Redshift as evidence of expansion

[budrap]

You seem to want to have your cake and eat it too. Isaac Newton supposed the universe was unbounded, but static because gravitational forces canceled out, (so Gauss's law didn't apply). You seem to want to take this universe, but then not have gravitation cancelling out when it affects light.

Hmm, I may have misunderstood something you said. Can you elaborate on the following, specifically with regard to how Gauss's law means that you can't have a static universe?

If Gauss's law applies then it's nonsense to think that you can explain the redshift by gravity in a static universe, since Gauss's law means that you can't have a static universe.

 

[budrap]

How so?

I'm treating the spherical wavefront (at the time of observation) as a Gaussian surface and therefore the matter within the sphere is of consequence and that without is not.

 

[chronon]

Hmm, I may have misunderstood something you said. Can you elaborate on the following, specifically with regard to how Gauss's law means that you can't have a static universe?

If you take a sphere of matter in a homogeneous universe then Gauss's law says that a body on the edge will experience a force towards the centre. Hence all matter will tend to collapse together.

 

[budrap]

If you take a sphere of matter in a homogeneous universe then Gauss's law says that a body on the edge will experience a force towards the centre. Hence all matter will tend to collapse together.

OK, that's kind of what I thought you meant. My response was poorly worded. Instead of:

I don't think that Gauss's law would apply, however, in a universe that is static yet unbounded.

It should have read:

I don't think that Gauss's law would apply, however, TO a universe that is static yet unbounded.

Gauss's law would apply within such a universe but it could not be applied to the entire universe because the extent of an unbounded universe is inherently indeterminate and therefore there is no defined edge or center and so no preferred direction around which a collapse could organize itself.

Another way to think about it would be to consider the wavefront analysis again and extend its radius to the Schwarzschild radius implied by the average mass density. At that point the wavefront would cease its outward expansion. In such a scenario (with every galaxy at the bottom of its own gravitational well) the extent of the universe is inherently indeterminate and the concept of a universal center or edge would have no meaning and therefore a collapse would have no place to start from or go to.

 

[Chalnoth]

I don't think that Gauss's law would apply, however, TO a universe that is static yet unbounded.

Gauss's law would apply within such a universe but it could not be applied to the entire universe because the extent of an unbounded universe is inherently indeterminate and therefore there is no defined edge or center and so no preferred direction around which a collapse could organize itself.

Huh? No, this is false. In such a universe, any point works as a focal point of the collapse. After rethinking my post very early in this thread, let me put it this way:

When using Gauss's Law, you can take any point you like as your reference. So, pick a spot in a homogeneous, unbounded universe. Any point will do, they're all the same. Then, draw a spherical surface around that point and use Gauss's law to see the force that the particles on that surface feel. With this picture, everything wants to collapse inward, with further objects wanting to collapse inward more quickly.

If you moved the reference point to somewhere else in the universe, by the way, you'd find that it would calculate all of the exact same motions for all of the particles, once you correct for the change in reference frame.

 

[salvestrom]

Huh? No, this is false. In such a universe, any point works as a focal point of the collapse. After rethinking my post very early in this thread, let me put it this way:

When using Gauss's Law, you can take any point you like as your reference. So, pick a spot in a homogeneous, unbounded universe. Any point will do, they're all the same. Then, draw a spherical surface around that point and use Gauss's law to see the force that the particles on that surface feel. With this picture, everything wants to collapse inward, with further objects wanting to collapse inward more quickly.

If you moved the reference point to somewhere else in the universe, by the way, you'd find that it would calculate all of the exact same motions for all of the particles, once you correct for the change in reference frame.

Here goes again:

The sphereical wavefront is being equated to the surface of a Gaussian sphere, causing gravitational redshift.

In a homogenous universe any single photon can be seen as being on the surface of an infinite number of surrounding Gaussian spheres, all of equal size and mass, such that neither direction has any overwhelming affect on the photon, and so redshift cannot occur.

Gravitational collapse occurs because the universe is not locally homogenous.

If the universe is not locally homogenous then the photon is acted upon in some direction, causing gravitational redshift.

I tried to keep the sentences short and to the point so we can determine at which point the logic goes out the window.

Is it me, or in an finite, unbound universe any particle is actually on the opposite sides of the largest possible Gaussian sphere at the same time?

 

[Drakkith]

I'm treating the spherical wavefront (at the time of observation) as a Gaussian surface and therefore the matter within the sphere is of consequence and that without is not.

Why would you ignore the material outside the sphere?
Or rather, what is the purpose of treating light this way?

 

[Chalnoth]

Is it me, or in an finite, unbound universe any particle is actually on the opposite sides of the largest possible Gaussian sphere at the same time?

The problem there is that you're assuming that the space-time is rigid. This isn't the case in General Relativity.

 

[budrap]

Why would you ignore the material outside the sphere?
Or rather, what is the purpose of treating light this way?

1) Because the effect of the matter outside the sphere sums to zero.
2) To demonstrate that it is possible to generate a redshift-distance relationship without invoking universal expansion.

 

[Drakkith]

1) Because the effect of the matter outside the sphere sums to zero.
2) To demonstrate that it is possible to generate a redshift-distance relationship without invoking universal expansion.

Perhaps I don't understand Gauss' law or whatever, but I don't see how this could be correct.

 

[Chalnoth]

Perhaps I don't understand Gauss' law or whatever, but I don't see how this could be correct.

Which part?

 

[Drakkith]

Which part?

How the matter inside and outside doesn't sum to zero, only the outside. Can someone elaborate a bit more perhaps?

 

[budrap]

In a homogenous universe any single photon can be seen as being on the surface of an infinite number of surrounding Gaussian spheres, all of equal size and mass, such that neither direction has any overwhelming affect on the photon, and so redshift cannot occur.

Well this is where the logic breaks down. I am talking about a clearly defined physical entity, an expanding spherical wavefront. All the photons on that sphere are interrelated because they have a common origin and share a common reference frame. All of your other theoretical Gaussian spheres have no such physical significance and are therefore irrelevant to the physics of the situation. Don't confuse the maths with the physics.

 

[salvestrom]

How the matter inside and outside doesn't sum to zero, only the outside. Can someone elaborate a bit more perhaps?

I can't seem to find the nice picture of a sphere and a little spaceman that I was looking at. The Gauss Law of gravity states that the gravitational strength of a region of r radius is determined only by the mass contained within the described circle, and that everything beyond the circle cancels out. The description was done to explain how the gravitational pull on an object varies beneath the surface of a planet. That's all I know about it. The wikipedia page has the usual inaccessible maths plastered all over. No idea where I read about it, then.

While searching the web I did find a forum post here that actually stated that Gauss's Law, while it works always, is hard to directly apply if there is no symmetry.

 

[salvestrom]

The problem there is that you're assuming that the space-time is rigid. This isn't the case in General Relativity.

A traveller through the universe would reach that point where the distance to their point of origin is equally distant ahead and behind (and up and down and so on). I then take that as the diameter of the largest possible, non-overlapping sphere.

I'm not sure what you mean by rigid, other than to say I'm treating space as if it isn't moving.

 

[Chalnoth]

How the matter inside and outside doesn't sum to zero, only the outside. Can someone elaborate a bit more perhaps?

Well, first of all, this is only the case if you have a symmetric system. Gauss's law states that if you take the integral over a surface of the gravitational field pointing out of that surface, that integral is equal to the mass enclosed within (with a constant adjustment factor to correct the units). If you don't have a highly symmetric system, this can be difficult to calculate, as the field across the surface could be pointing in all sorts of different directions, with different strengths and whatnot.

But if the system is a symmetric one, and you pick a shape of the surface that conforms to that symmetry, then your job is easy: if the surface was picked well, then the gravitational field across that surface is the same everywhere, so you can simply multiply the surface area times the gravitational field.

So, if we have a spherically-symmetric system, then if we draw a sphere, the gravitational field will have to be pointing inward at every point on the surface of the sphere with the exact same magnitude. So, using Gauss's law we can write:

$$A*\Phi = 4\pi G m$$

Here A is the surface area of the sphere, $\Phi$ is the gravitational field, such that the gravitational force $F_g = \Phi m$ for a mass $m$ in a gravitational field $\Phi$.

Since our area is a sphere, $A = 4\pi r^2$. So we can write:

$$\Phi = {G m \over r^2}$$

...exactly as expected. Whatever is going on outside the sphere is completely inconsequential, as long as it is spherically-symmetric.

 

[Drakkith]

Thanks Chalnoth and Salvestrom, I think I understand it a bit better now. However, I still have to wonder how you can use this for a large section of space. There is no "surface" like there is on a planet. You could simply use the surface of an imaginary sphere, but that doesn't seem reasonable to me. Am I incorrect?

 

[Chalnoth]

Thanks Chalnoth and Salvestrom, I think I understand it a bit better now. However, I still have to wonder how you can use this for a large section of space. There is no "surface" like there is on a planet. You could simply use the surface of an imaginary sphere, but that doesn't seem reasonable to me. Am I incorrect?

Yeah, no need for a real surface. You just need a symmetric system to make the calculations easy. If you have a symmetric system, you can draw any imaginary surface you like that obeys the symmetries of the system. Using these calculations, for example, it's relatively easy to compute the gravitational field of a long bar by using a cylindrical surface instead of a spherical one.

 

[Chalnoth]

I don't understand why you think there is no "surface" involved in the wavefront analysis. Do you think that the wavefront has no physical significance and therefore has no "real" surface, that it is some arbitrary mathematical construct? While the surface is not comprised of matter as in the case of a planet it is certainly comprised of energy. Why do you think we can't treat the two surfaces similarly?

What wavefront are you talking about?

 

[salvestrom]

What wavefront are you talking about?

He is referring to the wavefront of a photon, bringing us back where it started. Would gravity cause a redshift using Gauss's Law? More to the point, is this the cause of the cosmological redshift, rather than expansion?

Using the photon's point of emission as the center of the Gaussian sphere, we should be able to discard the gravity outside the sphere because of the law but also the gravity from beyond the observable universe (the one seen from the photon's frame of reference) because gravity won't have reached the photon from those locations.

The primary 'no' against the redshift seems to be that while the photon is redshifted leaving the area of a galactic cluster and entering deep space it will simply get blueshifted back again as it begins to enter the graitational field of another cluster.

 

[Chalnoth]

He is referring to the wavefront of a photon, bringing us back where it started. Would gravity cause a redshift using Gauss's Law? More to the point, is this the cause of the cosmological redshift, rather than expansion?

No. Gauss's Law doesn't apply so easily in General Relativity. This is fine for normal matter: you actually get the exact same equations of motion for a homogeneous, isotropic universe for normal matter in General Relativity or Newtonian gravity. But photons behave very differently between the two. So you can't just apply Gauss's Law. Instead, you have to use General Relativity, and in GR, well, there are a few ways to look at it, but the easiest is to just say that the expansion stretches the photons as well.

 

[salvestrom]

Instead, you have to use General Relativity, and in GR, well, there are a few ways to look at it, but the easiest is to just say that the expansion stretches the photons as well.

This was the original point, I believe. It is understood that expansion is part of GR and the redshift is attributed to this expansion. I believe the contention was that the redshift can be explained by another process. You both appear to be at loggerheads. I for one would be interested in your account of how Gauss's Law doesn't apply so well in GR, just out of curiosity.