Thomas precession is relatively easy to understand. Imagine you're a

gerbil running around a circular path in flat spacetime, carrying a

gyroscope with you. Now instead of going around a circular path,

substitute a regular polygonal path with a many (large number) sided

polygon.

For simplicity in visualization, assume the polygon is in a vertical

plane and the edge you are moving parallel along is horizontal. You

begin to approach a vertex. It looks like the foot of a slight hill.

The*steepness of this hill is frame dependent!* This is the key to

understanding! You should be able to figure it out from here, but I

will go on. In the laboratory frame the angle of this hill is 360/n

where n is the number of sides. However, because our little gerbil is

running*real fast*, the longitudinal direction is Lorentz contracted,

making the angle steeper.

In our little gerbil's frame, moving from one side of the polygon to

the next, he moves through an angle of theta'. In the lab frame we see

him move through an angle of theta.

Since our polygon has many sides (approaching infinity) the angles are

very small, so in the lab frame tan(theta) ~~ theta ~~ y/x. To our

little gerbil, tan(theta') ~~ theta' ~~ y'/x' = y/(gamma*x). Where

gamma = sqrt(1 - v^2/c^2)

In circumnavigating the polygon Mr. Gerbil rotates through an angle of

n*theta' = n*360/(n*gamma) = 360/gamma. So of course, he sees the

little gyroscope he is carrying with him rotate the same amount in the

opposite direction.

When he gets back to his starting point he stops! He sees that his

gyroscope has precessed (really just stood still, it's the gerbil that

turned) 360/gamma. In the lab frame Mr. Gerbil only circled 360

degrees. Since the gyroscope RECORDED a rotation of 360/gamma, and

since we only circled it 360 degrees, and since it is now stationary,

it must have precessed 360(1/gamma - 1) in the opposite direction in

the lab frame.

For gamma near "1" we have (1/gamma - 1) ~~ (1/2)*v^2/c^2.

If we specify an angular frequency w instead of v we get a precession

RATE of (for gamma near one):

precession rate = -w^3*r^2/(2*c^2)

-Bruce

gerbil running around a circular path in flat spacetime, carrying a

gyroscope with you. Now instead of going around a circular path,

substitute a regular polygonal path with a many (large number) sided

polygon.

For simplicity in visualization, assume the polygon is in a vertical

plane and the edge you are moving parallel along is horizontal. You

begin to approach a vertex. It looks like the foot of a slight hill.

The

understanding! You should be able to figure it out from here, but I

will go on. In the laboratory frame the angle of this hill is 360/n

where n is the number of sides. However, because our little gerbil is

running

making the angle steeper.

In our little gerbil's frame, moving from one side of the polygon to

the next, he moves through an angle of theta'. In the lab frame we see

him move through an angle of theta.

Since our polygon has many sides (approaching infinity) the angles are

very small, so in the lab frame tan(theta) ~~ theta ~~ y/x. To our

little gerbil, tan(theta') ~~ theta' ~~ y'/x' = y/(gamma*x). Where

gamma = sqrt(1 - v^2/c^2)

In circumnavigating the polygon Mr. Gerbil rotates through an angle of

n*theta' = n*360/(n*gamma) = 360/gamma. So of course, he sees the

little gyroscope he is carrying with him rotate the same amount in the

opposite direction.

When he gets back to his starting point he stops! He sees that his

gyroscope has precessed (really just stood still, it's the gerbil that

turned) 360/gamma. In the lab frame Mr. Gerbil only circled 360

degrees. Since the gyroscope RECORDED a rotation of 360/gamma, and

since we only circled it 360 degrees, and since it is now stationary,

it must have precessed 360(1/gamma - 1) in the opposite direction in

the lab frame.

For gamma near "1" we have (1/gamma - 1) ~~ (1/2)*v^2/c^2.

If we specify an angular frequency w instead of v we get a precession

RATE of (for gamma near one):

precession rate = -w^3*r^2/(2*c^2)

-Bruce