Integrating partial derivatives in a field equation

  • #1
Samson Ogaga Ojako
11
1
I am integrating the below:

\begin{equation}
\psi(r,v)=\int \left( \frac{\frac{\partial M(r,v)}{\partial r}}{r-2M(r,v)}\right)dr
\end{equation}

I am trying to write ψ in terms of M.

Please, any assistance will be appreciated.​
 
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Answers and Replies

  • #2
Delta2
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Is it good enough to write ##\psi## in terms of ##M## and ##\int \frac{M}{r^2}dr##?

If yes then just write ##\psi=\int ((\frac{\partial M}{\partial r}\frac{1}{r}-\frac{M}{r^2})+\frac{M}{r^2}-2M)dr## and notice that the first term is the derivative of ##\frac{M}{r}## so after all it will be $$\psi=\frac{M}{r}-2\int Mdr+\int \frac{M}{r^2}dr+C$$
 
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  • #3
Orodruin
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Is r a radius? (Or any other dimensionsl quantity?) If so, where did this integral come from?
 
  • #4
Samson Ogaga Ojako
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Is r a radius? (Or any other dimensionsl quantity?) If so, where did this integral come from?
The 'r' is a shell radius while the "v" is the time.
The question is related to spacetime.

Thank you for getting in touch
 
  • #5
Samson Ogaga Ojako
11
1
Is it good enough to write ##\psi## in terms of ##M## and ##\int \frac{M}{r^2}dr##?

If yes then just write ##\psi=\int ((\frac{\partial M}{\partial r}\frac{1}{r}-\frac{M}{r^2})+\frac{M}{r^2}-2M)dr## and notice that the first term is the derivative of ##\frac{M}{r}## so after all it will be $$\psi=\frac{M}{r}-2\int Mdr+\int \frac{M}{r^2}dr+C$$

This looks so good enough, Sir.
I must say, thank you very much.
Let me look at how you were able to simplify the integration first.
This is brilliant.
I will get back to you just now.

Cheers
 
  • #6
Orodruin
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The 'r' is a shell radius while the "v" is the time.
The question is related to spacetime.

Thank you for getting in touch
This does not answer where your integral came from. The reason I am asking is that if ##r## is dimensionful, then your expression is dimensionally inconsistent because ##(\partial M/\partial r)/r## does not have the same physical dimension as ##M##.
 
  • #7
Delta2
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This does not answer where your integral came from. The reason I am asking is that if ##r## is dimensionful, then your expression is dimensionally inconsistent because ##(\partial M/\partial r)/r## does not have the same physical dimension as ##M##.
Maybe there is some sort of constant (with the proper units) in front of the ##-2M## term, which he omitted.
 
  • #8
Samson Ogaga Ojako
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This does not answer where your integral came from. The reason I am asking is that if ##r## is dimensionful, then your expression is dimensionally inconsistent because ##(\partial M/\partial r)/r## does not have the same physical dimension as ##M##.
See the attached for details of how \psi, #M# and #r# are related.

Best regard.
 

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  • Solve.pdf
    302.3 KB · Views: 186
  • #9
Orodruin
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So you have not transcribed the problem correctly then. The integrand is
$$
\frac{\partial_r M}{r - 2M},
$$
not
$$
\frac{\partial_r M}{r} - 2M.
$$
 
  • #10
Samson Ogaga Ojako
11
1
Maybe there is some sort of constant (with the proper units) in front of the ##-2M## term, which he omitted.

I think he is unable to see the typing clearly because of the way the question is typed.

So, I've decided to attach a copy of the typed question to him for a better understanding of the question.

Thank you once again.
Cheers
 

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  • Solve.pdf
    302.3 KB · Views: 176
  • #11
Samson Ogaga Ojako
11
1
So you have not transcribed the problem correctly then. The integrand is
$$
\frac{\partial_r M}{r - 2M},
$$
not
$$
\frac{\partial_r M}{r} - 2M.
$$
Yeah
 
  • #12
Delta2
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Ah, then what I did in post # 2 is not correct...
 
  • #13
Samson Ogaga Ojako
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Ah, then what I did in post # 2 is not correct...

Oh ok.

You can take a second look at it again while I retype the question in LaTeX.

Cheers
 
  • #14
Delta2
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Well this look like some sort of assignment or homework. You have to show us some effort from your side.

All I can think right now is to transform the integral so that we get as end result ##-\frac{1}{2}\ln|r-2M|+\frac{1}{2}\int \frac{1}{r-2M} dr +C##
 
  • #15
Samson Ogaga Ojako
11
1
Well this looks like some sort of assignment or homework. You have to show us some effort from your side.

All I can think right now is to transform the integral so that we get as end result ##-\frac{1}{2}\ln|r-2M|+\frac{1}{2}\int \frac{1}{r-2M} dr +C##

It is not really a homework or assignment. This is part of my research finding. I've gotten to a stage where I have 3 field equations with 3 variables #M#, #\psi# and #phi# all are functions of #(r,v)#

\begin{equation}
\frac{\partial {m}}{\partial v} = 4\pi{r^{2}}\phi_{v} \left[e^{-\psi}\phi_{v} + \left( 1-\frac{2m}{r}\right) \phi_{r} \right] = m^{\prime}(v)
\end{equation}
\begin{equation}
\frac{\partial {\psi}}{\partial r} = 4\pi{r^{2}}\phi_{r}^2 = \psi^{\prime}(r)
\end{equation}
\begin{equation}
\frac{\partial {m}}{\partial r} = 2\pi{r^{2}}\phi_{r}^2 \left( 1-\frac{2m}{r}\right) = m^{\prime}(r)
\end{equation}

From equations 2 and 3, I was able to get what I shared online.

\begin{equation}
\psi(r,v)=\int \left( \frac{\frac{\partial M(r,v)}{\partial r}}{r-2M(r,v)}\right)dr
\end{equation}.

I have made some effort to get #\psi# in terms of #M# so that I can solve equation 1.

Please, you can help if you can.
 
  • #16
Delta2
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Can you have a look again at your equations (3) and (4) , because from those two I get that

$$\psi=\int \frac {2r\frac{\partial M}{\partial r}}{r-2M}dr$$
 
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  • #17
Samson Ogaga Ojako
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Can you have a look again at your equations (3) and (4) , because from those two I get that

$$\psi=\int \frac {2r\frac{\partial M}{\partial r}}{r-2M}dr$$
Oh ok.

Let me check again then.
 
  • #18
Samson Ogaga Ojako
11
1

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