David Mumford's* paper ('*Ruminations on Cosmology and Time*') in last November's __ Notices Of the American Mathematical Society__ features an entertaining first section headed ** A Trip by Space Ship.** Therein he describes a thought experiment (based on relativity and the assumption of a constant 1g acceleration) to the Sagittarius A (

*Sag A*) black hole at the center of the galaxy and back. He discovers:

"To my astonishment, I found that I could travel to the black hole in the center of the Milky Way galaxy, known as Sagittarius A*, and back, all in less than forty years. Meanwhile, tens of thousands of years will have elapsed on earth when it returns."

The following is almost word for word from his own derivation in his Sec. 1:

**Introduction:**

Let x, t be space time coordinates and let :

ds** ^{2 }**= dt

**-**

^{2 }**|**

**dx**

^{2}

**|**

**/ c**

^{ }

^{2}Be its special relativistic metric, with s measuring proper time along any time-like trajectory, with c the speed of light. Assume the trip takes place in the space y = z = 0.

The Lorentz group*:

Acts transitively on any hyperbola whose asymptotes are light rays, i.e. a curve given by: c**( 1 – A)**

^{2}**- (x – B)**

^{2}**= - R**

^{2 }

^{2 }The curves are then of
constant acceleration. Then a hyperbola through
the origin and tangent there to the time axis represents the trajectory of an
initially stationary space ship with constant acceleration and trajectory
expressed as:

𝑥⃗ = 𝑎.⃗
cosh(𝑠/𝜆),
𝑡
= 𝜆.
sinh(𝑠/𝜆)

Here, **|**** **𝑎.⃗**|**** **** ^{ }**= c

**/ g**

^{2}𝜆 = c / g

Where g = 32.174 ft/ sec ^{2 }

with s measuring proper time in
the space ship. Choosing ‘years’ and
light years for units, then c = 1. And also g
» 1.

Then more precisely, Mumford
computes exactly 31, 557,600 seconds in a year.
(The Julian year as used by the I.A.U.)
He further finds then that 3.104
x 10** ^{16}** ft. make
a light year, making g » 1.03 in year-light year units. The latest measurements show that SagA* is about 26,000 light years distant. Then if one sets c= 1, g » 1.03, s = 9.87 we find that in less than 10 subjective
years, the ship has traveled 26,000 light years. At the same time t is 13,000 yrs which means
13 millennia have passed on Earth. At
which point your speed is:

c **· **tanh (10.16)
or 99.99999985% of the speed of light.

Mumford writes in conclusion:

A few caveats: firstly, space-time is not flat, meaning you need some extra power to get out of the sun’s gravitational field and you will be accelerated/decelerated by the galaxies gravitation on the way in, respectively, way back. I think these are minor. When your own space ship hits its maximum speed midway, everything is normal inside your space ship but outside, stars and interstellar gas are rushing by extremely close to the speed of light. The usual formula shows that their masses increase by the factor:

(1
− ‖𝑑𝑥/𝑑𝑡‖ ** ^{2}**/𝑐

**)−1/2 = cosh(𝑔𝑠/𝑐) ≈ 13000.**

^{2}You’ll need good shielding. Finally, if
you use a photon drive, accelerating by emitting high energy
photons, and take the mass of the space ship to be the
same as that of the space shuttle (2 million kg), my calculations
show that the needed energy for this trip is
roughly 10^{18} kilowatt-hours, or roughly * **10,000 **years
*worth of the total energy
being generated today on
earth. This seems mean that we need some breakthroughs.

**See Also:**

Brane Space: Special Relativity Revisited (2) (brane-space.blogspot.com)

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*David Mumford is an emeritus professor of applied mathematics at Brown University.

** The Lorentz group is the group of all Lorentz transformations of Minkowski spacetime,

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