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Gravity with SR

  1. Jun 30, 2015 #1

    ChrisVer

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    Starting from a locked thread I tried to work the gravity of a body of mass ##M## on another body starting from infinity to some distance ##d## from the gravitating body.
    We have from the SR 2nd Newton law that:
    [itex]\gamma^3 a = \frac{GM}{r^2}[/itex]

    writting [itex]a= \frac{dv}{dt}= v \frac{dv}{dr} = \frac{1}{2} \frac{dv^2}{dr}[/itex]

    Naming [itex]v^2/c^2= x[/itex] the above relation becomes:

    [itex] \frac{dx}{(1-x)^{3/2}}= \frac{2GM}{c^2 r^2}dr[/itex]

    Integrating:

    [itex] \int_{0}^{x_d}\frac{dx}{(1-x)^{3/2}}= \int_{\infty}^{d} \frac{2GM}{c^2 r^2}dr[/itex]
    [itex]\frac{2}{\sqrt{1-x_d}}-2= - \frac{2GM}{c^2 d}[/itex]

    [itex]x_d=1- \Big( 1- \frac{GM}{c^2 d}\Big)^{-2}[/itex]

    And so:

    [itex]v(d)= c \bigg[ 1- \Big( 1- \frac{GM}{c^2 d}\Big)^{-2} \bigg]^{1/2} [/itex]

    I tried plotting this solution as [itex]v(d)[/itex], the good part is that [itex]v<c[/itex] for all distances however I don't understand why for [itex]d=GM/c^2[/itex] I'm obtaining an infinity (and worse- in the imaginary regime)? By the way, that's the Schwarzchild radius....

    Even worse, if I set [itex] \frac{GM}{c^2}=1[/itex] the plot of [itex] \beta = v/c = \bigg[ 1- \Big( 1- \frac{1}{d}\Big)^{-2} \bigg]^{1/2} [/itex] is given in my attachment...and doesn't seem to have a real solution away from 1?
     

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  3. Jun 30, 2015 #2

    A.T.

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    Are you trying to compute the falling speed relative to a local hovering observer? You cannot have one hovering at the Schwarzchild radius.
     
  4. Jun 30, 2015 #3

    ChrisVer

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    So the information of Schwarzchild radius singularity is contained within SR? I solved the problem in a mechanical way...
    But the main problem is that for larger radii from the source (distances ##d>R_s##), I am not getting a real solution for ##v##...
     
  5. Jun 30, 2015 #4

    Mentz114

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    Typo here maybe

    ##\frac{1}{2} \frac{dv^2}{dr}##
     
  6. Jun 30, 2015 #5

    ChrisVer

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    Eh?
    [itex] \frac{1}{2} \frac{dv^2}{dr}= \frac{1}{2} \big( \frac{dv}{dr} v + v \frac{dv}{dr}\big) = v \frac{dv}{dr}= \frac{dr}{dt} \frac{dv}{dr} = \frac{dv}{dt}=a[/itex]
     
  7. Jun 30, 2015 #6

    A.T.

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  8. Jun 30, 2015 #7

    ChrisVer

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    That looks like a GR conversation..:sorry:
     
  9. Jun 30, 2015 #8

    A.T.

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    Yes, for comparison of g under GR with the value you assume based on SR and Newton.
     
  10. Jun 30, 2015 #9

    ChrisVer

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    I think this is a mistake with the gravitational force I used...I think I should have used a minus sign: [itex]\gamma^3 a = - GM/r^2[/itex]...
    The main problem is that, as it is, there is no solution for [itex]d > R_s[/itex] ([itex]v^2<0[/itex]) and this makes physically no sense, since the force I used could as well be some other [itex]1/r^2[/itex] type force (like electromagnetism), with the change [itex]R_s \rightarrow D=\frac{k q_1 q_2}{c^2 m}[/itex].
     
  11. Jun 30, 2015 #10

    ChrisVer

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    The result for [itex]v[/itex] is then:

    [itex]v= c \sqrt{1- (1+\frac{GM}{c^2 d} )^{-2}}[/itex]
    With a plot (for [itex]R_s= \frac{GM}{c^2}=1[/itex]) like the attached, which makes "sense".
     

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  12. Jun 30, 2015 #11

    Mentz114

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    Oh, you mean ##\frac{d}{dr}v^2##. Your notation confused me.

    What is the plot of in your latest post ?
     
  13. Jun 30, 2015 #12

    ChrisVer

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    It is let's say [itex] \frac{v}{c} = \beta(r)[/itex]... it shows the change of the velocity of a mass that began from infinity with 0 velocity under the "newtonian" gravitational law of a point-like mass source [itex]M[/itex] (in terms of "schwarzchild radii" of the source body- of course it should not be confused with schwarzchild radii since I don't work in GR). So for example, when the body is at a distance [itex]x=2 \times \frac{GM}{c^2}[/itex] away from the source, it will have a velocity [itex]v \approx 0.75c[/itex]...
    In a closed thread someone said that the velocity of the object would eventually reach a value larger than [itex]c[/itex] due to acceleration. I just wanted to counter this idea via special relativity alone...

    I guess this could change for GR, but OK... It was more like a special relativistic application of [itex]F= \frac{dp}{dt}[/itex]. Showing that by applying special relativity alone, it doesn't matter for how long your body has travelled (starting from infinity up to reaching the source at 0), its velocity won't exceed the speed of light.
     
    Last edited: Jun 30, 2015
  14. Jun 30, 2015 #13

    ChrisVer

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    It saves the "classical"/non-relativistic prediction that:
    [itex] v= \sqrt{\frac{2GM}{r}}[/itex]
    Which allows for [itex]r[/itex] such that [itex]v>c[/itex], eg [itex]r= \frac{2GM}{9c^2}[/itex] which gives [itex]v=3c[/itex]
     
  15. Jun 30, 2015 #14

    Mentz114

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    The curve certainly makes sense. You can see those ##\gamma##'s kicking in.
     
  16. Jun 30, 2015 #15

    PeterDonis

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    SR can't be used to model gravity.

    No. You need to use GR to model gravity.

    So should this thread be. You can't model gravity using SR.
     
  17. Jun 30, 2015 #16

    PeterDonis

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    But for gravity, ##F = 0##. Gravity is not a force in relativity. A force in relativity is something that causes proper acceleration. Gravity doesn't. However the computations you are making work out formally, they are physically meaningless as far as gravity is concerned.
     
  18. Jun 30, 2015 #17

    Dale

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    Please do not reopen locked threads.
     
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