Lorentz coordinate transformation

In summary, the frames S and S' have a relative velocity u and at t=t'=0 they coincide. The distance x' of a light pulse from the origin of S' at time t' is given by x'^2 = c^2 t'^2. By using the relevant equations x' = \gamma(x - ut) and t' = \gamma(t - \frac{ux}{c^2}), we can transform this to an equation in x and t, which is x^2 = c^2 t^2.
  • #1
physicyst
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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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  • #2
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.
 

What is the Lorentz coordinate transformation?

The Lorentz coordinate transformation is a mathematical tool used in special relativity to convert measurements of time and space between different frames of reference that are moving at constant velocities relative to each other. It takes into account the effects of time dilation and length contraction.

Why is the Lorentz coordinate transformation important?

The Lorentz coordinate transformation is important because it allows us to accurately describe the behavior of objects moving at high speeds. It forms the basis of special relativity, which is a fundamental theory in physics that has been verified by numerous experiments.

What is the difference between the Lorentz coordinate transformation and the Galilean transformation?

The main difference between the Lorentz coordinate transformation and the Galilean transformation is that the Galilean transformation assumes that time and space are absolute, while the Lorentz transformation takes into account the fact that the speed of light is constant and the laws of physics are the same in all inertial frames of reference.

How do you use the Lorentz coordinate transformation?

To use the Lorentz coordinate transformation, you need to know the velocity of one frame of reference relative to another. Then, you can use the equations to convert measurements of time and space from one frame to the other. It is important to note that the transformation is only valid for objects moving at constant velocities.

Where did the Lorentz coordinate transformation get its name from?

The Lorentz coordinate transformation is named after the Dutch physicist Hendrik Lorentz, who first derived it in the late 19th century. It was later incorporated into Albert Einstein's theory of special relativity, which revolutionized our understanding of time, space, and motion.

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