5
$\begingroup$

I have read through this Wikipedia article on Dark Matter and it contains references to the methods used to demonstrate the existence of dark matter.

A galaxy rotation curve is a plot of the orbital velocities (i.e., the speeds) of visible stars or gas in that galaxy versus their radial distance from that galaxy's center. The rotational/orbital speed of galaxies/stars does not decline with distance, unlike other orbital systems such as stars/planets and planets/moons that also have most of their mass at the centre. In the latter cases, this reflects the mass distributions within those systems. The mass observations for galaxies based on the light that they emit are far too low to explain the velocity observations. The dark matter hypothesis accounts for the missing mass, explaining the anomaly.

My post is not to question anything that the article says, but rather to ask how can we judge the amount of gravitional force exerted by an arbitrary mass of dark matter, if we cannot use light or any other form of em radiation to estimate the volume or density involved?

If gravity is the only force that we can use to detect dark matter, how do we know it couples to ordinary matter with the same strength as ordinary matter does? I am not advancing any particular reason it should not, but would the density of the CDM support this assumption that dark matter attracts ordinary matter to the same degree as ordinary matter does?

Or does it depend on the nature of the particles involved with each particular model of what dark matter is composed of?

Improve this question
$\endgroup$
2
  • $\begingroup$ @RobJeffries has it right, we don't know. I think it is probably fair to say that the assumption that dark matter is similar to neutrinos, i.e. it's "ordinary matter that's just hard to detect" is probably the minimal hypothesis under Occam. If we won't be able to find one or several particles over the next decades, then it might be prudent to modify gravity to a larger extent than we have, so far. $\endgroup$
    – CuriousOne
    Jun 19 '16 at 20:49
  • $\begingroup$ A comparison to neutrinos must be qualified with the notion that Standard Model neutrinos themselves are far too light and as a result have far too fast average velocities. Any dark matter candidate must have a much slower average velociyt. $\endgroup$
    – ohwilleke
    Oct 14 '16 at 1:14
4
$\begingroup$

We assume that gravity couples to the stress-energy tensor i.e. the Einstein equation relates the curvature to the stress-energy tensor. In cosmology the only important terms in the stress-energy tensor are the diagonal terms $T_{00}$, $T_{11}$, $T_{22}$ and $T_{33}$. The $T_{00}$ term is the energy density i.e. how much stuff there is per unit volume - note that mass and energy are treated as related by Einstein's famous equation $E=mc^2$. The other three terms, $T_{11}$ to $T_{33}$, are the pressure.

Matter has a non-zero (and positive) density and a negligible pressure so it only contributes to $T_{00}$. Dark energy also has a non-zero (and positive) density but in addition it has a negative pressure. So dark energy contributes to all four diagonal terms.

The way stuff contributes to the stress-energy tensor dictates how we describe it. So for example cluster dynamics and galaxy rotation curves tell use there is something present that contributes only to $T_{00}$. Therefore we call that something matter. Dark matter is just matter defined in this way that we can't see. Measurements of the universe expansion rate tell us there is something present with a positive energy density contributing to $T_{00}$ and a negative pressure contributing to $T_{11}$ to $T_{33}$, and we call this dark energy.

So your question is the wrong way round. We define dark matter as matter because it contributes to the stress-energy tensor in the same way that the everyday matter around us does. So it couples to gravity in the same way ordinary matter does because that's the way it's defined.

If we had some alternative way to measure dark matter that wasn't based on measuring its gravity then it would be interesting to compare the two measurements. But right now gravity is the only way we have of measuring dark matter density.

Improve this answer
$\endgroup$
3
$\begingroup$

The answer is that we do not know. It is a working assumption that it does, and it is this assumption that leads to an estimate of the total mass and how that mass is distributed - essentially by the application of Poisson's equation for gravitation. $$ \nabla^2 \Phi(r) = 4 \pi G \rho(r) $$

If for some reason dark matter did not couple gravitationally in the same way as normal matter, then it could be distributed differently and it would be unclear what you actually meant by the mass of dark matter, since in this case we mean gravitational mass.

It certainly has been conjectured that dark matter and dark energy may couple non-gravitationally (e.g. Koyama et al. 2009, Murgia et al. 2016 and many others). This might make dark matter behave differently to baryonic matter - for instance the gravitational acceleration might not be independent of dark matter mass and it may subtly alter the relative clustering properties of dark and baryonic matter.

Improve this answer
$\endgroup$
0
$\begingroup$

The rotation curve exists because of he presence of dark matter and baryonic matter. And we can estimate it using the orbital velocity of the Sun and its distance from the Galactic Centre. The fact that we can do that demonstrates that we can estimate the interaction generated by Dark Matter. That is how its existence was initially suggested. We knew the interaction existed and couldn't find enough "ordinary" matter to explain the gravitational "force".

Improve this answer
$\endgroup$
0
$\begingroup$

My view on the constant speed of stars orbiting around the centre of the galaxy is this: the density of stars is such that a ball (same centre as the ghalaxy) with radius $r$ has to contain mass directly proportional to $r$. As long as the mass outside the ball is distributed evenly enough, that outside objects' mass's gravitational effects cancels themselves out. (Imagine being inside a hollow spheric object with uniform density, it's easy to see that the gravitational effect of that hollow object onto you will cancel itself out).

Such a distribution will cause the gravitational acceleration at a point within the galaxy with radial distance $r$ to be inversely proportional to $r$ because it's directly proportional to $m\over r^2$ with $m$ (mass inside the ball with radius $r$) being directly proportional to $r$. Centripetal acceleration inversely proportional to $r$, of course, means constant speed.

OK, the galaxy may end up being very bright if it is to contain enough stars, but that doesn't have to be the case. We just need some black holes scattered in there to provide enough mass, circling around the centre just like shining stars, their planets and whatever else, as long as the mass inside a ball is directly proportional to its radius.

So there is no need for dark energy to explain these speeds, just this simple mass distribution. However, regarding brightness, I don't have a clue what proportion of mass would have to be contained in black holes.

Improve this answer
$\endgroup$