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Which quantity is invariant-that is, has the same value-in all reference frames? a) time interval, \(\Delta t\) d) space-time interval, b) space interval, \(\Delta x\) \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\) c) velocity, \(v\)

Short Answer

Expert verified
a) Time interval, \(\Delta t\) b) Space interval, \(\Delta x\) c) Velocity, \(v\) d) Space-time interval, \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\) Answer: d) Space-time interval, \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\)

Step by step solution

01

Option a: Time Interval, \(\Delta t\)

The time interval measures the difference in time between two events. According to the theory of special relativity, time dilation occurs when an object moves relative to an observer. This means that two observers may measure different time intervals for the same event, depending on their relative velocities. Therefore, time intervals are not invariant under different reference frames.
02

Option b: Space Interval, \(\Delta x\)

The space interval measures the spatial distance between two points. In special relativity, length contraction occurs when an object moves relative to an observer. Due to length contraction, two observers may measure different space intervals for the same event, depending on their relative velocities. Hence, space intervals are not invariant under different reference frames.
03

Option c: Velocity, \(v\)

The velocity of an object is the rate of change of its position concerning time. Since both time intervals and space intervals can change depending on the relative velocities of observers, it is clear that the observed velocity of an object will also change depending on the observer's reference frame. Thus, velocities are not invariant under different reference frames.
04

Option d: Space-time Interval, \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\)

According to the theory of special relativity, the space-time interval is defined as the difference between the squared time interval multiplied by the speed of light squared (\(c^{2}(\Delta t)^{2}\)) and the squared space interval (\((, \Delta x)^{2}\)), given by the equation: \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\). This quantity remains unchanged for all observers, regardless of their relative velocities. The space-time interval is, therefore, invariant under different reference frames. In conclusion, the quantity that is invariant in all reference frames is:
05

Answer

d) Space-time interval, \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\)

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Key Concepts

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

Special Relativity
Special relativity, a fundamental theory established by Albert Einstein, revolutionized our understanding of space, time, and motion. The theory postulates that the laws of physics are the same for all non-accelerating observers, and that the speed of light in a vacuum is constant, regardless of the motion of the light source.

Special relativity introduces concepts such as time dilation and length contraction, which suggest that measurements of time and space are not absolute but depend on the relative motion of the observer and the object being observed. This goes against our everyday experiences, where time and space seem constant. Through its mathematical framework, special relativity provides tools for understanding how time and space intertwine—a concept known as the space-time continuum.

For example, if a spaceship flies past Earth at a significant fraction of the speed of light, an observer on Earth will perceive the spaceship's length as being shortened, a phenomenon known as length contraction. The same observer will also detect that time is passing more slowly for the people on the spaceship, illustrating time dilation.
Time Dilation
Time dilation is one of the most intriguing and counterintuitive predictions of special relativity. It describes how a moving clock ticks more slowly than a stationary one. Specifically, if two observers are in relative motion, each will perceive the other's clock as ticking at a slower rate than their own.

The classic 'twin paradox' provides a striking example of time dilation. If one twin travels on a spaceship close to the speed of light, while the other remains on Earth, upon return the spacefaring twin would be younger than their Earth-bound sibling. Time dilation is not merely a theoretical concept; it has practical implications and has been confirmed experimentally, such as by observing the decay rates of particles moving at high speeds or by accounting for it in the GPS satellite technology which would otherwise drift and provide inaccurate positioning data.
Space-Time Interval
The space-time interval is an essential concept in special relativity. It represents a quantity that combines measures of space intervals and time intervals into one invariant number for all observers, regardless of their state of motion. This invariance provides a way to describe events in a consistent manner.

The equation for the space-time interval, \(c^{2}(\Delta t)^{2}-(\Delta x)^{2}\), might appear daunting, but it's simply the subtraction of the square of the spatial distance between events from the square of the time distance between those same events, all multiplied by the square of the speed of light. This formula always gives the same result for all observers, which makes it a 'true' measure of the separation between two events in space-time.

Why does this matter? Because it’s a way to ensure that despite varying observations of time and space (due to, say, relative motion), there is something fundamental and unchanging in how different events are related across the universe. It champions the idea that the fabric of the universe is consistent and absolute, even if our individual perceptions are not.

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Most popular questions from this chapter

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to Earth's rest frame. Assume that the two reference frames have parallel axes and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by a distance \(l\) in the \(x^{\prime}\) -direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in Earth's reference frame.

What is the value of \(\gamma\) for a particle moving at a speed of \(0.800 c ?\)

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28 ), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, the observer sees that the particle is, in fact, moving toward the wire.)

A nucleus with rest mass \(23.94 \mathrm{GeV} / c^{2}\) is at rest in the lab. An identical nucleus is accelerated to a kinetic energy of \(10,868.96 \mathrm{GeV}\) and made to collide with the first nucleus. If instead the two nuclei were made to collide head on in a collider, what would the kinetic energy of each nucleus have to be for the collision to achieve the same center-of-mass energy?

A particle of rest mass \(m_{0}\) travels at a speed \(v=0.20 c .\) How fast must the particle travel in order for its momentum to increase to twice its original value? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

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