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If we intercept an electron having total energy 1533 \(\mathrm{MeV}\) that came from Vega, which is 26 ly from us, how far in lightyears was the trip in the rest frame of the electron?

Short Answer

Expert verified
In the electron's rest frame, the trip distance is about 0.00867 light-years.

Step by step solution

01

Restate the Problem

We need to find the distance the electron traveled from Vega to us in its own rest frame. We are given: - Total energy of electron: 1533 MeV - Distance from Vega to Earth: 26 light-years We will use the concept of relativistic effects for this calculation.
02

Understand the Energy-Mass Relationship

The total energy of a particle is given by the relation:\[E = \gamma mc^2\]where \(E\) is the total energy, \(m\) is the rest mass of the electron (0.511 MeV/c²), \(c\) is the speed of light, and \(\gamma\) is the Lorentz factor. We will use this equation to find \(\gamma\).
03

Calculate the Lorentz Factor \(\gamma\)

We know the total energy \(E = 1533 \mathrm{MeV}\) and the rest mass energy \(mc^2 = 0.511 \mathrm{MeV}\).Use the equation:\[\gamma = \frac{E}{mc^2} = \frac{1533 \mathrm{MeV}}{0.511 \mathrm{MeV}}\]Calculate \(\gamma\).
04

Calculate Lorentz Factor Value

Using the expression for \(\gamma\):\[\gamma = \frac{1533}{0.511} \approx 3000\]This means the electron is moving at a very high relativistic speed.
05

Understand Distance in Rest Frame

In the rest frame of the electron, due to the relativistic length contraction, the distance \(L'\) it travels to reach Earth is given by:\[L' = \frac{L}{\gamma}\]where \(L\) is the distance in the Earth frame (26 light-years).
06

Calculate Rest Frame Distance

Substitute \(L = 26 \text{ ly}\) and \(\gamma = 3000\) into:\[L' = \frac{26}{3000} \text{ ly}\]Perform the calculation.
07

Solution to Rest Frame Distance

Calculating:\[L' = \frac{26}{3000} \approx 0.00867 \text{ ly}\]Thus, the trip distance in the electron's rest frame is approximately 0.00867 light-years.

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Key Concepts

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

Lorentz Factor
The Lorentz factor, denoted as \( \gamma \), is crucial in relativity. It measures how much time, length, and relativistic mass appear to change for an object moving at a significant fraction of the speed of light. The formula for the Lorentz factor is:
  • \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
where \( v \) is the velocity of the object and \( c \) is the speed of light in a vacuum.

This concept becomes especially important when dealing with high speeds, as it dictates how time dilates and lengths contract from one frame of reference to another. When an object's speed approaches the speed of light, the Lorentz factor increases significantly. In the exercise, the calculated \( \gamma \) of approximately 3000 indicates that the electron moves at extremely high speeds, close to that of light, causing notable relativistic effects.
Total Energy in Relativity
In relativity, the total energy of an object is more complex than in classical physics. It's not just kinetic energy but includes its rest energy—a concept introduced by Einstein.

The total energy \( E \) is defined as:
  • \( E = \gamma mc^2 \)
where \( m \) is the rest mass, and \( c \) is the speed of light.

This equation shows that as an object's speed increases, its energy also increases due to the Lorentz factor, \( \gamma \). In the given problem, the total energy of 1533 MeV was used to find the Lorentz factor, demonstrating how the electron's high speed affects its energy state. The relationship highlights the intricate connection between energy and motion at relativistic speeds, differing significantly compared to typical non-relativistic scenarios.
Rest Frame
Instead of viewing motion from an external viewpoint, the rest frame allows you to look at it from the object's perspective. In this frame, the object is at rest, and the surroundings appear to move.

The concept of the rest frame is important for understanding "length contraction." This phenomenon is an essential part of special relativity, where the length of an object in motion appears contracted (or shorter) to an observer in a different frame. In the rest frame, an object experiences its full length as there are no effects of motion relative to itself.
  • Rest frame simplifies calculating distances affected by relativistic speeds.
  • In this exercise, calculating the electron’s journey in its rest frame reveals the trip was only 0.00867 light-years, showing significant length contraction.
This perspective aids in simplifying complex relativistic equations and understanding relativistic effects from a familiar vantage point, making the abstract more concrete.

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Most popular questions from this chapter

(Come) back to the future. Suppose that a father is \(20.00 \mathrm{y}\) older than his daughter. He wants to travel outward from Earth for \(2.000 \mathrm{y}\) and then back to Earth for another \(2.000 \mathrm{y}\) (both intervals as he measures them) such that he is then \(20.00\) y younger than his daughter. What constant speed parameter \(\beta\) (relative to Earth) is required for the trip?

A spaceship of rest length \(130 \mathrm{~m}\) races past a timing station at a speed of \(0.740 \mathrm{c} .\) (a) What is the length of the spaceship as measured by the timing station? (b) What time interval will the station clock record between the passage of the front and back ends of the ship?

A space traveler takes off from Earth and moves at speed \(0.9900 c\) toward the star Vega, which is \(26.00\) ly distant. How much time will have elapsed by Earth clocks (a) when the traveler reaches Vega and (b) when Earth observers receive word from the traveler that she has arrived? (c) How much older will Earth observers calculate the traveler to be (measured from her frame) when she reaches Vega than she was when she started the trip?

To eight significant figures, what is speed parameter \(\beta\) if the Lorentz factor \(\gamma\) is \(\begin{array}{lll}\text { (a) } 1.010 & 000 & 0, \text { (b) } & 10.000000, & \text { (c) } 100.00000 \text { , }\end{array}\) and (d) \(1000.0000\) ?

In a high-energy collision between a cosmic-ray particle and a particle near the top of Earth's atmosphere, \(120 \mathrm{~km}\) above sea level, a pion is created. The pion has a total energy \(E\) of \(1.35 \times 10^{5}\) \(\mathrm{MeV}\) and is traveling vertically downward. In the pion's rest frame, the pion decays \(35.0 \mathrm{~ns}\) after its creation. At what altitude above sea level, as measured from Earth's reference frame, does the decay occur? The rest energy of a pion is \(139.6 \mathrm{MeV}\).

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