Relativity, reference frames and Lorentz transformations

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SUMMARY

The discussion centers on the application of Lorentz transformations to determine the coordinates of a reference point in two inertial frames, S and S', moving relative to each other with velocity \(\upsilon\). The key equations used are the Lorentz transformations for time and space, specifically \(x'=\frac{x-\upsilon t}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}\) and \(t'=\frac{t-\frac{\upsilon}{c^2}x}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}\). The user attempts to solve for the condition where \(t=t'\) and derives the equation \(x=\frac{\upsilon t}{\sqrt{1-\frac{\upsilon^2}{c^2}}+1}\). This solution is confirmed as correct within the context of special relativity.

PREREQUISITES
  • Understanding of Lorentz transformations in special relativity
  • Familiarity with the concept of inertial reference frames
  • Basic knowledge of the speed of light constant, \(c\)
  • Ability to manipulate algebraic equations involving variables and constants
NEXT STEPS
  • Study the derivation and implications of Lorentz transformations in detail
  • Explore the concept of simultaneity in different reference frames
  • Learn about time dilation and length contraction effects in special relativity
  • Investigate applications of Lorentz transformations in modern physics scenarios
USEFUL FOR

Students of physics, particularly those studying special relativity, educators teaching the principles of reference frames, and anyone interested in the mathematical foundations of relativistic motion.

Petar Mali
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Homework Statement



The system [tex]S'[/tex] moves in relation to the system [tex]S[/tex] with velocity [tex]\upsilon[/tex] along the -[tex]x[/tex]- axis. At the time when the beginnings of the coordinate system are in the same point, clocks in both system shows [tex]t=t'=0[/tex]. Which coordinates will have a reference point during the motion in every of these systems, which has the property that in some next moment clocks in systems [tex]S[/tex], [tex]S'[/tex] shows the same time [tex]t=t'[/tex]. Determine the law of motion of motion of this point.



Homework Equations


Lorentz transformation

[tex]x'=\frac{x-\upsilon t}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]

[tex]y'=y[/tex]

[tex]z'=z[/tex]

[tex]t'=\frac{t-\frac{\upsilon}{c^2}x}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]



The Attempt at a Solution



I tried like this

[tex]t'=\frac{t-\frac{\upsilon}{c^2}x}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]

[tex]t=\frac{t'+\frac{\upsilon}{c^2}x'}{\sqrt{1-\frac{{\upsilon}^2}{c^2}}}[/tex]

[tex]t=t'[/tex]

[tex]t-\frac{\upsilon}{c^2}x=t'+\frac{\upsilon}{c^2}x'[/tex]

[tex]t-t'=\frac{\upsilon}{c^2}(x+x')[/tex]

[tex]0=\frac{\upsilon}{c^2}(x+x')[/tex]

and get

[tex]x=\frac{\upsilon t}{\sqrt{1-\frac{\upsilon^2}{c^2}}+1}[/tex]

Is this correct?
 
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Any answer?
 

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