Algebra/differentiation/trig expansions/ small approximation

[binbagsss]

Homework Statement

$\frac{\partial u}{\partial \phi}=-\frac{Me}{J^{2}}sin\phi + \frac{M^{3}}{J^{4}}(-\frac{1}{e}sin\phi+e^{2}sin2\phi+3e\phi cos \phi )$.

Assume this to be $0$ at $\phi=\pi+\epsilon$.
Find $\epsilon$?

Homework Equations

The only method I can think of is to expand out $cos(A+B), sin(A+B)$ terms and then use that for $x$ small $sin x = x$ and $cos x=1$ (so neglected terms of order $x^{2}$).

The Attempt at a Solution

Okay so doing the above I get $0=-\frac{Me}{J^{2}}sin(\pi + \epsilon) + \frac{M^{3}}{J^{4}}(-\frac{1}{e}sin(\pi+\epsilon)+e^{2}sin(2\pi+2\epsilon)+3e(\pi+\epsilon) cos (\pi+\epsilon)$

$=-\frac{Me}{J^{2}}(sin(\pi)cos( \epsilon) + cos(\pi)sin(\epsilon))+ \frac{M^{3}}{J^{4}}(-\frac{1}{e}(sin(\pi)cos( \epsilon) + cos(\pi)sin(\epsilon))+e^{2}(sin(2\pi)cos( 2\epsilon) + cos(2\pi)sin(2\epsilon))+3e(\pi+\epsilon) (cos \pi cos\epsilon-sin\epsilon sin\pi))$.

$=\frac{Me}{J^{2}} -\epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))$

The answer is however $\epsilon=\frac{3M^{2}}{J^{2}}\pi$.
(source An introduction to GR, L.P Hughston and K.P Tod).

Cheers in advance.

 

[wabbit]

Your solution is unduly complicated by using $\sin(\pi+\epsilon)=\sin(\pi)\cos(\epsilon)+\cos(\pi)\sin(\epsilon)$. This simplifies to the identity $\sin(\pi+\epsilon)=-\sin(\epsilon)$. What do you get when using that instead ?

 

[binbagsss]

Your solution is unduly complicated by using $\sin(\pi+\epsilon)=\sin(\pi)\cos(\epsilon)+\cos(\pi)\sin(\epsilon)$. This simplifies to the identity $\sin(\pi+\epsilon)=-\sin(\epsilon)$. What do you get when using that instead ?

the same? (after the appoximation $-\sin(\epsilon)=-\epsilon$)
or should i not be using this approximation?

 

[wabbit]

Should be the same yes, I am just too lazy to check the long formulas:)

One other thing, your first line has an unmatched patenthesis, it's not clear how it should be read.

 

[Mark44]

One other thing, your first line has an unmatched patenthesis, it's not clear how it should be read.

In fact, there are two unmatched parentheses.

$\frac{\partial u}{\partial \phi}=-\frac{Me}{J^{2}}sin\phi + \frac{M^{3}}{J^{4}})(-\frac{1}{e}sin\phi+e^{2}sin2\phi+3e\phi cos \phi$.

 

[binbagsss]

In fact, there are two unmatched parentheses.

Apologies, ta, edited.

 

[wabbit]

You write
$0=\frac{Me}{J^{2}} -\epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))$
But this should be
$0=\frac{Me}{J^{2}} \epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))$
But after that you don't try to solve for $\epsilon$ so I don't know what your solution is.

Also, what you give as the answer doesn't seem correct.

 

[binbagsss]

You write
$0=\frac{Me}{J^{2}} -\epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))$
But this should be
$0=\frac{Me}{J^{2}} \epsilon+ \frac{M^{3}}{J^{4}}(\frac{1}{e}(\epsilon)+e^{2}(2\epsilon)+3e(\pi+\epsilon))$
But after that you don't try to solve for $\epsilon$ so I don't know what your solution is.

Also, what you give as the answer doesn't seem correct.

Thats a multipication by $-\epsilon$ not an addition ! (due to the $cos(\pi)$.

Whilst I haven't tried to solve for $\epsilon$, it is clear that the error occurred in an earlier stage as the expression is not linear in $e$ and so it's not going to cancel- the actual solution is indepedent of $e$...

 

[wabbit]

Well I give up, you keep moving signs and parentheses around, and writing products as sums, I can't keep track.

 

[binbagsss]

bump.

 

[epenguin]

Have you tried inserting the given solution into the expression and seeing whether it works out as 0? If it does you have actually proved it, but comparison with your work might indicate where it went wrong.

Perhaps first check the book to be sure what you have reported as problem really is what they are saying?

Why would anyone change a variable φ to π + ε when that just changes sin φ into - sin ε etc. and doesn't look as if it can advance any argument? (You didn't need to use sine addition formulae!)

This is very easy to solve quite directly in a small angle approximation if φ is close to 0 or to (maybe was meant?) to π/2, and otherwise i think impossible to do so as a useful formula. Does not look right offhand.

If I understood right this is really an ordinary d.e. which is quite easy to solve?

 

[binbagsss]

Have you tried inserting the given solution into the expression and seeing whether it works out as 0? If it does you have actually proved it, but comparison with your work might indicate where it went wrong.

Im not looking for any old proof of it, I'm looking to show and explain it.

Why would anyone change a variable φ to π + ε when that just changes sin φ into - sin ε etc.

There's also a none sin/cos term in the expression?
..But the context here is GR comparing Newtonian orbit zero solution which occurs at $\pi$ and it is justified that we expect the zero solution in GR to occur close to this so plus a small term - $\epsilon$. Hence we plug into solve for this small term.

If I understood right this is really an ordinary d.e. which is quite easy to solve?

well i don't know how to solve it..which is why i posted it here.
But I thought small approximations sounds like the right thin to do in this situation rather than solving directly.
In fact,the book does a similar exercise a few pages on (for timelike geodesics instead of null) and so the equations differ very slightly, the mathematical method is the same, and by using small approximations in a similar way as discussed in the OP I could explain $\epsilon$.

 

[epenguin]

I have no chance of understanding anything from GR, but the math you present makes so little sense I expect only someone knowing GR could guess what the real problem was.

Im not looking for any old proof of it, I'm looking to show and explain it.

If it's so difficult you'd better use the possibility you do have of making sure it's right before spending any more effort on it.

There's also a none sin/cos term in the expression?
..

If LHS is zero then the equation gives φ = (A sin φ + B sin 2φ)/cos φ
and the RHS is unchanged when you change φ into π + φ, so it's a periodic function so if it's satisfied by π + ε so it is by ε as well as by nπ + ε, so there would have been no point in this change of variable.

well i don't know how to solve it..which is why i posted it here.

I see only two variables u and φ there which I think makes this an ode.
On the RHS you have only one very slightly difficult term φ cos φ which = d(φ sin φ)/dφ - sin φ, so you have on RHS all functions you know the integrals of.

I do not have the impression I have solved anything but that you have misdescribed or misunderststood the problemε,.

 

[wabbit]

As I understand the problem statement (it seems clear to me, not sure what I miss), there is no ODE here, just an equation of the form A×epsilon+B=0, and the solution is epsilon =-B/A. That is, after the correct approximations sin(pi+epsilon)=-epsilon etc. have been made, and if the equation is derived correctly without replacing at some stage a multiplication C×(-epsilon) by a difference (C-epsilon), which was OP's error.

 

[epenguin]

It is easy enough to do a small angle calculation of the given equation with LHS = 0. It gives φ = J/M X something, quite unlike the answer given, and π never comes into it, unless in the ± trivial way I mentioned before.
Maybe some brackets have been missed or misplaced?

The d.e. Is incidental to answering the question asked, but I said this looks like an ode and the OP did not deny it, but then it is so easily solved you wonder why they would do approximations, another reason to suspect misquotation.

This is enough exegesis to try and find out what the real problem was and question should have been (I am often doing it) and the OP should reproduce the relevant passages from the book to get any help.

 

[wabbit]

Nothing to add to that:) that was pretty much my sentiment in post#9.

 

[binbagsss]

In addition to post 12 on the context of the question, here is the book, pretty much word for word:

$\frac{\partial u}{\partial \phi}=-\frac{Me}{J^{2}}sin\phi+\frac{M^{3}}{J^{4}}(\frac{-1}{e}sin \phi + e^{2}sin 2\phi+3e\phi cos \phi)$ [1]

This is zero at $\phi=0$, and then the next zero is not at $\phi=\pi$, as it is Newtonian theory but at say $\phi=\pi+\epsilon$. Substituting into [1] we find $\epsilon=\frac{3M^{2}}{J^{2}}\pi$ **

My working:

Using:

$cos(\pi+\epsilon)=-cos(\epsilon)$
$sin(\pi+\epsilon)=-sin(\epsilon)$
$sin(\epsilon) = \epsilon$
$cos(\epsilon)=1$

The last two approximations for some small quantity.

$0= \frac{Me}{J^{2}}sin\epsilon-\frac{M^{3}}{J^{4}}(\frac{-1}{e}sin \epsilon + e^{2}sin 2\epsilon-3e( \pi+ \epsilon) cos \epsilon)) =\frac{Me}{J^{2}}\epsilon-\frac{M^{3}}{J^{4}}(\frac{-1}{e}\epsilon + e^{2} 2\epsilon-3e( \pi+ \epsilon) ))$

Which is not linear in $e$, I can't see any more approximations to use, and so don't get answer ** as the book.

 

[epenguin]

You have an e written small and written large. I have been assuming they are the same thing, are they?

A way it would begin to make sense would be if the second and third terms can be assumed negligible and then you do get the stated result, among others.
Difficult for (-1/e + 2e²) to be negligible (<< 3πe) it seems to me.

 

[wabbit]

@binbagsss, you say the equation you arrive at is not linear in $e$ - yes, but that doesn't matter : your problem statement says "Find $\epsilon$ ",so you are looking to solve for $\epsilon$ ,not $e$ . so what is the equation for $\epsilon$? Can you regroup the terms that have the same power of $\epsilon$? What do you get ?

You really need to think a bit more carefully, read your problem statement, derive your solution with care, and you will find that this is far easier than you expect. And if you do that and find an answer that is different from the book, check all your steps, make sure you didn't make a mistake, redo the calculation if you find a mistake, and then tell us what you found. We can tell you then if there is a mistake in the book or if you missed something. Show us your result, don't just say "I get something different."

 

[epenguin]

Mods, maybe this should be transferred to the S&G Relativity or Advanced Physics forums where there is surely someone familiar with what these authors are telling or similar things and what the point is?

Student who has made some effort should not be left stuck on one formula (and I wouldn't mind seeing the explanation, and is it true this is just a solvable ode?).

 

[binbagsss]

@binbagsss, you say the equation you arrive at is not linear in $e$ - yes, but that doesn't matter : your problem statement says "Find $\epsilon$ ",so you are looking to solve for $\epsilon$ ,not $e$ . so what is the equation for $\epsilon$? Can you regroup the terms that have the same power of $\epsilon$? What do you get ?

You really need to think a bit more carefully, read your problem statement, derive your solution with care, and you will find that this is far easier than you expect. And if you do that and find an answer that is different from the book, check all your steps, make sure you didn't make a mistake, redo the calculation if you find a mistake, and then tell us what you found. We can tell you then if there is a mistake in the book or if you missed something. Show us your result, don't just say "I get something different."

My point was that if if its not linear in $e$ they are not going to cancel and so the solution of $\epsilon$ will not be independent of $e$ as the actual solution is. Actually sorry, I could have made that clearer, I meant the same power of $e$

I didn't make the next step because from the above reasoning something has already gone wrong, and it's a very small step to get to the final wrong solution I'm getting,from #17, but here it is:$\epsilon(\frac{Me}{J^{2}}+\frac{M^{3}}{J^{4}e}-\frac{2e^{2}M^{3}}{J^{4}}+\frac{3eM^{3}}{J^{4}})=\frac{3e\pi M^{3}}{J^{4}}$

 

[epenguin]

You haven't noticed there is an unnecessary factor you can remove?

Now the e's seem to be all t he same. What are they, e of nat log fame? charge on electron?...?

Doh, I'm beginning to see now: removing the factor you get

ε = (M²/J²)(aε + 3π)

and as ε is small you get the quoted result.
(Though many others).

 

[wabbit]

My point was that if if its not linear in $e$ they are not going to cancel and so the solution of $\epsilon$ will not be independent of $e$ as the actual solution is. Actually sorry, I could have made that clearer, I meant the same power of $e$

I didn't make the next step because from the above reasoning something has already gone wrong, and it's a very small step to get to the final wrong solution I'm getting,from #17, but here it is:$\epsilon(\frac{Me}{J^{2}}+\frac{M^{3}}{J^{4}e}-\frac{2e^{2}M^{3}}{J^{4}}+\frac{3eM^{3}}{J^{4}})=\frac{3e\pi M^{3}}{J^{4}}$

OK that's close, you only got one sign wrong here (oops, sorry, two), this should be :
$\epsilon(\frac{Me}{J^{2}}+\frac{M^{3}}{J^{4}e}+\frac{2e^{2}M^{3}}{J^{4}}-\frac{3eM^{3}}{J^{4}})=\frac{3e\pi M^{3}}{J^{4}}$

Now this is not the result yet. You still need to express epsilon, I won't do that small easy step for you. Also, divide everything by Me/J^2, just to make it more readable.

Once you do that, how does this compare to the expression in the book ? Does the book say anything about the values of M, J, and e ?

 

[binbagsss]

ε = (M²/J²)(aε + 3π)

.

What the is $a$?

 

[epenguin]

What the is $a$?

An expression which is sum of several but the point is it's a constant. I thought wabbit might think I was spoonfeeding too much.

 

[binbagsss]

An expression which is sum of several but the point is it's a constant. I thought you might think I was spoonfeeding too much.

ok
I've done this and got $a=\frac{M^{2}}{J^{2}}-\frac{eM^{2}}{J^{2}}-\frac{M^{2}}{e^{2}J^{2}}$
But justifying removing this because $\epsilon$ is small doesn't make sense to me.
Why wouldn't you have considered these terms when making $\epsilon$ the subject

 

[binbagsss]

OK that's close, you only got one sign wrong here (oops, sorry, two), this should be :
$\epsilon(\frac{Me}{J^{2}}+\frac{M^{3}}{J^{4}e}+\frac{2e^{2}M^{3}}{J^{4}}-\frac{3eM^{3}}{J^{4}})=\frac{3e\pi M^{3}}{J^{4}}$

Now this is not the result yet. You still need to express epsilon, I won't do that small easy step for you. Also, divide everything by Me/J^2, just to make it more readable.

Once you do that, how does this compare to the expression in the book ? Does the book say anything about the values of M, J, and e ?

$e^{2}=1+h\frac{J^{2}}{M^{2}}$, $h$ is a constant positive or negative.
But there's a term containing $e$ only so you get $(1+h\frac{J^{2}}{M^{2}})^{1/2}$ which looks very messty.

 

[wabbit]

ok
I've done this and got $a=\frac{M^{2}}{J^{2}}-\frac{eM^{2}}{J^{2}}-\frac{M^{2}}{e^{2}J^{2}}$
But justifying removing this because $\epsilon$ is small doesn't make sense to me.
Why wouldn't you have considered these terms when making $\epsilon$ the subject

I agree with you here,

The equation $\epsilon=\frac{M^2}{J^2}(a\epsilon+3\pi)$ is equivalent to $(1-a\frac{M^2}{J^2})\epsilon=3\pi\frac{M^2}{J^2}$ so you still need to show that $a\frac{M^2}{J^2}\ll 1$ if you are to reach the book's solution.

 

[epenguin]

I may have expressed it badly, sufficient was my earlier assumption that M²/J² << 1 essentially. OK I thought I had said that.

Then maybe we should say we get ε = something small which is OK, but within that small something 3π is much bigger than the rest of the expression in brackets.

 

[wabbit]

@binbagsss do you have any other information about e maybe or h? You need to show that $\frac{1}{e^2}+2e-3\ll\frac{J^2}{M^2}=\frac{e^2-1}{h}$.

 

[vela]

In fact, the book does a similar exercise a few pages on (for timelike geodesics instead of null) and so the equations differ very slightly.

Is this problem considering a null geodesic? I figured it would be a timelike geodesic since you're comparing it to a Newtonian orbit.

I take it M=mass, J=angular momentum, e=eccentricity, and u=1/r. Is that right?

For what it's worth, I don't see how you're supposed to get the expected result without some other assumption that allows you to neglect more terms.

 

[binbagsss]

Is this problem considering a null geodesic? I figured it would be a timelike geodesic since you're comparing it to a Newtonian orbit.

I take it M=mass, J=angular momentum, e=eccentricity, and u=1/r. Is that right?

For what it's worth, I don't see how you're supposed to get the expected result without some other assumption that allows you to neglect more terms.

Apologies yes it's time-like, I said this the wrong way around.
Yes, with $e$ as given by #27.

In justifying neglecting the $u^{3}$ the book gives an idea of figures for the case of the Earth orbitting the Sun, these are:
$2Mu$ ~ $10^{-8}$
$J^{2}u^{2}$~$10^{-8}$
$2MJ^{2}u^{3}$~$10^{-16}$