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Wheeler's gravitation book exercise

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Has anyone solved the part (d) of 5.6 problem of that book?
    I am unable to solve it.
    It asks the reader to prove that the radius ##R## of a rotating cylinder (rotating around its symmetry axis) has to be greater or equal than ##\frac{|S|}{ M } ##, in other words,
    [tex] |S| ≤ R M [/tex]

    Where ##S## is the angular momentum respect to the center of mass of the system in the rest frame and ##M## is its rest mass.
    It has to be proven using the fact that ##T(u,u) ≥ 0## for any timelike 4-vector ##u## ( ##T## is the stress-energy tensor )

    Thanks in advance.
    Cesar

    2. Relevant equations

    Metric (-1,1,1,1).

    ## P^0 = M ## Total mass of the system in the system's rest frame ( that one with total momentum equal to zero ## P^j = 0 ## for ## j=1,2,3 ## )
    Angular momentum tensor about a 4-point "a" :
    [tex] J^{μβα} = (x-a)^μ T^{βα} - (x-a)^β T^{μα} [/tex]
    Total angular momentum on a hypersurface of constant time:
    [tex] J^{μ\nu} = \int J^{\mu\nu0}\,dx dy dz = \int (x-a)^μ T^{\nu0} - (x-a)^\nu T^{μ0}\, dx dy dz[/tex]
    location of the center of mass:
    [tex] x_{CM}^j = \frac{1}{M} \int x^j T^{00}\,d^3x[/tex]
    Intrinsic angular momentum ( ##S^{\mu\nu}## ) is defined as angular momentum about any event on the center of mass's world line, and works out to be:
    [tex] S^{\mu0} = S^{0\mu} =0, S^{ij} = \int (x-x_{CM})^i T^{j0} - (x-x_{CM})^j T^{i0}\, dV [/tex]
    As there are only three components, ## S^i ## is defined in the equation ## S^{jk} = \epsilon^{jki} S^i ##
    3. The attempt at a solution

    As it is a cylinder, ## T^{\mu\nu} = 0 ## for ## r > R ##. I chosed z as the rotating axis, therefore ## S^{jz} = S^{zj} = 0 ##, and ## |S| = |S^{xy}| ##.
    I can also choose that ## x_{CM}^\mu = 0 ##, and the time hypersurface is that having ## x^0 = 0 ##.
    Therefore,
    [tex] S^{xy} = \int ( x T^{y0} - y T^{x0}) \, dx dy dz = \int_{-z}^z \int_0^{2\pi} \int_0^R r T^{\theta0} r\,dr d\theta dz[/tex]
    where I am using x without superscript as the x component of ##x^\mu##, and y as the y component of ##x^\mu##. ## r,\theta ## are the normal polar coordinates and ## T^{\theta0} = - T^{x0} \sin(\theta) + T^{y0} \cos(\theta) ##.
    Here I start to get stuck, becuase I see no way of inserting the "## u⁰ > \|\vec{u^j}\| \Rightarrow T(u,u) ≥ 0 ##" condition.

    I try to choose a u vector with only time and ##\theta## components:
    [tex] T(u,u) = T^{00}(u_0)^2 - 2 T^{\theta0} u_\theta (-u_0) + T^{\theta\theta} (u_\theta)^2 ≥ 0 \Rightarrow 2 T^{\theta0} u_\theta (-u_0) ≤ T^{00}(u_0)^2 + T^{\theta\theta} |u_\theta|^2 [/tex]
    As ## u⁰ = (-u_0) > |u_\theta| ##,
    [tex] 2 T^{\theta0} u_\theta (-u_0) < T^{00}(u_0)^2 + T^{\theta\theta} (-u_0)^2 \Leftrightarrow 2 T^{\theta0} u_\theta < (-u_0) ( T^{00} + T^{\theta\theta} ) [/tex]
    Inserting it in the above equation (I can choose that ## u_\theta > 0 ##, don't I? ):
    [tex] S^{xy} = \int_{-z}^z \int_0^{2\pi} \int_0^R r^2 T^{\theta0}\,dr d\theta dz< \int_{-z}^z \int_0^{2\pi} \int_0^R r^2 \frac{1}{2 u_\theta} (-u_0) (T^{00} + T^{\theta\theta} ) \,dr d\theta dz [/tex]

    Therefore,
    [tex] |S^{xy}| < | \int_{-z}^z \int_0^{2\pi} \int_0^R r^2 \frac{-u_0}{2 u_\theta} (T^{00}+ T^{\theta\theta} ) \,dr d\theta dz | ≤ \int_{-z}^z \int_0^{2\pi} \int_0^R r^2 \frac{|u_0|}{2 |u_\theta|} |T^{00} + T^{\theta\theta} | \,dr [/tex]

    I would like to say:

    [tex] |S^{xy}| < \int_{-z}^z \int_0^{2\pi} \int_0^R r^2 |T^{00} + T^{\theta\theta} | \,dr d\theta dz < \int_{-z}^z \int_0^{2\pi} \int_0^R r^2 |T^{00} | \,dr d\theta dz < R \int_{-z}^z \int_0^{2\pi} \int_0^R |T^{00}| r \,dr d\theta dz < R M [/tex]

    but I can not since ## |u_0| > |u_\theta| ##.

    And I do not know where to go from now. I am probably missing something.

    If I try to use intuition in the simplest imaginable problem I can solve it: I imagine two particles of ##m_0## rest mass going in opposite directions in XY plane a distance r from the center. Say, for example that one's position is (0,R,0,0) and other's is (0,-R,0,0). And their 4-velocity are ##( u^0, 0, u^y, 0 ) ## and ##( u^0, 0, -u^y, 0 ) ## respectively. Then,

    [tex] |S^{xy}| = 2 R m_0 |u^y| < R 2 m_0 u⁰ = R M [/tex]

    But I did not use ## T(u,u) ≥ 0 ## condition. Just that ## u⁰ > |u^y| ##.
    And this is not a cylinder. In a real cylinder I would get radial tensions so that the particles do not go far away from each other. In the mathematical development I did not take into account any tensions, and I think they should be present.

    Thanks again for the help.
     
  2. jcsd
  3. Dec 31, 2013 #2

    TSny

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    Hello Cesarth.

    After looking at the statement of the problem in MTW, I have a somewhat different interpretation of part (d). It seems to me that they want you to consider an arbitrary isolated system of finite extent which has intrinsic angular momentum S. They then want you to show that in the rest frame of the system if you construct an imaginary cylinder "parallel to S" that contains the entire system, then the radius of the imaginary cylinder must be greater than or equal to |S|/M. So the physical system itself doesn't need to be a cylinder. Of course, you could consider the special case where the system is a rotating cylinder, but I think they want you to prove the theorem for any finite system.

    It also seems to me that the problem sort of guides you to the answer by having you first work parts (a), (b), and (c). I believe the answer to (d) follows without much calculation from (c).

    If I'm not mistaken, the condition ##T(u,u) \geq 0## is needed only to insure that the centroid as defined in part (a) lies within the region occupied by the system. If the energy density ##T(u,u)## could be negative for parts of the system according to some observer with 4-velocity u, then that observer might find that the centroid does not lie "within" the system. In that case I don't think that (d) would follow from (c).
     

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  4. Jan 2, 2014 #3
    Thank you TSny,
    I'll try this alternative path. It does not seem obvious for me either.
     
  5. Jan 11, 2014 #4
    Ok, Tsny.
    I've been working on this this whole week on my train trips to work, and I am not able to get it following your path. I have filled three pages and this is the best I've been able to come up with:

    Fist I simplify the problem supposing that ##\vec{v}## is perpendicular to ##\vec{S} = (0,0,S_z)## in the rest frame of the system. For example, ## \vec{v} = ( v_x,0,0)##. Therefore, as we know from the formula:

    [tex]\vec{\xi} = - \frac{ \vec{v} \times \vec{S} } { M } [/tex]

    that resulted in part c of the exercise, we have that ##\vec{\xi}=(0,\xi_y,0)##, and taking absolute value:

    [itex] |\xi_y| M = |v_x| |S_z| [/itex]

    Now, as you say, I try to prove that ##|\xi_y| < R ##. For that, I suppose that the origins of the system of the center of mass and the system that moves with velocity ##\vec{v}## match at ## t=x=y=z=t'=x'=y'=z'=0 ##. Primed coordinates are for the system moving with ##v_x## velocity from the center of mass.

    [tex]\xi_{y'} = \xi_y = \frac{1}{P^{0'}} \int_\Sigma y T^{0'0'} dx'dydz [/tex]

    Where ##\Sigma## is the region defined by ## t'=0, -\inf < z = z' <\inf, x²+y² = (u_0 x')²+y² < R ##. Being ##u_0## the 4-velocity's 0-th component associated with ##\vec{v}##.

    Using ## ∇·T = 0 ##, and knowing that outside the cylinder ##T=0## we can obtain:

    [tex] \xi_{y'} = \xi_y = \frac{1}{P^{0'}} \int_V y T^{0'0'} dxdydz [/tex]

    Where V is the region ## t=0, z\in(-\inf,\inf), x²+y²<R ##. As ## 0 < T^{0'0'} = u·T·u = u_0{}^2T^{00} + 2 u_0 u_x T^{0x} + u_x{}^2 T^{xx} ##,

    [tex] | \int_V y T^{0'0'} dxdydz | < R \int_V \mathbf{u·T·u} dxdydz = R u_0{}^2 M + 2 R u_0 u_x (P^x=0) + R u_x{}^2 \int_V T^{xx} dxdydz [/tex].

    If the last term were 0, (I am not sure, but I think it should be, because otherwise there would be a force applied over the whole system and that would mean that the system is not isolated ) we would have:
    [tex] | \xi_{y} | < \frac{ R u_0{}^2 M } { P^{0'} } = R u^0 [/tex]
    And then,
    $$ |v_x||S_z| = \frac{|u_x|} { u^0 } |S_z| = M |\xi_y| < M R u^0 → |u_x| |S_z| < M R (u^0{})^2 . $$
    As ## u⁰ > u_x ##, I can not prove it.
     
  6. Jan 11, 2014 #5

    TSny

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    OK. You have [tex] |\xi_y| M = |v_x| |S_z| [/tex] I think you can get the result for part (d) without much more calculation if you interpret the meaning of this equation.

    Note that ##v_x## is the x-component of velocity of some observer, Bob, moving relative to the rest frame of the system that has spin ## S_z## . Bob can calculate the location of the centroid of the system using equation (5.56) of the text. Call this “Bob’s centroid” of the system.

    Now, ##\xi_y## denotes the y-component of the displacement of Bob’s centroid of the system from the center of mass of the system as measured in the rest frame. (The "center of mass" is just the centroid as calculated in the rest frame.)

    What does the equation [itex] |\xi_y| M = |v_x| |S_z| [/itex] tell you about the maximum possible value for ##|\xi_y|##?
     
  7. Jan 11, 2014 #6

    TSny

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    To show ##|\xi_y| < R ##, try a proof by contradiction. Thus, suppose ##|\xi_y| \geq R##. This would imply that Bob's centroid lies outside the cylinder of radius R that contains all the material of the system.

    Now, suppose in Bob's frame you choose a coordinate system ##(\tilde{x},\tilde{y},\tilde{z})## with origin at Bob's centroid. We are assuming this origin lies outside the cylinder of radius R. Thus, we may orient the coordinate system such that the positive ##\tilde{x}## axis runs from the centroid through the center of the cylinder and all points within the cylinder have positive ##\tilde{x}## values. (Bob sees the cylinder as Lorentz contracted, but that shouldn't affect the argument.)

    Suppose Bob uses this coordinate system to calculate the ##\tilde{x}## coordinate of the centroid. Can you show that there is a contradiction if ##T^{00}## for Bob is ##\geq 0## everywhere?
     
  8. Jan 14, 2014 #7
    So, thanks to your guidance, now I have proven that ##|\xi_y|<R## easily. But,

    [tex] R>|\xi_y| = \frac{ |v_x| |S_z| }{ M } < \frac{ |S_z| }{ M } [/tex]

    because ##|v_x|<1 ## (natural units ). From there it can not be proven.
     
    Last edited: Jan 14, 2014
  9. Jan 14, 2014 #8

    TSny

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    But you can let ##|v_x|## be any speed you want (less than 1) and still you must have ##R>|\xi_y|## for that choice of ##v_x##. So, you can take the limit of ##|\xi_y|## as ##|v_x| \rightarrow 1##.
     
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