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The way I understand conservation laws - which I am asking you to correct - is that if I observe any slice of the universe perpendicular to the time axis and count up all the mass/energy, momentum, charge, etc., I should obtain the same sum as I would had I observed any other slice of the universe perpendicular to the time axis.

Stop me right there if you have to, but here is where I start scratching my head.

The general theory of relativity informs me that I can't observe a slice of the universe perpendicular to the time axis. The slice I observe is skewed based on my velocity. Yet worse, the special theory of relativity informs me that I can't even observe a flat slice of the universe if I'm accelerating. It's curved based on my acceleration and has all sorts of bumps and such around every massive object. Again, please re-inform me if I'm mistaken.

So let's say I make my observation and sum up some conserved quantity over the entire universe. The slice of space-time that I've integrated over is S1, and my total is Q1. I change my direction of motion but not my speed such that the slice of the universe I observe S2 is an affine transformation of S1 that is not equal to S1. I then integrate the same conserved quantity over S2 and call it Q2.

Does Q1 = Q2? If it doesn't, then how can we ever verify that conservation laws describe our universe, and how did we come up with them in the first place?

If Q1 does = Q2, would this not imply that all conserved quantities are distributed perfectly evenly, since I can tilt my observation to exclude any symmetry-ruining lump of conserved quantity from the total? Since that doesn't appear to be the case, who is the stupid: Einstein, the guy that came up the conservation laws, or me?

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  • $\begingroup$ The observable quantity is not preserved. Just the quantity itself. $\endgroup$
    – Martin Drautzburg
    Commented Jan 9, 2014 at 18:28
  • $\begingroup$ @MartinDrautzburg How can you tell? $\endgroup$
    – sqykly
    Commented Jan 9, 2014 at 18:31

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You can simplify your Gedankenexperiment even further. In a stationary universe assume a mass which is so distant from an obsever, that light from that mass had not enough time to travel to the observer. It is outside of the observer's horizon.

Now let time pass, and eventually that mass will be within the observer's horizon. You may interpret this as the sudden appearence of mass, apparently violating the conservation of mass/energy. (In reality the universe is expanding and the effect would be just the opposite - you see mass disappearing)

It appears you misconcieved the word observable. Not being observable is not the same as being non-existent. "Observable" is all about events and not about things. If an event is not observable, this means you cannot receive signals from this event, not that the thing experiencing the event simply isn't there.

Take for example a black hole, where the "inside" is beyond the horizon of any outside observer. This does not mean, that the mass of a black hole is undetectable from an outside observer.

I think you asked an interesting question, because conservation laws are hardly ever discussed in the presence of a horizon.

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  • $\begingroup$ Follow-up queries: This does mean that Alice and Bob may come up with different figures for the total conserved quantity at different times/speeds/etc. and Q1 =/= Q2, right? Does an event with mass/energy, momentum, and/or charge entering another system's horizon necessitate any change to that system without necessarily involving some physical force, like that ubiquitous demonstration of conservation of angular momentum with a bike wheel? Finally, does this have anything to do with black holes with a net angular momentum? $\endgroup$
    – sqykly
    Commented Jan 11, 2014 at 14:48
  • $\begingroup$ Even without a horizon, Alice and Bob will already see different kinetic energies of objects if they move relative to each other. I have no answer for the other follow-up questions though $\endgroup$
    – Martin Drautzburg
    Commented Jan 12, 2014 at 19:52

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