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As seen in frame \(\mathcal{S}\), two rockets are approaching one another along the \(x\) axis traveling with equal and opposite velocities of \(0.9 c .\) What is the velocity of the rocket on the right as measured by observers in the one on the left? [This and the previous two problems illustrate the general result that in relativity the "sum" of two velocities that are less than \(c \text { is always less than } c . \text { See Problem } 15.43 .]\)

Short Answer

Expert verified
The velocity is approximately \(0.9945c\).

Step by step solution

01

Identify the Given Information

We have two rockets approaching each other along the \(x\)-axis with velocities \(v = 0.9c\) and \(-v = -0.9c\). We need to find the velocity of one as seen from the other.
02

Apply the Velocity Addition Formula

In special relativity, the velocity addition formula is given by:\[v' = \frac{u + v}{1 + \frac{uv}{c^2}}\]where \(u\) and \(v\) are the velocities of the two objects. In this case, \(u = 0.9c\) and \(v = -0.9c\).
03

Substitute the Velocities

Substitute \(u = 0.9c\) and \(v = -0.9c\) into the formula:\[v' = \frac{0.9c + (-0.9c)}{1 + \frac{(0.9c)(-0.9c)}{c^2}}\]
04

Simplify the Expression

Calculate the numerator and the denominator:- Numerator: \(0.9c - 0.9c = 0\)- Denominator: \(1 - \frac{0.81c^2}{c^2} = 1 - 0.81 = 0.19\).Then:\[v' = \frac{0}{0.19} = 0\]
05

Interpret the Result

The calculated result of \(v' = 0\) implies zero velocity with respect to each other, which is incorrect due to a mistake. Correcting: Both velocities should be considered positive in opposite directions; re-calculate:\[v' = \frac{0.9c + 0.9c}{1 + \frac{(0.9c)(0.9c)}{c^2}} = \frac{1.8c}{1 + 0.81} = \frac{1.8c}{1.81} = 0.9945c\]
06

Conclude the Calculation

Therefore, the velocity of the rocket on the right as measured by observers in the left rocket is approximately \(0.9945c\).

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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 fundamental theory in physics proposed by Albert Einstein in 1905. It describes how the laws of physics are identical for all non-accelerating observers, and it introduced groundbreaking concepts about the nature of space and time. One of the main features of special relativity is that the speed of light in a vacuum is the same for all observers, regardless of the motion of the light source or observer. This universality leads to several surprising results, such as time dilation and length contraction, challenging our intuitive understandings.
Special relativity implies that time and space are intertwined into a single four-dimensional continuum known as spacetime. This theory fundamentally changes how we perceive velocity and movements at high speeds, particularly those approaching the speed of light. In this context, velocities don't add in the straightforward way they do at low speeds due to relativistic effects. This is where the velocity addition formula becomes crucial.
Velocity Addition Formula
The velocity addition formula in special relativity helps understand how to combine velocities in a way consistent with Einstein's propositions. When dealing with high velocities (often a significant fraction of the speed of light, denoted as \(c\)), this formula is essential.
Here is the equation: \[ v' = \frac{u + v}{1 + \frac{uv}{c^2}} \]
  • \(v'\) is the resultant velocity as observed from one reference frame.
  • \(u\) and \(v\) are the velocities of two objects as measured in another reference frame.
In classical physics, two velocities simply add together. However, in special relativity, the velocities are not just summed - they are modified by the factor \(1 + \frac{uv}{c^2}\). This ensures that the resulting velocity never exceeds the speed of light. It's a corrective mechanism for when objects move at significant fractions of \(c\).
This formula ensures that even with both velocities close to \(c\), the result remains below \(c\), evidencing how the nature of velocity addition is fundamentally altered under special relativity.
Speed of Light Limit
The speed of light limit is one of the most intriguing aspects of special relativity. According to Einstein's theory, nothing can travel faster than the speed of light in a vacuum, which is approximately 299,792,458 meters per second. This speed limit is not just a property of light but a fundamental feature of the universe, representing the maximum speed at which information or matter can travel.
A common misinterpretation before special relativity was the assumption that speeds could add indefinitely. However, as velocity addition formula implies, even when summing velocities close to the speed of light, the resulting velocity cannot surpass \(c\). This is due to the factor \(1 + \frac{uv}{c^2}\) in the denominator of the addition formula, ensuring the speed of light remains an upper limit.
  • The limitation ensures causality, meaning that cause always precedes effect.
  • It prevents any signal from traveling instantaneously, preserving the integral space-time structure.
This limit ensures compliance with the fundamental laws of physics, leading to unique outcomes, such as objects appearing to move slower or faster due to relativistic effects, depending on the observer's frame of reference.

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Most popular questions from this chapter

Consider two events that occur simultaneously at \(t=0\) in frame \(\mathcal{S},\) both on the \(x\) axis at \(x=0\) and \(x=a\). (a) Find the times of the two events as measured in a frame \(\mathcal{S}^{\prime}\) traveling in the positive direction along the \(x\) axis with speed \(V\). (b) Do the same for a second frame \(\mathcal{S}^{\prime \prime}\) traveling at speed \(V\) but in the negative direction along the \(x\) axis. Comment on the time ordering of the two events as seen in the three different frames. This startling result is discussed further in Section 15.10 .

(a) What is a mass of \(1 \mathrm{MeV} / c^{2}\) in kilograms? ( \(\mathbf{b}\) ) What is a momentum of \(1 \mathrm{MeV} / c\) in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s} ?\)

Show that any two zero-mass particles have a CM frame, provided their three- momenta are not parallel. [Hint: As you should explain, this is equivalent to showing that the sum of two forward light-like vectors is forward time-like, unless the spatial parts are parallel.]

A particle \(a\) traveling along the positive \(x\) axis of frame \(\mathcal{S}\) with speed 0.5c decays into two identical particles, \(a \rightarrow b+b,\) both of which continue to travel on the \(x\) axis. (a) Given that \(m_{a}=2.5 m_{b},\) find the speed of either \(b\) particle in the rest frame of particle \(a .\) (b) By making the necessary transformation on the result of part (a), find the velocities of the two \(b\) particles in the original frame S.

Prove that if \(x\) is time-like and \(x \cdot y=0,\) then \(y\) is space-like.

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