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   Greg Egan (gregegansf@mathstodon.xyz)'s status on Thursday, 19-Jun-2025 23:03:07 JST Greg Egan

   An intergalactic spacecraft cruises past a galaxy at 50% the speed of light, and without firing its engines, travels for so long before it passes another galaxy that the length scale of the universe has increased by 25%.

   At what speed does it pass the second galaxy?

   In an expanding universe energy and momentum aren’t conserved, but to the extent that the universe is homogeneous and isotropic on large scales, we can use *that* to answer our question. It turns out that the spatial symmetry of the universe lets us determine that the quantity:

   a(t) p(t)

   *is* conserved, where a(t) is the length scale of the universe, and p(t) is the momentum of a body in free fall, measured relative to the background of the overall expansion.

   So if a(t) has increased by a factor of 5/4, p(t) must have decreased by a factor of 4/5, and the spacecraft will pass the second galaxy at 40% of lightspeed.

   Why is a(t) p(t) conserved? This follows from a remarkably simple geometric fact, known as Killing’s Theorem.

   If we start with the geometry of the plane, suppose we draw two vectors, w_A and w_B, based at points on the straight line segment AB, which show the rate at which the points A and B would move if we rotated everything in the plane around some point.

   Then the difference, w_B-w_A, will always be at right angles to AB. Why? Because rotating everything in the plane won’t change the length of AB, so the motion can only turn it, not bring its endpoints closer together or further apart.

   This in turn means that the length of the projection of the two w vectors onto a unit-length vector tangent to the line segment at each point will be the same.

   • Embed this notice
     Anthk (anthk@paquita.masto.host)'s status on Thursday, 19-Jun-2025 23:04:31 JST Anthk

     in reply to

     @gregeganSF I've read something similar from a book written by Ian Stewart. I'm more a Math guy than a Physics one.

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     Greg Egan (gregegansf@mathstodon.xyz)'s status on Thursday, 19-Jun-2025 23:04:32 JST Greg Egan

     in reply to

     All of this is pretty obvious in the flat geometry of the plane, but it continues to hold true in the curved geometry of an expanding spacetime, with the following provisos:

     • Instead of a straight line AB, we have a geodesic AB, the closest thing to a straight path through spacetime, which is the worldline of a body in free fall.

     • Instead of the w vectors showing how a point would move if we rotated the plane, we consider a symmetry of the whole of spacetime that shifts everything in the direction in space that the spacecraft travelled. Because the overall size of spacetime scales with a(t), so must the vectors w, in order to slide everything along rigidly, without changing the geometry.

     The spacecraft’s relativistic energy-momentum vector P is a vector of a fixed length that is tangent to its worldline, and its momentum vector p is just the projection of P into the direction we are treating as “space”.

     But that is also the direction of the w vectors. So a(t) p(t) is the length of the projection of the w vectors onto a unit vector that points along the spacecraft’s worldline. And just as in the plane, it remains the same all along the geodesic!

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     Anthk (anthk@paquita.masto.host)'s status on Thursday, 19-Jun-2025 23:58:11 JST Anthk

     in reply to

     @gregeganSF yeah I found that missing

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     Greg Egan (gregegansf@mathstodon.xyz)'s status on Thursday, 19-Jun-2025 23:58:12 JST Greg Egan

     in reply to

     *Edit: Relativistic momentum isn’t proportional to velocity, so reducing the momentum by 4/5 actually changes the velocity to 41.9% of lightspeed.

     The exact calculation is:

     p = m v/√(1-v^2) in units where c=1.

     v = 1/2 gives p = m/√3
     p = 4m/(5√3) gives v = 4/√(91) ≈ 0.4193