How Do Lorentz Transformations Prove Light Pulse Symmetry?

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SUMMARY

The discussion centers on the application of Lorentz transformations to demonstrate the symmetry of light pulses in different inertial frames. The relevant equations include x' = (gamma)(x - ut) and t' = (gamma)(t - ux/c^2). The user attempts to derive the relationship x^2 = (c^2)(t^2) from the equation x'^2 = (c^2)(t'^2) but encounters algebraic errors. Key corrections involve ensuring the gamma factor is squared and properly accounting for the factor of 1/c^2 in the transformations.

PREREQUISITES
  • Understanding of Lorentz transformations
  • Familiarity with the concept of gamma factor in special relativity
  • Basic algebra skills for manipulating equations
  • Knowledge of the speed of light (c) and its implications in physics
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  • Study the derivation of Lorentz transformations in detail
  • Learn about the implications of the gamma factor in relativistic physics
  • Explore examples of light pulse symmetry in various inertial frames
  • Review algebraic techniques for solving physics problems involving transformations
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Students of physics, particularly those studying special relativity, educators teaching relativity concepts, and anyone interested in the mathematical foundations of light behavior in different frames of reference.

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Homework Statement



Frame S' has an x component of velocity u relative to the frame S and at t=t'=0 the two frames coincide. A light pulse with a spherical wave front at the origin of S' at t'=0. Its distance x' from the origin after a time t' is given by x'^2=(c^2)(t'^2). Transform this to an equation in x and t, showing that the result is x^2=(c^2)(t^2).


Homework Equations



I think the relevant equations are:

x'=(gamma)(x-ut) and
t' = (gamma)(t-[ux/c^2])

The Attempt at a Solution



x'^2=(gamma)([x^2]-(2xut)+([ut]^2))
t'^2=(gamma)([t^2]-(2tux/c^2)+([ux/c^2]^2))

plugged into the equation x'^2=(c^2)(t'^2) and reduced somewhat...

(x^2)-(2xut)+((ut)^2) = ((c^2)(t^2))-2tux+((c^2)((ux/c^2)^2)

I think my algebra is crud somewhere, but the farthest I can seem to get this down to is...

(x^2)(1-(u^2)) = (t^2)((c^2)-(u^2))

Thank you for looking at my problem! I hope I put enough clear information down, this is my first time here, please let me know if I need to clear anything up, and thanks again!
 
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physicyst said:

Homework Statement



Frame S' has an x component of velocity u relative to the frame S and at t=t'=0 the two frames coincide. A light pulse with a spherical wave front at the origin of S' at t'=0. Its distance x' from the origin after a time t' is given by x'^2=(c^2)(t'^2). Transform this to an equation in x and t, showing that the result is x^2=(c^2)(t^2).


Homework Equations



I think the relevant equations are:

x'=(gamma)(x-ut) and
t' = (gamma)(t-[ux/c^2])

The Attempt at a Solution



x'^2=(gamma)([x^2]-(2xut)+([ut]^2))
t'^2=(gamma)([t^2]-(2tux/c^2)+([ux/c^2]^2))
You forgot to square the gamma factor but it does not affect the rest of your calculation because this factor cancels out anyway.
plugged into the equation x'^2=(c^2)(t'^2) and reduced somewhat...

(x^2)-(2xut)+((ut)^2) = ((c^2)(t^2))-2tux+((c^2)((ux/c^2)^2)

I think my algebra is crud somewhere, but the farthest I can seem to get this down to is...

(x^2)(1-(u^2)) = (t^2)((c^2)-(u^2))

Thank you for looking at my problem! I hope I put enough clear information down, this is my first time here, please let me know if I need to clear anything up, and thanks again!
You forgot a factor of 1/c^2 on the left side (it should read [tex]x^2 (1 - \frac{u^2}{c^2}[/tex]). This is because you had [tex]c^2 \times (\frac{ux}{c^2})^2 = \frac{u^2 x^2}{c^2}[/tex]. With this correction, everything works out.
 

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