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Change volume V(t)

  1. Jun 20, 2010 #1
    Somebody knows as to prove the expression below? (V (t) is the volume of a sphere in free fall, following a geodesic one).

    (delta V(t)/ V(t) = -(1/2) (d^2 V / dt^2) (delta t)^2

    delta= [Tex]\delta[/Tex]
  2. jcsd
  3. Jun 20, 2010 #2


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    Hi, victorneto -- welcome to PF!

    To make your LaTeX code get rendered properly, you need to surround it with itex tags. To see how it's done, click on the Quote button for this post, where I've marked up your equation: [itex]\Delta V/V = -(1/2)(d^2 V/dt^2)\Delta t^2[/itex]

    First off, the equation is incorrect as written. The left-hand side is unitless, but the right-hand side has units of volume. It should be like this: [itex]\Delta V = -(1/2)(d^2 V/dt^2)\Delta t^2[/itex]

    The way to prove this is by starting from [itex]d^2 V/dt^2=constant[/itex] and integrating twice. The first constant of integration is gotten rid of by using [itex]\Delta V=V-V_o[/itex]. The second constant of integration is presumably zero because of the initial conditions. It's exactly equivalent to motion of a particle in one dimension with constant acceleration.
  4. Jun 20, 2010 #3
    Very grateful for the return.

    It forgives me, it had error of digitação, it lacked to the V in the denominator of segund member:

    [tex]\delta[/tex]V/V=(-1/2){(d2V/dt2)/V}(\delta t)2

    Not yet I learned to use its tags. The digitação above was optimum that I could make, so far. I will continue studying this system, new for me.

    Found the expression above in the study that I am making of an article of John C.Baez And Emory F.Bunn, " The Meaning of Einstein´s Equation" , pg. 9. (arXiv: gr.-qc/0103044v5 5 Jan 2006) Is about a very good work.

    It forgives me the English. It is that I am Brazilian, engineer and self-taught person in RG.
  5. Jun 20, 2010 #4
    In time:
    The correct expression is the following one, without the signal (-).

    [itex]\delta[/itex]V/V=(1/2)[(d2V/dt2)/V ](\delta t)2
  6. Jun 20, 2010 #5
    Hi Victorneto,

    click one of the equations below and wait. Eventually a small window pops up with the code for the equation. Copy and paste this equation (including the tex tags at each end in red) into your post. Edit as desired. Follow the link in the pop up for more information about formating tex codes.

    [itex]\delta V/V = (1/2)[(d^2 V/dt^2)/V](\delta t)^2[/itex]

    [tex]\frac{\Delta V}{V} = \frac{1}{2V}\frac{d^2 V}{dt^2}(\Delta t)^2[/tex]
    Last edited: Jun 20, 2010
  7. Aug 6, 2010 #6
    \frac{\Delta V}{V} = \frac{1}{2V}\frac{d^2 V}{dt^2}(\Delta t)^2
  8. Aug 6, 2010 #7


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    Did my #2 answer your question?
  9. Aug 6, 2010 #8

    Finally I found the way to prove the veracity of the equation below:


    \frac{\Delta V}{V} = \frac{1}{2V}\frac{d^2 V}{dt^2}(\Delta t)^2


    V=V (t).

    Thus, it follows the steps:
    1 - It makes an expansion of Taylor, preserving the values until second order;
    2- delta V = V (t) - V (0), although the acceleration it is not (as a body in the surface of the land: deltaV(t)=0, but g=10 m/s² !;
    3 - dV (t) /dt in t-0 is zero;
    4- now, divides the two sides for V.
    Note.: V (t) is the volume of of mass M in free fall, following a geodesic one.
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