Orbital velocities in the Schwartzschild geometry

[yossell]

Hi espen180

It seems you find the fact that a function of independent coordinates is set equal to a constant.

I'm not sure what you meant to say here - there's a missing word - `confusing'?

I resolution is that when such a restraint is set on the system, the coordinates are no longer fully independent. If you have s function of 4 variables, you may only choose 3 of them freely. The fourth is determined from the values of the other three.

Right - I think I see what you're saying here, but I'll write out what I think you're suggesting to make relevant distinctions explicit. I apologise for making this long-winded, but I think the standard abbreviations are causing confusion, and I want to clearly distinguish between functions, equations, points that satisfy an equation etc.

This is the kind of process you describe in your paragraph, fleshed out using your nice example from a few posts ago, where this starting function 3 place rather than 4 place though.

Begin with the *function* f(x, y, z) = ax + by + cz + d.

Consider those points <x, y, z> which, when plugged into the function f, make f output 0.

This defines a set of triples (or points of a 3-d manifold) <x, y, z>

Another way of defining the same set of points is by the equation: ax + by + cz + d = 0.

Because, for each x and y, there is one and only one z which satisfies this equation, we can rewrite the same set of points as <x, y, (- d - ax - by)/c>.

(note that we would have to proceed with more care if we had begun with a function h(x y z) = x^2 + y^2 + z^2 - d^2. The <x y z> which satisfy h = 0 define a sphere, and in this case, for each x and y, there are two values of z that solve the equation)

So the third value is clearly a function of the first two.

So we can define a NEW *two place* function g(x y) = (-d - ax - by)/c.

This two place function can now be differentiated as usual - with respect to x or with respect to y, to get -a/c and -b/c respectively.

Ok, very longwinded, sorry. But hopefully the worry is now clear: this isn't what happens in the example you present, and it doesn't permit the differentiations you make.

If L is just the function which equals ax + by + cz + d, then the differentiations you write make sense and follow from the definition of L, but you can't assert that the *function* L = 0.

If, on the other hand, you are implicitly defining a function along the (longwinded) lines above by considering the x,y,z that satisfy L = zero, then you need to make explicit which function you've chosen (x in terms of y and z - or y in terms of z and x - or z in terms of x and y) AND the differentiations you write do not follow. Firstly, there are only two variables in the function defined; secondly, as you see above, the answers you get for these functions are not simply a, b or c.

 

[espen180]

It's an invalid "counter-example". r(s) is a function of a single variable. Basic calculus tells you that r(s)=0 implies r&#039;(s)=0.

I was talking about your objection about differentiating the Lagrangian.

 

[Altabeh]

You, espen180 and Altabeh share the same basic misconception. It is clear that your misconception cannot be fixed.

Did you learn the Lagrangian procedure described in Hobson's book? If not, then stop following your wishful thinking and try reading books! I'm ready here to see if you're "able" to object that author's ideas or otherwise I have to reall that you've been declared "finished" finished since your post 251!

AB

 

[yuiop]

It's an invalid "counter-example". r(s) is a function of a single variable. This is the first error.
The second error is that you did not define a function, you defined an equation. It will become clearer to you when you correct your "counter example" to look like this f(x)=2x^2=1
You, espen180 and Altabeh share the same basic misconception. It is clear that your misconception cannot be fixed.

We are talking in the context of the Schwarzschild metric. Clearly the metric is a function of more than one variable.

 

[Altabeh]

H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}

From the above, it is clear that, contrary to your repeated fallacious claims, H_c is a function of r. Since H_c is a function of r, your whole differentiation of L as if H_c did not depend on r fails

Wow! I have to confess that I'm really starting to enjoy wittnessing someone making this much of nonsense claims that even do not cost a penny to be corrected. Look, H=r^2\dot{\phi} is a "constant of motion" and is to be considered in the equatorial plane so that by definition the spherical polar coordinate r is the same as the plane polar coordinate R=R(\phi). Thus the derivative of H wrt r is zero. By a similar reasoning we can get that K is an absolute constant. To learn all of this stuff, I want you to take a look at the pages 193-196 of this introductory book:

Ray D'inverno, Introducing Einstein's Relativity, 1998.

If you had a good knowledge in math, I would discuss this in a better way using the so-called "Killing vectors" but unfortunately you're really a rookie and have deep problems with basic calculus/algebra so I think it's good to keep it at a very low degree of difficulty!

As pointed out repeatedly to you in several posts in this thread, your derivation fails since it is based on a gross fallacy

Really? Go find another way to leak into kev's 100% "correct" method! You nonsense claims are now countless!

 

[starthaus]

I was talking about your objection about differentiating the Lagrangian.

Your objection is still wrong. See post 330.

 

[starthaus]

We are talking in the context of the Schwarzschild metric. Clearly the metric is a function of more than one variable.

No, we are talking about your inability (and espen180's and Altabeh) to understand the simple facts of calculus : f(x)=0=&gt;df/dx=0.

 

[Altabeh]

It's an invalid "counter-example". r(s) is a function of a single variable. This is the first error.
The second error is that you did not define a function, you defined an equation. It will become clearer to you when you correct your "counter example" to look like this f(x)=2x^2=1
You, espen180 and Altabeh share the same basic misconception. It is clear that your misconception cannot be fixed.

The second error is nothing but another nonsense of yours! The first CP has nothing to do with your old fallacy so get a clue on what the whole problem was! Nonetheless with all those nonsense claims I feel like you don't have access to books/papers to read and learm. So you can ask us if you want a book!

AB

 

[Altabeh]

No, we are talking about your inability (and espen180's and Altabeh) to understand the simple facts of calculus : f(x)=0=&gt;df/dx=0.

This has nothing to do with your old fallacy. Get a clue!

AB

 

[ZapperZ]

Closed, pending moderation.

Zz.

 

[ZapperZ]

Thread has been reopened, but it is still under moderation.

I would strongly recommend those participating to stick with the physics discussion and cease with the personal attack and trying to belittle other participants in here. There will be serious infractions forthcoming.

If you notice another member providing false information, REPORT IT. If you take it upon yourself to tackle it and it turns ugly, you bear the same responsibility in its escalation.

Zz.

 

[Altabeh]

Hi Altabeh,

Of your example, I don't see why the Lagrangian you write is, as you claim, either -1, 0 or 1. It looks to me like a function of position and velocity vectors, and will vary accordingly. What am I missing?

Hi my friend

This is simply because I've taken the affine parameter \tau to be the proper length s so that deviding each side of the metric

ds^2=g_{ab}dx^adx^b

by ds^2 we get

1=g_{ab}\dot{x}^a\dot{x}^b=L

where the over-dot here represents differentiation wrt s. In some textbooks, this often reads

-1=g_{ab}\dot{x}^a\dot{x}^b=L

in which the difference comes from the fact that for non-null geodesics the line-element conditions are replaced, i.e. for spacelike geodesics ds^2&gt;0 and for timelike geodesics ds^2&lt;0.

But for null geodesics we can't use s as affine parameter because simply ds=0. Hence we choose an arbitrary affine parameter e.g. \tau and set the Lagrangian as

L=g_{ab}\dot{x}^a\dot{x}^b=0

where now the over-dot is the derivative wrt the parameter \tau.

But why a constant Lagrangian seems to not have a zero derivative when differentiated in the Euler-Lagrange equations? Note that when dealing with Euler-Lagrange equations and the Lagrangian L=g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}, if \tau=s, we are not allowed to insert g_{ab}\frac{dx^a}{ds}\frac{dx^b}{ds}=1 into L; this condition is only valid along the actual worldline of particle where the geodesic parameter u coincides with the proper length s. The other nearby worldlines that must be considered in the variational principle have their own proper length distinct from \tau and therefore along them we generally have

g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau} not equal to 1.

Obviously by putting g_{ab}\frac{dx^a}{ds}\frac{dx^b}{ds}=1 carelessly into the equation of Lagrangian, we end up with a constant Lagrangian and then no equations of motion will be resulted. But When could this insertion be done? Writing the action as

I=\int^{P_2}_{P_1}{\sqrt{\frac{dx^a}{ds}\frac{dx^b}{ds}}}ds

along the actual worldline of a particle moving freely, the Lagrangian is given by

L=\sqrt{\frac{dx^a}{ds}\frac{dx^b}{ds}}ds.

Now from the Euler-Lagrange we obtain

\frac{d}{ds}(\frac{dx_a/ds}{L})=\frac{d}{ds}(\frac{dx_a/ds}{1})=0,

where we have inserted L=1 into the equation because now it simply refers to the actual worldline (\tau=s). You see that as expected from the classical physics, the equation of motion d(p_a)/ds=0 where p_a represents the covariant components of the 4-momentum, is extracted.

The important note here is that after calculating the Euler-Lagrange equations we have only this right to insert L=1 because generally this is not a constant so it has to be differentiated first.

I hope this helps.

AB

 

[yuiop]

H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}

From the above, it is clear that, contrary to your repeated fallacious claims, H_c is a function of r. Since H_c is a function of r, your whole differentiation of L as if H_c did not depend on r fails

You are completely missing the point here. It is irrelevant whether H_c depends on r or not. What is important is that L is dependent on r and we should avoid taking the partial differential of L with respect to \dot{r}. I obtain H_c by taking the partial differential of L with respect to (\dot{\phi}) and L is independent of \phi.

You may be unpleasantly surprised by the fact that K_c is also a function of r, so your "derivation" is based on a double fallacy.

Same applies. I obtain K_c by taking the partial differential of L with respect to \dot{t} and since L is independent of t, this is OK.

This a more formal explanation of how the Euler-Lagrange formalism works by way of clarification:

The Euler-Lagrange equations of the Schwarzschild metric with respect to proper time are:

\frac{d}{ds} \frac{\delta{L}}{\delta \dot{x}} =\frac{ \delta{L}}{\delta{x}}

where \dot{x} can be (dt/ds), (dr/ds) or (d\phi/ds) in the reduced metric with d\theta=0 and \theta = \pi/2.

Starting with:

L = (1-2M/r)(dt/ds)^2 - (dr/ds)^2/(1-2M/r) - r^2(d\phi/ds) = 1

--------------------------------------

For the (dt/ds) term we take the partial derivative of L wrt (dt/ds) and obtain:

\frac{\delta L}{\delta \dot{t}} = 2(1-2M/r) \dot{t}= K

Now from the Euler-Lagrange equation we get:

\frac{d}{ds} \frac{\delta{L}}{\delta \dot{t}} = \frac{ \delta{L}}{\delta{t}} = 0

because L does not explicitly depend on t. This proves that K=2(1-2M/r) \dot{t} is a constant.

--------------------------------------

For the (dr/ds) term we take the partial derivative of L wrt (dr/ds) and obtain:

\frac{\delta {L}}{\delta \dot{r}} = 2\dot{r}/(1-2M/r)

From the Euler-Lagrange equation we get:

\frac{d}{ds} \frac{\delta{L}}{\delta \dot{r}} = \frac{ \delta{L}}{\delta{r}} \ne 0

because L does explicitly depend on r. This shows we can NOT assume 2\dot{r}/(1-2M/r) to be a constant.

--------------------------------------

For the (d\phi/ds) term we take the partial derivative of L wrt (d\phi/ds) and obtain:

\frac{\delta {L}}{\delta \dot{\phi}} = 2r^2 \dot{\phi} = H

From the Euler-Lagrange equation we get:

\frac{d}{ds} \frac{\delta{L}}{\delta \dot{\phi}} = \frac{ \delta{L}}{\delta{\phi}} = 0

because L does not explicitly depend on \phi. This proves that H=2r^2 \dot{\phi} is a constant.

--------------------------------------

It can be seen from the above (largely obtained from a textbook) that the constants K and H contain the variable r (just like my constants K_c and H_c), but contrary to your claims this does not invalidate them from being constants.

Err, no. H_c is a function of r and you are incorrectly treating it as a constant. So, your differentiation attempt is a failure. You have persisted in this error over several threads now.

Hc and Kc do indeed contain the variable r (as do H and K in your blog) but it does not mean that these constants are a function of r, because these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. If a function remains unchanged for any value of r then the function is not a function of r. Let me give you a very simple example. Let us say we have a function f defined as f = r/r. This is not a function of r because the value of f is 1 for any value of r. This is an obvious example, but it is not always so obvious. Let us say we have another variable s defined as s=2r and a function g defined as g = 3+s/r. The function g is not a function of r because the function g always evaluates to 5 for any value of r and g is in fact a constant.

Your conviction that since the constants of motion for the Schwarzschild metric contain the variable r that they must therefore be functions of r, is major misconception and I have given you the counterproof in the paragraph above.

 

[starthaus]

You are completely missing the point here. It is irrelevant whether H_c depends on r or not. What is important is that L is dependent on r and we should avoid taking the partial differential of L with respect to \dot{r}.

]

This is false.

1. The lagrangian L depends on both r and \dot{r}.

The critical Euler Lagrange equation that I have shown to you repeatedly is, in fact:\frac{d}{ds} \frac{\delta{L}}{\delta \dot{r}} -\frac{ \delta{L}}{\delta{r}} =0

2. Your expression L depends on H_c and K_c. Since both H_c and K_c are clear function of r your attempt at differentiatin L as if it weren't a function of r is incorrect.

Hc and Kc do indeed contain the variable r (as do H and K in your blog) but it does not mean that these constants are a function of r, because these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r.

3. This is easily provable to be false .I have already shown that, according to your very own definition:H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}

So, your statement "these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. " is easily proven false.

 

[yuiop]

]This is false.

1. The lagrangian L depends on r

I said:

"What is important is that L is dependent on r..."

You said:

"This is false. 1. The lagrangian L depends on r"

Do you not see that we the substance of what we both said is the same?

 

[starthaus]

I said:

"What is important is that L is dependent on r..."

You said:

"This is false. 1. The lagrangian L depends on r"

Do you not see that we the substance of what we both said is the same?

You need to read point 1 and point 2 and point 3 n order to understand your mistake.

 

[atyy]

Pretty much what kev and Altabeh have been saying (I haven't checked algebra details, but certainly their big picture is correct, and from experience Altabeh makes very, very few algebraic errors in PF):

"It follows from the geodesic equations that L is constant. In fact, on the worldline of a particle in free fall, ds2 = gab.dxa.dxb, by definition, so L = gab.(dxa/ds)(dxb/ds) = 1. ... worldline of a photon is also given by the geodesic equations with gab.(dxa/ds).(dxb/ds)= 0. The parameter s does not here have the interpretation of time: it is called an affine parameter and can be replaced by any linear function of s." From p 27 of http://people.maths.ox.ac.uk/nwoodh/gr/gr03.pdf

 

[yuiop]

]
3. This is easily provable to be false .I have already shown that, according to your very own definition:


H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}

So, your statement "these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. " is easily proven false.

The term:

\sqrt{\frac{m}{r^3}}

is only true for circular motion when dr/dt=0 and d^2r/dt^2=0 and the radius r is constant.

See your own post here:

There is a third Euler-Lagrange equation, it is dependent on the other two, so it does not contain any extra information per se:

-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0

If you make r=R in the above, this means the cancellation of the terms in dr/ds and if you giving you

(d\phi/ds)^2=\frac{m}{R^3}

i.e.

\omega^2=\frac{m}{R^3}

To remind ourselves that r is constant in your equation I will use R to mean the constant radius of a particle in circular orbit and rewrite your equation as:

H_{cR}=\frac{R^2}{\alpha} \frac{d\phi}{dt}=\frac{R^2}{1-2M/R}*\sqrt{\frac{M}{R^3}}

Since both M and R are constants there are no variables on the RHS and H_{cR} must therefore also be constant. There is no need for any other variables to change to compensate for changes in R because R is a constant and is therefore not changing, in the expression you gave.

 

[espen180]

kev, since you calculated \frac{d^2r}{d\tau^2} for arbitrary orbits a few pages back. I'm aiming for the same thing, and found

\frac{d^2r}{d\tau^2}=-\frac{GM}{r^2}+\left(r-\frac{3GM}{c^2}\right)\left(\frac{d\phi}{d\tau}\right)^2

I'm currently working on \frac{d^2r}{dt^2}. I'll post my derivation once I'm done.

 

[yuiop]

Good , you finally realized your errors of misdirection, you could have accepted that my equation was correct as shown to you 200 posts ago, at post 53.
BTW, it isn't "quoted from a textbook", it is derived from scratch from the Euler-Lagrange equations. You should make the effort to learn the formalism sometimes.

Not quoted from a textbook eh? here is post #53 again:

If you want to obtain the lagrangian, then you shout divide by ds, not by dt. If you do this, you get the correct Lagrangian:

L=\alpha (dt/ds)^2-\frac{1}{\alpha}(dr/ds)^2-r^2(d\phi/ds)^2

Once you get the Lagrangian, you can get one of the Euler-Lagrange equations:

...

There is a third Euler-Lagrange equation, it is dependent on the other two, so it does not contain any extra information per se:

-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0

If you make r=R in the above, this means the cancellation of the terms in dr/ds and if you giving you

(d\phi/ds)^2=\frac{m}{R^3}

i.e.

\omega^2=\frac{m}{R^3}

I am quite sure that I have shown you this before as well.

Substitute d\phi/ds and dr=0 into the metric equation and you get:

ds^2=(1-3m/R)dt^2

The above makes sense only for R&gt;3m

Compare this to http://books.google.co.uk/books?id=...w#v=onepage&q=lagrange schwarzschild&f=false"

Your equations are the exactly the same as equations 11.31, 11.32 and 11.33 in Rindler's book, with the same odd use of parentheses, the same introduction of the variable \omega[/tex] even though it is never used later. All you have done is copied his equations in the same order. Your only original contribution is the introduction of the expression (d\phi/ds)^2=m/R^3 which you got wrong and which Rindler later gave correctly as (d\phi/dt)^2=m/R^3

 

[Altabeh]

Hc and Kc do indeed contain the variable r (as do H and K in your blog) but it does not mean that these constants are a function of r, because these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. If a function remains unchanged for any value of r then the function is not a function of r. Let me give you a very simple example. Let us say we have a function f defined as f = r/r. This is not a function of r because the value of f is 1 for any value of r. This is an obvious example, but it is not always so obvious. Let us say we have another variable s defined as s=2r and a function g defined as g = 3+s/r. The function g is not a function of r because the function g always evaluates to 5 for any value of r and g is in fact a constant.

I don't like the examples, but definitely your reasoning is completely correct!

AB

 

[yuiop]

Pretty much what kev and Altabeh have been saying (I haven't checked algebra details, but certainly their big picture is correct, and from experience Altabeh makes very, very few algebraic errors in PF):

"It follows from the geodesic equations that L is constant. In fact, on the worldline of a particle in free fall, ds2 = gab.dxa.dxb, by definition, so L = gab.(dxa/ds)(dxb/ds) = 1. ... worldline of a photon is also given by the geodesic equations with gab.(dxa/ds).(dxb/ds)= 0. The parameter s does not here have the interpretation of time: it is called an affine parameter and can be replaced by any linear function of s." From p 27 of http://people.maths.ox.ac.uk/nwoodh/gr/gr03.pdf

Thanks for the link atyy.

Looks like a LOT of useful material in that document. Are you the author?

 

[atyy]

Thanks for the link atyy.

Looks like a LOT of useful material in that document. Are you the author?

Nope, I am a clueless biologist. Woodhouse's notes are just a free source I have found useful in my own self-study. He gives lots of the orbits in a Schwarzschild spacetime later on in the same set of notes.

 

[espen180]

Thanks for the link atyy.

Looks like a LOT of useful material in that document.

I agree. The person who wrote this seems very pedagogically inclined.

 

[yuiop]

kev, since you calculated \frac{d^2r}{d\tau^2} for arbitrary orbits a few pages back. I'm aiming for the same thing, and found

\frac{d^2r}{d\tau^2}=-\frac{GM}{r^2}+\left(r-\frac{3GM}{c^2}\right)\left(\frac{d\phi}{d\tau}\right)^2

OK, this seems in agreement with what we obtained here:

Re-insert the full form of H back into the equation:

\frac{d^2r}{ds^2} = -\frac{m}{r^2} + \frac{d\phi^2}{ds^2}(r-3m)

so, so far so good!

I'm currently working on \frac{d^2r}{dt^2}. I'll post my derivation once I'm done.

It would be nice to see an alternative derivation for the general equation of the acceleration of a particle in freefall, in coordinate time.

 

[Altabeh]

]

This is false.

1. The lagrangian L depends on both r and \dot{r}.

Correct.

2. Your expression L depends on H_c and K_c. Since both H_c and K_c are clear function of r your attempt at differentiatin L as if it weren't a function of r is incorrect.

Completely nonsense and without a physical/mathematical basis. Read the sources given to you to get to see how H_c and K_c are both derived to be CONSTANTS and indeed they each correspond to a conserved quantity (respectively angular momentum and energy of particle):

m_0r^2\dot{\phi}=m_0H_c=const.

is, for instance, the angular momentum (m_0 being the mass of particle). But let's dig through the details of how to get K_c and why it is a constant.

From the invariance of Killing vector fields along a symmetry axis, or

\xi_au_a=const.

where \xi^a are the contravariant components of the Killing vector field and u_a is a geodesic tangent, for the Schwartzschild metric with the only two non-null normalized Killing vectors \xi=(1,0,0,0) and \eta=(0,0,0,1) which correspond altogether to time-independence and axial symmetry of the spacetime, we get (because of time-independence)

\xi^au_0=(1,0,0,0).(u_0,...,u_3)=u_0=const.

which means u_0=\dot{x_0}=g_{00}\dot{x}^0=const. Now let's take \dot{x}^0=ct and with g_00=1-2m/r one would immediately obtain

u_0=(1-2m/r)c\dot{t}=cK_c.

On the other hand, p_0=m_0\dot{x}_0=m_0u_0 where p_0 is the time component of the four-momentum and again m_0 is the mass of particle and in a flat spacetime it is obvious that p_0=E/c, with E being the energy. Thus

E=p_0c=m_0u_0c=km_0c^2.

And this is the total energy for motion in a Schwarzschild metric.

Please do not attempt to collect nonsense claims and rather read books and use information provided here to understand things sounding to be at a higher level than your knowledge. If you persist on nonsense, I'll have to report your inutile posts.

3. This is easily provable to be false .I have already shown that, according to your very own definition:


H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}

So, your statement "these constants contain other variables that are dedendent on r and change in such a way that the constants remain constant for any value of r. " is easily proven false.

The same nonsense as above which has no mathematical/physical knowledge behind. Read the following books and stick with physics not your own wishful thinking:

Hobson M., Efstathiou G., Lasenby A. General relativity.. an introduction for physicists (CUP, 2006, pp 205-209.

A. Papapetrou, Lectures on GR, 1974, pp 70-73.

AB

 

[starthaus]

The same nonsense as above which has no mathematical/physical knowledge behind. Read the following books and stick with physics not your own wishful thinking:

This is precisely the H_c definition used by kev in his derivation.

 

[starthaus]

Not quoted from a textbook eh? here is post #53 again:Compare this to http://books.google.co.uk/books?id=...w#v=onepage&q=lagrange schwarzschild&f=false"

Your equations are the exactly the same as equations 11.31, 11.32 and 11.33 in Rindler's book, with the same odd use of parentheses, the same introduction of the variable \omega[/tex] even though it is never used later.

<br /> <br /> Of course they are , I have been telling you this for 5 weeks since we started discussing thie subject in the thread dealing with orbital acceleration. I even cited the exact paragraph and equation numbers. With one notable exception, the Lagrangian (11.31) in Rindler is incorrect, so I corrected it in post 53.

 

[starthaus]

Correct.
Completely nonsense and without a physical/mathematical basis. Read the sources given to you to get to see how H_c and K_c are both derived to be CONSTANTS and indeed they each correspond to a conserved[/color] quantity (respectively angular momentum and energy of particle):

m_0r^2\dot{\phi}=m_0H_c=const.

...meaning that \frac{dH_c}{dt}=0 . Not that \frac{dH_c}{dr}=0, as you and kev incorrectly keep claiming.
The fact that H_c is a conserved quantity (and so is K_c) , does not in any way preclude them being functions of r as both quantities obviously are. You only need to look at their algebraic expressions.

 

[yuiop]

The same nonsense as above which has no mathematical/physical knowledge behind. Read the following books and stick with physics not your own wishful thinking:

Hobson M., Efstathiou G., Lasenby A. General relativity.. an introduction for physicists (CUP, 2006, pp 205-209.

AB

Hi Altabeh, I managed to find the book on google here: http://books.google.co.uk/books?id=...resnum=8&ved=0CDEQ6AEwBw#v=onepage&q&f=false"

On page 209 of the linked book they give equation (9.35) as:

\frac{dr^2}{ds^2} = c^2(K^2-1) +\frac{2GM}{r}

They then differentiate equation (9.35) with respect to s and then divide through by (dr/ds) (which is effectively the same as differentiating (dr/ds) by s) to obtain:

\frac{d^2r}{ds^2} = -\frac{GM}{r^2}

Clearly (contrary to the claims of Starthaus) they have treated K as a constant here. They can not directly differentiate the expression for (dr/ds) by s, because the variable s does not appear in the expression. Presumably they used implicit differentiation with respect to s but this is quite a lengthy procedure. A much simpler method is to differentiate (dr/ds)^2 with respect to r (instead of s) and divide the result by 2. The end result is the same and this demonstrates that K is not only independent of s but also of r, or an incorrect result would have been obtained.

Just in case anyone is wondering, I can provide a proof that:

\frac{d}{ds}\left(\frac{dr}{ds}\right) = \frac{1}{2}*\frac{d}{dr}\left(\frac{dr^2}{ds^2}\right)