Consider a photon on "mass" m, 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.
(a) Give an expression for the gravitational acceleration in the vertical direction in terms of M, b, and G.
The expression for the gravitational force between objects of mass M and m at a distance b from each other is given by the equation: \[F = ma = \frac{GMm}{r^2}\] The two smaller masses cancel, and we are left with: \[a = \frac{GM}{b^2}\]
(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.
We know that the photon is traveling at a speed c, \(3 * 10^{10} \text{ cm s}^{-1}\). As the formula for velocity, or speed, is given by speed = \(\frac{\text{distance}}{\text{time}}\), the formula for the photon's speed is given by \[c = \frac{2b}{\Delta t}\] Rearranging the equation, we have \[\Delta t = \frac{2b}{c}\]
(c) Solve for the change in velocity, \(\Delta v\), in the direction perpendicular to the original photon path, over this time of interaction.
If the acceleration of an object is constant, we know that the change in velocity of that object is given by the equation \(\Delta v = a(\Delta t)\), where a is the object's acceleration. As we know the acceleration of the object from part (a) and the \(\Delta t\) from part (b), we can solve for the \(\Delta v\) by plugging in our previous answers: \[\Delta v = (\frac{GM}{b^2})(\frac{2b}{c})\] \[\Delta v = \frac{2GM}{bc}\]
(d) Now solve for the deflection angle (\(\alpha\)) in terms of G, \(M_L\), b, and c using your answers from part (a), (b), and (c). This result is a factor of 2 smaller than the correct, relativistic result.
If the think of the photon's original speed as a horizontal vector of length c, and the photon's final velocity as a result of the gravitational acceleration as a vector of length \(v_{\text{final}}\), we know by addition of vectors that the change in velocity, \(\Delta v\), must be the vertical vector as shown in the diagram below:
From this diagram, we know that \[\text{tan}(\alpha) = \frac{\Delta v}{c}\] As \(\alpha\) is extremely small, we approximate \[ \text{tan}(\alpha) = \frac{\Delta v}{c} \approx \alpha\] \[\alpha \approx \frac{(\frac{2M_LG}{bc})}{c}\] \[\alpha = \frac{2M_LG}{bc^2}\]
Excellent!
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