center of mass relation

Why : From the Center of Mass to Barycentric Orbits

In a binary star, the two components orbit their common center of mass.

A standard relation is

(1)

where

• M₁ and M₂ are the stellar masses,
• a₁ is the semimajor axis of star 1 about the center of mass,
• a₂ is the semimajor axis of star 2 about the center of mass.

This equation is the basis of the mass-ratio relation

(2)

At first sight, Equation (1) looks like the familiar one-dimensional center of mass equation.

That intuition is correct, but one additional step is needed for an eccentric orbit.

For a circular binary, the distances from the center of mass are constant and are equal to the orbital radii.

For an eccentric binary, the instantaneous distances vary with time.

The reason Equation (1) still applies to the semimajor axes is that the complete barycentric orbits are constant scaled copies of the same relative Kepler orbit.

This derivation develops that result in three stages:

1. the one-dimensional center of mass,
2. the vector center of mass,
3. the scaling of complete Kepler ellipses.

1 Start with the one-dimensional center of mass

For two point masses on an x axis, the center of mass coordinate is

(3)

Choose the origin at the center of mass:

(1)

Then

(2)

Suppose star 1 lies to the right of the origin and star 2 to the left.

Write

x1	= +r1,	(3)
x2	= −r2,	(4)

where r₁ and r₂ are positive distances from the center of mass.

Substitution gives

(5)

Therefore

(4)

This is the basic balance relation.

2 Physical meaning of the one-dimensional relation

Equation (4) can be rearranged as

(5)

The distances from the center of mass are inversely proportional to the masses.

Therefore:

• the more massive star stays closer to the center of mass,
• the less massive star moves farther from the center of mass,
• equal masses place the center of mass halfway between the stars.

This is the same principle encountered in balances, levers, and two-particle mechanics.

3 Is the binary really a one-dimensional system?

No.

The orbit generally occurs in a two-dimensional orbital plane embedded in three-dimensional space.

The one-dimensional derivation works because, at any instant, the two masses and the center of mass lie on the same line.

One can temporarily choose an x axis along that instantaneous line.

The more fundamental relation is vectorial.

4 The vector center of mass

For two bodies,

(6)

where R is the center of mass position in an arbitrary inertial frame.

Choose a coordinate frame whose origin is at the center of mass.

Then

(6)

Equation (6) becomes

(7)

Therefore

(8)

The minus sign means that the two barycentric position vectors always point in opposite directions.

Taking magnitudes gives

(9)

Thus the one-dimensional result is simply the magnitude form of the full vector center of mass equation.

5 Introduce the relative position vector

Define the relative position

(10)

Let

(7)

From Equation (8),

(8)

Insert this into Equation (10):

r	= r1 + r1	(9)
	= r1	(10)
	= r1.	(11)

Therefore

(11)

Similarly,

(12)

These two equations are the key to understanding the semimajor-axis relation.

6 Why the scale factors are constant

The factors

(12)

and

(13)

contain only the stellar masses.

For an isolated binary whose masses are constant, these scale factors do not change with time.

Thus Equation (11) does not merely say that one instantaneous distance is a fraction of another.

It says that the entire time-dependent orbit of star 1 is a scaled copy of the relative orbit.

Likewise, Equation (12) says that star 2 follows a scaled copy rotated by 180^∘ about the center of mass.

7 Circular-orbit case

For a circular binary, the instantaneous barycentric distances are constant:

r1	= a1,	(14)
r2	= a2.	(15)

Equation (9) immediately becomes

(13)

So for circular motion, Equation (1) follows directly from the one-dimensional center of mass definition.

This is the simplest way to introduce the relation.

8 Why eccentric motion needs one more step

For an eccentric binary,

(16)

and

(17)

These instantaneous distances vary throughout the orbit.

In general,

(18)

and

(19)

Therefore one should not derive the eccentric-orbit semimajor-axis relation by simply replacing the instantaneous distances in Equation (9) with semimajor axes.

Instead, use the constant vector scaling in Equations (11) and (12).

9 Scaling an ellipse

Suppose the relative orbit is an ellipse described by

(20)

If every vector on that orbit is multiplied by a constant positive number c,

(21)

then every linear dimension of the ellipse is multiplied by c.

Therefore:

ac	= ca,	(22)
bc	= cb,	(23)
rp,c	= crp,	(24)
ra,c	= cra.	(25)

The eccentricity remains unchanged because it is dimensionless.

10 Apply the scaling to star 1

Equation (11) has the form

(26)

with

(27)

Therefore the semimajor axis of star 1 is

(14)

where a is the semimajor axis of the relative orbit.

11 Apply the scaling to star 2

Equation (12) contains a minus sign.

The minus sign reverses the direction by 180^∘ but does not change lengths.

Thus the semimajor axis of star 2 is

(15)

Both component orbits have the same eccentricity as the relative orbit.

12 Derive the semimajor-axis mass relation

Multiply Equation (14) by M₁:

(28)

Multiply Equation (15) by M₂:

(29)

The right sides are identical.

Therefore

(16)

This is Equation (15) of BIN01.

The derivation is valid for both circular and elliptical Keplerian binaries.

13 The relative semimajor axis

Add Equations (14) and (15):

a1 + a2	= a + a	(30)
	= a	(31)
	= a.	(32)

Thus

(17)

This is another important binary star relation.

The relative orbit measures the separation of one star from the other.

The barycentric orbits measure each star relative to the center of mass.

Their semimajor axes add to the relative semimajor axis.

14 Mass ratio

Equation (16) can be rearranged:

(33)

Divide by M₂a₁:

(18)

This inverse relation is physically intuitive.

The more massive star has the smaller barycentric orbit.

Define the mass ratio

(19)

Then

(20)

Different fields sometimes define q with the reciprocal convention, so the definition must always be stated.

15 Example 1: unequal masses

Suppose

M1	= 2M⊙,	(34)
M2	= 1M⊙,	(35)
a	= 12 AU.	(36)

The total mass is

(37)

Then

a1	= (12) = 4 AU,	(38)
a2	= (12) = 8 AU.	(39)

Check:

M1a1	= (2)(4) = 8,	(40)
M2a2	= (1)(8) = 8.	(41)

Therefore

(42)

16 Example 2: equal masses

If

(43)

then Equation (18) gives

(44)

Because

(45)

we obtain

(46)

The center of mass lies halfway between the two stars at every instant.

17 Connection to velocities

Differentiate Equation (11):

(21)

where

(47)

Similarly,

(22)

Therefore

(23)

This is simply conservation of momentum in the center of mass frame.

Taking corresponding velocity amplitudes gives another inverse mass-ratio relation.

18 Connection to spectroscopic binaries

For a double-lined spectroscopic binary, the radial-velocity semiamplitudes are

(48)

and

(49)

Because both stars share the same

• orbital period,
• eccentricity,
• inclination,

their radial-velocity amplitudes scale with their barycentric semimajor axes:

(50)

Therefore

(24)

Thus the same center of mass relation appears in both astrometry and spectroscopy.

19 Connection to total dynamical mass

Kepler’s third law gives the total mass:

(25)

The barycentric relation gives the mass ratio:

(26)

Together, total mass and mass ratio determine the individual masses.

Let

(51)

Then

(52)

Thus

(53)

so

(27)

and

(28)

This is why measuring both the total orbit and the barycentric division of that orbit is so powerful.

20 A geometric interpretation

Equation (8),

(54)

can be viewed as a balance of first moments about the center of mass.

The quantity

(55)

is the mass multiplied by its lever-arm distance from the center of mass.

The two first moments have equal magnitude and opposite direction.

This is exactly the same center of mass principle used in elementary mechanics.

Binary star orbital geometry is therefore a direct astronomical application of the familiar balance relation.

21 Why the center of mass is at a focus

Each star follows a Kepler ellipse under the gravitational interaction.

In barycentric coordinates, the common center of mass lies at a focus of each component ellipse.

The two ellipses:

• share the same eccentricity,
• share the same orbital period,
• are oppositely directed about the center of mass,
• have semimajor axes in the inverse ratio of the stellar masses.

The barycenter is not generally the geometric center of either ellipse.

This distinction becomes important when interpreting visual-binary orbit diagrams.

22 When the relation can change

The derivation assumes:

• an isolated two-body system,
• constant component masses,
• positions measured relative to the system center of mass.

If substantial mass is transferred or lost, the mass ratio can evolve.

If a third body is present, the motion can contain additional barycentric structure.

The instantaneous center of mass definition remains valid, but a simple fixed pair of Kepler ellipses may no longer describe the full motion.

23 Common mistakes

1. Calling the binary orbit fundamentally one dimensional.
2. Forgetting the sign difference between the two barycentric position vectors.
3. Replacing instantaneous radii with semimajor axes without explaining why the full orbits are scaled copies.
4. Assuming r_i = a_i at every point of an eccentric orbit.
5. Using a₁ or a₂ as the relative semimajor axis.
6. Forgetting that a = a₁ + a₂.
7. Reversing the mass-ratio relation.
8. Concluding that the more massive star has the larger barycentric orbit.
9. Forgetting that radial-velocity amplitudes carry the same inverse mass-ratio information.
10. Applying the fixed-Kepler-orbit relation without caution to systems undergoing strong mass transfer or third-body perturbations.

24 Practice exercises

1. Starting from the one-dimensional center of mass definition, derive M₁r₁ = M₂r₂.
2. Derive the vector relation M₁r₁ = −M₂r₂ in the barycentric frame.
3. Starting from the relative vector r = r₁ − r₂, derive Equations (11) and (12).
4. Explain why Equations (11) and (12) imply that the component orbits are scaled copies of the relative orbit.
5. Derive a₁ = M₂a∕(M₁ + M₂).
6. Derive M₁a₁ = M₂a₂ for an eccentric binary.
7. A binary has M₁ = 3M_⊙, M₂ = M_⊙, and a = 16 AU. Find a₁ and a₂.
8. An astrometric binary has a₁ = 2 AU and a₂ = 5 AU. Find M₁∕M₂.
9. An SB2 has K₁ = 35 km/s and K₂ = 70 km/s. Find M₁∕M₂.
10. Explain why the center of mass relation alone gives a mass ratio but not the total mass.

25 Summary

The one-dimensional center of mass relation is

(56)

Putting the origin at the center of mass gives

(57)

The full vector form is

(58)

Using the relative coordinate,

(59)

Because these scale factors are constant, the complete barycentric ellipses are scaled copies of the relative ellipse.

Therefore

(60)

and hence

(61)

Thus Equation (15) in BIN01 is fundamentally a center of mass relation extended from instantaneous positions to the semimajor axes of two geometrically similar Kepler orbits.

References

References