Do you accelerate through time when you stand still on Earth?

  • Context: Undergrad 
  • Thread starter Thread starter Grasshopper
  • Start date Start date
  • Tags Tags
    Accelerate Earth Time
Click For Summary
SUMMARY

The discussion centers on the concept of acceleration in the context of general relativity, specifically addressing the idea of "accelerating through time" while standing still on Earth. Participants clarify that while one experiences proper acceleration due to the normal force on the Earth's surface, this does not equate to motion through space. The equation presented, $$\frac{dr^2}{dτ^2}= a^r - Γ_{tt}^r (u^{t})^2$$, is derived from the Geodesic Equation in Schwarzschild coordinates, emphasizing that the time component of four-velocity is crucial in understanding the relationship between acceleration and motion in spacetime. The conversation also highlights the coordinate dependence of these concepts, asserting that acceleration does not imply absolute motion through space.

PREREQUISITES
  • Understanding of general relativity principles
  • Familiarity with the concept of four-velocity
  • Knowledge of Schwarzschild coordinates
  • Basic grasp of tensors and their differences from vectors
NEXT STEPS
  • Study the Geodesic Equation in detail, particularly in the context of Schwarzschild coordinates
  • Learn about the implications of proper acceleration in non-inertial frames
  • Explore the concept of gravitational time dilation and its effects in curved spacetime
  • Investigate the differences between coordinate systems in general relativity, including Eddington-Finkelstein coordinates
USEFUL FOR

Students and professionals in physics, particularly those focused on general relativity, cosmology, and theoretical physics, will benefit from this discussion. It is also valuable for anyone interested in the mathematical foundations of spacetime and acceleration concepts.

  • #31
cianfa72 said:
the scalar product between two 4-velocity tangent vectors

The tangent vector to a null worldline is not properly referred to as a "4-velocity". The term "4-velocity" implies a unit vector.

cianfa72 said:
shouldn't be related in some way to the 3-velocity relative speed between them ?

If both vectors are timelike, the scalar product is related to the 3-velocity, yes, since the scalar product is the relative ##\gamma## factor between them, and ##\gamma = 1 / \sqrt{1 - v^2 / c^2}##.

If one vector is null and the other is timelike, the scalar product is, as @vanhees71 has said, the frequency of the light ray (null vector) as measured in the rest frame of the timelike vector. So no, unfortunately, it is not related to the speed of the light ray. If you think about it, you will see that it can't possibly be related to the speed, since the speed of the light ray is ##c## regardless of which timelike vector you pick, but the scalar product is different for different timelike vectors. This is related to the fact that Lorentz transformations act differently on timelike vectors than on null vectors; they rotate timelike vectors in spacetime, but they dilate null vectors.
 
  • Like
Likes   Reactions: cianfa72, vanhees71 and PeroK

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 35 ·
2
Replies
35
Views
3K
  • · Replies 69 ·
3
Replies
69
Views
7K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K