- #1

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I know that an external magnetic field can create an anysotropy in the degeneracy pressure, but i don't know if the gas itself shows magnetic properties after been affected by a magnetic field.

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- Thread starter ChinoSupay
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- #1

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I know that an external magnetic field can create an anysotropy in the degeneracy pressure, but i don't know if the gas itself shows magnetic properties after been affected by a magnetic field.

- #2

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I know that an external magnetic field can create an anysotropy in the degeneracy pressure, but i don't know if the gas itself shows magnetic properties after been affected by a magnetic field.

First of all, since we know that Zeeman effect is real, then it is no longer questionable that a Fermi gas do interact with an external field. In fact, this is why we have electron paramagnetic resonance phenomenon. So yes, just like a magnetic dipole in a magnetic field, it can influence an external field.

I don't quite understand what you mean by "...

Zz.

- #3

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First of all, since we know that Zeeman effect is real, then it is no longer questionable that a Fermi gas do interact with an external field. In fact, this is why we have electron paramagnetic resonance phenomenon. So yes, just like a magnetic dipole in a magnetic field, it can influence an external field.

I don't quite understand what you mean by "...shows magnetic properties...."? Do you mean after the external field has been switched off?after been affected by a magnetic field

Zz.

TY

Not really. I just want to know if the gas increase the total magnetic field because i am studying magnetic fields in White Dwarf (that is actually degenerated matter). The ions due to a process of movement can generate a magnetic field. But still we can't reach the real values that are being mesured.

So if it is possible, do you know a kind of paper or research that i could see ??

Thank you so much again

- #4

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TY

Not really. I just want to know if the gas increase the total magnetic field because i am studying magnetic fields in White Dwarf (that is actually degenerated matter). The ions due to a process of movement can generate a magnetic field. But still we can't reach the real values that are being mesured.

So if it is possible, do you know a kind of paper or research that i could see ??

Thank you so much again

If you have a magnet behind a piece of metal, do you think a lot of the magnet's magnetic field will pass through that metal? Or do you think a lot of it will be shielded by the metal?

Relevance to your question, you may ask? Metals have what can be approximated as having a Fermi gas, i.e. the conduction electrons.

Zz.

- #5

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If you have a magnet behind a piece of metal, do you think a lot of the magnet's magnetic field will pass through that metal? Or do you think a lot of it will be shielded by the metal?

Relevance to your question, you may ask? Metals have what can be approximated as having a Fermi gas, i.e. the conduction electrons.

Zz.

But doesn't it depends on the magnetic properties of the metal? It could be para, dia or ferro

- #6

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But doesn't it depends on the magnetic properties of the metal? It could be para, dia or ferro

You can pick aluminum, or even copper, which doesn't get magnetized.

Zz.

- #7

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But not gold or silverYou can pick aluminum, or even copper, which doesn't get magnetized.

Zz.

- #8

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But not gold or silver

Since you want to know what happens to the Fermi gas, then you don't want the rest of metal ions to play a role. Picking a material such as, say, iron, is not a good idea since a lot of other factors than just the presence of conduction electrons can affect the field.

BTW, how significant is the effect that you're looking at? One would think that the induced, time-varying factor will dwarf any effect from the presence of just the Fermi gas affecting the external field. A static magnetic field against copper doesn't do a lot of interesting thing, but time-varying magnetic field produces effects that are considerably more significant than any "shielding" of the field.

Zz.

- #9

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Maybe the response to that is in Pauli Paramagnetism? do you know something about that?You can pick aluminum, or even copper, which doesn't get magnetized.

Zz.

- #10

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Since you want to know what happens to the Fermi gas, then you don't want the rest of metal ions to play a role. Picking a material such as, say, iron, is not a good idea since a lot of other factors than just the presence of conduction electrons can affect the field.

BTW, how significant is the effect that you're looking at? One would think that the induced, time-varying factor will dwarf any effect from the presence of just the Fermi gas affecting the external field. A static magnetic field against copper doesn't do a lot of interesting thing, but time-varying magnetic field produces effects that are considerably more significant than any "shielding" of the field.

Zz.

Yes it had to be very significant. Also the magnetic field is not constant at all, because is generated via plasma convection and it hve many turbulence and other phenomena

- #11

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My question is, somebody knows if a fermi gas (degenerate electrons) can increase or decrease de total magnetic field due to an influence of an external magnetic field

To know the total magnetic field requires calculating the magnetic susceptibility of the material, in this case that 'material' is a Fermi gas. The magnetic response of a fermi gas (ie: no correlations between electrons) in the presence of a magnetic field is well known. It will consist of two contributions. The first is Pauli paramagnetism. This is the spin response of the electrons. In terms of energy, Pauli paramagnetism is due to the external field producing a finite energy difference between quantized spin up and spin down states of the electrons. The Pauli exclusion principle requires that electrons in the Fermi sea ignore this perturbation. However, near the Fermi surface the electrons spins can polarize in response to this field. In a constant field the Pauli contribution to the magnetic susceptibility is given by:

$$\chi_{P} = \frac{3n\mu_{0}\mu_{B}^{2}}{2E_{F}}$$

The second contribution is Landau diamagnetism. This is the orbital response of the electrons. The simplest picture is that the magnetic field drives the electrons into orbits around the field direction. This is very much akin to the production of a current in a wire loop upon application of a magnetic field, and just like in the wire loop, the induced orbits will try and cancel the field. This is why the orbital contribution is a diamagnetic response. In terms of energy, the B=0 electron states produce a spherical volume that is discreetized into ##k_{x}##, ##k_{y}##, and ##k_{z}## boxes (ie: just our good old fashioned free electron gas model). When a field is applied, say along the z direction, the ##k_{x}## and ##k_{y}## momenta condense into degenerate nested cylinders oriented along the z-direction, so called Landau levels. The production of these levels changes the total energy of the system. Namely, the ##k_{x}## and ##k_{y}## values near the Fermi surface must adjust their energy to the nearest Landau level upon application of a magnetic field. The Landau contribution to the magnetic susceptibility in a constant field can be derived by working through the energetics of this picture and is given by:

$$\chi_{L} = -\frac{1}{3}\chi_{P}$$

Again, the results are for a free electron gas, if correlations between electrons must be taken into account than this will change things, enough so that I don't think a PhysicsForums post will give much traction, it will take a lot of reading and tailored application with your specific problem in mind; you will need band structure and so, at the very least I expect, you will also need DFT. I will state the simplest modification though. Correlations will renormalize the electrons effective mass ##m^{*}## and the net result is that the Landau susceptibility will now scale as

$$\chi_{L} = -\frac{1}{3}\left(\frac{m_{e}}{m^{*}}\right)^{2}\chi_{P}$$

Thus, if the effective mass is small enough, then it can cause the diamagnetic contribution to be larger than the paramagnetic and your total field will reduce.

Also the magnetic field is not constant at all, because is generated via plasma convection and it hve many turbulence and other phenomena

This is going to make your life much harder, because they do have an effect. Note that I was very careful to state that the above results were for a constant magnetic field. In the presence of a wavevector-dependent ##\mathbf q## and frequency-dependent ##\omega## magnetic field then the magnetic susceptibility becomes a tensor of the form:

$$ \mathbf\chi\left(\mathbf q, \mathbf k; \Omega, \omega\right)$$

where ##\mathbf k## and ##\Omega## are the Fourier domain of the spatial and temporal change of the electron magnetization in response to the applied field. Things can become a bit tough. How far down the rabbit hole you have to go is dependent on the problem at hand. Good luck.

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- #12

- 12

- 6

To know the total magnetic field requires calculating the magnetic susceptibility of the material, in this case that 'material' is a Fermi gas. The magnetic response of a fermi gas (ie: no correlations between electrons) in the presence of a magnetic field is well known. It will consist of two contributions. The first is Pauli paramagnetism. This is the spin response of the electrons. In terms of energy, Pauli paramagnetism is due to the external field producing a finite energy difference between quantized spin up and spin down states of the electrons. The Pauli exclusion principle requires that electrons in the Fermi sea ignore this perturbation. However, near the Fermi surface the electrons spins can polarize in response to this field. In a constant field the Pauli contribution to the magnetic susceptibility is given by:

$$\chi_{P} = \frac{3n\mu_{0}\mu_{B}^{2}}{2E_{F}}$$

The second contribution is Landau diamagnetism. This is the orbital response of the electrons. The simplest picture is that the magnetic field drives the electrons into orbits around the field direction. This is very much akin to the production of a current in a wire loop upon application of a magnetic field, and just like in the wire loop, the induced orbits will try and cancel the field. This is why the orbital contribution is a diamagnetic response. In terms of energy, the B=0 electron states produce a spherical volume that is discreetized into ##k_{x}##, ##k_{y}##, and ##k_{z}## boxes (ie: just our good old fashioned free electron gas model). When a field is applied, say along the z direction, the ##k_{x}## and ##k_{y}## momenta condense into degenerate nested cylinders oriented along the z-direction, so called Landau levels. The production of these levels changes the total energy of the system. Namely, the ##k_{x}## and ##k_{y}## values near the Fermi surface must adjust their energy to the nearest Landau level upon application of a magnetic field. The Landau contribution to the magnetic susceptibility in a constant field can be derived by working through the energetics of this picture and is given by:

$$\chi_{L} = -\frac{1}{3}\chi_{P}$$

Again, the results are for a free electron gas, if correlations between electrons must be taken into account than this will change things, enough so that I don't think a PhysicsForums post will give much traction, it will take a lot of reading and tailored application with your specific problem in mind; you will need band structure and so, at the very least I expect, you will also need DFT. I will state the simplest modification though. Correlations will renormalize the electrons effective mass ##m^{*}## and the net result is that the Landau susceptibility will now scale as

$$\chi_{L} = -\frac{1}{3}\left(\frac{m_{e}}{m^{*}}\right)^{2}\chi_{P}$$

Thus, if the effective mass is small enough, then it can cause the diamagnetic contribution to be larger than the paramagnetic and your total field will reduce.

This is going to make your life much harder, because they do have an effect. Note that I was very careful to state that the above results were for a constant magnetic field. In the presence of a wavevector-dependent ##\mathbf q## and frequency-dependent ##\omega## magnetic field then the magnetic susceptibility becomes a tensor of the form:

$$ \mathbf\chi\left(\mathbf q, \mathbf k; \Omega, \omega\right)$$

where ##\mathbf k## and ##\Omega## are the Fourier domain of the spatial and temporal change of the electron magnetization in response to the applied field. Things can become a bit tough. How far down the rabbit hole you have to go is dependent on the problem at hand. Good luck.

Thank you so much friend, you have been very useful <3

- #13

DrDu

Science Advisor

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- #14

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To know the total magnetic field requires calculating the magnetic susceptibility of the material, in this case that 'material' is a Fermi gas. The magnetic response of a fermi gas (ie: no correlations between electrons) in the presence of a magnetic field is well known. It will consist of two contributions. The first is Pauli paramagnetism. This is the spin response of the electrons. In terms of energy, Pauli paramagnetism is due to the external field producing a finite energy difference between quantized spin up and spin down states of the electrons. The Pauli exclusion principle requires that electrons in the Fermi sea ignore this perturbation. However, near the Fermi surface the electrons spins can polarize in response to this field. In a constant field the Pauli contribution to the magnetic susceptibility is given by:

$$\chi_{P} = \frac{3n\mu_{0}\mu_{B}^{2}}{2E_{F}}$$

The second contribution is Landau diamagnetism. This is the orbital response of the electrons. The simplest picture is that the magnetic field drives the electrons into orbits around the field direction. This is very much akin to the production of a current in a wire loop upon application of a magnetic field, and just like in the wire loop, the induced orbits will try and cancel the field. This is why the orbital contribution is a diamagnetic response. In terms of energy, the B=0 electron states produce a spherical volume that is discreetized into ##k_{x}##, ##k_{y}##, and ##k_{z}## boxes (ie: just our good old fashioned free electron gas model). When a field is applied, say along the z direction, the ##k_{x}## and ##k_{y}## momenta condense into degenerate nested cylinders oriented along the z-direction, so called Landau levels. The production of these levels changes the total energy of the system. Namely, the ##k_{x}## and ##k_{y}## values near the Fermi surface must adjust their energy to the nearest Landau level upon application of a magnetic field. The Landau contribution to the magnetic susceptibility in a constant field can be derived by working through the energetics of this picture and is given by:

$$\chi_{L} = -\frac{1}{3}\chi_{P}$$

Again, the results are for a free electron gas, if correlations between electrons must be taken into account than this will change things, enough so that I don't think a PhysicsForums post will give much traction, it will take a lot of reading and tailored application with your specific problem in mind; you will need band structure and so, at the very least I expect, you will also need DFT. I will state the simplest modification though. Correlations will renormalize the electrons effective mass ##m^{*}## and the net result is that the Landau susceptibility will now scale as

$$\chi_{L} = -\frac{1}{3}\left(\frac{m_{e}}{m^{*}}\right)^{2}\chi_{P}$$

Thus, if the effective mass is small enough, then it can cause the diamagnetic contribution to be larger than the paramagnetic and your total field will reduce.

This is going to make your life much harder, because they do have an effect. Note that I was very careful to state that the above results were for a constant magnetic field. In the presence of a wavevector-dependent ##\mathbf q## and frequency-dependent ##\omega## magnetic field then the magnetic susceptibility becomes a tensor of the form:

$$ \mathbf\chi\left(\mathbf q, \mathbf k; \Omega, \omega\right)$$

where ##\mathbf k## and ##\Omega## are the Fourier domain of the spatial and temporal change of the electron magnetization in response to the applied field. Things can become a bit tough. How far down the rabbit hole you have to go is dependent on the problem at hand. Good luck.

Hey! you give me an idea.

If a large effective mass leads to low the total magnetic field, then

To know the total magnetic field requires calculating the magnetic susceptibility of the material, in this case that 'material' is a Fermi gas. The magnetic response of a fermi gas (ie: no correlations between electrons) in the presence of a magnetic field is well known. It will consist of two contributions. The first is Pauli paramagnetism. This is the spin response of the electrons. In terms of energy, Pauli paramagnetism is due to the external field producing a finite energy difference between quantized spin up and spin down states of the electrons. The Pauli exclusion principle requires that electrons in the Fermi sea ignore this perturbation. However, near the Fermi surface the electrons spins can polarize in response to this field. In a constant field the Pauli contribution to the magnetic susceptibility is given by:

$$\chi_{P} = \frac{3n\mu_{0}\mu_{B}^{2}}{2E_{F}}$$

The second contribution is Landau diamagnetism. This is the orbital response of the electrons. The simplest picture is that the magnetic field drives the electrons into orbits around the field direction. This is very much akin to the production of a current in a wire loop upon application of a magnetic field, and just like in the wire loop, the induced orbits will try and cancel the field. This is why the orbital contribution is a diamagnetic response. In terms of energy, the B=0 electron states produce a spherical volume that is discreetized into ##k_{x}##, ##k_{y}##, and ##k_{z}## boxes (ie: just our good old fashioned free electron gas model). When a field is applied, say along the z direction, the ##k_{x}## and ##k_{y}## momenta condense into degenerate nested cylinders oriented along the z-direction, so called Landau levels. The production of these levels changes the total energy of the system. Namely, the ##k_{x}## and ##k_{y}## values near the Fermi surface must adjust their energy to the nearest Landau level upon application of a magnetic field. The Landau contribution to the magnetic susceptibility in a constant field can be derived by working through the energetics of this picture and is given by:

$$\chi_{L} = -\frac{1}{3}\chi_{P}$$

Again, the results are for a free electron gas, if correlations between electrons must be taken into account than this will change things, enough so that I don't think a PhysicsForums post will give much traction, it will take a lot of reading and tailored application with your specific problem in mind; you will need band structure and so, at the very least I expect, you will also need DFT. I will state the simplest modification though. Correlations will renormalize the electrons effective mass ##m^{*}## and the net result is that the Landau susceptibility will now scale as

$$\chi_{L} = -\frac{1}{3}\left(\frac{m_{e}}{m^{*}}\right)^{2}\chi_{P}$$

Thus, if the effective mass is small enough, then it can cause the diamagnetic contribution to be larger than the paramagnetic and your total field will reduce.

This is going to make your life much harder, because they do have an effect. Note that I was very careful to state that the above results were for a constant magnetic field. In the presence of a wavevector-dependent ##\mathbf q## and frequency-dependent ##\omega## magnetic field then the magnetic susceptibility becomes a tensor of the form:

$$ \mathbf\chi\left(\mathbf q, \mathbf k; \Omega, \omega\right)$$

where ##\mathbf k## and ##\Omega## are the Fourier domain of the spatial and temporal change of the electron magnetization in response to the applied field. Things can become a bit tough. How far down the rabbit hole you have to go is dependent on the problem at hand. Good luck.

A question, the effects are the same if the gas is degenerated?

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