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Magnetic Nuclear Resonance

  1. Sep 12, 2004 #1
    Hello,

    consider a dihydrogen molecule in a powerful magnetic field and an x-ray that is hitting this molecule.
    Many things could happen in this case I think:
    compton effect, ionisation, electrons changing orbitals, nuclear magnetic spin state changing....

    My question is: can I be certain that the nuclear magnetic spin state will change and that consequently the nucleus would emit a wave at the larmor frequency?

    Thanks a lot.
     
  2. jcsd
  3. Sep 15, 2004 #2
    Hello again,

    my question might be a little bit vague.

    If I want to put it in another way, I would ask:
    consider a proton in a strong magnetic field.
    can we change the proton spin state by giving him energy through a photon whose energy is greater than Larmor's energy?

    thanks again
     
  4. Sep 15, 2004 #3
    Yes cosmonova, it is possible in the way your last question is formulated. People studying the proton structure are especially interested in how the spin emerges from the constituents, and they do often use a lepton beam interacting with the proton through photon exchange. In that case, the energy transfer can be large.

    I could not reply to your first question, sorry. You would have to calculate also a Thomas term in the spin-orbit coupling.
     
  5. Sep 15, 2004 #4
    Isn't the probability that such a process occurs maximal if the energy of the incident photon equals the energy corresponding to the induced spinflip???

    I think, so, just wondering though

    regards
    marlon
     
  6. Sep 15, 2004 #5

    ZapperZ

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    This is still something very vague about this question, which is why I didn't address it in the first place. The vagueness come in by what you mean by a "photon" in this scenario.

    Let's use your example with a protons in a magnetic field. This external magnetic field is constant. So then, the protons can be in two different spin states, with an energy difference between these states since the degeneracy has been removed.

    Now, classically, the magnetization vector for each dipole spin will precess with the Larmour frequency about the direction of the external magnetic field. If you apply an external EM field (this is where the "photon" comes in), typically an RF field, the frequency of the RF field must be equal to the Larmour frequency. This is a simple application of the classical "resonance" effect - the most efficient transfer of energy is when the "transferer" is at the same frequency as the "transferee". Thus the word "resonance" in Nuclear Magnetic Resonance.

    Note that the above is the classical picture of the process. It turns out that the Larmour frequency from the classical picture is equivalent to the difference in the energy of the two spin states! So one can easily view the RF field as photons with energy equal to that energy difference. It is all consistent!

    Zz.
     
  7. Sep 16, 2004 #6
    thanks a lot for your replies.
    ZapperZ, what you have mentionned is exactly what my question meant.
    I am sorry for not taking my time to clarify it .

    But having mentionned it, I have more related questions:

    1- what is the reality of the resonance phenomena, i.e why does the maximal energy transfer we have talked about occur when the frequency of the wave is equal to the Larmor frequency?

    2- is it pure luck that the energy difference between the two spin states equal to the energy of the photon at the Larmor frequency?

    3- Can we change the spin state of the proton in the magnetic field by another process than the RF wave ? I am thinking about heat or another indirect process.

    4-I am still getting some difficulties to understand why this phenomena (proton magnetic resonance) is different in nature from another quantum phenomena involving discrete energy levels which involves the electron energy levels of the hydrogen atom.For the latter , if the energy difference between two states is equal to E, we could send a photon whose energy W is bigger than E and still have the shift between energy levels.(in this case , I think the remainder W-E will be emited as a photon by the hydrogen atom and that doesn't seem to happen in NMR)

    Thanks again.
     
  8. Sep 16, 2004 #7
    Yes, via heat one can dramatically change the magnetic interactions between neighboring atoms on a lattice-structure for example. I am referring to the transition of ferromagnetica, ferrimagnetica and anti-ferromagnetica to paramagnetica.

    ferromagnetica, ferrimagnetica and anti-ferromagnetica are specimen with specific magnetic interactions between two neigbours on the lattice. A rise in temperature weakens these interactions and this leads to more "chaos" of the different spin-states. For example in a ferromagnetic material two atoms next to eachother will have the "tendency" of aligning their spins in to same direction. This yields ORDER in the lattice of atoms. When temperature rises this spontaneous order will become weaker and weaker and two spins will have much lower probability to be aligned along the same direction.


    regards
    marlon
     
  9. Sep 16, 2004 #8
    Resonance is very logic. It basically states that an incident photon "must" (read : highest probability) have the exact same energy as the energy-difference of the induced spin-flip.

    So the probability that the spin-flip will occur is maximal when the energy of the incident photon is equal to the energy of the spin-flip. Now on a more intuitive note you could say that this process is most likely to occur when there is just enough energy available in order to make the spin-flip happen. if the foton has more energy it "might" happen that other interactions take place that would disturb or even prevent the spin-flip from happening


    regards
    marlon
     
  10. Sep 16, 2004 #9

    ZapperZ

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    Resonance means transfering energy at the right phase of an oscillation. This is a very general principle and not just confined to NMR (see, for example, a radio).

    No, it is not pure luck.

    You may apply heat if all you want to do is change the net magnetization of the bulk sample. However, keep in mind that heat tends to RANDOMIZE the spins. The BEST temperature to do any NMR measurement is usually close to zero kelvin (less noise).

    In an NMR, you are directly probing the nucleus of certain atoms in a material. This is because each nucleus of different atoms has different resonance frequency. You can probe that nucleus if you apply an interaction that has the same frequency. Because everything else other than that type of nucleus are at different frequencies, they are all immune to your probe. This means that in a hydrogen atom, the rest of the atom isn't directly affected. Only the proton (the nucleus of the H atom) is the one affected. So to probe the nucleus, you need not strip away all the surrounding electrons, or all the surrounding nucleus of different specie.

    [This is a highly simplified picture. There are higher order effects in which this isn't entirely true].

    Zz.
     
  11. Sep 17, 2004 #10
    Hello again,

    Okay, I think I understood what resonance is about.

    What is the physical interpretation of the T1 and T2 time constants usually invoked in texts about MRI?
    I have read somewhere that they corresponded respectively to “spin lattice” and “spin spin” relaxation times.
    Could someone please explain to me what these terms mean?

    One more question : when we consider the net magnetic vector (NMV), what do the angles 0,90,180 really mean?

    Thanks.
     
    Last edited: Sep 17, 2004
  12. Sep 17, 2004 #11
    Hi,

    Let me give you some examples...

    Spin lattice relaxation time : suppose you want to study certain atoms surrounded by a lattice of other atoms. An atom A in the centre of a cubic lattice with at the corners the atoms B for example.

    If you were to change something about the lattice then the atom A at the centre will feel this change indirectly via it's interaction with the surrounding B-atoms. The time needed for the atom A to "adapt" itself to the new conditions generated by the change in constitution of the lattice is the spin-lattice relaxation time.

    Spin-spin-relaxation time : Let me take the example of a ferromagnetic specimen. Ferromagnetic means that one spin will tell his neighbor to point itself into the same direction and along the same axis. So spins will align.
    Such a specimen can consist of two types of atoms A and B and the lattice of this specimen can be "divided" in two sub-lattices each containing one type of atom. So a sub-lattice with A-atoms at the corners and a sub-lattice with B atoms at the corners. The atoms of the A-sub-lattice will tell the ones of the B-sub-lattice to point in to same direction. Suppose that lattice A has atoms pointing down the x-axis and the B-lattices has atoms that point up the x-axis. Now, we bring the two lattices together so that after each A-atom there is a B-atom (think of this as you would "shove" the lattices into eachother). The A-atoms will tell the B-atoms to point down the x-axis and the time needed for the B-atoms to adapt to this new situation is the spin-spin-relaxation-time. It could also have been the other way around where the A atoms follow the ORDER of the B atoms. The system remains the same though

    Relaxation basically means equilibrium and the relaxation-time is the period needed for a certain physical state to evolve into this equilibrium that is caused due to interactions between atoms...

    regards
    marlon
     
  13. Sep 17, 2004 #12

    ZapperZ

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    T1 is defined as the spin-lattice relaxation time. When you have a nucleus in an external magnetic field, you will have spin states separated with some energy difference. If you apply an RF pulse, you can induce a transition from a lower spin state to a higher spin state. But this situation is unstable. Eventually, it will decay back to the lower spin state. The probability or rate which this decay occurs depends on two important factor: (1) the probability of that transition (governs by the Fermi Golden Rule), and (2) the probability that the energy given off during the decay can be taken up by the rest of the lattice. Thus the name spin-lattice relaxation.

    What this means is that with the (1) being relatively constant for the same type of nucleus in a constant magnetic field, then the only variation in T1 will be due to the surrounding structure or lattice. So T1 for a proton in water will be different than in human skin, etc., because the surrounding or lattice is different for each.

    T2 is the spin-spin relaxation time. In this case, what you are doing is measure how fast the individual spins in the bulk material randomize back to their original state. Typically what happens here is this. When you have the material in an external field, the spins in the material are alligned either along or opposite the direction of the applied field. However, the spin component perpendicular to this direction is undefined and randomized, resulting in a net value of zero. In an experiment to measure T2, an external RF or magnetic field is applied perpendicular to the original magnetic field. This causes some, if not all, of the spins to allign in this direction, resulting in a net magnetization in this perpendicular direction. When this perpendicular field is turned off, the time taken for the spins to go back to the randomize state is measured. T2 then measures how stongly the spins correlate with each other. Thus that's why it is called spin-spin relaxation.

    The angles you asked about are the angles that the net magnetization of the bulk have been rotated in a PULSE NMR. So a 90 degree pulse will cause the net magnetization to flip from along the external field to somewhere perpendicular to it. Various combinations of 90-90, and 180-90, etc. can be used to measure T1 and T2.

    Zz.
     
  14. Sep 17, 2004 #13
    Hello again,

    I think I got the whole picture. Correct me if I am wrong:

    When we consider a certain bulk of hydrogen atoms in a static magnetic field B, at the temperature T, there are two magnetic spin states for the proton. One along the main field axis and the other orthogonal to this axis.The latter has higher energy and is thus more unstable.
    In the absence of B, temperature practically randomizes these states and we have approximately the same number of protons in each state.
    When B is applied, the number of protons in the lower energy state is bigger statistically than the one in the higher energy state.The difference is the net magnetic vector (MNV)
    Because the protons are precessing around the B axis at the Larmor frequency , the most efficient way to give them energy for changing their state is by giving them a radiation at the Larmor frequency.
    When we do so, practically all of the protons (not only those of the MNV at that moment) shift to the highest energy state as long as the radiation pulse still exists. When the pulse disappears , the protons begin to get back to the lower energy state, and the spin-spin influence takes place, giving rise to an exponential decrease of the number of high energy protons to zero.
    After that, all the protons point in the B axis direction. And here the temperature randomization acts again via the spin-lattice interaction, which leads to another exponential decrease of the number of protons in lower energy state to the number of MNV.And it is obvious in this process that the time characteristic T1 would be larger than that of the spin-spin interaction T2.

    I hope I got that right.

    Thanks a lot really .
     
  15. Sep 17, 2004 #14

    ZapperZ

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    Not quite. The two spin states corresponds to spin parallel and anti-parallel to the external B field, not along and orthorgonal. At T=0, all of them will be alligned to B. At finite T, the state antiparallel to B will have some non-zero probability of occupation.

    In the absence of B, there are no spin states. The randomization is not just that the two states have equal number, because there are no "two states" to speak of. When B is not present, the spins are in random orientation all over space.

    Zz.

     
  16. Sep 18, 2004 #15
    Hi,

    When trying to compare my last visualization of the phenomenon to some explanations given by many textbooks or websites on this matter, I realize that something is not right. And this is confusing me a lot.

    I will resume these contradictions in the following points:

    1- I think that under a certain static magnetic field, protons have only two possible states. One parallel to the field and another anti-parallel to it. If that is correct, how can they say that under a 90 degree RF field, the protons rotate in the transverse plane (orthogonal position to the main magnetic field)?Shouldn’t they take the energy carried by the RF wave and take the anti-parallel position?
    2- In my scenario , I thought that T1 relaxation was associated with the decrease of the longitudinal part of the NMV to its maximal allowed value.It seems that it’s the opposite where this part goes from zero to its maximal allowed value under the considered temperature.

    3- By the way, I still don’t understand what they mean by the NMV angle? I tried to give it a mathematical formula and I found it should be as follows:
    If N is the number of protons in the lower energy state , S the number of protons in the higher energy state and M is the maximum value for N-S then the angle the NMV takes when we apply the RF wave should be equal to:
    theta=arccos((N-S)/M)
    Is that right?
    In this case the longitudinal and transverse magnetization they talk about are the respective projections of this vector along and orthogonal to the main magnetic field.


    4- I have difficulty in understanding why the two relaxation processes T1 and T2 are independent. If the proton, spinning in the transverse plane imposed by the 90 degree RF wave amplitude ,is relaxing and regaining his alignment with the main field, then any lost in his transverse magnetization is a gain in his longitudinal magnetization and the two phenomena are one. I still don’t get it folks. I also don’t understand when the spin-spin relaxation occurs and when the spin-lattice relaxation occurs.
    5- I think some of the confusion rises from mixing classical and quantum stuff together. I wish someone could give me a coherent, causal, chronological and detailed explanation of what happens at the molecular level when we apply to the proton the RF signal and when this signal stops.I prefer that this explanation forgets NMV and talks only about protons in two spin states.(even if it would be a more difficult approach)

    Thanks you for the inconvenience about my questions, but I really want to understand.
     
  17. Sep 18, 2004 #16

    ZapperZ

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    The problem here is that you're trying to mix the classical and the quantum picture at the same time. While they're consistent with each other, they shouldn't be mixed like that or they'll cause you indigestion.

    When you put a proton in a magnetic field, it will have two spin states - corresponding to up and down. Now, at ordinary room temperature, there will be more of these protons IN THE BULK material that will be in the parallel states than an antiparallel states. If you then switch over to the classical picture, you will see the bulk net magnetization FOR THE WHOLE SAMPLE having a magnetization vector along the external field.

    Now, if you apply an RF pulse that causes a 90 tilt, then you are now changing the number of occupation of each of the spin states. In fact, a precise 90 pulse will cause both spin states to have equal number of occupation! What this means is that the bulk material has ZERO net magnetization along the original external magnetic field.

    However, there is another possible complication here. When you apply a strong enough 90 field, then the original spin eigenstates are no longer the eigenstate of the new system (external field plus 90 pulse). So depending on how long you apply the 90 pulse, you can create a NEW set of spin states. This will get way too complicated for me to explain here.

    Zz.
     
  18. Sep 18, 2004 #17
    Hi ZapperZ,

    if the 90 degrees pulse made the Net Magnetization of the bulk equal to zero, I find it difficult to relate it with the classical view where the protons rotate 90 degrees and precess around the propagation axe of the RF wave(because classicly the RF wave is an electromagnetic wave which carries a magnetic field), thus forming a magnet orthogonal to the main field.
     
  19. Sep 18, 2004 #18

    ZapperZ

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    OK, classical picture. Each dipole moment in the material precesses around the direction of the B field. They each make a cone around this direction. So there is a net magnetization of the bulk material in the B direction when you sum up all of these individual dipoles.

    When you apply the 90 rf field, you flatten each cone, but keeping the vector length of each dipole moment constant. So each dipole moment are still precessing around the direction of the B field, but the cone is flattened until it is flat as a pancake, with the moment rotating as the radius of the circle. This produces a net magnetization of zero along the direction of the B field. The 90 pulse can do this because it is an RF field has the same frequency as the precession frequency of each moment (the Larmour freq).

    Zz.
     
  20. Sep 18, 2004 #19
    Good explanation ZapperZ. I posted a reply before but i was confused so I deleted my post, I´m sorry. Thanks for your advice
     
  21. Sep 19, 2004 #20
    I agree with altered-gravity. It is the best explanation I have been told so far.

    It should be very difficult to translate into quantum theory, and because of that I will stick for now with this explanation.

    Thanks again Zapper Z.

    Anyway, for those interested I have found a nice reference on spin dynamics and its application for NMR.
    http://www.cis.rit.edu/htbooks/nmr/chap-3/chap-3.htm
     
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