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

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The discussion explores the concept of acceleration in the context of general relativity, particularly regarding standing still on Earth. It highlights that while one experiences proper acceleration due to gravity, there is no movement through space in their own reference frame. The conversation delves into the mathematical representation of this phenomenon, including a specific equation related to four-velocity and spacetime curvature. Participants clarify that "accelerating through time" is a coordinate-dependent concept and lacks physical significance. Overall, the thread emphasizes the complexities of motion, acceleration, and reference frames in the framework of general relativity.
  • #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.
 
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