# I Still Don't Get General Relativity

I get objects bend the space around them. I have been told two things: these dents are attractive, and that they are not.

If they are not attractive, shouldn't objects slow down when they come near an object?
Think about an ant walking over a surface with a dent in in. The surface is 2d, but the dent is in 3d. If you look at it from the top, the ant will slow down (and change direction) over the dent.
http://img214.imageshack.us/img214/5752/1267964150095m.png [Broken]
The distance and time are both longer, but from the second dimension, the distance is the same, so it slows down.
Also, if the bending of space causes the change in direction, what causes things to fall when we release them from a height?

If they are attractive, wouldn't gravity propagate from a higher dimension?
http://img641.imageshack.us/img641/2696/1267964983741.png [Broken]

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Jonathan Scott
Gold Member
Two things to note:

1. The bending of space-time involved in everyday gravity is incredibly tiny - so much so that it is extremely difficult to observe. A light beam passing close to a massive body such as the sun or earth is bent by space and time curvature through a total angle of about (4GM/Rc^2) radians, where M is the mass of the body and R is its radius. For the sun this is about 8.5*10^-6 radians, and for the earth this is about 2.8*10^-9 radians.

2. It's space-time which is bent, not just space, otherwise slow-moving objects wouldn't be affected by gravity. Roughly speaking, as time elapses, the vector describing the displacement of the particle in space and time turns in the direction of the gravitational field. For objects moving near the speed of light, the bending of space and of space-time combine to double the overall acceleration relative to flat space.

A.T.
I get objects bend the space around them.
Space-time, not just space. This is crucial:
http://www.relativitet.se/spacetime1.html

If they are not attractive, shouldn't objects slow down when they come near an object?
"Attractive" is rather a vague term here. The path is bend towards the mass, so you can call it "attractive". But the effect on the speed observed from a distance depends on the speed of the object.

For fast objects: The curvature of space can cause light crossing region near a mass to appear slower, because there is "more space" to cross "in the dent".

For slow objects: But when you send a space ship through that region, the curvature of time has a much greater effect as it accelerates the ship to cross the region much faster.

Also, if the bending of space causes the change in direction, what causes things to fall when we release them from a height?
The bending of time:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

If they are attractive, wouldn't gravity propagate from a higher dimension?
All that matters in GR are distorted distances within the 4 space-time-dimensions. The extra dimensions are needed solely for visualization, and have no physical meaning.

Galileo
Homework Helper
The distance and time are both longer, but from the second dimension, the distance is the same, so it slows down.

Also, if the bending of space causes the change in direction, what causes things to fall when we release them from a height?
The path a freely falling object takes is a straight(est possible) line in spacetime, so you need to think of time in your space as having a coordinate axis too. The easiest way to picture this is with a worldline in a spacetime diagram (usually with time on the vertical axis).

The worldline of a particle is a geodesic in this spacetime. Without gravitation, spacetime is flat and the worldlines of objects are straight lines and the slope of the worldline tells you the velocity of the object (doesn't change for a straight line).
In a curved spacetime, a line that is initially vertically straight in spacetime will bend and have a different slope further down the line. I.e. an object that has no velocity initially will start to move if spacetime is curved.

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The basic principle of GR is that near a mass in empty space, Rij, the Ricci tensor with two indices, is zero, while the curvature tensor R with 4 indices is non zero. Weinberg discusses this in his 1972 book. I do not understand this logic. Can someone explain it?

bcrowell
Staff Emeritus
Gold Member
The basic principle of GR is that near a mass in empty space, Rij, the Ricci tensor with two indices, is zero, while the curvature tensor R with 4 indices is non zero. Weinberg discusses this in his 1972 book. I do not understand this logic. Can someone explain it?

The Ricci tensor measures the kind of curvature that is produced by masses that are right there in that region of space. The Riemann tensor R measures all kinds of curvature, including curvature that is produced by tidal forces of distant masses. Here's a more detailed explanation: http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.1 [Broken]

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Sorry, but I do not get your explanation of the Riemann tensor R. Look at the flea on a football. At any point on the surface we can imagine two perpendicular planes. The intersections of these planes are circles. The curvature at (x,y) is defined by the radii of these planes. Okay, this is not enough info for the flea to get to the tip. For this we need to know the curvature at all points on the surface. Mass will affect curvature at a point (x,y) which is defined by the radii of intersecting planes. I do not understand the meaning of R with 4 indices.

Sorry, but I do not get your explanation of the Riemann tensor R. Look at the flea on a football. At any point on the surface we can imagine two perpendicular planes. The intersections of these planes are circles. The curvature at (x,y) is defined by the radii of these planes. Okay, this is not enough info for the flea to get to the tip. For this we need to know the curvature at all points on the surface. Mass will affect curvature at a point (x,y) which is defined by the radii of intersecting planes. I do not understand the meaning of R with 4 indices.

Are we saying that General Relativity requires the use of a tensor describing a space-time point (x,y,z,t) to also specify the geometry of a different point? When we look at the change in geometry due to an external mass, we must also consider the geometry somewhere else? This cannot be, for how can we define this second point?

Again, what do we mean by the tensor R with 4 indexes?

I understand things a bit better, but am still puzzled.

en.wikipedia.org/wiki/Geodesic_deviation_equation

Here is how I understand the 4th order R (4 subscripts), the curvature tensor.

Let g be a geodesic (the site uses gamma, but I am writing in text mode and cannot use Greek).

g(t) defines a curve

g sub s defines different geodesics.

E.g., surface of a sphere.
g is a line from the north pole to the south pole.
The parameter t is latitude, from 0 to pi.
The parameter s is longitude.

Let a vector T be dg/dt, the tangent vector to g.
Let X be dg/ds, the normal vector to g.

We examine the relative acceleration of two particles on neighboring geodesics. The 4th order R defines this acceleration. Two of the parameters are for two geodesics. One parameter is for the connecting field X. The fourth is the free parameter.

We need four parameters to define curved space. Okay.

My question is what is the meaning of the trace, which is the second order Ricci tensor (2 subscripts).

http://myyn.org/m/article/ricci-tensor/ [Broken]

"In Riemannian geometry, the Ricci tensor represents the average value of the sectional curvature along a particular direction. "

I do not understand the Ricci tensor. I do not understand the significance of choosing the first and third parameters of the curvature tensor to sum and get the trace, the Ricci tensor. Why those parameters? What is the geometrical meaning of the Ricci tensor? It is extremely important to understand this, for textbooks, such as Weinberg 1972, postulate that Ricci must be zero. I do not understand the geometrical meaning of this statement.

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I do not understand the Ricci tensor. I do not understand the significance of choosing the first and third parameters of the curvature tensor to sum and get the trace, the Ricci tensor. Why those parameters? What is the geometrical meaning of the Ricci tensor? It is extremely important to understand this, for textbooks, such as Weinberg 1972, postulate that Ricci must be zero. I do not understand the geometrical meaning of this statement.

You know the example of drawing a triangle on the surface of the Earth such that the angle of all corners is 90° - and then you compare that triangle to a 2d triangle?

Ricci Tensor denotes the same kind of transform, except for 3 dimensions over 4, instead of 2 dimensions over 3.

e.g. We know what a type-G sun looks like when it's <8 min away. Compare one that's 80, 800, or 8000 light years away, and see how much it's warped. The difference between the two denotes the Ricci Tensor, which tells us what spacetime between ourself at the 8000lyr away sun looks like. If we do this for standard candles in the universe (supernovae, cepheid variables) we can determine what the spacetime looks like between them.

It might help if you review the Einstein Field equations and see where the Ricci tensor is implemented, and why it is important (i.e deriving the cosmological constant)
http://en.wikipedia.org/wiki/Einstein_field_equation

You know the example of drawing a triangle on the surface of the Earth such that the angle of all corners is 90° - and then you compare that triangle to a 2d triangle?

Ricci Tensor denotes the same kind of transform, except for 3 dimensions over 4, instead of 2 dimensions over 3.

e.g. We know what a type-G sun looks like when it's <8 min away. Compare one that's 80, 800, or 8000 light years away, and see how much it's warped. The difference between the two denotes the Ricci Tensor, which tells us what spacetime between ourself at the 8000lyr away sun looks like. If we do this for standard candles in the universe (supernovae, cepheid variables) we can determine what the spacetime looks like between them.

It might help if you review the Einstein Field equations and see where the Ricci tensor is implemented, and why it is important (i.e deriving the cosmological constant)
http://en.wikipedia.org/wiki/Einstein_field_equation

I am familiar with textbooks on GR, such as Weinstein 1972, hence my question.

I do not follow your logic explaining the geometrical meaning of the trace of the curvature tensor. What 2 coordinates are being summed over?

Now that you mentioned GR, why does Weinstein say that the Ricci tensor (the trace of the curvature tensor) =0 where there is no mass, yet <>0 where there is mass? I do not understand his logic. If you dive into a swimming pool, Ricci will be zero outside and nonzero in the water. I am a swimmer, but I do not follow this logic.

Time is slower on earth than on a satellite, in accordance to GR. This is critical for GPS. We can measure this slowness. How can we set up an experiment to detect the sudden change in Ricci as we dive in the water?

Patterns said:
Time is slower on earth than on a satellite, in accordance to GR.
This time difference is not related to curvature, but is a function of g00.

Coming back to curvature: Curvature is defined by the deficit angle of a vector transported around a loop. A difficult calculation shows that the rotation of the transported vector depends on the Riemann tensor. Similar to the geodesic deviation.

The time difference is due to the curvature tensor. This is the basis of general relativity. Read Weinstein's book on gravitation, 1972.

My question is why does Weinstein postulate that the trace of the curvature tensor, the Ricci tensor, is zero.

The time difference is due to the curvature tensor. This is the basis of general relativity. Read Weinstein's book on gravitation, 1972.

I suggest you read it a bit more carefully. Gravitational time dilation does not depend on the curvature tensor. This is from the Wiki page on 'Gravitational Time Dilation'
This has been demonstrated by noting that atomic clocks at differing altitudes (and thus different gravitational potential) will eventually show different times. The effects detected in such experiments are extremely small, with differences being measured in nanoseconds
also see the Pound-Rebka experiment.

My question is why does Weinstein postulate that the trace of the curvature tensor, the Ricci tensor, is zero.
Does Weinstein say that the trace of Rab is always zero ? Or only for certain space-times.

JesseM
You know, if matter formed a gravitational field. Wouldn't that violate energy conservation since this would mean that a mass, such as a grain of sand would be emitting energy since a field (G) is energy.
Apparently there are problems with defining the "energy of the gravitational field" in general relativity so many physicists prefer to just say that energy is not globally conserved in GR, see here and here for example.

The Ricci Tensor is set to zero...so we can solve the other equations! (?)

http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/WeylTensor.htm

Einstein's hypothesis is that the curvature of space-time is zero in the vacuum which is thus a flat space. This is true in two dimensions where the Gaussian curvature is zero. In higher dimensions, only the Ricci tensor is zero according to the Einstein equation. In matter, the Ricci tensor is different from zero. We shall not consider this case, here, but it should be considered to describe the universe which contains matter.

Page 305-306 explains a little as to why