# Simple gravity puzzle

1. Nov 16, 2005

### pervect

Staff Emeritus
Imagine that we have two masses m1 & m2 connected by a "rigid" but lightweight bar. The masses are attracting each other in essentially a Newtonian way (the masses are not so large as to need GR to calculate the force).

Picture:

m1
|
|<... bar of length d, cross sectional area A
|
m2

Imagine that the bar is oriented so its long direction lies along the z-axis. Then we can write the pressure in the bar

P_z = Force/Area = G m1 m2 / (d^2 A)

Now, imagine an observer comes by at some relativistic velocity v oriented in the x direction, perpendicular to the bar. We can no longer use Newtonian gravity to calculate the forces between the masses in this frame, but by the principles of covariance, we can convert the answers that we know in the rest frame of the bar to the moving frame.

This gives us the following set of questions, which I am going to defer answering for a few days. (I do have a simple answer, however my formulation of the answer requires some familiarity with the stress-energy tensor).

1) What is the pressure, P_z, in the bar in the moving frame?

2) What is the 3-force between the masses (i.e P_z * area) in the moving frame?

3) Can we give a hand-waving type explanation for the above result for the force between the masses?

Extra credit if you can say "oh, that old problem was discussed in such-and-such textbook or paper". (I'm sure it probably has been, but I'm not aware of where.)

2. Nov 23, 2005

### pervect

Staff Emeritus
No takers, so I'll post the solution I had in mind.

1) The pressure, P_z, does not change in the moving frame. There are several ways to see this, the easiest is the transformation properties of the stress-energy tensor, the mathematical construct that models pressure in relativity.

2) The pressure P_z does not change, but the cross-sectional area of the bar decreases by a factor of $\gamma=1/\sqrt{1-(v/c)^2}$. Thus the 3-force also decreases by a factor of $\gamma$. This is a general result for how forces transverse to the direction of motion transform in SR (they appear weaker in a moving frame).

If one is familiar with 4-forces, the 4-force in the transverse direction, $dP/d\tau$, does not change. (P here is the momentum). The 3-force is related to the 4-force by the equation $dP/dt = (dP/d\tau) \, (d\tau/dt)$, so the 3-force does change.

3) This is the interesting part. In the moving frame m1 has more energy (if you prefer, more "relativistic mass"), and m2 has more energy. Gravity in GR couples to energy (or relativistic mass), yet we are saying that the force between two masses has decreased as a result of the motion, not increased!

There are two factors to consider. The first is that Newton's law does not work in the moving frame to calculate the gravitational force. But a slightly more detailed answer can be given as well.

If we consdier that m1 is rapidly moving and passing by a _stationary_ m2, we do find that m1 exerts a larger force on m2 due to its larger energy, though details of how much larger depend on finicky questions about the coordinate system that is used. Also note that the "force" from the moving m1 is not spherically symmetric.

But when m1 and m2 are both moving, something else enters the picture - gravitomagnetism.

The gravitomagnetic forces create a repulsive force between m1 and m2 that more than counterbalances any increase. This is very similar in nature to the magnetic repulsion between two co-moving charges of opposite sign. (The electric charges are of opposite sign so that they attract, equivalent to two masses of the same sign, so that the force points in the same direction).

Last edited: Nov 23, 2005
3. Nov 23, 2005

### HallsofIvy

There is no such thing as a "rigid" bar in relativity- it is theoretically impossible.

4. Nov 24, 2005

### pervect

Staff Emeritus
The bar doesn't have to be perfectly rigid for this problem. If you like, replace it with a spring.

The bar has to be able to measure the force (a spring can do that), and it has to hold the bodies in an equilibrium condition a constant distance a part (a spring can do that, too).

5. Nov 25, 2005

### lightgrav

The Magnetic repulsion of opposite charges with parallel velocities is weaker than their Electrical attraction at v < c . The magnetic repulsion essentially compensates for not including time-dilation.

Why do you think it should work differently with gravity?

By the way, I was expecting these fast travelers to determine the bar's
composition and internal stress spectroscopically, which would give them
the same P as in the rest frame (presuming they can handle optical doppler shifts) ... they'd measure bar width (delta-y) and presume a circular cross-section [how can they measure delta-x?] , getting the same F, too. :shy:

6. Nov 25, 2005

### pervect

Staff Emeritus
It works very similarly for gravity as for electromagnetism, though not quite identically.

In the case of comparing co-moving charges with co-moving masses the similarities between E&m and gravity far outweight the differences. In _both_ cases, the total force must transform in a Lorentz invariant manner. (The only tricky part about doing this in GR is the issue of how to talk about graviy as a "force" - the mechanism of the "bar" or "spring" provides a convenient way about talking about the "force" of gravity).

The gravito-magnetic repulsive force is in fact smaller than the gravitational attractive force, I'm not sure why you think I said it was not. The masses still wind up with a net attraction. The attractive force, however, must scale so that the total force transforms in a Lorentz invariant manner - this is just as true for gravity as it is for E&M.

I'm not sure how you expect the traveller to measure the stress "spectroscopically" (??).

If you aren't familiar with the stress-energy tensor, you can, with more work, model the force due to stress as being due to bouncing particles. You will get the same result as you do with the stress-energy tensor approach.

Imagine the bar filled with particles moving left and right. The following is a "cross" section of the bar

----------------
-> -> -> ->
<- <- <- <-
----------------

The left moving particles transfer a certain amount of momentum P in a certain time to the left edge of the bar, the right moving particles transfer an identical amount of momentum P to the right edge of the bar in the same amount of time.

The amount of momentum P, transfered per unit time, is the force, F = dP/dt

We can calculate the total force F as
(amount of momentum transferred by each particle) * (number of particles / second)

If the momentum per each particle is +P, this will be

2*P*N/t_t

where P is the momentum of each particle (it changes from going left to going right, transfering a total amount of 2P to the wall)

N is the number of particles

t_t is the transit time it takes a particle to move from the left wall to the right.

Now do a Lorentz boost, towards the top of the page, with velocity V.

By time dilation, the particles in the moving frame are all moving more slowly. However, the momentum of each particle remains the same. (This happens because the momentum / velocity ratio of the particles, sometimes called the "relativistic mass", goes up).

The net result is that the rate at which momentum is transfered from the left end of the bar to the right (the force) goes down by a factor of "gamma".

Referring to the formula above, N and P remain constant, but t_t goes up by a factor of gamma in the "boosted" frame.

The number of particles stays fixed in this approach, so we do not have to compensate for the area change to calculate the total force. When we look at the force/unit area, we do need to include the area change, when we do so we find that the force/unit area (i.e. the pressure) stays constant.

There should be no confusion about the area of the bar decreasing, this is very basic SR

The cross section of the bar which appears like this
----
.
.
.
.
----

in the rest frame, get's squished like this

----
.
.
----

in the moving frame. The other dimensions of the bar remain unchanged by the boost, only the dimension of the bar in the direction of the boost is affected by length contraction.

Last edited: Nov 25, 2005