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.
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.
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