nmr units

Units for NMR Work PhL 3.28.08 I am "rusty" with SI units, and NMR people tend to use their own conventions, so now is time to get this mess cleared up a bit. 1. Preliminaries. I see I have already written units2.doc which goes into lots of detail with many digressions, but here let's look for something concise. That document eventually arrives at this point: divE = 4k1 curl E = - (1/)B/t (Faraday) k1 = c2 k2 div B = 0 curl B = 4 k2 J + ( k2/k1) t E where k1 and k2 and are arbitrary constants except we must have k1/k2 = c2 , speed of light squared. We usually have = 1, so let's eliminate that one right away, then we have divE = 4k1 curl E = - B/t (Faraday) k1 = c2 k2 div B = 0 curl B = 4k2 J + c-2 t E Simpler and more familiar equations are these, showing the same two constants k1 and k2: F = k1*qQ/r2 (1) / force between two charges dF/dx = k2*2iI/D (2) / force between two parallel wires separated by D Now, in the SI system we use k2 = (0/4) = 10-7 nt C-2 s2 // = kg m C-2 k1 = (1/40) = 10-7c2 9 x 109 nt C-2 m2 // = kg m3 C-2 s-2 Both of these constants "carry dimensions" or "have units" in the SI system. From equation (2) above we find that k2 has units nt C-2 s2 because that makes the units match on both sides of the equation. Then the fact that k1 = k2 c2 tells us that the units of k1 must be nt C-2 m2. We can regard this choice of constant k2 as "defining the Coulomb" in terms of other basic units. The alternate forms on the right go down to basic units, replacing nt = kg m s-2 from F = ma. The electric field E in any system of units is defined in terms of F = qE , so units of E are nt/C in SI. The potential V is defined as E = -V and this tells us that nt/C = volts/m where we thus define the "volt" as the unit of potential V. Thus, volt = nt m C-1. The magnetic field B from Biot-Savart ( = 1) is given by dB = k2* (IdX x r)/r3 . We conclude that: dB = ( nt C-2 s2 ) * C s-1 m-1 = nt C-1 s m-1 = tesla // = kg s-1 C-1 where here we define the tesla as shown. Plank's constant E = so = joules s = nt m s = kg m2 s-1 = J s . As shown a few lines below, these are also the units of angular momentum. The magnetic moment of a planar current loop of area A is given by m = I A where A points in the RHR sense based on the current direction. Therefore magnetic moment = C s-1 m2 /= joule/tesla from a few lines below! = J/T = Amp m2 Here is another example: the Bohr magneton is e = e/2me = C * kg m2 s-1 * kg-1 = C m2 s-1 which agrees with our magnetic moment dimensions., Now consider Levitt page 24 which says E = -B // Levitt page 24 and is a magnetic moment. We should then have joule = C s-1 m2 * tesla Let's check it out C s-1 m2 * tesla = C s-1 m2 * nt C-1 s m-1 = nt-m = joules // checks Our next item of interest is the gyromagnetic ratio: = s s = "the spin" See also Bleaney page 20 where the spin is called G. In Bleaney, G has units of angular momentum G = angular momentum = r x p = r x mv = kg m2 s-1 // = nt m s = J s The second form is more obvious from r x F = dJ/dt = N = torque. So what are the units of gyro ? = / G = C s-1 m2 / [kg m2 s-1 ] = C / kg / = s-1 tesla-1 Do this another way // above says = / G = J T-1 / [ J s] = s-1 T-1 where the last form just comes from the tesla units as noted above. Aside: what are the units of B? B = s-1 T- * T = s-1 = -units = Larmor frequency. Now, what is the angular momentum of a proton (ie, what is its spin): spin = 1/2 = angular momentum units = J s What is for a proton? First, from page 408 of Livesey, we are told that n = e/2Mp = "the nuclear magneton" = 5.05 x 10-27 A m2 But this is not the proton's actual magnetic moment, it is the scale used for same. From page 410 we find that P = 2.79270 N = 2.79270 * 5.05 x 10-27 A m2 We know that the proton's spin angular momentum is /2 which is (1/2) 1.0545 * 10-34 J s. Therefore we should find that P = P / spinP = 2.79270 * 5.05 x 10-27 * 2 / 1.0545 * 10-34 A m2 / [ J s] The units are J T-1 / [J s] = T-1 s-1 . The number is 2.79270 * 5.05 * 2 / 1.0545 * 107 = 26.748477952 x 107 = 267.48 x 106 T-1s-1 and this agrees with Levitt page 14, we can just ignore his "rad" thing. Now let's jump to page 204 of Levitt where we are talking dipole-dipole. We have b = - (0/4) 12 / r3 What are the dimensions of this "thing" ? { nt C-2 s2 } { s-1 T-1 }2 { nt m s } /m3 = nt2 C-2 s m-2 T-2 Let's replace T = nt C-1 s m-1 to get = nt2 C-2 s m-2 { nt C-1 s m-1}-2 = nt2 C-2 s m-2 nt-2 C2 s-2 m2 = s-1 So we conclude that b is a frequency of some sort, probably radians/second as he suggests. Basically, in this equation we have b = b and the I's are dimensionless. Now on page 149 Levitt tells us " to get the true Ham from the script Ham, mult by ". So the units of a script Ham are always units, in agreement with what we just found. Comments on Levitt page 204 on dipole-dipole: 1) the dimensions of script H are 2) the dimensions of b are 3) the spin operators are dimensionless. Now look at Levitt page 175 Equation 7.9 says = I. What units are implied in this equation ? If I were dimensionless, we woujld want to say physics = Idimless But what he has is this: NMR = Idimless So he is using NMR = physics / This then works right in equation (17.10) where we have H NMR = - NMR B So he never messes with the definition of itself, just and H.