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Homework Help: Help with differential equation

  1. Jan 3, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\frac{dr}{\sqrt{r^2- 2mr}}= \frac{d\rho}{\rho}[/itex]
    Solve for r.

    2. Relevant equations

    3. The attempt at a solution
    For those familiar with GR, this is an attempt to come up with the isotropic form of the Schwarzschild solution to the field equations, but I'm having trouble with the math, so I just need someone to check my math. The solution I am supposed to get is [itex]r=\rho(1+ \frac{1}{2}m/\rho)^2[/itex], but I am coming up with [itex]r=\rho(\frac{1}{2}+ m/\rho)^2[/itex].

    After integration, the equation I came up with is [itex]2\ln(\sqrt{r-2m}+ \sqrt{r})= \ln\rho[/itex]
    [itex]e^{2\ln(\sqrt{r-2m}+ \sqrt{r})}= e^{\ln\rho}[/itex]
    [itex]e^{\ln(\sqrt{r-2m}+ \sqrt{r})+ \ln(\sqrt{r-2m}+ \sqrt{r})}= e^{\ln\rho}[/itex]
    [itex]e^{\ln(\sqrt{r-2m}+ \sqrt{r})}e^{\ln(\sqrt{r-2m}+ \sqrt{r})}= e^{\ln\rho}[/itex]
    [itex](\sqrt{r-2m}+ \sqrt{r})^2= \rho[/itex]
    [itex]r- 2m+ 2\sqrt{r^2- 2mr}+ r= \rho[/itex]

    I moved all terms on the left side of the equal sign to the right except for the square root term and squared both sides.
    [itex]4(r^2- 2mr)= (\rho+ 2m- 2r)^2= \rho^2+ 4m\rho- 4r\rho+ 4m^2- 8mr+ 4r^2[/itex]

    The terms [itex]4r^2[/itex] and [itex]-8mr[/itex] cancelled, and then I moved the term [itex]-4r\rho[/itex] to the left side of the equal sign.
    [itex]4r\rho= \rho^2+ 4m\rho+ 4m^2[/itex]
    [itex]r\rho= (\frac{1}{2}\rho+ m)^2[/itex]
    [itex]r= \rho^2(\frac{1}{2}+ m/\rho)^2/\rho= \rho(\frac{1}{2}+ m/\rho)^2[/itex]

    Does anyone see a problem?
  2. jcsd
  3. Jan 3, 2012 #2


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    Homework Helper

    I see one. The integration will give you [itex]2\ln(\sqrt{r-2m}+ \sqrt{r})= \ln\rho +C[/itex]. There's an undetermined constant, C. I don't think you can assume it's zero. I think a choice of that constant will give you the books answer. Don't you have some way to determine the constant?
  4. Jan 4, 2012 #3
    Doing the math in my head, the constant would have to be [itex]\ln(\rho-m)[/itex]. The constant from integration of the left side of the equation allows for the rho term. I risk leaving the topic of this forum by asking this, but every book and article I have seen uses the specific transformation [itex]r=\rho(1+\frac{1}{2}m/\rho)^2[/itex] when deriving the isotropic form of the Schwarzschild solution. Why could I not use 0 as the constant of integration and have the transformation be [itex]r=\rho(\frac{1}{2}+m/\rho)^2[/itex]? It seems to me that also leads to a legitimate isotropic construct of the Schwarzschild solution.
  5. Jan 4, 2012 #4


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    [itex]\ln(\rho-m)[/itex] isn't a constant. You can't add that. You generally have some boundary conditions you want your coordinates to satisfy. In the books solution for large [itex]\rho, r[/itex] you have [itex]\rho \approx r[/itex]. Your version wouldn't have that.
  6. Jan 4, 2012 #5


    Staff: Mentor

    you could try differentiating your answer to see what factors got changed then go back thru your steps to see if you properly accounted for them.

    in undergrad we once did an eperiment to measure the speed of light via intereference and a rotating magnetic and got 1/2 the seepd of light. We never figured out why and the prof couldn't find anything wrong with what we did. After many years, I realized we may not have factered in that the mirror was double sided which meant it effectively spun at twice the speed. Oh well, that's why I went into programming that and the money even though physics was way more interesting.
  7. Jan 4, 2012 #6
    how did you get this? Seems you should have had
    \ln(\sqrt{(r-m)^2-m^2} + r-m) = \ln(C \rho)
    choosing [itex]C=2[/itex] you get your answer
  8. Jan 4, 2012 #7
    Bah, you're right, since [itex]\rho[/itex] is a function of r, I can't use it as a constant. My bad.

    Ha ha, I'm actually in IT myself, though I am not a hardcore programmer; I do some here and there. Physics is a very interesting subject to me, but a career in it, on the other hand, doesn't appeal to me too much.

    Well, to be honest, I cheated and used Wolfram to compute the integral for the left side. I thought it might be a bad idea, but I looked into using substitution and integration by parts and couldn't get anywhere using those techniques. Yes, differentiating your solution does indeed return [itex]\frac{dr}{\sqrt{r^2-2mr}}[/itex]. I am curious to know how you came up with that.

    Also, I was not aware you would get a constant multiplied by [itex]\rho[/itex] on the right side. I don't remember learning about that in calculus, but it has been 7 years since I took it, so maybe I just completely forgot.
  9. Jan 4, 2012 #8
    Ok, first you notice that [itex]r^2-2mr = (r-m)^2-m^2[/itex] and then you substitute [itex]x\equiv r-m[/itex], and [itex]dr=dx[/itex]. The integral then becomes
    \int\frac{dx}{\sqrt{x^2-m^2}} = \ln(\sqrt{x^2-m^2}+x)
    Second, on the right hand side you have
    \int\frac{d\rho}{\rho}=\ln\rho + C_1
    Let's write [itex]C_1=\ln C[/itex] so that we have pure [itex]\ln[/itex]s on both sides
    \ln\rho + C_1 = \ln(C\rho)
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