Is the Cosmological Principle Limited to Space Only?

[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.