Algebra /derivatives/ chain rule/

[binbagsss]

Homework Statement

$J=r^{2}\dot{\phi}$ [1]
$\dot{r^{2}}=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}$. [2]
(the context is geodesic equation GR, but I'm pretty sure this is irrelevant).

where $u=r^{-1}$

Question: From these two equations to derive $(\frac{du}{d\phi})^{2}=\frac{1}{J^{2}}(E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r})$

Homework Equations

As above.

The Attempt at a Solution

[/B]
I know that $(\frac{du}{d\phi})^{2}=(\frac{\dot{u^{2}}}{\dot{\phi^{2}})}$.

The $J^{2}$ in the denominator is throwing me, I see from [1] that $\dot{\phi}=Jr^{-2}$ so there's some relation between $\dot{\phi}$ and $J$, but what about a $r^{-4}$ term with the $J^{2}$, since the long term in brackets remains from [2].

I'm guessing I'm missing something simple but I've spent a long time looking at it and no luck.

Cheers.

 

[HallsofIvy]

Homework Statement

$J=r^{2}\dot{\phi}$ [1]
$\dot{r^{2}}=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}$. [2]
(the context is geodesic equation GR, but I'm pretty sure this is irrelevant).

Question: From these two equations to derive $(\frac{du}{d\phi})^{2}=\frac{1}{J^{2}}(E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r})$

There is NO "u" in either of the two equations, so your result makes no sense.

2. Homework Equations

As above.

The Attempt at a Solution

[/B]
I know that $(\frac{du}{d\phi})^{2}=(\frac{\dot{u^{2}}}{\dot{\phi^{2}})}$.

The $J^{2}$ in the denominator is throwing me, I see from [1] that $\dot{\phi}=Jr^{-2}$ so there's some relation between $\dot{\phi}$ and $J$, but what about a $r^{-4}$ term with the $J^{2}$, since the long term in brackets remains from [2].

I'm guessing I'm missing something simple but I've spent a long time looking at it and no luck.

Cheers.

 

[binbagsss]

There is NO "u" in either of the two equations, so your result makes no sense.

cheers, edited.

 

[binbagsss]

Bump. Anyone please, I'm still stuck?

 

[Fredrik]

I don't understand this problem. $r$ and $\phi$ are coordinates, right? So r isn't an invertible function, and 1/r isn't a function of $\phi$. So what do you mean by $r^{-1}$? I think you will need to include a lot more information.

 

[binbagsss]

I don't understand this problem. $r$ and $\phi$ are coordinates, right? So r isn't an invertible function, and 1/r isn't a function of $\phi$. So what do you mean by $r^{-1}$? I think you will need to include a lot more information.

Original post, first lst line under the attempt at the solution
We have both $\phi$ and $r$ differentiated with respect to some parameter, denoted by a dot, so dividing these two has gave $du / \ d \phi$

 

[Fredrik]

Original post, first lst line under the attempt at the solution
We have both $\phi$ and $r$ differentiated with respect to some parameter, denoted by a dot, so dividing these two has gave $du / \ d \phi$

All you're saying there, is that if $u$ is a function of $\phi$, and $\phi$ is a function of $t$, then
$$\frac{du}{dt}=\frac{du}{d\varphi}\frac{d\varphi}{dt}.$$ This is just the chain rule. But you haven't posted a meaningful definition of $u$. You said that $u=r^{-1}$, but the map that takes $t$ to $r$ isn't invertible so did you really mean the inverse of that non-invertible function? If you meant $1/r$, then $r$ has to be a function of $\phi$ (if it's not, then $u$ isn't, and the chain rule can't be applied as written).

You need to post the full problem statement.

 

[binbagsss]

All you're saying there, is that if $u$ is a function of $\phi$, and $\phi$ is a function of $t$, then
$$\frac{du}{dt}=\frac{du}{d\varphi}\frac{d\varphi}{dt}.$$ This is just the chain rule. But you haven't posted a meaningful definition of $u$. You said that $u=r^{-1}$, but the map that takes $t$ to $r$ isn't invertible so did you really mean the inverse of that non-invertible function? If you meant $1/r$, then $r$ has to be a function of $\phi$ (if it's not, then $u$ isn't, and the chain rule can't be applied as written).

You need to post the full problem statement.

I'm looking at Tod and Hughston introductuction to GR and it says introducing a new variable $u=r^{-1}$ from [1] and [2] we can solve for $u$ as a function of $\phi$.

This is all the book says

 

[Mark44]

All you're saying there, is that if $u$ is a function of $\phi$, and $\phi$ is a function of $t$, then
$$\frac{du}{dt}=\frac{du}{d\varphi}\frac{d\varphi}{dt}.$$ This is just the chain rule. But you haven't posted a meaningful definition of $u$. You said that $u=r^{-1}$, but the map that takes $t$ to $r$ isn't invertible so did you really mean the inverse of that non-invertible function? If you meant $1/r$, then $r$ has to be a function of $\phi$ (if it's not, then $u$ isn't, and the chain rule can't be applied as written).

I'm pretty sure that u = r^-1 means u = 1/r in this context.

$J=r^{2}\dot{\phi}$ [1]
$\dot{r^{2}}=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}$[2].

Let u = 1/r, so r = 1/u
From this we have dr/dt = -1/u² ( du/dt)

Dispensing with the dot notation, equation [2] is $(\frac{dr}{dt})^2=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}$
Hence with the u substitution, we have $(\frac{-1}{u^2})^2(\frac{du}{dt})^2=E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r}$
Multiply both sides by u⁴ to get
$(\frac{du}{dt})^2=u^4(E^{2}-1-\frac{J^{2}}{r^{2}}+\frac{2MJ^{2}}{r^{3}}+\frac{2M}{r})$
Multiply both sides by $(\frac{dt}{d\phi})^2$ and the desired result can be quickly obtained.

From the OP, I am inferring that both $\phi$ and r are differentiable functions of t, otherwise $\dot{\phi}$ and $\dot{r}$ wouldn't make sense.

 

[Fredrik]

OK, if u=1/r, and there's a constraint that ensures that if you know the value of one of the variables $r$, $\phi$ and $t$, you can calculate the values of the other two, then the proof is a straightforward calculation. The word "geodesic" was mentioned, so I guess these variables must be sufficiently constrained by the assumption that the curve we're dealing with is a geodesic. I haven't tried to prove that, but it sounds reasonable when you think about straight lines in $\mathbb R^2$ for example.

I would do this calculation by proving that $\frac{du}{dt}$ is equal to both $-u^2\dot r$ and $\frac{du}{d\phi}Ju^2$. This implies that $\big(\frac{du}{d\phi}\big)^2=\frac{\dot r^2}{J^2}$.