Light deflection and geodesics

[VladZH]

It is known that light beam bends near massive body and the object sendind deflected the beam is seen in shifted position.

Now about spacetime curvature. As I undestand the things are like that:
http://i11.pixs.ru/storage/3/3/4/pic2png_7037348_21446334.png
The question is why are geodesics in left side not the same as in right? So I'd expect this situation
http://i10.pixs.ru/storage/4/5/6/pic3png_9235518_21446456.png
So that we can see objects before the body without shifting

 

[PeroK]

You mean that everything must move "horizontally" in flat spacetime?

 

[VladZH]

I mean that after passing curved region light has to return to flat region with the same straight geodesic as it was before the curved one. I depict the red horizontal line as one geodesic and it just bends in near the body and has to get straight again in flat region. Where am I wrong?

 

[Orodruin]

Where am I wrong?

A geodesic is a straight line, the line you have drawn is significantly curved in the flat region. In the flat region, the line is straight in both cases in the first two figures.

 

[PeroK]

I mean that after passing curved region light has to return to flat region with the same straight geodesic as it was before the curved one. I depict the red horizontal line as one geodesic and it just bends in near the body and has to get straight again in flat region. Where am I wrong?

Why in the same direction as before? Why can't it just continue in a straight line?

If a particle is moving in a straight line and you deflect it, it moves in a new straight line. It doesn't curve back to its original direction.

 

[pixel]

There can be many straight line geodesics in flat space.

 

[VladZH]

A geodesic is a straight line, the line you have drawn is significantly curved in the flat region. In the flat region, the line is straight in both cases in the first two figures.

I am sorry.There must be no curved

 

[VladZH]

Thanks, guys for your replies. So when there is no curvature the path between two points i.e. geodesic is straight line

If we place a massive body in center the geodesic will be curved

But actually we see that light appears in point C

And we see an object that is in A as if it was above point A.
So where is misconception?

 

[Orodruin]

Your curved line in the middle figure (and in the beginning of the lower one) does not follow how the curvature actually bends the light. Part of your misconception probably stems from that your "curved" region seems like a slab rather than a spherically symmetric region.

Edit: Also, it is not as easy as declaring a part of your space-time to be flat and another to be curved - there is a gradual change from small curvature to large curvature.

 

[A.T.]

So where is misconception?

You seem to misunderstand how geodesics on curved surfaces work. Have a look at section 2 in this link:

http://demoweb.physics.ucla.edu/content/10-curved-spacetime

Note that this spatial curvature explains only half of the light bending. The other half involves the time dimension. But the pictures under 2 give you a good idea how you can model geodesics, by approximating the curved surface with flat pieces.

 

[VladZH]

You seem to misunderstand how geodesics on curved surfaces work. Have a look at section 2 in this link:

http://demoweb.physics.ucla.edu/content/10-curved-spacetime

So from the second picture in your link it appears that if there is curvature instead of light moving to green line:

it moves more under the green line:

So it seems that there is a break of geodesic
Or another situation when right flat region moves down and there is no break with green line

But is that movement possible?
Sorry guys for inaccurate pictures

 

[jbriggs444]

So from the second picture in your link it appears that if there is curvature instead of light moving to green line:
it moves more under the green line:

No. It is not that the light is offset downwards. It is curved downwards. It does not retain its original direction.

To be more precise, it does always retain its direction at every infinitesimal step along its path. However, the curvature of space-time means that the incrementally straight path seems to curve when judged against the asymptotically flat space-time far from the region of curvature.

 

[stevendaryl]

I mean that after passing curved region light has to return to flat region with the same straight geodesic as it was before the curved one.

But that's not true. It won't return to the "same" geodesic. Here's an exaggerated picture of the bending of light around the sun:

Passing near the sun causes the light ray to change directions.

 

[VladZH]

But that's not true. It won't return to the "same" geodesic. Here's an exaggerated picture of the bending of light around the sun:

Passing near the sun causes the light ray to change directions.

And the direction change is caused by spatial or time curvature?

 

[A.T.]

Sorry guys for inaccurate pictures

Your pictures have no intrinsic curvature, because they are flat. Try to find a real surface with a bump (or dent). Then take adhesive tape and stick it along one side of the bump, without stretching or folding the tape edges (keep it locally straight like a geodesic is).

 

[A.T.]

And the direction change is caused by spatial or time curvature?

Both, but you cannot show that many dimensions in one picture.

 

[PeroK]

And the direction change is caused by spatial or time curvature?

Apart from the fact that it's the path of light being bent, this is no different from classical gravitation. The reasons for the gravity are different, but the paths followed by particles are essentially the same. Geodesics in flat spacetime are straight lines: all straight lines. Forget about A' for the moment, the path A-B is just a curved path about a gravitational body that straightens out as the gravitational attraction reduces. If that were an asteroid following the path A-B, there is nothing that is going to change its direction at B back to the original direction. Would you really expect an asteroid to do a 90° turn at B to get itself back on its original course?

I think you're getting yourself all confused about something that is just elementary geometry.

 

[VladZH]

Your pictures have no intrinsic curvature, because they are flat. Try to find a real surface with a bump (or dent). Then take adhesive tape and stick it along one side of the bump, without stretching or folding the tape edges (keep it locally straight like a geodesic is).

Do I misunderstand something? Intrinsic curvature needs no higher dimension so curvature of 2d space can be shown on 2d space

Apart from the fact that it's the path of light being bent, this is no different from classical gravitation. The reasons for the gravity are different, but the paths followed by particles are essentially the same. Geodesics in flat spacetime are straight lines: all straight lines. Forget about A' for the moment, the path A-B is just a curved path about a gravitational body that straightens out as the gravitational attraction reduces. If that were an asteroid following the path A-B, there is nothing that is going to change its direction at B back to the original direction. Would you really expect an asteroid to do a 90° turn at B to get itself back on its original course?

Well, as far as I see if you are getting farther from B to the left the shifted picture A' will be getting closer to A?

 

[PeroK]

Well, as far as I see if you are getting farther from B to the left the shifted picture A' will be getting closer to A?

I have no idea what that means.

 

[stevendaryl]

And the direction change is caused by spatial or time curvature?

The meaning of "curvature" is exactly that the notion the direction of vectors change as you move around. In the picture below, I've drawn two different paths that take you from point B to point A. Starting at point B, we pick a direction, as shown by the little arrow near B. If you follow the bold path, and keep the little arrow pointing what you think of as "the same direction" all along the path, you end up with arrows pointing in two different directions, depending on what path you took. That's what curvature means, mathematically: that the notion of two arrows pointing in the "same direction" is path dependent.

 

[jbriggs444]

Do I misunderstand something? Intrinsic curvature needs no higher dimension so curvature of 2d space can be shown on 2d space

Look at a map of the Earth. Yes, a curved 2-d space (the surface of the earth) can be displayed on a 2-d space (the map in front of you). On this map, straight lines (great circle paths) will appear to be curved.

If you want to consider trajectories, you now have two dimensions of space and one of time. Trying to present that, complete with curvature, on a static two dimensional flat map is more challenging.

 

[A.T.]

Intrinsic curvature needs no higher dimension so curvature of 2d space can be shown on 2d space

Yes, the adhesive tape stuck geodesically onto a intrinsically curved surface never takes off from the 2d surface into the embedding 3d space. And if you connect three points on that surface with three geodesic tape segments you get a triangle with a angle sum different from 180°, indicating intrinsic currvature .

 

[Janus]

It is known that light beam bends near massive body and the object sendind deflected the beam is seen in shifted position.

Now about spacetime curvature. As I undestand the things are like that:
http://i11.pixs.ru/storage/3/3/4/pic2png_7037348_21446334.png
The question is why are geodesics in left side not the same as in right? So I'd expect this situation
http://i10.pixs.ru/storage/4/5/6/pic3png_9235518_21446456.png
So that we can see objects before the body without shifting

You have light coming in from the left leaving flat space-time and entering curved space-time as shown in the image below. There are an infinite number of geodesics that intersect any given point, and this is true for the point where the light enters curved space-time. Some possible ones are shown as the red curved lines. But only one of those geodesics "lines up" properly with the incoming light ray and is the only allowed path for the light upon entering light ray.
Upon leaving curve space-time, there are also an infinite number of straight line geodesics in flat space-time that intersect that point. Some are shown by the red straight lines. But again, only one of them (the white line) lines up properly with the curved space-time geodesic the light is now following, and it is the only allowed path the light can take upon leaving curved space-time. This new flat space-time geodesic will be a different direction to the original straight line path.

 

[VladZH]

View attachment 98818

And another question. Will there be light deflection if there is no time curvature?

 

[Orodruin]

And another question. Will there be light deflection if there is no time curvature?

This statement makes no sense because what is "space" and what is "time" is frame dependent. You would first have to define what you consider to be "space" and what you consider to be "time" and then define what you mean by curvature belonging to "space" or "time". There is space-time curvature.

 

[VladZH]

This statement makes no sense because what is "space" and what is "time" is frame dependent. You would first have to define what you consider to be "space" and what you consider to be "time" and then define what you mean by curvature belonging to "space" or "time". There is space-time curvature.

By absence of time curvature I imply the invariance of dt for all observers. By space curved regions I mean ones where the smallest path between A and B is not straight line. Here I wonder how the light would behave only in curved space and Newtonian time or there is no even hypothetical answer in physics?

 

[A.T.]

And another question. Will there be light deflection if there is no time curvature?

A puerly spatial geodesic would be deflected too by the spatial part of the Schwarzshild geometry.

 

[Orodruin]

By absence of time curvature I imply the invariance of dt for all observers.

dt is not invariant for all observers, not even in special relativity (where it transforms as $dt' = \gamma(dt - v\, dx)$ under the Lorentz transformation in standard configuration).

A puerly spatial geodesic would be deflected too by the spatial part of the Schwarzshild geometry.

Note that this is a more precise statement than the one made by VladZH and restricts the situation to the Schwarzschild geometry with global time $t$ as a singled out coordinate.

 

[VladZH]

dt is not invariant for all observers, not even in special relativity (where it transforms as $dt' = \gamma(dt - v\, dx)$ under the Lorentz transformation in standard configuration).

OK, I mean absolute (Newtonian) time. So suppose we have such spacetime (only with space curvature) can we describe light deflection with GR equations or something?

 

[stevendaryl]

By absence of time curvature I imply the invariance of dt for all observers.

Technically, there is no such thing as "time curvature". The way curvature is defined is in terms of parallel transport.

• You have two different events (points in spacetime): $e_1$ and $e_2$.
• You have two different paths $\mathcal{P_1}$ and $\mathcal{P_2}$ connecting those events (a path being a curve through spacetime).
• You have a vector (direction in spacetime) $V^\mu$ defined at $e_1$.
• You move along path $\mathcal{P_1}$ from $e_1$ to $e_2$, and "parallel transport" $V^\mu$ along the path to get a vector $V^\mu_1$ defined at point $e_2$
• You move along path $\mathcal{P_2}$ from $e_1$ to $e_2$, and "parallel transport" $V^\mu$ along the path to get a vector $V^\mu_2$ defined at point $e_2$
• If $V^\mu_1$ is different from $V^\mu_2$, then spacetime is curved.

Note that the paths $\mathcal{P_1}$ and $\mathcal{P_2}$ enclose a 2-dimensional surface in spacetime. So you can't really talk about curvature for a single coordinate, such as $t$. Curvature necessarily must involve at least $2$ coordinates.

 

[VladZH]

Technically, there is no such thing as "time curvature". The way curvature is defined is in terms of parallel transport.

• You have two different events (points in spacetime): $e_1$ and $e_2$.
• You have two different paths $\mathcal{P_1}$ and $\mathcal{P_2}$ connecting those events (a path being a curve through spacetime).
• You have a vector (direction in spacetime) $V^\mu$ defined at $e_1$.
• You move along path $\mathcal{P_1}$ from $e_1$ to $e_2$, and "parallel transport" $V^\mu$ along the path to get a vector $V^\mu_1$ defined at point $e_2$
• You move along path $\mathcal{P_2}$ from $e_1$ to $e_2$, and "parallel transport" $V^\mu$ along the path to get a vector $V^\mu_2$ defined at point $e_2$
• If $V^\mu_1$ is different from $V^\mu_2$, then spacetime is curved.

Note that the paths $\mathcal{P_1}$ and $\mathcal{P_2}$ enclose a 2-dimensional surface in spacetime. So you can't really talk about curvature for a single coordinate, such as $t$. Curvature necessarily must involve at least $2$ coordinates.

OK, let's suppose there is only space, without t. That will be the same way to define curvature?

 

[stevendaryl]

OK, let's suppose there is only space, without t. That will be the same way to define curvature?

Yes, you can talk about purely spatial curvature. It's the same definition: use parallel transport to define curvature.

A particularly simple example is the surface of the Earth, which is a curved 2-D surface. Imagine standing on the equator, at the point of $0^o$ longitude. Take a spear (representing your vector) and point it parallel to the ground pointing straight north. Now, walk straight north until you get to the North Pole, trying not to twist your spear. Now at the North Pole, your spear is pointing south, along the line $180^o$ longitude. Now, go back to where you started, at the equator, at $0^o$ longitude. Instead of going straight north, you go east to the point $90^o$ east longitude, keeping your spear pointing in the same direction (north). Now you go straight north until you reach the north pole. Your spear will now be pointing south, along the line of $90^o$ west longitude. So even though you tried to keep your spear pointing in the same direction at all times, the direction it is pointing when you get to the North Pole depends on the path you took. That's what spatial curvature means.

In the case of the bending of starlight by the sun, I haven't done the calculation, so I'm not sure how much of the effect is due to pure spatial curvature.

 

[A.T.]

I'm not sure how much of the effect is due to pure spatial curvature.

I think it's half of the effect for light:
http://mathpages.com/rr/s8-09/8-09.htm

 

[PeterDonis]

I mean absolute (Newtonian) time.

There is no such thing.

can we describe light deflection with GR equations or something?

Certainly. Most GR textbooks treat this problem. A decent online treatment is here:

http://lacosmo.com/DeflectionOfLight/index.html