Kerr Metric Confusion: Problem 1 on Page 138

[latentcorpse]

Hi.

I'm trying problem 1 on p138 of this
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf

Now when I try and get the Euler Lagrange equation for $\phi$ I get
(the Kerr metric in BL coordinates can be found at the bottom of p77)

$\frac{\partial L}{\partial \phi} = \frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{\tau}} \right)$

$0= \frac{d}{d \tau} \left( 2 \frac{(r^2+a^2)^2- \Delta a^2 \sin^2{\theta}}{\Sigma} \sin^2{\theta} \frac{d \phi}{d \tau} - \frac{2a \sin^2{\theta} ( r^2 + a^2 - \Delta ) }{\Sigma} \frac{dt}{d \tau} \right)$

But the question says we are in the equatorial plane and so when I impose $\theta=\frac{\pi}{2}$ all the sine terms cancel and I'm left with $0=\frac{d}{d \tau} (0)$ which is useless - what am I doing wrong here?

 

[fzero]

In what way do the sine terms cancel?

 

[latentcorpse]

sorry i meant they vanish (not cancel) and that leads to the equation 0=0 which is useless...

 

[fzero]

They don't vanish either, $$\sin(\pi/2)=1$$.

 

[latentcorpse]

Ooops!
Sorry!

But anyway, I still find the EL equations for t give

$\frac{2 \Delta - 2a^2}{\Sigma} \frac{dt}{d \tau} + \frac{2ar^2 + 2 a^3 -2a \Delta}{\Sigma} \frac{d \phi}{d \tau} = E$

and for $\phi$ they give

$\frac{2(r^2+a^2)^2-\Delta a^2}{\Sigma} \frac{d \phit}{d \tau} - \frac{2a(r^2+a^2-\Delta)}{\Sigma} \frac{dt}{d \tau} = h$

Now by analogy with what we do in the Schwarzschild case (where we rearrange to get an expression for $\frac{dt}{d \tau}$ in terms of E and an expression for $\frac{d \phi}{d \tau}$ in terms of h), we want to try and find $\frac{dt}{d \tau}$ and $\frac{d \phi}{d \tau}$.
Is that correct? How would I go about this? I can't seem to solve these equations simultaneously?

 

[fzero]

You have 2 equations in 2 unknowns, so it's easy to solve these equations. The algebra is quite a bit simpler if you don't substitute expressions for the metric components $$g_{\mu\nu}$$.

 

[latentcorpse]

You have 2 equations in 2 unknowns, so it's easy to solve these equations. The algebra is quite a bit simpler if you don't substitute expressions for the metric components $$g_{\mu\nu}$$.

I seem to be getting overcomplicated though:

$\mathcal{L}=-g_{\mu \nu} \frac{dx^\mu}{d \tau} \frac{d x^\nu}{d \tau}=-g_{tt} \dot{t}^2 - g_{t \phi} \dot{t} \dot{\phi} - g_{\phi \phi} \dot{\phi}^2 - g_{rr} \dot{r}^2$ where we used $\dot{\theta}=0$

E-L for t gives $2g_{tt} \dot{t} + g_{t \phi} \dot{\phi}=E$ A

and for $\phi$ the E-L give $g_{t \phi} \dot{t} + 2 g_{\phi \phi} \dot{\phi} = h$ B

Now consider $2g_{\phi \phi} A - g_{t \phi} B$ this gives $\dot{t}=\frac{2 g_{\phi \phi} E - g_{t \phi} h}{4 g_{tt} g_{\phi \phi} - g_{t \phi}^2}$

and consider $g_{t \phi} A -2 g_{tt} B$ this gives $\dot{\phi} = \frac{g_{t \phi} E - g_{tt} h}{g_{t \phi}^2 -4g_{\phi \phi} g_{tt}}$

Then to get the radial equation we use the fact that

$g_{\mu \nu} \frac{d x^\mu}{d \tau} \frac{dx^\nu}{d \tau}=-\sigma$
$\Rightarrow g_{tt} \dot{t}^2 + g_{t \phi} \dot{t} \dot{\phi} + g_{\phi \phi} \dot{\phi}^2 + g_{rr} \dot{r}^2 + \sigma=0$

Substituting gives

$g_{tt} \left( \frac{2g_{\phi \phi} E - g_{t \phi} h}{4 g_{tt} g_{\phi \phi} - g_{t \phi}^2} \right)^2 + g_{t \phi} \left( \frac{g_{\phi \phi} E - g_{t \phi} h}{4g_{tt} g_{\phi \phi} - g_{t \phi}^2} \right) \left( \frac{ g_{t \phi} E - 2 g_{tt} h}{g_{t \phi}^2 - 4 g_{\phi \phi} g_{tt}} \right) + \left( \frac{g_{t \phi} E - 2 g_{tt} h}{g_{t \phi}^2 - 4 g_{\phi \phi}g_{tt}} \right)^2 g_{\phi \phi} + g_{rr} \dot{r}^2 + \sigma=0$

$g_{tt} \left( \frac{ 4g_{\phi \phi}^2 E^2 - 4 g_{\phi \phi} g_{t \phi} Eh + g_{t \phi}^2 h^2}{16 g_{tt}^2 g_{\phi \phi^2} - 8 g_{tt} g_{\phi \phi} g_{t \phi}^2 + g_{t \phi}^4} \right) + g_{t \phi} \left( \frac{2 g_{\phi \phi} g_{t \phi} E^2 - g_{t \phi}^2 Eh - 4 g_{tt} g_{\phi \phi} Eh + 2 g_{tt} g_{t \phi} h^2}{ 16 g_{tt}^2 g_{\phi \phi}^2 - 8 g_{tt} g_{\phi \phi} g_{t \phi}^2 + g_{t \phi}^4} \right) + \left( \frac{g_{t \phi}^2 E^2 - 4 g_{tt} g_{t \phi} Eh + 4 g_{tt}^2 h^2}{16 g_{tt}^2 g_{\phi \phi}^2 - 8 g_{tt} g_{\phi \phi} g_{t \phi}^2 + g_{t \phi}^4} \right) g_{\phi \phi} + g_{rr} \dot{r}^2 + \sigma=0$

Then collect the $E^2 , E h$ and $h^2$ terms to get

$\left( \frac{ ( 4g_{tt}g_{\phi \phi}^2 + 3 g_{t \phi}^2 g_{\phi \phi} ) E^2 - ( 8 g_{\phi \phi} g_{t \phi} + 4 g_{tt} g_{ \phi \phi} + g_{t \phi}^2) Eh + ( g_{t \phi}^2 + 2 g_{tt} g_{t \phi} + 4 g_{tt}^2 ) h^2}{16 g_{tt}^2 g_{\phi \phi}^2 - 8 g_{tt} g_{\phi \phi} g_{t \phi}^2 + g_{t \phi}^4} \right) + g_{rr} \dot{r}^2 + \sigma=0$

But it still seems far to complicated to even attempt substituting for the metric components?

 

[fzero]

I seem to be getting overcomplicated though:

$\mathcal{L}=-g_{\mu \nu} \frac{dx^\mu}{d \tau} \frac{d x^\nu}{d \tau}=-g_{tt} \dot{t}^2 - g_{t \phi} \dot{t} \dot{\phi} - g_{\phi \phi} \dot{\phi}^2 - g_{rr} \dot{r}^2$ where we used $\dot{\theta}=0$

This should be

$$-g_{tt} \dot{t}^2 - 2g_{t \phi} \dot{t} \dot{\phi} - g_{\phi \phi} \dot{\phi}^2 - g_{rr} \dot{r}^2$$

That factor of 2 leads to some cancellations later on.

Then collect the $E^2 , E h$ and $h^2$ terms to get

$\left( \frac{ ( 4g_{tt}g_{\phi \phi}^2 + 3 g_{t \phi}^2 g_{\phi \phi} ) E^2 - ( 8 g_{\phi \phi} g_{t \phi} + 4 g_{tt} g_{ \phi \phi} + g_{t \phi}^2) Eh + ( g_{t \phi}^2 + 2 g_{tt} g_{t \phi} + 4 g_{tt}^2 ) h^2}{16 g_{tt}^2 g_{\phi \phi}^2 - 8 g_{tt} g_{\phi \phi} g_{t \phi}^2 + g_{t \phi}^4} \right) + g_{rr} \dot{r}^2 + \sigma=0$

But it still seems far to complicated to even attempt substituting for the metric components?

If you fix the factors of 2, you'll find something like

$\left( \frac{ g_{\phi \phi} E^2 + 2 g_{t \phi} Eh + g_{tt} h^2}{ g_{tt} g_{\phi \phi} - g_{t \phi}^2 } \right) + g_{rr} \dot{r}^2 + \sigma=0.$

 

[latentcorpse]

This should be

$$-g_{tt} \dot{t}^2 - 2g_{t \phi} \dot{t} \dot{\phi} - g_{\phi \phi} \dot{\phi}^2 - g_{rr} \dot{r}^2$$

That factor of 2 leads to some cancellations later on.



If you fix the factors of 2, you'll find something like

$\left( \frac{ g_{\phi \phi} E^2 + 2 g_{t \phi} Eh + g_{tt} h^2}{ g_{tt} g_{\phi \phi} - g_{t \phi}^2 } \right) + g_{rr} \dot{r}^2 + \sigma=0.$

Where's the factor of 2 from?

 

[fzero]

Where's the factor of 2 from?

When you sum over indices, both $$g_{t\phi}$$ and $$g_{\phi t}$$ terms contribute as usual.

 

[latentcorpse]

When you sum over indices, both $$g_{t\phi}$$ and $$g_{\phi t}$$ terms contribute as usual.

Here's what I've got so far:

E-L for t gives $g_{tt} \dot{t} + g_{t \phi} \dot{\phi} =E$
and for $\phi$ we get $g_{t \phi} \dot{t} + g_{\phi \phi} \dot{\phi} =h$

These can be solved to get

$\dot{t} = \frac{g_{\phi \phi} E - g_{t \phi} h}{g_{\phi \phi} g_{tt} - g_{t \phi}^2} , \quad \dot{\phi} = \frac{g_{t \phi} E - g_{tt} h}{g_{t \phi}^2 - g_{tt} g_{\phi \phi}}$

Radial equation also needs a missing factor of 2 from $g_{\mu \nu} \frac{dx^\mu}{d \tau} \frac{dx^\nu}{d \tau}=-\sigma$

and so we get

$g_{tt} \dot{t}^2 + 2 g_{t \phi} \dot{t} \dot{\phi} + g_{\phi \phi} \dot{\phi}^2 + g_{rr} \dot{r}^2 + \sigma=0$

Subbing in and expanding out the squares we get

$g_{tt} \left( \frac{g_{\phi \phi}^2 E^2 + g_{t \phi}^2 h^2 - 2 g_{\phi \phi} g_{t \phi} Eh}{g_{\phi \phi}^2 g_{tt}^2 + g_{t \phi}^4 - 2 g_{\phi \phi} g_{tt} g_{t \phi}^2} \right) + 2g_{t \phi} \left( \frac{ g_{\phi \phi} g_{t \phi} E^2 + g_{t \phi} g_{tt} h^2 - ( g_{t \phi}^2 + g_{tt} g_{\phi \phi}) Eh}{g_{\phi \phi} g_{tt} g_{t \phi}^2 - g_{t \phi}^4 -g_{tt}^2 g_{\phi \phi}^2 -g_{t \phi}^2 g_{tt} g_{\phi \phi}} \right) + g_{\phi \phi} \left( \frac{ g_{t \phi}^2 E^2 + g_{tt}^2 h^2 -2g_{t \phi} g_{tt} Eh}{g_{t \phi}^4 + g_{tt}^2 g_{\phi \phi}^2 - 2g_{tt} g_{\phi \phi} g_{t \phi}^2} \right) + g_{rr} \dot{r}^2 + \sigma=0$

Now if you look at the denominator of the 2nd term in that last line, it will reduce to the denominator that you have in your answer but the denominators of the other two terms remain a problem as they don't cancel?

 

[fzero]

You have a sign error in the last term of that denominator. The denominators of all 3 terms are equal to the square of the same determinant up to a sign. When you collect terms in the numerator, you cancel one factor of the determinant.

 

[latentcorpse]

You have a sign error in the last term of that denominator. The denominators of all 3 terms are equal to the square of the same determinant up to a sign. When you collect terms in the numerator, you cancel one factor of the determinant.

Ok. So I got to that result.

Now for the substitutions, I find

$g_{\phi \phi} = \frac{(r^2+a^2)^2-\Delta a^2}{r^2} =\frac{r^4+r^2a^2+2Mra^2-e^2a^2}{r^2}$

$g_{tt}=-( \frac{\Delta - a^2}{r^2} ) = \frac{2Mr-r^2-e^2}{r^2}$

$g_{t \phi} = -2a ( \frac{r^2+a^2-r^2+2Mr-a^2-e^2}{r^2})=\frac{2ae^2-4Mra}{r^2}$

$g_{rr}=\frac{r^2}{\Delta}=\frac{r^2}{r^2}=1$

Do these look correct? Are there any simplifications I have missed?

I then rearranged our equation to give

$(g_{\phi \phi} g_{tt} - g_{t \phi}^2)(g_{rr} \dot{r}^2+\sigma}) = - g_{\phi \phi} E^2 -g_{tt} h^2 + 2 g_{t \phi} Eh$

Does this look ok to go ahead with substitution now?

Thanks.

Do these look correct? (I'm supposed to be getting an $E^2$ term on its own on the other side from $\dot{r}^2$ and I don't see how that's going to happen - it looks to me like it will always be attached to some factor?)

 

[fzero]

Ok. So I got to that result.

Now for the substitutions, I find

$g_{\phi \phi} = \frac{(r^2+a^2)^2-\Delta a^2}{r^2} =\frac{r^4+r^2a^2+2Mra^2-e^2a^2}{r^2}$

$g_{tt}=-( \frac{\Delta - a^2}{r^2} ) = \frac{2Mr-r^2-e^2}{r^2}$

$g_{t \phi} = -2a ( \frac{r^2+a^2-r^2+2Mr-a^2-e^2}{r^2})=\frac{2ae^2-4Mra}{r^2}$

$g_{rr}=\frac{r^2}{\Delta}=\frac{r^2}{r^2}=1$

Do these look correct? Are there any simplifications I have missed?

I really haven't taken the calculation this far, but I can see you've got a factor of 2 wrong in one place, since

$g_{t \phi} = -a ( \frac{r^2+a^2-r^2+2Mr-a^2-e^2}{r^2})=\frac{ae^2-2Mra}{r^2}$

The other terms are probably ok.

I then rearranged our equation to give

$(g_{\phi \phi} g_{tt} - g_{t \phi}^2)(g_{rr} \dot{r}^2+\sigma}) = - g_{\phi \phi} E^2 -g_{tt} h^2 + 2 g_{t \phi} Eh$

Does this look ok to go ahead with substitution now?

You probably want to solve for $$\dot{r}^2$$ before making substitutions.

Thanks.

Do these look correct? (I'm supposed to be getting an $E^2$ term on its own on the other side from $\dot{r}^2$ and I don't see how that's going to happen - it looks to me like it will always be attached to some factor?)

It's probably just massaging of the coefficient. It's likely that you can obtain that term by adding and subtracting 1 in the right place.

 

[latentcorpse]

I really haven't taken the calculation this far, but I can see you've got a factor of 2 wrong in one place, since

$g_{t \phi} = -a ( \frac{r^2+a^2-r^2+2Mr-a^2-e^2}{r^2})=\frac{ae^2-2Mra}{r^2}$

The other terms are probably ok.



You probably want to solve for $$\dot{r}^2$$ before making substitutions.



It's probably just massaging of the coefficient. It's likely that you can obtain that term by adding and subtracting 1 in the right place.

Well we have $\dot{r}^2=\frac{g_{\phi \phi} E^2 + 2 g_{t \phi} Eh + g_{tt} h^2}{g_{t \phi}^2-g_{tt} g_{\phi \phi}} - \sigma$

Now from the form of the final answer it looks like we want to evaluate the E^2, Eh and h^2 terms separately.

But let's consider first that tricky denominator:

$g_{t \phi}^2 = ( \frac{ae^2-2Mra}{r^2})^2 = \frac{a^2e^4 + 4 M62 r^2 a^2 - 4Mra^2e^2}{r^4}$
$g_{\phi \phi} g_{tt} = ( \frac{r^4+r^2a^2 + 2Mra^2 -e^2a^2}{r^2})(\frac{2Mr-r^2-e^2}{r^2})=\frac{2Mr^5+2Mr^3a^2+4M^2r^2a^2-2Mre^2a^2-r^6-r^4a^2-2Mr^3a^2+r^2e^2a^2-r^4e^2-r^2a^2e^2-2Mra^2e^2+e^4a^2}{r^4}=\frac{2Mr^5+2Mr^3a^2+4M^2r^2a^2-4Mre^2a^2-r^6-r^4a^2-2Mr^3a^2+-r^4e^2+e^4a^2}{r^4}$
$g_{t \phi}^2-g_{\phi \phi}g_{tt}=\frac{r^6+r^4a^2+r^4e^2-4Mr^3a^2-2Mr^5}{r^4}$ after much cancellation

But then suppose we try and work out the first term of the answer (the h^2 term), it should have coefficient $(1-\frac{2M}{r}+\frac{e^2}{r^2}$ but if we put $g_{tt}$ over the above denominator, it doesn't simplify down?

 

[fzero]

I find

$$g_{\phi \phi}g_{tt}-g_{t \phi}^2= \Delta,$$

which ends up canceling against the factor of $$\Delta$$ in $$g_{rr}$$. You should leave $$\Delta$$ unsubstituted to make this calculation easier.

 

[latentcorpse]

I find

$$g_{\phi \phi}g_{tt}-g_{t \phi}^2= \Delta,$$

which ends up canceling against the factor of $$\Delta$$ in $$g_{rr}$$. You should leave $$\Delta$$ unsubstituted to make this calculation easier.

Ok. I find that $g_{\phi \phi}g_{tt}-g_{t \phi}^2=-\Delta$ actually.
This is because $g_{\phi \phi}g_{tt}=\frac{(r^2+a^2)^2}{r^2}\frac{a^2-\Delta}{r^2}$ and if you take the $r^2^2$ from the first term and the $-\Delta$ from the second you get $-r^4 \Delta$ which reduces to $-\Delta$ when you divide by the $r^4$ on the bottom.But our rearranged formula for $\dot{r}$ is

$\dot{r}^2=\frac{1}{g_{rr}} \left( \frac{2g_{t \phi} Eh - g_{\phi \phi} E^2 - g_{tt} h^2}{g_{\phi \phi} g_{tt} - g_{t \phi}^2} \right) - \frac{1}{g_{rr}} \sigma$

Now the $\frac{1}{g_{rr}}=\frac{1}{\frac{\Sigma}{\Delta}}=\frac{\Delta}{\Sigma}=\frac{\Delta}{r^2}$ and so as you said the $\Delta$'s will cancel.

Consider the $h^2$ term now. We have

$\frac{1}{g_rr} \frac{-g_{tt}}{-\Delta}h^2=\frac{g_{tt}}{\Delta}\frac{\Delta}{r^2}h^2=\frac{g_{tt}}{r^2}h^2 = \frac{a^2-\Delta}{r^2} \frac{h^2}{r^2} =\frac{a^2-r^2+2Mr-a^2-e^2}{r^2} \frac{h^2}{r^2} = ( \frac{2M}{r} - 1 - \frac{e^2}{r^2} ) \frac{h^2}{r^2}$
which if you look at the answer is out by a factor of -1 which means you must have been correct with the sign of the denominator that we worked out. But I don't see how? I clearly get a $-\Delta$ when I do that calculation?

 

[fzero]

Ok. I find that $g_{\phi \phi}g_{tt}-g_{t \phi}^2=-\Delta$ actually.

That's right, I found my sign mistake.

Consider the $h^2$ term now. We have

$\frac{1}{g_rr} \frac{-g_{tt}}{-\Delta}h^2=\frac{g_{tt}}{\Delta}\frac{\Delta}{r^2}h^2=\frac{g_{tt}}{r^2}h^2 = \frac{a^2-\Delta}{r^2} \frac{h^2}{r^2} =\frac{a^2-r^2+2Mr-a^2-e^2}{r^2} \frac{h^2}{r^2} = ( \frac{2M}{r} - 1 - \frac{e^2}{r^2} ) \frac{h^2}{r^2}$
which if you look at the answer is out by a factor of -1 which means you must have been correct with the sign of the denominator that we worked out. But I don't see how? I clearly get a $-\Delta$ when I do that calculation?

There's a - sign put in when $$V_\text{eff}$$ is defined.

 

[latentcorpse]

That's right, I found my sign mistake.



There's a - sign put in when $$V_\text{eff}$$ is defined.

Oh yeah. So I now get the answer except for the $Eh$ term where I'm out by a minus sign since

$\dot{r}^2$ has a $\frac{\Delta}{r^2} \frac{2 g_{t \phi}Eh}{-\Delta}=-\frac{2g_{t \phi}Eh}{r^2}$ term

Now if we sub for the metric component using $g_{t \phi}=-\frac{a(r^2+a^2-\Delta)}{r^2}=\frac{a(\Delta-r^2-a^2)}{r^2}=\frac{a(e^2-2mr)}{r^2}$

we get

$-\frac{2g_{t \phi}Eh}{r^2}=-\frac{2a(e^2-2Mr)}{r^2} \frac{Eh}{r^2} = - \frac{2aEh}{r^3} ( \frac{e^2}{r}-2M)$

but then when we take the minus at the front out when we define $V_{\text{eff}}$ we find that the term that contributes to the effective potential is $\frac{2aEh}{r^3} ( \frac{e^2}{r}-2M)$ whereas it's meant to be $\frac{2aEh}{r^3} ( 2M-\frac{e^2}{r})$

 

[fzero]

You're missing a minus sign in front of that term. You should have $\dot{r}^2=\frac{1}{g_{rr}} \left( \frac{-2g_{t \phi} Eh - g_{\phi \phi} E^2 - g_{tt} h^2}{g_{\phi \phi} g_{tt} - g_{t \phi}^2} \right) - \frac{1}{g_{rr}} \sigma$

 

[latentcorpse]

You're missing a minus sign in front of that term. You should have $\dot{r}^2=\frac{1}{g_{rr}} \left( \frac{-2g_{t \phi} Eh - g_{\phi \phi} E^2 - g_{tt} h^2}{g_{\phi \phi} g_{tt} - g_{t \phi}^2} \right) - \frac{1}{g_{rr}} \sigma$

I don't see how

if you follow the contributions from the radial eqn, we had a

$g_{tt} \dot{t}^2 = g_{tt} ( \frac{g_{\phi \phi} E - g_{t \phi} h}{g_{\phi \phi} g_{tt} - g_{t \phi}^2})^2$
the cross term here will be $\frac{-2 g_{tt}g_{\phi \phi} g_{t \phi} Eh}{(g_{\phi \phi} g_{tt} - g_{t \phi}^2)^2}$

we also had a $g_{\phi \phi} \dot{\phi}^2=g_{\phi \phi} ( \frac{g_{ t \phi} E - g_{tt} h}{g_{t \phi}^2 - g_{tt} g_{\phi \phi}} )^2$
the cross term will be $\frac{-2g_{tt} g_{\phi \phi} g_{t \phi} Eh}{(g_{\phi \phi} g_{tt} - g_{t \phi}^2)^2}$

and we also had a $2g_{t \phi} \dot{t} \dot{\phi} = 2 g_{t \phi} ( \frac{g_{\phi \phi} E - g_{t \phi} h}{g_{\phi \phi} g_{tt} - g_{t \phi}^2} ) ( \frac{g_{t \phi} E - g_{tt} h}{ g_{t \phi}^2 - g_{tt} g_{\phi \phi}})$
the Eh contribution will be $\frac{-2g_{t \phi} ( g_{t \phi}^2 + g_{\phi \phi} g_{tt})Eh}{(g_{\phi \phi} g_{tt} - g_{t \phi}^2)^2}$Putting all these contributions together we get

$\frac{-2 g_{\phi \phi} g_{tt} g_{t \phi} + 2g_{t \phi}^2 + 2 g_{\phi \phi} g_{tt} g_{t \phi} - 2 g_{\phi \phi} g_{tt} g_{t \phi}}{(g_{\phi \phi} g_{tt} - g_{t \phi})^2} Eh$

$=\frac{2 g_{t \phi} ( g_{t \phi} - g_{tt} g_{\phi \phi})}{( g_{\phi \phi} g_{tt} - g_{t \phi})^2}Eh$

$=\frac{-2g_{t \phi} Eh}{g_{\phi \phi} g_{tt} - g_{t \phi}}$

and then when we take that over to the other side so that we have $\dot{r}^2 = \dots$, it will become positive, no?

 

[fzero]

I found another mistake and agree with your sign now. I haven't been able to match the sign of the quoted answer.