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GRE Subject Test · GRE Physics الموضوع 42 5:44 سرد باللغة الإنجليزية · ترجمة مدمجة بالإنجليزية + الصينية
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Two observers can disagree about which of two events happened first, and both can be right.
两个观察者可能对两个事件谁先发生意见不同，而且两人都可能是对的。
This topic transforms the separations between events, separates proper length from a moving rod's measured length, adds velocities relativistically, and uses the interval to decide whether events can be causally connected.
本主题变换事件之间的间隔，区分固有长度与运动杆的测量长度，用相对论方法叠加速度，并用时空间隔判断事件之间能否有因果联系。
Use the Lorentz transformation with declared frame motion.
使用洛伦兹变换并说明参考系的运动。
If frame S prime moves at plus v along x relative to S, then x prime equals gamma times x minus v t, and t prime equals gamma times t minus v x over c squared, with gamma equal to 1 over the square root of 1 minus v squared over c squared.
若 S 撇系相对 S 系沿 x 方向以正 v 运动，则 x 撇等于 γ 乘以 x 减 vt，t 撇等于 γ 乘以 t 减 vx 除以 c 平方，其中 γ 等于 1 除以根号下 1 减 v 平方除以 c 平方。
Apply it to the event separations delta x and delta t, not only to single coordinates.
要把变换用于事件间隔 Δx 和 Δt，而不只是单个坐标。
Two events simultaneous in S, with delta t equal to zero, at different places have delta t prime equal to minus gamma v delta x over c squared; the event further in the positive x direction happens earlier in S prime.
在 S 系中同时（Δt 等于零）但位于不同地点的两个事件，其 Δt 撇等于负 γ v Δx 除以 c 平方：沿正 x 方向更远的事件在 S 撇系中更早发生。
Simultaneity is frame-dependent, not a bookkeeping error.
同时性依赖于参考系，而不是记账错误。
A moving rod's length must be read from endpoint positions at the same lab time, with delta t equal to zero.
运动杆的长度必须在同一实验室时刻（Δt 等于零）读取两端位置。
Then delta x prime equals gamma delta x gives L equals L nought over gamma, with L nought the rest-frame, or proper, length.
此时 Δx 撇等于 γ Δx，得出 L 等于 L0 除以 γ，其中 L0 为静止系中的固有长度。
Reading the endpoints simultaneously in the rod frame instead does not produce the lab length; the two reading conditions answer different questions.
若改为在杆的参考系中同时读取两端，得到的不是实验室长度；两种读数条件回答的是不同的问题。
State the frame and the reading condition before quoting any length.
引用任何长度之前，先说明参考系和读数条件。
The interval s squared equals c squared delta t squared minus delta x squared, in the mostly-minus convention declared once, is invariant.
时空间隔 s 平方等于 c 平方 Δt 平方减 Δx 平方（采用一次声明的“主负”约定），是不变量。
If s squared is positive, the interval is timelike; a frame exists with the events co-located, and delta tau equals s over c is the proper time between them.
s 平方为正时为类时间隔：存在一个参考系使两事件同地发生，Δτ 等于 s 除以 c，是两者之间的固有时。
If s squared is negative, it is spacelike; a frame exists with the events simultaneous, and the square root of minus s squared is the proper distance.
s 平方为负时为类空间隔：存在一个参考系使两事件同时发生，负 s 平方的平方根是固有距离。
If s squared is zero, it is lightlike.
s 平方为零时为类光间隔。
Timelike order is the same in every frame; spacelike order can swap, and no signal can join the events because the size of delta x over delta t exceeds c.
类时事件的先后顺序在所有参考系中相同；类空事件的顺序可以颠倒，且没有信号能连接这两个事件，因为 Δx 除以 Δt 的绝对值大于 c。
Collinear velocities add as u equals u prime plus v, over 1 plus u prime v over c squared; the denominator keeps u below c for any u prime and v below c, and u prime equal to c gives u equal to c exactly.
共线速度相加公式为 u 等于 u 撇加 v，除以 1 加 u 撇 v 除以 c 平方。 分母确保对于任意小于 c 的 u 撇和 v，u 均小于 c；且当 u 撇等于 c 时，u 恰好等于 c。
Energy and momentum form a four-vector with the invariant E squared minus p squared c squared equals m squared c to the fourth; compute it in any convenient frame.
能量与动量构成四维矢量，其不变量为 E 平方减 p 平方乘 c 平方等于 m 平方乘 c 的四次方：可在任意方便参考系中计算。
Check each result in the nonrelativistic limit, where the addition rule reduces to u equals u prime plus v and the invariant reduces to the rest energy.
请在非相对论极限下检验各结果，此时速度相加规则简化为 u 等于 u 撇加 v，不变量则简化为静止能量。
Two lab events are 900 metres and 2 microseconds apart.
两个实验室事件相距 900 米和 2 微秒。
c delta t is only 600 metres, less than delta x, so the interval is spacelike.
c Δt 只有 600 米，小于 Δx，所以间隔是类空的。
Setting delta t prime to zero gives v equals c squared delta t over delta x: two thirds of c.
令 Δt′ 为零，得到 v 等于 c² Δt 除以 Δx：三分之二 c。
In that frame the events are simultaneous, separated by the square root of 900 squared minus 600 squared: 670.8 metres.
在该参考系中两事件同时发生，间距为 900 的平方减 600 的平方再开方：670.8 米。
No frame can put these spacelike events at the same place.
没有任何参考系能让这两个类空事件发生在同一地点。
Two events are separated by 300 metres and 2 microseconds in the lab.
在实验室中，两个事件的空间间隔为300米，时间间隔为2微秒。
Find the velocity of a frame in which they occur at one position, and their time separation there.
求一个使这两个事件发生在同一位置的参考系的速度，以及该参考系中的时间间隔。
Can a frame reverse their order?
是否存在一个参考系能反转这两个事件的先后顺序？
Set delta x prime equals gamma times the quantity delta x minus v times delta t, and set that equal to zero. So v equals delta x divided by delta t, which is 0.5 c.
令 Δx' = γ(Δx − vΔt) = 0，因此 v = Δx/Δt = 0.5c。
The proper time is the square root of the quantity delta t squared minus the quantity delta x over c squared, which equals the square root of 3 microseconds.
固有时为 √(Δt² − (Δx/c)²) = √3 微秒。
They are timelike separated, and every time-oriented inertial frame retains their order.
它们是类时间隔的，且所有时间定向惯性系均保持其先后顺序。
Award velocity condition, speed, proper time, and causal classification or order.
评分标准：速度条件、速率、固有时以及因果分类或顺序。
Quoting L equals L nought over gamma without the simultaneous-endpoint condition, or adding velocities as u equals u prime plus v at relativistic speeds.
常见错误：引用 L 等于 L0 除以 γ 却不满足两端同时读取的条件，或在相对论速度下直接用 u 等于 u 撇加 v 相加速度。
A simultaneity difference is physical, not a mistake.
同时性的差异是物理事实，而不是错误。
Spacetime interval, 时空间隔, is the invariant combination c squared delta t squared minus delta x squared under the declared sign convention, whose sign classifies an event separation.
时空间隔：在约定的符号下，不变组合 c 平方 Δt 平方减 Δx 平方，其符号用于对事件间隔分类。
Proper length, 固有长度, is the endpoint separation measured in the object's rest frame; any other frame must read both endpoints simultaneously in its own time.
固有长度：在物体静止系中测得的两端间距；其他参考系必须在自己的时间里同时读取两端。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不看定义的情况下回忆相关含义，然后改变原始变换中的一个参数并检验极限情况。