Dark matter energy loss mechanisms

[Xomena]

Hi, I'm new, so this may be on the wrong board, but I thought it was more particle-energy-stuff than astro.

So to the point - dark matter is non interacting with EM fields, and can't emit photons. Yes?

OK, given this, how could dark matter lose energy. For example, when normal matter is accelerated in a circle we can get synchrotron radiation. If you could set up something similar (but non-em induced) then dark matter can't lose energy by photon emission. Maybe it would be some kind of thermal emission.. but does that make any sense either..?

I am unaware of any kind of dark matter energy loss mechanisms having been discussed, and I was just wondering if anyone could answer this or point me to a paper where this kind of idea is investigated.

 

[jambaugh]

"thermal emission" is typically electromagnetic but otherwise not specific. The "thermal" qualifier just means randomized e.g. with maximized entropy.

Now as far as dark matter goes, there are two context of this phrase. Originally "dark matter" simply referred to conventional matter which was not in the form of stellar objects and so invisible to telescopes. Calculations of galactic masses gave excesses beyond the mass of visible stars within them.

More recently there are speculations that some of this dark matter is in the form of "exotic dark matter" i.e. weakly interacting massive particles (WIMPs). But until and unless a specific theoretical type is suggested I don't know how to answer your question. Look for example at known particles which do not interact electromagnetically, e.g. neutrinos and neutrons. They interact through the weak force (and neutrons also through the strong force) and so can loose energy through scattering interactions which exchange weak bosons (and or gluons).

Ultimately however all particles interact via gravity so they can loose energy that way.

 

[Vanadium 50]

jambaugh has it exactly right - they lose energy through gravity. I wish I had a more straightforward argument, but this is the best I can do:

1) Gravitationally bound objects have negative specific heats. That is, as they lose energy, their temperature goes up. You can prove this, but maybe an example is better - as interstellar gas clouds collapse, they heat up.

2) Dark matter particles in orbit around the center of the galaxy have velocities of about 1/1400th of the speed of light - the same as everything else in orbit. For a 100 GeV particle, that means a kinetic energy of around 25 keV, which corresponds to a temperature of about 300 million K. So galactic dark matter is hot.

3) Primordial dark matter particles would be expected to have a temperature comparable to the CMB, a few kelvins.

Conclusion - gravity has caused dark matter to warm up, which means it has lost energy.

 

[kaksmet]

1) Gravitationally bound objects have negative specific heats. That is, as they lose energy, their temperature goes up. You can prove this, but maybe an example is better - as interstellar gas clouds collapse, they heat up.

How is collapsing equivalent to loosing energy?

2) Dark matter particles in orbit around the center of the galaxy have velocities of about 1/1400th of the speed of light - the same as everything else in orbit. For a 100 GeV particle, that means a kinetic energy of around 25 keV, which corresponds to a temperature of about 300 million K. So galactic dark matter is hot.

3) Primordial dark matter particles would be expected to have a temperature comparable to the CMB, a few kelvins.

Conclusion - gravity has caused dark matter to warm up, which means it has lost energy.

So, you are saying that a particle with less energy would be warmer? Let the 100 GeV particle loose its kinetic energy so that it decreases to 2.5 keV, should then have a higher temperature?

 

[jambaugh]

How is collapsing equivalent to loosing energy?

The system is losing potential energy. Some of it goes into heating and indeed most of it is lost through radiation of that heat.

Let's start simple. Consider a small body on a highly eccentric orbit about a larger mass. As it comes closer it looses potential energy and gains kinetic energy... zero net change in energy... except as the body is not a perfectly rigid mass it will deform under tidal stresses, warm up due to friction of that deformation, and its orbit will decay slightly.

Assume the object has zero emissivity and the total energy is still unchanged but some of the kinetic plus potential energy has been converted to heat. Now allow the body to radiate and that energy is lost to the environment.

 

[jal]

I am unaware of any kind of dark matter energy loss mechanisms having been discussed, and I was just wondering if anyone could answer this or point me to a paper where this kind of idea is investigated.

I could not find anything. So instead I tried to find out “How is light made?”.

See my blog.
https://www.physicsforums.com/blog.php?b=3039
“How is light made?”

jal

 

[Vanadium 50]

An object in a gravitational potential has kinetic energy equal to half the potential energy, or total energy equal to minus the kinetic energy. Temperature is proportional to kinetic energy, so E ~ -T: i.e. the heat capacity is negative.

This is true for each particle in an ensemble, so it's true for the entire ensemble.

 

[kaksmet]

An object in a gravitational potential has kinetic energy equal to half the potential energy, or total energy equal to minus the kinetic energy. Temperature is proportional to kinetic energy, so E ~ -T: i.e. the heat capacity is negative.

And since you can choose the zero of your potential energy to be at any point the particle can have any temperature?

 

[Vanadium 50]

No, this is true with the conventional definition of gravitational potential energy, i.e. zero at infinity.

 

[smith345]

An important property of all dark matter is that it behaves like and is modeled like a perfect fluid, meaning that it does not have any internal resistance or viscosity.

 

[jambaugh]

An important property of all dark matter is that it behaves like and is modeled like a perfect fluid, meaning that it does not have any internal resistance or viscosity.

How do you know?