I've now started chapter 4 on Gravitation just in time for 2020. Very exciting!
Carroll first states two formulas of Newtonian Gravity his 4.1 and 4.2$$
\mathbf{a}=-\nabla\Phi
$$where ##\mathbf{a}## is the acceleration of a body in a gravitational potential ##\Phi##. And Poisson's differential equation for the potential in terms of the matter density ##\rho## and Newton's gravitational constant ##G##:$$
\nabla^2\Phi=4\pi G\rho
$$I had a long pause thinking about the various formulas for the Laplacian ##\nabla^2## here.
How to these tie up with the old-fashioned laws? Newton's law of gravity is normally stated as$$
F=G\frac{m_1m_2}{r^2}
$$which combined with Newton's second law ##F=m\mathbf{a}## gives us the acceleration of a mass in the presence of another as$$
\mathbf{a}=G\frac{M}{r^2}
$$In exercise 3.6 we were given 'the familiar Newtonian gravitational potential'$$
\Phi=-\frac{GM}{r}
$$A bit of rough reasoning shows these are equivalent.
At his 4.4 Carroll states that the next equation gives the path of a particle subject to no forces$$
\frac{d^2x^i}{d\lambda^2}=0
$$If we solve it in polar coordinates for ##r,\theta## instead of ##x,y## Carroll says we get a circle and he cheekily suggests that we might think free moving particles follow that path. But the solution is $$
r=m\theta+k
$$where ##m,k## are constants. We can plot that and, obviously if ##m=0## we get a circle of radius ##k## but if ##m\neq0## we get other more interesting lines which are equally wrong. See above. Another error by Carroll, but only minor 😏. The next one is a corker.
Then we examine the equations in a near Newtonian environment and equation 4.13 ##g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}## is wrong. The actual equation is obviously$$
g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}
$$Properly 4.13 might be
$$
g^{\mu\nu}=\eta^{\mu\nu}-h_{\rho\sigma}\eta^{\mu\sigma}\eta^{\nu\rho}
$$which is true to first order and gives ##h^{00}=-h_{00}## which is used in the next section. If one accepts the approximation that ##\eta## can be used to raise and lower indices on any object of order ##h## then that also gives us$$
g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}
$$which says ##h^{\mu\nu}=0##. Oops! But it turns out it turns out that the sign on ##h^{\mu\nu}## is immaterial in this section. There is a full analysis in the pdf.
We soon arrive at the conclusion that in the near Newtonian environment we have the time, time component of the metric is $$
g_{00}=-1-2\Phi
$$which is also what we were given in Exercise 3.6.
 |
| Fools straight line |
\mathbf{a}=-\nabla\Phi
$$where ##\mathbf{a}## is the acceleration of a body in a gravitational potential ##\Phi##. And Poisson's differential equation for the potential in terms of the matter density ##\rho## and Newton's gravitational constant ##G##:$$
\nabla^2\Phi=4\pi G\rho
$$I had a long pause thinking about the various formulas for the Laplacian ##\nabla^2## here.
How to these tie up with the old-fashioned laws? Newton's law of gravity is normally stated as$$
F=G\frac{m_1m_2}{r^2}
$$which combined with Newton's second law ##F=m\mathbf{a}## gives us the acceleration of a mass in the presence of another as$$
\mathbf{a}=G\frac{M}{r^2}
$$In exercise 3.6 we were given 'the familiar Newtonian gravitational potential'$$
\Phi=-\frac{GM}{r}
$$A bit of rough reasoning shows these are equivalent.
At his 4.4 Carroll states that the next equation gives the path of a particle subject to no forces$$
\frac{d^2x^i}{d\lambda^2}=0
$$If we solve it in polar coordinates for ##r,\theta## instead of ##x,y## Carroll says we get a circle and he cheekily suggests that we might think free moving particles follow that path. But the solution is $$
r=m\theta+k
$$where ##m,k## are constants. We can plot that and, obviously if ##m=0## we get a circle of radius ##k## but if ##m\neq0## we get other more interesting lines which are equally wrong. See above. Another error by Carroll, but only minor 😏. The next one is a corker.
Then we examine the equations in a near Newtonian environment and equation 4.13 ##g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}## is wrong. The actual equation is obviously$$
g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}
$$Properly 4.13 might be
$$
g^{\mu\nu}=\eta^{\mu\nu}-h_{\rho\sigma}\eta^{\mu\sigma}\eta^{\nu\rho}
$$which is true to first order and gives ##h^{00}=-h_{00}## which is used in the next section. If one accepts the approximation that ##\eta## can be used to raise and lower indices on any object of order ##h## then that also gives us$$
g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}
$$which says ##h^{\mu\nu}=0##. Oops! But it turns out it turns out that the sign on ##h^{\mu\nu}## is immaterial in this section. There is a full analysis in the pdf.
g_{00}=-1-2\Phi
$$which is also what we were given in Exercise 3.6.
More at Commentary 4.1 Physics in curved spacetime.pdf (5 pages)
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