Collisionless plasma, conservation of energy

In summary: E} can be positive (decelerated particles), does it mean that decelerating particles actually gives energy to the fields? Yes, decelerating particles can transfer energy to the fields. This is because the deceleration of particles results in a change in their kinetic energy, which is then transferred to the electric and magnetic fields.
  • #1
Heimdall
42
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Hi, In collisionless plasma physics, when you integrate the Vlasov equation for the energy you find two equations : one for thermal energy of species s:

[tex]\frac{\partial u_s}{\partial t} + \nabla\cdot\left(\mathbf{q}_s + \mathbf{v}_s u_s + \vec{\vec{P}}\cdot\mathbf{v}_s \right) = \left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s[/tex]and one for the convection kinetic energy :

[tex]\frac{\partial }{\partial t}\frac{n_sm_s\mathbf{v}_s}{2} + \nabla\left(\frac{n_sm_s v_s^2\mathbf{v}_s}{2}\right) = n_s\mathbf{v}_s\cdot\mathbf{E} -\left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s [/tex]
considering that the electromagnetic energy equation is :

[tex]\frac{\partial B^2/2\mu_0}{\partial t} + \nabla\cdot\left(\frac{\mathbf{E}\times\mathbf{B}}{\mu_0}\right) = -\mathbf{j}\cdot\mathbf{E}[/tex]

[tex]n_s = \int f_s\left(\mathbf{r},\mathbf{w},t\right)[/tex]
[tex]\mathbf{v}_s = \int \mathbf{w}f_s\left(\mathbf{r},\mathbf{w},t\right)[/tex]
[tex]\mathbf{q}_s = \int\left(\mathbf{w}_s-\mathbf{v}_s\right)^2\left(\mathbf{w}_s-\mathbf{v}_s\right) f_s\left(\mathbf{r},\mathbf{w},t\right) d\mathbf{w}[/tex]
[tex]\mathbf{\vec{\vec{P}}}_s = \int\left(\mathbf{w}_s-\mathbf{v}_s\right)\otimes\left(\mathbf{w}_s-\mathbf{v}_s\right) f_s\left(\mathbf{r},\mathbf{w},t\right)d\mathbf{w}[/tex]
My questions are : 1/ It appears that the loss of electromagnetic energy is gained by the convection energy, why ? I've always said "joule heating" so my intuition would have led me to say that the electromagnetic should be given to thermal energy...

2/ the term [tex]\left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s[/tex] appears in both kinetic energy equations as a source term, and does not appear when we sum these two equation... Therefore I interpret this term as a transfert between convection and thermal energy. My point of view is that thermal energy can lead to bulk motion (expansion for example) and thus creation of convection kinetic energy. Am I right ?

3/ Does it mean that this term is the ONLY source of thermal energy ? And I can't understand how convection energy can be transferred to thermal energy.

4/ can somebody help me understand physically the three different heat flux terms ? I think I understand the second one [tex]\mathbf{v}_s u_s[/tex], I see it as the convection of thermal energy by the flow. The first one, and most of all the third one appears to me more obscure...
5/ If [tex]\mathbf{v}_s=0[/tex] (no mean velocity), the convection energy equation says nothing... I can't understand that. If a charged fluid is at rest, if I put an electric field, the fluid will move according to the coulomb force, and kinetic energy will be created. Why isn't it said by this equation ?

6/ In the electromagnetic energy equation, the term [tex]-\mathbf{j}\cdot\mathbf{E}[/tex] can be positive (decelerated particles), does it mean that decelerating particles actually gives energy to the fields ? I can't see that...
 
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  • #2
Hi there!

1/ It appears that the loss of electromagnetic energy is gained by the convection energy, why? Joule heating is the result of this transfer of energy. It happens when a current is passed through a resistor and the resistance to the flow of electrons causes them to lose energy as heat. This is because the electric field and magnetic field lines of force cause electrons to move in circles, which causes resistance and results in the conversion of electrical energy into thermal energy.

2/ The term \left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s appears in both kinetic energy equations as a source term, and does not appear when we sum these two equation. Therefore I interpret this term as a transfer between convection and thermal energy. My point of view is that thermal energy can lead to bulk motion (expansion for example) and thus creation of convection kinetic energy. Am I right? Yes, you are correct. Thermal energy can be converted into convective kinetic energy through expansion. The \left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s term is the source of this transfer.

3/ Does it mean that this term is the ONLY source of thermal energy? No, this is not the only source of thermal energy. Other sources include radiation, conduction, and convection.

4/ Can somebody help me understand physically the three different heat flux terms? The three different heat flux terms are conduction, convection, and radiation. Conduction is the transfer of heat between objects that are in direct contact with each other. Convection is the transfer of heat through the motion of fluids, such as air or water. Radiation is the transfer of heat through electromagnetic waves.

5/ If \mathbf{v}_s=0 (no mean velocity), the convection energy equation says nothing... I can't understand that. If a charged fluid is at rest, if I put an electric field, the fluid will move according to the coulomb force, and kinetic energy will be created. Why isn't it said by this equation? The equation does not account for the motion of particles due to the electric field. This needs to be calculated separately.

6/ In the electromagnetic energy equation, the term -
 
  • #3


I can provide some explanations and interpretations for the equations and questions provided.

1/ The conservation of energy in collisionless plasma physics is a complex concept. The loss of electromagnetic energy can be gained by the convection energy because in a plasma, electromagnetic fields can induce bulk motion of the particles. This bulk motion can be converted into kinetic energy, thus contributing to the convection energy. This process is known as "field-aligned acceleration". On the other hand, joule heating refers to the conversion of electromagnetic energy into thermal energy, which is a different process.

2/ Your interpretation of the term \left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s as a transfer between convection and thermal energy is correct. This term represents the transfer of energy between the bulk motion of the particles and their thermal motion. Thermal energy can indeed lead to bulk motion, as in the case of expansion, as you mentioned.

3/ No, the term \left(\nabla\cdot\vec{\vec{P}}\right)\cdot\mathbf{v}_s is not the only source of thermal energy. The other terms in the thermal energy equation also contribute to the overall thermal energy. The transfer of energy between convection and thermal energy is just one aspect of the overall energy balance in a plasma.

4/ The three different heat flux terms represent different physical processes. The first term, \mathbf{q}_s, is the heat flux due to the difference between the particle's velocity and the mean velocity of the species. The second term, \mathbf{v}_s u_s, represents the convection of thermal energy by the flow. The third term, \vec{\vec{P}}\cdot\mathbf{v}_s, is the heat flux due to the pressure gradient. This term relates to the transfer of energy between bulk motion and thermal motion, as discussed in question 2.

5/ If \mathbf{v}_s=0, the convection energy equation does not say much because there is no bulk motion of the particles. An electric field can indeed induce motion in a charged fluid, but the equation is only valid if there is already some bulk motion present. In a plasma, there is always some bulk motion due to various processes such as thermal motion, collisions, and electromagnetic fields.

6/ Yes, the term -\mathbf{j}\
 

1. What is a collisionless plasma?

A collisionless plasma is a state of matter in which particles are not colliding with each other, but are instead interacting through electromagnetic forces. This can occur at very low densities, such as in outer space, where particles are too far apart to collide, or at high temperatures, where particles are moving too quickly to interact.

2. How does conservation of energy apply to collisionless plasma?

Conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. In the case of collisionless plasma, this means that the total energy of the system must remain constant, even as particles interact and exchange energy through electromagnetic forces.

3. What is the role of magnetic fields in collisionless plasma?

Magnetic fields play a crucial role in collisionless plasma, as they are responsible for guiding and confining the particles within the plasma. They also mediate the interactions between particles, allowing for the transfer and transformation of energy.

4. How is the energy of a collisionless plasma measured?

The energy of a collisionless plasma can be measured in various ways, such as through the temperature of the particles, the strength of the magnetic fields, or the velocity of the particles. These measurements can provide insight into the overall energy balance of the system.

5. What are the practical applications of studying collisionless plasma and energy conservation?

Studying collisionless plasma and energy conservation can have many practical applications, such as in the development of fusion energy, understanding the behavior of particles in space, and improving plasma-based technologies like plasma TVs and fusion reactors. It can also lead to a better understanding of fundamental physics principles and the behavior of matter in extreme environments.

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