#### Neddy Bate

**Valued Senior Member**

Mike,

Somewhere online you should be able to find a GR solution to the SR twin 'paradox' using pseudo-gravity, and the equivalence principle, rather than just SR. Gravitational potential depends on distance, so it should give the same answer, I would think.

But why? You already know how to draw Minkowski diagrams, so just look at the rules for drawing lines of simultaneity:

When the velocity changes from v=+0.866 to v=-0.866 the simultaneity lines should look symmetrical, showing that there is no absurd and inconsistent asymmetry as you have described above.

Another way is to simply look at the SR equations rather than starting with the GR equations and having to work backwards from there.

Here is a simple SR equation:

t = γ(t' + (vx' / c²))

The above equation is the

For simplicity, choose the time t'=0 and use units where c=1:

t = γ(0 + (vx' / 1))

t = γvx'

This demonstrates that for the case t'=0, the times (t) are directly related to their location as measured by the x' axis. For a constant v, clearly t varies with x'. Gamma is always positive, but velocity can be positive or negative. So for t'=0 and a positive velocity, t is increasingly positive on the positive x' axis, and increasingly negative on the negative x' axis. But for t'=0 and a negative velocity, that is reversed, so t is increasingly positive on the negative x' axis, and increasingly negative on the positive x' axis. The asymmetry you describe is not there, so you have made a mistake someplace.

(Note: negative t values here only mean that they were an earlier time than the time assigned t=0.)

Somewhere online you should be able to find a GR solution to the SR twin 'paradox' using pseudo-gravity, and the equivalence principle, rather than just SR. Gravitational potential depends on distance, so it should give the same answer, I would think.

But why? You already know how to draw Minkowski diagrams, so just look at the rules for drawing lines of simultaneity:

When the velocity changes from v=+0.866 to v=-0.866 the simultaneity lines should look symmetrical, showing that there is no absurd and inconsistent asymmetry as you have described above.

Another way is to simply look at the SR equations rather than starting with the GR equations and having to work backwards from there.

Here is a simple SR equation:

t = γ(t' + (vx' / c²))

The above equation is the

*given*SR relationship between a time in an unprimed frame (t) and the time in a primed frame (t') at various locations along the x' axis.For simplicity, choose the time t'=0 and use units where c=1:

t = γ(0 + (vx' / 1))

t = γvx'

This demonstrates that for the case t'=0, the times (t) are directly related to their location as measured by the x' axis. For a constant v, clearly t varies with x'. Gamma is always positive, but velocity can be positive or negative. So for t'=0 and a positive velocity, t is increasingly positive on the positive x' axis, and increasingly negative on the negative x' axis. But for t'=0 and a negative velocity, that is reversed, so t is increasingly positive on the negative x' axis, and increasingly negative on the positive x' axis. The asymmetry you describe is not there, so you have made a mistake someplace.

(Note: negative t values here only mean that they were an earlier time than the time assigned t=0.)

Last edited: