Characteristic time sqrt(R_e/g) of Earth's gravity in General Relativity

In summary, the time t=sqrt(R/g) characterizes the Earth's gravity through the concepts of gravitational time dilation and geodesic deviation. This equation can be used to understand the slower passage of time in a stronger gravitational field and the change in distance between initially parallel geodesics.
  • #1
makehhh
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Homework Statement



Hiya,

I was looking over a General Relativity paper today, and at the end of a question on Geodesic Deviation, was "Give a brief statement of a physical effect or thought-experiment which illustrates how the time sqrt(R/g) characterises the Earth’s gravity." (R is the radius of the earth)


Homework Equations



t=sqrt(R/g)


The Attempt at a Solution



Well, I have been looking through chapters of GR texts on Schwarzschild thought experiments, Bondi thought experiments etc., and can't find anything related to this quantity. Obviously any expression containing g characterises Earth's gravity, but I simply don't know what physical effect or thought-experiment the question could be referring to. If anyone could help it would be very much appreciated.

Thanks a lot.
 
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  • #2




Hello,

The time t=sqrt(R/g) characterizes the Earth's gravity through the concept of gravitational time dilation. This is the phenomenon where time appears to pass slower in a stronger gravitational field. In this case, R represents the radius of the Earth and g represents the gravitational acceleration at the surface of the Earth.

To understand this concept, we can use the thought experiment of a clock placed at the surface of the Earth and another clock placed at a higher altitude, where the gravitational acceleration is lower. According to the equation t=sqrt(R/g), the clock at the surface of the Earth will experience a slower passage of time compared to the clock at a higher altitude. This is because the gravitational acceleration is stronger at the surface of the Earth, causing time to pass slower.

Another way to think about this is through the concept of geodesic deviation. Geodesics are the paths that objects follow in the presence of a gravitational field. In this case, the geodesic deviation refers to the change in the distance between two initially parallel geodesics. The equation t=sqrt(R/g) can be used to calculate the amount of geodesic deviation that occurs due to the Earth's gravity.

I hope this helps to clarify the physical effect or thought-experiment that the question is referring to. Let me know if you have any further questions.
 

1. What does the characteristic time sqrt(R_e/g) represent?

The characteristic time sqrt(R_e/g) represents the time it takes for a free falling object to travel the radius of the Earth, assuming Earth's gravity follows the predictions of General Relativity.

2. How is the characteristic time sqrt(R_e/g) derived?

The characteristic time sqrt(R_e/g) is derived from the Schwarzschild radius (R_e) and the acceleration due to gravity (g) at the Earth's surface. It is calculated using the formula sqrt(R_e/g).

3. How does the characteristic time sqrt(R_e/g) differ from the classical formula for free fall time?

The characteristic time sqrt(R_e/g) differs from the classical formula for free fall time because it takes into account the effects of General Relativity on gravity. This results in a slightly longer free fall time compared to the classical formula.

4. Why is the characteristic time sqrt(R_e/g) important in General Relativity?

The characteristic time sqrt(R_e/g) is important in General Relativity because it helps to explain the effects of gravity on objects in space. It also plays a role in understanding the curvature of spacetime and how it affects the motion of objects.

5. How does the characteristic time sqrt(R_e/g) change for different celestial bodies?

The characteristic time sqrt(R_e/g) will vary for different celestial bodies depending on their mass and radius. The larger the mass and radius of a celestial body, the longer the characteristic time sqrt(R_e/g) will be. This is because the effects of gravity will be stronger on larger objects.

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