Sunday, September 27, 2015

Blog #10, Worksheet #4.1, Problem 1, Microlensing

1) Mass bends space-time! This is a prediction of general relativity, but fortunately we can heuristically derive the effect (up to a factor of 2) using Newtonian mechanics and some simplifying assumptions. Consider a photon of “mass” mγ 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).

(a) Give an expression for the gravitational acceleration in the vertical direction in terms of M, b and G.

To solve for the vertical acceleration we simply use our standard gravitational force equation \[F_g = \frac {GMm}{b^2} = ma\]Where b is the distance between the photon and our larger mass of the lens. We then simplify and solve for \(a\) giving us \[\frac{GM}{b^2} = a\]

(b) Consider the time of interaction, ∆t. Assume that most of the influence the photon feels occurs in a horizontal distance 2b. Express ∆t in terms of b and the speed of the photon.

We know the equation for distance with respect to velocity and time is \[\Delta d = \Delta t \cdot v\]Rearranging and solving for \(\Delta t\) we get \[\Delta t = \frac{\Delta d}{v}\] and we know that \(\Delta d = 2b\). So we end up with \[\Delta t = \frac{2b}{c}\] where \[c = v_{photon} = 3x10^8 \frac{m}{s}\]

(c) Solve for the change in velocity, ∆v, in the direction perpendicular to the original photon path, over this time of interaction.

To solve for \(\Delta v\) we simply have to combine the previous two parts using the equation \[\Delta v = a \Delta t\]Plugging in we get \[\Delta v = \frac{GM}{b^2} \cdot \frac{2b}{c}\] which we simplify as \[\Delta v = \frac{2GM}{bc}\]

(d) Now solve for the deflection angle (α) 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.

We just solved for the vertical velocity of the photon as \[\Delta v = \frac{2GM}{bc}\]and we know that a photon must have a horizontal velocity of simply the speed of light, c. We use these two velocities to create a triangle with the two shorter sides as the velocities and find our deflection angle. We find that \[tan (\alpha) = \frac{\frac{2GM}{bc}}{c}\]Simplifying and using the fact that for small angles \(tan(\alpha) \thickapprox \alpha\), we end up with \[\alpha = \frac{2GM_L}{bc^2}\] which is in fact a factor of 2 smaller than the general relativism answer of \[\alpha = \frac{4GM_L}{bc^2}\]

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