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Decay rates 
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#1
Oct1907, 03:42 PM

PF Gold
P: 2,892

In our exchange previous to my guest post, Dorigo suggested me to speak of decay rates, but I did not take the courage to extend about this topic. Still, it is an interesting one for a whole thread here, and see what are the ideas of the people of PF.
to start with, I think some dimensional analisis can be interesting: what can we say about decays of massive particles depending of the quantity of dimensional parameters in the theory. For instance, consider electromagnetism, where photon is massless and the fine structure constant is adimensional. Another question I was pondering was the decay of extended objects (strings, branes, superstrings, (superbranes?) etc) against the decay of point like objects. What differences can we expect? 


#2
Oct1907, 04:45 PM

PF Gold
P: 2,892

OK, first thinking: decay rate (or decay width, if you prefer) has units of mass. So if you have no scale, the only dependence you can get is
[tex]\Gamma(M)= g_M * M[/tex] with g a constant depending only of the combinatorial details of the decay. In this kind of theories, the only scale you can get in principle comes from decay products, if the decay is to a particle of mass m. Still, in the limit [tex]m\to 0[/tex], one should expect to recover the linear rule. So we can expect rules of the kind [tex]\Gamma_{M\to m} = f(m/M) * (M \lambda m)^j * M^k[/tex] with j+k=1, and more generally sums of this kind of terms. 


#3
Oct2007, 07:09 AM

PF Gold
P: 2,892

Suppose now you have a theory with a dimensional constant, for instance fermi [tex]G_F[/tex], and consider the decay of a masive particle into massless or almost massless particles, so that the mass of these particles is not so relevant... in this way we avoid a lot of pole and analyticity analysis.
We will have [tex] \Gamma= M* f(G_F M^2) [/tex] and while the naivest idea is to have cubic dependence on the mass, in the concrete case of Fermi interaction the coupling appears in the amplitude and then it is squared for the decay width, so the result is a quintic dependence of the mass. We could also have dependence on [itex](M^2  \lambda G_F)[/itex], should we? Other puzzling set of decays are the "long lived strings", where the dimensional coupling is used to reverse the depencence, and decay width goes inverse proportional to the mass. It is kind of amusing. 


#4
Oct2107, 04:45 PM

PF Gold
P: 2,892

Decay rates
One point about decay rates is that there are two components in the calculation: the amplitude for the probability of changing state and the size of the phase space, the momentum and energy distribution, for the decay products. Pretty obviously this part of the calculation will depend on the number of dimensions of space (or extra uncompactified dimensions, because the whirling about the compact dimesions is considered part of the final state). Is this dependence calculated explicitly in some webpage or could it be worthwhile to do here the exercise?



#6
Oct2207, 03:57 AM

PF Gold
P: 2,892

(When I noticed the cubic scaling on electromagnetic decays someone, I do not remember who, told that it was "just kinematics". Of course the whole point was that, after removing kinematics, the extant factor coincided within the 1sigma error for most of the known decays. But still, it will be interesting to see how the kinematics arise. ) 


#7
Oct2307, 03:11 PM

PF Gold
P: 2,892

It is interesting to consider what happens in an electroweak decay when the mass of the decaying particle evolves from less than M_W to more than M_W: the decay rate changes from quintic "Sargent's rule" dependence to cubic, and while the former decay was into three bodies, this later one is into two bodies, because the W particle can survive onshell.
It is a pretty way to understand the meaning of "unification" and "symmetry breaking" in the electroweak GWS theory. References I found: http://www.slac.stanford.edu/spires/...=PHRVA,D30,947 http://www.slac.stanford.edu/spires/...PHLTA,B181,157 http://www.slac.stanford.edu/spires/...PHRVA,D37,2676 


#8
Oct2507, 05:23 AM

PF Gold
P: 2,892

With this notation, the paper defines also [itex]Q_0=(m^2+Q^2\epsilon^2)/2m[/itex] and [itex]\vec Q^2=Q_0^2Q^2[/itex] and then [tex] \Gamma={G^2_F m^5 \over 24 \pi^3} \int_0^{(m\epsilon)^2} dQ^2 {M^4_W \vec Q \over (Q^2M_W^2)^2 + M^2_W \Gamma_W^2} (2 \vec Q^2 + 3 Q^2 (1{Q_0\over m})) [/tex] I still fail to see how units match, on one side this [itex]Q[/itex] should be adimensional, and on the other hand it seems to have dimensions of mass. (EDIT: given that the overall integral has dimension of M^5 too, I am inclined to think that the m^5 in the prefactor is a "typo"). The paper Phys.Lett.B181:157,1986. takes more care with dimensions and you get a formula you could actually plot. First they define an adimensional function f(a,b,c) and then they give (we allow ourserves to put Vqq equal unity as the previous paper) [tex] \Gamma= {G^2_F m^5 \over 192 \pi^3} f(\frac {m^2} {m^2_W}, \frac {\epsilon^2} {m^2}, \frac {\Gamma_W^2} {m^2_W}) [/tex] The function f appears in the paper, eq (4), without to me noticeable typos. Ok, lets copy it for completeness: [tex] f(\rho,\mu,\gamma)= 2 \int_0^{(1\sqrt \mu)^2} {dx \over ((1x \rho)^2+\gamma^2)^2} ((1\mu)^2+(1+\mu)x 2x^2) \sqrt{1 + \mu^2 + x^2 2 (\mu +\mu x +x)} [/tex] You can notice that these formulae do not take the Standard Model as granted, ie no relationship yet between W, its mass and its width, and G_F. We can contemplate the change from quintic to cubic dependence on m and still keep some intrigue about the role of fermi constant (although I can not understand how they have got to put W inside in the first place if they are still playing with fermi beta decay). PS: note an overall factor 9 is needed if you want to account for all the decays of W and not only a partial width. 


#9
Oct2507, 09:04 AM

PF Gold
P: 2,892

via gordon watts, this tv series informs us that Vtb is near unity, as said above
http://www.tvguide.com/tvshows/bigb...hotos/288041/4 


#10
Oct2507, 12:03 PM

PF Gold
P: 2,892

[tex] \Gamma={G^2_F \over 24 \pi^3} {m^5 \over 4} \int_0^{1} {dx \over (1  x \rho )^2 + \gamma} (1 +x x^2) \sqrt {1  2x +x^2} [/tex] To be compared with the preprint of Phys.Lett.B181:157 [tex] \Gamma= {G^2_F m^5 \over 192 \pi^3} 2 \int_0^{1} {dx \over ((1x \rho)^2+\gamma^2)^2} (1+x 2x^2) \sqrt{1 + x^2 2 x} [/tex] It almost works! Except for the BreitWignerlike denominator, which has a different power. In this case, it seems the mistake is in the second formula; note I have checked only the preprint, not the published version. 


#11
Oct2707, 08:35 AM

PF Gold
P: 2,892

Well, I think I can confirm that there was also a typo in the Phys.Lett.B181:157, and that this second ^2 exponent was most probably a bad interpretation of the instruction "remove this 2" to edit out the other ^2 over the gamma. Albeit in more recent papers, the authors prefer to keep the first ^2 and to redefine gamma as the square root of the one we are using here.
So the tree level formula for a weak charged decay into massless particles is [tex] \Gamma= 2 {G^2_F m^5 \over 192 \pi^3} \int_0^{1} {dx \over ((1x \rho)^2+\gamma)} (1+x 2x^2) \sqrt{1 + x^2 2 x} [/tex] where [itex]\rho=m^2/M_W^2[/itex], [itex]\gamma=\Gamma^2_W/M_W^2[/itex] and you can still put a factor 9 if you want to account for the nine different decays of the W. (EDIT: question: but the [itex]\Gamma_W[/itex] in the formula is the total decay width, isn't it?) Ah, there are more recent calculations of this formula in the literature, some of them even including the first QCD correction. http://www.slac.stanford.edu/spires/...hepph/9302295 http://www.slac.stanford.edu/spires/...=PRLTA,66,3105 http://www.slac.stanford.edu/spires/...j=NUPHA,B314,1 http://www.slac.stanford.edu/spires/...=NUPHA,B320,20 Well, the next step, standard also in the literature, is to get rid of the integral. They are two different limits where you can do it: 1) consider the "narrow width limit", using the representation of the dirac delta as a limit of cauchy distribution. 2) consider the limit where m is a lot smaller than M_W. 


#12
Oct2707, 05:16 PM

PF Gold
P: 2,892

I'd believe we are all using free software, so we can try maxima or wxmaxima



#13
Oct2807, 07:31 AM

PF Gold
P: 2,892

Really! The bad interface between numerics and analytics in maxima, 20 years later, is disgusting. I first used Macsyma, the VAX VMS version, in 1989.
Anyway:
Observe the second attached plot, and consider what should happen if the mass of W were smaller, while you start still from the top mass scale... the cubic line should prolongue, the change to quintic happening to lower energy. Ideally, a limit can be taken where the mass of W is null and no change to quintic happens; in this limit, isospin symmetry is restored, as the elders told us in the school. For an excursion into the unknown, compare the third plot here with the one discussed in http://dorigo.wordpress.com/2006/09/...dthezwidth/. Note that here we are using GeV, while that plot uses MeV. Note also we are not including QCD, neither here not in the other plot, and that here we are including the factor 9 of the different decays as if all the masses were zero except the top. The coincidence with the cubic line discussed in that entry is stronger if this factor 9 is not used. EDIT: If you want to pursue further the numerical investigation, quad_qag will be better than quad_qags. 


#14
Oct2907, 07:08 PM

PF Gold
P: 2,892

[QUOTE=arivero;1484081]
For an excursion into the unknown, compare the third plot here with the one discussed in http://dorigo.wordpress.com/2006/09/...dthezwidth/. Note that here we are using GeV, while that plot uses MeV. Note also we are not including QCD, neither here not in the other plot, and that here we are including the factor 9 of the different decays as if all the masses were zero except the top. The coincidence with the cubic line discussed in that entry is stronger if this factor 9 is not used. What happens is that the fork here sits nice and mystically about a factor ten inside of the fork there. Thus the use of this fork to calculate the breaking of isospin in decay widths of a isomultiplet works nicely in the qualitative (dependence on M^2) but fails about one or two orders of magnitude in the quantitative. The quintic leg would meet exactly the muon cross if the factor nine were progressively removed as the energy crosses the thresholds for the decay of virtual W into cs (3), taunu (1), ud (3) and muon itself (1). The exact calculation becomes lengthy because of the CKM mixing. But there is not a clear way for the cubic leg here to reach the experimental cubic leg. If we enable QCD in the formula here, the decay width decreases (see Jezabek and Kuhn 1993). In any case, note that the decays of the electrically neutral isospin partners of charged mesons are into photons, so it had been very surprising to get a match with only SU(2) weak and without the U(1) EM group. Still, I do not know yet how to enter it into the Z0 mystery line, so I guess that the detour stops here and we can go back to regular schedule. 


#15
Oct3007, 05:55 PM

PF Gold
P: 2,892

I still would like to derive this formula straight from the Feynman rules step by step. Meanwhile, lets attack the limits
[tex] \Gamma=N_f* 2 * {G^2_F m^5 \over 192 \pi^3} {1 \over 1+\gamma} \int_0^{1} dx (1+x 2x^2) \sqrt{1 + x^2 2 x} = N_f * {G^2_F m^5 \over 192 \pi^3} * {1 \over 1+{\Gamma^2_W/M_W^2}} [/tex] and remember that for the standard model [itex]{\Gamma^2_W/M_W^2}=0.00071[/itex], so when the number of available particles for the decay is only[itex]N_f=1[/itex] we practically hit the usual result of muon decay (besides a correctness check, it is a realistic approx, but sort of inconsistent touch because we have set all the other quark or leptons masses to zero so it is really =9 always) 


#16
Oct3007, 06:33 PM

PF Gold
P: 2,892

(1) is almost the case [itex]\gamma \to 0 [/itex]. We do not want to go so far because in such case the integral gets an infinity. But if we are near enough then we can substitute
[tex] \Gamma=N_f* 2 * {G^2_F m^5 \over 192 \pi^3} \int_0^{1} dx {\pi \over \sqrt \gamma} \delta(1x \rho) (1+x 2x^2) \sqrt{1 + x^2 2 x}= [/tex] [tex] =N_f* 2 * {G^2_F m^5 \over 192 \pi^3} \int_0^{1} dx {\pi \over \rho \sqrt \gamma} \delta(\frac 1 \rho x) (1+x 2x^2) \sqrt{1 + x^2 2 x}=[/tex] [tex] =N_f* 2 * {G^2_F m^5 \over 192 \pi^2} { M_W^2 \over m^2 \sqrt {\Gamma^2_W/M_W^2}} (1+{ M_W^2 \over m^2 } 2{ M_W^4 \over m^4 } ) \sqrt{1 + { M_W^4 \over m^4 } 2 { M_W^2 \over m^2 } }=[/tex] [tex] =N_f* 2 * {G^2_F m^3 \over 192 \pi^2} { M_W^3 \over \Gamma_W} (1+{ M_W^2 \over m^2 } 2{ M_W^4 \over m^4 } ) \sqrt{1 + { M_W^4 \over m^4 } 2 { M_W^2 \over m^2 } }[/tex] If furthermore (lets call it case 1') we consider that m is bigger than the mass of W, we get rid of the corrections and we get [tex] \Gamma=N_f* 2 * {G_F \ m^3 \over 192 \pi^2} { G_F M_W^3 \over \Gamma_W} [/tex] And if we want to believe that [tex]\Gamma_W = \frac 32 { G_F M_W^3 \over \pi \sqrt 2 }[/tex] then [tex] \Gamma=9* 2 * {G_F \ m^3 \over 192 \pi^2} {2 \pi \sqrt 2 \over 3 } = 12 \sqrt 2 * {G_F \ m^3 \over 192 \pi} = {G_F \ m^3 \over 8 \pi \sqrt 2 } [/tex] and yes, it is the standard approximation for the decay of the top quark. 


#17
Oct3007, 06:52 PM

PF Gold
P: 2,892

A qualitatively right estimate but still too small in the quantitative is the quotient between decay rates of two isospin partners, say neutral pion vs charged pion
[tex] (N_f* 2 * {G_F \ m^3 \over 192 \pi^2} { G_F M_W^3 \over \Gamma_W}) / (N_f * {G^2_F m^5 \over 192 \pi^3} * {1 \over 1+{\Gamma^2_W/M_W^2}}) [/tex] it simplifies to [tex] ({ 2 M_W^3 \over \Gamma_W}) / ( { m^2 \over \pi} * {1 \over 1+{\Gamma^2_W/M_W^2}}) [/tex] lets call it q [tex] q= {2 \pi \over m^2} * {M_W^3 \over \Gamma_W} * (1+{\Gamma^2_W/M_W^2} ) \approx {4 \pi^2 \over 3 m^2} { \sqrt 2 \over G_F } [/tex] The failure is due to the neutral particles, whose decay seems to scale empirical and unexplainablely as the cubic law for Z0 decay, and not the cubic law for the top quark. 


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