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

Relativistic apple

  1. Jan 22, 2006 #1
    Hi, everybody. Can anyone settle this question that I've been puzzling about?

    If an object with a small rest-mass, say, an apple, were to whiz very close past you in space with a relative velocity so great that its mass, as measured in your own inertial frame, was comparable to the rest-mass of the sun, would you experience any gravitational effects as it passed by?

    Considered in its rest frame it is obvious that such a small mass has a negligible effect on the geometry of space-time. Which at first inclined me to think, even though it seems slightly paradoxical, that you would feel nothing as the object went past you. But then I had doubts. I'm pretty good at special relativity, and have made a start at general relativity, but solving even simple problems in GR is a little beyond me at the moment.

  2. jcsd
  3. Jan 22, 2006 #2


    User Avatar
    Homework Helper

    Sure it would, although it's Schwarzchild radius at that mass would be a few kilometers, so it wouldn't look like an apple, and you wouldn't want it to pass very close to you. In it's rest frame, everything else, like you and the earth, would have a much greater mass, and so it would be pulled towards those things with a large force, while you would see those things (and you) being pulled toward it. Those are clearly consistent observations.
    Last edited: Jan 22, 2006
  4. Jan 23, 2006 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    What you would measure would depend on the direction that the small mass was travelling.

    First of all, I should point out that what you can measure in a coordinate independent manner is not the "gravitational acceleration", but the tidal forces - the acceleration per unit length. Basically, an accelerometer will not directly measure any acceleration, because you will be in a locally inertial frame - but you can easily measure the change in acceleration per unit distance with, for example, a "Forward mass detector" aka a "Forward gravity gradiometer".

    It would be possible in principle to measure your distance to some distant object, but this is a more difficult measurement than it might appear. The massive body will be distorting the path that light takes while you are trying to perform your distance measurement. The tidal acceleration, though, is a purely local measurement that doesn't need a complete coordinate system.

    Anyway, the tidal forces you will measure will not change if the body is moving directly towards or away from you, no matter how rapidly. (This is explained for instance in MTW's "Gravitation", I could look up the page number if anyone is interested). This point actually needs some further clarification (a diagram).

    #1----> massive body

    Spaceship #1 is moving towards (or away) from the massive body at high speeds. Spaceship 2 is at rest. The massive body is at rest. Spaceship 1 and 2 are at the same point in space and time.

    By the principle of relativity, it doesn't matter whether it is the massive body that is moving, or the spaceship, in the latter case we consider #1 to be stationary and both #2 and the massive body to be in the same frame (co-moving), which is the way you asked the question. (But it is easier to compute the answer using a frame in which the massive body is stationary, though it doesn't matter to the end result.)

    In any event, in the diagram above, both spaceship #1 and #2 measure the same tidal forces.

    What is unfortunately not in the textbooks is what happens if spaceship is moving in another direction, i.e. "up the page". So I had to calculate this case myself.

    In this case, spaceship #1, now moving up the page, experiences a greater tidal force than spaceship #2 which is co-moving with the heavy mass. (In my diagram the heavy mass and spaceship 2 are again both stationary. Again, this is for the ease of calculation, it doesn't matter to the result).

    Somewhere or other in this forum I have the detailed calculations on the exact tidal forces, I calculated, and the details of the calculation. They are also quite technical, so I won't post more unless someone is interested in them.

    I'll summarize the approach briefly by saying that one computes the Riemann tensor using an orthonormal basis of one-forms (not the coordinate version!), which calculation can be found in textbooks, and then boosts the result for the moving spaceship (which result is not in the textbooks for non-radial motion), and that one then computes the tidal forces from the Riemann and a unit time-like vector (also boosted) via the geodesic deviation equation.
  5. Jan 23, 2006 #4
    Thanks guys. I get the gist of it. Someday I hope I'll be able to re-read Pervect's reply and understand it. I'm working though track 2 of Wheeler, Misner and Thorne at the moment. It's been mostly only differential geometry so far.

    In the meantime here's another one I've long wondered about.

    We all know how to use Gauss's theorem to show, on the basis of Newtonian gravity, that the acceleration towards a uniform, infinite plane of matter is always directed towards the plane, and independent of the distance from it.

    I'm curious to know whether the same result is found if the calculation is performed using GR.
  6. Jan 23, 2006 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Check out pg 821-822. You may not understand the details yet, but it calculates the tidal forces on someone falling into a black hole.

    You can construct the metric for an infinte plane of matter by "gluing together" two Rindler metrics.

    The resulting metric is
    ds^2 = (1+g|z|)^2 dt^2 - dx^2 - dy^2 - dz^2

    You should find a similar metric in MTW's gravitation on the section labelled "accelerated observers" without the absolute value sign.

    Ths metric represents matter being in the x-y plane, and the acceleration being in the 'z' direction. The stress-energy tensor is zero everywhere, except for a singular disk of matter on the x-y plane.

    You can use the Komar mass intergal to show that there is singular sheet of matter present in the x-y plane associated with the above metric. Unfortunately MTW doesn't talk much about the Komar mass intergal, and I haven't found anything on-line that explains it in simple terms. It is discussed in Wald's "General Relativity", but the notation may be somewhat daunting :-(.

    This above metric, without the absolute value sign, is often called a "uniform gravitational field", however the acceleration required to hold station (the "felt" acceleration required to keep z constant) is _not_ constant, but a function of z.

    Using local clocks and rulers, I get that the acceleration required to "hold station" is g / (1+ g|z|) - always directed away from the x-y plane.

    The local clocks and rulers are defined by the following orthonormal basis of one-forms:

    (1+g|z|)dt; dx; dy; dz;
    Last edited: Jan 23, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Relativistic apple
  1. Relativistic mass (Replies: 28)

  2. Relativistic Gravity (Replies: 28)

  3. Relativistic Molecules (Replies: 3)