# Potential energy of a particle within a dark matter halo

1. Jun 14, 2009

### florian101

Hay
I would like to calculate the potential energy of a particle which is within a big dark matter halo. The Newton shell theorem says that if I have a particle within a spherical mass distribution (let us say at radius r) the mass with R>r exert no net Force to the particle. But there is a constant potential...

phi(r) = -G*M/R

M is the mass outside r and R is the radius of the dark matter halo. My first problem is that the dark matter halo does not have a sharp edge so what is R? Furthermore the density is not constant -> rho(r) and therefore I assume that the formula above should be modified in my case to

phi(r) = -G *int^infinity_r M(r)/r dr

is this correct???

The potential within r should be

phi(r) = -G*M(r)/r

is this correct?
So I just have to add both components and this should give the net potential?
I am thankful for any advice or comment
best regards
florian

2. Jun 14, 2009

### Wallace

You can't define 'the potential' of a particle until you define where zero is on your potential energy scale, so it is not clear what you are asking? Note that you can't define it with reference to infinity, because a dark matter halo does not sit within an otherwise empty universe. The same goes for the mass of the halo, you can't integrate to an infinite radius. Indeed many of the formalism that people have developed to describe the radial density profile of halos integrate to infinite mass if you integrate to infinity (i.e. the integral does not converge).

Conventionally, the mass of a dark matter halo is taken to be the mass within the 'virial radius', which is defined in a number of different ways but typically (and sometimes exactly) it is the radius that encloses 200 times the mean density of the Universe.

On the other hand, all of the above assumes that the Halo is spherical, which is usually a poor assumption, given the results of N-body simulations that show most halo's have significant 'tri-axiality' and sub-structure.

If you give some more details of exactly what you are trying to find out that would make it easier to give you a better answer.

3. Jun 14, 2009

### florian101

In principle I would like to see a lindblad plot of all dark matter particles in my dark matter halo. This means a plot with total energy vs. angular momentum. To calculate the total energy I need the potential to calculate the potential energy.
My question is therefore whether

E = 1/2*m*v^2 - grav_const*m*M(r)/r

is the correct formula for the total energy. M(r) is the mass within r (radius of the DM particle) and m is the mass of the particle and v its velocity.
Of course a spherical symmetric DM halo is not 100% realistic but this is just a model (first I have to understand the simple case).

I attached a plot which I get with the formula explained above... the Y axis is total energy/mass of the DM particle and the x-axis is total angular momentum/mass of the DM particle
Actually I expected to have negative energies for the particles???? But for almost all particles the kinetic energy is much larger than the potential energy what means in my understanding that this particles will leave the halo??? What can't be because the halo is stable...
thank you very much
best regards
florian

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4. Jun 15, 2009

### Wallace

Not quite, the mass of the Halo appearing in your formulas should not have a variable r. That is to say, even if *at this instant* the shell theorem tells you the gravitational attraction of the particle to the halo depends on it's radial position, if you are thinking about whether the particle is bound you need to consider that it is moving, hence it's r co-ordinate is changing.

The mass of the halo needs to be fixed, and there is no one right way to do this. Usually, as I say, the mass is taken to be that within the virial radius, but there is no one right way to define this. It looks like you are working with an N-body simulation snapshot? There are a lot of tools out there for analysing halo's from N-body shots, so you could look to see if there is something already available to do what you want to do?

5. Jun 15, 2009

### florian101

I thought because the force within r (position of the particle) is not zero and outside r it is zero I have two different contributions to the potential. Outside it should be constant because the gradient of the potential (=force) has to be zero and within r it should depend on r...
If I use the mass of the whole halo instead of M(r) (I used your definition of the viral radius) then I get negative total energies, but because of my argumentations above I am not sure whether this is correct... (remember I want to calculate the energy of one DM particle not of the whole halo)

E = 1/2*m*v^2 - grav_const*m*M(r_viral)/r

plot is attached
thanks for help
florian

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6. Jun 15, 2009

### Wallace

At any instant yes, but you are not interested in the instantaneous force on a particle, you want to know whether the particle is bound, so you need to consider the total mass of the halo. Think about it in terms of escape velocity. By your thinking, the escape velocity of a particle at r=0 is zero, which is clearly not the case. In fact the escape velocity does not depend on the instantaneous position of any given particle, it is a property of the whole object (assuming spherical symmetry).

Yep, negative energy means that a particle is bound. I'm not necessarily saying your plot is 100% correct (I'm not saying it's wrong either) but it is true that bound particles should have negative energy, and therefore you plot implies the particles are all bound. The large gap in energy between your highest energy particles and zero is somewhat odd. The problem is that you haven't given us much information. Where have you got these numbers from? Has someone given you the details of a Halo, derived from a simulation? If so then the method that they used to define what is and isn't part of the Halo should be considered in your analysis. You need to use the same definition of the virial radius as they have, otherwise your results will not mean very much.

7. Jun 15, 2009

### florian101

I did the simulation by myself. The halo mass within the viral radius (r_viral = 32.4kpc) is 2e09 solar masses. The scale radius is 0.81kpc and the concentration parameter is 40.
Just to be sure about the formula

E = 1/2*m*v^2 - grav_const*m*M(r_viral)/r

r is still the position of the particle, so the potential is depending on the particle position (not like the escape velocity).
regards
florian

8. Jun 15, 2009

### Wallace

Right, the potential of the particle depends on the position, but the mass of the DM halo doesn't depend on the particles position (hence the use of M(r_virial) ). The formula above is exactly as you would use to work out the escape velocity.

If you did you simulation, you must have already decided which particles are part of the halo and which are not by some criteria, so you should be using that in your analysis. By the way, from memory your concentration parameter is very very high, and the halo is quite a low mass, not normally one that would be well characterised by a DM simulation? Normally you would look at halos much below 10e10-10e11 or so, unless you are including gas physics.

9. Jun 15, 2009

### florian101

ok I had a chat to my prof and he told me that the potential should go with ln(r) so the equation we are using is wrong...:-(
florian

10. Jun 20, 2009

### florian101

hay
ok if anybody is interested I think I found the solution...

d phi = - GdM/r

Therefore

r<r': phi_1 = -GM(r')/r'
r>r': phi_2 = int^infinity_r' rho(R)*R dR

phi = phi_1 + phi_2

The result shows a small discrepancy which I do not understand so far but I think this is a numerical problem
thanks for any attempts to help
best regards
florian