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能量不守恒及经典薛定谔猫? 2021-03-08 06:26:56

戴榕菁

物理学家Sabine Hossenfelder在她的视频中指出宇宙膨胀导致宇宙内的总能量不守恒,原因是宇宙膨胀使得行进在宇宙中的光的波长增加,因而能量减少,而这部分减少的能量没有参与到任何能量交换的过程,因此这部分能量就白白地从宇宙中消失了。

按照她的这种说法(听起来挺有道理的,我现在也找不出她这种说法有什么毛病来),其实可以得出结论说不需要膨胀宇宙中的总能量也不守恒,因为在任何两个相互离开的物体之间都会产生红移,而在任何两个相互趋近的物体之间都会产生蓝移,其中红移损失的能量就白白地没有了,而蓝移增加的能量也是凭白地产生的。试想你在宇宙飞船上射出一束蓝光,它的能量是hf,其中h是普朗克常数而f是光的频率。这时后面一艘与你同速的飞船接收到了你的这束光,它仍然是蓝光;然后,后面那艘飞船突然减速,这时你又射出一束蓝光,仍然被那艘船接收到了,但是它已经没有那么蓝了,而这由于红移损失掉的能量并没有转换成任何其它的能量,而是就这么白白地不见了。

如果你说在整个宇宙中红移产生的能量与蓝移产生的能量可以抵消掉因而总能量不变的话,那也已经不是我们现在所说的能量守恒的意义了,而成为在统计意义上的能量守恒了!

不仅如此,由于红移和蓝移的能量损失是由光的接收体决定的,因此一束光打出后,它的能量到底是会因为红移而损失,还是因为蓝移而增多,或是保持不变,那是要等到它被一个物体接受到的那一瞬间才决定的。但是,这里的要害在于,红移损失的能量或蓝移增加的能量本身却不是由接受物造成的,而是在光的传播过程中出现的,因此,这里的能量变化就如同是经典意义(即不涉及量子意义)的薛定谔猫,而且是无时不在的真实的薛定谔猫,它的实际状态是由观察者的状态决定,但它的状态却是在被观察之前就已经注定了的。

物理学出现了悖论???


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作者:慕容青草 留言时间:2021-03-17 06:47:05

下面是在网上与英文读者就本文英文版的对话 (https://www.academia.edu/s/49453d40aa#comment_770918):

Eduardo D Dib23 hrs ago

Doesn't have the velocity of spaceship B to be modified in order to change the shift? That requires energy, possibly compensating the loss due to the shift...

rongqing Dai23 hrs ago

The energy changes is irrelevant to the change of the wave length of the photon....actually you might even say the total of the red shifts and blue shifts in the universe might cancel each other to keep the total energy the same...but that would no longer be the same sense of the energy conservation as we have been assumed, because that would be the energy conservation in a statistical sense....Thanks

rongqing Dai22 hrs ago

Put it this way: When the mass of spaceship B changes, the energy that used to change its speed would change, but the loss of the energy of the laser beam would not change, which is only related to the amount of the speed change...so that energy cost for changing the speed of spaceship B does not come into the equation of energy conservation (or nonconservation) of the laser beam at all...

Eduardo D Dib12 hrs ago

However, the amount of the speed change itself is related to the amount of energy used to produce it. So, there is not a covariant relation linking the energy loss of the beam with the specific amount of energy used to produce that specific speed change? The more energy used in decelerate B, the more energy lost in the shift. Nothing random there...

rongqing Dai12 hrs ago

the deceleration is the illusion in my layout of the scenario for the sake of comparing "no change" and "loss"....we don't need the deceleration...the whole point is the relative departure of two spaceships...look at it this way: we have three spaceships: A, B, C...and John in A shoots a blue laser beam to B which is received by Jack on B without any energy loss because those two spaceships moving at the same speed, but then he shoots another blue beam laser to C, and received by Joe on C which is so much slower than A, and what Joe received is red laser beam.....the energy loss goes to nowhere, but just completely lost...The weirdest thing here is that when John shoots out the laser beam, nobody knows whether the energy would be conserved or changed until it hits the retina of Joe...

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作者:慕容青草 留言时间:2021-03-09 06:56:04

光能的守恒性其实是一个相当严肃的问题:

1)在相对论的推导中光能E=hf起着很重要的作用;

2)所有有温度的物体都无时不刻地在辐射着红外光也在接收着其它物体的红外辐射。

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作者:慕容青草 留言时间:2021-03-08 08:55:14

Sabine的视频链接:https://youtu.be/ZYM6HMLgIKA

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