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How does SR quantify the direction relative properties of inertia?

  1. Apr 11, 2013 #1
    Pre-theoretically, we notice that objects have some property responsible for their varying dispositions to resist being pushed around in space, or to resist changes in motion given applied forces ("inertia"). So we introduce the term "inertial mass" to apply to the most physically interesting property responsible for such dispositions.

    We then, naturally, try to get precise about the exact relationship between this "mass" property, and resulting accelerations given applied forces. Testing at low relative velocities, we quantify the mass / inertia relationship with F=MA.

    But then we realise that spacetime is structured so as to prevent particles from accelerating past the speed of light. Consequently, objects have more inertia, more resistance to acceleration, in the direction of their velocity: objects are harder to "push around" in the direction of their velocity. So if we want to keep 'mass' close to inertia (the original rationale for the term's introduction), then it looks like we had better introduce direction relative masses, Transverse mass and Longitudinal mass. Quantifying the mass / inertia relationship then yields F= ##\gamma^3##MA = MLA; and F= ##\gamma##MA= MTA.

    That all makes a lot of sense to me. What doesn't make sense to me, is modern SR formalism, which purports to be able to describe this phenomenon, without introducing direction relative quantities.

    According to wikipedia's entry on mass in SR, we don't need to distinguish direction relative equations, we just need to give up on F=MA and instead work with F=d(MRV)/dt. I do not understand this suggestion.
    I thought the reason that F=MA is no good is because the applied force can alter the mass - mass is not constant in SR. So we had better put 'M' into the time derivative to its right to get F=d(MRV)/dt or F=d(M0##\gamma##V)/dt. But if we work with one of the two latter equations, haven't we lost crucial information about inertial behaviour? Where has the cubed gamma gone? How can one derive the phenomenon of direction relative inertia from this equation?

    Any suggestions would be most welcome.
     
    Last edited: Apr 11, 2013
  2. jcsd
  3. Apr 11, 2013 #2

    Dale

    Staff: Mentor

    The usual way that this is handled in relativity is to introduce the four-momentum and the four-force.
    http://en.wikipedia.org/wiki/Four-force
    http://en.wikipedia.org/wiki/Four-momentum

    The particle's mass is the norm of the four-momentum, which is a frame invariant quantity equal to the rest mass. So with this approach the whole concept of relativistic mass as well as the transverse or longitudinal mass can be discarded. This is the usual approach today.
     
  4. Apr 11, 2013 #3
    Thanks. Fair enough that longitudinal, transverse and relativistic mass have been discarded in favour of invariant mass. I see many reasons for this. My only worry is that I don't see how SR (without such concepts) can give any transparent explanation of direction relative inertia.

    Is there a simple way of stating, what it is about four-force (as opposed to three-force), that makes clear why the four-force equations don't need to be split in two, one for force parallel to velocity, one for force perpendicular? Is there a simple way of stating an explanation of direction relative inertia in terms of four force equations?
     
  5. Apr 11, 2013 #4

    Dale

    Staff: Mentor

    Direction relative inertia doesn't exist in four-vectors. It is an artifact of the three-vector representation.

    I guess that you could derive the three-vector representation from the four-vectors and show that the derivation produces the direction relative artifact.
     
  6. Apr 11, 2013 #5
    Good, this is what I'm after. I'm after an explanation of an observable phenomenon. Our perceptual system is roughly a three-vector representation. So an explanation of the observable phenomenon in terms of four-vector representation, is effectively equivalent to, a derivation of the three-vector representation (of direction relative inertia) in terms of some four-vector representation.

    So I should go and study more about how three-vector formalism is derived from four-vector formalism. But what would the relevant four-vector representation be? If anyone has any hints to get me started, for the specific case of deriving F=##\gamma##3MA + ##\gamma##MA; I would be very interested.
     
  7. Apr 11, 2013 #6

    Dale

    Staff: Mentor

    I disagree with this pretty much completely. In the four vector approach observables are frame invariant scalars, not coordinate dependent three vectors. You never need to invoke any three vectors at all in order to explain any observation.
     
  8. Apr 11, 2013 #7
    Here's an observation: (i) we observe that objects are harder to push or accelerate (i.e. have more inertia) in the direction of their velocity.

    You said: (ii) "Direction relative inertia doesn't exist in four-vectors. It is an artifact of the three-vector representation".

    And you also said: (iii) "You never need to invoke any three vectors at all in order to explain any observation".

    These three claims don't seem logically compatible to me. For (iii) entails that you don't need three vectors to explain (i). But (ii) suggests that you DO need three vectors to explain (i)--for how can four-vector representation explain something that "doesn't exist" in four vector representation?

    Can you please clarify what you mean?
     
  9. Apr 11, 2013 #8

    Dale

    Staff: Mentor

    I don't think that is an actual observation. I don't know of any measurement device which produces a number which corresponds to the above statement.

    An observation is a measurement. There is some measuring device that produces some specific measured outcome. Those are represented by scalars and don't need three-vectors.
     
  10. Apr 11, 2013 #9
    If there are true equations that entail that the exact coefficient between force and acceleration are direction dependent, then there must be observations/measurements that confirm that these equations are true.

    According to wiki: "The precise relativistic expression relating force and acceleration for a particle with non-zero rest mass moving in the x direction with velocity v are [click here for the three equations].

    Wiki provides different equations for different directions. The coefficient between force and acceleration in the direction of the particle's velocity (x direction) is different from the coefficient between the force and acceleration for force in different direction to velocity. If these equations are true then there must be SOME observations that confirm them. In that case, those are the observations I'm referring to.
     
  11. Apr 11, 2013 #10

    Bill_K

    User Avatar
    Science Advisor

    In a circular accelerator like the LHC, a certain collinear E field within the rf cavities is required to accelerate the particles, while a certain transverse B field is required to keep them in orbit. The correct functioning relies on the difference between what used to be called the "transverse mass" and "longitudinal mass".
     
  12. Apr 11, 2013 #11
    Why do you think this is the case?
    Are you sure you do not mix up coordinate and proper acceleration?

    Proper acceleration in relativity can be represented by a frame invariant 3-vector.

    With respect to force the equation F = mα also applies where α is the proper acceleration, or in other form F = m (yL3/yT) a where a is the coordinate acceleration.
     
    Last edited: Apr 11, 2013
  13. Apr 11, 2013 #12

    Dale

    Staff: Mentor

    I don't know what those observations may be, but any of them can be described without reference to any 3 vectors.

    When a measurement device produces some number the number it produces depends only on the state of the device and the thing being measured. You can predict what that measurement should be based on some physical theory, and in the process of calculating that predicted number most of the time you will choose some coordinate system convenient for ease of calculation. However, the actual number obtained cannot depend on your choice of coordinate system.

    In relativity, the way that is done is through the use of scalars, which are coordinate independent quantities. All actual observables are coordinate independent scalars which can be directly calculated from other scalars, four-vectors, or tensors. If you have a specific measurement in mind, then I can help you calculate it using relativistic quantities.
     
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