Difference between revisions of "4-momenta"

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<center><math>\mathbf P \cdot \mathbf P = \frac{E^2}{c^2}-E^2+m^2</math></center>
 
<center><math>\mathbf P \cdot \mathbf P = \frac{E^2}{c^2}-E^2+m^2</math></center>
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<center><math>\mathbf P \cdot \mathbf P = m^2</math></center>

Revision as of 19:56, 8 June 2017

4-momenta

As was previously shown for the space-time 4-vector, a similar 4-vector can be composed of momentum. Using index notation, the energy and momentum components can be combined into a single "4-vector" [math]\mathbf{p^{\mu}},\ \mu=0,\ 1,\ 2,\ 3[/math], that has units of momentum(i.e. E/c is a distance).

[math]\mathbf{P} \equiv \begin{bmatrix} p^0 \\ p^1 \\ p^2 \\ p^3 \end{bmatrix}= \begin{bmatrix} E/c \\ p_x \\ p_y \\ p_z \end{bmatrix}[/math]


As shown earlier,


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


Following the 4-vector of space-time for momentum-energy,


[math]\mathbf P \cdot \mathbf P = p_0^2-(p_1^2+p_2^2+p_3^2)[/math]


[math]\mathbf P \cdot \mathbf P = \frac{E^2}{c^2}-\vec p\ ^2[/math]


Using the relativistic equation for energy


[math]E^2=\vec p\ ^2+m^2[/math]


[math]\mathbf P \cdot \mathbf P = \frac{E^2}{c^2}-E^2+m^2[/math]


[math]\mathbf P \cdot \mathbf P = m^2[/math]