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Edit - Maybe formulated differently: Does it make sense to talk about the top mass at energies below $m_t$, although in all processes the corresponding energy scale is above $m_t$, because of the rest mass energy of the top quark?


In what sense is it or isn't it meaningful to speak about the top quark mass at energy scales below the top quark mass? Using an effective field theory approach, the top quark decouples at energies below the top quark mass and therefore has no influence on the mass running of the other fermions. Nevertheless, it is possible to compute how the top quark mass $m_t$ changes at energies below $m_t$, for example, because of the energy dependence of the gauge coupling constants. For example, in Updated Estimate of Running Quark Masses the authors compute

enter image description here

but write

We also illustrate the behavior of the heavy quark masses m q ( μ ) ( q = c,b,t ) in Fig. 3. Exactly speaking, the word “the running mass value m Q ( μ )” of a heavy quark Q at a lower energy scale μ than $μ = m_Q ( m_Q )$ loses the meaning. For example, the effective quark flavor number $n_q$ is three at μ = 1 GeV, so that the value of $m_t( μ )$ at $ μ = 1$ GeV has not the meaning. However, for reference, in Fig. 3, we have calculated the value of m Q ( μ ) ( Q = q N ) at μ n ≤ μ < μ n +1 ( n < N ) by using the relation m Q ( μ ) = c m Q R ( N ) ( μ ) [not m Q ( μ ) = c m Q R ( n ) ( μ )].

What do they mean by "loses the meaning"?

Similarly, the authors in Updated Values of Running Quark and Lepton Masses list in table II the top quark mass at various energy scales, for example $m_t(\mu= 2$ GeV) $ \approx 384.4$ GeV, which is considerably different from its value at the $Z$ scale $m_t(\mu= 91$ GeV) $ \approx 171.7$ GeV.

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    $\begingroup$ As they state at 1 GeV the flavor number is 3, I.e the top quark (and bottom and charm) does not even exist in the theory in this region. Hence it becomes less clear what its mass and other properties mean. $\endgroup$ – Your Majesty Feb 22 '16 at 8:16
  • $\begingroup$ @Faq Thanks for your comment! Yes of course the top gets integrated out at low energies and it is unclear what its mass means at low energies. Thus I asked a new question physics.stackexchange.com/questions/238989/… assuming a stable top, which therefore would be present at low energies. $\endgroup$ – jak Feb 22 '16 at 8:30
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Here thinking about effective theory approach and decoupling unnecessarily complicates the issue. Just think about full standard model lagrangian. Top quark mass is just another running/sliding parameter, depending on a
referent scale $\mu$. The choice of $\mu$ is free, and you are allowed to take it $\mu = 2\,{\rm GeV}$, and $m_t(\mu)$ makes perfect sense even then. Note, that $\mu$ is just a referent scale used to define renormalization procedure and, in principle, nothing should depend on its value.

In practice, you would keep the $\mu$ close to $m_t$ to avoid the appearance of large logarithms (which is the main reason to introduce running parameters in the first place), but you are not forbidden to define, use, and calculate with $m_t(2\,{\rm GeV}$.

Also note that in quantum electrodynamics, you can work with renormalization scale $\mu = 0$ and electron mass is just fine with that.

In practice, at low scales, you would work within effective theory, decouple top quark, which then gets absorbed in other parameters, and doesn't exist any more in the theory, so then trivially its mass also becomes meaningless.

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    $\begingroup$ Thanks a lot for your answer! There is one thing that confuses me a bit about the omnipresent statement "$\mu$ can be chosen arbitrarily": In Grand Unified Theories, such as $SU(5)$ a famous result is that "at the $SU(5)$ scale" we have $m_b (SU(5))= m_\tau(SU(5))$, which becomes through the RGE running of the Yukawa couplings at the "electroweak scale" $m_b (M_Z)= 3 m_\tau(M_Z)$. Now, if we compute the decay of a bottom at a very high scale, we get that it cannot decay into a tau because $m_b \approx m_\tau$. However if we choose a low renormalization scale the decay is easily possible... $\endgroup$ – jak Feb 24 '16 at 9:23
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    $\begingroup$ Note that for the kinematics of the decay it is not the running mass that is relevant, but a pole mass, which doesn't depend on scale, and is what would be usually quoted as the physical mass of the particle. (Here I ignore the confinement of quarks which makes things more coplicated.) Also note that it doesn't make sense to talk about "decay at some scale". You can always go to the rest frame of the particle and then it decays at rest and scale zero. $\endgroup$ – kkumer Feb 25 '16 at 9:49
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    $\begingroup$ Also, I think that some of your confusion comes from the fact that renormalization scale is usually set to be equal to the typical scale of the process to make the problematic logarithms small, but physical scale $Q$ of the process and arbitrary renormalization scale $\mu$ are in general two different things. $\endgroup$ – kkumer Feb 25 '16 at 9:49
  • $\begingroup$ Wow thanks, that's very helpful! Two little things: Given a Yukawa coupling as a function of scale $\mu$ and a vev, $m_R(\mu)= Y(\mu) \langle \phi \rangle$, how can I compute the corresponding pole mass? A relation I was able to find is $m_p = m_R(0) − Σ(m_P )$ (page 39 in lup.lub.lu.se/luur/…), where $m_p$ is the pole mass and $ m_R(0)$ the running mass at $\mu=0$. Thus essentially $m_p \approx m_R(0)$, assuming $ Σ(m_P )$ is a small correction? $\endgroup$ – jak Feb 25 '16 at 12:14
  • $\begingroup$ And secondly, I always though that the rest frame meant for a decay $\mu = M$, where $M$ is the mass of the decaying particle. However you wrote the rest frame corresponds to scale zero? Do you have any reference for this? $\endgroup$ – jak Feb 25 '16 at 12:15
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An answer can be found at page 176 of Srednicki's book:

"If the particle mass is nonzero, this process stops at $µ ∼ m$. This is because the minimum value of $s$ is $4m^2$, and so the factor of $ln( \frac{s}{µ^2} )$ becomes an unwanted large $log$ for $µ ≪ m$. We should therefore not use values of $µ$ below $m$.

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  • $\begingroup$ Doesn’t it mean we should somehow resum the large logs? $\endgroup$ – innisfree Nov 18 '16 at 13:32

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