• Prof. Robert Stewart

Bending of light in a Gravitational Field


Yes, we were being succinct in case the questioner wasn't a professional Physicist. All that you stated was correct. We'll add the bending of light in a gravitational field.

For the Newtonian or Special Relativistic calculation, it's easiest to treat the light as a particle, although a wave description would also work. The light is traveling at speed c. Say that the gravitational field g is at right angles to the velocity. (That's the part of g that gives curvature.) The radius of curvature (this is just simple geometry) is then c2/g. In other words, for each little distance dx traveled, the direction changes by an angle dθ = gdx/c2 (in radians).

To get the total curvature as the light goes by the sun (say with closest approach distance L from the center) we have to integrate dθ over the whole path, taking into account the changing distance from the sun and the changing angle between the acceleration and the velocity. So long as you make the (excellent) approximation that the total curvature is very small you can just add the curvatures on the nearly un-curved path. If I've done the integral correctly, that gives θ=2GM/Lc2 for the total curvature.

Now what happens in GR? I'm a little more comfortable with a wave picture here. That part of the curvature we just calculated basically corresponds to the change in kinetic energy when something falls in a field. Energy is just the same thing as frequency, quantum mechanically. So what that part consists of is just that the part of the wave closer to the sun is higher frequency, with the wavefronts spaced closer together. Try drawing some wavefronts spaced a little closer together on one side and farther apart on the other, remembering that the wave is propagating at right angles to the fronts. You'll see the wave curve.

Why did I go through that wave exercise? In GR, the gravitational distortion affects not only the time part of spacetime (the part we just treated) but also the space part. In a standard coordinate choice, there's extra space, more path length, for the parts of the wave nearer the sun. Projection of that curved space onto a flat plane makes waves which have equal spacing nearer and farther from the sun look like they are more tightly spaced near the sun, since the flat space doesn't have that extra stretch in there. So that's the source of the extra bending. This second bending is just as big for massive particles as for light, but for particles moving slowly compared to c, the first component of the bending is so much bigger that it's hard to see this second part.

Why are the two components of equal importance for light? In dividing the effect up into the two parts, we've implicitly assumed a conventional coordinate system in which the speed of light is isotropic and everywhere constant. [The previous answer I'd given here was messed up./mw] I can now do the calculation, and it works out. It looks like the time-like part of the effect comes mainly from the region where the light gets closest to the sun. The space-like part seems to come mostly from two regions, where the light is approaching and leaving that region of closest approach. The reason for that is that the peculiar lengths (in this choice of coordinates) are only along the radial direction, not along the tangential direction.


The above was in response to the gentleman below......


Bill C. RiemersPhysicist, Mathematician, Software Engineer, Science Fiction Fan Works at Red Hat2006-present Ph.D. Experimental High Energy Physics, Purdue UniversityGraduated 1994 Lives in Canada2001-present 2m answer views101.1k this month Top Writer2018 Bill C. Riemers, High Energy Experimental Physicist (Not Practicing) Answered Oct 20, 2018 · Author has 2.5k answers and 2m answer views Robert Stewart is pretty good. The only correction I would make is that all rest mass is invariant mass, but not all invariant mass is rest mass. The difference is context.Since you brought up the topic of light, we’ll use that explicitly. The energy of a photon is:E=hfE=hf E = hf This equation works regardless if the photon is in a vacuum or passing through a medium. When a light wave enters a medium such as glass it slows down. Our energy equation is based on the concept that E energy and momentum form a four dimensional vector. We could write this in vector notation:(Ec,ipx,ipy,ipz)(Ec,ipx,ipy,ipz) ( \dfrac{E}{c}, i p_x,i p_y, i p_z) i≡−1−−−√i≡−1 i\equiv \sqrt{-1} We use ii i for the spacial coordinates as relativity teaches us time and space have a hyperbolic relationship. cc c is needed to put everything in the same units. Four-momentum - WikipediaPythagoras’ Theorem tells us the length of this vector is:E0c=E2c2−p2−−−−−−−√E0c=E2c2−p2 \dfrac{E_0}{c} = \sqrt{\dfrac{E^2}{c^2}-p^2} E0=E2−p2c2−−−−−−−−√E0=E2−p2c2 E_0 = \sqrt{E^2-p^2 c^2} In relativity, velocity is a Lorentz transformation, which just means it is a rotation across the hyperbolic coordinates. The length of a vector does not change when you rotate. So we call the length of the vector Lorentz Invariant.The equation E=mc2E=mc2 E=mc^2 is an equivalence. It tells us mass and energy are the equivalent, except the unit conversion constant c2c2 c^2 , but it does not put any restrictions on what type of energy or what type of mass, just that they are the same type. So if we consider m0m0 m_0 rest mass, we can apply this our equation and say:m0c2=E2−p2c2−−−−−−−−√m0c2=E2−p2c2 m_0 c^2 = \sqrt{E^2-p^2 c^2} So far so good. Now most physicists agree the only mass value that is useful to publish is rest mass. As such, we can go ahead and drop the subscript of 0, because it is implied.Now frequently we want our equation in terms of velocity, not momentum. In which case we have to say:p=Evc2p=Evc2 p = \dfrac{E v}{c^2} Our Newton’s equation of p=mvp=mv p = mv won’t work, because we just defined mm m to be m0m0 m_0 . So now when we are talking about relativistic mass, we need to say Ec2Ec2 \dfrac{E}{c^2} instead. This is a mass, but it is more useful to call it energy, so we don’t need an extra variable. And since mass and energy are equivalent, this works just fine.So now I have my equation:m0c2=E2−E2v2c2−−−−−−−−−√m0c2=E2−E2v2c2 m_0 c^2 = \sqrt{E^2-E^2 \dfrac{v^2}{c^2}} m0c2=E1−v2c2−−−−−−√m0c2=E1−v2c2 m_0 c^2 = E \sqrt{1 - \dfrac{v^2}{c^2}} But wait we agreed to call m0m0 m_0 just mm m . So this becomes:m=Ec21−v2c2−−−−−−√m=Ec21−v2c2 m = \dfrac{E}{c^2}\sqrt{1 - \dfrac{v^2}{c^2}} m=Ec2γ−1m=Ec2γ−1 m = \dfrac{E}{c^2}\gamma^{-1} where γ−1≡1−v2c2−−−−−−√γ−1≡1−v2c2 \gamma^{-1} \equiv \sqrt{1-\dfrac{v^2}{c^2}} Now it is important to remember, this equation is not say our original equation was wrong.Our original equation E=mc2E=mc2 E = m c^2 is for EE E is any time type of energy, and mm m is the same type of mass.Our new equation, which just happens to use the same variable is for EE E is total energy, and mm m is invariant mass.For this revised equation this second equation we are assuming potential energy is 0. That is a far different, and much more specific than our original equation. It was derived from our original equation, but we changed the meaning of the variables along the way.Now back to the story of light, since that is what you asked about. The equation we started out with E=hfE=hf E = hf also used our EE E for our more specific equation. So we can directly plug that in:m=hfc2γ−1m=hfc2γ−1 m = \dfrac{hf}{c^2}\gamma^{-1} Now for light in a vacuum his we have:γ−1=1−c2c2−−−−−−√=0γ−1=1−c2c2=0 \gamma^{-1} = \sqrt{1 - \dfrac{c^2}{c^2}} = 0 So the invariant mass of a photon in a vacuum is:m=0m=0 m = 0 But what happens when the light enters glass with an index of refraction of 1.5 ? That means the speed of light in the glass is:v=c1.5v=c1.5 v = \dfrac{c}{1.5} γ−1=1–11.52−−−−−−√≈0.75γ−1=1–11.52≈0.75 \gamma^{-1} = \sqrt{1–\dfrac{1}{1.5^2}} \approx 0.75 m≈0.75hfc2m≈0.75hfc2 m \approx 0.75 \dfrac{hf}{c^2} Which means my photon passing through a medium has an invariant mass associated with it.Now if I start to going around telling people light has rest mass, I’ll be a laughing stock. And rightly so. This is an invariant mass. And it is emergent property of the system. I can calculate a total invariant mass for any system of particles, and again ignoring potential energy it is always calculated as:mc2=(∑E)2−(cp→)2−−−−−−−−−−−−−−−−√mc2=(∑E)2−(c∑p→)2 mc^2 = \sqrt{(\sum{E})^2-(c\sum{\overrightarrow{p}})^2} If I’m talking about a contained system, such as a gas cloud, it is reasonable to call this a rest mass. But lets say my system is star A on one side of the galaxy, and star B on the opposite, it is unreasonable to call that rest mass when trying to determine what happens to star C somewhere else in the galaxy. It is an invariant mass (the length of a vector), but not a rest mass in that context.Invariant mass is conserved. A photon doesn’t just suddenly gain invariant mass, and then lose it. Even after the photon leaves the medium, than invariant mass still exists, somewhere, and eventually you would probably expect it to be somewhere outside the glass. But there is nothing where look for that invariant mass. No particle at all. This is just a description of an emergent property of the system, and fails to carry any physical meaning to one close enough to tell the two are objects are not in the same approximate location.However, while the photon is in the glass we can meaningfully treat the invariant mass associated with the photon as if it was a rest mass for the photon.So in summary, all rest mass, is an invariant mass. But not all invariant mass is rest mass. And indeed light has a relativistic mass. Light can also have an invariant mass associated with it. But light never has a rest mass. Rather any rest mass associated with a photon is an emergent property of a more complex system of more than just a photon.

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