Đồng thời, đo độ dài và biến đổi vận tốc
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| Lorentz transformation/ˈlɒrents trænsfɔːˈmeɪʃn/ | The relativistic rule converting event coordinates between inertial frames. |
| simultaneity/ˌsɪməltəˈneɪɪti/ | Two events happening at the same time in a given frame. |
| spacelike/ˈspeɪslaɪk/ | Too far apart in space for light to connect the two events. |
Who went first?
- Two lightning strikes are simultaneous on the ground. A fast train passenger sees the forward one first — and both observers are right.
- GRE relativity tests Lorentz transformations 洛伦兹变换 of event pairs, simultaneity 同时性, length measurement, the interval and velocity addition.
Transform separations
- $S'$ moving at $+v$: $\Delta x' = \gamma(\Delta x - v\Delta t)$, $\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$.
- Simultaneous in $S$ ($\Delta t = 0$) but apart: $\Delta t' = -\gamma v\Delta x/c^2$ — the event farther along $+x$ happens first in $S'$.
Two events are simultaneous in $S$, with B farther along $+x$. In a frame moving at $+v$,
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Δt′ = −γvΔx/c² < 0 for B − A.
Measuring a moving rod
- Length means both ends read at the same time in your frame. Then $L = L_0/\gamma$, with $L_0$ the rest length.
- At $0.6c$ ($\gamma = 1.25$) a 2 m rod is 1.6 m long in the lab.
A rod of rest length 3 m moves at $0.8c$. Its lab length in m?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
At $0.8c$, $\gamma = 5/3$; the moving rod contracts to $3/(5/3) = 1.8$ m.
The invariant interval
- $s^2 = c^2\Delta t^2 - \Delta x^2$ is the same in all frames. $s^2 > 0$ timelike (proper time $s/c$, order fixed); $s^2 < 0$ spacelike 类空 (order can swap; no signal connects them); $s^2 = 0$ lightlike.
$\Delta x = 900$ m, $c\Delta t = 600$ m: spacelike. A frame moving at $v = c^2\Delta t/\Delta x = \tfrac23c$ sees them simultaneous, $\sqrt{900^2 - 600^2} \approx 671$ m apart.
Timelike, spacelike or lightlike?
Classify event pairs by the sign of the interval.
Two events have $\Delta x = 1200$ m and $c\Delta t = 1500$ m. Proper time between them in μs ($c = 3\times10^8$ m/s)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
√(1500² − 1200²) = 900 m; 900/c.
The time order of two spacelike-separated events can differ between inertial frames.
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Only timelike order is fixed.
Velocity addition
- $u = \dfrac{u' + v}{1 + u'v/c^2}$. A $0.5c$ probe from a $0.5c$ ship: $u = 0.8c$. With $u' = c$, $u = c$.
- At low speeds it reduces to $u' + v$.
A probe leaves a $0.6c$ ship forwards at $0.6c$ relative to it. Lab speed in units of $c$ (three decimal places)?
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Relativistic velocity addition: $(0.6 + 0.6)/(1 + 0.36) = 1.2/1.36 = 0.882c$.
Recap
- Transform $\Delta x$ and $\Delta t$ together; simultaneity is relative.
- $L = L_0/\gamma$ needs endpoints read simultaneously in the measuring frame.
- Interval sign decides causal order; add velocities relativistically.
- Lorentz transformation: The relativistic rule converting event coordinates between inertial frames.
- simultaneity: Two events happening at the same time in a given frame.
- spacelike: Too far apart in space for light to connect the two events.