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Chapter 2: Problem 18

A light bulb at point \(x\) in the frame of reference of \(O\) blinks on and off at intervals \(\Delta t=t_{2}-t_{1} .\) Observer \(O^{\prime}\) moving relative to \(O\) at speed \(u\), measures the interval to be \(\Delta t^{\prime}=t_{2}^{\prime}-t_{1}^{\prime} .\) Use the Lorentz transformation expressions to derive the time dilation expression relating \(\Delta t\) and \(\Delta t^{\prime}\)

Short Answer

Expert verified
\[ \triangle t' = \frac{\triangle t}{\big(1 - \frac{u^2}{c^2}\big)^{1/2}} \]

Step by step solution

01

Identify Given Variables

Start by identifying the given variables. The intervals in the time frame of observer O are given by \(\triangle t = t_2 - t_1\). For observer O', the intervals are given by \(\triangle t' = t_2' - t_1'\). The relative speed between the observers is \(u\).
02

Lorentz Transformation for Time

According to Lorentz transformations, the time coordinates in two reference frames moving relative to each other are related by: \[ t' = \frac{t - ux/c^2}{\beta} \] where \[ \beta = \frac{1}{\big(1 - \frac{u^2}{c^2}\big)^{1/2}} \] where \(c\) is the speed of light.
03

Apply Lorentz Transformation to Both Time Coordinates

Substitute the Lorentz transformation for each time coordinate: \[ t_1' = \frac{t_1 - ux/c^2}{\beta} \] \[ t_2' = \frac{t_2 - ux/c^2}{\beta} \]
04

Calculate \(\triangle t'\)

To find \(\triangle t'\), subtract the equation for \(t_1'\) from \(t_2'\): \[ \triangle t' = t_2' - t_1' \] \[ \triangle t' = \frac{t_2 - ux/c^2}{ \beta} - \frac{t_1 - ux/c^2}{\beta} \] Combine the terms, noting that the position \(x\) is constant during the blink and cancels out: \[ \triangle t' = \frac{t_2 - t_1}{\beta} \] Hence, \[ \triangle t' = \frac{\triangle t}{\beta} \]
05

Relate Time Dilation

Express the time dilation relationship: \[ \triangle t' = \frac{\triangle t}{\big(1 - \frac{u^2}{c^2}\big)^{1/2}} \]

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Key Concepts

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

Lorentz Transformation
When discussing relativity, one crucial element is the Lorentz transformation. It enables us to connect the coordinates (both time and space) of events as seen from two different inertial frames of reference moving relative to each other.

Consider an observer at rest in a reference frame O and another observer moving at a constant speed, denoted as O'. The Lorentz transformations for time can be expressed as follows:

\( t' = \frac{t - ux/c^2}{\beta} \)

Here:
  • \(t'\) represents the time in the moving frame O'
  • \(t\) represents the time in the original frame O
  • \(u\) is the relative velocity between the two frames
  • \(x\) is the position
  • \(c\) is the speed of light
  • \(\beta = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}\)


      This transformation helps us understand how time and space coordinates change under the relative motion, giving rise to phenomena such as time dilation.
Relativity
Relativity encompasses the idea that the laws of physics are the same for all non-accelerating observers. This forms the foundation of Einstein's theory of Special Relativity.

Special relativity introduces two key postulates:
  • The laws of physics are invariant (identical) in all inertial frames of reference.
  • The speed of light in a vacuum is the same for all observers, regardless of their motion relative to the light source.


      This led Einstein to conclude that time is not absolute. Instead, time can stretch and contract, depending on the relative velocity between observers. This phenomenon is quantified through time dilation.

      In the context of the provided problem, relativity allows us to relate the time intervals measured by two observers moving at different relative speeds via the Lorentz transformations. It鈥檚 important to realize that these transformations are not just mathematical tricks but have real implications on space and time as observed in high-speed contexts.
Time Interval
Understanding time intervals is essential in studying time dilation within the framework of special relativity. A time interval is simply the difference in time between two events.

In the provided problem, two observers measure the interval between a light bulb blinking on and off:
  • \(\Delta t = t_2 - t_1\) for observer O
  • \(\Delta t' = t_2' - t_1'\) for observer O'


      Using Lorentz transformations, we derive the time dilation formula:

      \( \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{u^2}{c^2}}} \)

      This equation shows how time intervals between two events differ when measured by observers in different frames of reference.

      The term \(\sqrt{1 - \frac{u^2}{c^2}}\) (often denoted as \(\gamma\) in relativity) is crucial here. It accounts for the relative speed (u) between observers. For speeds much less than the speed of light, \(\gamma\) approaches 1, meaning the time interval remains similar. However, as velocity increases, time intervals measured in the moving frame (O') become significantly longer, demonstrating the real-life impact of time dilation.

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Most popular questions from this chapter

According to observer \(O,\) a blue flash occurs at \(x_{b}=\) \(10.4 \mathrm{m}\) when \(t_{\mathrm{b}}=0.124 \mu \mathrm{s},\) and a red flash occurs at \(x_{\mathrm{r}}=23.6 \mathrm{m}\) when \(t_{\mathrm{r}}=0.138 \mu \mathrm{s} .\) According to observer \(O^{\prime},\) who is in motion relative to \(O\) at velocity \(u,\) the two flashes appear to be simultaneous. Find the velocity \(u\).

An electron moving at a speed of \(v_{\mathrm{i}}=0.960 \mathrm{c}\) in the positive \(x\) direction collides with another electron at rest. After the collision, one electron is observed to move with a speed of \(v_{1 f}=0.956 c\) at an angle of \(\theta_{1}=9.7^{\circ}\) with the \(x\) axis. \((a)\) Use conservation of momentum to find the velocity (magnitude and direction) of the second electron. (b) Based only on the original data given in the problem, use conservation of energy to find the speed of the second electron.

For what range of velocities of a particle of mass \(m\) can we use the classical expression for kinetic energy \(\frac{1}{2} m v^{2}\) to within an accuracy of 1 percent?

A particle of rest energy \(m c^{2}\) is moving with speed \(v\) in the positive \(x\) direction. The particle decays into two particles, each of rest energy \(140 \mathrm{MeV}\). One particle, with kinetic energy \(282 \mathrm{MeV},\) moves in the positive \(x\) direction, and the other particle, with kinetic energy \(25 \mathrm{MeV}\), moves in the negative \(x\) direction. Find the rest energy of the original particle and its speed.

You are piloting a small airplane in which you want to reach a destination that is \(750 \mathrm{km}\) due north of your starting location. Once you are airborne, you find that (due to a strong but steady wind) to maintain a northerly course you must point the nose of the plane at an angle that is \(22^{\circ}\) west of true north. From previous flights on this route in the absence of wind, you know that it takes you 3.14 h to make the journey. With the wind blowing, you find that it takes \(4.32 \mathrm{h}\). A fellow pilot calls you to ask about the wind velocity (magnitude and direction). What is your report?
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