Exploring Lorentz Transformation Matrix: A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}

In summary, the conversation discusses the importance of the Lorentz transformation matrix, which is symmetric and has a trace of 2\gamma+2. It is also shown how this matrix relates to four-vectors and antisymmetric tensors. The participant has questions about the trace and diagonalizing the matrix, and it is explained that the trace is important for conserving volume in spacetime and the determinant must equal 1. The easiest way to diagonalize the matrix is also discussed, with the final result being a diagonal form with the trace and determinant remaining the same.
  • #1
Petar Mali
290
0
[tex]\alpha=\left(\begin{array}{cccc}
\gamma& 0&0& -\beta\gamma\\
0&1& 0 & 0\\
0 & 0 & 1 & 0\\
-\beta\gamma & 0 & 0 & \gamma \end{array} \right)[/tex][tex]x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}[/tex]

[tex]\alpha[/tex] is Lorrentz transformation matrix. Can I see something more about it? . It's symmetric. That is important.

[tex]Tr(\alpha)=2\gamma +2[/tex]

Is this value of trace important for something?

We can also say for four vectors

[tex]A'^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\nu}}A^{\nu}[/tex]

using this relation and

[tex]x'^{\mu}=\alpha^{\mu}_{\nu} x^{\nu}[/tex]

we get

[tex]A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}[/tex]

So

[tex]A'^{1}=\alpha^{1}_{\nu}A^{\nu}=\alpha^{1}_{1}A^{1}+\alpha^{1}_{4}A^{4}=\gamma A^{1}-\beta\gamma A^4=\gamma(A^1-\beta A^4)[/tex]

[tex]A'^{2}=\alpha^{2}_{\nu}A^{\nu}=A^2[/tex]

[tex]A'^{3}=A^3[/tex]

[tex]A'^{4}=\alpha^{4}_{\nu}A^{\nu}=\alpha^{4}_{1}A^{1}+\alpha^{4}_{4}A^{4}=-\beta\gamma A^{1}+\gamma A^4=\gamma(A^4-\beta A^1)[/tex]

Look now in antisymmetric tensors

[tex]A^{\nu\mu}=-A^{\mu\nu}[/tex]

[tex]A'^{12}=\alpha^1_{\rho}\alpha^2_{\sigma}A^{\rho\sigma}[/tex]

Nonzero terms are terms in which [tex]\rho[/tex] takes values [tex]1,4[/tex] and [tex]\sigma[/tex] takes [tex]2[/tex].

I have a question for component

[tex]A'^{14}=\alpha^1_{\rho}\alpha^4_{\sigma}A^{\rho\sigma}=\alpha^1_{1}\alpha^4_{1}A^{11}+\alpha^1_{1}\alpha^4_{4}A^{44}+\alpha^1_{4}\alpha^4_{4}A^{44}+\alpha^1_{4}\alpha^4_{1}A^{41}=-\beta\gamma^2A^{11}+\gamma^2A^{14}-\beta^2\gamma^2A^{14}-\beta\gamma^2A^{44}[/tex]

Can I say if I look antisymmetric tensors [tex]A^{11}=A^{44}=0[/tex]?

Then I will get

[tex]A'^{14}=A^{14}[/tex]

Thanks for your answers!
 
Last edited:
Physics news on Phys.org
  • #2
Interesting question about the trace.

The determinant does have a nice interpretation. The determinant equals one, and this means that the Lorentz transformation conserves volume in spacetime. This is plays a prominent role in my favorite derivation of the Lorentz transformation. There is a purely geometrical argument, based only on the symmetries of spacetime, that shows that the Lorentz transformation must conserve volume. If you then demand causality, you can prove that there is an invariant velocity c, with Galilean relativity corresponding to the special case where c approaches infinity. So in this approach the existence of an invariant velocity is a theorem rather than an assumption (as Einstein took it to be in 1905).
 
  • #3
Yes its interesting me to trace question. What is easiest way to diagonalize matrix

[tex]
\alpha=\left(\begin{array}{cccc}
\gamma& 0&0& -\beta\gamma\\
0&1& 0 & 0\\
0 & 0 & 1 & 0\\
-\beta\gamma & 0 & 0 & \gamma \end{array} \right)
[/tex]

?

When we diagonalize this matrix trace must stay the same so

[tex]Tr(\alpha)=2(\gamma+1)[/tex]

and determinant which will be product of the numbers in main diagonal must be equal [tex]1[/tex] also.
 
  • #4
To diagonalize it, you'll have an easier time if you just work in 1+1 dimensions. Then the diagonal form has to have the form diag(a,b). The trace and determinant give [itex]a+b=2\gamma[/itex] and ab=1. Solve for a and b.
 

1. What is the Lorentz transformation matrix?

The Lorentz transformation matrix, denoted by A'^{\mu}=\alpha^{\mu}_{\nu}A^{\nu}, is a mathematical tool used in special relativity to describe how the coordinates of an event in one frame of reference are related to the coordinates of the same event in another frame of reference. It takes into account the effects of time dilation and length contraction, and is crucial in understanding the behavior of objects moving at high speeds.

2. How is the Lorentz transformation matrix derived?

The Lorentz transformation matrix is derived from the Lorentz transformation equations, which were first proposed by Dutch physicist Hendrik Lorentz in 1904. They were further developed by Albert Einstein in his theory of special relativity. The equations describe how time and space coordinates in one inertial reference frame are related to those in another inertial reference frame that is moving at a constant velocity relative to the first frame.

3. What are the elements of the Lorentz transformation matrix?

The Lorentz transformation matrix \alpha^{\mu}_{\nu} is a 4x4 matrix that contains 16 elements. The first three rows and columns represent the spatial coordinates (x, y, z) and the fourth row and column represent the time coordinate (t). The elements of the matrix are determined by the velocity of the moving frame relative to the stationary frame and the speed of light.

4. What is the significance of the Lorentz transformation matrix?

The Lorentz transformation matrix is significant because it allows us to make precise predictions about the behavior of objects moving at high speeds. It is a fundamental tool in the theory of special relativity and has been confirmed by numerous experiments. It also forms the basis for the more general theory of relativity, which includes the effects of gravity.

5. Can the Lorentz transformation matrix be used for any type of motion?

No, the Lorentz transformation matrix is only applicable to objects moving at constant velocities in a straight line. It cannot be used for objects that are accelerating or moving in curved paths. In these cases, more complex mathematical tools, such as the general theory of relativity, must be used to accurately describe the motion of the objects.

Similar threads

  • Special and General Relativity
Replies
1
Views
688
  • Special and General Relativity
Replies
1
Views
534
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
2
Views
931
  • Special and General Relativity
Replies
2
Views
772
  • Special and General Relativity
Replies
6
Views
1K
  • Special and General Relativity
Replies
5
Views
1K
  • Special and General Relativity
Replies
1
Views
619
  • Special and General Relativity
Replies
34
Views
3K
Back
Top