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Chapter 38: Problem 36

An evacuated tube at rest in the laboratory has a length \(3.00 \mathrm{~m}\) as measured in the laboratory. An electron moves at speed \(v=0.999987 \mathrm{c}\) in the laboratory along the axis of this evacuated tube. What is the length of the tube measured in the rest frame of the electron?

Short Answer

Expert verified
The length of the tube in the electron鈥檚 rest frame is approximately 0.0153 meters.

Step by step solution

01

Identify the Given Information

The length of the tube in the laboratory frame is given as \(L_0 = 3.00 \mathrm{~m}\). The speed of the electron is given as \(v = 0.999987 \mathrm{c}\), where \(c\) is the speed of light.
02

Recall the Length Contraction Formula

The length contraction formula in special relativity is given by \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\), where \(L\) is the length measured in the moving frame (rest frame of the electron), \(L_0\) is the proper length (length in the laboratory frame), \(v\) is the velocity of the moving object (electron), and \(c\) is the speed of light.
03

Substitute the Known Values

Substitute \(L_0 = 3.00 \mathrm{~m}\), \(v = 0.999987 \mathrm{c}\), and \(c\) into the length contraction formula to get \(L = 3.00 \mathrm{~m} \sqrt{1 - (0.999987)^2}\).
04

Simplify the Expression Inside the Square Root

Simplify the term inside the square root: \(1 - (0.999987)^2\). This equals approximately \(2.599\text{e-5}\).
05

Calculate the Contraction

Now compute \(\sqrt{2.599\text{e-5}}\), which is approximately \(0.0051\).
06

Determine the Contracted Length

Finally multiply \(3.00\mathrm{~m}\) by \(0.0051\). This gives us the length \(L\) in the electron's rest frame \(L = 3.00 \mathrm{~m} \times 0.0051 \approx 0.0153 \mathrm{~m}\).

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Key Concepts

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

Special Relativity
Special relativity is a theory proposed by Albert Einstein that describes the physics of objects moving at high speeds, close to the speed of light. One of its key principles is that the laws of physics are the same for all non-accelerating observers. It also states that the speed of light in a vacuum is constant and does not change regardless of the motion of the source or observer.

Special relativity leads to several extraordinary effects, including time dilation and length contraction. In our exercise, we are dealing with length contraction, which occurs when an object is moving close to the speed of light relative to an observer. The object appears shorter along the direction of motion than it would if it were at rest.
Lorentz Transformation
The Lorentz transformation is a set of equations that relate the space and time coordinates of two systems moving at a constant velocity relative to each other. These transformations are essential for understanding and calculating the effects predicted by special relativity.

The length contraction formula can be derived from the Lorentz transformations. This formula expresses how an object's length changes when measured from different reference frames: \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}}\).

Here, \(L_0\) is the proper length, which is the length of the object in the frame where it is at rest. \(L\) is the length measured in the moving frame, \(v\) is the relative velocity between the observer and the object, and \(c\) is the speed of light. By using this formula in our exercise, we see how the high speed of the electron causes the length of the tube to contract significantly in the electron's rest frame.
Proper Length
Proper length is a specific term in special relativity referring to the length of an object as measured by an observer at rest relative to the object. In our example, the proper length (\(L_0\)) of the tube is its length as measured in the laboratory, which is at rest relative to the tube. This is given as \(3.00 \text{ meters}\).

When we switch to the frame of reference moving with the electron, we need to use the length contraction formula to find how the length appears in this moving frame. As derived in the exercise, substituting the values into the formula: \(L = 3.00 \text{ meters} \sqrt{1 - (0.999987)^2}\), we find that the observed length (\(L\)) in the electron's rest frame is dramatically shorter. This illustrates how objects moving at speeds close to the speed of light are significantly contracted in the direction of motion. Proper length helps us ground these calculations in a frame of reference where the object is not in motion, making it easier to understand length contraction in relative motion.

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Most popular questions from this chapter

Evelvn Brown does not approve of our latticework of rods and clocks and the use of a light flash to synchronize them. (a) "I can synchronize my clocks in any way I choose!" she exclaims. Is she right? (b) Evelyn wants to synchronize two identical clocks, called Big Ben and Little Ben, which are at rest with respect to one another and separated by one million kilometers in their rest frame. She uses a third clock, identical in construction with the first two, that travels with constant velocity between them. As her moving clock passes Big Ben, it is set to read the same time as Big Ben. When the moving clock passes Little Ben, that outpost clock is set to read the same time as the traveling clock. "Now Big Ben and Little Ben are synchronized," says Evelyn Brown. Is Evelyn's method correct? (c) After Evelyn completes her synchronization of Little Ben by her method, how does the reading of Little Ben compare with the reading of a nearby clock on a latticework at rest with respect to Big Ben (and Little Ben) and synchronized by our standard method using a light flash? Evaluate in milliseconds any difference between the reading on Little Ben and the nearby lattice clock in the case that Evelyn's traveling clock moved at a constant velocity of 500000 kilometers per hour from Big Ben to Little Ben. (d) Evaluate the difference in the reading between the EvelynBrown- synchronized Little Ben and the nearby lattice clock when Evelyn's synchronizing traveling clock moves 1000 times as fast as the speed given in part (c).

An unstable high-energy particle is created in a collision inside a detector and leaves a track \(1.05 \mathrm{~mm}\) long before it decays while still in the detector. Its speed relative to the detector was \(0.992 c\). How long did the particle live as recorded in its rest frame?

(a) Can a person, in principle, travel from Earth to the center of our galaxy, which is 23000 ly distant, in one lifetime? Explain using either length contraction or time dilation arguments. (b) What constant speed with respect to the galaxy is required to make the trip in \(30 \mathrm{y}\) of the traveler's lifetime?

(a) Find an equation for the unknown mass \(m\) of a particle if you know its momentum \(p\) and its kinetic energy \(K\). Show that this expression reduces to an expected result for nonrelativistic particle speeds. (b) Find the mass of a particle whose kinetic energy is \(K=55.0 \mathrm{MeV}\) and whose momentum is \(p=\) \(121 \mathrm{MeV} / \mathrm{c}\). Express your answer as a decimal fraction or multiple of the mass \(m_{\mathrm{e}}\) of the electron.

You see a sudden eruption on the surface of the Sun. From solar theory you predict that the eruption emitted a pulse of particles that is moving toward the Earth at oneeighth the speed of light. How long do you have to seek shelter from the radiation that will be emitted when the particle pulse hits the Earth? Take the light-travel time from the Sun to the Earth to be 8 minutes.
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