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국제바칼로레이트 · IB Diploma · Physics · SL 주제 19 3:35 영어 내레이션 · 영어 + 중국어 자막 burned-in
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Two stars can have the same apparent brightness while having different luminosities.
两颗恒星可能具有相同的视亮度，但光度不同。
Distance changes the flux 辐射通量密度 reaching an observer.
距离会改变到达观测者的辐射通量密度。
Luminosity is the total power emitted.
光度是发射的总功率。
Flux is the power received per unit area.
通量是单位面积接收到的功率。
A stellar spectrum carries information about surface temperature and composition.
恒星光谱携带有关表面温度和成分的信息。
Fusion transfers energy as light nuclei combine.
聚变通过轻核结合传递能量。
For isotropic emission without absorption, flux follows an inverse-square relationship with distance.
在无吸收的各向同性辐射中，通量与距离的平方成反比。
Observed brightness alone therefore cannot establish luminosity.
因此，仅凭观测亮度无法确定光度。
In the lab: Keep distance units consistent, identify which quantities are intrinsic to the star, and distinguish observational evidence from a model of stellar evolution.
在实验室中，请保持距离单位一致。 识别哪些量是恒星本身的固有属性。
Do not confuse a red giant stage with every possible final remnant.
区分观测证据与恒星演化模型。 不要将红巨星阶段与所有可能的最终残骸混淆。
Known: distance doubles while luminosity remains fixed.
已知：距离加倍，而光度保持不变。
Use F proportional to 1 over d squared.
使用公式 F 正比于 1 除以 d 的平方。
Flux ratio equals 1 over 2 squared, which is 1 over 4.
通量比值等于 1 除以 2 的平方，即 1/4。
A flux of 12 units becomes 3 units.
原本为 12 单位的通量变为 3 单位。
This assumes no change in absorption or source output.
此结论假设吸收率和光源输出均无变化。
Your turn: A source gives flux 18 units.
轮到你了。 一个光源的辐射通量为18个单位。
Find flux at three times the distance.
求距离变为三倍时的辐射通量。
Pause the video and work it out.
请暂停视频，自己算一算。
Check: three times the distance divides the flux by 9, so 18 becomes 2 units.
核对：距离变为 3 倍，通量除以 9，18 变为 2 个单位。
The Sun is not expected to become a supernova.
太阳预计不会成为超新星。
Redshift can support cosmological expansion; it does not mean every nearby object must move away from every observer.
红移可以支持宇宙膨胀，但这并不意味着每个邻近天体都必须远离每个观测者。
Not true.
不正确。
Apparent brightness alone does not always determine a star's luminosity.
仅凭视亮度并不能总是确定恒星的光度。
Apparent brightness depends on distance as well as luminosity, F equals L divided by 4 pi d squared.
视亮度既取决于距离，也取决于光度，F 等于 L 除以 4 π d 平方。
A dim-looking star may be very luminous but far away, so apparent brightness alone cannot fix luminosity without a distance.
一颗看起来暗淡的恒星可能实际上光度很高，只是距离遥远，因此如果没有距离信息，仅凭视亮度无法确定其光度。
Luminosity is total power emitted.
光度是发射的总功率。
Flux is power received per area.
通量是单位面积接收到的功率。
For isotropic emission without absorption, flux follows an inverse-square relationship with distance.
在无吸收的各向同性辐射中，通量与距离遵循平方反比关系。
Watch for: the Sun is not expected to become a supernova.
请注意：太阳预计不会成为超新星。
A source gives flux 32 units.
一个光源的辐射通量为32个单位。
Find the flux at twice the distance.
求距离变为两倍时的辐射通量。
Apparent brightness alone does not give luminosity.
仅凭视亮度无法得出光度。
The result is a flux of 8 units.
结果是通量为8个单位。
A star’s power spreads over a sphere of area four pi d squared.
恒星的功率分布在面积为4πd²的球面上。
At twice the distance the same power covers four times the area, so the flux is one quarter.
距离加倍时，同样的功率覆盖四倍的面积，所以通量为四分之一。
Identical sources have fluxes 8 and 2 units.
两个相同的辐射源通量分别为8和2个单位。
Find their distance ratio.
求它们的距离之比。
What ambiguity arises if luminosity is not known?
如果光度未知，会产生什么歧义？
The fainter source is twice as far under inverse-square propagation and equal absorption.
在平方反比传播和吸收相等的条件下，较暗的源距离是两倍。
Different luminosity or absorption can also lower flux; flux alone is insufficient without calibration.
不同的光度或吸收也会降低通量。 若无校准，仅凭通量是不够的。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。