\subsubsection{Reparametrization Invariance and Residual Mass Term}
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Full QCD has Lorentz invariance as well as chiral symmetry, which ensures
that no mass renormalization can occur in the massless limit. 
We shall discuss how much of these invariances survives once we go to 
the heavy mass limit and integrate out the small component fields. 

Lorentz invariance as it is usually implemented in a quantum field 
theory is not present any more in the heavy mass limit, since the choice
of the vector $v$ corresponds to a choice of a specific coordinate frame. 
Only if we would transform the vector $v$ ``by hand'' we would obtain 
again invariance. Another way of dealing with this problem is to 
``integrate in'' more degrees of freedom by summing over all possible 
velocities as suggested in \cite{GeorgiHQET} 

On the other hand, when we started from the full QCD Lagrangian, there 
was no dependence on the vector $v$ and even when $v$ appeared explicitly
in the equations everything was still independent of $v$ as long as all 
orders in $1/m_Q$ were taken into account. In other words, a dependence on 
$v$ emerges at that point when we truncate the $1/m_Q$ expansion. 

Hence we expect that the $1/m_Q$ expansion exhibits an invariance under 
infinitesimal changes of the vector $v$ and that this so called 
reparametrization invariance \cite{repara} results in relations between 
coefficients of different orders in the $1/m_Q$ expansion. 
 
Full QCD depends only on the momentum of 
the heavy quark $P$, and the splitting of this momentum in an on-shell 
part $m_Q v$ and a residual momentum $k$ corresponding to the covariant 
derivative acting on the heavy static field $h_v$ is arbitrary. Formally 
this means that the Lagrangian does not depend on the velocity $v$, if 
all orders of the $1/m_Q$ expansion are included. 

In order to explore the consequences of reparametrization invariance 
it is convenient to use  the representation (\ref{HarvLag}) and 
(\ref{HarvField}), since we may obtain closed expressions for all orders 
in the $1/m_Q$ expansion. In fact performing an infinitesimal shift of 
the velocity, combined with the corresponding shift of the covariant 
derivative and the fields one explicitly checks that 
the Lagrangian is invariant under the transformation 
\begin{eqnarray} \label{repara}
&& v \to v + \delta v \qquad v \cdot \delta v = 0 \\
&& h_v \to h_v + \frac{\delta v \hskip -0.75em / \hskip 0.25em }{2} 
\left(1+ P_- \frac{1}{2m_Q + ivD} i \fmslash{D} \right) h_v \nonumber \\
&& iD \to iD - m_Q \, \delta v \nonumber .
\end{eqnarray}
This invariance is the so-called reparametrization invariance, 
which has non-trivial consequences, since it relates terms of different 
orders of the $1/m_Q$ expansion. 

One of the key points is that reparametrization invariance in fact 
survives renormalization, which means that the relations between 
coefficients of the Lagrangian as derived from (\ref{repara}) hold 
true beyond tree level. As we shall see when we discuss QCD radiative 
corrections it is reparametrization invariance which implies that 
the kinetic energy operator $\bar{h}_v (iD)^2 h_v$ is not renormalized 
to all orders and furthermore one may extract relations between the 
renormalization constants of the second order terms in the Lagrangian. 

Another point related to the way the heavy mass limit is taken is the 
so called residual mass term \cite{ResMass}. 
The mass of a quark is not a physical 
quantity and one has in principle some freedom of how to define the 
heavy quark mass. In the derivation of the heavy mass limit from QCD 
we have chosen a phase in such a way that the mass term (with the mass 
as it appears as a parameter in the Lagrangian) is compensated. However, 
one could as well choose a different mass $m_Q^\prime = m_Q + \delta m_Q$ 
and split the momentum of the heavy quark using $m_Q^\prime$
\begin{equation} 
p_Q = m_Q^\prime v + k^\prime 
    = (m_Q - \delta m_Q ) v + (k + \delta m_Q v) 
\end{equation}
The heavy quark fields obtained from the two different definitions 
$m_Q$ and $m_Q^\prime$ differ by a phase factor, such that
\begin{equation}
h_v = e^{i \delta m_Q vx} h_v^\prime \mbox{ and }
iD h_v = e^{i \delta m_Q vx} (iD - \delta m_Q v) h_v^\prime
\end{equation}
The Lagrangian may be expressed in terms of the field $h_v^\prime$ and 
reads to leading order
\begin{equation}
{\cal L} = \bar{h}^\prime_v v(iD - \delta m_Q v) h_v 
\end{equation}
yielding an equation of motion of the form
\begin{equation}
(iD - \delta m_Q v) h_v^\prime = 0
\end{equation}
The quark mass is not a physical quantity, in contrast to the mass of a 
meson. In the heavy mass limit, the mass of a hadron is given by
\begin{equation}
M_H = m_Q + \bar\Lambda 
\end{equation}
and so a redefinition of the heavy quark mass by inserting a residual mass term
amounts to a redefinition of $\bar\Lambda$. The appearance of a residual mass
term is connected to the appearance of renormalon ambiguities in the 
definition 
of the heavy quark mass \cite{renormalons}, 
and the question of which mass to chose is 
a matter of 
convention. As we shall see below, a particular choice (as e.g. the 
pole mass or
a short distance definition of the mass such as the $\overline{\rm MS}$-mass) 
does not change the predictions of the heavy mass expansion as far as physical 
quantities are concerned, since the nonperturbative matrix elements parameterizing
the correction terms to order $1/m_Q$ and higher will depend on the definition
of the mass, in such a way that the net result is independent of the definition 
chosen for the mass.   
