# Geodesics and stretched strings

Consider this real situation: Light from a far-background galaxy is observed to be lensed by a closer massive galaxy. Individual photons that graze the closer galaxy travel to the observer along various geodesic paths, as prescibed by GR.

Now consider a very hypothetical situation. Suppose a point on the background galaxy is connected to the observer by a weightless stretched string. Does this string lie along a geodesic?

Or, more simply: in principle, do laser pointers and builder's chalk lines define "straight lines" equally well? If so, why should they?

## Answers and Replies

George Jones
Staff Emeritus
Gold Member
Consider this real situation: Light from a far-background galaxy is observed to be lensed by a closer massive galaxy. Individual photons that graze the closer galaxy travel to the observer along various geodesic paths, as prescibed by GR.

Now consider a very hypothetical situation. Suppose a point on the background galaxy is connected to the observer by a weightless stretched string. Does this string lie along a geodesic?

Or, more simply: in principle, do laser pointers and builder's chalk lines define "straight lines" equally well? If so, why should they?
A (timelike or lightlike) geodesic is a 1-dimensional worldline in spacetime. A string traces out a 2-dimensional worldsheet in spacetime.

I am going to try and formulate two versions of your question. You might have something totally different in mind, and I don't really know the answer to either version.

Version 1: If a magic marker is used to put a small red dot on the string at an point on the string, is the worldline of the dot a geodesic in spacetime?

If there are (non-uniform?) stresses in the string, probably not.

On to the second version.

Define space as a 3-dimensional hypersurface in spacetime given by t = constant, where t is some appropriate timelike (i.e., $\partial / \partial t$ is a future-directed timelike vector field) coordinate. In other words, space is a hypersurface of "simultaneity".

The intersection of the string's worldsheet with space is a 1-dimensional curve in space. Restricting the spacetime metric to space naturally defines a positive-definite 3-metric on space.

Version 2: Is the string's 1-dimensional curve in space a geodesic in space with respect to this spatial metric?

A (timelike or lightlike) geodesic is a 1-dimensional worldline in spacetime. A string traces out a 2-dimensional worldsheet in spacetime.

I am going to try and formulate two versions of your question. .....
.... the second version.

Define space as a 3-dimensional hypersurface in spacetime given by t = constant, where t is some appropriate timelike (i.e., $\partial / \partial t$ is a future-directed timelike vector field) coordinate. In other words, space is a hypersurface of "simultaneity".

The intersection of the string's worldsheet with space is a 1-dimensional curve in space. Restricting the spacetime metric to space naturally defines a positive-definite 3-metric on space.

Version 2: Is the string's 1-dimensional curve in space a geodesic in space with respect to this spatial metric?
Thanks for recasting my naive question so clearly. Version 2 is the one I'd like answered..... I think. The answer may well be "yes" , but I really don't know, since the "1-dimensional curve" of the stretched string is shaped ultimately by interactions that are not gravitational (strong and electroweak).

Fredrik
Staff Emeritus
Gold Member
I think the answer is almost always no. Consider e.g. a planet in a circular orbit around its star. The orbit is a geodesic in space-time, but not in space.

I think the answer is almost always no. Consider e.g. a planet in a circular orbit around its star. The orbit is a geodesic in space-time, but not in space.
You are correct about a planet, which is a body freely falling along a spacetime geodesic. But the stretched string I'm asking about is not a body in free fall, so the answer you give is inapplicable. But thanks for the interest!

Perhaps one should think of a static situation, like the Schwarzchild geometry. Consider first this geometry, with no central mass, where the geodesics followed by light rays and occupied by stretched strings are what we call "straight lines" in space sections.

Now turn on the central mass, quasi-statically, as if you were slowly creating the sun. Consider a grazing light ray, say from a distant star, as in the 1918 expedition. This light becomes perceptibly deviated as you turn the sun on. Would a string stretched from the star to Eddington also become bent along a grazing geodesic? Or not?

Fredrik
Staff Emeritus
Gold Member
the stretched string I'm asking about is not a body in free fall, so the answer you give is inapplicable.
You're right. I didn't read enough of your post and George's reply to understand the question. How about this then? Imagine two very tall buildings some distance from each other. Tie one end of the string to the top of one of them and the other end to the top of the other. The string will have a curved shape. If we increase the string tension (i.e. make the string shorter), the shape will be less curved. How "stretched" do you want your string to be? Are we talking about the limit where string tension goes to infinity? In that case, it seems obvious that the shape of the string goes towards a straight line, i.e. a geodesic in space, but I suppose we'd have to do some calculations to know for sure.

Jorrie
Gold Member
... Now turn on the central mass, quasi-statically, as if you were slowly creating the sun. Consider a grazing light ray, say from a distant star, as in the 1918 expedition. This light becomes perceptibly deviated as you turn the sun on. Would a string stretched from the star to Eddington also become bent along a grazing geodesic? Or not?
I think even your unphysical 'massless, stretched string' will sag towards the Sun in the gravitational field, unless it consists out of massless particles moving along the 'string' at c, in which case it is the same as a grazing light beam!

Jorrie

You're right. I didn't read enough of your post and George's reply to understand the question. How about this then? Imagine two very tall buildings some distance from each other. Tie one end of the string to the top of one of them and the other end to the top of the other. The string will have a curved shape. If we increase the string tension (i.e. make the string shorter), the shape will be less curved. How "stretched" do you want your string to be? Are we talking about the limit where string tension goes to infinity? In that case, it seems obvious that the shape of the string goes towards a straight line, i.e. a geodesic in space, but I suppose we'd have to do some calculations to know for sure.
In a strong gravitational field a light beam shone from one building to the other will follow a curved path a bit like the trajectory of a cannon ball rising at first and then falling ( a parabola?). The string on the other hand starts with a sag and tends towards a straight line as the tension tends towards infinity, so it would seem the string with extreme tension would not approach the path of the light no matter how much tension is applied to it. In other words a string under tension does not follow a null geodesic which is the answer to the OP of this thread IMHO. I am assuming of course that the tension does not cause an antigravity effect that causes it to curve upward, but that seems unlikely.

In another thread https://www.physicsforums.com/showthread.php?t=238428 (post #14 onwards) there is currently an argument as to why light curves in a gravitational field. Is it because the light has inertial and gravitational mass due to its energy, or is it because gravity is simply a geometrical effect that causes object to follow paths determined completely by their velocity irrespective of whether they have any form of mass or not? The answer to the question in that thread reflects on the answer to the question in this thread.

Another approach is this. Light passing from a vacuum to a medium with a refractive index not equal to unity will follow a bent path that minimises the time for light to get from one point to another. On the other hand, a string under tension follows a path that minimises the distance from one point to another. This point of view suggests that path of string under tension and the path of a photon is not the same.

......seems obvious that the shape of the string goes towards a straight line, i.e. a geodesic in space, .
I was thinking of a massless string. I think the answer is this: a light ray grazing the sun traces a geodesic path though spacetime which is not a "straight line" in the space section, near the sun. A massless stretched string would also lie along this geodesic, which is as straight a line as you could hope to define in the curved part of the space section. But maybe I'm wrong?

I think even your unphysical 'massless, stretched string' will sag towards the Sun in the gravitational field, unless it consists out of massless particles moving along the 'string' at c, in which case it is the same as a grazing light beam!

Jorrie
It might melt, but won't sag! On reflection I think that in any case it would appear to lie along the grazing light beam, just like a stick appears bent when poked into water. Dankie, Jorrie.

......The answer to the question in that thread reflects on the answer to the question in this thread.......
I don't see how -- the mass of the string in this situation is a red herring mistakenly introduced by me, by not specifying an (imaginary) massless string. I wish I hadn't done this!

Jorrie
Gold Member
It might melt, but won't sag!
If the solar wind and other such phenomena are left out of the equation, how can any physical line not sag under the gravitational acceleration? Consider a Schwarzschild black hole with the string passing the hole at r=3M, where light can orbit. If there were something like a static massless particle from which your string could be made, those particles would have to be forced out of their spacetime geodesics (at closest approach) by a radial force component, which you do not have in a straight string - you need some sag for that!

Anyway, how would one measure the sag in such a situation?

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......how can any physical line not sag under the gravitational acceleration?
First, I was considering only an (imaginary) massless string, not a physical line. No mass, no gravitational acceleration! Second, if a physical string is strong enough, and stretched tightly enough, the sag needed to balance gravitational forces can, in principle at least, be made as small as you like. A massless string seems a reasonable model in a practical case, such as that of sun-grazing rays, where melting seems the greater hazard. I agree that in the case of a physical string grazing a black hole at r = 3M sags might be big, but you perhaps have stellar-massed black holes in mind? What about really massive black holes (10^ 6 to 9 solar masses)? Aren't the forces then smaller at r =3M?

It might melt, but won't sag! On reflection I think that in any case it would appear to lie along the grazing light beam, just like a stick appears bent when poked into water. Dankie, Jorrie.
The apparent bent path of a straight stick poked into water does not follow the bent path of the light rays from the stick.

Have a look at the attached diagram. The red line is the straight stick lying on the physical path ABC. The blue line is the apparent position of the stick lying on the path DBC. The average path followed by the light rays is on the path AEC.

Substitute a gravitational field for the water and the path of the photons is AEC and the path of the massless string unaffected by gravity (or water) is along ABC and the apparent path of the string is along DBC. So the answer to your OP is no.

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Here is another diagram. It shows the geodesics of a low velocity particle, a light ray, a low tension string with mass and a high tension string with mass compared to the path of a massless string. In a gravitational field none of the paths lie on the path of the massless string. Switch off the gravitational field and all the paths follow the path of the massless string.

I think I was pretty much "on the money" when I stated:

1) A particle with motion follows a path that minimises the time between two points.

2) An object with tension follows a path that minimises the distance between two points.

A particle with motion follows a path that tends towards a straight line as the velocity tends towards to infinity but it is limited by the speed of light, so it can never follow a straight horizontal path in a gravitational field.

An object with tension follows a path that tends towards a straight line as the tension tends towards infinity but presumably no real material can have infinite strength so this path is never truely a straight horizontal line in a gravitational field.

Only an object with no motion and no mass (a hypothetical massless string) can follow a true horizontal straight line in a gravitational field.

In the absense of readily available massless string in the hardware stores, if you want to build a perfectly straight garden wall you will have to take the average of the builders chalk line and laser leveling device.

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Jorrie
Gold Member
First, I was considering only an (imaginary) massless string, not a physical line. No mass, no gravitational acceleration!
I'm not convinced. It is surely true in Newtonian mechanics, but I can't agree for relativistic mechanics. Spacetime geodesics (even null-geodesics) don't care about the proper mass of entities. They all 'fall' in the same way, AFAIK.

I agree that in the case of a physical string grazing a black hole at r = 3M sags might be big, but you perhaps have stellar-massed black holes in mind? What about really massive black holes (10^ 6 to 9 solar masses)? Aren't the forces then smaller at r =3M?
Yes, the radial gravitational acceleration at r=3M decreases linearly with M, but only approaches zero as M => infinity. In such a case the acceleration will approach zero, because the event horizon circumference will approach infinite length. In all other cases, the radial acceleration will be non-zero.

I'm not sure this leads anywhere, because static massless 'strings' (as per everyday meaning) cannot exist anyway. Practical strings will surely sag, not so?

Jorrie

The apparent bent path of a straight stick poked into water does not follow the bent path of the light rays from the stick.

Have a look at the attached diagram. The red line is the straight stick lying on the physical path ABC. The blue line is the apparent position of the stick lying on the path DBC. The average path followed by the light rays is on the path AEC.

Substitute a gravitational field for the water and the path of the photons is AEC and the path of the massless string unaffected by gravity (or water) is along ABC and the apparent path of the string is along DBC. So the answer to your OP is no.
Yes, you're quite right. Nice diagram -- thanks for all the trouble you've taken. And I agree with your "no", as being the answer to both my posts, 1 and 5. I was too hasty in drawing conclusions from the similarities between the situation I was considering in say post #5 to that of a stick in water.

Your second post seems to address some points raised by Jorrie rather than the questions of my posts 1 and 5, so I'll give it a bit more thought before replying more fully.

Yes, you're quite right. Nice diagram -- thanks for all the trouble you've taken. And I agree with your "no", as being the answer to both my posts, 1 and 5. I was too hasty in drawing conclusions from the similarities between the situation I was considering in say post #5 to that of a stick in water.

Your second post seems to address some points raised by Jorrie rather than the questions of my posts 1 and 5, so I'll give it a bit more thought before replying more fully.
Thanks ,I was just going to post a link to a googled image but believe it or not I couldn't find one which is surprising for such a basic often quoted optical illusion!

A new question also occured to me. How do we know that the apparent bending of a rod poked into water is an optical illusion and that the rod does not "really" bend and repair itself when withdrawn. The answer is that we could measure the stress on the rod and note that the sides of the rod do not experience tension or compression on its sides. This may give an method to define a straight line in a gravitational field. Start with a straight rod far away from a gravitational field and when the rod is lowered into a gravitational field support it in such away that it does not experience any stresses. The supported stress free rod would aproximate the hypothetical massless string that you suggested.

Practical strings will surely sag, not so?

Jorrie
Yes, I agree, but the questions I asked in posts #1 and #5 weren't about the sagging of strings at all. Kev has given an interesting and (I now think after making mistakes) correct answer. This is that starlight grazing the sun follows path that is different from string imagined to be somehow stretched between the star and an observer here on earth.

This seems true whether you imagine the string to have mass and sag, or be massless and not sag. It's the difference that is significant, not how we choose to imagine what the string is made of, how tightly it's stretched or how to tie it to the star! I'll discuss this difference after some more thought, in another post.

Thanks for the clarification of how gravitational acceleration ar r = 3M varies with M. I suppose tidal forces vary faster than linearly?

Jorrie
Gold Member
Kev has given an interesting and (I now think after making mistakes) correct answer.
Yes, I also now think Kev has it spot-on! I originally had problems with the 'sag' of a hypothetical massless string, but I think Kev's definition of the 'thing' in his later post clarified that.

Thanks for the clarification of how gravitational acceleration ar r = 3M varies with M. I suppose tidal forces vary faster than linearly?
Yep, where gravitational acceleration at r=3M is inversely proportional to the mass of the hole, tidal gravity is approximately proportional to the inverse square of the mass. It is roughly proportional to the inverse third power of r, which in itself is directly proportional to mass; hence the inverse square law.

Jorrie
Gold Member
This may give an method to define a straight line in a gravitational field. Start with a straight rod far away from a gravitational field and when the rod is lowered into a gravitational field support it in such away that it does not experience any stresses. The supported stress free rod would aproximate the hypothetical massless string that you suggested.
Interesting definition, Kev! Almost like the Born-rigid (lengthwise) accelerated rod that is kept stress free. Shall we call your's the Kev-rigid rod?

It is probably possible to calculate the local propulsion force required at any point on your rod by noting that the local gravitational acceleration of a free, momentarily static particle is given in geometric units by:

$$\ddot{r} = \frac{-M}{r^2\sqrt{1-2M/r}}$$

-J

Jorrie
Gold Member
Kev-rigid Rod

Removed duplicate post by deleting content.

.....which is surprising for such a basic often quoted optical illusion!
I'm glad you mentioned optical illusions, because this provides an opportunity to take us into deeper water, as it were.

The bent-stick-in-water is one of a myriad illusions that the refraction and reflection of light can confuse us with. In this case we are able to distinguish between "reality" (whatever that is -- leave it to the philosophers) and what we are deluded into observing by using the machinery of optical physics, ranging from Snell's law to the dispersion of wave packets in optical media. The diagram you drew is an application of this method that I should have remembered. And, as you have carefully explained, there is no doubt in our minds that the stick does not "really" bend and unbend as you poke it in and out of the water.

When we turn to describe gravity with GR it is easy to become confused by a similar distinction between "what is really there" and what we observe. Take starlight deviated when it grazes the sun as an example. The star does not "really" move relative to other stars when its starlight happens to graze the sun. For us, its apparent shift of position is indeed an optical illusion, which could in principle be revealed with measuring tapes (a.k.a. calibrated strings) stretched between stars. Readings of these tapes would not change when the star's light happened to graze the sun.

It then seems fair to ask: is the "curvature of space" near the sun is "real" or just an artefact of the mathematical machinery we use to describe gravity, namely GR? It seems to me likely that the curvature of space sections, or of spacetime, which folk often discuss as if curvature in an abstract entity like spacetime were somehow a physical reality, is nothing more than a mathematical device we use to describe gravity and its effects. And I doubt if it has any bearing on the geometry and action of the other non-gravitational interactions we know of, viz: strong and electroweak, which
keep stretched strings straight.

If this is so, then it is no wonder that Physics has so far failed to describe these interactions and gravity with a common dialect!

This may give an method to define a straight line in a gravitational field. Start with a straight rod far away from a gravitational field and when the rod is lowered into a gravitational field support it in such away that it does not experience any stresses.......
The snag here is that gravity is a body force (in non-GR terminology) that acts on all the particles of matter in a rod. So even a uniform gravitational field, like the one you're in, creates internal stresses in your body, no matter how you sit in your chair. That's why you evolved bones! But floating in water helps!

Dale
Mentor
2020 Award
I am no GR expert and I have not been following this thread closely, but it seems to me that what you really want is a physical interpretation of spacelike geodesics. We have easy physical interpretation of timelike geodesics (worldlines of objects in free fall) and null geodesics (worldlines of massless particles), but what about spacelike geodesics? What is their physical interpretation?

In general, I would not expect there to be a spacelike geodesic that differed from a null geodesic only by its timelike coordinate any more than I would expect the same for satellites and photons. In other words, I would be very surprised to find that "laser pointers and builder's chalk lines define 'straight lines' equally well". Or at least, not the same straight lines.

...
The snag here is that gravity is a body force (in non-GR terminology) that acts on all the particles of matter in a rod. So even a uniform gravitational field, like the one you're in, creates internal stresses in your body, no matter how you sit in your chair. That's why you evolved bones! But floating in water helps!
Calling the rod "stress free" is not entirely accurate. The horizontal rod will always be slightly compressed or even crushed across it's width when suported. It would be slightly more accurate to say the rod has equal tension along the length of its top and lower sides. If it bends, one side is stretched while the other side is compressed.

It seems a horizontal straight line as observed by a stationary observer is one of the hardest things to measure or define in a gravitational field. (It is easier to define for a free falling observer in a uniform gravitational field because all light rays appear to travel in straight lines according to him) In Newtonian physics, an object with no forces acting on it, is said to remains stationary or contiue with constant motion in a straight line. Einstein pointed out that an object in freefall has no forces acting on it, and follows a geodesic according to its velocity and position in the gravitational field. This, in a way, redefines a straight line as the path of shortest time between two points rather than the path of shortest distance between two points that we normally think of.

For example if you want to get from A to B as quickly as possible and there is a swamp between A and B it might be quicker to take a road that goes around the perimeter of the marsh to get from A to B. This is basically how light acts when travelling through mediums of various refractive indexes. When examing the bent path that light takes when it goes from A in air to B in water, it turns out that there is no possible quicker path for light to take when it is recognised that the light slows down in the medium with greater refractive index (the water). In the distorted spacetime around a massive body, the imaginary medium near the body has a greater refractive index near the surface of the body, than the efffective refractive index of the spacetime further away from the body. A light ray going from point C to D on the surface of the body, initially climbs upward away from the body, where the effective refractive index is lower. Higher up the light can move faster before descending back down to D. This curved path followed by the light ray is the fastest possible route light can take between C and D or the path of least time rather than the path of least distance.

Visualising the spacetime around a gravitational body as an imaginary medium with refractive index increasing as you get closer to the body, makes it easy to visualise gravitational lensing by massive bodies. However, unlike meduims we normally think of, this imaginary spacetime medium has the unusual property that it bends or slows down all wavelengths of light equally and also that it slows down the motion and bends the path of particles with rest mass in the same way. The smoke and mirrors behind the magic of the imaginary spacetime medium is the distortion of the relationship of time and distance by a gravitational field. The higher the refractive index the slower the rate of time so everything moves slower, distances seem longer and the local speed of light appears constant everywhere to any local observer.

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