Lecture 11.tex

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\title{{\Huge General Relativity}\\{\Large{Class 11 --- February 14, 2020}}} %replace with class number
\author{Rabia Husain}

\emailAdd{r.husain@utexas.edu} %replace with your email
\begin{document}
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\part{More on One Forms}

    \section{One Forms and the Gradient}
    In 3D, vectors can be decomposed via the Helmholtz decomposition into:
        \begin{equation}
            \begin{aligned}
            \bar{V} &= \bar{\nabla} \Phi + \bar{\nabla} \cross \bar{A}
            \end{aligned}
        \end{equation}
        \sn{(1.1) Where $\Phi$ is some scalar function and $\bar{A}$ is some vector field.}
    So, one forms are not uniquely represented by the gradient of some function since there is a component that is the curl of a field. However, we do not yet have a definition for the cross product in higher dimensions. \\
    An example of a one form is the gradient:
        \begin{equation}
            \begin{aligned}
            \undertilde{\omega} &= \undertilde{d}f
            \end{aligned}
        \end{equation}
    This can be used to build a more general definition for a one form:
        \begin{equation}
            \begin{aligned}
            \undertilde{\omega} &= &= \omega_\mu dx^\mu 
            \end{aligned}
        \end{equation}
    in which dx$^\mu$ is the coordinate basis.
    Now, consider if $\omega_\mu = \partial_\mu f$. Then,
        \begin{equation}
            \begin{aligned}
            \partial_\mu \omega_\nu - \partial_\nu \omega_\mu &= \partial_\mu \partial_\nu f - \partial_\nu \partial_\mu f &= 0
            \end{aligned}
        \end{equation}
    This shows that not all one forms can be written as the gradient of a function.
    
    \section{Kinds of One Forms}
    \begin{itemize}
    \item
    A "rotation-free" one form is defined as:
        \begin{equation}
            \begin{aligned}
            \undertilde{\omega} &= h \undertilde{d}f
            \end{aligned}
        \end{equation}
    for h and f as some other functions on the manifold. This type of one form can no longer be written as a gradient of a function. It is possible to show that $\omega_{[\mu} \partial_\nu \omega_{\rho]} = 0$, meaning that all indices are fully antisymmetrized
    \sn{Recall the definition of antisymmetrization:\\*
            $A_{[\mu \nu]} = \frac{1}{2}(A_{\mu\nu} - A_{\nu\mu})$\\*
            $A_{[\mu \nu \rho]} = \frac{1}{6}(A_{\mu\nu\rho} + A_{\rho\mu\nu} + A_{\nu\rho\mu} - A_{\rho\nu\mu} - A_{\mu\rho\nu} - A_{\nu\mu\rho})$\\*
        and so on.
        }
    , where $\omega^\mu$ is perpendicular to some hypersurface.\\
    \begin{figure} [H]
        \begin{center}

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\end{tikzpicture}
        \caption{A vector field (in black) normal to some weirdly shaped surface (in red). It is possible to check if a vector field is normal to a hypersurface using $\omega_{[\mu} \partial_\nu \omega_{\rho]}$}
            \end{center}
    \end{figure}
    
    \item
    A "curl-free" one form is a special type of one form defined as:
        \begin{equation}
            \begin{aligned}
            \undertilde{\omega} &= \undertilde{d}f
            \end{aligned}
        \end{equation}
    If a one form is "curl-free", then it obeys equation (1.4). So, does every one form which obeys equation (1.4) have the form $\omega_\mu = \partial_\mu f$? 
    The answer is no, not always. This depends on the space that you are in. If you are looking locally, meaning in the neighborhood of the point, then yes, $\omega_\mu = \partial_\mu f$ holds. We call this a closed form, which implies that the form is exact. If you are looking globally, however, $\omega_\mu = \partial_\mu f$ does not hold, meaning that in some spaces, not all closed forms are exact.
    \end{itemize}

\newpage

\part{The Metric Tensor}

    The metric tensor in curved spacetime is:
        \begin{equation}
            \begin{aligned}
            \textsl{g} &= g_{\mu\nu} \undertilde{d}x^\mu \otimes \undertilde{d}x^\nu
            \end{aligned}
        \end{equation}
    where $g_{\mu\nu}$ represents the components of \textsl{g}. This should not be confused with $\eta_{\mu\nu}$, which was used previously to represent the metric for flat space. While it is appropriate to use $g_{\mu\nu}$ as the metric for flat space as well as for curved space, $\eta_{\mu\nu}$ is reserved solely for flat space. 
    \sn{Some properties of $g_{\mu\nu}$:\\*
            $g_{(\mu\nu)} = g_{\mu\nu}$ (symmetric)\\*
            $g_{\mu\nu}$ has 10 degrees of freedom.\\*}
\
\begin{figure} [h!]
                \begin{centering}

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    \caption{The manifold $\mathcal{M}$ on which we have a point $x^\mu$. The metric $g_{\mu\nu}$ at that point is defined by $g_{\mu\nu}(x^\mu)$.}
                \end{centering}
\end{figure}
\\
    We can define the determinant of the metric tensor as the following.
        \begin{equation}
            \begin{aligned}
            det(g_{\mu\nu}) = g
            \end{aligned}
        \end{equation}
    If $g \neq 0$, then the inverse matrix of $g_{\mu\nu}$, called $g^{\mu\nu}$, exists. $g^{\mu\nu}$ has the property that $g^{\mu\alpha} g_{\alpha\nu} = \tensor{\delta}{^\mu_\nu}$. 

\newpage

    \section{Raising and Lowering Indices}
    Since the metric tensor forms an invertible map between one forms and vectors, we can now define a way to raise and lower the indices of one forms and vectors, 
        \begin{equation}
            \begin{aligned}
            \omega^\mu &= g^{\mu\alpha} \omega_\alpha
            \end{aligned}
        \end{equation}
         \begin{equation}
            \begin{aligned}
            V_\mu &= g_{\mu\alpha} V^\alpha
            \end{aligned}
        \end{equation}
    Equation (3.1) defines a way to raise indices, going from one forms to vectors. Equation (3.2) defines a way to lower indices, going from vectors to one forms.
    
    \section{The Dot Product}
    We can also use the metric tensor to define the dot product. 
        \begin{equation}
            \begin{aligned}
            \bar{U} \cdot \bar{V} &= \textsl{g}(\bar{U}, \bar{V}) = g_{\mu\nu} U^\mu V^\nu
            \end{aligned}
        \end{equation}
    Using this definition of the dot product, we can define the magnitude of vectors. 
        \begin{equation}
            \begin{aligned}
            V_\mu V^\mu &= \begin{cases} 
                            < 0 & "timelike" \\
                            > 0 & "spacelike" \\
                            = 0 & "null"/"lightlike" 
                            \end{cases}
            \end{aligned}
        \end{equation}
    \sn{This is similar to our previous definitions of "timelike separated", "spacelike separated", and "lightlike separated" when we were discussing the spacetime interval $\Delta s^2$ in special relativity.  }
    for some $\bar{V}(f) = \frac{df}{d\lambda}$ where $\bar{V} = \frac{\bar{d}}{d\lambda}$. $\bar{V}$ is a directional derivative that can act on a function. It is fine to think about $\bar{V} = V^\mu \bar{e}_{(\mu)}$, but $\bar{e}_{(\mu)}$ is just a directional derivative now. For the one form $\undertilde{w}$: $T_{p} \to \mathbb{R}$ acting on $\bar{V}$, $\undertilde{w}(\bar{V}) = w_\mu V^\mu$. This one form is acting via contraction to map a derivative to the real numbers. 
    
    \section{Distance}
    We can also write down a definition for distance using the metric. For a spacelike curve, we define distance as proper distance s.
        \begin{equation}
            \begin{aligned}
            s &= \int_{\lambda_a}^{\lambda_b} \sqrt{g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d\lambda
            \end{aligned}
        \end{equation}
        
        \begin{figure} [H]
                \begin{centering}
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\caption{The manifold $\mathcal{M}$ on which we have a spacelike curve from points a to b. The curve is parameterized by $\lambda$.}
\end{centering}
\end{figure}
    
    For a timelike curve, we define distance as proper time $\tau$.
        \begin{equation}
            \begin{aligned}
            \tau &= \int_{\lambda_a}^{\lambda_b} \sqrt {-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d\lambda
            \end{aligned}
        \end{equation}
        
    \begin{figure} [H]
        \begin{centering}
    

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    \caption{The manifold $\mathcal{M}$ on which we have a timelike curve from points a to b. The curve is parameterized by $\lambda$.}
        \end{centering}
    \end{figure}
    
    
    In both cases, we have assumed nothing about the parameterization $\lambda$. It is important to note that the proper time and proper distance along a specified curve between two points is invariant under coordinate changes.
     \begin{equation}
            \begin{aligned}
            V^\mu V_\mu = V^{\mu'} V_{\mu'}
            \end{aligned}
        \end{equation}
    A convenient shorthand for distance, ds, can be written as follows:
        \begin{equation}
            \begin{aligned}
            ds^2 &= g_{\mu\nu} \undertilde{d}x^\mu \otimes \undertilde{d}x^\nu\\*
            &= g_{\mu\nu} dx^\mu dx^\nu
            \end{aligned}
        \end{equation}
    While the second line of equation (5.4) is an abuse of notation, this is safe to use if one acts as though $dx^\mu$ and $dx^\nu$ are just differential objects.
    
    \begin{example}
	Let us consider cartesian coordinates in 3D flat space. In this space, 
	    \begin{equation}
            \begin{aligned}
            g_{ij} &= \begin{pmatrix}
                        1  &  0  &  0\\
                        0  &  1  &  0\\
                        0  &  0  &  1\\
                        \end{pmatrix}
            \end{aligned}
        \end{equation}
    We can write the distance as:
        \begin{equation}
            \begin{aligned}
            ds^2 &= \undertilde{d}x \otimes \undertilde{d}x + \undertilde{d}y \otimes \undertilde{d}y + \undertilde{d}z \otimes \undertilde{d}z \\
            &= dx^2 + dy^2 + dz^2
            \end{aligned}
        \end{equation}

    This is just the Pythagorean theorem! 
    
   \begin{figure} [H]
        \begin{centering}
        

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        \caption{Two points a and b separated by a distance $ds^2$ in 3D flat space.}
        \end{centering}
    \end{figure}
    
    Now, we will transform coordinates from (x, y, z) $\to$ (r, $\theta$, $\phi$). Let us denote (x, y, z) with $x^i$ and (r, $\theta$, $\phi$) with $x^{i'}$. We will perform this transformation using the Jacobian transformation matrices $\frac{\partial x^{i'}}{\partial x^{i}}$. 
    For i = 1:
        \begin{equation}
            \begin{aligned}
            \frac{\partial x^{1'}}{\partial x^1} &= \frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2 + y^2 + z^2} = \frac{x}{r}
            \end{aligned}
        \end{equation}
    This same process can be done for i = 2 and i = 3. The distance in spherical polar coordinates is as follows:
        \begin{equation}
            \begin{aligned}
            ds^2 &= dr^2 + r^2d\theta^2 + r^2 \sin^2{\theta}d\phi^2
            \end{aligned}
        \end{equation}
    Now, we can write the metric and its inverse for this coordinate system.
        \begin{equation}
            \begin{aligned}
            g_{ij} &= \begin{pmatrix}
                        1  &  0  &  0\\
                        0  &  r^2  &  0\\
                        0  &  0  &  r^2\sin^2{\theta}\\
                        \end{pmatrix}
            \end{aligned}
        \end{equation}
        \begin{equation}
            \begin{aligned}
            g^{ij} &= \begin{pmatrix}
                        1  &  0  &  0\\
                        0  &  \frac{1}{r^2}  &  0\\
                        0  &  0  &  \frac{1}{r^2\sin^2{\theta}}\\
                        \end{pmatrix}
            \end{aligned}
        \end{equation}
    \end{example}
    
    \begin{example}
    
    Another example that we can consider is the 4D Friedmann–Lemaître–Robertson–Walker (FLRW) metric.
        \begin{equation}
            \begin{aligned}
            ds^2 &= -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2)
            \end{aligned}
        \end{equation}
    This looks like our spacetime interval $\Delta s^2$ for flat 4D spacetime except for the $a^2(t)$ term, which is called the "scale factor".
    
    \begin{figure} [H]
        \begin{centering}

\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        

\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]
%uncomment if require: \path (0,381); %set diagram left start at 0, and has height of 381

%Shape: Axis 2D [id:dp8310412399811793] 
\draw  (274,211.6) -- (451.5,211.6)(291.75,73) -- (291.75,227) (444.5,206.6) -- (451.5,211.6) -- (444.5,216.6) (286.75,80) -- (291.75,73) -- (296.75,80)  ;
%Shape: Axis 2D [id:dp06756589499718757] 
\draw  (309.5,211.51) -- (132,212.38)(292.43,350.2) -- (291.67,196.2) (139.03,217.35) -- (132,212.38) -- (138.98,207.35) (297.39,343.17) -- (292.43,350.2) -- (287.39,343.22)  ;
%Shape: Ellipse [id:dp3232775634506875] 
\draw  [fill={rgb, 255:red, 0; green, 0; blue, 0 }  ,fill opacity=1 ] (337,212) .. controls (337,210.34) and (338.46,209) .. (340.25,209) .. controls (342.04,209) and (343.5,210.34) .. (343.5,212) .. controls (343.5,213.66) and (342.04,215) .. (340.25,215) .. controls (338.46,215) and (337,213.66) .. (337,212) -- cycle ;
%Shape: Ellipse [id:dp39951307098776034] 
\draw  [fill={rgb, 255:red, 0; green, 0; blue, 0 }  ,fill opacity=1 ] (288.5,211.6) .. controls (288.5,209.94) and (289.96,208.6) .. (291.75,208.6) .. controls (293.54,208.6) and (295,209.94) .. (295,211.6) .. controls (295,213.26) and (293.54,214.6) .. (291.75,214.6) .. controls (289.96,214.6) and (288.5,213.26) .. (288.5,211.6) -- cycle ;
%Straight Lines [id:da31362189603752033] 
\draw [line width=1.5]    (340.25,212) -- (291.75,211.6) ;

% Text Node
\draw (293,59) node   [align=left] {ct};
% Text Node
\draw (463,209) node   [align=left] {x};
% Text Node
\draw (315,226) node   [align=left] {dx};
% Text Node
\draw (281,200) node   [align=left] {a};
% Text Node
\draw (352,200) node   [align=left] {b};


\end{tikzpicture}
        \caption{Two points in 4D spacetime separated by a distance in space, dx.}
        \end{centering}
    \end{figure}
    For Figure 6 above, $ds^2 = a^2(t)dx^2 = a^2dx^2 \to ds = adx$. This means that observers at fixed coordinates see the distance between themselves grow or shrink in time depending on the value of $a$, the scale factor.
    
    \end{example}
    
    \section{Transformations}
    For a tensor T, the transformation of its components is defined as follows:
        \begin{equation}
            \begin{aligned}
           T_{\mu ' \nu '} &= \frac{\partial x^\mu}{\partial x^{\mu '}} \frac{\partial x^\nu}{\partial x^{\nu '}} T^{\mu \nu}
            \end{aligned}
        \end{equation}
    The components of the tensor change under transformation, however the tensor itself is invariant under coordinate transformation.
    \begin{example}
    Let us consider an example which is $\boldsymbol{not}$ a tensor: $\Tilde{\epsilon}_{\mu\nu\rho\sigma}$.
        \begin{equation}
            \begin{aligned}
           \Tilde{\epsilon}_{\mu '\nu '\rho '\sigma '} &= \frac{\partial x^\mu}{\partial x^{\mu '}} \frac{\partial x^\nu}{\partial x^{\nu '}} \frac{\partial x^\rho}{\partial x^{\rho '}} \frac{\partial x^\sigma}{\partial x^{\sigma '}} \Tilde{\epsilon}_{\mu\nu\rho\sigma}\\
           &= det \begin{vmatrix}
                    \frac{\partial x^\mu}{\partial x^{\mu '}}  \\
                    \end{vmatrix} \Tilde{\epsilon}_{\mu\nu\rho\sigma}
            \end{aligned}
        \end{equation}
    There is a "scaling factor", $\begin{vmatrix}
                    \frac{\partial x^\mu}{\partial x^{\mu '}}  \\
                    \end{vmatrix}$
    , multiplying $\Tilde{\epsilon}_{\mu\nu\rho\sigma}$ which shows that $\Tilde{\epsilon}_{\mu\nu\rho\sigma}$ is not invariant under coordinate transformation. Thus, $\Tilde{\epsilon}_{\mu\nu\rho\sigma}$ is not a tensor. 
    \end{example}
\end{document}