Difference between revisions of "4-vectors"

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<center><math>\textbf{\underline{Navigation}}</math>
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<center><math>\underline{\textbf{Navigation}}</math>
  
[[Relativistic_Frames_of_Reference|<math>\vartriangleleft </math>]]
+
[[Relativistic_Units|<math>\vartriangleleft </math>]]
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
+
[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
[[Phase_space_Limiting_Particles|<math>\vartriangleright </math>]]
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[[4-momenta|<math>\vartriangleright </math>]]
  
 
</center>
 
</center>
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<center>Since <math>ds^2 </math> is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship.</center>
+
Since <math>ds^2 </math> is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship. Following the rules of matrix multiplication, the dot product of two 4-vectors should follow the form:
  
 
+
<center><math>(ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}</math></center>
<center><math>(ds)^2\equiv \mathbf{x_{\mu}} \mathbf{x^{\nu}}</math></center>
 
  
  
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\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
  
 +
This gives the desired results as expected.
  
 
<center><math>(ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}</math></center>
 
<center><math>(ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}</math></center>
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from the contravarient term
 
from the contravarient term
  
<center><math>\mathbf{x^{\nu}}=
+
<center><math>\mathbf{x^{\mu}}=
 
\begin{bmatrix}
 
\begin{bmatrix}
 
dx^0  \\
 
dx^0  \\
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<center><math>\eta_{\mu \nu}=
+
<center><math>\eta_{\mu \mu}=
 
\begin{bmatrix}
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
1 & 0 & 0 & 0  \\
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<center><math>ds^2 \equiv \eta_{\mu \nu} \mathbf{dx^{\mu}} \mathbf{dx^{\nu}}=d\mathbf{R} \cdot d\mathbf{R}</math></center>
+
<center><math>ds^2 \equiv \eta_{\nu}^{ \mu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}</math></center>
 
 
  
  
<center><math>\mathbf R \mathbf R = s = \eta_{\mu \nu} \mathbf{x^{\mu}} \mathbf{x^{\nu}}</math></center>
 
  
 +
<center><math>\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}</math></center>
  
<center><math>\mathbf R \mathbf R = x_0^2-(x_1^2+x_2^2+x_3^2)</math></center>
 
  
 +
<center><math>\mathbf R_1 \cdot \mathbf R^1 = x_0^2-(x_1^2+x_2^2+x_3^2)</math></center>
  
This is useful in that it shows that the scalar product of two 4-vectors is an invariant since the time-space interval is an invariant.
 
  
  
 +
Similarly, for two different 4-vectors,
  
  
Using the Lorentz transformations and the index notation,
 
  
<center><math>
+
<center><math>\mathbf R_1 \cdot \mathbf R^2 = x_{0_1}x^{0_2}-(x_{1_1}x^{1_2}+x_{2_1}x^{2_2}+x_{3_1}x^{3_2})</math></center>
\begin{cases}
 
t'=\gamma (t-vz/c^2) \\
 
  
x'=x' \\
 
  
y'=y' \\
+
This is useful in that it shows that the scalar product of two 4-vectors is an invariant since the time-space interval is an invariant.
 
 
z'=\gamma (z-vt)
 
\end{cases}
 
</math></center>
 
 
 
 
 
<center><math>\begin{bmatrix}
 
x'^0 \\
 
x'^1 \\
 
x'^2\\
 
x'^3
 
\end{bmatrix}=
 
\begin{bmatrix}
 
\gamma (x^0-vx^3/c)  \\
 
x^1 \\
 
x^2 \\
 
\gamma (x^3-vx^0)
 
\end{bmatrix}
 
=
 
\begin{bmatrix}
 
\gamma (x^0-\beta x^3)  \\
 
x^1 \\
 
x^2 \\
 
\gamma (x^3-vx^0)
 
\end{bmatrix}</math></center>
 
 
 
 
 
 
 
<center>Where <math>\beta \equiv \frac{v}{c}</math></center>
 
 
 
This can be expressed in matrix form as
 
 
 
<center><math>\begin{bmatrix}
 
x'^0 \\
 
x'^1 \\
 
x'^2\\
 
x'^3
 
\end{bmatrix}=
 
\begin{bmatrix}
 
\gamma & 0 & 0 & -\gamma \beta  \\
 
0 & 1 & 0 & 0 \\
 
0 & 0 & 1 & 0 \\
 
-\gamma \beta & 0 & 0 & \gamma
 
\end{bmatrix}
 
\cdot
 
\begin{bmatrix}
 
x^0  \\
 
x^1 \\
 
x^2 \\
 
x^3
 
\end{bmatrix}</math></center>
 
 
 
 
 
Letting the indices run from 0 to 3, we can write
 
 
 
<center><math>x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu})x^{\nu}</math></center>
 
 
 
 
 
<center>Where <math>\Lambda</math> is the Lorentz transformation matrix for motion in the z direction.</center>
 
 
 
 
 
The Lorentz transformations are also invariant in that they are just a rotation, i.e. Det <math>\Lambda=1</math>. The inner product is preserved,
 
 
 
 
 
 
 
<center><math>\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu}=\eta_{\nu}^{\mu}</math></center>
 
 
 
 
 
<center><math>
 
\begin{bmatrix}
 
\gamma & 0 & 0 & -\gamma \beta  \\
 
0 & 1 & 0 & 0 \\
 
0 & 0 & 1 & 0 \\
 
-\gamma \beta & 0 & 0 & \gamma
 
\end{bmatrix}\cdot
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 &-1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}\cdot
 
\begin{bmatrix}
 
\gamma & 0 & 0 & -\gamma \beta  \\
 
0 & 1 & 0 & 0 \\
 
0 & 0 & 1 & 0 \\
 
-\gamma \beta & 0 & 0 & \gamma
 
\end{bmatrix}=
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 &-1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}^T</math></center>
 
 
 
 
 
 
 
<center><math>
 
\begin{bmatrix}
 
\gamma^2-\beta^2 \gamma^2 & 0 & 0 & 0  \\
 
0 & -1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -\gamma^2+\beta^2 \gamma^2
 
\end{bmatrix}=
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 &-1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}</math></center>
 
 
 
 
 
 
 
<center><math>
 
\begin{bmatrix}
 
\gamma^2(1-\beta^2) & 0 & 0 & 0  \\
 
0 & -1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -\gamma^2(1-\beta^2)
 
\end{bmatrix}=
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 &-1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}</math></center>
 
  
 
 
<center>Where <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}</math></center>
 
 
 
<center><math>
 
\begin{bmatrix}
 
\frac{\gamma^2}{\gamma^2} & 0 & 0 & 0  \\
 
0 & -1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -\frac{\gamma^2}{\gamma^2}
 
\end{bmatrix}=
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 &-1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}</math></center>
 
  
  
  
  
<center><math>
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 & -1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}=
 
\begin{bmatrix}
 
1 & 0 & 0 & 0  \\
 
0 &-1 & 0 & 0 \\
 
0 & 0 & -1 & 0 \\
 
0 & 0 & 0 & -1
 
\end{bmatrix}</math></center>
 
 
----
 
----
  
  
  
<center><math>\textbf{\underline{Navigation}}</math>
+
<center><math>\underline{\textbf{Navigation}}</math>
  
[[Relativistic_Frames_of_Reference|<math>\vartriangleleft </math>]]
+
[[Relativistic_Units|<math>\vartriangleleft </math>]]
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
+
[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
[[Phase_space_Limiting_Particles|<math>\vartriangleright </math>]]
+
[[4-momenta|<math>\vartriangleright </math>]]
  
 
</center>
 
</center>

Latest revision as of 18:47, 15 May 2018

[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]

4-vectors

Using index notation, the time and space coordinates can be combined into a single "4-vector" [math]\mathbf{x^{\mu}},\ \mu=0,\ 1,\ 2,\ 3[/math], that has units of length(i.e. ct is a distance).

[math]\mathbf{R} \equiv \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}= \begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix}[/math]


We can express the space time interval using the index notation

[math](ds)^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2[/math]


[math](ds)^2\equiv (dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}= (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2[/math]


Since [math]ds^2 [/math] is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship. Following the rules of matrix multiplication, the dot product of two 4-vectors should follow the form:

[math](ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}[/math]


[math](ds)^2\equiv \begin{bmatrix} dx_0 & -dx_1 & -dx_2 & -dx_3 \end{bmatrix} \cdot \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix}[/math]

This gives the desired results as expected.

[math](ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}[/math]


The change in signs in the covariant term,

[math]\mathbf{x_{\mu}}= \begin{bmatrix} dx_0 & -dx_1 & -dx_2 & -dx_3 \end{bmatrix}[/math]

from the contravarient term

[math]\mathbf{x^{\mu}}= \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix} [/math]


Comes from the Minkowski metric


[math]\eta_{\mu \mu}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}[/math]


[math]ds^2= \begin{bmatrix} dx_0 & dx_1 & dx_2 & dx_3 \end{bmatrix}\cdot \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\cdot \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix} [/math]


[math]ds^2 \equiv \eta_{\nu}^{ \mu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}[/math]


[math]\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}[/math]


[math]\mathbf R_1 \cdot \mathbf R^1 = x_0^2-(x_1^2+x_2^2+x_3^2)[/math]


Similarly, for two different 4-vectors,


[math]\mathbf R_1 \cdot \mathbf R^2 = x_{0_1}x^{0_2}-(x_{1_1}x^{1_2}+x_{2_1}x^{2_2}+x_{3_1}x^{3_2})[/math]


This is useful in that it shows that the scalar product of two 4-vectors is an invariant since the time-space interval is an invariant.





[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]