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Solve Velocity 4-Vector Homework
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[QUOTE="MikeLizzi, post: 5628110, member: 104390"] I think I got it! It took a couple days to learn latex but PeroK's hint and seeing the formulas clearly made me realize what I was doing wrong. Here's the latex by the way (still working on format) The Lorentz Transformation is defined as $$\lambda= \begin{bmatrix} \gamma & -\gamma\beta_x & -\gamma\beta_y & -\gamma\beta_z \\ -\gamma\beta_x & 1+\frac{(\gamma -1)\beta_x^2}{\beta^2} & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & \frac{(\gamma -1)\beta_x \beta_z}{\beta^2} \\ -\gamma\beta_y & \frac{(\gamma -1)\beta_x\beta_y}{\beta^2} & 1+\frac{(\gamma -1)\beta_y^2}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} \\ -\gamma\beta_z & \frac{(\gamma -1)\beta_x\beta_z}{\beta^2} & \frac{(\gamma -1)\beta_y\beta_z}{\beta^2} & 1+\frac{(\gamma -1)\beta_z^2}{\beta^2} \end{bmatrix} $$ Where v is the velocity of S with respect to S' and $$\beta_x = v_x/c$$ $$\beta_y = v_y/c$$ $$\beta_y = v_y/c$$ $$\beta^2 = \beta_x^2 + \beta_y^2 + \beta_z^2$$ $$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$ The velocity 4-vector in reference frame S' is defined as $$u'_\mu= \begin{bmatrix} \gamma_{u'} \\ \gamma_{u'} u'_x \\ \gamma_{u'} u'_y \\ \gamma_{u'} u'_z \end{bmatrix} $$ Where u' is the 3-vector velocity of an object with respect to the S' reference frame and gamma is the gamma value using u'. The velocity 4-vector in reference frame S is defined as $$u_\mu= \begin{bmatrix} \gamma_{u} \\ \gamma_{u} u_x \\ \gamma_{u} u_y \\ \gamma_{u} u_z \end{bmatrix} $$ Where u is the 3-vector velocity of an object with respect to the S reference frame and gamma is the gamma value using u. Thanks PeroK ! [/QUOTE]
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