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tallal hashmi
What is the value of curvature which caused by earth(geodetic effect) on space-time?
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..."
ChrisVer said:This sounds like a very easy thing to verify... judging from the precision of our clocks.
Chestermiller said:the world line of the object would have to have a radius of curvature of about 1 light-year.
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.
ChrisVer said: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) ?
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.
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?PeterDonis said:Yes, but this is the curvature of the object's worldline, not the curvature of spacetime. They're two different things.
A.T. said:You don't need to go to the Moon for this:
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?
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.tallal hashmi said:Still don't have the exact value.
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?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.
Chestermiller said:the radius of curvature of spactime is several orders of magnitude smaller, implying considerably greater curvature?
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.
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.
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.
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.
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.