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Lorentz contraction and gravitational field

  1. Sep 28, 2012 #1
    Here's a scenario:

    A rod shaped object 1m in diameter and 300m in length is moving through space at a velocity of .999994444429013c (picked that velocity arbitrarily for the 300:1 length contraction). The Lorentz factor is 300 for this problem. So the equation to figure length contraction is:

    Δx'=Δx/300=Δ300/300=1

    So, to a remote observer, the apparent length of the rod would be 1m. Assume the rod has sufficient density to exert a measurable gravitational field.

    Question 1 - When the rod is at rest, would its gravitational field be non-spherical? (I already assume it would not be, the geometry of the field would be something akin to the shape of the rod itself I think assuming that the rod is of uniform density.)

    Question 2 - When the rod is traveling at the velocity above so that it is subject to Lorentz contraction, would the remote observer, given proper equipment, detect the gravitational field as "rod shaped" or spherical (not truly spherical unless the rod was shaped in such a way that when subject to Lorentz contraction, the apparent shape was perfectly spherical.)

    Or am I making incorrect assumptions about one part of this or another?
     
  2. jcsd
  3. Sep 28, 2012 #2

    Bill_K

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    Question 1 - Yes

    Question 2 - Sort of. But before plunging into GR, take a step backwards and consider first a simpler problem, which involves SR only. What's the corresponding situation for electromagnetism? Suppose the rod is uniformly charged. What does the E field of the rod look like in the frame in which it is moving? Is the E field "spherical"?
     
  4. Sep 28, 2012 #3
    Hrmm, I would think that in the rod's frame of reference, the E field would be roughly rod shaped also?
     
  5. Sep 28, 2012 #4

    Bill_K

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    What about the frame in which the rod is Lorentz contracted?

    E & M provides a very useful analogy with gravity, and is much more familiar to us. Almost always, it's a good idea to think it through with E & M first, and then say well, gravity is just like that, except...
     
  6. Sep 28, 2012 #5
    Honestly, I don't know how or if an EM field would change in those circumstances. I need to break it down even simpler :)

    If the rod was a giant bar magnet and you compressed it in some way so that it went from 300m in length to 1m in length and retained the same 1m diameter, and you had a detector 301m away from the center point of the magnet (making sure the detector is placed directly on the x-axis), the reading wouldn't change when you compressed the magnet. If you got closer in to the magnet, the field strength would increase but the total nature of the field would stay the same?

    I don't know, what calculations do I need to do to solve this?
     
  7. Sep 28, 2012 #6

    Bill_K

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    I'll keep this non-mathematical. Let me know if you want to see some math!
    (Don't forget that a moving charge produces a B field as well.) Quoting the E & M result from Jackson,

    "At high speeds the peak transverse E field becomes γ times its nonrelativistic value, while the duration is decreased by a factor 1/γ. For v → c the observer sees nearly equal transverse and mutually perpendicular E and B fields, indistinguishable from the fields of a pulse of plane polarized radiation."

    In other words, just as you proposed, the field surrounding the object is Lorentz contracted in the direction of motion. If the object was a point to begin with, the moving field becomes pancake-shaped. If the object had been rod-shaped, the moving field would at first become spherical. Even higher velocities would lead again to a pancake. Such a field resembles a pulse of EM radiation.

    Same for gravity. The field grows in the transverse direction. It becomes Lorentz contracted, and at even greater velocities comes to resemble a pulsed gravitational wave.
     
  8. Sep 28, 2012 #7
    Heck yeah, show me the math! :) I'd like to see at what speed it would become spherical.
     
  9. Sep 29, 2012 #8
    You seem to be asking if length contraction due to speed would be denied by gravitation measurements, if such could be done. GR does not disagree with SR on that point.

    But perhaps you want to know the predicted field intensity as function of distance and angle? That's much more difficult, and Bill made a good start with giving a reasonable answer.
     
    Last edited: Sep 29, 2012
  10. Sep 29, 2012 #9
    I don't think I asked either one of those :) Length contraction can't be denied, whether you observe Lorentz contraction or don't just depends on what frame you're in.

    My basic question is, does a field whether it be EM or gravitational compress along with the object generating it? Bill gave a great answer earlier. It seems to follow the same principle as the compression of the rod itself, in the rod's frame it doesn't happen, in the observer's frame it does. Am I explaining this properly?
     
  11. Sep 30, 2012 #10

    zonde

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    Seems kind of obvious that field should depend on distribution of charge.
    But more difficult part seems to be the change of field due to the motion of charge i.e. partial transformation of polar (vector) field into axial (vector) field. And because of that part the word "compress" does not seem appropriate.

    Hmm and sufficient distance away any localized distribution of charge will look like point charge anyway so the field wouldn't look compressed.

    So it does not seem right to say that "field compress along with the object generating it".
     
  12. Sep 30, 2012 #11

    Bill_K

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    Electromagnetism

    A static charge distribution produces a Coulomb potential φ satisfying ∇2φ = 4πρ and an electric field E = -∇φ. What happens when we view the field in a frame moving with velocity v along the x axis?

    Naive guess: There is a Lorentz contraction by a factor γ in the x direction, which compresses the field and increases the gradient of the potential in that direction. Therefore Ex increases by a factor γ while Ey stays the same. Both of these conclusions are wrong.

    In the first place, φ is not a Lorentz scalar, it's the fourth component of a 4-vector Aμ = (φ, A) where A is the magnetic vector potential. Under a Lorentz transformation (A = 0 for a charge distribution), A0 → Λ00 A0, implying φ → γφ. The Coulomb potential is larger in the moving frame by a factor γ. Thus Ey ≡ - ∂φ/∂y → γEy.

    Now we might naively suppose that Ex → γ2Ex, one γ from the growth in φ, and one from the compression. This is also wrong.

    E is part of an antisymmetric tensor Fμν: F0i = - Fi0 = Ei. Under a Lorentz transformation, F0y → Λ00Λyy F0y, implying Ey → γEy as before. But, F0x → Λ00Λxx F0x + Λ0xΛx0 Fx0, implying Ex → γ2(1 - v2/c2)Ex = Ex. Ex does not grow at all, because of a cancellation. The result is Ex → Ex, Ey → γEy.

    The inclination of the field lines increases by a factor γ, and the entire field compresses in the direction of motion like a pancake.

    Gravity

    A static mass distribution produces a Newtonian potential φ satisfying ∇2φ = 4πρ, and a gravitational field G = ∇φ (different sign convention). What happens when we view the field in a frame moving with velocity v along the x axis?

    The naive version is similar.

    Again, φ is not a Lorentz scalar. It's the 00 component of a symmetric tensor φμν, where ∇2φμν = 16πTμν, Tμν being the stress-energy tensor. (Note additional factor of 4.) From this, the Newtonian potential is larger in the moving frame by a factor γ2, and we might suppose Gy = γ2Gy, while Gx is compressed to produce one more γ, Gx = γ3 Gx. The former is True, the latter is False.

    There is a partial cancellation in the gravitational case too, but it happens rather differently. In the weak field approximation, gμν ≈ ημν + hμν, where hμν = φμν - ½ημν φαα. Therefore h00 = hxx = hyy = hzz = ½φ00. The relativistic gravitational force field is given by the Christoffel symbol, Γμνσ = ½gσλ(gλν,μ + gμλ,ν - gμν,λ) ≈ ½(hσν,μ + hμσ,ν - hμν,σ). For a static mass distribution the Newtonian gravitational force is Gi = Γ00i. Under a Lorentz transformation, Γ00y → Λ00Λ00ΛyyΓ00y, implying Gy → γ2 Gy, that part is right. But Γ00x → Λ00Λ00ΛxxΓ00x + Λ0xΛ0xΛxxΓxxx, implying Gx → γ3(1 - v2/c2) Gx, or Gx → γGx.

    Just as in the electromagnetic case, the inclination of the field lines increases by a factor γ, and the field compresses in the direction of motion like a pancake. Unlike the electromagnetic case, the field grows by an additional factor of γ.
     
    Last edited: Sep 30, 2012
  13. Sep 30, 2012 #12
    That is true if just applying a Lorentz transformation at some particular field location. At large distance from a charged rod of finite length, it will present a field essentially like that of a point charge. It is usual to compare the fields of such a charge, stationary vs moving, at a fixed instantaneous axial displacement in the one reference frame. The fields are then as shown in (8) here Then see the comments in (9), (10a), (10b). Notice Ex for moving charge in this setting is reduced by a factor γ2.
     
  14. Sep 30, 2012 #13

    Bill_K

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    Very good. Regardless of what you consider "usual", I am just comparing the field of a charge or mass at the same point, in two different frames.
     
  15. Sep 30, 2012 #14
    Ok, so for an EM field, Ex → Ex, Ey → γEy says that the field in the y-axis is increased by a factor of γ. Does the field expand in radius or is it just the field strength that increases? Or do both happen?
     
  16. Sep 30, 2012 #15

    jtbell

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    The field still extends to infinity in all directions. It's the magnitude (strength) at a given point that changes.
     
  17. Sep 30, 2012 #16
    I think I answered my own question by looking at Q-reeus' link.

    "The field is "pancaked" -- it's stronger out to the sides, and weaker in front and in back of the charge. If we view the field as a "bubble" around the charge, we can view the "bubble" as being Lorentz contracted. However, Gauss's law still holds and the integral of the field strength over the surface of a sphere around the charge is the same in all inertial frames. So, if the "bubble" is contracted fore-to-aft, it must bulge out to the sides, which isn't quite the same as physical Lorentz contraction."

    So for an external observer, the rod itself would appear to be denser because of Lorentz contraction. But the gradient of the EM field generated by the rod wouldn't change, it would just shorten along the x-axis and widen along the y-axis?
     
  18. Sep 30, 2012 #17
    Sorry, that was bad vocabulary, I do understand that fields are infinite. I was thinking of a limit of measurement, say the point where measured field magnitude dropped below some arbitrary amount.
     
  19. Sep 30, 2012 #18

    Bill_K

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    That's precisely the issue. The original post concerned the field of a charged rod. That's what my answer was geared to. Somewhere along the line we've changed questions, and now talk instead about what happens to the field of a point charge.

    The field of a rod is rather general, and what you can do is sit still at a single point and say the field strength increases/decreases. In fact we saw it increases uniformly, just picking up a factor of γ or γ2.

    For the field of a point charge, on the other hand, you're given the radial dependence and so you can compare field values at different points. You can rescale r and say: this field here came from that field over there because it expanded/contracted.
     
  20. Sep 30, 2012 #19
    Thanks for the insightful answers Bill, I think I have a general understanding of what happens in this scenario and I copied the calculations down. I'm definitely ignorant in many basic areas. I started a calculus course ~2 months ago that covers everything up to vector calculus so hopefully soon I'll have the requisite skills to actually work out some of the basic math that describes EM and other fundamental forces.

    I do think I'll ignore GR in the beginning and focus on understanding SR. SR does seem to be more straightforward and demonstrable. Thanks again for humoring a novice.
     
  21. Sep 30, 2012 #20

    pervect

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    You can also represent electromagnetic fields via simple geometrical objects, though the theoretical justification is rather advanced.

    See for instance http://125.71.228.222/wlxt/ncourse/DCCYDCB/web/condition/9.pdf "TEACHING ELECTROMAGNETIC FIELD THEORY USING DIFFERENTIAL FORMS".

    Then the electric field of a point charge can be represented by the geometric object in figure 12. Squash it, and you get the electric field of a moving point charge. Of course you have to add in the magnetic field, too, to keep things covariant.

    Gravity is much trickier...
     
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