Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravity exerted by a fast moving object versus stationary object?

  1. Apr 25, 2013 #1
    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. jcsd
  3. Apr 25, 2013 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  4. Apr 25, 2013 #3
    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.
     
  5. Apr 25, 2013 #4

    Bill_K

    User Avatar
    Science Advisor

    The Olson-Guarino paper is available online here. It looks pretty believable. :approve:
     
  6. Apr 26, 2013 #5
    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.
     
  7. Apr 26, 2013 #6

    Mentz114

    User Avatar
    Gold Member

    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.
     
  8. Apr 26, 2013 #7

    Bill_K

    User Avatar
    Science Advisor

    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.
     
  9. Apr 26, 2013 #8

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.

    Start with Newton's equations

    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 [itex]\gamma[/itex] and not [itex]\beta[/itex]

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

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

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

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

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

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

    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).

    [tex]
    ds^2 = -f(r)\, dt^2 + g(r)\, dr^2 + h(r)\, d\phi^2
    [/tex]

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

    [tex]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)
    [/tex]

    standard Schwarzschild is

    [tex]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[/tex]

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

    The radial term gives us:

    [tex]
    \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
    [/tex]

    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
    [tex]\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} [/tex]
     
    Last edited: Apr 27, 2013
  10. Apr 26, 2013 #9
    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?
     
  11. Apr 27, 2013 #10
    Classically you have:

    [tex]\frac{d}{dt}(m\bar{v})=-\frac{GMm}{r^2}\hat{r}[/tex]

    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:

    [tex]\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}[/tex]

    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.
     
  12. Apr 27, 2013 #11

    Mentz114

    User Avatar
    Gold Member

    The link is that any massless energy propagation will travel at c. c is the fundamental constant, not permeability nor permittivity.
     
  13. Apr 27, 2013 #12

    Mentz114

    User Avatar
    Gold Member

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

    ( I like 'perihelium' for 'perihelion' ).
     
  14. Apr 27, 2013 #13
    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.
     
  15. Apr 27, 2013 #14

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  16. Apr 27, 2013 #15
    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.
     
  17. Apr 27, 2013 #16

    Bill_K

    User Avatar
    Science Advisor

    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.
     
  18. Apr 27, 2013 #17
    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!
     
  19. Apr 27, 2013 #18

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    OK, the equations of motion one gets when substiting [itex]\beta = \gamma = 0[/itex] leads to a metric like:

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


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

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

    where E and L are "costants of motion"

    We have one final equation needed to compute the orbits:

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

    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
  20. Apr 27, 2013 #19
    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.
     
  21. Apr 30, 2013 #20
    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?
     
  22. Apr 30, 2013 #21

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I would say the types of references to look at are for ADM mass and Bondi mass. The former is conserved in (and only defined for) asymptotically flat spacetimes. For these, nothing about 'mass measured from infinity' changes no matter the dynamics of objects coalesing or BH mergers. Meanwhile, Bondi mass decreases for BH mergers or BH + star merger. This reflects that Bondi mass excludes the GW radiated to infinity. Thus, a 1kg mass (measured 'far' from a BH), when absorbed by a BH, will generally add less than 1kg to the BH mass due to essentialy inevitable GW.
     
  23. Apr 30, 2013 #22

    Bill_K

    User Avatar
    Science Advisor

    Well guess what, in GR there is nothing complicated going on, it works exactly the same way it does in the Newtonian physics which you're already familiar with (hopefully!) When the bodies are far apart, the PE and KE are both zero and the total energy of the system is 2M. As they fall together the KE becomes positive, the PE becomes negative, while the sum remains the same. For example for a test particle falling into a Schwarzschild field, E = energy/mass is one of the conserved quantities attached to the geodesic.
     
  24. Apr 30, 2013 #23

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not sure if it will help anyone, but I'll give one other approach for getting the correct/accepted equations of motion, without direclty mentioning the geodesic equations.

    You start out with the Lagrangian formulation of special relativity. In free space, this is just [itex]L = \int d \tau = \int \sqrt{1-v^2/c^2} dt [/itex].

    You use variational principles and the Euler-Lagrange equations as usual to find the equations of motion.

    [add]To spell this out more:

    You define the Lagrangian L

    [tex]
    L(t, r, \phi, \dot{r}, \dot{\phi})[/tex]
    where
    [tex]\dot{r} = \frac{dr}{dt} \quad \dot{\phi} = \frac{d \phi}{dt}[/tex]

    Then you use Lagrange's equations, which are the solution from extremizing the Lagrangian:

    [tex]\frac{\partial L}{\partial r} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = 0 \quad \frac{\partial L}{\partial \phi} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right) = 0 [/tex]


    Note that one way of putting this is that SR extremizes proper time.

    How to incorporate gravity into the Lagrangian isn't at all obvious. But the GR inspired approach is simple: you don't really need to do anything. GR extremizes proper time too! So you don't even need to add a potential function to the free-space Lagrangian.

    What you do need is to be able to compute proper time, you unfortunately can't extremize it without being able to compute it. The obvious way to do this is via a metric. Various possible metrics of interest have been discussed - there's the exact solution that general relativity predicts, of course, the "Newtonian" metric which is the PPN metric with beta = gamma = 0, and the PPN metric with beta = gamma = 1.

    Of course, extremizing the proper time turns out to lead directly to the same geodesic equations presented earlier. But this is to be expected, it's just a different route to getting to the same end mathematical result. The appeal is for people who have some familiarity with Lagrangian mechanics and less familiarity with geodesics.

    I'm afraid I don't have any good "motivators" for people without Lagrangian mechanics - other than to highly recommend learning it, it makes even Newtonian mechanics a lot more error free, and it's a gateway into more advanced realms of physics as well.
     
    Last edited: Apr 30, 2013
  25. May 1, 2013 #24
    I am familiar with the convention that PE is negative but this did not get me out of the mess I got myself in. A system with it's mass spread out (eg a spherical cloud of gas) has more energy than a system of identical mass but more clumped together (eg a more compacted spherical cloud of gas) - because the latter system would require energy input to lift out the gas to covert it to the state of the former system.

    And if it has more energy then it has more mass!.

    I got into trouble when I tried to calculate how much extra energy there was in the former system because this depended on how compact one chose to make the latter. If the latter was very dense eg a nuetron star the energy advantage of the former was so great that it would add serveral percent to the mass. Infinitely worse if the latter system was a BH singularity. Clearly this is rediculous since the mass of the gas is what it is and can't depend on how compact it is now and certainly not on how compact it might become in the future!

    The convention that PE is negative must have a deeper meaning that I am currently grasping.

    I shall read up on ADM and bondi as this looks promising.
     
  26. May 1, 2013 #25

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Have you considered that a collapsed cloud is much hotter, initially, until it radiates and cools? Alternatively, that a cloud is radiating energy as it collapses? In short, a sufficiently hot collapsed cloud will have the same mass measured at infinity, as a larger, cooler cloud; and that the difference in energy between a collapsed versus large cloud of the same temperature lies precisely in the energy that needs to radiate away to achieve same temperature in the collapsed cloud.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook