# NMR spectroscopy question

1. Nov 27, 2015

### BobP

Hi.

I know that the chemical shift (Hz) of a particular chemical will vary with B0. As I understand (although this may be wrong) ppm is frequency independent so the peaks should always be in the same position

There are two things I do not understand though

1) Why does increasing the frequency increase the separation of peaks on a ppm curve if ppm is frequency indepndenant? (my lecturer said it did)

2) When trying to localise the position of a spectral signal in 3D space why can we NOT use a frequency encoded gradient but we can use a slice or phase encoded gradient (I assume this links with the previous question).

Thank you very much

2. Nov 27, 2015

### Staff: Mentor

Increasing the frequency (or, more appropriately, the magnetic field) doesn't change the separation of the peaks in ppm, but it does allow for a better resolution (the peaks are narrower).

3. Nov 27, 2015

### BobP

Thanks. Why does resolution increase?
Has it anything to do with a longer FID decay time? If so, how does this happen?

Don't think so. I am talking about acquiring a signal from a specific voxel in the brain, for example, using spectroscopy in a clinical setting. My lecturer said we cannot use frequency gradients but we can use other gradients. His reasoning was that ppm is affected by frequency but clearly it isn't so I am wondering why we cannot use them.

thanks

Last edited: Nov 27, 2015
4. Nov 27, 2015

### Staff: Mentor

There are two reasons for that. One reason is simply SNR. Since you have higher SNR at higher field strengths two peaks that are blurred out by noise at los field strengths may be separable at higher field.

The second is due to shimming. A poor shim causes blurring in the spectrum. Say you can shim down to 10 Hz at both fields, then that is a smaller ppm blurring at the higher field.

You can, but the math gets substantially more complicated. Andrew Maudsley at University of Miami and Stefan Posse at New Mexico have each developed their own approaches for doing that.

Are you familiar with the k space approach for understanding encoding?

5. Nov 27, 2015

### BobP

Why does SNR increase with B0? thanks

Sadly not :( Is it possible to explain why it is more complicated without going into it?

6. Nov 27, 2015

### Staff: Mentor

As the field strength increases you get more net longitudinal magnetization in the fully relaxed state. I don't remember the numbers exactly, but it is something like 6 excess protons per million at 1.5 T vs 12 excess protons per million at 3.0 T.

7. Nov 27, 2015

### BobP

Ah of course! thanks

On a slightly unrelated note, I have been told that the seperation of J-coupled peaks increases as B0 increases. Please could you suggest why this is, again, as ppm is unrelated to B0

8. Nov 27, 2015

### Staff: Mentor

I had to think a bit about this. Basically, what makes the Fourier transform easy is that it is separable. A 2D FFT is just a bunch of 1D FFT in one direction followed by a bunch of 1D FFT in the other direction. But this relies on the data being laid out in a nice rectilinear grid in k space.

But the corresponding "gradient" for spectroscopy is just time, which you cannot turn off or rewind. So if you do spatial encoding and spectroscopic encoding you wind up going diagonally through k space which complicates things.

9. Nov 27, 2015

### Staff: Mentor

I think that this is purely due to SNR, but I am more on the imaging side than the spectroscopy side. So I am not certain.

10. Nov 27, 2015

### Staff: Mentor

Be careful here. The chemical shift is independent of B0, but the calculation of ppm depends on the operating frequency, as it is basically
$$\frac{\nu_\mathrm{sample} - \nu_0}{\nu_0} \times 10^6$$
Since the transition frequency due to the chemical shift ($\nu_\mathrm{sample} - \nu_0$) and the reference frequency have the same variation with respect to B0, the field cancels out and the chemical shift, expressed in ppm, is independent of B0.

However, the J-coupling comes from spin-spin interaction, which is independent of B0. Since it induces a fixed shift in $\nu_\mathrm{sample}$, in terms of ppm it decreases as B0 increases.:
$$J\ \mathrm{Hz} = \delta\ \mathrm{ppm} \times \nu_0\ \mathrm{MHz}$$

11. Nov 27, 2015

### BobP

I see. thanks

12. Nov 27, 2015

### BobP

thank you. To clarify, did you mean However, the J-coupling comes from spin-spin interaction, which is DEPENDANT on B0.

13. Nov 27, 2015

### Staff: Mentor

No. The spin-spin coupling is intrinsic to the molecule, thus is independent of any external field.

I've stolen the following at http://orgchem.colorado.edu/Spectroscopy/nmrtheory/splitting.html