# What is the value of curvature which caused by earth

• tallal hashmi
This is because tidal gravity is not the same thing as "acceleration due to gravity". In summary, the value of curvature caused by the Earth (geodetic effect) on space-time is not a single number, but rather a complex concept that includes factors such as the proximity to the Earth's gravitational well and the velocity of an object through spacetime. The radius of curvature of the object's worldline is not the same as the radius of curvature of spacetime, as tidal gravity is not the same as "acceleration due to gravity".
tallal hashmi
What is the value of curvature which caused by earth(geodetic effect) on space-time?

I need the value of curvature as a length,distance,area e.tc.

There isn't a single number that describes the curvature of spacetime due to the Earth. You'll need to be more specific about what you're interested in.

I think I have an idea what you are asking. If 4D space-time were Euclidean, rather than the non-Euclidean entity that it actually is, to produce what you would consider an acceleration of 1 g, the world line of the object would have to have a radius of curvature of about 1 light-year. This is because the velocity of the object through fictitious 4D Euclidean space-time would be so high (the speed of light c) that even the slightest curvature to the world line would be enough to produce this significant an acceleration. I hope this doesn't offend the sensibilities of the true experts like Peter too much, but, as an engineer, this is my feeble way of getting an intuitive notion of what is happening.

Chet

DaveC426913 said:
"...proximity to the Earth's gravitational well will cause a clock on the planet's surface to accumulate around 0.0219 fewer seconds over a period of one year than would a distant observer's clock..."

This sounds like a very easy thing to verify... judging from the precision of our clocks.

ChrisVer said:
This sounds like a very easy thing to verify... judging from the precision of our clocks.

It's not enough just to have precise clocks; we would have to have a precise clock out at infinity to compare with our precise clocks on Earth.

We can (and have) verified gravitational time dilation over much smaller changes of altitude (which requires much more accurate clocks than verifying 0.0219 seconds per year would--fortunately we have them). But we haven't verified the time dilation on Earth as compared with infinity, and won't be able to any time soon.

Chestermiller said:
the world line of the object would have to have a radius of curvature of about 1 light-year.

Yes, but this is the curvature of the object's worldline, not the curvature of spacetime. They're two different things.

PeterDonis said:
It's not enough just to have precise clocks; we would have to have a precise clock out at infinity to compare with our precise clocks on Earth.

Well I wouldn't mean infinity... but our clocks can achieve a precision of ns... so even if it wasn't 0.022s but 0.000001s, we would be able to detect it.

I wonder... if you go can make the two measurements one on the Earth and one on the Moon...wouldn't you be able to see a difference (due to different gravity potentials on each surface) ?

ChrisVer said:
if you go can make the two measurements one on the Earth and one on the Moon...wouldn't you be able to see a difference (due to different gravity potentials on each surface) ?

If you could get the two clocks to be at rest relative to each other--but that's going to be tough since the Moon is moving relative to the Earth.

PeterDonis said:
If you could get the two clocks to be at rest relative to each other--but that's going to be tough since the Moon is moving relative to the Earth.

So you would measure a difference in clocks: $\Delta t = \delta t_{relat} + \delta t_{grav}$?

PeterDonis said:
Yes, but this is the curvature of the object's worldline, not the curvature of spacetime. They're two different things.
Yes. I was thinking in terms of the acceleration of an object, as reckoned in Special Relativity, in a non-gravity situation. However, for my own edification, if a body is stationary relative to massive object in curved space-time, if it is experiencing a force of approximately 1 g (say on the surface of a planet), doesn't that imply approximately the same curvature of space-time as the curvature of the world line in the non-gravity situation?

Chet

A.T. said:
You don't need to go to the Moon for this:

Well in a talk given by David Jeffery Wineland (I think concerning trapping atoms), I heard they even considered General Relativistic corrections in a setup the concerns atoms within a laboratory environment (so some meters)...

Chestermiller said:
if a body is stationary relative to massive object in curved space-time, if it is experiencing a force of approximately 1 g (say on the surface of a planet), doesn't that imply approximately the same curvature of space-time as the curvature of the world line in the non-gravity situation?

No. Spacetime curvature is not "acceleration due to gravity"; it's tidal gravity, which, roughly speaking, is the rate at which the magnitude and/or direction of "acceleration due to gravity" changes with position.

For example, the Riemann tensor components in the vacuum region outside a spherically symmetric gravitating mass (which we can approximate the Earth to be for this discussion) are equal to either ##M / R^3## in geometric units (so ##M## here is ##GM / c^2## in conventional units), or twice that (depending on whether we are looking at tangential or radial components--also the signs are different but we are just looking at magnitudes here). The "radius of curvature" of spacetime is then the inverse square root of this. The "acceleration due to gravity" is ##M / R^2## in geometric units (in the Newtonian approximation, the fully relativistic formula has an extra factor of ##\sqrt{1 - 2M / R}## in the denominator--note that this factor is not present in the Riemann tensor components), and the radius of curvature of the stationary object's worldline is the inverse of this.

Plugging in values for Earth, we get:

Radius of curvature of stationary worldline: ##9.1 \times 10^{15}## meters.

Radius of curvature of spacetime (tangential): ##2.4 \times 10^{11}## meters.

As you can see, the two results are very different.

Still don't have the exact value.

A few comments on the OP's question. As others have pointed out, curvature as we define it in GR is a rather complicated mathematical entity , known as a tensor, that is a bit like a matrix, except that it has even more dimensions than a matrix does. It's full name is the Riemann curvature tensor.

To further confuse things, in mathematics there are several different things that might be called "curvature". In the context of General relativity, though, the sort of curvature we are interested in is called the Riemann curvature, a form of intrinsic curvature.

The good news is that if you consider a specific two dimensional surface, you can represent the curvature of that surface by a singe number. Wikki has a good background on it, at http://en.wikipedia.org/wiki/Curvature#Gaussian_curvature, that also describes the difference between extrinsic and intrinsic curvature, and a few techniques for measuring the Gaussian curvature. Only one of the techniques so described is intrinsic, that's the technique that involves examining the circumference of small circles.

Applying this insight to GR, you could ask what the curvature was in any 2-dimensional plane you specify, for instance the r-t plane, though you can't ask for "the" curvature at a point, there is a value of cuvature for every plane containing that point. The most important aspects of space-time curvature turn out to be physically interpretable as tidal forces, these involve planes in which one of the dimensions is timelike. Space is also curved in GR, so if you specify a pair of spatial axes, you will typically get nonzero values for the Gaussian curvature as well, which is why we sometimes say that "space is curved".

tallal hashmi said:
Still don't have the exact value.

That's because there isn't one "exact value", as I pointed out in post #2. But the "radius of curvature of spacetime" number I gave in post #15 is a reasonably representative value.

tallal hashmi said:
Still don't have the exact value.
To paraphrase Pervect, there is no one answer to this. There are 64 numbers in the Riemann tensor, although some of them must be the same (I think there are only 24 independent values). So you need to be specific about which number you want to know before you can get an answer.

PeterDonis said:
No. Spacetime curvature is not "acceleration due to gravity"; it's tidal gravity, which, roughly speaking, is the rate at which the magnitude and/or direction of "acceleration due to gravity" changes with position.

For example, the Riemann tensor components in the vacuum region outside a spherically symmetric gravitating mass (which we can approximate the Earth to be for this discussion) are equal to either ##M / R^3## in geometric units (so ##M## here is ##GM / c^2## in conventional units), or twice that (depending on whether we are looking at tangential or radial components--also the signs are different but we are just looking at magnitudes here). The "radius of curvature" of spacetime is then the inverse square root of this. The "acceleration due to gravity" is ##M / R^2## in geometric units (in the Newtonian approximation, the fully relativistic formula has an extra factor of ##\sqrt{1 - 2M / R}## in the denominator--note that this factor is not present in the Riemann tensor components), and the radius of curvature of the stationary object's worldline is the inverse of this.

Plugging in values for Earth, we get:

Radius of curvature of stationary worldline: ##9.1 \times 10^{15}## meters.

Radius of curvature of spacetime (tangential): ##2.4 \times 10^{11}## meters.

As you can see, the two results are very different.
These results are really quite dazzling. So, I was approximately correct about the radius of curvature of the stationary worldline, but the radius of curvature of spactime is several orders of magnitude smaller, implying considerably greater curvature?

Chet

Chestermiller said:
the radius of curvature of spactime is several orders of magnitude smaller, implying considerably greater curvature?

Yes.

## 1. What is curvature caused by the earth?

The curvature caused by the earth refers to the slight bending or rounding of the earth's surface due to its spherical shape. This curvature is most noticeable at large scales, such as viewing the horizon from a high point, and is a result of the earth's gravity pulling towards its center.

## 2. How is the value of curvature calculated?

The value of curvature caused by the earth is calculated using the equation: C = (R^2)/(R+h), where C is the curvature, R is the radius of the earth, and h is the height or distance above the surface. This equation can be used to calculate the curvature at different points on the earth's surface.

## 3. What is the unit of measurement for curvature?

The unit of measurement for curvature caused by the earth is typically expressed in degrees or radians. This is because curvature is measured as an angle, representing the amount of deviation from a straight line on the earth's surface.

## 4. How does curvature affect everyday life?

Curvature caused by the earth is not typically noticeable in everyday life as it is a gradual change. However, it does play a crucial role in navigation and surveying, where accurate measurements of distance and direction must take into account the earth's curvature.

## 5. Can curvature caused by the earth be seen from space?

Yes, curvature caused by the earth can be seen from space. Astronauts in orbit around the earth can see the curvature of the earth's surface, and images taken from satellites also show the curvature of the earth's horizon. However, this curvature is more noticeable at higher altitudes and is not visible from the surface.

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