What happens when the photons are there but the star is not

We are bounded in a nutshell of Infinite space: Blog Post #10, Worksheet # 4.1, Problem #1: What happens when the photons are there but the star is not

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” \(m_\gamma\) 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).

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

(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.

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

(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.

(a) First off, let us give a brief review of the nature of light and gravity, as described by Albert Einstein in his theory of General Relativity. Relativity describes the nature of the Four-dimensional world we inhabit, these dimensions being length, width, height, and time, 3 space dimensions and a time dimensions. Working off this knowledge, Einstein developed a description of all space as truly being a sort of fabric, a space-time fabric, whose perception is changed based on individual reference frames (from where you observe the event(s) ). Adding in the nature of the speed of light in a vacuum, the one absolute constant in the universe, and the nature of gravity and how it warps space time, bending the fabric, we can understand how light is affected by gravity.

Essentially, gravity changes the curvature of space time, and light will always travel in a straight line in concord with the space-time fabric, so if space time is bent, light is bent as well. This is what happens in Microlensing and Gravitational lensing events, the gravity of large celestial objects has the power to bend the light that comes from a star behind it in our field of view and thus redirect more light in our direction and we thus see the far star more clearly.

Now then, understanding the role gravity has on light and space-time curvature, we should start this problem with the classical equations for accelerations and gravity. Coming from Isaac Newton’s Second Law, we have \[F=ma ,\] describing how a force exists when a mass has acceleration, and we also have: \[F_{Gravity} = \frac {GM_1 M_2}{R^2} ,\] where G is the gravitational constant, the M’s are the masses of the two interacting objects, and R is the distance between these objects.

Now taking the diagram for the Lens and the path of light from the original source, we can equate both equations for force to find a and substitute R for b, the distance between the photon path and the Lens.

\[ ma = \frac {GM_L m_{photon}}{b^2} ,\] the masses cancel out:

\[ a = \frac {GM_L}{b^2} ,\] and we now have an equation describing the acceleration in a vertical direction.

(b) The time of interaction is relatively simple to solve for. Already, the problem establishes that the distance the photons feel a gravitational interaction from the Lens is a distance 2b (see image), and we previously established that the speed of light is a constant (c) (whose actual value is approximately \( 2.998 \times 10^10 \frac {cm}{s}\). Now then, we also know the general equation for any speed, \[v = \frac{Distance}{Time} ,\] and now just plugging in the values of the speed of light and the distance it will be affected by gravity, we have: \[c = \frac{2b}{\delta t} .\] And solving for \(\delta t\) we now have our resulting equation: \[ \delta t = \frac{2b}{c} .\]

(c) Now that we have the two previous equations, we can begin to solve for the general change in velocity of the photons. The general equation for any change in velocity is \[\delta v = a (\delta t) ,\] and, as you might have inferred, we have just solved for both a and \(\delta t\) in the last two problems. Having the previous two problems’ results:  \[ \delta t = \frac{2b}{c} ,\] \[ a = \frac {GM_L}{b^2} ,\], these can now be plugged into the general change in velocity equation. \[\delta v = a (\delta t) ,\] \[\delta v = \frac {GM_L}{b^2} (\frac{2b}{c}) ,\] which yields the equation for change in velocity in the direction perpendicular to the original photon path: \[\delta v = \frac {2GM_L }{bc}.\]

(d) Furthermore, if we now take these last few equations, we can start to see the practical effect the lensing has, changing the angle of the path the light follows. This angle, seen in the image as \(\alpha\), hels make a triangle composed of the distance gravity affects the photon (2b), the new trajectory, and a third side which is actually the change in velocity \(\delta v\). But, you are correct to think there a piece missing, for distances and velocities do not have the same units and cannot be considered in a similarity, but we must also recognize that the distance 2b is being traveled at the speed of light c. These two values, c and \(\delta v\) can certainly be used together in an equation to find the angle. The change in the velocity has the characteristic equation we just saw in the previous problem, and c is a constant. Knowing a bit of trigonometry, we know c and \(\delta v\) are the adjacent and opposite sides (respectively) to \(\alpha\), which can now be turned into a trigonometric equation: \[ tan(\alpha) = \frac{\delta v}{c} .\] This equation would serve us well, but in the case of this problem, the angle is infinitesimally small, so the small angle theorem (the trigonometric function of a sufficiently small angle is that angle) simplifies the equation into:  \[ \alpha = \frac{\delta v}{c} , \] when speaking of such incredible distances. Plugging in the previous equations we had: \[ \alpha = \frac{\delta v}{c} , \] \[ \alpha = \frac{\frac {2GM_L }{bc}}{c} , \] and the total equation for the angle in Newtonian mechanics (which is off by a factor of two once actual relativity is used in the equation) is: \[ \alpha = \frac {2GM_L }{bc^2} .\]

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