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Conservation of momentum a universal truth

  1. May 14, 2010 #1
    Wherever I have read, the conservation of momentum is a universal truth. This is the basis, as I undertsand it, to the mass equation:

    m = m0/SQRT(1 - v2/c2) in which m0 = the mass in the rest frame and m is the mass in the moving frame relative to the rest frame moving at v.

    The idea behind it, as I understand it, that in order to conserve momentum, one must increase the mass to keep the mv product the same, as the "moving frame" is time dilated and v is functionally less than the v in the rest frame.

    Does this automatically happen?. The interesting part about this is that the v in the moving frame is "slowed down" by time dilation yet the v is perpendicular to the path of the moving frame. I guess time dilation must occur in all three axes (x, y, z) as it would be impossible for an object to exist at one time in x and "simultaneously" other times in y and z.
     
    Last edited: May 14, 2010
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  3. May 15, 2010 #2
    At speeds we are used to everyday, momentum is a reduction of that formula to p = mv.
    Normally we only consider v to be in the x direction.
    v in several axes at once gets very confusing.
    while momentum is a handy measure for doing calcs of energy distribution it is not in itself very useful.
    The kinetic energy of a moving body, relative to the rest frame is, as I'm sure you know, is Ek = 1/2mv^2
    v is not 'slowed down' the time dilation refers to any processes on the moving body as seen from the rest frame. The observers on the moving body see the rest frame as being slower. this is the basis of relativity.
     
  4. May 15, 2010 #3
    I know, v is not slowed down but momentum is "slowed down" so to speak which requires the gamma kick on the mass to up it back to where it should be.

    Ain't life grand?
     
  5. May 15, 2010 #4


    Energy-Momentum conservation has nothing to do with "relativistic mass", nor does it have anything to do with the definition of relativistic momentum as [tex]\vec{p}=\gamma m_0\vec{v}[/tex].
     
  6. May 15, 2010 #5

    Mentz114

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    Stevmg, energy and momentum are not invariant under Lorentz transformations.
     
  7. May 15, 2010 #6
    Well, I guess you are both saying the same thing so it would appear that momentum DOES change when looking at it from different frames of reference.

    Back to AP French and to try to decipher what the ---- this is all about. The preservation of momentum was presented to me in a math blog - Karl's Calculus Tutor just to let you know that I really wasn't on Pluto when I came up with this.

    I've attached the Karl's Calculus Tutor reply to my question.
     

    Attached Files:

  8. May 15, 2010 #7
    It may be useful to consider the prerelativity, Galilean transformation. The momentum of a particle is not the same for two observers in relative motion. It continues to hold true in relativity. The conserved quantity in special relativity is E2- c2p2.
     
  9. May 15, 2010 #8
    Let's try a practical example.
    two meteorites collide in space.
    This collision is observed from 3 different planets. Thus the meteorites have 3 different sets of relative velocities.
    each planetary observer sees total momentum before the collision equal to total momentum after the collision.
    Thus the 3 frames of reference all see momentum conserved, regardless of the relative speeds of the meteorites.
    But these are three Different Values of momentum. because they are measured from 3 different frames of reference.
    See: Modern Physics by Krane.
     
  10. May 15, 2010 #9
    You are comingling two DIFFERENT terms:

    -frame invariance
    -conservation

    we are telling you that:

    -momentum is NOT frame-invariant
    -momentum IS conserved
     
  11. May 15, 2010 #10

    Dale

    Staff: Mentor

    Starthaus is correct. And the same applies to energy. It is not frame invariant but it is conserved.

    The quantity Phrak mentioned is both frame invariant and conserved.
     
  12. May 16, 2010 #11
    "Momentum is not frame invariant..." [Double negative here...]

    Does that mean it varies according to frame?

    "Momentum is conserved." Does that mean within each frame it is "preserved?" Or does it mean that momentum is preserved across all frames.

    Got to get on the same page.

    The article from Karl's Calculus Tutor is trying to show that by conserving momentum across frames, the mass must go up (by gamma) to preserve the momentum.

    When I used the term relativistic mass above I meant [tex]\gamma[/tex]m0. Maybe that is the wrong term.
     
  13. May 16, 2010 #12

    DrGreg

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    I will quote myself from another thread (click on the little blue arrow to see it in context):
     
  14. May 16, 2010 #13
    I would recommend that you read carefully DrGreg's excellent post above mine. I will put DrGreg's post in a mathematical format.

    Yes, momentum, being dependent on velocity, varies from frame to frame. If you have two frames , F and F' moving wrt each other , the velocity of a partcle is [tex]\vec{u}[/tex] in F and [tex]\vec{u'}[/tex] in F'. Thus , the momentum is [tex]\gamma(u)m_0 \vec{u}[/tex] in F and [tex]\gamma(u')m_0 \vec{u'}[/tex] in F'. So, momentum is not frame invariant. Neither is total energy [tex]\gamma(u)m_0c^2[/tex].
    Yet, the energy-momentum [tex]E^2-(\vec{p}c)^2[/tex] is frame invariant for a system of particles.

    Neither. It means exactly what it means in Newtopnian mechanics, the variation of momentum of a system is equal to the sum of the external forces applied to the system:

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

    If [tex]\Sigma \vec{F}=0[/tex] then [tex]\vec{P}=constant[/tex].
    If total momentum is conserved in one frame, it is conserved in all frames.
     
  15. May 16, 2010 #14
    Maybe my use of "relativistic mass" was incorrect to describe what I was trying to say. the equation of m = m0[itex]\gamma[/itex] I used m0[itex]\gamma[/itex] as "relativistic mass."

    Karl, from Karl's Calculus Tutor tries to explain this that one must use whatever you want to call this above equation to preserve momemntum over different frames.

    The question was, how did we derive m = m0[itex]\gamma[/itex]? What is the basis of it? Why does mass increase in the moving frame in relation to the orignal "rest frame?"
     
    Last edited: May 16, 2010
  16. May 16, 2010 #15
    Exellent post. Makes AP French more "palatable."

    But, the question is - what makes this rest mass in different frames change according to the frame? As you said, if momentum is preserved in one frame it is preserved in all. Is that what "forces" the rest mass to vary according to what frame you are in? According to Karl, it does, but he is a mathematician and not a physicist.
     
  17. May 16, 2010 #16
    "We" didn't. Einstein did, long ago. See here

    The usage of "relativistic mass" is in disfavor these days. We use instead the relativistic energy [tex]\gamma(u) m_0c^2[/tex] and the relativistic momentum[tex]\gamma(u)m_0\vec{u}[/tex].
    If you want to learn how we arrived to these definitions, you need to get a good book, like "Spacetime Physics" by Taylor and Wheeler and start studying. Cobbling info from websites like "Kurt Calculus" will not do, sorry.
     
  18. May 16, 2010 #17
    That's what I have read ("relativistic mass" is in disfavor.)

    I have seen that book Spactime Physics somewhere recently. Not a cheap one so I will get it from the library, even if it takes three months to get it. Will take a lot of energy to get into.

    Don't knock Karl - he does OK but he is not a physicist.

    Will get back later on it as this is NOT number one on my priority list right now. This is mental exercising. There is a real world out there.
     
  19. May 16, 2010 #18
    I am not knocking Karl, I am criticizing your approach to learning. Spacetime Physics can be bought (used) for about 15$ from Amazon.com. I highly recommend it.
     
  20. May 16, 2010 #19
    Greg, you have to modify this in such a way that the conservation of momentum can also be included in there in general. That is just correct for the conservation of energy; as we all know the momentum has to remain invariant over the spatial translations not time translations in order for it to be conserved.

    AB
     
  21. May 16, 2010 #20
    I am 67 years old, hold a college degree, a MA, a MD degree, am board certified in 2 specialties and have been practicing medicine for longer than you are alive. I have read countless medical books, journals, monographs and have outstanding board scores in both my specialties. I think my approach to learning has spoken for itself. This venture into spacetime physics, etc. is to satisfy a curiosity I have had for years but never pursued because I was reading medical articles (I didn't want to kill anybody through ignorance) rather than "free-lancing." Fredrik and others have told me to continue to write "simplistic" questions on this forum (I did ask them for a another site so as to not clog up the forum with "trivial" questions, but they said "no, continue to write." ) I have learned more by this method in the past two months than ever before on this subject and I have accompanied the questions by reading textbooks suggested by DaleSpam and JesseM. Spacetime Physics was one of them but it was too advanced for me but the AP French Special Relativity book was more in line with what I knew and my level of comprehension of physics.

    I wish I could find Taylor's Spacetime Physics book for $15 but it isn't there. I have looked and looked. It is not on Amazon, that's for sure. If you hear of a site please let me know at stevmg@yahoo.com as I will be most grateful.

    Verstehst du? Caspisci? Entiendes?
     
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