Rotational motion and special relativity

Question

I was trying to solve two questions from problem book on relativity and gravitation by lightman.Questions are

1. Calculate the nonzero components in an inertial frame S of the stress-energy tensor for the give system: A ring of N similar particles of mass m rotating counter- clockwise in the x-y plane about some point fixed in S at a radius $a$ and angular velocity $\omega$. (The width of the ring is much less than a.) Do not include the stress-energy of whatever forces keep them in orbit. Assume N is large enough that one can treat the particles as being continuously distributed. (problem 5.1)
2. An infinitesimally thin rod of length 2a has a point mass m at each of its ends. The center of the rod is fixed in the laboratory and the rod rotates about this point with a relativistic angular velocity $\omega$. (i.e. $\omega l$ is comparable with c). Assume the rod is massless. What is $T^{\mu\nu}$ for the rod and particle system? (problem 5.7)

The solutions of the first problem is give as: Let the circle of motion be in the $x$ -y plane. Consider a con- tinuum of rest mass density $\rho_{0}$ at the point $x=0, y=a$ moving with velocity $\beta=\omega$ a. At $x=0, y=a$ we have $T^{00}=\rho_{0} \gamma^{2}$ etc. Now if we consider a ring of matter, all points on the ring are equivalent so that in polar coordinates the nonzero components of $\mathrm{T}^{\mu \nu}$ are

\begin{equation} \ \mathrm{T}^{00}=\rho_{0} \gamma^{2}, \quad \mathrm{T}^{0 \hat{\phi}}=\rho_{0} \gamma^{2} \beta=\mathrm{T}^{\hat{\phi} 0}, \quad \mathrm{T}^{\hat{\phi} \hat{\phi}}=\rho_{0} \gamma^{2} \beta^{2} \tag{1}\end{equation} In this case, $\rho_{0}=\mathrm{Nm} \delta(\mathrm{r}-\mathrm{a}) \delta(\mathrm{z}) / 2 \pi \mathrm{a} \gamma$

While the solution for second problem is given as: Let S' be the instantaneous rest frame of an element of the rod. In this rest frame the only nonzero components are $\mathrm{T}^{0^{\prime} 0^{\prime}}=\rho, \quad \mathrm{T}^{\mathrm{x}^{\prime} \mathrm{x}^{\prime}}=\mathrm{p}$ . If we Lorentz transform to the lab frame we find the nonzero components: $T^{x x}=p, \quad T^{y y}=y^{2} \beta^{2} \rho$ $\mathrm{T}^{00}=\gamma^{2} \rho, \quad \mathrm{T}^{0 \mathrm{y}}=\gamma^{2} \beta \rho, \quad$ where $\beta=\omega \mathrm{r}$ and $\gamma \equiv\left(1-\beta^{2}\right)^{-\frac{1}{2}} .$ If spherical polar coordinates are used the nonzero components are \begin{equation}\mathrm{T}^{\mathrm{rr}}=\mathrm{p}, \quad \mathrm{T}^{\phi \phi}=\gamma^{2} \beta^{2} \rho / \mathrm{r}^{2}, \quad \mathrm{T}^{0 \phi}=\gamma^{2} \beta \rho / \mathrm{r}, \quad \mathrm{T}^{00}=\gamma^{2} \rho\tag{2}\end{equation} In this case, $\rho=m \delta(r-a) \delta(\cos \theta)[\delta(\phi-\omega t)+\delta(\phi-\omega t-\pi)] / r^{2}$

Can anyone explain why equation 1 and equation 2 are not same?(Notice the extra factor of r in equation 2)

Answer 1

You need to consider the coordinates you are using and the matter density, in the first example you have polar cylindrical and in the second they are polar spherical coordinates. Thus Dirac's Delta Function should be written, respectively as $$ \delta(\vec{r} - \vec{r}_o) = \frac{1}{2\pi r} \delta(r - r_o)\delta(z - z_o), $$since de symmetry over $\phi,$ and $$\delta(\vec{r} - \vec{r}_o) = \frac{1}{r^2} \delta(r - r_o)\delta(\theta - \theta_o) \delta(\phi - \phi_o).$$

Check some reference for that, for example, http://www.fen.bilkent.edu.tr/~ercelebi/mp03

After this you need to realize that in the first example you have the $0\mu$-th components are related to energy-momentum density as in $T^{0\mu} \sim P^{\mu},$ the $\phi \phi$ component goes as kinetic energy and given the coordinates and metric you need to make sure $\nabla_{\mu} T^{\mu\nu}= 0.$ In the second example it should have the same idea but they start from a rest frame and change after to a moving one to facilitate the calculation.

Answer 2

My solution of question 2 in cylindrical co-ordinate: Let S' be the instantaneous rest frame of an element of the rod. In this rest frame the only nonzero components are $\mathrm{T}^{0^{\prime} 0^{\prime}}=\rho, \quad \mathrm{T}^{\mathrm{x}^{\prime} \mathrm{x}^{\prime}}=\mathrm{p}$ . If we Lorentz transform to the lab frame we find the nonzero components: $T^{x x}=p, \quad T^{y y}=y^{2} \beta^{2} \rho$ $\mathrm{T}^{00}=\gamma^{2} \rho, \quad \mathrm{T}^{0 \mathrm{y}}=\gamma^{2} \beta \rho, \quad$ where $\beta=\omega \mathrm{r}$ and $\gamma \equiv\left(1-\beta^{2}\right)^{-\frac{1}{2}} .$ If cylindrical coordinates are used the nonzero components are \begin{equation}\mathrm{T}^{\mathrm{rr}}=\mathrm{p}, \quad \mathrm{T}^{\phi \phi}=\gamma^{2} \beta^{2} \rho / \mathrm{r}^{2}, \quad \mathrm{T}^{0 \phi}=\gamma^{2} \beta \rho / \mathrm{r}, \quad \mathrm{T}^{00}=\gamma^{2} \rho\tag{2}\end{equation}

To find $\mathrm{p}(\mathrm{r})$ we can use the $\mathrm{r}$ component of the equation of motion

\begin{aligned} 0 &=T_{; \nu}^{r \nu}=T_{, r}^{r r}+T^{\phi \phi} \Gamma_{\phi \phi}^{r}+T^{r r}\left[\log (-g)^{\frac{1}{2}}\right]_{, r} \\ &=\left(T^{r r} \right)_{, r}-r T^{\phi\phi} +(1/r) T^{rr}\\ &=(rT^{rr})_{, r}-\gamma^{2} \beta^{2} \rho \end{aligned} Let, $\rho=m \delta(r-a) \delta(z)[\delta(\phi-\omega t)+\delta(\phi-\omega t-\pi)] / r$

We can now integrate the equation of motion with the boundary condition p = 0 for r > a, to find p. Upon doing so we get $$T^{rr}=\frac{m}{1-{\omega}^2a^2}\delta(z)[\delta(\phi-\omega t)+\delta(\phi-\omega t-\pi)] (\frac{{\omega}^2a}{r})$$ Which then upon integrating over total volume gives $$\frac{m{\omega}^2a}{1-{\omega}^2a^2}$$ which is same if we had used spherical coordinates