Summary:
Is the image of the background radiation fixed through out most of the universe life? Can we use it as the ultimate compass?
The image of the background radiation unlike galaxies and super galactic clusters are fixed through out time yes? Is this true only to a certain extent?

Also can the radiation it self be used as a coordinate system since it's unchanging in our observable universe and presumingly fixed to the super galactic coordinate system and compass?

In fact I would think that the supergalactic clusters would move behind the background of the radiation which are more distant and fixed compared to the super clusters yes?

Why don't we have a new coordinate system based on the radiation image?

And why don't we have any images of the background radiation with celestial coordinates overlayed?

Lastly why is it that the image is in oval shaped format instead of a rectangular shape standard for wraping around a sphere? I am a software engineer and am wrting astronomy software. I want to have the background radiation image wraped around a sphere surounding the camera of the software so the user is surounded by the image. But I need a rectangular image.

David

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Staff Emeritus

is better than this?

bhobba, Motore and PeroK
Bandersnatch
The image of the background radiation unlike galaxies and super galactic clusters are fixed through out time yes? Is this true only to a certain extent?
It's not fixed.
The the temperature fluctuations that you can see on the pictures are the result of matter collapsing on itself and bouncing back at the time and place of emission. It's as if you had a pot of boiling water, and you took a snapshot of bubble density along a single plane going through the volume of water.
With passing time, the fluctuations evolve, as each consecutive snapshot is taken from a slice a bit further away than the last. So the fluctuations grow and shrink over time (this takes too long to be observable with currently existing instruments).
Should you allow for intergalactic travel, observer located at significantly distant points would also see different fluctuations, as - again - the slice would go through different bubbles.
For this reason it's not a good basis for a universal map.

One can use it as a sort of universal standard of rest, as is regularly done in cosmology, since all you need for this purpose is knowing your state of motion w/r to the CMBR.

The oval, as cheekily alluded to above, is just one possible projection of a sphere onto a 2D surface (here: Mollweide).

bhobba and mfb
View attachment 290782
is better than this?
View attachment 290783
I can use the second one in my code by wrapping it as a texture around a sphere, I cannot do so with the first.

Check above youtube link. Tell me why we can't use this as a fixed reference frame superior to the supergalactic plane. In our life time galactic clusters might move a tiny bit, but less so for the background radiation yes?

David

It's not fixed.
The the temperature fluctuations that you can see on the pictures are the result of matter collapsing on itself and bouncing back at the time and place of emission. It's as if you had a pot of boiling water, and you took a snapshot of bubble density along a single plane going through the volume of water.
With passing time, the fluctuations evolve, as each consecutive snapshot is taken from a slice a bit further away than the last. So the fluctuations grow and shrink over time (this takes too long to be observable with currently existing instruments).
Should you allow for intergalactic travel, observer located at significantly distant points would also see different fluctuations, as - again - the slice would go through different bubbles.
For this reason it's not a good basis for a universal map.

One can use it as a sort of universal standard of rest, as is regularly done in cosmology, since all you need for this purpose is knowing your state of motion w/r to the CMBR.

The oval, as cheekily alluded to above, is just one possible projection of a sphere onto a 2D surface (here: Mollweide).
Ok so nothing is fixed except maybe the accelerating expansion of space it self to calculate a fixed reference frame. But comparing the supergalactic clusters to background radiation, the background radiation is in order of magnitudes more "still" than the galaxy clusters yes? Why havent we devealoped a coordinate system for it instead since it fluctuates so much less than the galaxies move?

It wont be useful for intergalactic travelers but it would be useful for Earth observers still!

David

Bandersnatch
In our life time galactic clusters might move a tiny bit, but less so for the background radiation yes?
But comparing the supergalactic clusters to background radiation, the background radiation is in order of magnitudes more "still" than the galaxy clusters yes?
In our lifetime, neither will move in any appreciable sense. On very large time-scales, you'd have more trouble with CMBR than clusters. I haven't put the number to work, but offhand I'd say you have the orders of magnitude the wrong way around.
With clusters, the confounding bit when building a map is the tangential (i.e. across the line of sight) motion, as the radial movement doesn't affect the relative position of reference points. This is both small by itself (much less than c) and hidden by distance (parallax).
With changes in CMBR you're not dealing as much with local motion of any material thing (as is the case with clusters), but with images of different things from further away. The limiting factor on the rate of change of the image is not local motion, but the speed of light; and there's no parallax along the radial direction.
Over the next few dozens of billions of years, as it gets more redshifted and dimmer, the CMBR will indeed appear to gradually 'freeze'. But then you'll have that dimming issue to confound your map-making efforts. To an extent all of this is true of clusters as well.

In either case, changes are negligible for all practical purposes. You can treat both clusters and CMBR as static. So you're free to use either as a basis of an impromptu coordinate system. Same is true for coordinate systems based on the galactic structure, or even relative positions of local stars. You'll find various such systems used across various subsets of astronomy, based pretty much on what is most convenient given the purpose at hand. I don't see why a CMBR-based system should be universally preferred.

darkdave3000 and PeroK
In our lifetime, neither will move in any appreciable sense. On very large time-scales, you'd have more trouble with CMBR than clusters. I haven't put the number to work, but offhand I'd say you have the orders of magnitude the wrong way around.
With clusters, the confounding bit when building a map is the tangential (i.e. across the line of sight) motion, as the radial movement doesn't affect the relative position of reference points. This is both small by itself (much less than c) and hidden by distance (parallax).
With changes in CMBR you're not dealing as much with local motion of any material thing (as is the case with clusters), but with images of different things from further away. The limiting factor on the rate of change of the image is not local motion, but the speed of light; and there's no parallax along the radial direction.
Over the next few dozens of billions of years, as it gets more redshifted and dimmer, the CMBR will indeed appear to gradually 'freeze'. But then you'll have that dimming issue to confound your map-making efforts. To an extent all of this is true of clusters as well.

In either case, changes are negligible for all practical purposes. You can treat both clusters and CMBR as static. So you're free to use either as a basis of an impromptu coordinate system. Same is true for coordinate systems based on the galactic structure, or even relative positions of local stars. You'll find various such systems used across various subsets of astronomy, based pretty much on what is most convenient given the purpose at hand. I don't see why a CMBR-based system should be universally preferred.
So in order of frozen-ness this is the correct order?:
0. Expansion of the universe(Since we know the acceleration rate of the universe expanding we can work out a stationary coordinate system in theory).
1. Supergalactic coordinate system(Those galaxies drift across our sky real slow , nothing is slower than this).
2. Cooling CMB(That background image is going to change color maybe faster than those galaxies move).
3. Galactic Coordinate System(Earth moves faster around the galaxy compared to changes in above 2)
4. Ecliptic (precession of the ap and per will throw this system off over time)
5. Celestial Coordinates (Earth's 24 hour spin is not exact and is slowing down faster than above)

Have I got the right picture in my head now?

Regards, David

Can anyone help me find CMB mapping image with any kind of astronomical coordinate metadata or overlay so I can orientate it correctly relative to the stars and other celestial coordinate systems?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I can use the second one in my code by wrapping it as a texture around a sphere, I cannot do so with the first.
Of course you can. You’re just using the wrong wrapping function.

Bandersnatch
Can anyone help me find CMB mapping image with any kind of astronomical coordinate metadata or overlay so I can orientate it correctly relative to the stars and other celestial coordinate systems?
Have you seen the image here?
https://planck.ipac.caltech.edu/image/planck13-002a
It's centred around galactic coordinates (mid point is 0,0; plane horizontal). It's even in the projection you wanted. Take note of the bit in the description about mapping it onto the inside of a sphere.

So in order of frozen-ness this is the correct order?:
I wouldn't entirely sign off on that. E.g. a full precession of Earth's axis is far faster than any appreciable slowing of rotation, and I think the time scale of 3 and 2 is comparable enough to require checking first. But then again what is exactly meant by frozenness would need clarifying to compare these meaningfully.
Still, as a rough picture it looks ok to me.

darkdave3000
Of course you can. You’re just using the wrong wrapping function.
Um... how? Are you an OpenGL programmer?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
It is not a question of using OpenGL or not. It is a simple mathematical fact.

It is not a question of using OpenGL or not. It is a simple mathematical fact.
Is there a mapping conversion I can use for the oval shaped image that I can use for the inside of a sphere?

https://planck.ipac.caltech.edu/image/planck13-002a

David

mfb
Mentor
The expansion of space is uniform and isotropic on large scales, it doesn't lead to any sort of universal coordinate system.

The largest fluctuations of the CMB are somewhat stable over 100 million years. Using many galaxies far away is probably more reliable. The motion will average out.

bhobba
The expansion of space is uniform and isotropic on large scales, it doesn't lead to any sort of universal coordinate system.

The largest fluctuations of the CMB are somewhat stable over 100 million years. Using many galaxies far away is probably more reliable. The motion will average out.
I disagree about the expansion not leading to a universal coordinate system, the reason is that it indirectly leads to what we all can agree to be a stationary frame of reference if we let computers do the work for us by oberving the motions of stars and factorinig in the expasion of space for their motion.

Because of the expansion of space-time all objects are having a tendancy to slow down to a universal stationary frame of reference. Imagine putting dots all over an inflating balloon. Each of those dots are stationary to that balloon(universe). As the dots become farther apart because space is added between them objects between them "slow down". We don't have such dots in our universe but we can calculate their ideal redshift based on their position since we know how fast they should move if they are stationary. This is something I want to simulate in my software suite eventually. Virtual stationary reference nodes for our universe. We can then have absolute speed.

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Staff Emeritus
Because the oval distorts less. Look at the maps. Is Greenland really bigger than Africa? Is Antarctica bigger than all the other continents put together?

mfb
Mentor
@darkdave3000: You can go to the reference frame where the dipole moment of the CMB vanishes, but that doesn't give you any direction reference.

collinsmark
Homework Helper
Gold Member
Is there a mapping conversion I can use for the oval shaped image that I can use for the inside of a sphere?
I would like to know why oval projection is superior yes, I dislike the inconsistency horizontally.

All map projections have distortions. Choosing one map projection over another is simply a matter of which distortions you're willing to accept, and which distortions you are not.

Most cylindrical projections (which end up looking like squares or rectangles), including Mercator, Web Mercator, Equirectangular, Miller, Behrmann, Gall stereographic, and others, suffer from the distortion that if you slide around a small, fixed area on the map/projection, the area on the sphere to which it corresponds is not constant. So if you're trying to compare the sizes of features on the map, these projections are not good. On the other hand, some of them, such as the Mercator maintain consistency of direction (i.e., what Northwest is), so if you're a sailor, the Mercator projection is good for choosing your heading.

The Mollweide (ellipse) projection is good because it maintains equal area. The downside is that it does not maintain consistent direction (i.e., what Northwest is).

If you're starting with a Mollweide (elliptical) projection, and you want to map that to a rectangular projection that you can use in a UV map that is used in game engines and whatnot, use the inverse transformations shown from this link on the Mollweide projection wiki: https://en.wikipedia.org/wiki/Mollweide_projection.

$$\varphi = \arcsin \frac{2 \theta + \sin 2 \theta}{\pi}$$
$$\lambda = \lambda_0 + \frac{\pi x}{2 R \sqrt{2} \cos \theta}$$
where $\theta$ can be found by the relation
$$\theta = \arcsin \frac{y}{R \sqrt{2}}$$
and where $x$ and $y$ correspond to the Mollweide map coordinates, where $x = y = 0$ is the center of the map; $x$ varies from [$-2R \sqrt{2}, 2R \sqrt{2}$], and $y$ ranges from [$-R\sqrt{2}, R\sqrt{2}$]; $\varphi$ is the latitude (ranging from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$), $\lambda$ is the longitude (ranging from $-\pi$ to $\pi$); $\lambda_0$ is the central meridian; and $R$ is the radius of the globe to be projected.

Bandersnatch