Problem 2: Schutz chap 5

1.

Redshift from gravity. Fire a photon up. Convert into mass and let mass drop. If energy conserved then you gain mgh of energy where m = \hbar \omega', the freqeuncy at the top. Hence you have \hbar \omega = m+mgh=\hbar \omega (1+gh).

The frequency at the top is reduced by (1+gh), i.e. it is redder.

2. Uniform gravity raises no tides? That’s odd. naively I would think that the water would flow towards the lowest energy configuration, which would be at the bottom of the external potential. On the total gravitational potential (+ centrifugal force) constant surface. Which does get distorted by an external field.

Oh. I see. Looking at the text, the earth would be freely falling as well. ¬†Wow. That’s an important point. The earth is falling towards the moon at exactly the rate of free fall. It’s the differences of force that cause the tides. A little stronger on the front, a little weaker on the back. This is sort of the same thing as a draping cloth in a falling elevator.

7.

Can’t raise an lower \Lambda using the metric. They are coordinate transformations. I believe upper indices are components of vectors and lower indices are components of 1-forms. dx^\mu \partial_\mu and \partial_\mu \phi

x = r\cos(\theta)

y = r\sin(\theta)

This matrix will convert the gradient into new coordinates

\begin{bmatrix} \cos{\theta} & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{bmatrix}

\theta = \arctan(\frac{y}{x})

r = \sqrt{x^2+y^2}

This one will convert vectors into new coordinates

\begin{bmatrix} x/r& y/r\\ -y/r^2 &x/r^2\end{bmatrix}

Multiplying the two should give an identity matrix, since inner product should stay invariant under coordinate change. Quite miraculously they do. Try multiplying the two matrices. Nice.

 

Leave a Reply

Your email address will not be published. Required fields are marked *