Following a light ray through a curved spacetime.

[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)

 

[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.

 

[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.

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

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?

 

[Dale]

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?

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 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?

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.

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)

Same as above.

 

[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?

 

[Lavabug]

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.

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.

Let's say, first, for simplicity, the target star location were fixed relative to your departure 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?

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.

 

[pervect]

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)

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 let's 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 traveling 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 traveling 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.

 

[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.

 

[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.

 

[WannabeNewton]

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.

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?

 

[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 traveling 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 traveling 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?

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.

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.

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.

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

 

[WannabeNewton]

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.

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.

 

[Naty1]

For v∞<c it is always larger than the Newtonian value but not exactly twice it.

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...

 

[Bill_K]

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.

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.

 

[Dale]

The "reasoning" is fallacious

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.

 

[Bill_K]

you can geometrize Newtonian gravity and get some deflection.

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.

 

[nitsuj]

Einstein calculated that the curvature of a light path due to GR is exactly twice the curvature predicted by Newtonian physics.

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

 

[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.

 

[Bill_K]

If you take Maxwell's equations in a Newtonian inertial frame then you might expect no deflection... You can also treat it as a massless particle and get some deflection.

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

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.

 

[A.T.]

In Newtonian physics, light is an electromagnetic wave.

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

Nevertheless people seem to enjoy the simple idea that General Relativity's result differs from Newtonian theory by a factor of two.

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

 

[Bill_K]

Newton himself saw light as particles:
http://en.wikipedia.org/wiki/Corpuscular_theory_of_light#Sir_Isaac_Newton

So what? Today we see it differently.

And the idea that gravity affects light is certainly older than General Relativity:
http://en.wikipedia.org/wiki/Black_hole#History

So what? GR was preceded by many ideas, some right, some wrong.

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

Again, so what. Today, one hundred years later, we (hopefully) have a better understanding.

 

[A.T.]

Today we see it differently.

You were talking about Newton, not about today.

Today, one hundred years later, we (hopefully) have a better understanding.

Yes, and we can compare the old and new understanding.

 

[pervect]

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.

That's a fair statement, I see I skipped over and/or misstated essential parts of my thought process on review.

So let's try again:

Start with a PPN formalism as a description of gravity on a "test theory" basis. (For the unfamiliar, http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism).

Start at the most simple level of approximation Set beta=gamma=0. This is called the "Newtonian order" of approximation.

At this level of approximation, one can think of the metric as having only a time component, ie. g_00 = -1 + 2U, and that the spatial derivatives of U are "gravitational forces".

Unfortunately, this level of approximation is not sufficient to explain the bending of light accurately. And if we want to discuss what happens when a spaceship follows the path of a light beam, we probably want to have an accurate description of how light deflects.

What PPN parameters do we need to predict the paths of light beams? Basically, traditional light defelction experiments are only affected by gamma, so we know we need to include that parameter. The betas, at least on first glance, don't seem to be significant. (Perhaps a more detailed analysis will show they are needed, I don't know for sure, I'm assuming we don't need to worry about them for now).

When we introduce gamma, we introduce "spatial curvature" into the metric, the spatial components of the metric g_ij are no longer diagonal, some of the off diagonal elements become non-zero.

Casting around for words to describe the mathematics, we stumble upon the much-maligned "bolwling ball on a sheet" picture of curved space.

As a "mixed metaphor" for those struggling to understand the predictions of GR, I am suggesting that they may continue to think of gravity at the Newtonian order as being due to the traditional Newtonian forces (rather than space-time curvature, though of course the later is really better!)

Then, on top of these traditional "forces", we have "curved space", the much maligned bowling ball image, that explains the effects of gamma. We can see some of the potential for confusion here - if an object if following a spatial geodesic (the spatial path of shortest possible distance) but space itself is curved, does it experience a "force" or not? And how do we calculate it's momentum? I won't attempt to answer such confusions except to say that if you really want to understand how it all fits together, it's a good idea and easier in the long run to abandon the notions of "gravitaitonal forces" altogether and express EVERYTHING in terms of "space time curvature".

When one does this, one explains all the effects by the same metaphor. The hybrid metaphor approach attempts to combine two different metaphors for how gravity works, making it a bit unsatisfying. But I think it still might be helpful to some readers.

A second point I'm trying to make is that we may not even have to get into the complexities of curved space to get an answer to the OP's question. Adding these complexities will give a more accurate answer, but I think even the crudest model will show that an object following the spatial path of a light beam (but slower!) will have a nonzero proper acceleration as measured by an onboard accelerometer.

Going back to definitions, for the mathemeticians among us I'll point out that what I mean by the "following the path of a light beam" is following the same spatial path (excluding time), and that to define this path, we need to slice space-time into space-time slices. I assume we have a static spacetime, then we can do this slice by the usual method of making the space-like slices orthogonal to the time-like Killing vectors.

 

[Dale]

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.

It doesn't violate the correspondence principle if you geometrize Newtonian gravity. It is not fallacious.

 

[Dale]

I have no idea of a unique way to answer any question about light in this theory.

I agree, I don't know of a unique way to answer the question, which makes the various possible answers unclear rather than fallacious.

 

[Dale]

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.

Just as using a Lagrangian to solve a pendulum is still under the purview of Newtonian physics, so Newton Cartan gravity is still Newtonian gravity. They are just different mathematical formalisms for calculating the same experimental results.

 

[Naty1]

Lavabug: Is your question answered?

Lavabug:

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.

[Unfortuntely, it IS complicated...simple answer from this non expert is, "NO". ]I am now on the same general track as pervect, here...

I won't attempt to answer such confusions {his examples} except to say that if you really want to understand how it all fits together, it's a good idea and easier in the long run to abandon the notions of "gravitaitonal forces" altogether and express EVERYTHING in terms of "space time curvature".

but even that leads to interpretational issues:So, Lavabug, if this is the kind of thing 'bugging' you, pun intended:

So then tell me how does an observer observe the effects of
A) spatial curvature
B) temporal curvature

I think you will find this past discussion interesting...if not entirely resolved...

from the original question in the thread by passionflower:

Apparently if someone goes fast (whatever that might mean in relativity) curvature is more in the time direction while if someone goes slow (again whatever that might mean) curvature is more in the spatial direction.

[The quote above may be be a better answer than my "NO"...you can decide...]

Check out my attempt at a summary of that discussion here, post #37, page 3...
or better, just read the whole discussion...good insights from experts...

Is Spacetime Curvature Observer and
Coordinate independent
https://www.physicsforums.com/showthread.php?t=596224&page=3including what I took to be the final insurmountable obstacle:

Miller,Thorner, Wheeler:

… nowhere has a precise definition of the term “gravitational field” been
given --- nor will one be given. Many different mathematical entities are
associated with gravitation; the metric, the Riemann curvature tensor, the
curvature scalar … Each of these plays an important role in gravitation
theory, and none is so much more central than the others that it deserves the name “gravitational field.”

 

[Lavabug]

I still don't know if the last incarnation of my question (above post) is answered, I am still absorbing much of what's gone on in the past few posts.

Might try posing the question to one of the GR experts at my faculty if I run into them.

Clearly my single course in GR has left me with a bunch of IMO hard conceptual gaps (I have a few more questions like these but nowhere near as thought out, verbally), I'll take the previous post as a reminder to yank MTW from the library over the holidays for some exploration.

 

[Naty1]

Apparently it took the greatest minds of the 1920's and 1930's to fully understand the implications of GR...so don't feel too bad...some two decades before GR was widely accepted...