I have a fundamental mis-understanding about gravity.

I have a layman understanding about how the curvature of space describes the motions of planets and other large celestial bodies. What I don't understand is how curved space makes say; an apple fall to the ground. Any help appreciated. Thanks.
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 Quote by Dr Ninja I have a layman understanding about how the curvature of space describes the motions of planets and other large celestial bodies. What I don't understand is how curved space makes say; an apple fall to the ground. Any help appreciated. Thanks.
The principle is the same. The apple is following the 'curvature' of spacetime that is caused by the Earth. All objects follow General Relativity, it's just that deviations from Newtonian gravity are only seen at the large celestial body scale.

 Quote by Dr Ninja I have a layman understanding about how the curvature of space describes the motions of planets and other large celestial bodies. What I don't understand is how curved space makes say; an apple fall to the ground. Any help appreciated. Thanks.
It's not only curved space, but curved space-time. The time dimension is crucial to understand how objects at rest in space are affected at all. This links might help:
http://www.physics.ucla.edu/demoweb/...spacetime.html
http://www.relativitet.se/spacetime1.html

Note that planets are also mainly affected by this "time curvature". It's only for very fast objects like photons, that the purely spatial curvature causes a significant fraction of the total path bending.

I have a fundamental mis-understanding about gravity.

 Quote by A.T. Note that planets are also mainly affected by this "time curvature". It's only for very fast objects like photons, that the purely spatial curvature causes a significant fraction of the total path bending.
Could you show the math to back this statement up?
Because I am not buying it.

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 Quote by A.T. Note that planets are also mainly affected by this "time curvature". It's only for very fast objects like photons, that the purely spatial curvature causes a significant fraction of the total path bending.
It's true. For example, you can see it from this metric
$$d\tau^2=g_{00}dt^2-\frac{1}{c^2}g_{ab}dx^adx^b,$$
where a,b,c are spatial indexes. The spatial contributions to the proper interval are all divided by c2 so g00 (time curvature?) dominates. The convention of setting c=1 hides this a bit, but doesn't change it.

 Quote by Mentz114 It's true. For example, you can see it from this metric $$d\tau^2=g_{00}dt^2-\frac{1}{c^2}g_{ab}dx^adx^b,$$ where a,b,c are spatial indexes. The spatial contributions to the proper interval are all divided by c2 so g00 (time curvature?) dominates. The convention of setting c=1 hides this a bit, but doesn't change it.
Why do you think that formula is relevant?

We are talking about why planets 'bend' inwards when they orbit a mass right?

In general you first have to define a map with a space-like and time-like component and then you have to take into consideration over which range this map is valid. Only then can you even begin to talk about how much curvature there is in the time-like and space-like parts. An observer's 'four legs' can point in any angle in the 'directions' of spacetime. Thus different observers can attribute different amounts of curvature in the spatial and temporal direction.

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 Quote by Passionflower Why do you think that formula is relevant? We are talking about why planets 'bend' inwards when they orbit a mass right? In general you first have to define a map with a space-like and time-like component and then you have to take into consideration over which range this map is valid. Only then can you even begin to talk about how much curvature there is in the time-like and space-like parts. An observer's 'four legs' can point in any angle in the 'directions' of spacetime. Thus different observers can attribute different amounts of curvature in the spatial and temporal direction.
Of course it's relevant. The curvature is made up of first and second derivatives of the metric, including the 1/c2 factors, so the contributions from the spatial curvature will be much less than that from the temporal components.

Since we're talking about orbits, I'll try to come up with something for the Hagihara frame that might convince you. The problem is that c is set to 1 in all my scripts ...
 Recognitions: Gold Member Consider a plot of x against ct, starting at rest, so the line starts by rising parallel to the t axis. If there is a Newtonian gravitational field g along the x-axis, the line starts to curve towards the source with radius of curvature c2/g, which is a very large radius for normal gravitational accelerations because c is so large, but again because c is so large this also results in the velocity changing at rate g. This is effectively the "curvature of space with respect to time", that corresponds to the Newtonian acceleration. The linear curvature of space has a similar radius but only affects moving objects, proportionally to v2/c2. (In General Relativity the term "curvature" is often used to denote the intrinsic curvature of space-time created by mass-energy, which roughly corresponds to the sort of curvature on the surface of a ball that results in angles adding up differently from the way they do in flat space. In this case, I'm using the word "curvature" to describe the ordinary curvature of lines).

 Quote by Mentz114 It's true. For example, you can see it from this metric $$d\tau^2=g_{00}dt^2-\frac{1}{c^2}g_{ab}dx^adx^b,$$ where a,b,c are spatial indexes. The spatial contributions to the proper interval are all divided by c2 so g00 (time curvature?) dominates. The convention of setting c=1 hides this a bit, but doesn't change it.
 Quote by Passionflower Why do you think that formula is relevant?
It is relevant because the word-line of a free-faller is maximizing the proper-time interval. And since for a planet those dx/c are relatively small compared to dt, the g00 dominates. The two extremes are:
- An object at rest in space (dx=0) is not affected at all by spatial curvature.
- For photons (dx=c*dt) the contribution of spatial and temporal components is the same.

For math on those two cases see:
http://www.mathpages.com/home/kmath409/kmath409.htm
http://www.mathpages.com/rr/s6-03/6-03.htm

The planets are closer to rest than c.

 Quote by Passionflower Thus different observers can attribute different amounts of curvature in the spatial and temporal direction.
I was considering the rest frame of the gravitation source.

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I think posts #8 and #9 wrap this up. But to answer this

 Quote by Passionflower An observer's 'four legs' can point in any angle in the 'directions' of spacetime. Thus different observers can attribute different amounts of curvature in the spatial and temporal direction.
For the stationary observer in the Schwarzschild vacuum the components of the 'fourleg' (cobasis) are
$$p_0= -c\sqrt{1-2m/r}\ dt,\ \ p_1=dr/\sqrt{1-2m/r},\ \ p_2= r\ d\theta,\ \ p_3=r\sin(\theta)\ d\phi$$
from which we can see cdt dominates for large r. For the orbiting frame the timelike cobasis vector is a mixture of the coordinate dt and dr directions
$$h_0 = -(1-2m/r)/(\sqrt{1-3m/r})\ dt-\sqrt{mr}/(c\sqrt{1-3m/r})\ dr$$
and again the spatial part is divided by c. The same pattern is shown by the new $\phi$ cobasis vector but I'm too tired to tex it in.

I think this shows that for these frames the new time direction takes only a small contribution from the coordinate spatial directions compared to the contributions from the coordinate time direction.
 Recognitions: Science Advisor Staff Emeritus Going back to the original poster here, i.e. Dr Ninja, has the question been answered sufficiently. Without getting into the technical details of how to specify curvature or some of the fine technical distinctions, the main point is that it's space-time that's curved, not just space. One can consider a worldline as being a path through space-time, as a curve parameterized by proper time, tau. You can think of an object as "moving through" spatial coordinates and time coordinates as proper time advances. Because space-time is curved, this "motion" through time (which is actually the change in coordinate time as proper time changes) causes relative accelerations relative to other worldlines of other objects, which is what we describe as gravity.

 Quote by Mentz114 I think posts #8 and #9 wrap this up. But to answer this For the stationary observer in the Schwarzschild vacuum the components of the 'fourleg' (cobasis) are $$p_0= -c\sqrt{1-2m/r}\ dt,\ \ p_1=dr/\sqrt{1-2m/r},\ \ p_2= r\ d\theta,\ \ p_3=r\sin(\theta)\ d\phi$$ from which we can see cdt dominates for large r. For the orbiting frame the timelike cobasis vector is a mixture of the coordinate dt and dr directions $$h_0 = -(1-2m/r)/(\sqrt{1-3m/r})\ dt-\sqrt{mr}/(c\sqrt{1-3m/r})\ dr$$ and again the spatial part is divided by c. The same pattern is shown by the new $\phi$ cobasis vector but I'm too tired to tex it in. I think this shows that for these frames the new time direction takes only a small contribution from the coordinate spatial directions compared to the contributions from the coordinate time direction.
A small contribution to the accumulation of proper time?
OK, I agree but so what?
What has that to do with spatial curvature?
What has the rate of the clock in a satellite to do witch spatial curvature?

Take one circumference of a satellite with a circular orbit of length r and compare this to a satellite going in a straight line. We can easily see how much the satellite was displaced by spatial curvature.

You are saying that it was displaced more by temporal curvature?
Show it!

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 Quote by Passionflower Still what is your point? A small contribution to the accumulation of proper time? OK, I agree but so what? What has that to do with spatial curvature? What has the rate of the clock in a satellite to do witch spatial curvature? Take one circumference of a satellite with a circular orbit of length r and compare this to a satellite going in a straight line. We can easily see how much the satellite was displaced by spatial curvature. You are saying that it was displaced more by temporal curvature? Show it!
The numbers I've shown have nothing to do with clock rates. They show the time cobasis vector of the orbiting observers frame. The time direction basis vector is only tilted by a small amount from the coordinate time direction. In other words the contribution of the spatial curvature is much smaller than the contribution of the time curvature. Read the equation.

The Newtonian orbit equations work very well, despite ignoring spatial curvature. Only tiny relativistic effects have been unexplained like the perihelion precession of Mercury.

 We can easily see how much the satellite was displaced by spatial curvature.

We are hijacking this thread. Start a new thread if you think it necessary.

 Quote by Mentz114 In other words the contribution of the spatial curvature is much smaller than the contribution of the time curvature. Read the equation.
So you agree with AT that the curvature of spacetime depends on how fast one travels through it (whatever how fast might even mean)? If you go slow (again whatever that might mean) spacetime is curved more in the spatial direction and if you go fast (idem for fast) it curves more in the temporal direction?

Sorry but I think that is not even wrong.

Perhaps someone can point to a credible textbook that makes such statements.

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 Quote by Passionflower So you agree with AT that the curvature of spacetime depends on how fast one travels through it (whatever how fast might even mean)? If you go slow (again whatever that might mean) spacetime is curved more in the spatial direction and if you go fast (idem for fast) it curves more in the temporal direction? Sorry but I think that is not even wrong. Perhaps someone can point to a credible textbook that makes such statements.
No, that is not what I'm saying. You raised the point the the axes of the local frame should be considered and I've answered that question for the orbiting observer. You obviously don't understand what I'm saying so let us just agree to disagree.

 Quote by Mentz114 They show the time cobasis vector of the orbiting observers frame. The time direction basis vector is only tilted by a small amount from the coordinate time direction. In other words the contribution of the spatial curvature is much smaller than the contribution of the time curvature. Read the equation. The Newtonian orbit equations work very well, despite ignoring spatial curvature. Only tiny relativistic effects have been unexplained like the perihelion precession of Mercury.
If the planetary orbits are straight lines in curved spacetime doesn't their manifest ellipse shape in space result from and describe that curvature.

Is there any way you could conceptually describe how time effectuates this if the spatial curvature is negligible???
Thanks
 Recognitions: Science Advisor Staff Emeritus Imagine drawing your space-time diagram on a curved surface. For a very simple example,, you might draw a space-time diagram on a sphere. http://www.pitt.edu/~jdnorton/teachi....html#Geodesic illustrates how initially parallel geodesics on a sphere converge. So, "curved space-time" can be visualized as "drawing space-time diagrams on a curved surface". As previously mentioned, in all of the space-time diagrams one can visulaize the static wordline as progressing through time by imagining a point on the worldline moving along it's length. Because time always "flows", the space-time diagram of an object is a world-line, not a world-point. One other note - the underlying geometry is still Lorentzian, i.e. the standard flat space-time diagram in SR transforms via the Lorentz transform. So, if you really want to understand GR well, you need to understand SR first, you need to be familiar with the SR space-time diagrams and how they transform via the Lorentz transform.