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Dec 6, 2023

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1. According to general rela0vity, 0me dila0on occurs in gravita0onal fields. The strength of the gravita0onal field depends on the distance from the source of gravity. In this case, the clock on the ground experiences a slightly stronger gravita0onal field than the clock on top of your head. According to the principle of general rela0vity, 0me passes more slowly in stronger gravita0onal fields. Therefore, the clock on the ground would run slightly slower than the clock on top of your head. The 0me difference between the two clocks is rela0vely small and can be calculated using the gravita0onal 0me dila0on formula: \[ \Delta t = \Delta t_0 \sqrt{1 - \frac{2GM}{rc^2}} \] where: - \(\Delta t\) is the 0me interval experienced by the clock in the stronger gravita0onal field (on the ground), - \(\Delta t_0\) is the 0me interval experienced by the clock in the weaker gravita0onal field (on top of your head), - G is the gravita0onal constant, - M is the mass of the Earth (assuming the gravita0onal field is due to the Earth), - r is the distance from the center of the Earth to the clocks, - c is the speed of light. Assuming the clocks are at a typical height of around 1.7 meters, and neglec0ng other factors like the Earth's rota0on, the difference in 0me between the clock on the ground and the clock on top of your head would be extremely small, on the order of nanoseconds or even less. Therefore, you would have to wait an imprac0cally long 0me for the two clocks to differ by one second. The effect is minuscule in everyday scenarios and is usually only significant in extreme condi0ons, such as near massive celes0al bodies like black holes. 1.4 To complete the table, we can use the concept of the space0me interval, which is an invariant quan0ty in special rela0vity. The space0me interval (\(ds\)) is defined as: \[ ds^2 = c^2 \Delta t^2 - \Delta x^2 \] where: - \(ds\) is the space0me interval, - \(c\) is the speed of light, - \(\Delta t\) is the 0me separa0on, - \(\Delta x\) is the space separa0on.

In special rela0vity, the space0me interval is the same for all observers, regardless of their rela0ve mo0on. This allows us to relate the measurements made by different observers. Given the laboratory observer's measurements (0me lapse and space separa0on) and the rocket observer's measurements (0me lapse), you can use the space0me interval to find the missing quan00es. Here's how you can do it: 1. For each row, you can use the laboratory observer's measurements to calculate the space0me interval (\(ds\)) using the formula: \[ ds^2 = c^2 \Delta t_{\text{laboratory}}^2 - \Delta x_{\text{laboratory}}^2 \] 2. Once you have \(ds\), you can use the rocket observer's 0me lapse (\(\Delta t_{\text{rocket}}\)) to find the missing space separa0on (\(\Delta x_{\text{rocket}}\)) using the rearranged formula: \[ \Delta x_{\text{rocket}} = \sqrt{c^2 \Delta t_{\text{rocket}}^2 - ds^2} \] Now, let's apply this approach to the given table: ``` SPACE AND TIME SEPARATIONS 20 29 21 a ? 10.72 5.95 b 20 ? 99 c 66.8 72.9 ? d ? 8.34 6.58 e 21 22 ? ``` 1. For row a: \[ ds^2 = c^2 \0mes (10.72)^2 - (5.95)^2 \] Calculate \(ds^2\) and then use it to find the missing rocket distance (\(\Delta x_{\text{rocket}}\)). 2. For row b: \[ ds^2 = c^2 \0mes (20)^2 - (99)^2 \] Calculate \(ds^2\) and then use it to find the missing rocket distance (\(\Delta x_{\text{rocket}}\)). 3. For row c: \[ ds^2 = c^2 \0mes (72.9)^2 - (66.8)^2 \] Calculate \(ds^2\) and then use it to find the missing rocket distance (\(\Delta x_{\text{rocket}}\)).

4. For row d: \[ ds^2 = c^2 \0mes (8.34)^2 - (6.58)^2 \] Calculate \(ds^2\) and then use it to find the missing rocket distance (\(\Delta x_{\text{rocket}}\)). 5. For row e: \[ ds^2 = c^2 \0mes (22)^2 - (21)^2 \] Calculate \(ds^2\) and then use it to find the missing rocket distance (\(\Delta x_{\text{rocket}}\)). Note: Make sure to use consistent units (e.g., seconds and light-seconds) throughout the calcula0ons. 1.7 To compute the space0me interval between two events (1 and 2, 1 and 3, 1 and 4, and 1 and 5), we use the space0me interval formula in special rela0vity: \[ ds^2 = c^2 \Delta t^2 - \Delta x^2 \] where: - \( ds \) is the space0me interval, - \( c \) is the speed of light, - \( \Delta t \) is the 0me separa0on between the events, - \( \Delta x \) is the space separa0on between the events. Given the space0me map, we can read the values of \( \Delta t \) and \( \Delta x \) directly from the graph. ### a) Between Event 1 and Event 2: \[ \Delta t_{12} = 2 \, \text{meters} \] \[ \Delta x_{12} = 0 \, \text{meters} \] \[ ds_{12}^2 = c^2 \cdot (2 \, \text{meters})^2 - (0 \, \text{meters})^2 \] \[ ds_{12}^2 = 4c^2 \] ### b) Between Event 1 and Event 3: \[ \Delta t_{13} = 4 \, \text{meters} \] \[ \Delta x_{13} = 4 \, \text{meters} \] \[ ds_{13}^2 = c^2 \cdot (4 \, \text{meters})^2 - (4 \, \text{meters})^2 \] \[ ds_{13}^2 = 12c^2 \] ### c) Between Event 1 and Event 4:

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\[ \Delta t_{14} = 6 \, \text{meters} \] \[ \Delta x_{14} = 2 \, \text{meters} \] \[ ds_{14}^2 = c^2 \cdot (6 \, \text{meters})^2 - (2 \, \text{meters})^2 \] \[ ds_{14}^2 = 32c^2 \] ### d) Between Event 1 and Event 5: \[ \Delta t_{15} = 8 \, \text{meters} \] \[ \Delta x_{15} = 0 \, \text{meters} \] \[ ds_{15}^2 = c^2 \cdot (8 \, \text{meters})^2 - (0 \, \text{meters})^2 \] \[ ds_{15}^2 = 64c^2 \] ### e) Rocket moving from Event 1 to Event 2: Since the events 1 and 2 occur at the same place in the rocket frame, \( \Delta x_{12} = 0 \) and \( \Delta t_{12} \) is the 0me interval recorded on the rocket clock. Therefore, \( \Delta t_{12} = 2 \) meters. These are the values for the space0me intervals between the specified events on the space0me map. 2.2 Bouncing on a trampoline while standing on a bathroom scale can provide an interes0ng insight into the rela0onship between weight, gravita0onal force, and accelera0on. Let's analyze the situa0on: 1. **At the Bolom of the Bounce:** - When you are at the bolom of the bounce, your feet are momentarily at rest, and you experience maximum weight. The scale reads your actual weight, which includes both your mass and the gravita0onal force ac0ng on you. 2. **During the Jump Up:** - As you start moving up, the scale reading decreases. This is because the force exerted by the scale is not enough to counteract the gravita0onal force pulling you downward. The scale will s0ll show a posi0ve value but less than your actual weight. 3. **At the Top of the Bounce:** - At the highest point of your bounce, your velocity momentarily becomes zero, and you experience weightlessness. The scale reads zero because there is no normal force ac0ng on you. You are in free fall under the inﬂuence of gravity, and there is no contact force between you and the scale. 4. **During the Descent:** - As you start coming back down, the scale reading becomes nega0ve. This is because, during the descent, the scale is providing an upward normal force that is greater than the gravita0onal

force, resul0ng in a net force upward. The nega0ve reading indicates that the force from the scale is greater than your weight. **Longest Part of the Cycle for Free Float:** During the part of the bounce when you are in free fall, which occurs at the top of the jump, you can consider yourself to be in a free-ﬂoat frame. This is the longest part of the cycle where you experience weightlessness, and the scale reads zero. It's important to note that neglec0ng air resistance assumes an idealized scenario. In reality, air resistance would have some effect on the mo0on, and the actual readings on the scale might be inﬂuenced by various factors like the elas0city of the trampoline and the precision of the scale. 2.5 Let's break down the problem into parts: ### Part a: **Given:** - Speed of the elementary par0cle, \(v = 0.96 c\) (where \(c\) is the speed of light) - Width of the spark chamber, \(d = 1\) meter - Accelera0on due to gravity, \(g = 10 \, \text{m/s}^2\) **1. Time in Transit Through the Spark Chamber:** The 0me (\(\Delta t\)) for the par0cle to travel through the spark chamber can be found using the 0me dila0on formula in special rela0vity: \[ \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}} \] where \( \Delta t_0 \) is the proper 0me (0me measured in the par0cle's frame). Since the par0cle is moving perpendicular to the direc0on of mo0on through the spark chamber, \( \Delta t_0 \) is the same as the laboratory 0me. \[ \Delta t = \frac{1}{\sqrt{1 - 0.96^2}} \] **2. Time the Experiment is "In Progress":** This is the laboratory 0me, so it's the same as the 0me in transit through the spark chamber. **3. Distance of Fall for a Separate Test Par0cle:** Use the equa0on of mo0on for free fall: \[ h = \frac{1}{2} g t^2 \] where \( h \) is the distance of fall, \( g \) is the accelera0on due to gravity, and \( t \) is the 0me of free fall.

\[ h = \frac{1}{2} \0mes 10 \0mes \leu(\frac{1}{\sqrt{1 - 0.96^2}}\right)^2 \] ### Part b: **Given:** - Sensi0vity of detec0on, \( \Delta x = 5 \0mes 10^{-9} \) meters (500 nanometers) - Wavelength of visible light, \( \lambda = 5 \0mes 10^{-7} \) meters (500 nanometers) - Accelera0on due to gravity, \( g = 10 \, \text{m/s}^2 \) **1. Time to Fall the Distance of One Wavelength:** Use the equa0on of mo0on for free fall: \[ \Delta t_{\text{fall}} = \sqrt{\frac{2 \Delta x}{g}} \] **2. Distance the Elementary Par0cle Moves in that Time:** Use the rela0vis0c velocity equa0on: \[ \Delta x_{\text{par0cle}} = v \Delta t_{\text{fall}} \] **3. Maximum Width of the Spark Chamber for Detec0on Sensi0vity:** The width of the spark chamber can be determined by mul0plying the speed of light by the laboratory 0me (\(\Delta t\)). \[ d_{\text{max}} = c \0mes \frac{1}{\sqrt{1 - 0.96^2}} \] These calcula0ons should give you the values you need for both parts a and b of the problem.

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