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Gravity caused by curved space - am I right?

  1. Feb 8, 2004 #1
    gravity caused by curved spacetime - am I right?

    I went over this with someone here a while ago and I think they became exhausted with me. Anyway, I went away to think about it. Here is what I have so far.

    I was told that gravity is caused by warped space time (according to, is it GR or SR). I was told that when you move into a curve, you accelerate. This is what was really confusing me. I tried to picture something moving faster because it was going from a straigh path to a curved one but I just wasn't seeing it. How does moving along a curve translate to increasing speed? Then it dawned on me that one thing I could see was that if the curve was getting tighter and tighter, as 'twere, one thing that does increase and that is the rate at which the object in question is changing directions! So the picture I formed in my head is this: Imagine the earth by itself in space (to avoid distractions). Now pop something into existence, say, a person, some ways away from the Earth. Now this person, having just popped into existence, instead of just floating in space will start to move towards the earth, slowly at first and then faster and faster, making a straight path. Now, this straight line being traced is what we see, but the reason the person is accelerating is because this is not the whole picture. What we see as a straight line path is actually a person changing directions because they are actually travelling in a curve and the reason they are accelerating is because the curve they are caught on is getting more and more curved. In other words, the faster speed we see is the increased rate the person is changing direction on the ever tightening curve.

    Could someone please tell me if this is what the whole deal is about or have I descended into madness? Many many thanks.
    Last edited: Feb 8, 2004
  2. jcsd
  3. Feb 8, 2004 #2


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    Yes, it is true that in General Relativity, gravity is pictured as a "warping" of space. In classical, Newtonian, physics, all objects move in a straight line unless they are affected by a "force". General Relativity replaces the straight line with the "geodesic" of a curved surface. The "force" of gravity is replaced by the fact that geodesics are no longer straight lines.

    "I was told that when you move into a curve, you accelerate. This is what was really confusing me. I tried to picture something moving faster because it was going from a straigh path to a curved one but I just wasn't seeing it."

    In physics, "accelerate" does not mean "move faster". "Accelerate" means "change velocity". Slowing down (in common parlance "decelerate") is also a form of acceleration.

    In fact, since velocity is a "vector" quantity (having both speed and direction) any time you change direction, you are changing your velocity and so are "accelerating". If you move around a circle, with constant speed, since you are constantly changing direction, you are accelerating. Since going around a curve necessarily means changing direction, going around a curve necessarily means "accelerating" but not "going faster".
  4. Feb 8, 2004 #3
    Re: gravity caused by curved spacetime - am I right?

    You were told wrong. Gravity is not caused by curved spacetime. Curved spacetime is just another term for tidal force and tidal force is not the cause of gravity. In fact you can have a region of space where there is a gravitational field where there are no tidal forces in that region. That's what a uniform gravitational field is and that is why Einstein refered to it in his principle of equivalence, i.e. Eintein said
    You wrote
    When you're sitting in your chair in front of your computer then you're at rest in a gravitational field. Being at rest in a gravitational field is (locally) equivalent to accelerating with respect to an inertial frame of reference. Do not confuse the curved spatial path of a particle with spacetime curvature. You can have a spatially curved path in flat spacetime.

    Geodesics should not be thought of as straight lines. Think of them as lines of extremal lenght. Or if you wish think of it as a "straightest possible line." But a path, even a geodesic, on a curved surface surface is not a straight line. However if one resides within such a surface then geodesics would appear as straight lines in that sense.

    Keep in mind that geodesics are the "straightest possible lines" not in space but in space*time*
    Last edited: Feb 8, 2004
  5. Feb 8, 2004 #4

    Not quite, you must be very careful. Objects in free fall always follow geodesics. The fact that spacetime is curved doesn't mean that the line is no longer straight. It is, in fact, a straight line on a curved surface (the same way lines of latitude and longitude on a sphere are straight lines). They "look" curved, because we like to associate the term "straight" with flat space(time) only. But it is important to remember that they are straight.

    The fact that spacetime is curved and geodesics may not behave the way they do in flat spacetime is what manifests itself as a deflection from an otherwise straight path.
  6. Feb 8, 2004 #5


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    the same way lines of latitude and longitude on a sphere are straight lines

    Sartor resartus. Lines of latitude aren't geodesics on the sphere, except for one, the equator. Lines of longitude are geodesics ("great circles"); the use of the term "straight line" for them is, well,...specialized.
  7. Feb 8, 2004 #6
    I know. I was told that a curved path causes acceleration. So I though, ok, a change in velocity, it's either slowing down or speeding up. I don't see how moving in a curved path makes you speed up or slow down, so I don't see how following a curving path in spacetime causes you to accelerate towards the earth.

    Right, and when I read that I thought, ok, the direction is changing, but the speed isn't. Same speed, different direction which now turns out, according to my research, to be the definition of velocity. What this sounds like is that I can move at a uniform speed, change direction and announce, "look, I'm accelerating". Kinda confusing.

    So I'm back to, why do you speed up as you approach the earth?
  8. Feb 8, 2004 #7
    Re: Re: gravity caused by curved spacetime - am I right?

    Whoa. Tidal forces? First time I've heard that expression! I know that accelerating with respect to an inertial frame is equivalent to standing on the earth. I know that a geodesic is the shortest path between two points on a dome shaped surface. What I don't know is why I move faster and faster as I fall towards the Earth. Someone on another thread suggested, I thought, that it's because I'm actually tracing a curved path in spacetime and that because I'm residing in that curved space time it appears to be a straight line to me and since I'm on this spacetime geodesic, I'm accelerating, which I took to mean getting faster and faster as I approach landing only now I find out that the definition of acceleration is merely changing direction. So while I'm changing direction on this curve I'm not actually accelerating (as oppose to decelerating) and so I'm left with my question which is why is my speed getting faster and faster.

    Hmm, what I'm sort of seeing is that residing on this geodesic means I'm changing direction and I take it this is why I started moving towards the earth in the first place. But you say that this isn't so and mentioned something called "tidal force"...

    I'll leave it there.
  9. Feb 8, 2004 #8
    Re: Re: gravity caused by curved spacetime - am I right?

    ummm...what are you talking about tidal forces? Mass warps spacetime...this is gravity...not sure what you mean in your post...in fact i ahve no clue where tidal forces came from. PLease explain.
  10. Feb 9, 2004 #9
    Re: Re: Re: gravity caused by curved spacetime - am I right?

    Tidal forces are just another way to say "spacetime curvature" = they mean the exact same thing. Spacetime curvature refers to the geodesic deviation, i.e. two geodesics which start out parallel do not remain parallel. Physically this manifests itself as two particles in free-fall accelerating relative to each other. Thus two particles will cross paths even when they start out moving parallel to each other. That's why Kip Thorne wrote
    If there is mass at a point X then spacetime is curved at that point. However if there is a point Y which is different than X which is near X but for which there is no matter at Y may still have a gravitational field at Y.

    Here is a Newtonian example: If there is mass at point X then the Newtonian tidal force tensor does not vanish at X. But that doesn't mean that the gravitational field in nearby points must have a tidal gradient there. Here is such an example


    Here is a derivation of tidal accelerations in Newtonian language

    Here is a derivation of tidal accelerations in Einsteinian language
  11. Feb 9, 2004 #10
    Re: Re: Re: gravity caused by curved spacetime - am I right?

    Nobody else knows why either. Einstein did not explain why. He just described it differently and much more accurately. But general relativity does not exlain why things accelerate in free-fall in a gravitational field. But to see all this clearly keep in mind the difference between a gravitational field and spacetime curvature. Picture yourself in an accelerating frame of referance. You will be unable to distinguish it from a uniform gravitational field. Yet there is no spacetime curvature since spacetime curvature refers to the non-uniformities in a gravitational field.
  12. Feb 9, 2004 #11


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    Re: Re: gravity caused by curved spacetime - am I right?

    Actually, without Riemannian spacetime curvature one is left with the spacetime of special relativity where the affine connections can be globally transformed away. This leaves one with no gravitational forces anywere in the universe. On the other hand if one has this spacetime curvature the affine connections can not be globally transformed away and one always haves gravitational forces somewhere in the universe. This is sufficient to demonstrate a causal relation.

    No, the Riemann tensor depends on both first order derivatives of the affine connections and products of the connections without derivatives. Even if one has a nonzero Riemann tensor one can always choose a frame for which the derivative terms vanish even though the product terms do not.

    He was referring to the Newtonian field, not general relativity's field of spacetime curvature.
  13. Feb 9, 2004 #12
    Re: gravity caused by curved spacetime - am I right?

    There has been an unfortunate trend in GR where the deflection of particles was attempted to have been explained in geometrical terms. However GR has never been more geometrical than any other theory. The analogy with curved surfaces has led people to the erroneous belief that it was the actual cause of gravity rather than a different way to describe gravity. But GR is just as geometrical as EM. As Einstein said
    That is in a letter to Lincoln Barnett dated 1948.
  14. Feb 9, 2004 #13


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    Re: gravity caused by curved spacetime - am I right?

    If the world-line of a particle is curved, then acceleration exists for this particle.

    Consider a 1-D space. The worldline is drawn in 2-D space-time, one D for space, and the other D for time. The slope of the world-line is the speed. So, if the world-line is curved, then its slope changes, therefore the speed changes.

    For higher D space, even if the slope of spatial displacement wrt time displacement is constant, the world-line can have a change in the dx/dy slope, for instance. As you pointed out, this would be a change in direction, which also cannot happen in the absence of acceleration. Just don't forget about the space to time slope.

    In other words:
    The dx/dy, dy/dz, and dz/dx slopes are direction. If they change, then there is acceleration by virture of changing direction.
    The dl/dt slope is speed (dl2 = dx2 + dy2 + dz2). If it changes, then there is acceleration by virtue of changing speed.
    Last edited: Feb 9, 2004
  15. Feb 9, 2004 #14
    Re: Re: gravity caused by curved spacetime - am I right?

    If the worldline is a geodesic then the acceleration exists for the observer, not the parricle. Whether a worldline is curved or not depends on one's choice of coordinates. I.e. consider a free-particle in flat spacetime. In an inertial frame of referance the worldline in a spacetime diagram will be straight if the spatial coordinates are Cartesian. However switch to a coordinate system which corresponds to a frame of reference which is accelerating relative to the first. If the spatial coordinates are chosen to be Cartesian then the worldline will now be curved. This is equivalent to plotting a straight line in the xy-plane and plottint the same line in the r-theta (i.e. polar) coordinate system when r and theta are plotted on linear orthogonal axes.
  16. Feb 10, 2004 #15
    Re: Re: gravity caused by curved spacetime - am I right?

    Well, you responded to someone else's response using grammar and terms I couldn't follow in the least, so I'm replying again in hopes that you'll have another go, this time keeping in mind that I am merely a dilettant.

    I don't know how many times I've shouted this from the mountain tops, but I KNOW that if I accelerate out into space inside an elevator that it will feel the same way as standing on the earth and things I let go of will drop etc. etc.! I've never thought of as geodesics as "straight lines" except in so far as the appear straight to those residing in the curved space and in so far as they are the shortest path between two points on a domed surface.

    So here is what I've gotten from you so far. Spacetime curves are sometimes called tidal forces. Gravity, which I understand to be the pull exerted by a mass (what every particle does to every other particle in the universe), is not the result of moving through curved spacetime. Spacetime is merely a way to describe the phenomenon of being pulled towards a mass (don't quite know what that means). This pull that masses exert, which we call gravity, can exist without the presence of curved spacetime. Umm, do we happen to know of a planet or star or something out there which has gravity but does not curve spacetime?
  17. Feb 10, 2004 #16
    In GR there is no need for forces.
    In fact the basics of GR ar simpel and clear.
    The curved spacetime defines how particels move (read mass/energy) and mass/energy defines the curvature of spacetime.
  18. Feb 10, 2004 #17
    The gravitational force still exists in GR. What Einstein did was to place the gravitational force as being equivalent to inertial forces which thus raised inertial forces to the level of "real" forces.

    Particles moving through curved spacetime is not an explanation of gravity - it is merely a description of tidal effets in geometrical terms
  19. Feb 10, 2004 #18
    Re: Re: Re: gravity caused by curved spacetime - am I right?

    I'm sorry. I'll try again. Short and sweet definition of Tidal force - That which can cause a change in shape of an extended body from "normal".

    The normal shape of a drop of water in free-fall in the absense of gravitational tides is a sphere. Now place that drop of water in a region of spacetime where there are tidal gradients present and the drop of water will distort and no longer be spherical.

    Let me start with a statement of fact. It'll be a bit more mathematical for the mathematically inclined who are also reading this but I'll explain it all below without the math.

    Let Xu be the components of the seperation 4-vector, X, of two nearby particles, each of which is in free-fall in a gravitational field. X is also called the deviation 4-vector. The tail of the vector originates on one particle and the tip of the arrow ends on another nearby particle. The quantity

    [tex]\mathbf{A} = \frac {d\mathbf{X} }{d\tau}[/tex]

    is called the tidal acceleration 4-vector and has components Au. It is related to the spacetime curvature through the Riemann curvature tensor, R, also known as the tidal force 4-tensor (the relativistic version of the Newtonian tidal force 3-tensor) and the curvature tensor. The relationship is

    [tex]A^{\mu} = -R^{\mu}_{\beta\sigma\alpha}U^{\beta}X^{\sigma}U^{\alpha}[/tex]

    where U is the 4-velocity of a referance particle, i.e. it's the 4-velocity of the particle at the tail of the deviation 4-vector. The quantity on the left is called the tidal acceleration of two nearby particles. It is for this reason that the Wikipedia Encyclopedia states that (http://en.wikipedia.org/wiki/Schwarzschild_metric)
    The Newtonian equivalent is described here

    Tidal acceleration should not be confused with differences in coordinate accelerations. In General Relativity they are different quantities altogether.

    There is a nice discussion of tidal forces here
    http://www.fys.uio.no/~sigbjorh/gravity.pdf [Broken]

    On to a less mathy description ---------------------
    No. That was not what I meant exactly. Do not confuse the curved paths of particles in a gravitational field with spacetime curvature. If you're in that accelerating elevator and you see a beam of light zip across from one side of the elevator to the other and thus deflecting then that is a gravitational effect. In this case a uniformly accelerating frame of reference is equivalent to a uniform gravitational field. So the spatial part of the path is curved as you measure it in your elevator frame. However that is not an indication of spacetime curvature.

    Tidal force - The term Tidal force comes from ocean tides. I.e. it is that force which, if there were a planet with an ocean on it, would cause tides. In GR it refers to the relative acceleration of particles in free-fall as measured by a free-fall observer. It also refers to the tidal-force 4-vector.

    Tidal force in Newtonian gravity is a result of the gravitational force varying with position throughout space. In GR its a bit different since things which held in Newtonian gravity don't hold in GR. Thus in GR tidal accelerations are the relative accelerations of particles as measured by an obsever in free-fall. It refers to a 4-vector which has an absolute existance rather than a relative existance like the gravitational force. The gravitational force is an inertial force and not a 4-force and as such it is not represented byu a 4-vector. For example: The force on a particle with mass m due to a planet whose mass is M has a magnitude of

    [tex]F = \frac {GMm}{r^{2}}[/tex]

    If you're in free-fall then you will experience no gravitational force in a small region surrounding you. It will seem as if you're in an inertial frame of reference and that the gravitational field has been transformed away. That is called a locally inertial frame of reference.

    The difference in the magnitude of the gravitational force can be detected by such an observer in free-fall by the fact that objects which are near him will experience accelerations relative to him. By this I mean that if you're falling radially inward towards the center of the Earth (in vacuum) and you place a ball near you then that ball will accelerate relative to you. The further away from you that you place it the greater this relative acceleration. If you place it along the same line that you're falling along then it will accelerate away from you. If you place it off to the side of the line it will accelerate towards you. This is how you can tell if you're in an elevator accelerating in the absense of gravity or if you're in the gravitational field of a planet etc. Drop two balls and see if they accelerate towards each other. See diagram here


    Each particle is moving along a geodesic and they accelerate relative to each other. I.e. if their paths start out parallel then they will not remain parallel. If your geodesic were to remain parallel to the geodesic of a one of the balls which is near you then they would not accelerate relative to you. Now picture this in geometrical terms. If you and a buddy are standing on the Equator and each of you are facing North and you each start walking North then your paths started out parallel. But each of you will wind up at the North pole and therefore there is a relative acceleration between you and your buddy. This is the same kind of phenomena as you and the falling ball. Since the surface of the Earth is curved we call the tidal phenomena "curved spacetime". Notice that the curvature that is being spoken of is not really the curavture of your paths but the curvature of the surface.

    Now notice that this will happen with all planets and stars etc. Any spherical body in fact. And it happens due to the fact that particles have an acceleration which is independant of the mass of that particle. However if the field is uniform then all of what I said above no longer appplies.

    For an example of a gravitational field which is perfectly uniform in linearized GR see

    The field inside the cavity is uniform for all practical purposes.

    Hope that helps
    Last edited by a moderator: May 1, 2017
  20. Feb 10, 2004 #19


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    Re: Re: Re: Re: gravity caused by curved spacetime - am I right?

    No, Riemannian spacetime curvature means that the Riemann tensor is not zero. You can have the derivative terms in the expression for it vanish without the tensor vanishing and you can have all terms add to zero without the derivative terms by themselves being zero.

    It is quite a streatch to make that mean what you seem to think it means. Either way, no one is arguing that one can't have a nonzero field at a seperate location from the source either in the Newtonian acceleration field sense or in the modern relativistic Riemannian spacetime curvature field concept sense.

    Snipped Newtonian example as its irrelevent to relativity.

    These are parts of your site, not references, and your words are not Einstein's. Instead see-

    Actually it does explain why in terms of more fundamental premisis.

    The field for modern relativity is the spacetime curvature. Don't mix up the Newtonian gravitational acceleration field with modern relativity's spacetime curvature field.

    Actually you should say you won't "locally" be able to distinguish whether you are accelerating with respect to an inertial frame Vs standing stationary with respect to a source in the presence of its "Newtonian" gravitational field. It is the Newtonian field that is the acceleration field.

    Actually it refers to modern relativity's field itself as that IS Riemannian spacetime curvature. A nonzero Riemann tensor is irrelevent in this local limit anyway. Even if it wasn't zero it is the affine connection alone, not the Riemann tensor that shows up in the equation of motion describing relativistic local dynamics.

    This isn't unfortunate, nor merely attempted. This is how such dynamics is correctly explained.

    It isn't erroneous and Einstein didn't say what you seem to think. He was saying that other physics was also fundamentally geometrical. This isn't mere coincidence and maybe at that time he didn't see the full ramifications of his general relativity for the future of physics. But, this common geometry is the manefestation of an underlaying principle and sense general relativity gives us the correct geometry it is the skelleton of all modern classical mechanics. The power of general relativity and the spacetime geometry it describes is not merely in a description of gravitation, but in providing a templet for modeling and deriving all other physics for macroscopic scales.

    The coordinate acceleration you should say.

    No absolutely not. In general relativity the gravitational force is a force of affine connection which is the same thing as any other inertial force, a fictitious force. What it did was the other way around. It described gravity as a fictitious force. It does NOT describe inertial forces as "real" i.e. four-vector forces. For gravitation alone, the real force or four vector force acting on a test particle is zero. This reduces the relativistic equation of motion [tex]F^\lambda = \frac{DP^\lambda}{d\tau} = mA^\lambda[/tex] to four vector acceleration equaling zero which is the geodesic equation.

    Actually general relativity does explain gravity in terms of more fundamental premisis, unlike what your claim here.

    You are referring to the Newtonian gravitational field. In modern relativity the spacetime curvature is the field. The coordinate acceleration is not.

    No, that would be a useless definition as not all planets have oceans. Instead the total tidal effect is the the large scale result of variance in affine connections.

    The rest of your arguments are based on Newtonian equations and premisis which have no validity in relativistic scenarios and includes plugs for your site's sections which you try to pass of as references. Instead see-
    Last edited: Feb 10, 2004
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