empirical observations that seem to indicate it, and philosophical reasons to expect it. But, we must also be prepared for surprises.

Yes, but one of the nice things is that there's a very simple way to check: continue improving the precision and accuracy of our measurements. If something is wrong about one of our assumptions, then it is highly likely to show up as a set of observations that do not agree with one another.

There are, since the advent of the Lambda-CDM model, no clear indications of this to date, with all apparent discrepancies in areas where the systematic errors are not under adequate control.

 

[AWA]

Yes, but one of the nice things is that there's a very simple way to check: continue improving the precision and accuracy of our measurements. If something is wrong about one of our assumptions, then it is highly likely to show up as a set of observations that do not agree with one another.

Unless the model is built in such a way that whenever an observation that doesn't agree with the assumptions does show up (say, like faint SNaeIa) it can be integrated by changing the parameters of the model. So it's not so simple, the assumptions are alway right it seems, or would you give me an example of an observation that would make us reconsider some fundamental assumption?

 

[Chronos]

Unless the model is built in such a way that whenever an observation that doesn't agree with the assumptions does show up (say, like faint SNaeIa) it can be integrated by changing the parameters of the model. So it's not so simple, the assumptions are alway right it seems, or would you give me an example of an observation that would make us reconsider some fundamental assumption?

A high redshift object superimposed over a lower redshift object, an impossible orbital velocity in a system whose distance has been determined by parallax . . .

 

[Chalnoth]

Unless the model is built in such a way that whenever an observation that doesn't agree with the assumptions does show up (say, like faint SNaeIa) it can be integrated by changing the parameters of the model. So it's not so simple, the assumptions are alway right it seems, or would you give me an example of an observation that would make us reconsider some fundamental assumption?

Well, the difficulty is that if the observations we have really are consistent with the standard model anyway, then merely contemplating other models won't help you discover that any other model is actually better.

Basically, the only time that it becomes really useful for the progress of science to propose a new idea is if that new idea leads to the development of new experiments/observations that would not have been performed otherwise. The problem right now in Cosmology is that there isn't any clear direction for deviations from the current model, so the most reasonable course of action is to just continue to test the current standard model more and more accurately (well, that and engage in low-cost experiments that test alternative hypotheses, but unfortunately in cosmology that's a bit challenging).

As a final point, it isn't just a matter of fitting the supernova data, but instead of fitting combinations of data, from supernovae to galaxy cluster counts to CMB data to baryon acoustic oscillations. They all have to add together and point in the same direction, or something is wrong. Most often that turns out to be some sort of systematic error, but if the same discrepancy keeps popping up again and again, that will eventually become a clear indication of where we should move from the standard model.

That, right now, be our best bet for progress in cosmology. Our second best bet is for new discoveries at the LHC to provide us with new ideas about the nature of dark matter (don't hold your breath, though: the LHC isn't very good at making dark matter particles, even if those particles have the right sort of properties).

 

[AWA]

A high redshift object superimposed over a lower redshift object

Well that observation seems to have been made but discredited on grounds of impossibility according to our model, and statistical irrelevance. So it's a good example of what I am saying, discordant observations are either integrated, dismissed as irrelevant for statistical reasons, or completely ignored, I'm just wondering if such a systematic approach to observations discordant with the standard model can be called science, since it provides us with a perfect excuse for never questioning the initial assumptions.
But perhaps this is not about science at all.

 

[Chalnoth]

Well that observation seems to have been made

Uh, what? No, this observation certainly has not been made. What we have seen are high redshift objects that closely align with low redshift objects. But we have not seen any that show indications of these high redshift objects actually being in the foreground.

The way you test this, by the way, is by looking at absorption lines. Intervening gas blocks light preferentially at specific wavelengths, and so we can see both intervening gas, and its redshift, by looking for such absorption lines.

So what you see is the background object emits at some frequencies and absorbs in others, while the foreground object emits at other frequencies (emitting light at some wavelengths the background object did not emit light), while absorbing light at other frequencies where the background object did emit light. Because the foreground object basically erases the background object's absorption lines, absorption lines tell us primarily about the foreground object.

And those absorption lines, whenever there is such an alignment, come from the low-redshift object, not the high-redshift one.