# How Do You Derive the Lorentz Transformation for Frame S''?

• brainpushups
In summary, the problem involves applying the Lorentz transformation twice to find the coordinates of an event in terms of the coordinates in a different frame of reference. This is shown to be equivalent to using the standard Lorentz transformation with the velocity V given by the relativistic sum of the velocities of the two frames. Some algebraic manipulation may be required to simplify the final equation.
brainpushups

## Homework Statement

Frame S' travels at speed V1 along the x-axis of frame S. Frame S'' travels at speed V2 along the x' axis of frame S'. Apply the Lorentz transformation twice to find the coordinates x'', y'', etc of any event in terms of x, y, z, t. Show that this is the same as the standard Lorentz transformation with velocity V given by the relativistic sum of V1 and V2

## Homework Equations

$\x'=\gamma(x-Vt)$
$\t'=\gamma(t'-\frac{Vx}{c^{2}})$
$\v_{x}'=\frac{v_{x}-V}{1-\frac{v_{x}V}{c^{2}}}$

## The Attempt at a Solution

No problem with the first part. I'll just put my answer for the x'' coordinate for reference:

$\x''=\gamma_{1}\gamma_{2}(x(1+\frac{V_{1}V_{2}}{c^{2}})-(V_{1}+V_{2})t)$

The different gamma factors come from transforming between S and S' and then again between S' and S''

For showing that this is equivalent to applying the Lorentz transformation using the relativistic sum:

$\x''=\gamma(x-Vt)=\gamma(x-\frac{V_{1}+V_{2}}{1+\frac{V_{1}V_{2}}{c^{2}}}t)$

This is almost in the correct form. If I kill the fraction I can get

$\x''(1+\frac{V_{1}V_{2}}{c^{2}})=\gamma(x(1+\frac{V_{1}V_{2}}{c^{2}})-(V_{1}+V_{2})t)$

The only differences between this and the expression I am looking for is the extra factor by x'' and the fact that the factor gamma should be the product of two gamma factors. I figured I could write the gammas explicitly in terms of V1 and V2 in my first equation for x’’ and then do the same in the second equation by substituting the relativistic sum for V and then rearrange things until I got the extra factor next to x’’ in the second equation to cancel. I worked through the algebra for a little while, but it didn't seem like it was going to happen. I just want to make sure I'm not overlooking something before I continue with the algebraic gymnastics. If I’m on the right track please confirm and I’ll continue to work at it.

Everything looks OK. I would suggest that you now try to simplify ##\sqrt{1-\frac{V^2}{c^2}}## where V is the relativistic sum of V1 and V2.

TSny said:
Everything looks OK. I would suggest that you now try to simplify ##\sqrt{1-\frac{V^2}{c^2}}## where V is the relativistic sum of V1 and V2.
I cannot get it. I try to calculate
γ1γ2 with V1 = V2 = 0.8c and its relativistic speed.
It found that the equation could not hold.

xiaozegu said:
I cannot get it. I try to calculate
γ1γ2 with V1 = V2 = 0.8c and its relativistic speed.
It found that the equation could not hold.

Everything should work out. I'm not sure exactly what you have done. Can you show us the details of your work?

brainpushups said:
I worked through the algebra for a little while, but it didn't seem like it was going to happen. I just want to make sure I'm not overlooking something before I continue with the algebraic gymnastics. If I’m on the right track please confirm and I’ll continue to work at it.

It is just algebraic gymnastics. However, I'd suggest just working on the γ factor for the combined velocities first. And, perhaps, working on ##\frac{1}{\gamma ^2}## makes it slightly easier.

## 1. What is the Lorentz transformation for special relativity?

The Lorentz transformation is a mathematical formula used in the theory of special relativity to describe how measurements of space and time change for different observers moving at constant speeds relative to each other. It allows for the reconciliation of different measurements of space and time made by different observers, and is a fundamental concept in understanding the effects of time dilation and length contraction.

## 2. How is the Lorentz transformation derived?

The Lorentz transformation is derived from the postulates of special relativity, which state that the laws of physics are the same for all non-accelerating observers and that the speed of light in a vacuum is constant for all observers. By applying these postulates to the equations of motion for objects moving at different speeds, the Lorentz transformation can be derived.

## 3. Why is the Lorentz transformation important?

The Lorentz transformation is important because it forms the basis of special relativity, which is a crucial theory in modern physics. It has been confirmed by numerous experiments and has led to groundbreaking discoveries such as the equivalence of mass and energy (E=mc^2) and the concept of space-time. The Lorentz transformation also plays a key role in technologies such as GPS and particle accelerators.

## 4. How does the Lorentz transformation affect our perception of time and space?

The Lorentz transformation shows that time and space are not absolute, but are relative to the observer's frame of reference. This means that an event may appear to happen at different times and locations for different observers, depending on their relative speeds. This concept challenges our everyday perception of time and space as fixed and independent, and has led to a deeper understanding of the nature of the universe.

## 5. Are there any limitations to the Lorentz transformation?

The Lorentz transformation is only applicable to objects moving at a constant velocity in a straight line. It does not take into account the effects of acceleration or gravity, which require the more complex mathematics of general relativity. Additionally, the Lorentz transformation assumes a flat space-time geometry, and may not accurately describe extreme scenarios such as near black holes. However, for most everyday situations, the Lorentz transformation is a highly accurate and useful tool for understanding the effects of special relativity.

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