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Hydrostatic Equilibrium in a Star

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello.
    In my homework I was given to find a mistake in a derivation of hydrostatic equilibrium of a star.
    In contrast to "regular" derivation, where a cylindrical segment is chosen, in HW a segment of sphere is used:

    49PAJX3.png

    2. Relevant equations
    Derivation is given as following:
    a. Pressure on the outer surfase:
    [itex]P(r+dr)=P(r)+\frac{dP}{dr}dr[/itex]

    b. Area of the outer surface:
    [itex]A(r+dr)=d\Omega \left [ r+dr \right ]^2=r^2d\Omega+2rdrd\Omega+\left (dr \right )^2d\Omega[/itex]

    c. Therefore, force on the outer surface is:
    [itex]F(r+dr)=P(r+dr)A(r+dr)=\left [ P(r)+\frac{dP}{dr}dr \right ]\left [r^2d\Omega+2rdrd\Omega+\left (dr \right )^2d\Omega \right ]=[/itex]
    [itex]=r^2P(r)d\Omega+2rP(r)drd\Omega+P(r)\left (dr \right )^2d\Omega+r^2\frac{dP}{dr}drd\Omega+2r\frac{dP}{dr}\left ( dr \right )^2d\Omega+\frac{dP}{dr}\left (dr \right )^3d\Omega=[/itex]
    [itex]=r^2P(r)d\Omega+2rP(r)drd\Omega+r^2\frac{dP}{dr}drd\Omega+O\left ( \left ( dr \right )^2 \right )[/itex]

    d. The force on the inner surface is:
    [itex]F(r)=P(r)A(r)=P(r)A(r)=P(r)r^2d\Omega[/itex]

    e. Mass of the segment:
    [itex]dm=\rho r^2drd\Omega[/itex]

    f. Then, equation for equilibrium:
    [itex]gdm=F(r)-F(r+dr)[/itex]
    [itex]g\rho r^2drd\Omega=-2rP(r)drd\Omega-r^2\frac{dP}{dr}drd\Omega[/itex]
    [itex]g=-\frac{2}{\rho}\frac{P(r)}{r}-\frac{1}{\rho}\frac{dP}{dr}[/itex]

    But we know that equation for hydrostatic equilibrium is:
    [itex]g=-\frac{1}{\rho}\frac{dP}{dr}[/itex]

    3. The attempt at a solution
    After going and checking the solution step by step I found only 2 places where error can occur:
    1. The origin of the additional term is from the non-equal areas of the segment in [itex]r[/itex] and [itex]r+dr[/itex]. This makes sense, since in "regular" derivation of the equilibrium equation a cylinder is used, and there both areas are equal. In addition, if we make [itex]r[/itex] to be very big (in limit - [itex]r\rightarrow \infty[/itex]) - then areas of outer and inner surfaces will be, indeed, identical, and as supposed the additional term ([itex]-\frac{2}{\rho}\frac{P(r)}{r}[/itex]) will be zero.
    I told this to my teacher, but he said that this is not an origin of the problem.

    2. So I checked the derivation again and found that in line "c", the term [itex]2r\frac{dP}{dr}\left ( dr \right )^2d\Omega[/itex] cannot be included inside [itex]O\left ( \left ( dr \right )^2 \right )[/itex]. But if I include this term in the derivation, I get:
    [itex]g=-\frac{2}{r \rho} P(r)-\frac{1}{\rho}\frac{dP}{dr}-\frac{2}{r \rho}\frac{dP}{dr}dr[/itex]
    which gets even worse than was before.

    Can someone help me with that? What I'm missing here?
    Thanks in advance.
     
  2. jcsd
  3. Mar 8, 2014 #2
    Why are the forces on the other surfaces of the element ignored?
     
  4. Mar 8, 2014 #3
    Only radial components are taken:
    [itex]F(r+dr)[/itex] and [itex]F(r)[/itex] are vectors in [itex]\hat{r}[/itex] (radial direction).
    Gravitational force is of course in [itex]\hat{r}[/itex] direction too.
     
  5. Mar 8, 2014 #4
    Why do you think the other surfaces do not contribute a net radial component?
     
  6. Mar 8, 2014 #5
    Because 2 other surfaces are in [itex]\hat{θ}[/itex] and 2 additional surfaces are in [itex]\hat{\phi}[/itex] directions and do not contribute to the [itex]\hat{r}[/itex] components. Am I wrong?
     
  7. Mar 8, 2014 #6
    Consider a simpler case. Take a string (say, a shoelace) and wrap it around your leg so that it forms a complete circle. Then tighten it gently. Observe that tension is circumferential everywhere. Does that mean you leg won't feel any radial force?
     
  8. Mar 8, 2014 #7
    So the forces in [itex]\hat{θ}[/itex] direction:
    [itex]F(r, θ)=P(r, θ ) A(r, θ) =P(r, θ )\ d \phi dr [/itex]
    [itex]F(r, θ + d θ)=P(r, θ + d θ ) A(r, θ + d θ) =-P(r, θ + d θ ) d \phi dr [/itex]
    Sum of all forces equals zero:
    [itex]F(r, θ) + F(r, θ + d θ)=0 [/itex]
    [itex]P(r, θ )\ d \phi dr -P(r, θ + d θ ) d \phi dr =0[/itex]
    [itex]P(r, θ ) =P(r, θ + d θ ) [/itex]
    No gravitational force contributes, since no component of gravitational force is in [itex]\hat{θ}[/itex].
    Same with [itex]\hat{ \phi}[/itex] direction.

    In your example with the shoelace I will fill the radial force, but only because I tighten the shoelace, i.e. put the external force in [itex]\hat{θ}[/itex] direction. This is not the case.
     
  9. Mar 8, 2014 #8
    There is no single ##\theta## direction. The "left" and the "right" surfaces have different ##\theta## directions, and the resultant force is in the radial direction. Ditto for the other non-radial directions.
     
  10. Mar 8, 2014 #9
    As for the shoelace, you are mistaken. The role of the external force is solely to induce tension in the shoelace. You can replace the shoelace with a tightly fitting rubber band or something like that. There is no external force, but there is still tension (which is similar to pressure) and the resultant force has a radial component everywhere.
     
  11. Mar 8, 2014 #10
    I still don't understand.

    bwROV6H.png

    [itex]F(\theta)\hat {\theta}=P(\theta)\cdot A(\theta)=P(\theta)\cdot \left [\left (d \phi \hat {\phi} \right ) \times \left (d r \hat {r} \right ) \right ]=P(\theta)\cdot d \phi dr \hat {\theta}[/itex]
    [itex]F(\theta+ d \theta)\hat {\theta}=P(\theta+ d \theta)\cdot A(\theta+ d \theta)=P(\theta+ d \theta)\cdot \left [\left (d \phi \hat {\phi} \right ) \times \left (d r \hat {r} \right ) \right ]=P(\theta+ d \theta)\cdot d \phi dr \hat {\theta}[/itex]

    Still no radial components.
     
  12. Mar 8, 2014 #11
    You keep writing the same vector ##\hat \theta## for both sides, while you have been already told they are different vectors. No surprise that you keep getting "no radial components".

    The angle between the force on the left surface and the force on the right surface ##\pi - 2 d \theta ##, so the resultant is obviously radial and proportional to ##2 d \theta##.
     
  13. Mar 8, 2014 #12
    Oops, sorry. There should not be ##2## in either of ##2 d \theta ## above.
     
  14. Mar 8, 2014 #13
    I think I got your point:

    MpnvWiv.png

    So,
    [itex]F(\theta)=P(r)\cdot A(\theta)=P(r) d \phi dr[/itex]
    [itex]F(\theta + d \theta)=P(r)\cdot A(\theta + d \theta)=P(r) d \phi dr[/itex]

    This contributes additional force in radial direction:
    [itex]F\left ( \theta \right )sin\left (\frac{1}{2}d\theta \right )+F\left (\theta + d\theta \right )sin\left (\frac{1}{2}d\theta \right )\approx \frac{1}{2}d\theta F\left ( \theta \right )+\frac{1}{2}d\theta F\left (\theta + d\theta \right )=[/itex]
    [itex]= \frac{1}{2}d\theta P ( r ) dr d \phi +\frac{1}{2}d\theta P (r) dr d \phi=P(r)d \theta dr d \phi[/itex]

    Same will be with [itex] \phi [/itex] direction, so to the sum of all forces equation in the my first post I need to add additional force in radial direction:
    [itex]F_{add}=2P(r)d \theta dr d \phi=2P(r)dr d \Omega[/itex]

    Did I understand it right?
     
  15. Mar 8, 2014 #14
    Yes, exactly correct.
     
  16. Mar 8, 2014 #15
    voko!
    Thank you very much for help!
     
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