Redshift as evidence of expansion

[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.