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On The Definition Of An Inertial Reference Frame

  1. Dec 14, 2003 #1
    There are various definitions of an inertial reference frame out there, but only one is really accepted by the physics community.

    In some places, you will see an inertial reference frame defined as a a reference frame in which Newton's law of inertia is valid.

    In some places, you will see an inertial reference frame defined simply as a reference frame which isn't accelerating.

    In some places, you will see an inertial reference frame defined as a reference frame in which all three of Newton's laws are valid.
    (This is the standard definition)

    In some places, you will see an inertial reference frame defined as follows:

    An inertial reference frame is defined as a reference frame in which an object at rest will remain at rest, and an object in motion will remain in motion in a straight line at a constant speed, and if a repulsive force acts between two bodies of the same mass, they will acquire equal velocities in equal amounts of time.

    What I would like to do in this thread, is discuss inertial reference frames. The reason is this: If someone does not know exactly what an inertial reference frame is, then they certainly don't understand the theory of special relativity, precisely because its fundamental postulate is that the speed of light is c in any inertial reference frame .

    To begin with, consider Newton's laws:

    Law of inertia: An object at rest will remain at rest, and an object in motion will remain in motion in a straight line at a constant speed, unless acted upon by an outside force.

    Newton's second law: An external force will accelerate an object, and the change in the momentum of the object will be directly proportional to the applied force, and in the direction of the applied force.

    Newton's third law: Every action is accompanied by an equal and opposite reaction. In other words, if some object in the universe is acted upon by an external force of magnitude F, then some other object in the universe is simultaneously acted upon by a force of the same magnitude F, but in the opposite direction.

    Now, Newton's first and second law can be combined into a single mathematical equation which is this:

    [tex] F = dP/dt [/tex]

    In the above equation, F is a vector quantity, and P = momentum is also a vector quantity. The definition of momentum is as follows:

    P = momentum = mass times velocity = mV

    where velocity is a vector quantity. The magnitude of an object's velocity, is its speed in a frame.

    This is all you need in order to understand the definition of an inertial reference frame.

    An interesting consequence of the definition of an inertial reference frame, is that if frame 1 is an inertial reference frame, and frame 2 is moving at a constant speed relative to frame 1, then frame 2 is also an inertial reference frame, but if frame 2 is accelerating with respect to frame 1, then frame 2 is a non inertial reference frame.

    The first issue which I wish to address, is how is it shown that Newton's first and second laws are contained in the single mathematical statement given above?
     
    Last edited by a moderator: Dec 14, 2003
  2. jcsd
  3. Dec 14, 2003 #2

    Hurkyl

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    I'm not sure where the problem is...

    Newton's second law says:

    [tex]\frac{d\vec{p}}{dt} = \kappa \vec{F}[/tex]

    If we select units where [itex]\kappa=1[/itex], then Newton's second law is [itex]\vec{F} = d\vec{p}/dt[/itex].


    And because constant momentum implies constant velocity (for an object whose mass is unchanging), the first law is simply a special case of the second law...


    P.S. are you using 3-vectors or 4-vectors? I'm presuming 4-vectors, but I just want to make sure...
     
  4. Dec 14, 2003 #3
    First of all, I have already concluded that [tex] \kappa = 1 [/tex], so that carrying [tex] \kappa [/tex] through a long mathematical argument is fruitless in the end. That is really why the other thread is so important, because its whole purpose is to be the very beginning of a long argument. Proving that simultaneity is absolute right off the bat, simplifies mathematical work later.

    Next, I am most definitely not using four vectors, and provided simultaneity is absolute, four vectors are not needed to express the laws of physics in an inertial frame.

    P.S. I was hoping someone would show me how to write vector symbols. Thanks Hurkyl.

    Let me try my question again, and see what happens. Consider the following mathematical formula:

    [tex]\vec{F}= \frac{d\vec{P}}{dt} [/tex]

    My question is this:

    Given that [tex] \kappa = 1 [/tex] in Hurkyl's formula, how can we derive Newton's law of inertia from the formula above. In other words, one simple mathematical formalism 'contains' two of Newton's laws. How can one equation 'contain' two laws.

    What I am after is this. A mathematical argument using a rectangular coordinate system, in which velocity is defined using vector notation. Position vectors must be defined, etc. The goal will be to show that if we set F to zero, that Newton's law of inertia naturally comes out of the math.
     
    Last edited by a moderator: Dec 14, 2003
  5. Dec 14, 2003 #4

    Hurkyl

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    F = 0 -> dp/dt = 0 -> p is constant -> v is constant



    I would like to object, then, to [itex]\vec{p} = m\vec{v}[/itex]. It is known to be only a low velocity approximation to the actual (3-vector) equation:

    [tex]
    \vec{p} = \frac{m \vec{v} } {\sqrt{1 - |v|^2/c^2}}
    [/tex]

    (in the sense that it has been both theoretically deduced and experimentally verified)
     
    Last edited: Dec 14, 2003
  6. Dec 14, 2003 #5

    selfAdjoint

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    Hurkyl, in the inertial frame, v/c = 0.
     
  7. Dec 15, 2003 #6
     
    Last edited by a moderator: Dec 15, 2003
  8. Dec 15, 2003 #7

    Hurkyl

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    If you finish with the calculus, you'll see that if the LHS and RHS point opposite directions, thus they must both be zero.

    You don't need calculus, though; for a given momentum you can algebraically solve for the unique velocity that yields that momentum. (assuming rest mass is fixed)
     
    Last edited: Dec 15, 2003
  9. Dec 16, 2003 #8
    Hurkyl, please finish the analysis.

    We have: m dv = - v dm, for a particle in an inertial reference frame. So how do we now reach the conclusion that if the particle is at rest then it will remain at rest, and if the particle is in motion it will continue moving in a straight line at a constant speed?
     
  10. Dec 16, 2003 #9

    Hurkyl

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    Suppose 0 = F = dp/dt. By elementary calculus, this tells us that p is constant. We can write v as a function of p, so v is constant. QED.


    Do you really need to see it proved by brute force?
     
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