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

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##.

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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