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Consider two events that occur at positions \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) and times \(t_{1}\) and \(t_{2} .\) Let \(\Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}\) and \(\Delta t=t_{2}-t_{1} .\) Write down the Lorentz transformation for \(\mathbf{r}_{1}\) and \(t_{1}\), and likewise for \(\mathbf{r}_{2}\) and \(t_{2},\) and deduce the transformation for \(\Delta \mathbf{r}\) and \(\Delta t\). Notice that differences \(\Delta \mathbf{r}\) and \(\Delta t\) transform in exactly the same way as \(\mathbf{r}\) and \(t\). This important property follows from the linearity of the Lorentz transformation.

Short Answer

Expert verified
\( \Delta \mathbf{r} \) and \( \Delta t \) transform the same way as \( \mathbf{r} \) and \( t \) due to linearity of Lorentz transformation.

Step by step solution

01

Express the Lorentz Transformation

The Lorentz transformation relates two frames of reference moving relative to one another at a constant velocity. For an event in one frame located at \( \mathbf{r}_1 = (x_1, y_1, z_1) \) and a time \( t_1 \), the transformations to another frame \((x_1', y_1', z_1', t_1') \) moving at velocity \( v \) along the x-axis are:\[ x_1' = \gamma (x_1 - vt_1) \] \[ y_1' = y_1 \] \[ z_1' = z_1 \] \[ t_1' = \gamma \left( t_1 - \frac{vx_1}{c^2} \right) \]where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) and \( c \) is the speed of light.
02

Apply the Lorentz Transformation to the Second Event

Apply similar transformations for the second event located at \( \mathbf{r}_2 = (x_2, y_2, z_2) \) and time \( t_2 \):\[ x_2' = \gamma (x_2 - vt_2) \] \[ y_2' = y_2 \] \[ z_2' = z_2 \] \[ t_2' = \gamma \left( t_2 - \frac{vx_2}{c^2} \right) \]
03

Calculate Differences in Position and Time

Calculate the differences in spatial coordinates and time in both frames. In the original frame, they are:\[ \Delta \mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1 = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \] \[ \Delta t = t_2 - t_1 \]
04

Apply Differences to the Lorentz Transformation

Use the differences found in Step 3 to find their transformations:\[ \Delta x' = x_2' - x_1' = \gamma ((x_2 - vt_2) - (x_1 - vt_1)) = \gamma (\Delta x - v \Delta t) \]\[ \Delta y' = y_2' - y_1' = y_2 - y_1 = \Delta y \]\[ \Delta z' = z_2' - z_1' = z_2 - z_1 = \Delta z \]\[ \Delta t' = t_2' - t_1' = \gamma \left( (t_2 - \frac{vx_2}{c^2}) - (t_1 - \frac{vx_1}{c^2}) \right) = \gamma (\Delta t - \frac{v \Delta x}{c^2}) \]
05

Conclusion

We found that \( \Delta \mathbf{r}' \) and \( \Delta t' \) transform exactly like \( \mathbf{r}' \) and \( t' \). This confirms that differences \( \Delta \mathbf{r} \) and \( \Delta t \) also obey the Lorentz transformation just like position and time coordinates because of the linear nature of the Lorentz transformation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Event Transformation
When we talk about event transformations in the context of physics, particularly in the special theory of relativity, we refer to how events are observed differently in various frames of reference. An event is essentially a point in spacetime, specified by four coordinates: three for space and one for time. The Lorentz transformation is the key to understanding these differences.

Imagine two observers, each in their own spaceship flying past each other. Each sees events in their own specific way based on their respective frames of reference. The Lorentz transformation explains exactly how the coordinates of an event seen by one observer translate into what another observer sees. For instance, if a firework explodes in space, the coordinates of this explosion, such as position and time, will be different from one spaceship's perspective compared to the other.

The ability to transform these coordinates is particularly important in high-speed contexts, such as those encountered in particle physics, where classical Newtonian approaches fall short. Here, events are transformed using the Lorentz equations, ensuring that all observers, irrespective of their relative motion, can still agree on the nature of the spacetime structure. This is crucial for the integrity of physical laws across different frames.
Spacetime Coordinates
Spacetime coordinates are essential to describe the position and time of any event in the universe. Unlike in classical mechanics, where we can treat space and time independently, relativity combines these into a four-dimensional framework. To completely describe any event in this relativistic universe, we use four coordinates: three spatial (usually denoted as \( x \) , \( y \) , and \( z \)) and one temporal (\( t \)).

In our everyday experiences, time ticks uniformly for everyone, and spaces remain consistent; however, that's not the case when dealing with significant speeds or gravitational fields. Here, time can slow down or speed up, and space can stretch or compress, depending on the observer's relative motion. That's where the Lorentz transformation becomes vital, as it helps to recalculate these coordinates in another moving frame, preserving the form of physical laws.

The transformation showcases the linked nature of space and time, forming the fabric of our universe, known as spacetime. Understanding spacetime coordinates is fundamental for comprehending how events link to each other. For instance, when considering two events happening at different times and locations, the differences in their coordinates (\

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Most popular questions from this chapter

(a) Show that if a body has speed \(v < c\) in one inertial frame, then \(v < c\) in all frames. [Hint: Consider the displacement four-vector \(d x=(d \mathbf{x}, c d t),\) where \(d \mathbf{x}\) is the three-dimensional displacement in a short time \(d t .]\) (b) Show similarly that if a signal (such as a pulse of light) has speed \(c\) in one frame, its speed is \(c\) in all frames.

As an observer moves through space with position \(\mathbf{x}(t),\) the four- vector \((\mathbf{x}(t), c t)\) traces a path through space-time called the observer's world line. Consider two events that occur at points \(P\) and \(Q\) in space-time. Show that if, as measured by the observer, the two events occur at the same time \(t,\) then the line joining \(P\) and \(Q\) is orthogonal to the observer's world line at the time \(t\); that is, \(\left(x_{P}-x_{Q}\right) \cdot d x=0,\) where \(d x\) joins two neighboring points on the world line at times \(t\) and \(t+d t\).

Prove that if \(T\) and \(a\) are respectively a four-tensor and a four-vector, then \(b=T \cdot a=T G a\) is a four-vector; that is, it transforms according to the rule \(b^{\prime}=\Lambda b\)

A traveler in a rocket of proper length 2d sets up a coordinate system \(\mathcal{S}^{\prime}\) with its origin \(O^{\prime}\) anchored at the exact middle of the rocket and the \(x^{\prime}\) axis along the rocket's length. At \(t^{\prime}=0\) she ignites a flashbulb at \(O^{\prime} .\) (a) Write down the coordinates \(x_{\mathrm{F}}^{\prime}, t_{\mathrm{F}}^{\prime}\) and \(x_{\mathrm{B}}^{\prime}, t_{\mathrm{B}}^{\prime}\) for the arrival of the light at the front and back of the rocket. (b) Now consider the same experiment as observed from a frame \(\delta\) relative to which the rocket is traveling with speed \(V\) (with \(\delta\) and \(S\) ' in the standard configuration). Use the inverse Lorentz transformation to find the coordinates \(x_{\mathrm{F}}, t_{\mathrm{F}}\) and \(x_{\mathrm{B}}, t_{\mathrm{B}}\) for the arrival of the two signals. Explain clearly in words why the two arrivals are simultaneous in \(\mathcal{S}^{\prime}\) but not in \(\mathcal{S} .\) This phenomenon is called the relativity of simultaneity.

Show that the four-velocity of any object has invariant length squared \(u \cdot u=-c^{2}\)

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