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Validity of Newton's Laws on a MCRF

by Zag
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Zag
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Jan31-13, 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 stress-energy 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:

Let us in particular look at it in the MCRF (Momentarily Comoving Reference Frame), where there is no bulk flow of the fluid element, and no spatial momentum in the particles.
Afterwards, the author starts analizing the spatial components of the general stress-energy tensor. I reproduce below that part of the text:

The spatial components of the stress-energy tensor, [itex]T^{ij}[/itex]. By definition, [itex]T^{ij}[/itex] is the flux of [itex]i[/itex] momentum across the [itex]j[/itex] surface. Consider (Fig. 4.6) two adjacent fluid



elements, represented as cubes, having the common interface [itex]\ell[/itex]. In general, they exert forces on each other. Shown in the diagram is the force [itex]F[/itex] exerted by A on В (В of course exerts an equal and opposite force on A). Since force equals the rate of change of momentum (by Newton's law, which is valid here, since we are in the MCRF where velocities are zero), A is pouring momentum into В at the rate [itex]F[/itex] per unit time. Of course, В may or may not acquire a new velocity as a result of this new momentum it acquires; this depends upon how much momentum is put into В by its other neighbors. Obviously B's motion is the resultant of all the forces. Nevertheless, each force adds momentum to B. There is therefore a flow of momentum across [itex]\ell[/itex] from A to В at the rate [itex]F[/itex]. If [itex]\ell[/itex] has area [itex]\zeta[/itex], then the flux of momentum across [itex]\ell[/itex] is [itex]F/\zeta[/itex]. If if is a surface of constant [itex]x^{j}[/itex], then [itex]T^{ij}[/itex] for fluid element A is [itex]F^{i}/\zeta[/itex].
Having that in mind, the problem I've been facing is related to the bolded sentece above: "Since force equals the rate of change of momentum (by Newton's law, which is valid here, since we are in the MCRF where velocities are zero) (...)"

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?
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elfmotat
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Jan31-13, 03:07 PM
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First of all, welcome to PF!

Quote Quote by Zag View Post
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?
But that's not true. For example, consider (in ordinary Newtonian physics) throwing a ball up in the air near the Earth's surface: the ball traces out a parabola. At the top of the parabola the ball's velocity is zero w.r.t. Earth's surface, so we are (at that instant) in a MCRF with the ball. Does the fact that the ball's velocity is momentarily zero mean there's no force/acceleration on it? Of course not, because it has a constant acceleration of 9.8 m/s2 applied to it at all times: F=dp/dt≠0.
PeterDonis
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Jan31-13, 03:14 PM
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Quote Quote by elfmotat View Post
For example, consider (in ordinary Newtonian physics) throwing a ball up in the air near the Earth's surface: the ball traces out a parabola. At the top of the parabola the ball's velocity is zero w.r.t. Earth's surface, so we are (at that instant) in a MCRF with the ball. Does the fact that the ball's velocity is momentarily zero mean there's no force/acceleration on it? Of course not, because it has a constant acceleration of 9.8 m/s2 applied to it at all times: F=dp/dt≠0.
This is a bad example for two reasons:

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 acceleration--i.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.

elfmotat
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Jan31-13, 03:17 PM
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Validity of Newton's Laws on a MCRF

Quote Quote by PeterDonis View Post
This is a bad example for two reasons:

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 acceleration--i.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.
You're right. I was trying to think of a simple example where something's velocity was zero and its acceleration was nonzero.
PeterDonis
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Jan31-13, 03:19 PM
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Quote Quote by Zag View Post
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?
Because, as the passage you quoted says, the acceleration of the fluid element as a whole is determined by the *net* force on it, which might be zero. But you could still measure a force on each face of the fluid element--imagine putting a pressure gauge on each face, for example. It would read nonzero, indicating that a force is being exerted on it, even though the fluid element isn't moving.

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.
bcrowell
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Jan31-13, 03:26 PM
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Welcome to PF!

Let us in particular look at it in the MCRF (Momentarily Comoving Reference Frame), where there is no bulk flow of the fluid element, and no spatial momentum in the particles.
I don't have the context, but I don't think this necessarily says the particles are not moving. I think it says that when you average over a finite region, such as one of the boxes in the figure, the average momentum of the particles is zero.

In any case, Newton's second law is equally valid in the special case v=0, a=0, F=0.
Zag
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Feb1-13, 07:10 AM
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Thank you very much for the welcoming messages, everyone! And thank you also for replying with your ideas!

Quote Quote by PeterDonis View Post
Because, as the passage you quoted says, the acceleration of the fluid element as a whole is determined by the *net* force on it, which might be zero. But you could still measure a force on each face of the fluid element--imagine putting a pressure gauge on each face, for example. It would read nonzero, indicating that a force is being exerted on it, even though the fluid element isn't moving.

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.
Thank you PeterDonis, I think I understand what you mean. After thinking for some time, I came up with a mental picture of how it would be to look at the fluid element from the point of view of a MCRF. In fact, I found it much easier to think about this situation in terms of particles exchanging momentum through the boundaries of the element, as you suggested. From this point of view it becomes clear that momentum is indeed being transfered through the boundaries of the element, and there is actually no need to mention Newton's Second Law here.

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?

Quote Quote by bcrowell View Post
Welcome to PF!



I don't have the context, but I don't think this necessarily says the particles are not moving. I think it says that when you average over a finite region, such as one of the boxes in the figure, the average momentum of the particles is zero.

In any case, Newton's second law is equally valid in the special case v=0, a=0, F=0.
Thank you bcrowell, I think you are right when it comes to the motion of the particles. Even though the author stated the opposite, I believe he is making the situation too simple and carrying the abstraction too far, which really doesn't help when the reader tries to come up with an interpretation of what is happening. It is much easier to think in a real general situation.
PeterDonis
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Feb1-13, 09:50 AM
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Quote Quote by Zag View Post
I don't think the rate at which momentum is being transfered it necessarily given by Newton's Second Law.
If you look at the level of individual molecules, it is.

Quote Quote by Zag View Post
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.
Yes; the MCRF is freely falling, but the fluid element itself is not. That's why the "M" (for "momentarily") is there in "MCRF".

Quote Quote by Zag View Post
Therefore, momentarily, you would be able to apply Newton's Second Law - but this analysis would not be allowed in extended periods of time.
This is not quite the way to put it. Newton's Second Law is a local law, in both space *and* time: it relates the time derivative of momentum to force at a single point of space, at a single instant of time. To do an analysis over extended periods of time, you aren't actually using a single instance of Newton's Second Law: you're using a continuous series of such instances, one in each (different) MCRF at each different point of space and instant of time on the fluid element's trajectory. The law applies just fine at each point: what changes is the MCRF you evaluate it in from point to point.
Zag
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Feb1-13, 10:07 AM
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Quote Quote by PeterDonis View Post
If you look at the level of individual molecules, it is.

(...)

To do an analysis over extended periods of time, you aren't actually using a single instance of Newton's Second Law: you're using a continuous series of such instances, one in each (different) MCRF at each different point of space and instant of time on the fluid element's trajectory. The law applies just fine at each point: what changes is the MCRF you evaluate it in from point to point.
I agree! I think I haven't expressed myself very well when I adressed this question. In other words, just like you calculate an interval of time measured in an accelerated frame by integrating the contributions on a set of MCRFs, you would have to integrate small contributions of these Newton's Law instances to get the result in an extended period of time. Correct? :)
PeterDonis
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Feb1-13, 10:14 AM
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Quote Quote by Zag View Post
just like you calculate an interval of time measured in an accelerated frame by integrating the contributions on a set of MCRFs, you would have to integrate small contributions of these Newton's Law instances to get the result in an extended period of time. Correct? :)
Yes.


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