Is the Cosmological Principle Limited to Space Only?

[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:
$$\begin{array}{rrrr} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}$$

(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:

$$\begin{array}{rrrr} \rho & 0 & 0 & 0\\ 0 & -\rho & 0 & 0\\ 0 & 0 & -\rho & 0\\ 0 & 0 & 0 & -\rho \end{array}$$

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:
$$\begin{array}{rrrr} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}$$

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:

$$A' = UAU^T$$

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):

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

 

[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:

$$A' = UAU^T$$

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):

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

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,

$$\begin{pmatrix} \cosh \phi & -\sinh \phi \\ -\sinh \phi & \cosh \phi \end{pmatrix}$$​

and the inverse Lorentz Transform would be

$$\begin{pmatrix} \cosh( -\phi) & -\sinh (\phi) \\ -\sinh (-\phi) & \cosh (-\phi) \end{pmatrix}$$​

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:

$$\eta'_{\mu\nu} = \Lambda_\mu^\alpha \Lambda_\nu^\beta \eta_{\alpha\beta}$$

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:

$$d'= \Lambda d$$

..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:

$$s^2 = d^T A d = d'^T A d'$$

Some quick math reveals:

$$d'^T A d' = d^T \Lambda^T A \Lambda d$$

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:

$$\eta'_{\mu\nu} = \Lambda_\mu^\alpha \Lambda_\nu^\beta \eta_{\alpha\beta}$$

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.

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

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:

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

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:

$$d'= \Lambda d$$

..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:

$$s^2 = d^T A d = d'^T A d'$$

Some quick math reveals:

$$d'^T A d' = d^T \Lambda^T A \Lambda d$$

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:

$$v'_\mu = \Lambda_\mu^\alpha v_\alpha$$

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:

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

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:

$$v'_\mu = \Lambda_\mu^\alpha v_\alpha$$

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:

$$v'^\alpha = \Lambda^\alpha_\beta v^\beta$$

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:

$$m^2 = p^\alpha p^\beta \eta_{\alpha\beta} = p'^\alpha p'^\beta \eta_{\alpha\beta}$$

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:
$$\begin{array}{rrrr} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}$$

(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:

$$\begin{array}{rrrr} \rho & 0 & 0 & 0\\ 0 & -p & 0 & 0\\ 0 & 0 & -p & 0\\ 0 & 0 & 0 & -p \end{array}$$

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.

 

[JDoolin]

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?

Lessons learned from physics 101. If the data doesn't fit your equations, change the data.

Is that true? It seems to me that faint SN would imply a longer distance--and hence a smaller Hubble Constant, hence an earlier event.

 

[AWA]

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.

Uh, what? No, this observation certainly has not been made.

I meant that the observation has been claimed, not entering on whether the claim is right, I was using it just as an example since Chronos brought it up.

The core of my reflection is more general, and it's been generally responded by saying that noone's come up with anything better than what we have (the standard model), and that is actually true, and that everybody would be delighted and thrilled to find something that solved the standard model problems or misteries, even if it meant changing some apparently obvious assumptions ,and that is probably true too (but I'm less sure about this).

But IMO we must get rid of some circular reasonings that are often used here and that don't do any good to the standard model nor to science as a whole, like justifying expansion because there is spatial homogeneity, and spatial homogeneity because of expansion, this alone explains nothing, some other ingredient is needed, for instance redshift.

 

[Chalnoth]

But IMO we must get rid of some circular reasonings that are often used here and that don't do any good to the standard model nor to science as a whole, like justifying expansion because there is spatial homogeneity, and spatial homogeneity because of expansion, this alone explains nothing, some other ingredient is needed, for instance redshift.

Except that the reasoning isn't circular at all when you combine the results of multiple, independent observations that rely differently upon these assumptions. As I've shown previously, such detailed measurements can and do rule out inhomogeneous cosmologies:
http://arxiv.org/abs/1007.3725

 

[twofish-quant]

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?

Just off the top of my head...

1) If we find any star or galaxy with less that 20% of helium.
2) If we find evidence that the H, He, D, Li abundances change in any radical way by direction
3) If we find a highly evolved red dwarf or any black dwarf or anything else that is obviously more than 13 billion years old
4) If we find any evidence of heavy elements in the era of the CMB
5) If we find any reason to suspect that GR is wrong from any local experiment
6) Any new particles at CERN may cause reconsideration of LCDM. If we find another generation of quarks that would cause a rethink
7) If any of particle parameter goes out of certain bounds we'd have something to figure out. For example if it turns out that neutrinos are heavier than we think they are then this could cause a rethink
8) If we go for another decade and we can't pin down exactly what dark matter is, then we should probably rethink what is going on
9) Any sort of systematic asymmetry or anisotropy in the CMB or galaxy counts. For example, if someone points to a direction in space and finds five times as many galaxies in that direction, then we got some explaining

Also those are the things we could find now. There are about another dozen things that we could have found that would have killed LCDM, but we didn't find them.

 

[twofish-quant]

But IMO we must get rid of some circular reasonings that are often used here and that don't do any good to the standard model nor to science as a whole, like justifying expansion because there is spatial homogeneity, and spatial homogeneity because of expansion, this alone explains nothing, some other ingredient is needed, for instance redshift.

Except that's not what is going on.

Spatial homogeneity of galaxies is an observation. There's nothing to justify. You point your telescope and that's what you see. If we find any sort of direction in space in which there are more galaxies than in other directions, then the universe is not homogenous.