# Gravity exerted by a fast moving object versus stationary object?

1. Apr 25, 2013

### bcrelling

Take two scenarios:

1) A 2kg mass at rest

2) A 1kg mass accelerated to a speed such that its relatavistic mass is 2kg (0.866C)

Which excerts more gravity?

Now the obvoious answer is that they excert the same gravity as they have the same relativistic mass. The reason I ask, is that time dilation is affected in an inverse relationship to mass dilation. If gravity is affected by time dilation(i.e. gravity waves emitted at a lower rate) this would exactly counter the increased gravity due to the mass dilation resulting in no net change.

2. Apr 25, 2013

### pervect

Staff Emeritus
IT's unclear how one defines "more gravity". Note that the field of the moving source is NOT spherically symmetric.

If one accepts the definitions in Olson, D.W.; Guarino, R. C. (1985). "Measuring the active gravitational mass of a moving object", the moving mass generates "more gravity" than the stationary one, by roughly a factor of 2.

From the abstract:
Other approaches to "quantifying" the "amount of gravity" give different results.

3. Apr 25, 2013

### bcrelling

Thanks, yeah it makes sense that it's not spherically symmetrical as there's length contraction at play too- I guess the gravitational field would be a somewhat squashed in the direction of travel.

The scenario could be simplified if the mass was considered to be orbiting a much larger one at 0.866C, then at least the distance, angle and acceleration are unchanging.

4. Apr 25, 2013

### Bill_K

The Olson-Guarino paper is available online here. It looks pretty believable.

5. Apr 26, 2013

### bcrelling

Thanks man, it'll take me a while to digest it all.

BTW, I think I stumbled upon an aditional proof that gravitaty does increase for moving objects.
Consider the perihelion orbit of Mercury, its trajectory can be explaned that as mercury nears the Sun, its velocity increases and hence is mass and gravity also increase. This would cause a sling shot effect putting the eliptical orbit on a new trajector every time it passes.

6. Apr 26, 2013

### Mentz114

The orbit of Mercury is exactly solved in GR, and it comes out in terms of Weierstrass's elliptic function $\wp$.

The solution is here http://128.84.158.119/abs/astro-ph/0305181v3.

7. Apr 26, 2013

### Bill_K

I'd be very cautious about "explanations" like this. The mathematics has the final word, and does not lead easily to such a simplistic interpretation. The advance of the perihelion seems to be adequately explained by the geometry surrounding the central mass rather than the properties of the particle orbiting it.

8. Apr 26, 2013

### pervect

Staff Emeritus
I've seen this claim before, and it baffles me. Where did you read it?

Working things out for myself, I get a totally different answer.

F = GmM/r^2 = mv^2/r

If we substitute the "relativistic mass" blindly in for m on both sides of the equation, we conclude that nothing happens!

This is obvious and sensible - it says that things fall at the same rate, regardlelss of mass. If the mass of our particle changes with its velocity, it doesn't matter as long as our quasi-Newtonian-made-up-on-the-spot "gravitational mass" matches our quasi-Newtonian-made-up-on-the-spot "inertial mass".

If we substitute it on one side, and not the other, we are violating the conservation of momentum, the principle that every action has an equal and opposite reaction.

I don't think this even turns out to correctly predict the magniutde of the precession even if we take it seriously, and it's really ugly. As well as poorly motivated.

As far as the GR explanation goes, the majority of the precession can be explained by the PPN parameter gamma, which as other posters have remarked is due to the distortion of space.

There is also an affect from the PPN parameter beta, this effect actually goes in the opposite direction from the gamma effect.

This makes precession a more complex topic than light bending, or the Shapiro effect, both of which depend only on $\gamma$ and not $\beta$

I.e. from MTW's gravitation, pg 1110

$$\delta \phi_0 = \frac {\left( 2 - \beta + 2\gamma \right) }{3} \frac {6 \pi M_{sun}}{a \left( 1 - e^2 \right) }$$

Here $\beta = \gamma = 1$ are PPN parameters
$M_{sun}$ is the mass of the sun
a is the semi-major axis of the orbit
e is the eccentricity.
$\delta \phi_0$ is the perihelion shift.

So we see that $\gamma$ over-explains the precession, and $\beta$ fights this over-explanation, giving the right answer.

$\gamma$ models spatially curvature. $\beta$ is a second order term in the expression for gravitational time dilation, i.e.

$g_{00} = (1 - 2M/r + 2 \beta M^2 / r^2 )$

It might be instructive to sketch how we actually find the orbits in GR:

We start with the metric in the equatorial plane (we can use the whole metric if we want, but we don't need the non-equatorial terms, it's slightly simpler without them).

$$ds^2 = -f(r)\, dt^2 + g(r)\, dr^2 + h(r)\, d\phi^2$$

We can work it out in a couple of different coordinate systems, the PPN system uses

$$f = c^2 \left( 1 - \frac{2GM}{c^2 r} + \frac{2 G^2 M^2}{c^4 r^2} \right) \quad g = \left( 1 + \frac{2GM}{c^2 r} \right) \quad h = r^2 \left( 1 + \frac{2GM}{c^2 r} \right)$$

standard Schwarzschild is

$$f = c^2(1 -\frac{2 G M}{c^2 r}) \quad g = 1 / (1 -\frac{2 G M}{c^2 r}) \quad h = r^2$$

In either case, we apply the geodesic equations, http://en.wikipedia.org/wiki/Solving_the_geodesic_equations

The radial term gives us:

$$\frac{d^2r}{d \tau^2} + \Gamma^r{}_{tt} \left( \frac{dt}{d \tau} \right) ^2 + \Gamma^r{}_{rr} \left( \frac{d r}{d \tau} \right)^2 + \Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$

We need two more equations (this is one of three geodesic equations we need to solve, the one that's formally similar to the Newtonian radial force equation.)

The funky-looking Chrsitoffel symbols are well defined in the literature - they're a pain to compute by hand, but you can compute them directly from the metric coefficeints.

In particular
$$\Gamma^r{}_{tt} = \frac{(\frac{df}{dr})}{ 2g} \quad \Gamma^r{}_{rr} = \frac{(\frac{dg}{dr})}{ 2g} \quad \Gamma^r{}_{\phi \phi} = - \frac{(\frac{dh}{dr})}{ 2g}$$

Last edited: Apr 27, 2013
9. Apr 26, 2013

### Trenton

The paper is brilliant and I shall read it more until I fully grasp it. But it prompted a question which I hope will not sound too silly.

We all assume that gravity travels at C. I don't doubt this but ever since learning of the relationship between the speed of light and the permittivity and permeability of vacuum, I have been wondering what the gravitational equivelent might be.

To clarify the above, I can see why an electromagnetic wave would travel at a speed set by the permittivity and permeability. But I can't see why a gravitational wave which obtensively has nothing to do with permittivity and permeability, would travel at the same speed.

There must be a link somewhere. What is it?

10. Apr 27, 2013

### Agerhell

Classically you have:

$$\frac{d}{dt}(m\bar{v})=-\frac{GMm}{r^2}\hat{r}$$

using the "relativistic mass" from special relativity instead of a constant mass on both sides of the equation above and solving for $d\bar{v}/dt$ gives:

$$\frac{{\rm d}\bar{v}}{{\rm d}t}=-\frac{GM}{r^2}(\hat{r}\cdot\hat{v})\left(1-\frac{v^2}{c^2}\right)\hat{v}+\frac{GM}{r^2}(\hat{r}\times\hat{v}) \times \hat{v}$$

However, this new expression will only be able to explain ont third of the so called "anomalous perihelion shift". It is possible to let the mass of the orbiting body vary not only with the velocity, but also with the position within the gravitational field, and get the perihelium precession right using the gravitational force as formulated by Newton but that would be inventing new physical laws.

11. Apr 27, 2013

### Mentz114

The link is that any massless energy propagation will travel at c. c is the fundamental constant, not permeability nor permittivity.

12. Apr 27, 2013

### Mentz114

Interesting, but unnecessary, given that GR gives the correct answer in the weak field and the exact models.

( I like 'perihelium' for 'perihelion' ).

13. Apr 27, 2013

### dipole

I think "relativistic mass" is a concept which should be avoided. It's better to think in terms of energy and to understand that energy is a source of gravitation, so an object moving with a lot of kinetic energy is going to have a stronger gravitational field.

I once did a calculation like the one you're describing in the weak-field limit and I believe the answer is that the gravitational field is directional and is stronger in the forward direction, but I'd have to dig up my old HW to be sure.

14. Apr 27, 2013

### pervect

Staff Emeritus
That's very interesting - I can see where substituting relativistic mass on both sides is wrong. But I have the feeling I'm still missing something :-(. I shall have to think on it, but I can explain the problem.

The expression from MTW suggests that we should have 2/3 the perihelion shift with beta=gamma=0. You suggest that only 1/3 of it comes from what we've been calling the slingshot effect. So either there's still an error in the formulation of the effect (possibly relating to the relativistic / transverse mass formulation), or there's ANOTHER factor of 1/3 out there. Or my textbook reference has a typo, perhaps, but that seems like a low probability.

15. Apr 27, 2013

### Trenton

Taking the above quote as a good starting point we can rewrite the problem by reversing the reference frame. The fast moving mass becomes stationary (but retains it relativistic mass) and the test particle is moving at near light speed. Thinking of it this way makes me rather doubtful that the effect of gravity on the test particle is doubled. The paper draws a parallel between this doubling and the 'famous factor of 2' betwen the Newtonian and GR values for light deflection. I can't be sure but I am getting the feeling a slight of hand has been pulled.

16. Apr 27, 2013

### Bill_K

Yes, I think this is the same factor of 2. It's not really the difference between Newton and GR, it's the difference between scalar gravity and tensor gravity. A particle sitting still feels only the scalar (Newtonian) potential. A moving particle feels also the vector component (~v) and the tensor component (~v2). Hence the 1 + β2 factor.

17. Apr 27, 2013

### Trenton

Bill_k I think you could be right there, the factor of 2 is the difference between scalar and tensor gravity so the paper's claim would appear to be valid. When either the large mass or the test particle have relativistic speed relative to each other, the factor of 2 will be approached.

GR never seems to lose it's capacity to confuse! I noted that the paper pointed out that while in theory a massive object could aqquire enough relativistic mass to become a black hole, if it were to do so it would be a black hole in all frames of reference. Whilst I would have to agree with that I find the paper a reminded of the care one must take choosing frames of reference.

In my post where I swapped the frames, I suggested that the large mass should be treated as still possesing it's relativistic mass. This is quite a horrible step as it has overtones of absoluteness!

18. Apr 27, 2013

### pervect

Staff Emeritus
OK, the equations of motion one gets when substiting $\beta = \gamma = 0$ leads to a metric like:

$$-c^2 (1-2GM/c^2r) dt^2 + dr^2 + r^2 d \phi^2 = -c^2 d \tau^2$$

Setting G = c = 1 through an appropriate units choice (this is known as geometric units) we get the geometric equations of motion

$$\left( 1 - 2M/r \right) \frac{dt}{d\tau} = E$$
$$r^2 \frac{d\phi}{d\tau} = L$$

where E and L are "costants of motion"

We have one final equation needed to compute the orbits:

$$\left( \frac{dr}{d\tau} \right)^2 = \frac{E^2}{1-2M/r} - 1- \frac{L^2}{r^2}$$

It's reasonably obvious that this is different than the equations of motion produced by the "relativistic mass" concept, even after one throws all the units back in (or takes them out of the relativistic mass eq's to compare).

The details of making the "relativistic mass" concept give the same answers will be left to the reader who actually uses the concept (I'm not one of them), but I'll note for starters that the concept of unifying gravitational time dilation with SR's time dilation has not been addressed at all via the "relativistic mass" approach, and it's something that needs to be considered.

The approach above does define it implicitly through the metric and could be used as a guide for anyone interested enough to proceed to find the differences between the two approaches.

I will add that using the relativistic mass approach basically adds work when attempting to understand the problem - it doesn't appear to "make things easier".

Also, for my purposes I'll count anything that doesn't give the same results as standard textbooks as "wrong" rather than as "some diferent theory of gravitation" - as I am asuming that the basic idea is to understand General Relativity "as currently practiced by professionals" rather than to ome up with one's own personal theory that hasn't been put to experimental test.

Last edited: Apr 27, 2013
19. Apr 27, 2013

### Trenton

I shall consider this further and you are right, unification of the two time dilations needs to be addressed. I would say also that for the most part I too regard anything that does not give the same results as the textbooks as 'wrong' and have little time for exotic thinking such as MOND. So far as I have learned GR I find satisfaction when I finally am able to agree with the textbooks and remain troubled when I can't. That said it is easy to get the wrong end of the stick when learning GR and then write something that appears to support an exotic theory.

20. Apr 30, 2013

### Trenton

Whilst on the subject of relativistic mass, how does matter having relativistic mass as a result of being in a gravity well fit in? If one is standing on the surface of a non collapsed dense object one's clocks are dilated to the same extent that they would be if one were travelling through space at the escape velocity at said surface. If your clocks are dilated then you must also have aquired relativistic mass - Unless I am hopelsssly mistaken.

This is a miniscule factor if one is standing on Earth, rather more noticible on a white dwarf and absolutely critical on a neutron start approaching the Tolman-Oppenheimer-Volkoff limit. The thing that baffles me (and I am looking for the text book answer here), is how to resolve the apparent creation of mass/energy that gravitational fields seem to present.

The total energy of a 'system' should remain constant unless energy is put in or taken out and would be measured by the total mass + energy in the form of potential and kinetic energy etc. To make things simple assume we are examining a system cold enough for temperature to be neglected.

If the system consisted of a single body you just measure its mass to give the mass/energy value. But if it consisted of two identical bodies the value would not be 2M but 2M + the PE between them and any KE. There is then an mass/energy value that depends on spatial separation ie on space. But the PE and so the total mass/energy is much greater if the system has the capacity to collapse to much smaller distances. In the case of collapse to a BH this mass/enegy 'creation' seems to run away with itself.

I have heard that this assesment is wrong and I certainly feel this is wrong. I have been given examples such as 'throwing 1Kg at a black hole can only add 1Kg to the mass of the BH'

Can someone state what the proper text book explanation, accepted by the leading academics on GR, as to what is going on here?