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Gravitational Field & Reference Frames

  1. Jul 22, 2009 #1
    In Newtonian physics gravity is a force and in relativity it's curved spacetime if I understand it correctly. So my question is, does the gravitational field of an object look different to different observers? What I mean is, does one reference frame see the strength of the field to be greater or lesser depending on it's relative motion?

    My physics professor said yes, but a quote from "Exploring Black Holes" by Wheeler and Taylor has me wondering. It says "Special relativity uses laboratory and rocket frames as different vantage points to get an insight into flat spacetime that exists independent of any reference frame. In the same way we use alternative reference systems around a star to get insight into curved spacetime-a curved geometry that exists independent of any frame of reference."

    My apologies if this has been asked before, or if the answer should be blatantly obvious.
     
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  3. Jul 22, 2009 #2

    diazona

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    It's possible that your physics professor is talking about accelerated motion. If you are accelerating, you'll perceive a different strength of the gravitational field than someone who's not accelerating.

    Imagine being in a closed, perfectly insulated box - let's say, a space capsule with no windows. Now consider three different situations:
    • When the capsule is sitting on the ground not moving, you perceive the Earth's normal gravitational field.
    • When your capsule is being pushed up by the rocket during launch, you'll feel a very strong gravitational field, something like 8 or 9 times Earth's gravity if I remember right. Of course, someone on Earth would say that most of that field is "actually" acceleration, not gravity, but GR says that it's just as valid for you to claim that you're feeling more gravity. (Maybe gravity just got really strong all of a sudden.)
    • When your capsule is falling back to Earth during reentry, you'll feel virtually weightless - that's no gravitational field. Or someone on Earth would say that your acceleration downwards is "canceling out" the Earth's gravity.
    So those are some different reference frames (not all inertial reference frames, of course) that see the gravitational field strength to be of different values.

    But: if you really want to get insight into curved spacetime, you need to somehow correct for your acceleration relative to some reference point, which you can't do while you're locked up in your little capsule. You'd need to open a window or something to be able to see the reference point (i.e. the Earth).
     
  4. Jul 22, 2009 #3
    That makes sense. The class was actually Modern Physics & we were working with special relativity. The reasoning behind what the physics professor said I believe was, each force is measured differently for different observers traveling at different speeds. This is a result of different measurements of velocity, and hence different measurements of acceleration (F=ma). But it was the whole idea that relativity doesn't really treat gravity as a force that made me question the idea. I'm just now getting into the basics of general relativity with this book. So if we're dealing only with uniform motion, would it then be to safe to say that each individual would measure the gravitational field of say the Earth to be the same?
     
  5. Jul 23, 2009 #4

    zonde

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    Yes, up to 2 times grater when relative motion approaches c. That's the reason why light is bend twice stronger than predicted by Newton's gravity law.
    Can not offer any comment about geometry part as that is beyond me.
     
  6. Jul 23, 2009 #5

    A.T.

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    This has nothing to do with seeing a stronger field when approaching c. Newton and GR are just different theories predicting different things. GR predicts, additionally to time curvature which causes Newton's gravity, a spatial curvature. For fast moving objects like photons the influence of the spatial curvature gains relatively more influence on the path, so the difference to Newton gets bigger.

    Back to the OP-question: When moving fast relative to the source of gravitation, it's field is Lorentz-contracted, so you could argue that it is "stronger" or concentrated on less space.
     
  7. Jul 23, 2009 #6

    zonde

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    What is field if not it's influence on the path?
    Anyways the pure fact is that difference in predictions from Newtonian gravity and GR grow bigger with bigger relative motion.

    When moving fast relative to the source of gravitation:
    - gravity field (as seen from perspective of gravitating body) is Lorentz-contracted;
    - your trajectory (as seen from perspective of gravitating body) is Lorentz-contracted;
    - your timing (as seen from perspective of gravitating body) is Lorentz-time-dilated;
    - mass of gravitating body increases;
    - position of gravitating body is altered due to aberration.
    Not to mention that you have to define some global coordinate system (probably with coordinate light speed) where you will perform all these manipulations.
    So it's hardly trivial.
     
  8. Jul 23, 2009 #7

    A.T.

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    But the influence on the path doesn't get bigger with high speed as you claimed. The opposite is the case: The trajectories of photons are bend less than the trajectories of slower objects. It is just the relative contribution of spatial curvature to the path bending that is higher.
    I think the question was about the influence of the relative speed within GR. Not the difference between Newton and GR.

    Do these two effects add up, or is it the same thing?

    Not sure about this. It is a field, not radiation like light. In the frame of the mass aberration changes the direction you see the light coming from. But the pull of the gravitational field is still towards the mass, not the aberrated position.

    .
     
  9. Jul 23, 2009 #8

    zonde

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    It's a bit hard to follow you. It seems obvious to me that the faster the object goes the bigger influence is need to bend it's path by the same amount.
    The other thing is that as I understand in GR there is space-time curvature. I can somewhat interpret what you mean by time curvature - I understand it as time dilation effect. But what do you mean by spatial curvature?

    In opening post there is "In Newtonian physics gravity is a force and in relativity it's curved spacetime if I understand it correctly."
    So I assume that field refers to Newtonian gravitational force field.

    Well for one thing Lorentz-contraction happens only in one direction where gravitation field goes in all directions from gravitating body. So it's obviously not the same thing.

    But the field propagates at c just the same.
    As we talk about frame of moving body in that frame gravitating mass is in motion but the body (observer) is at rest.
    So will the pull of the gravitational field be toward the position of gravitating mass where you see it or will it be at some angle from that position?
    If you say it's towards the position where it is in gravitating body frame then the pull is at angle from visible position and the field propagates from source with this angle "incorporated" (as seen in observers frame).
     
  10. Jul 23, 2009 #9

    A.T.

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    Space curvature is what contributes 50% of the light bending and causes orbits to precess. Time vs. space curvature are explained here nicely:
    http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html
    Together they are called space-time curvature.
    The Newtonian gravitational accelerates everything equally, independent of it's speed. Neither Newtons field nor the GR-gravity affects fast object more than slow objects, if by "affecting" we mean the apparent acceleration. If by "affecting" we mean how much the trajectory is bend, then slow objects are affected more in both models. But I don't see how you would define "affecting" so that fast objects are affected more in any of the two models.
    Not the same thing but Lorentz-contracting the space-time metric could already account for the increase of the dynamic mass That was my question.
    It will be at some angle.
     
  11. Jul 24, 2009 #10
    i would like to raise some related topis, the gravitional waves. A free falling observer has all the right to considere his frame inertia, that is simple to understood. But suppose some abrupt change happens in the gravitonal mass, like sudden change in its total mass or so..according to GR, there must be gravitional waves initiated starting from the center of the field and proceed outward in the speed of the light. So the free falling observer should feel this gravitional waves. However, because his frame is inertial, he should not appreciate this waves because his frame will remian also inertial after that abrupt chages of the sourse. in both situation the free falling frame of that observer will remain inertial. So this is a pardox, because in one side the observer should feel the wave and on another hand he should not feel anything because his frame is inertial???
     
  12. Jul 24, 2009 #11

    A.T.

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    He will not experience proper acceleration, but could measure periodical distance changes or tidal effects.
     
  13. Jul 24, 2009 #12

    zonde

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    Point 1. in this link is clear but from point 2. follows that length of circumference drawn around gravitating object (with object at center) will be shorter than 2Pi*radius.
    Have you any other reference apart from this visualization that supports this view?

    If that was true then GR would not be able to explain perihelion precession of Mercury.

    What exactly do you mean with phrase "Lorentz-contracting the space-time metric"?
    I am familiar with Lorentz-contraction of distances but have not heard about Lorentz-contraction of space-time metric.
     
  14. Jul 24, 2009 #13

    A.T.

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    Correct
    The spatial curvature is described here:
    http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid

    I don't see why you think so. My statement just means that gravity is an inertial force in GR, which accelerates everything equally. The perihelion precession of Mercury is explained in GR by the spatial curvature discussed above.
    Just contraction of the gravitational field. In GR it is described by a metric, like the Schwarzschild metric linked above.
     
    Last edited: Jul 24, 2009
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