Calculating Four-Velocity and Four-Momentum Using Schutz's Method

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Discussion Overview

The discussion revolves around calculating the four-velocity and four-momentum of a particle using Schutz's method as presented in "A First Course in General Relativity." Participants explore the definitions and transformations involved in these calculations, particularly focusing on the application of Lorentz transformations in different reference frames.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions the derivation of the four-velocity components from Schutz's text, specifically how the transformation leads to the expression for U^{\alpha}.
  • Another participant suggests expressing the problem as a matrix equation to clarify the transformation process, providing a specific matrix representation and definitions for \beta and \gamma.
  • A third participant reiterates the initial question about the four-velocity and emphasizes the need to apply Lorentz transformations to find the components in a moving reference frame.
  • A later reply expresses gratitude for the matrix explanation, indicating a preference for that format over Einstein notation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivation process, as there are differing levels of understanding and preferences for notation. The discussion remains open with various approaches being explored.

Contextual Notes

Some participants express difficulty with Einstein notation, which may affect their understanding of the transformations discussed. The discussion does not resolve the mathematical steps involved in the derivation of the four-velocity and four-momentum.

schwarzschild
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From Schutz's A First Course in General Relativity

"A particle of rest mass m moves with velocity v in the x direction of frame O. What are the components of the four-velocity and four-momentum?"

By definition \vec{U} = \vec{e}_{\bar{0}

However, I don't see how he gets U^{\alpha} = \Lambda^{\alpha}_{\bar{\beta}}(\vec{e}_{\bar{0}})^{\bar{\beta}} = \Lambda^{\alpha}_{\bar{0}}

Where \vec{U} is the four-velocity vector, and U^{\alpha} are its components.
 
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Does it help to write it all out explicity as a matrix equation?

\begin{bmatrix}<br /> \gamma &amp; \beta \gamma &amp; 0 &amp; 0\\ <br /> \beta \gamma &amp; \gamma &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{bmatrix}<br /> <br /> \begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix} = \begin{bmatrix}\gamma\\ \beta \gamma \\ 0\\ 0 \end{bmatrix}

where

\beta = \frac{v}{c}

and

\gamma = \left ( 1 - \beta^2 \right )^{-1/2}.

The components of U^\alpha in reference frame O are the entries of the first column of the transformation matrix.
 
Last edited:
schwarzschild said:
From Schutz's A First Course in General Relativity

"A particle of rest mass m moves with velocity v in the x direction of frame O. What are the components of the four-velocity and four-momentum?"

By definition \vec{U} = \vec{e}_{\bar{0}

However, I don't see how he gets U^{\alpha} = \Lambda^{\alpha}_{\bar{\beta}}(\vec{e}_{\bar{0}})^{\bar{\beta}} = \Lambda^{\alpha}_{\bar{0}}

Where \vec{U} is the four-velocity vector, and U^{\alpha} are its components.

In the reference frame in which the mass is at rest, its four-velocity is simply e0, i.e. the components are (1, 0, 0, 0). Now all you have to do is find the components of this same vector in a reference frame which is moving in the opposite direction with speed v. This is done by applying an appropriate Lorentz transformation.
 
Last edited:
Thanks for the help guys! I completely understand it in matrix form - for some reason I struggle with Einstein notation.
 

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