Confusion in notation of Lorentz Transformations

[Hamiltonian]

$\bar{\mathcal{O}}$ is moving with a velocity $v$ relative to $\mathcal{O}$ along $x^{1}$
The Lorentz transformations between a Frame $\mathcal{O}$ and $\bar{\mathcal{O}}$ is given by:
$$\Delta x^{\bar{0}} = \gamma\left(\Delta x^0 - v\Delta x^1\right)$$
$$\Delta x^{\bar{1}} = \gamma\left(\Delta x^1 - v\Delta x^0 \right)$$
$$\Delta x^{\bar{2}} = \Delta x^{2}$$
$$\Delta x^{\bar{3}} = \Delta x^{3}$$

where $\gamma = 1/\sqrt{1- v^2}$
and $\mu \in [0, 1, 2, 3]$

In MIT 8.962, the prof. writes the Lorentz transformations as:
$$\Delta x^{\bar{\mu}} = \sum_{\nu = 0}^{3} \wedge_{\nu}^{\bar{\mu}} \Delta x^{\mu}$$

what does $\wedge_{\nu}^{\bar{\mu}}$ represent and how is the above equation equivalent to the Lorentz transformations?

 

[DrGreg]

$\Lambda_{\nu}^{\bar{\mu}}$ represents the 16 components of the matrix$$
\Lambda = \begin{pmatrix}
\gamma & -\gamma v & 0 & 0 \\
-\gamma v & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 &0 & 0 & 1
\end{pmatrix}
$$

 

[Ibix]

It's a capital \Lambda, not a wedge. He's just converted a system of linear equations into a matrix form (which Dr Greg beat me to writing out). If you write the coordinate differences as column vectors, you'll find that $\vec{\Delta x'}=\Lambda\vec {\Delta x}$, and that the prof's notation is equivalent to that matrix multiplication, which is equivalent to your four equations.

 

[Hamiltonian]

It's a capital \Lambda, not a wedge. He's just converted a system of linear equations into a matrix form (which Dr Greg beat me to writing out). If you write the coordinate differences as column vectors, you'll find that $\vec{\Delta x'}=\Lambda\vec {\Delta x}$, and that the prof's notation is equivalent to that matrix multiplication, which is equivalent to your four equations.

The prof. in his second lecture says that its very obvious to think of $\Lambda^{\bar{\mu}}_{\nu}$ as a matrix and the coordinate differences as column vectors, but we should going forward(in our studies of GR) not think of them as that. I didn't understand why he said this?

 

[Ibix]

I would have said that the Lorentz transform is a special case of the Jacobian matrix, to be honest. But it's true that vectors (and tensors, more generally) in differential geometry don't obey the same rules of grammar as matrices do. For example, it's tempting to say that a two index tensor like the metric, $g_{ab}$, is a 4×4 tensor. But it's perfectly legal to write $g_{ab}v^au^b$ (where sums over $a$ and $b$ are implied) and if you try to write that as a matrix equation it's illegal.

You can work around it with a sufficiently flexible definition of "matrix", but it's certainly not as simple as you might guess.

 

[PeroK]

The prof. in his second lecture says that its very obvious to think of $\Lambda^{\bar{\mu}}_{\nu}$ as a matrix and the coordinate differences as column vectors, but we should going forward(in our studies of GR) not think of them as that. I didn't understand why he said this?

A Lorentz Transformation, technically, has a physical description relating to a transformation from one IRF to another.

The Lorentz Transformations form a mathematical group, called the Lorentz Group. This means essentially that the composition of two Lorentz Transformations is another Lorentz transformation. I.e. if $L_1$ and $L_2$ are Lorentz Transformations, then so is $L_2 \circ L_1$. Note that this is saying something important physically about the nature of inertial frames and inertial motion. Think about it!

If we stick with Cartesian coordinates and a certain convention for alignment of axes between frames, then each Lorentz Transformation may be written as a 4x4 matrix. This is more generally called a representation of the group. I think it's too early in your studies to be worried about this. That said, group representations appear all over physics so you will eventually have to grasp the concept. That's probably why your Professor cautioned about getting too attached to one particular representation of the Lorentz Group.

In this case, we have a representation of the Lorentz Group as a set of 4x4 matrices with the following property:

$\Lambda$ is a Lorentz Transformation if
$$\eta_{ab}\Lambda^a_c \Lambda^b_d = \eta_{cd}$$Where $\eta$ is the diagonal matrix
$$
\eta = \begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 &0 & 0 & 1
\end{pmatrix}
$$This means that a Lorentz Transformation leaves the metric unchanged. Note that an alternative, equivalent definition is that$$\Lambda^T\eta \Lambda = \eta$$An exercise, if you want, is to show that the example of a Lorentz Transformation (boost of $v$ in the x-direction) given by @DrGreg above satisfies this definition.

PS and, if you want a bit of practice at abstract algebra, you could show that this does indeed define a group of matrices.

 

[cianfa72]

Just to avoid confusion maybe it would be better to use two different symbols for the same Lorentz transformation: ${\Lambda^{a}}_b$ for index notation and $\hat \Lambda$ for matrix notation.