Orbital velocities in the Schwartzschild geometry

[Altabeh]

kev,
Is it the case that your L is a constant function? It appears to be - Am I mistaken in this? If it is a constant function, then taking its derivative does, it seems to me, result in the zero function.

It is not your fault to be thinking of this nonsense that has been dragged into our consideration by starthaus. See the source introduced in my post 305 to get familiar with the Lagrangian procedure for geodesics. It's really helpful and pithy at the same time.

AB

 

[Altabeh]

Correct, I told kev this about 200 posts ago. He (and Altabeh) still don't get the difference.

Do you get the difference between "nonsense" and "correct"? All over this thread you have left behind a nonsense of yours and the last being the Lagrangian and its derivative. I have to add this to the list of your hacks I sent in a private message to you. LOL!

Besides, his "solution" is incorrect.

It behooves oneself to give reasons for claiming incorrectness of a solution otherwise the whole reasoning is nonsense as is yours right now!

I don't know why you have a mental block with such a basic thing, so I'll simplify it further for you:

Did you take those introductory courses in everything? If not, then stop playing the role of tutor here. With all those nonsense claims your knowledge won't be empty of leaks EVER.

-if $$y(x)=0$$ for all x then $$\frac{dy}{dx}=0$$ for all x

Basic stuff that doesn't have anything to do with your old fallacy. You've been declared as finished in your post 251.

-if you have a differential equation in $$y$$ and its derivatives, then, you are not allowed to put in by hand $$y=0$$ without implying immediately that the higher derivatives of $$y$$ are also null.

Another nonsense. The whole problem which was completely explained to you long time ago is that the initial conditions are all point-wise as is the proper velocity of the momentarily at rest particles. Go read the textbooks I gave you when discussing this. Please do not go for hacks/twists when you're stuck, or better say, blocked.

Basic stuff

Then why don't you understand it, huh?

AB

 

[Altabeh]

I have derived them in my paper. If you disagree with them, you should state the correct ones, or you are making groundless arguments and breaking PF rules.

Here I think mentors would come to check these mind-made theories of starthaus and don't let him publish wrong ideas/nonsense claims like the "old fallacy" $$dr/ds=0\Rightarrow d^2r/ds^2=0$$. We are doing our best we can to stop such fallacies to be made but as far as I know, individuals here are not allowed to disperse their own ideas specially when they are full of hacks.

AB

 

[yossell]

Hi Altabeh,

thanks for your comment in 316 - I had wondered at your post 305, but I admit it was too pithy for me to be sure I understood. I couldn't get hold of the reference - though the book's in Amazon, the preview that was available to me didn't have the reference.

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?

 

[espen180]

What "partial derivative of L"? You resurrected your old hack of attempting to "calculate" partial derivatives of a... constant. You just declared $$L=1$$ a few lines above, when will you stop with the ugly hacks?

Here's a nice counter-example.

Take the equation of a plane:

$$L=ax+by+cz+d=0$$ (letter L chosen for comparative purposes)

It is obvious that L is a constant (it is equal to zero), but the partial derivatives are

$$\frac{\text{d}L}{\text{d}x}=a$$

$$\frac{\text{d}L}{\text{d}y}=b$$

$$\frac{\text{d}L}{\text{d}z}=c$$All the equation above says is that

$$\text{d}L=\left(\frac{\text{d}L}{\text{d}x}\right)_{yz}\text{d}x+\left(\frac{\text{d}L}{\text{d}y}\right)_{xz}\text{d}y+\left(\frac{\text{d}L}{\text{d}z}\right)_{xy}\text{d}z=0$$

And equally, for the lagrangian, it is true that

$$\text{d}L=\sum_{i=0}^N \left(\frac{\text{d}L}{\text{d}q_i}\right)_{q_j\,,\,j\neq i}\text{d}q_i=0$$

where $$q_i$$ are the independent coordinates. The equation says nothing about the individual partial derivatives, it only restricts them as a group.

 

[yossell]

Hi espen180,

thanks for your example. Let me try and express my worries about over this example.

One can define a function of three variables, L(x, y, z) as:
L(x, y, z) = ax + by + cz + d

More explicitly: for every triple <x, y z>, the value of the function at this triple is ax + by + cz + d.

It is typical to avoid an explosion of notation hide the variables, keep them explicit, and just write L for the function - but we should remember they are there.

It makes perfect sense to take partial derivatives of this function wrt x, y and z, as you do in lines 5-7 and get the answers you say.

But it does not make sense to say: let this function be the zero function. You can't do it: you've already defined the function, as above, and it's just not the zero function.

When defining a plane, it *does* make sense to talk about the set of points <x y z> which are such that ax + by + cz + d = 0. This, of course, defines a plane in 3 dimensions. But I don't know what it means to start differentiating a plane, or differentiating anything like ax + by + cz + d = 0.

I may of course be very wrong - but I hope this worry makes sense.

 

[yuiop]

But if it's agreed that the derivative of constant function is zero, then I guess I do feel worried about starthaus' objection to your derivation in post 277.

Is it the case that your L is a constant function? It appears to be - Am I mistaken in this? If it is a constant function, then taking its derivative does, it seems to me, result in the zero function.

Here is a simple demonstration that although the derivative of a constant function is zero, it does not necessarily follow that the partial derivative of a constant function is zero.

Consider the function $f(x,y) = 2y^2 - x^2 = 1$

Since f(x,y) = 1, the function f is a constant function.

Now the partial derivative of f with respect to x is:

$$\frac{\delta}{\delta x} f(x,y) = \frac{\delta}{\delta x} (1) = \frac{\delta}{\delta x} (2y^2 - x^2) = 2x$$

The partial derivative of f with respect to y is:

$$\frac{\delta}{\delta y} f(x,y) = \frac{\delta}{\delta y} (1) = \frac{\delta}{\delta y} (2y^2 - x^2) = 4y$$

Clearly the partial differentiation of the constant function f(x,y)=1 with respect to x or y is not zero, except at the point where x=0 or y=0. It is easy to demonstrate that x or y can take other values.

Solve the original equation for x:

$$2y^2 - x^2 = 1$$

$$\Rightarrow x = \sqrt{2y^2-1}$$

When y=1, x = sqrt(2*1^2-1) = 1.

When y=5, x = sqrt(2*5^2-1) = sqrt(2*25-1) = sqrt(50-1) = sqrt(49) = 7.

Obviously x can take on many other values depending on the value of y.

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

 

[espen180]

When defining a plane, it *does* make sense to talk about the set of points <x y z> which are such that ax + by + cz + d = 0. This, of course, defines a plane in 3 dimensions. But I don't know what it means to start differentiating a plane, or differentiating anything like ax + by + cz + d = 0.

Think about it this way:

A line is given by

$$ax+by+c=0$$ (every line can be represented in 2 dimensions by a coordinate transformation)

By the equation, ax+by+c is always zero. Can be still talk about the slope of the line? Sure! The slop is equal to $$\frac{\text{d}x}{\textd}y}$$ .

All surfaces can be represented in the form $$f(x,y,z,...)=0$$. We often want to know the change of some coordinates when we change others. For that we have to differentiate f wrt. the coordinates. An example is the gradient of a manifold, which shows the direction of the steepest "slope".

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

Isn't r an independent coordinate? (This might just be a typo. I'm no Lagrange guru.).

 

[yuiop]

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

Isn't r an independent coordinate? (This might just be a typo. I'm no Lagrange guru.).

It would have perhaps been better for me to say something like: When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those variables that the metric is independent of. This is probably still not close to a formal statement. Perhaps I can explain it like this:

Consider the Schwarzschild metric:

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

Only the variables r and $\theta$ appear in their "raw" form so the metric is directly dependent of those variables, but the variables s, t, and $\phi$ only appear in the ds, dt, $d/phi$ forms respectively and it said that the metric is "independent" of s,t and $\phi$ in their "raw" from.

 

[yossell]

kev,
thanks. My worry (or mental block) with what you say is pretty much the same as my worry about espen180's post.

`consider the function $f(x,y) = 2y^2 - x^2 = 1$' is problematic.
You can consider the two-place function of x and y given by f(x y) = 2y^2 - x^2;
you can consider the two-place (constant) function of x and y given by f(x y) = 1
you can consider the set of points <x y> that satisfy the equation: 2y^2 - x^2 = 1
But you can't say that the whole function f(x y) is 1 for all x and y, and that it's 2y^2 - x^2 for all x and all y.

And are you really sure you want to say that the derivative of the constant function f(x y) = 1 is not zero (apart from at x = y = 0)? I thought we'd agreed that the derivatives of constant functions *were* zero everywhere. Since the function doesn't change, the limit of
f(x +dx, y) - f(x, y)/dx is the limit of k - k/dx (as the function is constant) = 0.

espen180, thanks for your response. It's true that a line is given by ax + by + c = 0. This is because it corresponds to a 1-place function f(x) = {c - ax}/b, and then *this* function is what is differentiated to find the slope of the line. Again, there is a notational issue - but it's another notational issue: the fact that in Leibniz notation, y is used sometimes as the name of a function, sometimes as a variable.

Anyway, thanks again for the posts and the spirit of your discussion with me. I see I'm in a minority here, so it could well be my own mental block, and I'll get some sleep now and study your posts again.

 

[espen180]

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

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.

 

[sf222]

My solution agrees with equations (4) and (5) of mathpages http://www.mathpages.com/rr/s6-07/6-07.htm...

That web page deals with purely radial motion, explaining how the acceleration is expressed in terms of Schwarzschild coordinates, but another page from the very same site gives a simple derivation of the equations of motion for general motion, which seems to be what you guys are interested in: http://www.mathpages.com/rr/s6-02/6-02.htm

 

[starthaus]

Here's a nice counter-example.

Take the equation of a plane:

$$L=ax+by+cz+d=0$$ (letter L chosen for comparative purposes)

It is obvious that L is a constant (it is equal to zero), but the partial derivatives are

$$\frac{\text{d}L}{\text{d}x}=a$$

$$\frac{\text{d}L}{\text{d}y}=b$$

$$\frac{\text{d}L}{\text{d}z}=c$$All the equation above says is that

$$\text{d}L=\left(\frac{\text{d}L}{\text{d}x}\right)_{yz}\text{d}x+\left(\frac{\text{d}L}{\text{d}y}\right)_{xz}\text{d}y+\left(\frac{\text{d}L}{\text{d}z}\right)_{xy}\text{d}z=0$$

And equally, for the lagrangian, it is true that

$$\text{d}L=\sum_{i=0}^N \left(\frac{\text{d}L}{\text{d}q_i}\right)_{q_j\,,\,j\neq i}\text{d}q_i=0$$

where $$q_i$$ are the independent coordinates. The equation says nothing about the individual partial derivatives, it only restricts them as a group.

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

 

[starthaus]

Here is a simple demonstration that although the derivative of a constant function is zero, it does not necessarily follow that the partial derivative of a constant function is zero.

Consider the function $f(x,y) = 2y^2 - x^2 = 1$

Since f(x,y) = 1, the function f is a constant function.

Now the partial derivative of f with respect to x is:

$$\frac{\delta}{\delta x} f(x,y) = \frac{\delta}{\delta x} (1) = \frac{\delta}{\delta x} (2y^2 - x^2) = 2x$$

The partial derivative of f with respect to y is:

$$\frac{\delta}{\delta y} f(x,y) = \frac{\delta}{\delta y} (1) = \frac{\delta}{\delta y} (2y^2 - x^2) = 4y$$

Clearly the partial differentiation of the constant function f(x,y)=1 with respect to x or y is not zero, except at the point where x=0 or y=0. It is easy to demonstrate that x or y can take other values.

Solve the original equation for x:

$$2y^2 - x^2 = 1$$

$$\Rightarrow x = \sqrt{2y^2-1}$$

When y=1, x = sqrt(2*1^2-1) = 1.

When y=5, x = sqrt(2*5^2-1) = sqrt(2*25-1) = sqrt(50-1) = sqrt(49) = 7.

Obviously x can take on many other values depending on the value of y.

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

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.

 

[starthaus]

Moving on to the general case for coordinate acceleration as promised, this is the derivation based on the one I started in post #48.

Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that $\theta = \pi/2$ and $d\theta = 0$

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

Divide both sides by $\alpha dt^2$ and rearrange so that the constant (1) is on the LHS:

$$L = 1 =\frac{1}{\alpha}\frac{ds^2}{dt^2} +\frac{1}{\alpha^2}\frac{dr^2}{dt^2}+\frac{r^2}{\alpha}\frac{d\phi^2}{dt^2}$$

The metric is independent of $\phi$ and t, so there is a constant associated with coordinate angular velocity $(H_c)$ which is obtained by finding the partial derivative of L with respect to $d\phi/dt$

$$\frac{\delta{L}}{\delta(d\phi/dt)} = \frac{r^2}{\alpha} \frac{d\phi}{dt} = H_c$$

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

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

The metric is independent of s and t, so there is a constant associated with time dilation $(K_c)$ which is obtained by finding the partial derivative of L with respect to $ds/dt$

$$\frac{\delta{L}}{\delta(ds/dt)} = \frac{1}{\alpha} \frac{ds}{dt} = K_c$$

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.

Substitute these constants into the equation for L

$$1 =\alpha K_c^2 +\frac{1}{\alpha^2}\frac{dr^2}{dt^2}+\frac{\alpha}{r^2}H_c^2}$$

and solve for (dr/dt)^2:

$$\frac{dr^2}{dt^2} = (1-2M/r)^2 -(1-2M/r)^3 (K_c^2 + H_c^2/r^2)$$

Differentiate the above with respect to r and divide by 2 to obtain the general coordinate radial acceleration of a freefalling particle in the metric:

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.

$$\frac{d^2r}{dt^2}= \frac{\alpha^2 H_c^2}{r^4}(r-5M) +\frac{M}{r^2}(2\alpha-3K_c^2\alpha^2)$$

Re-inserting the full forms of $H_c$ and $K_c$ back in gives:

$$\frac{d^2r}{dt^2}= \frac{d\phi^2}{dt^2} (r-5M) +\frac{M}{r^2}(2\alpha-3 \frac{ds^2}{dt^2} )$$

Now the (ds/dt)^2 term is a little inconvenient, but we can find an alternative form by solving the Schwarzschild metric to directly obtain:

$$(ds/dt)^2 = (\alpha - dr^2/(\alpha dt^2) - r^2d\phi^2/dt^2)$$

and substituting this form into the equation above it to obtain:

$$\frac{d^2r}{dt^2}= \frac{d\phi^2}{dt^2} (r-5M) +\frac{M}{r^2}(2\alpha-3 (\alpha - dr^2/(\alpha dt^2) - r^2d\phi^2/dt^2) )$$

which after a bit of algebra simplifies to:

$$\Rightarrow \frac{d^2r}{dt^2}= \alpha\left( r \frac{d\phi^2}{dt^2}-\frac{M}{r^2}+ \frac{3M}{\alpha^2 r^2} \frac{dr^2}{dt^2}\right)$$

This is the fully general form of acceleration in Schwarzschild coordinates in terms of coordinate time and gives Espen something to compare his results against in his latest document.


.

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

 

[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 $$r'(s)=0$$.

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 : $$f(x)=0=>df/dx=0$$.

 

[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 : $$f(x)=0=>df/dx=0$$.

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 $$ds^2>0$$ and for timelike geodesics $$ds^2<0.$$

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 $$R>3m$$

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 [itex]\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 $$(d\phi/dt)^2=m/R^3$$