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>]]
+
[[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>, that has units of length(i.e. ct is a distance).
+
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).
  
 
<center><math>\mathbf{R} \equiv  
 
<center><math>\mathbf{R} \equiv  
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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 x_{\mu} 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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The change in signs in the covariant term,  
 
The change in signs in the covariant term,  
<center><math>x_{\mu}= \begin{bmatrix}
+
<center><math>\mathbf{x_{\mu}}= \begin{bmatrix}
 
dx_0  & -dx_1 & -dx_2 & -dx_3  
 
dx_0  & -dx_1 & -dx_2 & -dx_3  
 
\end{bmatrix}</math></center>
 
\end{bmatrix}</math></center>
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from the contravarient term
 
from the contravarient term
  
<center><math>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=\Eta_{\mu \nu} x^{\mu} x^{\nu}</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>
 
 
 
 
This is useful in that it shows that the "dot product" of two 4-vectors is an invariant since the time-space interval is an invariant.
 
 
 
 
 
 
 
 
 
Using the Lorentz transformations and the index notation,
 
  
<center><math>
 
\begin{cases}
 
t'=\gamma (t-vz/c^2) \\
 
 
x'=x' \\
 
 
y'=y' \\
 
 
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><math>\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}</math></center>
  
<center>Where <math>\beta \equiv \frac{v}{c}</math></center>
 
  
This can be expressed in matrix form as
+
<center><math>\mathbf R_1 \cdot \mathbf R^1 = x_0^2-(x_1^2+x_2^2+x_3^2)</math></center>
  
<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
+
Similarly, for two different 4-vectors,
  
<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>
+
<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>
  
  
 +
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>
+
<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

Navigation_

4-vectors

Using index notation, the time and space coordinates can be combined into a single "4-vector" xμ, μ=0, 1, 2, 3, that has units of length(i.e. ct is a distance).

R[x0x1x2x3]=[ctxyz]


We can express the space time interval using the index notation

(ds)2c2dt2dx2dy2dz2=c2dt2dx2dy2dz2


(ds)2(dx0)2(dx1)2(dx2)2(dx3)2=(dx0)2(dx1)2(dx2)2(dx3)2


Since ds2 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:

(ds)2dxμdxμ


(ds)2[dx0dx1dx2dx3][dx0dx1dx2dx3]

This gives the desired results as expected.

(ds)2(dx0)2(dx1)2(dx2)2(dx3)2=(dx0)2(dx1)2(dx2)2(dx3)2


The change in signs in the covariant term,

xμ=[dx0dx1dx2dx3]

from the contravarient term

xμ=[dx0dx1dx2dx3]


Comes from the Minkowski metric


ημμ=[1000010000100001]


ds2=[dx0dx1dx2dx3][1000010000100001][dx0dx1dx2dx3]


ds2ημνdxμdxμ=dRdR


RR=s=ημνxμxμ


R1R1=x20(x21+x22+x23)


Similarly, for two different 4-vectors,


R1R2=x01x02(x11x12+x21x22+x31x32)


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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