H.W. 01 관련

• 1. Q-1.10
• 2. Q-1.13
• 3. Q-1.20(b)
• 4. Q-1.22
• 5. Q-1.24
• 6. Q-1.37
• 7. Q-1.41
• 8. Q-1.44

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1. Q-1.10

Suppose that $A'$, $B'$, and $C'$ are at rest in frame $S'$, which moves with respect to $S$ at speed $v$ in the $+x$ direction. Let $B'$ be located exactly midway between $A'$ and $C'$. At $t'=0$, a light flash occurs at $B'$ and expands outward as a spherical wave.

(a)

According to an observer in $S'$, do the wave fronts arrive at $A'$ and $C'$ simultaneously?

(a) - Answer

According to $S'$ frame, the wave fronts arrive simultaneously at $A'$, and $C'$.

(b)

According to an observer in $S$, do the wave fronts arrive at $A'$ and $C'$ simultaneously?

(b) - Answer

According to $S'$ frame, the wave fronts DO NOT arrive simultaneously at $A'$, and $C'$.

(c)

If you answered no to either (1) or (2), what is the difference in their arrival times and at which point did the front arrive first?

(c) - Answer

$$\begin{align*} \Delta{t}&=t_2-t_1,&&\begin{cases}t_2:&\text{time wavefront hits }C'\\t_1:&\text{time wavefront hits }A'\end{cases}\\ \end{align*}$$

$$\begin{align*} \Delta{t}&=\left(\frac{d_2}{v_2}\right)-\left(\frac{d_1}{v_1}\right)\\ &=\frac{\left(B'C'\right)}{\left(c-\mathbf{V}\right)}-\frac{\left(B'A'\right)}{\left(c+\mathbf{V}\right)}, & &\left<\begin{matrix}\text{wavefront arrives at A' first}\\\text{ according to }S'\text{ frame}\end{matrix}\right>\\ &=d\left(\frac{(1)}{c-\mathbf{V}}-\frac{(1)}{c+\mathbf{V}}\right), &&\left<B'C'=B'A'=d'\right>\\ &=d\left(\frac{(c+\mathbf{V})-(c-\mathbf{V})}{(c-\mathbf{V})(c+\mathbf{V})}\right)\\ &=2d\left(\frac{\mathbf{V}}{c^2-\mathbf{V}^2}\right); \end{align*}$$

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2. Q-1.13

Suppose that an event occurs in inertial frame $$S$$ with coordinates $x=75\:m$, $y=18\:m$, $z=4.0\:m$ at $t=2.0\times10^{-5}\:s$. The inertial frame $S'$ moves in the $+x$ direction with $v=0.85c$. The origins of $S$ and $S'$ coincided at $t=t'=0$.

(a)

What are the coordinates of the event in $S'$?

(a) - Answwer

$$\begin{align*} \underset{\text{Lorentz Transformation}}{ \begin{cases} x'=\gamma(x-\mathbf{V}t);\\ y'=y;\\ z'=z;\\ t'=\gamma\left(t-\frac{\mathbf{V}}{c^2}x\right) \end{cases} } ,&&\left<\gamma=\frac{1}{\sqrt{1-\left(\frac{\mathbf{V}}{c}\right)^2}}=\frac{1}{\sqrt{1-\left(\frac{0.85c}{c}\right)^2}}=1.8983;\right> \end{align*}$$

$$\begin{align*} x'&=\gamma(x-\mathbf{V}t)\\ &=(1.8983)\left((75)-(0.85)\left(3.0\times10^8\right)\left(2.0\times10^{-5}\right)\right)\\ &=-9.538\times10^3\:\text{m}\\ y'&=y=18\:\text{m}\\ z'&=z=4\:\text{m}\\ t'&=\gamma\left(t-\frac{\mathbf{V}}{c^2}x\right)\\ &=(1.8983)\left((2.0\times10^{-5})-\left(\frac{0.85c}{c^2}\right)(75)\right)\\ &=3.756\times10^{-5}\:\text{s}\\\\ \end{align*}$$

$$\therefore \begin{cases} x'=-9.538\times10^{3}\:\text{m};\\ y'=18\:\text{m};\\ z'=4\:\text{m};\\ t'=3.756\times10^{-5}\:\text{s} \end{cases}$$

(b)

Use the inverse transformation on the results of (1) to obtain the original coordinates.

(b) - Answwer

$$\begin{align*} \underset{\text{Inverse Lorentz Transformation}}{ \begin{cases} x=\gamma(x'+\mathbf{V}t);\\ y=y';\\ z=z';\\ t=\gamma\left(t'+\frac{\mathbf{V}}{c^2}x'\right) \end{cases} } ,&&\left<\gamma=\frac{1}{\sqrt{1-\left(\frac{\mathbf{V}}{c}\right)^2}}=\frac{1}{\sqrt{1-\left(\frac{0.85c}{c}\right)^2}}=1.8983;\right> \end{align*}$$

$$\begin{align*} x&=\gamma(x'-\mathbf{V}t')\\ &=(1.8983)\left((-9.538\times10^3)-(0.85)\left(3.0\times10^8\right)\left(3.756\times10^{-5}\right)\right)\\&=75.8\:\text{m}\\ y'&=y=18\:\text{m}\\ z'&=z=4\:\text{m}\\ t&=\gamma\left(t'+\frac{\mathbf{V}}{c^2}x'\right)\\ &=(1.8983)\left((3.756\times10^{-5})-\left(\frac{0.85c}{c^2}\right)(-9.538\times10^3)\right)\\&=2.0\times10^{-5}\:\text{s}\\\\ \end{align*}$$

$$\therefore \begin{cases} x=75.8\:\text{m};\\ y=18\:\text{m};\\ z=4\:\text{m};\\ t=2.0\times10^{-5}\:\text{s} \end{cases}$$

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3. Q-1.20(b)

Question

Use the binomial expansion to derive the following results for values of $v\ll{c}$ and use when applicable in the problems that follow in this section.

$$\frac{1}{\gamma}\approx1-\frac{1}{2}\frac{v^2}{c^2}$$

Answer

$$\frac{1}{\gamma}=\sqrt{1-\left(\frac{\mathbf{V}}{c}\right)^2};$$

BINOMIAL EXPENSION:

$$\begin{align*} f(\epsilon)&=f(0)+\frac{f'(0)\epsilon^1}{1!}+\frac{f''(0)\epsilon^2}{2!}+\cdots\\ &\left<\frac{1}{\gamma}=f(\epsilon)=\left[1-\epsilon\right]^{\frac{1}{2}},\right>\\ &\left<\epsilon=\frac{\mathbf{V}^2}{c^2},\:\:\:\:\epsilon'\approx\epsilon''\approx\epsilon\right>\\ \end{align*}$$

$$\left\{\begin{align*} f(0)&=\left[1-(0)\right]^{\frac{1}{2}}=1;\\ f'(\epsilon)&=\left(\frac{1}{2}\right)\left(1-(\epsilon)\right)^{-\frac{1}{2}}(-1);&f'(0)&=\left(\frac{1}{2}\right)(-1)=-\frac{1}{2};\\ f''(\epsilon)&=\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(1-(\epsilon)\right)^{-\frac{3}{2}}(-1);&f''(0)&=\left(-\frac{1}{4}\right)(-1)=\frac{1}{4};\\ \vdots& \end{align*}\right\}$$

$$\begin{align*} \gamma&\approx1+\frac{\left(-\frac{1}{2}\right)}{1!}\epsilon+\frac{\left(\frac{1}{4}\right)}{2!}\epsilon^2+\cdots+\\ &=1-\frac{1}{2}\epsilon+\frac{1}{8}\epsilon^2+\cdots,&\left<\epsilon=\frac{\mathbf{V}^2}{c^2}\right>\\ &=1-\frac{1}{2}\left(\frac{\mathbf{V}^2}{c^2}\right) +\frac{1}{8}\left(\frac{\mathbf{V}^2}{c^2}\right)^2+\cdots\\ &\approx1-\frac{1}{2}\left(\frac{\mathbf{V}^2}{c^2}\right) \end{align*}$$

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4. Q-1.22

A nova is the sudden, brief brightening of a star (see Chapter 13). Suppose Earth astronomers see two novas occur simultaneously, one in the constellation Orion (The Hunter) and the other in the constellation Lyra (The Lyre). Both nova are the same distance from Earth, $2.5 \times10^3\:c\cdot{y}$, and are in exactly opposite directions from Earth. Observers on board an aircraft flying at $1000\:\text{km}/\text{h}$ on a line from Orion toward Lyra see the same novas but note that they are not simultaneous.

(a)

For the observers on the aircraft, how much time separates the novas?

(a) - Answer

(b)

Which one occurs first? (Assume Earth is an inertial reference frame.)

(b) - Answer

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5. Q-1.24

The proper mean lifetime of $\pi$ mesons (pions) is $2.6\times10^{-8}\:\text{s}$. Suppose a beam of such particles has speed $0.9c$.

(a)

What would their mean life be as measured in the laboratory?

(a) - Answer

(b)

How far would they travel (on the average) before they decay?

(b) - Answer

(c)

What would your answer be to part (3) if you neglected time dilation?

(c) - Answer

(d)

What is the interval in spacetime between creation of a typical pion and its decay?

(d) - Answer

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6. Q-1.37

Question

Einstein used trains for a number of relativity thought experiments since they were the fastest objects commonly recognized in those days. Let's consider a train moving at $0.65c$ along a straight track at night. Its headlight produces a beam with an angular spread of $60^{\circ}$ according to the engineer. If you are standing alongside the track (rails are $1.5\:\text{m}$ apart), how far from you is the train when you see its approaching headlight suddenly disappear?

Answer

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7. Q-1.41

A friend of yours who is the same age as you travels to the star Alpha Centauri, which is $4c\cdot{y}$ away, and returns immediately. She claims that the entire trip took just 6 years.

(a)

How fast did she travel?

(a) - Answer

(b)

How old are you when she returns?

(b) - Answer

(c)

Draw a spacetime diagram that verifies your answer to (a) and (b).

(c) - Answer

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8. Q-1.44

A burst of $p+$ mesons (pions) travels down an evacuated beam tube at Fermilab moving at $\beta=0.92$ with respect to the laboratory.

(a)

Compute $\gamma$ for this group of pions.

(a) - Answer

(b)

The proper mean lifetime of pions is $2.6\times10^{-8}\:\text{s}$. What mean lifetime is measured in the lab?

(b) - Answer

(c)

If the burst contained 50,000 pions, how many remain after the group has traveled $50\:\text{m}$ down the beam tube?

(c) - Answer

(d)

What would be the answer to (c) ignoring time dilation?

(d) - Answer

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