# Following a light ray through a curved spacetime.

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1. Nov 20, 2013

### Lavabug

Here goes a conceptual question that has been bugging me:

Consider the famous eclipse experiment that shows the Sun's gravitational lensing effect, allowing a star that would otherwise be obscured by the Sun to be visible from Earth.

Say an observer wanted to travel to the star from Earth and pointed his spaceship directly towards the star in question (at a speed v<<c).

If the observer did not look at the rest of the background stars and just did everything necessary to keep the spaceship pointed towards the target star, is there any way he/she would realize that the path being taken is not straight but in fact curved?

I presume if you had manual control over the ship's direction, you would progressively have to reorient the ship as the star changed position as you performed a close encounter with the Sun's gravitational field.

If the observer didn't know any GR/know that space was curved would he reasonably conclude that any steering he did was to correct any deflection from the Sun's gravity, and that the path was apparently straight?

What if the observer and ship were practically massless (but somehow still moved at v<<c, not sure if this is a non-sequitur on classical grounds), would he/she then be able to tell that the path was curved? (again without looking at background star motions)

2. Nov 20, 2013

### WannabeNewton

Local experiments done by a single freely falling observer reduce to experiments in SR. If you want to detect curvature locally you need more than one observer (you need a swarm of observers).

EDIT: Let me emphasize again that measuring the curvature of space-time is not the same as measuring the curvature of a worldline. You are interchanging the two as if they were equivalent but they are not. The curvature of a worldline is the proper acceleration of the worldline and this an observer can measure locally using an accelerometer (a freely falling observer would of course measure a value of zero). Space-time curvature is a higher order quantity and requires a swarm of observers to measure (via the equation of geodesic deviation).

As a side note, the scattering angle for light rays passing by stars is given by in terms of the impact parameter for null geodesics so the measurement already relies on quantities evaluated at spatial infinity.

Last edited: Nov 20, 2013
3. Nov 20, 2013

### A.T.

I don't understand your question. Just because the ship points towards the star’s visual position, doesn't mean it moves in that direction. The star emits light rays in different directions.

What would "Sun's gravity" mean to him, if he doesn't know GR? Newtonian gravity? Are you asking about the differences between GR and Newton for navigation near large masses?

4. Nov 20, 2013

### Staff: Mentor

In GR curved paths in spacetime are characterized by proper acceleration, which is the kind of acceleration which is measured by an accelerometer. Whenever the spaceship fires its rockets the onboard accelerometers will show non zero acceleration and they will know that their path is in fact curved.

I don't know what such a person would conclude other than that Newtonian gravity seems to give the wrong deflection for the star's light.

Same as above.

5. Nov 20, 2013

### Naty1

Hey Lavabug, It's difficult for me to pick just what aspect of gravity and curvature you are wondering about.....

In general relativity, matter, energy and pressure cause spacetime to bend, stretch or compress...to 'warp' or 'curve' to use two common terms. Light rays and coasting spaceships follow these curves which we call geodesics through this bent, stretched or compressed spacetime. The warping of spacetime curves the paths of the light rays and the trajectory of the spaceship.

In addition, to these rather static curvature conditions, your example also includes a target star moving relative to your spaceship. Since light has a finite speed, as you approach the star you'd need to adjust the spaceship course to account for the smaller and smaller error in your visual location of the target, right?

[Generally, you can consider relativistic corrections for the curvature of light to be significant, roughly double the Newtonian effect, while Newtonian approximations for slow speed space ships is neglibible. I've read in these forums it's the latter approximation that is used to plan space exploration probes.]

Let's say, first, for simplicity, the target star location were fixed relative to your depature point; Do you think if you were coasting though space, the spaceship would get you to the star without any course correction? In other words, do you think your spaceship would follow the same geodesic as the light, or would you follow a different one yet arrive at the star, or neither?

6. Nov 20, 2013

### Lavabug

Huh? Correct me if I'm wrong: Newtonian gravity predicts no deflection of light at all, since it treats space as flat, and any optical path between two points in vacuum would just be the Euclidean distance.

AT: I think your dilemma can be resolved if you just assume the star to be a tiny point source, not an extended one (paraxial optics). By saying the observer knows nothing about GR I mean his default knowledge of Sun's gravity is that it's just an attractive force-field and nothing more, he/she does not know the Sun is introducing curvature.

WBN: So like Naty says, you would say an effective acceleration needs to take place along the path to the star? Would this allow the observer to say he/she did not travel in a straight line in space, or would he/she assume it was only a compensation for passing through the Sun's gravitational field? By "path" I always mean the path in R3, what any single inertial observer will measure, not the worldline in R4.

If the spaceship+observer are considered massless, I don't see why they can't go freely down the same geodesic as light without having to accelerate.

I guess my question in the most precise and realistic fashion is: can that -massive- observer make a unique distinction between accelerative corrections needed to compensate for gravitational deflection or for the curvature of spacetime? I hope I'm not making this more complicated than it needs to be.

Last edited: Nov 20, 2013
7. Nov 20, 2013

### pervect

Staff Emeritus
I haven't done a really detailed calculation, but it seems to me that if we have the following diagram

B-----------------Sun-----------------A

For convenience lets assume A and B are equally distant from the sun

If we draw the space-time geodesics (representing natural motion with no forces other than gravity) from A to B, it seems that the will be different depending on the velocity, the higher velocity trajectories passing closer to the sun, the lower velocity trajectories passing further away.

The paths are obviously different - the part of the argument to check is that the spacetime region is small enough that there is only one spacetime geodesic between two points. I believe this is correct, because the path is half an orbit, and I believe a whole orbit represents the first time we see two different space-time geodesics between the same two points. But if it's really important, it might be a good idea to double check this seemingly plausible argument.

This is true even without relativistic effects. So I would think that an observer travelling with the "wrong" velocity over the spatial curve (it's same curve in space, but different in time obviously thus it is not the same curve in space-time) would experience a proper acceleration as measured by an onboard accelerometer, because the force-free path is a different path.

There isn't really any such thing as an almost massless observer. A massive observer (travelling a timelike geodesic) could, by moving fast enough, travel in *almost* the same geodesic path as a massless observer (who must travel a null geodesic). Of course, it can't be truly identical - a null geodesic must always be different than a timelike one. But the curves should approach each other in the limit as v approaches c.

But you specified that v<<c, so this case of v~c is outside of the example you requested. Any massive observer, no matter how small their mass, should measure the same proper acceleration when travelling along the specific path you suggest. The acceleration readings will depend on the speed of the observer, not the mass.

Relativistic effects would be important for getting the details of the proper acceleration right, bu not for its presence or absence.

When I say relativistic effects I mean using Newton-Cartan theory as the non-relativistic test theory, to compare to the fully relativistic predictions of GR.

Last edited: Nov 20, 2013
8. Nov 20, 2013

### gabriel.dac

Well, light would get less curved than the rocked. That is because light travels much, much faster. The rocket would be in Sun's gravitational field for a longer time than the light. So the spaceship will have to be constantly adjusting it's direction to keep pointed towards the star. If it could travel as fast as light it could simply point itself to the star and go, no adjusts needed.

9. Nov 20, 2013

### nitsuj

the experiment you refer demonstrates a geodesic (null) path for a very specific reason, light travels at c / is massless.

A geodesic path is the shortest path through both space AND time. Because light travels at c it must follow this shortest path through spacetime. If you consider how c is a maximum speed yet the photon can still seemingly (coordinate) accelerate by curving around the sun, might make it a bit more clear how a photon is so unique with respect to motion in comparison to a spaceship.

In either case the scenarios you presented are not at all similar, WannabeNewton made the distinction clear.

Last edited: Nov 20, 2013
10. Nov 20, 2013

### WannabeNewton

Your question is still not clear to me. Just to clarify, when I speak of the curvature of a worldline this refers to the 4-acceleration of the worldline. If an observer is freely falling then gravitational scattering won't change the fact that he/she is freely falling. As such the 4-acceleration will always remain zero along the worldline of the observer. To wit, the path in space-time will always locally be a straight line.

You on the other hand are speaking of straight lines in "space" and acceleration in "space". Even locally, "space" has no meaning unless you specify a coordinate system so what coordinate system are you using? Fermi-Normal coordinates along the worldline of a static observer hovering at the point of closest approach of the freely falling observer about which gravitational scattering is measured? Or locally inertial coordinates along the worldline of the freely falling observer?

11. Nov 20, 2013

### yuiop

A lot depends on what you mean by straight and curved paths. Take a look at the attached sketch of two unfeasibly tall towers on the surface of the Earth. The curved red path is the path of projectile travelling between tower A and B. Since the projectile is in freefall between launch and landing, it experiences no proper acceleration and so this path is considered straight in GR. The orange curve is the path of a laser beam. The blue sky bridge is constructed far out in flat space and designed to be as straight as possible. Stress sensors are placed all over the sky bridge so that any deformations can be detected and then it is put in place between the two towers and supported in such a way that it is not deformed. The blue path might be considered straight in Newtonian physics (or geometrically straight in Euclidean geometry) but an object travelling along the blue path experiences proper acceleration due to gravity and so this path is considered curved in GR. (As a side note, a trolley at A on the blue bridge would accelerate towards the centre of the bridge before decelerating gradually to a stop at B, under the influence of gravity alone.) In other words, the definitions of straight and curved in Euclidean geometry and GR and almost exactly opposite to each other. Which definition are you using?

The rate that an object falls in Newtonian physics s independent of its mass. By this reasoning, a light particle should fall in Newtoniian physics. Einstein calculated that the curvature of a light path due to GR is exactly twice the curvature predicted by Newtonian physics. I might be wrong, but it is possible that your observer trying to keep a straight heading would need to expend twice as much effort than predicted by Newtonian predictions and thus detect the GR curvature.

They will only follow the same geodesic as light if they travel at the speed of light. The path is speed dependent.

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12. Nov 20, 2013

### WannabeNewton

The deflection angle for both light rays and massive particles is given by $\delta \phi = \frac{2GM}{bv_{\infty}^2}(1 + \frac{v_{\infty}^2}{c^2})$ where $b$ is the impact parameter and $v_{\infty}$ is the velocity as measured by an observer at infinity. You can see that for $v_{\infty} = c$ we get twice the Newtonian value of $\frac{2GM}{bv_{\infty}^2}$. For $v_{\infty} < c$ it is always larger than the Newtonian value but not exactly twice it. The derivation of $\delta \phi$ for massive particles is entirely analogous to the standard derivation of $\delta \phi$ via perturbation theory for light rays.

13. Nov 20, 2013

### Naty1

Yes, and for those interested in more derivational detail related to Wannabe's post, the last time I was in a thread of this sort, somebody posted the following link:

Deviation of Light {path}near the Sun in General Relativity
http://www.lacosmo.com/DeflectionOfLight/index.html

In a 20 second or so scan I couldn't tell if the link text actually gives the same result as Wannabe's....

14. Nov 20, 2013

### Bill_K

The "reasoning" is fallacious, and so the fact that it leads to the wrong answer should not be surprising. In Newtonian physics, light is an electromagnetic wave. Maxwell's Equations do not contain any reference to gravity, and a light wave is unaffected by it. Nevertheless people seem to enjoy the simple idea that General Relativity's result differs from Newtonian theory by a factor of two.

15. Nov 20, 2013

### Staff: Mentor

No, it isn't clearly correct, but it certainly isn't fallacious.

If you take Maxwell's equations in a Newtonian inertial frame then you might expect no deflection. On the other hand, since inertial and passive gravitational mass are equal in Newtonian gravity, you can geometrize Newtonian gravity and get some deflection. You can also treat it as a massless particle and get some deflection.

It is not clear, but it isn't fallacious either.

16. Nov 20, 2013

### Bill_K

Newtonian gravity does not come "geometrized." If you choose to do that then it's no longer standard Newtonian physics you're talking about, it's your own extension.

17. Nov 20, 2013

### nitsuj

Well, after he had second look and more significantly just prior to presenting the theory.

18. Nov 20, 2013

### PAllen

Well, you can say Newton-Cartan completely geometrizes Newton's theory. However, I have no idea of a unique way to answer any question about light in this theory. Since Galilean invariance is to NC as Lorentz invariance is to GR, you have big problem with Maxwell's equations. If you try to go the massless particle route, you certainly don't get properties associated with light - you would readily expect speeds much greater than c.

19. Nov 20, 2013

### Bill_K

The reason I said it's fallacious is that it violates the Correspondence Principle. The Correspondence Principle says that classical and quantum mechanics must agree in the limit of high quantum numbers. You can't have a theory in which photons fall but electromagnetic waves do not.

20. Nov 20, 2013

### A.T.

Newton himself saw light as particles:
http://en.wikipedia.org/wiki/Corpuscular_theory_of_light#Sir_Isaac_Newton
And the idea that gravity affects light is certainly older than General Relativity:
http://en.wikipedia.org/wiki/Black_hole#History

Einstein himself initially designed GR to match the Newtonian acceleration, via gravitational time dilation. Then he realized that, unlike in Newtons theory, space is curved, which doubles the deflection of light:
http://mathpages.com/rr/s8-09/8-09.htm