Inertial frames of references cannot be when 2 frames are accelerating relative to each other. (In circular motion there is acceleration)
CloseThe other postulate was that the laws of physics are the same in all inertial frames of references.
CloseThe time interval on Earth is given by: \[ \Delta t = \gamma \Delta t_0, \] where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\). For \( v = 0.6c \): \[ \gamma = \frac{1}{\sqrt{1 - (0.6)^2}} = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} = 1.25. \] \[ \Delta t = \gamma \Delta t_0 = 1.25 \times 2 \, \text{s} = 2.5 \, \text{s}. \]
CloseThe contracted length is given by: \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}}. \] For \( L_0 = 5.0 \, \text{m} \) and \( v = 0.8c \): \[ L = 5.0 \sqrt{1 - (0.8)^2} = 5.0 \sqrt{1 - 0.64} = 5.0 \sqrt{0.36} = 5.0 \cdot 0.6 = 3.0 \, \text{m}. \]
CloseThe time difference between clocks in the Earth frame, as measured in the spaceship's frame, is given by the relativity of simultaneity: \[ \Delta t = \frac{v \Delta x}{c^2}, \] where \( v = 0.9c \), \( \Delta x = 2 \, \text{light-seconds} \). Substituting: \[ \Delta t = \frac{(0.9c)(2 \, \text{light-seconds})}{c^2} = \frac{1.8c^2}{c^2} = 1.8 \, \text{s}. \]
CloseThe total time in the observer's frame is: \[ t = \frac{\text{distance}}{\text{velocity}} = \frac{200 \, \text{m}}{0.8c}. \] Substituting \( c = 3 \times 10^8 \, \text{m/s} \): \[ t = \frac{200}{0.8 \times 3 \times 10^8} = \frac{200}{2.4 \times 10^8} \approx 8.33 \times 10^{-7} \, \text{s}. \] The proper time on the rod is given by \( \Delta \tau = \frac{t}{\gamma} \), where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \). For \( v = 0.8c \): \[ \gamma = \frac{1}{\sqrt{1 - (0.8)^2}} = 1.67. \] Thus: \[ \Delta \tau = \frac{8.33 \times 10^{-7}}{1.67} \approx 5 \times 10^{-7} \, \text{s}. \]
CloseThe relativistic velocity addition formula is: \[ v_{\text{relative}} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}, \] where \( v_1 = 0.6c \) and \( v_2 = 0.8c \). Substituting: \[ v_{\text{relative}} = \frac{0.6c + 0.8c}{1 + \frac{(0.6c)(0.8c)}{c^2}} = \frac{1.4c}{1 + 0.48} = \frac{1.4c}{1.48} \approx 0.946c. \]
CloseThe cause of the difference in timing is the relativity of simultaneity, which states that events that are simultaneous in one frame are not necessarily simultaneous in another frame moving relative to the first.
CloseThe time dilation formula is: \[ \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}, \] where \( \Delta t_0 = 1 \, \text{hour} \), \( v = 0.85c \), and \( \Delta t = 5 \, \text{hours} \) in the observer's frame. First, calculate \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - (0.85)^2}} = \frac{1}{\sqrt{1 - 0.7225}} = \frac{1}{\sqrt{0.2775}} \approx 1.8974. \] Now, use the time dilation formula: \[ \Delta t_0 = \frac{\Delta t}{\gamma} = \frac{5 \, \text{hours}}{1.8974} \approx 2.63 \, \text{hours}. \]
CloseWe can calculate the time dilation using the formula: \[ \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}, \] where \( \Delta t_0 = 10 \, \text{years} \) is the proper time (time measured by the astronaut), and \( v = 0.9c \). First, calculate the Lorentz factor \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - (0.9)^2}} = \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}} \approx 2.294. \] Now, calculate the time observed on Earth: \[ \Delta t = \Delta t_0 \times \gamma = 10 \times 2.294 = 22.94 \, \text{years}. \]
CloseThe event is represented at the same position in the moving frame, with \( t' = 0 \), indicating that the event occurs at the origin of the moving frame at that instant in time. In the original reference frame, this event will occur at a different position, since it happens at the same time in the moving frame but is offset in space according to the relative velocity.
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