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In Special Relativity, acceleration i.e. a changing velocity 4-vector results in time dilation, that is asymmetric aging of observers.

In General Relativity, the 4-vector does not change along a geodesic. Or rather, the co-ordinate components of the four-vector do change along a geodesic. But we have defined the covariant derivative such that the velocity-four vector differentiates to 0 along a geodesic.

Consequently, the acceleration along a geodesic is zero. And hence, it seems like the the GR time-dilation is of a different nature from the SR time dilation. The GR one being attributed to a varying metric while the SR one attributed to a non-zero acceleration.

However, what if we re-interpret the four-velocity in GR? What if we simply say that the four-velocity along a geodesic IS changing? We would do this by using partial derivatives instead of covariant derivatives. If we did this, wouldn't the GR time dilation get attributed to a non-zero acceleration just like in SR?

Does this mean that both time dilations are of the same underlying nature, and that attributing the GR time dilation to the metric is just a matter of semantics?

And how does all of this relate to "time dilation due to a graviton"? I've seen people argue that the reason graviton is special is that it somehow affects the metric of spacetime. If we simply re-interpret a changing metric as a non-zero acceleration, then the graviton is no longer special in this regard.

Qmechanic
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Rain Deer
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1 Answers1

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This answer is not in terms of metrics and geodesics. Instead, it gives a mathematical distinction between the two types of time dilation. Gravitational time dilations and special relativity time dilations are different transformations. Special relativity time dilations result from space-time parallelepiped strains. Schwarzschild time dilations (eg: near a mass) are space-time squash strains, and just to complete the story, gravitational waves do space-space squash strains ($h_+$) and space-space parallelepiped strains ($h_X$).

Consider the group of all invertible 4x4 matrices M with det(M)=1. These are linear transformations that can be done to a 4-vector $(x,y,z,ct)$. Now label all these matrices near the identity matrix I. $$ M=I+\Theta $$ where all elements of $\Theta <<1$ then

$ \\ \Theta =radians= \begin{bmatrix} 0 & \theta_3 & -\theta_2 & \phi_1 \cr -\theta_3 & 0 & \theta_1 & \phi_2 \cr \theta_2 & -\theta_1 & 0 & \phi_3 \cr -\phi_1 & -\phi_2 & -\phi_3 & 0 \end{bmatrix}_{Asym} + $ $ \begin{bmatrix} h+_1 & hx_3 & hx_2 & \lambda_1 \cr hx_3 & h+_2 & hx_1 & \lambda_2 \cr hx_ 2 & hx_1 & h+_3 & \lambda_3 \cr \lambda_1 & \lambda_2 & \lambda_3 & h+_4 \end{bmatrix}_{Sym} $

where det(M)=1 implies $(h+_1+h+_2+h+_3+ h+_4=0)$. The M for larger $\Theta$ is obtain from $$ M=e^{\Theta}=I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +… $$

You recognize the rotation transformation angles $\vec{\theta}$ and the Lorentz Boost transformation parameters $\vec{\lambda}$. Rotations (orthogonal transformations) are done by the elements of the antisymmetric matrix, and strains (don't leave Euclidian length invariant) are done by all the elements of the symmetric matrix. All the relativity stuff are strains.

If you apply an M made from any of the off diagonal elements of the symmetric matrix to the four corner points of a square, the square will become a parallelepiped. The on diagonal elements do a squash or expansion in one dimension. For Schwarzschild $~h+_4=\frac{GM}{r}~$ so $~t'=(1+h+_4)t=(1+\frac{GM}{r})t~$ and $~h+_1=-\frac{GM}{r}~$ (which does the space squash if r is in the 1 direction). For special relativity time dilation vanishes to first order for $\lambda\ll 1$, but for larger boosts $~t'=cosh(\lambda)t=\gamma t$.

Gary Godfrey
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