Special relativity - frames of reference

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SUMMARY

The discussion focuses on the derivation of Lorentz transformations for two frames of reference, K and K', where K' moves with a velocity vector \(\vec{v} = v [\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}]\). The transformations are defined as follows: \(t' = \frac{t - (\vec{v} \circ (x, y))/c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\), \(x' = \frac{x - v_x t}{\sqrt{1 - \frac{v^2}{c^2}}}\), and \(y' = \frac{y - v_y t}{\sqrt{1 - \frac{v^2}{c^2}}}\), with \(v_x = v_y = \frac{v}{\sqrt{2}}\). The transformations are crucial for understanding how time and space coordinates change between moving frames in special relativity.

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  • Knowledge of vector notation and operations
  • Basic grasp of the speed of light constant (c)
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  • Explore the implications of time dilation and length contraction
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Homework Statement


We have two frames of reference: K (x,y,t) and K' (x',y',t') such that initially x=x'=y=y'=t=t'=0. Now let K' move with a velocity [tex]\vec{v} = v [\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}][/tex]
Write Lorentz transformations in such a case.

Homework Equations





The Attempt at a Solution


My try is:
[tex]t' = \frac{t - (\vec{v} \circ (x, y))/c^2}{\sqrt{1 - \frac{v^2}{c^2}}}, \; x' = \frac{x - v_x t}{\sqrt{1 - \frac{v^2}{c^2}}}, y' = \frac{y - v_y t}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

where [tex]v_x = v_y = \frac{v}{\sqrt{2}}[/tex]
 
Last edited:
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Looks good to me.
 
Great, thank you!
 

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