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>
  
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[[Relativistic_Frames_of_Reference|<math>\vartriangleleft </math>]]
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[[Relativistic_Units|<math>\vartriangleleft </math>]]
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[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
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[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
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[[Phase_space_Limiting_Particles|<math>\vartriangleright </math>]]
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[[4-momenta|<math>\vartriangleright </math>]]
  
 
</center>
 
</center>
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Using index notation, the time and space coordinates can be combined  into a single "4-vector" <math>x^{\mu},\ \mu=0,\ 1,\ 2,\ 3</math>
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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).
  
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<center><math>\begin{bmatrix}
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<center><math>\mathbf{R} \equiv
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x_0 \\
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\begin{bmatrix}
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x_1 \\
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x^0 \\
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x_2 \\
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x^1 \\
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x_3
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x^2 \\
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x^3
 
\end{bmatrix}=
 
\end{bmatrix}=
 
\begin{bmatrix}
 
\begin{bmatrix}
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z
 
z
 
\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
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We can express the space time interval using the index notation
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<center><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></center>
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<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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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:
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<center><math>(ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}</math></center>
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<center><math>(ds)^2\equiv
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\begin{bmatrix}
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dx_0  & -dx_1 & -dx_2 & -dx_3
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\end{bmatrix} \cdot
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\begin{bmatrix}
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dx^0  \\
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dx^1 \\
 +
dx^2 \\
 +
dx^3
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\end{bmatrix}</math></center>
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This gives the desired results as expected.
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<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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The change in signs in the covariant term,
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<center><math>\mathbf{x_{\mu}}= \begin{bmatrix}
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dx_0  & -dx_1 & -dx_2 & -dx_3
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\end{bmatrix}</math></center>
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from the contravarient term
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<center><math>\mathbf{x^{\mu}}=
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\begin{bmatrix}
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dx^0  \\
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dx^1 \\
 +
dx^2 \\
 +
dx^3
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\end{bmatrix}
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</math></center>
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Comes from the Minkowski metric
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<center><math>\eta_{\mu \mu}=
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\begin{bmatrix}
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1 & 0 & 0 & 0  \\
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0 &-1 & 0 & 0 \\
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0 & 0 & -1 & 0 \\
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0 & 0 & 0 & -1
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\end{bmatrix}</math></center>
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<center><math>ds^2=
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\begin{bmatrix}
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dx_0  & dx_1 & dx_2 & dx_3
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\end{bmatrix}\cdot
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\begin{bmatrix}
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1 & 0 & 0 & 0  \\
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0 &-1 & 0 & 0 \\
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0 & 0 & -1 & 0 \\
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0 & 0 & 0 & -1
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\end{bmatrix}\cdot
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\begin{bmatrix}
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dx^0  \\
 +
dx^1 \\
 +
dx^2 \\
 +
dx^3
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\end{bmatrix}
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</math></center>
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<center><math>ds^2 \equiv \eta_{\nu}^{ \mu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}</math></center>
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<center><math>\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}</math></center>
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<center><math>\mathbf R_1 \cdot \mathbf R^1 = x_0^2-(x_1^2+x_2^2+x_3^2)</math></center>
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 +
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Similarly, for two different 4-vectors,
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<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>
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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.
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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>]]
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[[Phase_space_Limiting_Particles|<math>\vartriangleright </math>]]
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[[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]

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