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You see a galaxy far away. That galaxy is attracting you with a certain amount of gravity. I'm wondering if the gravity influence of the galaxy on you, as measured by you, always ends up being what you see and measure from your frame. For examples consider the cases below:

  1. If that galaxy is moving towards you with relativistic speed, you see it blue-shifted and therefore you perceive it as having more energy. Do you feel more gravity from it proportional to the blue-shift?

  2. Now you start moving towards it instead of it towards you with the same speed. You see it as blue-shifted but also at a shorter distance due to length contraction. Do you feel more gravity from both these effects proportional to the blue-shift + the effects of the reduced distance?

  3. Now you are in a strong gravitational field where time is flowing at half the rate of most of the space between you and the galaxy. You have no relative velocity with respect to the galaxy. You "see" the galaxy as being at about half the distance an observer far in space would measure because you would time a 2 way light signal to return in half the time due to your slower clock. To you the galaxy is about half as many light years away. On top of that shorter distance, the galaxy is blue-shifted purely due to the difference in clock rates. Would you feel an increase in gravity from both these effects combined?

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4 Answers 4

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Unfortunately the whole premise of your questions is wrong. You seem to be trying to apply Newtonian gravity in a relativistic context, and that just doesn't work. You need general relativity to describe gravity in those situations.

In GR the curvature of spacetime is described by a tensor. All observers will agree on the geometry, but they use different coordinates to describe it. The physical results of any experiment are the same for everyone (and everyone will agree on the predictions for such experiments, even though they use different coordinates). So in analyzing any situation you can use whatever reference frame is most convenient.

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  • $\begingroup$ Is the premise wrong or simply a non standard way of looking at the problem? No part of the question fundamentally implies disagreement with any result you would get using GR. If an observer wants to calculate how strongly an object is affecting him gravitationally at any given moment, could he simply calculate the energy/mass and distance he perceives the object to be in his frame? If this would fail I'm looking to understand why. $\endgroup$
    – Zach
    Commented Jul 4 at 16:53
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    $\begingroup$ Gravity isn't a force in GR, and objects don't attract each other, they bend spacetime. This bending isn't caused by mass or energy alone, it's from the stress-energy-momentum tensor (in which the momentum terms loosely speaking cancel out kinetic energy terms). You can do the GR calculation in any frame you like, the results will be the same. $\endgroup$
    – Eric Smith
    Commented Jul 4 at 17:07
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    $\begingroup$ In the case of the example you gave, the calculation is simplest in the frame in which the galaxy is at rest (because its influence on spacetime is far more significant than the rocket's). If you want to use a frame in which the rocket is at rest and the galaxy is moving you can of course do so, but you cannot ignore the momentum terms of the stress energy tensor. If you do the calculations carefully you'll get the same answer either way. $\endgroup$
    – Eric Smith
    Commented Jul 4 at 17:14
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    $\begingroup$ @EricSmith Then what does the "same answer" look like? What observable effects would the observer expect, in each of the three scenarios that the OP mentioned? $\endgroup$
    – wzkchem5
    Commented Jul 5 at 14:51
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    $\begingroup$ @wzkchem5 Scenarios 1 and 2 appear to be the same (rocket and galaxy moving towards one another). The rocket's contribution to the situation is negligible so you might as well treat the galaxy as stationary. In that frame the rocket is attracted to the galaxy in roughly the same way it would be in ordinary Newtonian gravity. For the third scenario, if you're in a gravitational potential that causes that much time dilation then for all practical purposes that's the only gravity you'll feel, the distant galaxy may be ignored. $\endgroup$
    – Eric Smith
    Commented 2 days ago
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An easy way to see why this would not work: if you are moving rapidly, aberration makes most of the mass content of the Universe appear to lie in front of you. If gravity were based on apparent positions, the pull from the rest of the Universe would accelerate you in the direction of your motion. That acceleration would further aberrate the mass distribution, resulting in further acceleration. Objects within the Universe would thus tend to accelerate continuously without limit.

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  • $\begingroup$ It isn't obvious to me that that would be a problem. Aberration and other effects would only be considerable at extreme speeds. In the "rest frame" of the rest of the universe, that traveler's acceleration to c would take infinite time. In the traveler's frame as he reaches c, there is no longer matter in front of him accelerating him passed speed c because space is infinitely contracted. But realistically he would likely get obliterates by the blue-shifted radiation facing him. $\endgroup$
    – Zach
    Commented 5 hours ago
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The answer to the question in the title is "No." For a mass moving with constant velocity, the Newtonian approximation for gravity points toward the current position of the mass, not where it was when the photons you are viewing left it. This holds to second order; if the mass is under a constant acceleration, you are still attracted to where it is, not to where it was.

This is important in orbital mechanics. Jupiter's orbit around the Sun wouldn't be stable if they each were attracted to where the other one had been 40 minutes earlier.

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  • $\begingroup$ You seem to be implying that gravity travels faster than the speed of light. For example, if I possessed the technology to reverse Jupiter's obit, by your reasoning you could detect that instantly on Earth from the change in gravitational pull, before the light has time to reach me so that I can see it directly. In other words, you're describing an ansible. Obviously, that can't be correct. $\endgroup$
    – JBentley
    Commented 2 days ago
  • $\begingroup$ @JBentley it's correct though. There is no faster-than-light communication, but nature manages to "forecast" the position of the source to remarkable precision. See arxiv.org/abs/gr-qc/9909087 for a detailed discussion. $\endgroup$
    – Sten
    Commented 2 days ago
  • $\begingroup$ @JBentley you're correct that in that case it will not point to where Jupiter actually is, because you are changing the acceleration of Jupiter during the time you're reversing its orbit. It will point to where Jupiter would have been without your intervention, for the next 40 minutes (give or take 8 minutes depending on where the Earth is). No FTL signaling. $\endgroup$
    – BaddDadd
    Commented yesterday
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  1. The contraction of distance and the contraction of the gravity field cancel out. The additional energy quite obviously causes additional force. I do not know how much additional force exactly.

  2. Force between galaxy and neutron star can be calculated by Newton's gravity law. I will measure double that force when standing on the neutron star, when my time dilation factor is 2.

1 . Force is reduced because the length contraction of the gravity field has bigger effect than the increase of energy. I do not know how much the force is reduced exactly.

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