التزامن، قياس الطول وتحويل السرعة
مُقدَّم| English | المعنى |
|---|---|
| proper length/ˈprɒpə leŋθ/ | الطول الذاتي |
| spacetime interval/ˈspeɪstaɪm ˈɪntəvl/ | الفاصل الزمكان |
A decision before an answer
- Two observers can disagree about which of two events happened first, and both can be right.
- Your goal: Transform event separations and identify proper length versus simultaneous endpoint readings.
Transform separations
- Use the Lorentz transformation with declared frame motion: if frame S′ moves at +v along x relative to S, then x′=γ(x−vt) and t′=γ(t−vx/c²) with γ=1/sqrt(1−v²/c²). Apply it to event separations Δx and Δt, not only to single coordinates.
- Two events simultaneous in S (Δt=0) at different places have Δt′=−γvΔx/c²: the event further in the +x direction happens earlier in S′. Simultaneity is frame-dependent, not a bookkeeping error.
To obtain the contracted length of a moving rod, a lab observer must record the two endpoint positions:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Length in a frame is the endpoint separation at equal time in that frame; Δt=0 in the lab gives L=L0/γ.
Measure length in one frame
- A moving rod length must be read from endpoint positions at the same lab time (Δt=0). Then Δx′=γΔx gives L=L0/γ with L0 the rest-frame (proper) length.
- Reading the endpoints simultaneously in the rod frame instead does not produce the lab length; the two reading conditions answer different questions. State the frame and the reading condition before quoting any length.
A ship moves at 0.5c and launches a probe forward at 0.5c relative to the ship. The probe has lab speed:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
(0.5+0.5)/(1+0.25)=0.8; the denominator keeps the result below c.
Classify the interval
- The interval s²=c²Δt²−Δx² (mostly-minus convention, declared once) is invariant. s²>0 is timelike: a frame exists with the events co-located, and Δτ=s/c is the proper time between them.
- s²<0 is spacelike: a frame exists with the events simultaneous, and sqrt(−s²) is the proper distance; s²=0 is lightlike. Timelike order is the same in every frame; spacelike order can swap, and no signal can join the events because |Δx/Δt|>c.
Events A and B are 900 m apart in the lab and 2 μs apart in time (cΔt=600 m using c=3×10⁸ m/s). Since 600<900 the interval is spacelike; the simultaneity frame moves at v=c²Δt/Δx=(2/3)c and the proper distance is sqrt(900²−600²)=670.82 m. A 0.5c probe from a 0.5c ship has lab speed (0.5+0.5)c/(1+0.25)=0.8c, where γ=5/3. E=5 GeV with p=3 GeV/c gives rest energy sqrt(25−9)=4 GeV.
Two lab events have Δx=900 m and Δt=2 μs. Using c=3×10⁸ m/s, the frame in which they are simultaneous moves at ____ c (give the exact fraction).
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Set Δt′=0 in t′=γ(t−vx/c²): v=c²Δt/Δx=(600/900)c=2c/3.
Add velocities and check invariants
- Collinear velocities add as u=(u′+v)/(1+u′v/c²); the denominator keeps u<c for any u′,v<c, and u′=c gives u=c exactly.
- Energy and momentum form a four-vector with invariant E²−p²c²=m²c⁴: compute it in any convenient frame. Check the nonrelativistic limit, where the addition rule reduces to u=u′+v and the invariant reduces to the rest energy.
Quoting L=L0/γ without the simultaneous-endpoint condition, or adding velocities as u=u′+v at relativistic speeds. A simultaneity difference is physical, not a mistake.
أي إجابة تتناسب مع هذه الحالة؟
تحويل انفصال الأحداث وتحديد الطول الصحيح مقابل قراءات النقاط الزمنية المتزامنة للنهايات
If two events are spacelike separated, every inertial frame agrees on which event happened first.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Spacelike order can swap between frames; only timelike or lightlike order is invariant.
Keep the distinctions
- spacetime interval 时空间隔 — The invariant combination c²Δt²−Δx² under the declared sign convention, whose sign classifies an event separation.
- proper length 固有长度 — The endpoint separation measured in the object's rest frame; any other frame must read both endpoints simultaneously in its own time.
- Transform event separations and identify proper length versus simultaneous endpoint readings.
- Apply relativistic velocity addition and check a four-vector invariant.
- Classify the interval sign to order events and rule out causal contact.
Match each term with its precise meaning in this lesson.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
Difficulty 3/5 · 200 coins + a small bonus on first completion today
The order follows the stated process; check each stage before the next.