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Adding Gravitational Mass?

  1. Jan 28, 2009 #1
    This is just a question out of curiosity.

    When things are traveling at a faster relative speed they are supposed to have greater mass which makes it harder and harder to increase their speed relative to the observer due to if being harder to overcome momentum. (Correct that if it has mistakes).

    Mass and gravitational pull are related I thought.

    So does this greater mass due to higher relative movement also present a stronger gravitational pull?
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  3. Jan 28, 2009 #2


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    Hi gonegahgah ! :smile:

    Yes, higher relative movement presents a stronger gravitational pull (as measured by the stationary observer).

    Everything else you say is correct too, except:
    "overcome momentum"?

    No, momentum is just as easy to increase at any speed.

    Rather, it's that the same increase in momentum produces less and less increase in speed. :smile:
  4. Jan 28, 2009 #3


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    Could you explain exactly what you mean by this? I think you are wrong, but I may be misunderstanding what you mean.

    Say a test particle going .99c relative to the earth passes through the earth's gravitational field. The curvature of the test particle's path is the same whether it is the earth that is stationary and the particle moving at .99c or if it is the particle that begins stationary and the earth is moving at .99c. So in what sense does the gravitational pull increase when the earth is moving at .99c?
  5. Jan 28, 2009 #4
    Replaced by post #5.
    Last edited: Jan 28, 2009
  6. Jan 28, 2009 #5
    My apologies Tim. Rather than momentum, I meant that because of the greater mass - when having a higher relative speed to the observer - it requires greater force to accelerate it faster relative to the observer. Momentum = mass x velocity so as velocity increases the mass increases, rather than remaining constant, so the momentum increase is non-linear with velocity. Is that correct? Maybe I should have referred to inertia instead of momentum?

    Back to the other question... Dale you are saying no I think and Tim you are saying yes I think at the present. So I am still curious to know which is right.

    Newton's formula is:
    F = G(m1 x m2)/r2

    So under this the increase in mass due to higher relative velocity would suggest a higher gravitational force between the two masses.

    Even in GR the increase in mass would cause a greater curvature in the space-time causing the objects to deviate further.

    So there-in lies my question. Does the increase in mass due to increase in relative speed (given as the reason for the light speed limit) also translate to a corresponding increase in the gravitational effect (ie. increased force (Newton) or greater space-time curvature (GR))?
  7. Jan 28, 2009 #6


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    But that's the same situation being observed in two different ways …*for the question to be meaningful, we need different situations being observed in the same way :wink:

    I automatically assumed that the original question envisaged both the particle and the Earth being observed by an independent inertial goldfish …

    if the goldfish sees the test particle in two identical situations, with the Earth having the same (instantaneous) positions but different velocities,

    then the gravitational force on the test particle, as measured by the goldfish, will be greater when the Earth's speed is greater. :smile:

    Same if the Earth is rotating … extra energy = extra mass = extra gravity, from the equivalence of gravitational mass and inertial mass. :wink:
    Yes … "inertia" would have cured it. :biggrin:
    Yes, all correct … from the equivalence of gravitational mass and inertial mass, extra mass (even extra "relativistic" mass) = extra gravity. :smile:
  8. Jan 28, 2009 #7


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    There are two problems with this. The first is that modern physicists don't use the concept of relativistic mass. It is redundant, just another name for energy, and frame variant. Instead, the word "mass" is generally taken to mean the "invariant mass" or "rest mass", which is different from energy.

    The second, and more important reason, is that regardless of the name you use both rest mass and relativistic mass are scalars (one component) whereas the source of gravity in GR is the stress-energy tensor (16 components). Obviously, a scalar cannot equal a tensor. For a typical planet at rest the energy is predominantely in the mass and only the time-time component of the stress energy tensor is significant, so you can use the Newtonian approximation and your usual reasoning. However, as you examine the planet in reference frames where it is moving relativistically then the momentum components become significant also. You cannot ignore them, and you certainly cannot use the Newtonian approximation. Basically, you need to determine, not only how the energy (relativistic mass) warps spacetime, but also how the momentum does also.
  9. Jan 28, 2009 #8
    Thanks Tim. Very straight forward answers; though Dale still disagrees does he?

    Thanks Dale for your answers too (though not really getting to a quantifiable conclusion).

    So how much does the momentum warp space-time? Does it add, subtract, cancel?

    And does it play its corresponding part in setting the light speed limit with equivalence?
  10. Jan 28, 2009 #9


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    Can you justify this claim at all? I think this is wrong.

    Scenario A) planet, goldfish, and test particle all at rest wrt each other. (How does goldfish measure the force on the test particle?)

    Scenario B) planet moving relativistically, goldfish and test particle at rest. If I understand your argument, you are saying that the planet has more energy and therefore more relativistic mass and therefore exerts a greater force on the test particle.

    Scenario B') planet stationary, goldfish and test particle both moving relativistically with the opposite velocity as the planet in B. Here the planet has no extra energy so there is clearly no extra force. My argument is that since there is clearly no extra force in B' then, by the principle of relativity, there must be no extra force in B (unless the method you have in mind of measuring force is frame-dependent).

    No, most definitely not. If relativistic mass gave extra gravity than a sufficiently fast object of any mass would form a black hole, and http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html" [Broken].
    Last edited by a moderator: May 3, 2017
  11. Jan 29, 2009 #10
    Cool. Thanks Dale (as long as that is the prevailing view).

    It was just originally a question out of curiosity but it does lead me to ask:

    So it is okay to treat it as increased mass when giving that as a reason why it takes more force to accelerate it relatively faster (and why it can't reach the speed of light) but it is not okay to treat it as increased mass as far as gravity goes?
  12. Jan 29, 2009 #11


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    Hi DaleSpam! :smile:
    He plots its acceleration, and multiplies by its mass. :wink:
    Yes, I say that the goldfish measures a greater acceleration.

    There is an extra force in B without an extra force in B' (in which the planet is at rest, while the goldfish and test particle move relativistically) …

    the forces are related by the standard Lorentz equations for force, so the gravitational force from a moving mass changes in the same way as the electromagnetic force from a moving charge.
    No, as I read that link, John Baez is disagreeing with you … increased gravitational mass due to relative motion cannot be plugged into the 2GM/c2 formula for an event horizon to prove a physical contradiction. :wink:
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  13. Jan 29, 2009 #12


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    There are no forces in GR.
    The Lorentz transformation only works in GR between freely falling frames with constant relative velocity ( ie almost none ).

    If you are talking about forces you must be using a Newtonian approximation with SR added. That does not work, as others are telling you.
  14. Jan 29, 2009 #13


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    Hi Mentz114! :smile:
    I'm using a linear approximation in which acceleration (and therefore force) can be measured in the pre-GR way.

    The linear approximations with different local "zero" velocities are related locally by the Lorentz transformation … aren't they?

    The test particle, or the goldfish, don't need all the paraphernalia of GR to work out the distance and speed of the planet.

    They can work out that a moving (pure electric) charge has an electric force increased in the transverse directions by 1/√(1 - v2/c2) (and also a small transverse magnetic force, of course, though no increase of force in the longitudinal direction)

    and that a moving mass has a gravitational force increased in the transverse directions by 1/√(1 - v2/c2) also.

    A rotating star will have higher gravitational attraction than a non-rotating star of the same rest-mass for the same reason.
  15. Jan 29, 2009 #14


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    Hi Tiny,
    first what I said earlier about the LT in GR isn't right. Red face.

    I understand you're relating to weak field approximations. That's OK as long as nothing behaves relativistically. So the approximations may fail if your planet moves too fast. I'm not certain but I'd be concerned.

    The beauty of the LT in electrodynamics is that it ensures that all inertial observers see the charged particle going through the same hoops, so to speak. The magnetic fields that may be infered by a moving observer are just there to ensure this.

    The test of your treatment is this - will all observers see compatible things ?
  16. Jan 29, 2009 #15
    Seems like a good question!
  17. Jan 29, 2009 #16


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    It is probably not OK to give increased mass as a reason why the body will not reach c. But even if you do, it would still be wrong to attribute an increase in gravitational strength to an increase in mass because momentum gravitates and this might explain things without resorting a mass increase.

    Newtonian gravity does not mix well with relativity until it's been cooked into GR. Technically Newtonian gravity is a weak field approximation of GR where nothing is moving relativistically - an assumption this treatment breaks.
  18. Jan 29, 2009 #17


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    The norm of the electromagnetic http://en.wikipedia.org/wiki/Four-force" [Broken] on a moving charge is the same in both frames. Each frame will disagree about if the source of the force is due to the electric field or the magnetic field, but they will agree on all experimentally measurable values including the force.

    Similarly here, both frames should agree on any kind of frame-invariant measurement of "force" although some will attribute it to the energy of the planet and others will attribute it to the momentum of the planet.

    I agree, the increased energy cannot be plugged into 2GM/c², nor can you plug it into a=GM/r². You have to consider the momentum terms of the stress-energy tensor also, which you are ignoring. Baez mentions this when he says

    "gravity does not only couple to mass as it does in the newtonian theory of gravity. Gravity also couples to momentum and momentum flow"

    The Newtonian approximations you keep thinking about and refering to are simply not relevant when you have a relativistically moving mass.
    Last edited by a moderator: May 3, 2017
  19. Jan 29, 2009 #18
    That was the reason that was given I thought? If not then what is the hurdle to a mass reaching the speed of c or greater (apart from loads of fuel)? I was lead to believe that the non-linear increase in mass as the velocity approached c was attributable directly to the increase in inertia. An increase in inertia due to the greater mass meant that a greater force was required to accelerate the greater mass further. Approaching c meant that mass approached - I assume infinity - so that we approached the requirement of infinite force (an impossibility) to push that last increase to the speed of c. So what is the real barrier then?

    That's nice but there just happens to be a corresponding increase in mass as the momentum increases so why would we not look for some sort of equivalence around the subject of mass increases?
  20. Jan 30, 2009 #19


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    If you use Newton's a=F/m to explain what you see when someone accelerates away from you, then it's easy to say that what you are seeing is an increase in m or a decrease in F. Either of these will account for a decrease in a. This fine, but there's no reason to choose one over the other. It's formula juggling - with the wrong formula !

    Newton+SR is supreceded by GR, so GR must be used to investigate these things.

    The actual, real, underlying, physical ( call it what you will) mechanism is unknown. Even using the word 'mechanism' is probably wrong here.
  21. Jan 30, 2009 #20
    Okay thanks. That's what I wanted - a result - whatever it was; hopefully it is in agreement with the general prevailing view.

    So there is no understanding of why it happens only that it is observed to happen. (It was a brave man to suggest it before we could work out how to make observations of it). That is what I get from your last post.

    Interestingly, if the observations are correct, it means that if something is moving at 9.999999c away from you and falls into a gravity well at a distance it can not increase its speed at the local acceleration rate (due to the curvature) relative to you despite the space-time curvature. So it will fall into the gravity well no significantly faster than when it was not in the gravity well.

    So two objects falling into the gravity well at a distance - one moving at a low relative speed to you and the other at 9.999999c - will appear to accelerate at different rates with the faster object not following the space-time curvature correctly and falling with less acceleration; while the slow object falls at the expected acceleration rate.

    That has to hold true given what we are told does it not?
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