Consider a photon of "mass" $m_\gamma$ passing near an object of mass $M_L$; we'll call this object a "lens" (the 'L' in $M_L$ stands for "lens", which is the object doing the bending). The closest approach ($b$) of the photon is known as the impact parameter. We can imagine that the photon feels a gravitational acceleration from this lens, which we imagine is vertical (see diagram below:
(a) Give an expression for the gravitational acceleration in the vertical direction in terms of $M$, $b$, and $G$.
Okay, so we know that we are using basic Newtonian physics to solve for the distortion of space-time. Therefore, we can use basic Newtonian equations to solve for the gravitational acceleration in the vertical direction.
Recall that the relationship between force and acceleration can be described by Newton's Second Law of Motion:
\begin{align}
F = ma
\end{align}
Well in this case, we can use the force of gravitational acceleration, $F_g = \frac{GM_1M_2}{r^2}$, where the two masses, $M_1$ and $M_2$ are the mass of the "lens", $M_L$ and the "mass" of the photon, $m_{\gamma}$, respectively. The radius between the two bodies, $r$ is described by the variable $b$. Knowing this, we can rearrange Newton's second law and substitute in these variables for gravitational force, $F_g = \frac{G M_L M_\gamma}{b^2}$ in order to solve for the gravitational acceleration, $a$ as follows:
\begin{align}
F_g &= m_\gamma a\\
a &= \frac{F_g}{m}\\
a &= \frac{\frac{G M_L M_\gamma}{b^2}}{m_\gamma}\\
a &= \frac{G M_L}{b^2}\\
\end{align}
(b) Consider the time of interaction $\Delta t$. Assume that most of the influence the photon feels occurs in a horizontal distance $2b$. Express $\Delta t$ in terms of $b$ and the speed of the photon.
In order to solve for the "time of interaction", $\Delta t$, let's think about the direction that the photon was travelling. Since the photon was traveling in the horizontal direction, it feels the effects of gravity of $M_L$ the most over the horizontal distance, $2b$.
Since we know that the photon is traveling a distance of $2b$, and we are solving for the $\Delta t$, we can use the rudimentary knowledge of classical mechanics to relate the distance and time using the equation for velocity. Since velocity is measured as the distance traveled over a specified time, and we know that a photon travels at the speed of light, $c$, we can use the velocity equation to solve for $\Delta t$ as follows:
\begin{align}
v_{photon} &= \frac{\text{distance}}{\Delta t}\\
\Delta t &= \frac{\text{distance}}{v_{photon}}\\
\Delta t &= \frac{2b}{c}
\end{align}
(c) Solve for the change in velocity, $\Delta v$, in the direction perpendicular to the original photon path, over this time of interaction.
We are solving for the "change in velocity... over time of interaction", which sounds very much like the definition of acceleration ($a = \frac{\Delta v}{\Delta t}$)! Since the change in velocity, $\Delta v$, is happening over the distance perpendicular to the original photon path, we can use the acceleration derived in part (a) to solve for the change in velocity $\Delta v$. And since we are only measuring the "change in velocity", $\Delta v$ over the "time of interaction", $\Delta t$, we have all the necessary information to use the acceleration equation.
Recall that the acceleration in part (a) was $a = \frac{G M_L}{b^2}$, and the time of interaction, $\Delta t$ calculated in part (b) was $\Delta t = \frac{2b}{c}$. Using this information, we can rearrange the acceleration equation to solve for $\Delta v$ as follows:
\begin{align}
a &= \frac{\Delta v}{\Delta t}\\
\Delta v &= a \Delta t\\
\Delta v &= \frac{G M_L}{b^2} \times \frac{2b}{c}\\
\Delta v &= \frac{2G M_L}{bc}
\end{align}
(d) Now solve for the deflection angle ($\alpha$) in terms of $G$, $M_L$, $b$, and $c$ using your answer from part (a), (b), and (c). This result is a factor of 2 smaller than the correct, relativistic result.
According to the diagram above, the deflection angle $\alpha$ seems to create a right triangle with the horizontal and vertical components of the velocity of the path the photon is trying to take. Therefore, we can model the deflection angle as follows:
Using trigonometry, we can determine the following:
\begin{align}
\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{\Delta v}{c}
\end{align}
However, since the change in angle is very minute, we can use the small angle approximation to say that:
\begin{align}
\tan \alpha \approx \alpha = \frac{\Delta v}{c}
\end{align}
Since we can ignore the tangent function, we can solve for $\alpha$ using the equation above and the value calculated for $\Delta v = \frac{2G M_L}{bc}$ as follows:
\begin{align}
\alpha &= \frac{\Delta v}{c} \\
\alpha &= \frac{\frac{2G M_L}{bc}}{c} \\
\alpha &= \frac{2G M_L}{bc^2}
\end{align}
So now, we have the deflection angle, $\alpha$ using classical Newtonian mechanics. However, as the problem states, this answer is smaller than the actual answer obtained by general relativity by a factor of 2. Therefore, the correct relativistic deflective angle $\alpha$ is:
\begin{align}
\alpha = \frac{4G M_L}{bc^2}
\end{align}
Great work!
ReplyDelete5