Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Review: Cauchy Tetrahedron.

Referring to figure (\ref{fig:continuumL5:continuumL5fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig1}
\caption{Cauchy tetrahedron direction cosines.}
\end{figure}

recall that we can decompose our force into components that refer to our direction cosines

Or in tensor form

We call this the traction vector and denote it in vector form as

Constitutive relation.

Reading: section 2, section 4 and section 5 from the text [1].

We can find the relationship between stress and strain, both analytically and experimentally, and call this the Constitutive relation. We prefer to deal with ranges of distortion that are small enough that we can make a linear approximation for this relation. In general such a linear relationship takes the form

Consider the number of components that we are talking about for various rank tensors

latex 0^\text{th}$ rank tensor} & \mbox{ components} \\ \mbox{ rank tensor} & \mbox{ components} \\ \mbox{ rank tensor} & \mbox{ components} \\ \mbox{ rank tensor} & \mbox{ components}\end{array}\end{aligned} \hspace{\stretch{1}}(3.7)$

We have a lot of components, even for a linear relation between stress and strain. For isotropic materials we model the constitutive relation instead as

For such a modeling of the material the (measured) values and (shear modulus or modulus of rigidity) are called the Lam\’e parameters.

It will be useful to compute the trace of the stress tensor in the form of the constitutive relation for the isotropic model. We find

or

We can now also invert this, to find the trace of the strain tensor in terms of the stress tensor

Substituting back into our original relationship 3.8, and find

which finally provides an inverted expression with the strain tensor expressed in terms of the stress tensor

Special cases.

Hydrostatic compression

Hydrostatic compression is when we have no shear stress, only normal components of the stress matrix is nonzero. Strictly speaking we define Hydrostatic compression as

i.e. not only diagonal, but with all the components of the stress tensor equal.

We can write the trace of the stress tensor as

Now, from our discussion of the strain tensor recall that we found in the limit

allowing us to express the change in volume relative to the original volume in terms of the strain trace

Writing that relative volume difference as we find

or

where is called the Bulk modulus.

Uniaxial stress

Again illustrated in the plane as in figure (\ref{fig:continuumL5:continuumL5fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig2}
\caption{Uniaxial stress.}
\end{figure}

Expanding out 3.12 we have for the element of the strain tensor

or

where is Young’s modulus. Young’s modulus in the text (5.3) is given in terms of the bulk modulus . Using we find

FIXME: figure (\ref{fig:continuumL5:continuumL5fig3}) reference?

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig3}
\caption{stress associated with Young’s modulus}
\end{figure}

We define Poisson’s ratio as the quantity

Note that we are still talking about uniaxial stress here. Referring back to 3.12 we have

Recall (3.20) that we had

Inserting this gives us

so

We can also relate the Poisson’s ratio to the shear modulus

These ones are (5.14) in the text, and are easy enough to verify (not done here).

Appendix. Computing the relation between Poisson’s ratio and shear modulus.

Young’s modulus is given in 3.21 (equation (43) in the Professor’s notes) as

and for Poisson’s ratio 3.24 (equation (46) in the Professor’s notes) we have

Let’s derive the other stated relationships (equation (47) in the Professor’s notes). I get

or

For substitution into the Young’s modulus equation calculate

and

Putting these together we find

Rearranging we have

This matches (5.9) in the text (where is used instead of ).

We also find

References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

Motivation.

In a classical mechanics lecture (which I audited) Prof. Poppitz made the claim that an infinitesimal rotation in direction of magnitude has the form

where

I believe he expressed things in terms of the differential displacement

This was verified for the special case and . Let’s derive this in the general case too.

With geometric algebra.

Let’s temporarily dispense with the normal notation and introduce two perpendicular unit vectors , and in the plane of the rotation. Relate these to the unit normal with

A rotation through an angle (infinitesimal or otherwise) is then

Suppose that we decompose into components in the plane and in the direction of the normal . We have

The exponentials commute with the vector, and anticommute otherwise, leaving us with

In the last line we use and . Making the angle infinitesimal we have

We have only to confirm that this matches the assumed cross product representation

Taking the two last computations we find

as desired.

Without geometric algebra.

We’ve also done the setup above to verify this result without GA. Here we wish to apply the rotation to the coordinate vector of in the basis which gives us

But as we’ve shown, this last coordinate vector is just , and we get our desired result using plain old fashioned matrix algebra as well.

Really the only difference between this and what was done in class is that there’s no assumption here that .

I’d be curious to know what the rationale for that choice was? It seems reasonable to me to be able to use an instruction like this to do an atomic set bit or clear bit operation.

“what exactly has the kinetic energy of the second rocket’s two burns been transformed into?”

My response to my dad was:

There’s lots of things that the burns have been turned into. He demonstrates an impressive lack of understanding of not only physics, but chemistry. In the chemical reactions that expel the exhaust particles, there’s lots of heat and light produced, both of which carry out energy from the ship. The other thing that is glaringly obvious is that he doesn’t even consider the exhaust itself.

Rockets are a very primitive devices and they actually propel themselves by tossing bits of themselves out the back. You could do this if you could put yourself on a very frictionless surface, along with a very heavy mass. If you toss the mass in one direction, you’ll end up moving in the other direction because momentum of you plus the mass have to be conserved. Perhaps you could actually do this experiment. Get a big rock and a dolly and put the dolly on ice with you and the rock, and then toss the rock. You should move in the opposite direction. Rockets expel most of their mass doing this, with chemical reactions making that expelled mass move much faster so that they gain they momentum (in the opposite direction) of the exhaust they expel. It’s the rocket, plus its exhaust, plus all the light and heat changes that occur in the chemical reactions that have to be considered if you are looking at conservation of energy.

Basically, the guy who authored this page appears to be talking out of his ass.  I’d like to try the experiment that I outlined for my dad.  Since it is winter time, perhaps I could just put on my ice skates and find something big to throw in front of me.  I should move backwards a bit in response.

http://mathematica.stackexchange.com/

Whether or not it ends up as a permanent stackexchange site will depend on what sort of usage it gets, so if you are an active mathematica users, here’s a chance to build up a useful community.

Spherical tensor.

To perform the derivation in spherical coordinates we have some setup to do first, since we need explicit representations of all three unit vectors. The radial vector we can get easily by geometry and find the usual

We can get by geometrical intuition since it the plane unit vector at angle rotated by . That is

We can get by utilizing the right handedness of the coordinates since

and find

That and some Mathematica brute force can be used to calculate the differential strain element, and we find

A manual derivation.

Doing the calculation pretty much completely with Mathematica is rather unsatisfying. To set up for it let’s first compute the unit vectors from scratch. I’ll use geometric algebra to do this calculation. Consider figure (\ref{fig:qmTwoExamReflection:continuumL2fig5})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig5}
\caption{Composite rotations for spherical polar unit vectors.}
\end{figure}

We have two sets of rotations, the first is a rotation about the axis by . Writing for the unit bivector in the plane, we rotate

Now we rotate in the plane spanned by and by . With , our vectors in the plane rotate as

(with since ).

Now, these are all the same relations that we could find with coordinate algebra

There’s nothing special in this approach if that is as far as we go, but we can put things in a nice tidy form for computation of the differentials of the unit vectors. Introducing the unit pseudoscalar we can write these in a compact exponential form.

To summarize we have

Taking differentials we find first

Summarizing these differentials we have

A final cleanup is required. While is a vector and has a nicely compact form, we need to decompose this into components in the , and directions. Taking scalar products we have

Summarizing once again, but this time in terms of , and we have

Now we are set to take differentials. With

we have

Squaring this we get the usual spherical polar line scalar line element

With

our differential is

We can add to this and take differences

For each we have

and plugging through that calculation is really all it takes to derive the textbook result. To do this to first order in , we find

Collecting terms we have the result of the text in the braces

It should be possible to do the calculation to second order too, but to include all the quadratic terms in is again really messy. Trying that with mathematica gives the same results as above using the strictly coordinate algebra approach.

References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Stress tensor.

Reading: Portions of this lecture cover section 2 from the text [1].

For the stress tensor

a second rank tensor, the first index defines the direction of the force, and the second index defines the surface.

Observe that the dimensions of is force per unit area, just like pressure. We will in fact show that this tensor is akin to the pressure, and the diagonalized components of this tensor represent the pressure.

We’ve illustrated the stress tensor in a couple of 2D examples. The first we call uniaxial stress, having just the element of the matrix as illustrated in figure (\ref{fig:continuumL4:continuumL4fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig1}
\caption{Uniaxial stress}
\end{figure}

A biaxial stress is illustrated in figure (\ref{fig:continuumL4:continuumL4fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig2}
\caption{Biaxial stress.}
\end{figure}

where for our tensor takes the form

In the general case we have

We can attempt to illustrate this, but it becomes much harder to visualize as shown in figure (\ref{fig:continuumL4:continuumL4fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig3}
\caption{General strain}
\end{figure}

In equilibrium we must have

We can use similar arguments to show that the stress tensor is symmetric.

In 3D we have three components of the stress tensor acting on each surface, as illustrated in figure (\ref{fig:continuumL4:continuumL4fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig5}
\caption{Strain components on a 3D volume.}
\end{figure}

We have three unique surface orientations and three components of the force for each of these, resulting in nine components, but these are not all independent. For an object in equilibrium we must have (FIXME: justify?). Explicitly, that is

Diagonalization

We’ll look at the two dimensional case in some detail, as in figure (\ref{fig:continuumL4:continuumL4fig6})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig6}
\caption{Area element under strain with and without rotation.}
\end{figure}

Under this coordinate transformation, a rotation, the diagonal stress tensor is taken to a non-diagonal form

How do the stress tensor and the force relate

We form a Cauchy tetrahedron as in figure (\ref{fig:continuumL4:continuumL4fig7})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig7}
\caption{Cauchy tetrahedron}
\end{figure}

We write in terms of the direction cosines

Here

or

Force balance on direction, matching total external force in this direction to the total internal force () as follows

Similarily

and

We can therefore write these force components like

but these ratios are really just the projections of the areas as illustrated in figure (\ref{fig:continuumL4:continuumL4fig8})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL4fig8}
\caption{Area projection.}
\end{figure}

where an arbitrary surface with area can be decomposed into projections

utilizing the direction cosines. We can therefore write

or in matrix notation

This is just

This force with components is also called the traction vector

References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. Physics Today, 13:44, 1960.

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Introduction.

Mechanics could be defined as the study of effects of forces and displacements on a physical body

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig1}
\caption{Physical body.}
\end{figure}

In continuum mechanics we have a physical body and we are interested in the internal motions in the object.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig2}
\caption{Control volume elements.}
\end{figure}

For the first time considering mechanics we have to introduce the concepts of fields to make progress tackling these problems.

We will have use of the following types of fields

\begin{itemize}
\item Scalar fields. components. Examples: density, Temperature, …
\item Vector fields. components. Examples: Force, velocity.
\item Tensor fields. components. Examples: stress, strain.
\end{itemize}

We have to consider objects (a control volume) that is small enough that we can consider that we have a point in space limit for the quantities of density and velocity. At the same time we cannot take this limiting process to the extreme, since if we use a control volume that is sufficiently small, quantum and inter-atomic effects would have to be considered.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig3}
\caption{Mass and volume ratios at different scales.}
\end{figure}

Stress and Strain definitions.

\begin{definition}
\emph{(Stress)}

Measure of the Internal force on the surfaces.
\end{definition}

\begin{definition}
\emph{(Strain)}

Measure of the deformation of the body.
\end{definition}

Strain Tensor.

This follows [1] section 1 very closely.

Utilizing summation convention consider a set of small internal displacements to the coordinates so that the transformation is related by

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig4}
\caption{Differential change to the object.}
\end{figure}

(ie: is a function of all the initial coordinates, as are the displacements ).

or

with

we have

We write

where we define the \emph{strain tensor} as

Here is a matrix in Cartesian coordinates

We see from 3.11 that is symmetric, so we have

Because any real symmetric matrix can be diagonalized we can write in some coordinate system

If our changes are small enough we can also write approximately, taking the first order term in the square root evaluation

We are also free to define a volume element

So the change of volume is given by the trace

Strain Tensor in cylindrical coordinates.

At the end of the section in the text, the formulas for the spherical and cylindrical versions (to first order) of the strain tensor is given without derivation. Let’s do that derivation for the cylindrical case, which is simpler. It appears that use of explicit vector notation is helpful here, so we write

where

Since and are functions of position, we will need their differentials

but these are just scaled basis vectors

So for our and differentials we find

and

Putting these together we have

For the squared magnitude’s difference from we have

Expanding this out, but dropping all the terms that are quadratic in the components of or its differentials, we have

Grouping all terms, with all the second order terms neglected, we have

From this we can read off the result quoted in the text

Observe that we have to introduce factors of along with all the ‘s, when we factored out the tensor components. That’s an important looking detail, which isn’t obvious unless one works through the derivation.

Note that in class we retained the second order terms. That becomes a messier calculation and I’ve cheated using the symbolic capabilities of mathematica to do it

As with the first order case, we can read off the tensor coordinates by inspection (once we factor out the various factors of and ). The next logical step would be to do the spherical tensor calculation. That would likely be particularily messy if we attempted it in the brute force fashion. Let’s step back and look at the general case, before tackling there sphereical polar form explicitly.

Strain Tensor for general coordinate representation.

Now let’s dispense with the assumption that we have an orthonormal frame. Given an arbitrary, not neccessarily orthonormal, position dependent frame , and its reciprocal frame , as defined by

Our coordinate representation, with summation and dimensionality implied, is

Our differentials are

and

Summing these we have

Taking dot products to form the squares we have

and

Taking the difference we find

To evaluate this, it is useful, albeit messier, to group terms a bit

Here is used to denote summation over the pairs just once, not neccessarily any numeric ordering. For example with , this could be the set .

Cartesian tensor.

In the Cartesian case all the partials of the unit vectors are zero, and we also have no need of upper or lower indexes. We are left with just

However, since we also have , this is

This essentially recovers the result 3.11 derived in class.

Cylindrial tensor.

Now lets do the cylindrical tensor again, but this time without resorting mathematica brute force.

First we recall that all our basis vector derivatives are zero except for the derivatives, and for those we have

If we write

We have for all the partials

latex \alpha = x^\mu = r$ or } \\ 0 & \quad \mbox{otherwise}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(3.56)$

We are now set to evaluate the terms in the sum of 3.50 for the cylindrical coordinate system and shouldn’t need Mathematica to do it. Let’s do this one at a time, starting with all the squared differential pairs. Those are, for the value of

For both and all our unit vectors have zero derivatives so we are left respectively with

and

For the term we have

Now, on to the mixed terms. The easiest is the term, for which all the unit vector derivatives are zero, and we are left with just

Now we have the two messy mixed terms. For the , term we get

Finally for the , term we have

To summarize we have, including both first and second order terms,

Factors of have been pulled out so that the portions remaining in the braces are exactly the cylindrical tensor elements as given in the text (except also with the second order terms here). Observe that the pre-calculation of the general formula has allowed an on paper expansion of the cylindrical tensor without too much pain, and this time without requiring Mathematica.

Spherical tensor.

FIXME: TODO.

References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. Physics Today, 13:44, 1960.

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Strain.

Strain is the measure of stretching. This is illustrated pictorially in figure (\ref{fig:continuumL3:continuumL3fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig1}
\caption{Stretched line elements.}
\end{figure}

where is the strain tensor. We found

Why do we have a factor two? Observe that if the deformation is small we can write

so that we find

Suppose for example, that we have a diagonalized strain tensor, then we find

so that

Observe that here again we see this factor of two.

If we have a diagonalized strain tensor, the tensor is of the form

we have

so

Observe that the change in the volume element becomes the trace

How do we use this? Suppose that you are given a strain tensor. This should allow you to compute the stretch in any given direction.

FIXME: find problem and try this.

Stress tensor.

Reading for this section is section 2 from the text associated with the prepared notes [1].

We’d like to consider a macroscopic model that contains the net effects of all the internal forces in the object as depicted in figure (\ref{fig:continuumL3:continuumL3fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig2}
\caption{Internal forces.}
\end{figure}

We will consider a volume big enough that we won’t have to consider the individual atomic interactions, only the average effects of those interactions. Will will look at the force per unit volume on a differential volume element

The total force on the body is

where is the force per unit volume. We will evaluate this by utilizing the divergence theorem. Recall that this was

We have a small problem, since we have a non-divergence expression of the force here, and it is not immediately obvious that we can apply the divergence theorem. We can deal with this by assuming that we can find a vector valued tensor, so that if we take the divergence of this tensor, we end up with the force. We introduce the quantity

and require this to be a vector. We can then apply the divergence theorem

where is a surface element. We identify this tensor

and

as the force on the surface element . In two dimensions this is illustrated in the following figures (\ref{fig:continuumL3:continuumL3fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig3}
\caption{2D strain tensor.}
\end{figure}

Observe that we use the index above as the direction of the force, and index as the direction normal to the surface.

Note that the strain tensor has the matrix form

We will show later that this tensor is in fact symmetric.

FIXME: given some 3D forces, compute the stress tensor that is associated with it.

Examples of the stress tensor

Example 1. stretch in two opposing directions.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig4}
\caption{Opposing stresses in one direction.}
\end{figure}

Here, as illustrated in figure (\ref{fig:continuumL3:continuumL3fig4}), the associated (2D) stress tensor takes the simple form

Example 2. stretch in a pair of mutually perpendicular directions

For a pair of perpendicular forces applied in two dimensions, as illustrated in figure (\ref{fig:continuumL3:continuumL3fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig5}
\caption{Mutually perpendicular forces}
\end{figure}

our stress tensor now just takes the form

It’s easy to imagine now how to get some more general stress tensors, should we make a change of basis that rotates our frame.

Example 3. radial stretch

Suppose we have a fire fighter’s safety net, used to catch somebody jumping from a burning building (do they ever do that outside of movies?), as in figure (\ref{fig:continuumL3:continuumL3fig6}). Each of the firefighters contributes to the stretch.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig6}
\caption{Radial forces.}
\end{figure}

FIXME: what form would the tensor take for this? Would we have to use a radial form of the tensor? What would that be?

References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. Physics Today, 13:44, 1960.

I went looking for a decompiler for Linux amd64 code. I remember once using IDA pro in an ethical hacking course I took, but that’s not available in IBM’s standard licensed software offerings I can ask for. There’s a couple of free decompilers that I found listed on wikipedia. The only one that appeared to have any sort of amd64 support looked like it was backerstreet’s reverse engineering compiler. I happened to have a ubuntu VM around that I tried this on, but it crashed even on a 32-bit executable (tried command prompt: ‘open /bin/bash’), so I don’t really expect it to behave better on a cross target executable.

Since the code in question is not too big (~300 lines of disassembly), I was wondering if I could hack together something that at least removed all the addresses from the disassembly that weren’t jump targets.

Another thing that was required to make the disassembly sensible was the use of the linked output, nor just the .o files, as the objdump source. Otherwise I end up with all the function calls left unresolved, like:

     272:       e8 00 00 00 00          callq  277
     277:       48 83 c4 20             add    $0x20,%rsp

So, from the shared lib, I ran ‘objdump -d –no-show-raw-insn’ and filtered that with a simpler re-labler

#!/usr/bin/perl

my @lines = () ;
my %addressMap = () ;
my $lableCount = 0 ;

while (<>)
{
   chomp ;

   if ( /\tj\S+\s+(\S+)/ )
   {
      unless ( defined $addressMap{$1} )
      {
         $addressMap{$1} = "L$lableCount" ;

         $lableCount++ ;
      }
   }

   push( @lines, $_ ) ;
}

my @addrs = ( keys %addressMap ) ;

foreach my $line (@lines)
{
   foreach ( @addrs )
   {
      $line =~ s/\t(j\S+)  # example: <tab>je
                 \s+
                 $_
                 \s.*
                /printf("\t%-6s $addressMap{$_}", $1)/xe ;

      $line =~ s/^ $_:/ $addressMap{$_}:/ ;
   }

   $line =~ s/^ [0-9a-f]+:// ;

   print "$line\n" ;
}

So, now instead of a mess like:

 3639ff3:       test   %rbp,%rbp
 3639ff6:       mov    (%rax),%r14
 3639ff9:       je     363a02a
 3639ffb:       mov    0x788(%r14),%rbx
 363a002:       test   %rbx,%rbx
 363a005:       je     363a02c
 363a007:       mov    $0x1,%edi

I get something like:

   test   %rbp,%rbp
   mov    (%rax),%r14
   je     L21
   mov    0x788(%r14),%rbx
   test   %rbx,%rbx
   je     L31
   mov    $0x1,%edi

(with the lables in the jump targets also renamed, and retained).

The next logical step would be to implement a register renamer. Since I’ve now got all the basic blocks identified, it should be possible to figure out any time a general purpose register is clobbered and give the register a new name at any clobber point. For instance in the following BB these two pairs of rdx variables are logically different:

   mov    0x128(%rsp),%rdx
   inc    %rdx
   test   %rbx,%rbx
   je     L191
   mov    %rdx,0x128(%rsp)
   mov    0x78(%rbx),%rdi
   test   %rdi,%rdi
   je     L191
   mov    0xc88(%rdi),%rsi
   test   %rsi,%rsi
   je     L191
   decq   0xcb0(%rdi)
   mov    0x78(%rbx),%r8
   mov    0xcb0(%r8),%rdx
   test   %rdx,%rdx

The first rdx use above could be renamed without trouble since it is clobbered in the same BB, so you know that its use is purely local to that block. However, to rename registers intelligently in general you’d also have to identify what register dependencies exist between the basic blocks.

Identifying the dependencies would be extra messy on amd64 since we have different aliases for the same registers too, depending on the size of the access to the register (ie: rdx, edx, dx, …)

In the end I’ve come to the conclusion that taking this any further would really be too much work. Perhaps I can spot a difference just by inspection. Without some preprocessing the assembly is fairly hard to read though.