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An energy exchange (i.e. absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse (your original question) happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.

The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.

An energy exchange (which you could call absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse (your original question) happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.

The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.

An energy exchange (i.e. absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.

The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.

An energy exchange (i.e. absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse (your original question) happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.

The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.

An energy exchange (i.e. absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber.

An energy exchange (i.e. absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.

The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.