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Validity of Newton's Laws on a MCRF 
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#1
Jan3113, 02:48 PM

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Hey guys, I am new here and hope to be able to contribute with some of the discussions in Relativity. I will start my participation with a conceptual question that arised during my studies and has been boggling me since then.
I have been studying Relativity using the book "A First Course in General Relativity" by B. F. Shcutz. Right now I am reading Chapter 4 on Perfect Fluids in Special Relativity, more specifically I am studying the section that introduces the stressenergy tensor, [itex]T^{\alpha\beta}[/itex], and its properties. The author makes an important remark before starting the more detailed discussion about the fluid elements, he says and I quote: So here comes what has been bothering me: Since there is no bulk motion and no spatial momentum on the particles, how is it possible to use Newton's law in this argument since the MCRF sees no motion at all  and consequently no acceleration  in that fluid element? Also, going deeper into the problem, is it possible to ascertain that Newton's Laws are always valid from the point of view of a MCRF? 


#2
Jan3113, 03:07 PM

P: 260

First of all, welcome to PF!



#3
Jan3113, 03:14 PM

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PF Gold
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First, the ball's velocity is only zero momentarily; the velocity of a fluid element in a static configuration is zero indefinitely, even though it continues to have forces exerted on it. Second, the "force" of gravity, in relativity, is not really a force. The relativistic version of Newton's Laws only applies to forces that cause proper accelerationi.e., acceleration you can measure with an accelerometer or feel as weight. Gravity doesn't do that; the ball's proper acceleration is zero (if we neglect air resistance) all during its flight, so its dp/dt in an MCRF is zero. There is no force. 


#4
Jan3113, 03:17 PM

P: 260

Validity of Newton's Laws on a MCRF



#5
Jan3113, 03:19 PM

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If you really want to dive deep enough into this, you will end up looking at the individual molecules of the fluid; momentum exchange across the boundary of a fluid element then resolves into individual collisions between molecules, which do change the trajectories of the individual molecules. In a static situation where the fluid element as a whole does not move, it just happens that the trajectory changes cancel each other out: for each molecule that gets knocked one way by a collision, there's another molecule that gets knocked the opposite way, so when you average over all the molecules you get zero net momentum. But momentum is still being transferred at the molecular level, and things like pressure gauges can measure that. 


#6
Jan3113, 03:26 PM

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Welcome to PF!
In any case, Newton's second law is equally valid in the special case v=0, a=0, F=0. 


#7
Feb113, 07:10 AM

P: 43

Thank you very much for the welcoming messages, everyone! And thank you also for replying with your ideas!
I still find it weird, though, that the author mentioned the validity of Newton's Law in this situation. As I just said, it is obvious that momentum is being transfered through the boundaries of the element, but at the same time, I don't think the rate at which momentum is being transfered it necessarily given by Newton's Second Law. Here is a thought: maybe it is true for an instant in time because the MCRF (which is really a collection of inertial frames that "follow" the fluid element) does not experience the same acceleration that the fluid element does. Therefore, momentarily, you would be able to apply Newton's Second Law  but this analysis would not be allowed in extended periods of time. Would this be correct? 


#8
Feb113, 09:50 AM

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#9
Feb113, 10:07 AM

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#10
Feb113, 10:14 AM

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PF Gold
P: 6,149




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