I Schwarzschild metric not dependent on time

[stevendaryl]

Two of the Killing fields are easy to know, it follows directly from the independence of the metric on $t$ and $\varphi$ that $(-(2m - r), 0,0,0)$ and $(0,0,0, (r \sin \vartheta)^{-2})$ are the vectors components of the vectors of the two fields at any point $(t,r,\vartheta,\varphi)$.

But I'm afraid the vector components of the remaining vector fields can only be found by solving the system of PDE resulting from the Killing equations. And this is really hard.

Or perhaps we can split the metric into parts and compute only the necessary (unknown) terms for the Killing equations?

Actually, $\phi$ represents the angle for rotation about the z-axis. You can, from symmetry, immediately get the other two: $\phi_x$ representing rotation about the x-axis, and $\phi_y$ representing rotation about the y-axis. So rather than making a question of solving a differential equation, it becomes a geometric problem of how to compute $\phi_x$ and $\phi_y$ in terms of the usual $\theta, \phi$.

For illustration: The picture on the left shows the $\phi$, representing rotations around the z-axis (the line from north pole to south pole), the middle picture shows $\phi_y$ representing rotations about the y-axis (the line from the point on the equator at 90 degrees west longitude to the antipodal point at 90 degrees east longitude) and the third picture shows $\phi_x$, representing rotations about the x-axis (the line from the point on the equator at 0 degrees longitude to the antipodal point at 180 degrees longitude). I don't know an easy way to calculate $\phi_x$ and $\phi_y$ in terms of the usual latitude and longitude.

 

[stevendaryl]

Actually, $\phi$ represents the angle for rotation about the z-axis. You can, from symmetry, immediately get the other two: $\phi_x$ representing rotation about the x-axis, and $\phi_y$ representing rotation about the y-axis. So rather than making a question of solving a differential equation, it becomes a geometric problem of how to compute $\phi_x$ and $\phi_y$ in terms of the usual $\theta, \phi$.

It's actually very simple, using symmetry arguments.

The usual $\phi$ relates to the cartesian coordinates via:

$\phi = tan^{-1}(\frac{y}{x})$

The other two are:

$\phi_x = tan^{-1}(\frac{z}{y})$

$\phi_y = tan^{-1}(\frac{x}{z})$

 

[kent davidge]

But the relations you gave relate the three parameters (angles) of the spherical coordinates with the Cartesian parameters. I was talking about expressing the remaining Killing vector fields in Schwarszschild coordinates. They will have four components. For example, the (dual of the) timelike one will be $g_{tt} (1,0,0,0) = -(1 / (1 - 2m / r),0,0,0)$.

 

[PeterDonis]

I remind myself by thinking of Minkowski spacetime whose spacelike Killing fields are the translations and rotations, making six.

Actually it's somewhat more complicated than that. The "space-time rotation" Killing fields, which are a three-parameter group (three independent ones) are usually thought of as being the boosts, which are timelike; but actually they include all "hyperbolic rotations", which include spacelike ones such as the hyperbolic rotation whose integral curves are hyperbolas with $t^2 - x^2 = \text{const}$ in a particular frame (i.e., in the upper and lower "wedge" instead of the left and right "wedge"). So the "space-time rotation" Killing fields are sometimes timelike and sometimes spacelike (and sometimes null, on the Killing horizons). In this respect they are like the fourth Killing field in Schwarzschild spacetime (the easiest way to see the correspondence is to look at a Kruskal diagram).

 

[stevendaryl]

But the relations you gave relate the three parameters (angles) of the spherical coordinates with the Cartesian parameters. I was talking about expressing the remaining Killing vector fields in Schwarszschild coordinates. They will have four components. For example, the (dual of the) timelike one will be $g_{tt} (1,0,0,0) = -(1 / (1 - 2m / r),0,0,0)$.

Well, the Schwarzschild coordinates are $r, \theta, \phi, t$. Let $x$ be $r sin(\theta) cos(\phi)$, $y$ be $r sin(\theta) sin(\phi)$ and $z$ be $r cos(\theta)$. Then in terms of $x, y, z$: we can write down 3 alternative coordinate systems:

1. $r, \theta, \phi, t$ where $\theta = tan^{-1} (\frac{\sqrt{x^2 + y^2}}{z})$, and $\phi = tan^{-1} (\frac{y}{x})$
2. $r, \theta_x, \phi_x, t$ where $\theta_x = tan^{-1} (\frac{\sqrt{z^2 + y^2}}{x})$, and $\phi_x = tan^{-1} (\frac{z}{y})$
3. $r, \theta_y, \phi_y, t$ where $\theta_y = tan^{-1} (\frac{\sqrt{x^2 + z^2}}{y})$, and $\phi_y = tan^{-1} (\frac{x}{z})$

So the 4 Killing fields are

1. $e_t$, which is the tangent vector to the path where $\frac{dt}{ds} = 1, \frac{dr}{ds} = 0, \frac{d\theta}{ds} = 0, \frac{d\phi}{ds} = 0$
2. $e_\phi$, which is the tangent vector to the path where $\frac{dt}{ds} = 0, \frac{dr}{ds} = 0, \frac{d\theta}{ds} = 0, \frac{d\phi}{ds} = 1$
3. $e_{\phi_x}$, which is the tangent vector to the path where $\frac{dt}{ds} = 0, \frac{dr}{ds} = 0, \frac{d\theta_x}{ds} = 0, \frac{d\phi_x}{ds} = 1$
4. $e_{\phi_y}$, which is the tangent vector to the path where $\frac{dt}{ds} = 0, \frac{dr}{ds} = 0, \frac{d\theta_y}{ds} = 0, \frac{d\phi_y}{ds} = 1$

So Killing vector field $e_{\phi_x}$, for example, is the vector $V$ with components $V^r = V^t = V^{\theta_x} = 0, V^{\phi_x} = 1$. That's in the coordinate basis for the coordinate system $r, t, \theta_x, \phi_x$. To translate back into the Schwarzschild basis, you use the rule for translating vectors:

$V^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^\mu} V^\mu$

So in our case, $V^\mu = 0$ unless $\mu = \phi_x$, in which case it is equal to 1. So that simplifies things:

$V^{\mu'} = \frac{\partial x^{\mu'}}{\partial \phi_x}$

So you need to compute $\frac{\partial \phi}{\partial \phi_x}, \frac{\partial \theta}{\partial \phi_x}$ ($t$ and $r$ don't depend on $\phi_x$. So our answer is:

$V^r = V^t = 0$
$V^\phi = \frac{\partial \phi}{\partial \phi_x}$
$V^\theta = \frac{\partial \theta}{\partial \phi_x}$

Since $\phi = tan^{-1}(\frac{y}{x})$, we can in the second coordinate system write: $x = r cos(\theta_x), y = r sin(\theta_x) cos(\phi_x), z = r sin(\theta_x) sin(\phi_x)$. So

$\phi = tan^{-1}(\frac{sin(\theta_x) cos(\phi_x)}{cos(\theta_x)}) = tan^{-1} (tan(\theta_x) cos(\phi_x))$

We can similarly compute $\theta$ in terms of $\phi_x$ and $\theta_x$. Then take derivatives to get the components of $e_{\phi_x}$ in the Schwarzschild coordinate basis.

 

[kent davidge]

Although I (think) understand what you're doing, I'm afraid that's not what we should get. To get a feeling of what I mean, I have computed the (non trivial) Killing Vectors for a simpler metric $$g = -\bigg(1 - \frac{2m}{r} \bigg)dt^2 + \frac{1}{1 - \frac{2m}{r}}dr^2 \\ \xi^{(1)} = (1,0) \\ \xi^{(2)} = \bigg(0, \sqrt{1 - \frac{2m}{r}} \bigg)$$
Those are the kind of expressions one should obtain when computing them using the full metric. Your expressions will not involve elements such as $m$, $r$...

 

[PAllen]

Although I (think) understand what you're doing, I'm afraid that's not what we should get. To get a feeling of what I mean, I have computed the (non trivial) Killing Vectors for a simpler metric $$g = -\bigg(1 - \frac{2m}{r} \bigg)dt^2 + \frac{1}{1 - \frac{2m}{r}}dr^2 \\ \xi^{(1)} = (1,0) \\ \xi^{(2)} = \bigg(0, \sqrt{1 - \frac{2m}{r}} \bigg)$$
Those are the kind of expressions one should obtain when computing them using the full metric. Your expressions will not involve elements such as $m$, $r$...

No, the killing vector fields are just a congruence of curves in specified coordinates. They do not involve metric components - they are curves along which the metric doesn’t change. So you should NOT see m appearing in their expression.

 

[PeterDonis]

I have computed the (non trivial) Killing Vectors for a simpler metric

There is only one Killing field for this metric (the first one you list). I don't know why you think your $\xi^{(2)}$ is a Killing field.

 

[kent davidge]

I don't know why you think your $\xi^{(2)}$ is a Killing field.

The Killing equations present it as a solution.

 

[stevendaryl]

The Killing equations present it as a solution.

I had to work through this myself a number of times, but I convinced myself of the following:

Vector field $K$ is a killing vector field in some region if and only if there exists some local coordinate system $x^\mu$ such that in that coordinate system, there is some coordinate $x^{mu_0}$ where:

1. $K = e_{\mu_0}$ (That is, $K$ is one of the basis vectors in the coordinate basis)
2. $\frac{\partial g_{\mu \nu}}{\partial x^{\mu_0}} = 0$ (the metric tensor components don't depend on the coordinate $x^{\mu_0}$)

(Note that 1. means that $K^{\mu_0} = 1$ and for all other indices $\mu$, $K^\mu = 0$. That means that $\partial_{\lambda} K^\mu = 0$. $K$ is constant in this coordinate system.)

The implication in one direction is sort of straight-forward. If $K = e_{\mu_0}$, then that means that $\frac{\partial g_{\mu \nu}}{\partial x^{\mu_0}} = K^\lambda \partial_\lambda g_{\mu \nu}$

So we have:

1. $K^\lambda \partial_\lambda g_{\mu \nu} = 0$
   
2. $K^\lambda \nabla_\lambda g_{\mu \nu} = 0$ (that's always true, the metric tensor is covariantly constant)
   
3. $\nabla_\lambda g_{\mu \nu} = \partial_\lambda g_{\mu \nu} - \Gamma^\tau_{\lambda \nu} g_{\mu \tau} - \Gamma^\tau_{\lambda \mu} g_{\tau \nu}$ (that's the way the covariant derivative works)
4. $K^\lambda [ \Gamma^\tau_{\lambda \nu} g_{\mu \tau} + \Gamma^\tau_{\lambda \mu} g_{\tau \nu} ] = 0$ (putting 1, 2, and 3 together)
5. We also have, again from the way covariant derivatives work, $\nabla_\nu K^\tau = \partial_\nu K^\tau + \Gamma^\tau_{\nu \lambda} K^\lambda$ and $\nabla_\mu K^\tau = \partial_\mu K^\tau + \Gamma^\tau_{\mu \lambda} K^\lambda$
6. So using the fact that $K$ is constant in this coordinate system, 5 simplifies to: $\nabla_\nu K^\tau = \Gamma^\tau_{\nu \lambda} K^\lambda$ and $\nabla_\mu K^\tau = \Gamma^\tau_{\mu \lambda} K^\lambda$
7. Plugging 6 into 4 gives: $(\nabla_\nu K^\tau) g_{\mu \tau} + (\nabla_\mu K^\tau) g_{\tau \nu} = 0$
8. Since the metric tensor is covariantly constant, we can bring it inside the scope of $\nabla$ with no change, so 7 can be rewritten as: $\nabla_\nu (K^\tau g_{\mu \tau}) + \nabla_\mu (K^\tau g_{\tau \nu}) = 0$
9. Since $g$ lowers indices, 8 can be rewritten as: $\nabla_\nu K_\mu + \nabla_\mu K_\nu = 0$. Which is the Killing equation.

So, in my opinion, the most intuitive way to find Killing vector fields is to look for coordinates under which the metric doesn't change. Sometimes it might be difficult to find those coordinate systems.

 

[stevendaryl]

The Killing equations present it as a solution.

The Killing equation is $\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0$ In your case:

$K^t = 0$
$K^r = \sqrt{1-\frac{2M}{r}}$

To get the lower-index version, you use the metric tensor. $g_{rr} = \frac{1}{1 - \frac{2m}{r}}$, $g_{tt} = -(1 - \frac{2m}{r})$. That implies that the lower-index version of $K$ is:

$K_t = 0$
$K_r = \frac{1}{\sqrt{1-\frac{2M}{r}}}$

I haven't worked out the coefficients $\Gamma^\mu_{\nu \lambda}$, but it sure doesn't look likely that $\nabla_r K_r = 0$

 

[kent davidge]

Thank you for pointing this out to me. The components were indeed wrong. Here are the correct ones. This time I checked them against the Killing equations.
$$K_t = \sqrt{1- \frac{2m}{r}} \frac{mt}{r^2} \\ K_r = \sqrt{ \frac{1}{1- \frac{2m}{r}}}$$

 

[kent davidge]

I should say that the above solution is non trivial. The trivial one is $$K_t = -(1 - 2m / r) \\ K_r = 0$$ which follows from the independence of the metric on $t$. I also checked it using the Killing equations before posting it on here.

 

[PeterDonis]

The components were indeed wrong. Here are the correct ones.

No, these are wrong.

I should say that the above solution is non trivial. The trivial one is

$$
K_t = -(1 - 2m / r) \\ K_r = 0
$$

which follows from the independence of the metric on $t$.

No, your $K_t$ here is now not a Killing vector field. You had that one right the first time: $K_t = (1, 0, 0, 0)$.

Also, your $K_r$ is not a vector field at all. In your simplified metric, which is just the $t-r$ plane of Schwarzschild spacetime, there is only one Killing vector field, $K_t = (1, 0, 0, 0)$. If your calculations are giving you some other answer, your calculations are wrong.

I also checked it using the Killing equations before posting it on here.

Then you need to check your calculations, because your answer is wrong. Feel free to post them here if you need help.

 

[PeterDonis]

In your simplified metric, which is just the $t-r$ plane of Schwarzschild spacetime, there is only one Killing vector field, $K_t = (1, 0, 0, 0)$.

Just to be clear, this is a vector field, i.e., with an upper index, so it should really be written as $(K_t)^{\mu} = (1, 0, 0, 0)$, to make it clear that the $t$ is just a label, not an index.

 

[kent davidge]

Then you need to check your calculations, because your answer is wrong. Feel free to post them here if you need help.

No, your $K_t$ here is now not a Killing vector field. You had that one right the first time: $K_t = (1, 0, 0, 0)$.
Also, your $K_r$ is not a vector field at all.

I'm going to check the last one I posted.
The non vanishing connection coefficients are $$\Gamma^t{}_{rt} = \Gamma^t{}_{tr} = \frac{-m}{2mr - r^2} \\ \Gamma^r{}_{tt} = \frac{m(r - 2m)}{r^3} \\ \Gamma^r{}_{rr} = \frac{m}{2mr - r^2}$$ The Killing equations are $$0 = \nabla_r K_r = \partial_r K_r - \Gamma^t{}_{rr} K_t - \Gamma^r{}_{rr} K_r = 0 \\ 0 = \nabla_t K_t = \partial_t K_t - \Gamma^t{}_{tt} K_t - \Gamma^r{}_{tt} K_r = 0 \\ 0 = \nabla_t K_r + \nabla_r K_t = \partial_t K_r + \partial_r K_t - 2 \Gamma^r{}_{tr} K_r - 2 \Gamma^t{}_{tr} K_t = \frac{-2m}{r^2} - \frac{2m}{2mr -r^2} \bigg(1 - \frac{2m}{r} \bigg) = 0$$

 

[PeterDonis]

The non vanishing connection coefficients are

These look right. But I think it's helpful (as will be seen below) to rewrite the first one as

$$
\Gamma^t{}_{rt} = \Gamma^t{}_{tr} = \frac{m}{r \left( r - 2m \right)}
$$

and similarly for the third, with the sign flipped.

The Killing equations are

You're leaving out a crucial step here: you have to lower the index on the Killing vector field, since we've been writing them with upper indexes. So these equations should be rewritten as (note that we're using the fact that the metric is diagonal, otherwise these would be even more complicated):

$$
\partial_r \left( g_{rr} K^r \right) - \Gamma^t{}_{rr} \left( g_{tt} K^t \right) - \Gamma^r{}_{rr} \left( g_{rr} K^r \right) = 0
$$

$$
\partial_t \left( g_{tt} K^t \right) - \Gamma^t{}_{tt} \left( g_{tt} K^t \right) - \Gamma^r{}_{tt} \left( g_{rr} K^r \right) = 0
$$

$$
\partial_t \left( g_{rr} K^r \right) + \partial_r \left( g_{tt} K^t \right) - 2 \Gamma^r{}_{tr} \left( g_{rr} K^r \right) - 2 \Gamma^t{}_{tr} \left( g_{tt} K^t \right) = 0
$$

But some of these terms vanish because they contain connection coefficients which vanish, so we are left with:$$
\partial_r \left( g_{rr} K^r \right) - \Gamma^r{}_{rr} \left( g_{rr} K^r \right) = 0
$$

$$
\partial_t \left( g_{tt} K^t \right) - \Gamma^r{}_{tt} \left( g_{rr} K^r \right) = 0
$$

$$
\partial_t \left( g_{rr} K^r \right) + \partial_r \left( g_{tt} K^t \right) - 2 \Gamma^t{}_{tr} \left( g_{tt} K^t \right) = 0
$$

Now consider $K^\mu = (1, 0, 0, 0)$. The first and second equations are trivially satisfied. The third equation becomes

$$
\partial_r \left[ - \left( 1 - \frac{2m}{r} \right) \right] - 2 \frac{m}{r \left( r - 2m \right)} \left[ - \left( \frac{r - 2m}{r} \right) \right] = 0
$$

which becomes

$$
- \frac{2m}{r^2} + \frac{2m}{r^2} = 0
$$

which is valid, so $K^\mu = (1, 0, 0, 0)$ is a Killing vector field. This calculation should also make it evident that no other vector with just a $t$ component can satisfy the above equations: the second equation tells you that $K^t$ cannot depend on $t$, and the third equation tells you that $K^t$ cannot depend on $r$ (because the two terms that cancel in the equation above will still be there, just multiplied by $K^t$, and there will be a third term $g_{tt} \partial_r K^t$ which can't be canceled by anything, because there's nothing else there, so it must vanish).

I'll refrain for now from showing how the other possibilities are eliminated, leaving $K^\mu = (1, 0, 0, 0)$ as the only Killing vector field. (Note that you can rescale this by multiplying it by a constant, but that doesn't count as a separate Killing vector field; it just corresponds to rescaling the coordinates.)

 

[PeterDonis]

your $K_t$ here is now not a Killing vector field...

Also, your $K_r$ is not a vector field at all.

It occurs to me that I might have been misreading this; you might have intended $K_t$ and $K_r$ to be components of $K_\mu$, the covector field corresponding to $K^\mu$. If that is the case, yes, these are the correct covector components, corresponding to $K^\mu = (1, 0)$ (if we only include the $t$ and $r$ components).

 

[PeterDonis]

leaving $K^\mu = (1, 0, 0, 0)$ as the only Killing vector field.

Or $K^\mu = (1, 0)$ if we just include the $t$ and $r$ components, which I keep forgetting we are restricting to.

 

[kent davidge]

I don't understand, you just rewrote my previous post in another way, to conclude the same as I have concluded.

It occurs to me that I might have been misreading this; you might have intended $K_t$ and $K_r$ to be components of $K_\mu$, the covector field corresponding to $K^\mu$. If that is the case, yes, these are the correct covector components, corresponding to $K^\mu = (1, 0)$ (if we only include the $t$ and $r$ components).

Yes.

the second equation tells you that $K^t$ cannot depend on $t$

I don't think so. You are arriving at this conclusion by assuming $K^r$ vanishes, which is not a priori thing.

I'll refrain for now from showing how the other possibilities are eliminated, leaving $K^\mu = (1, 0, 0, 0)$ as the only Killing vector field. (Note that you can rescale this by multiplying it by a constant, but that doesn't count as a separate Killing vector field; it just corresponds to rescaling the coordinates.)

But the non trivial field of my post #42 satisfies the Killing equations.

 

[PeterDonis]

You are arriving at this conclusion by assuming KrK^r vanishes, which is not a priori thing.

Yes, here I am only considering the case where $K^t$ is the only nonzero component. As I said at the end of that post, I'm going to refrain for now from explicitly considering the other cases and showing how they are eliminated.

By checking the non trivial field of my post #42 you will see that it satisfies the Killing equations.

Ok, let's check. Since you wrote it with lower indexes, we can leave out the index lowering. So we have:

$$
K_t = \sqrt{1- \frac{2m}{r}} \frac{mt}{r^2}
$$

$$
K_r = \sqrt{ \frac{1}{1- \frac{2m}{r}}}
$$

The third Killing equation becomes

$$
\partial_t K_r + \partial_r K_t - 2 \Gamma^t{}_{tr} K_t = 0 + \sqrt{ \frac{1}{1- \frac{2m}{r}}} \left( \frac{m}{r^2} \right) \frac{mt}{r^2} - \sqrt{1- \frac{2m}{r}} \frac{mt}{r^3} - 2 \frac{m}{r^2 \left( 1 - \frac{2m}{r} \right)} \sqrt{1- \frac{2m}{r}} \frac{mt}{r^2}
$$

This does not all cancel out to zero, so the Killing equation is not satisfied.

I strongly suggest that you very, very carefully check your work. The fact that $(1, 0)$ is the only Killing vector field in the $t-r$ plane of Schwarzschild spacetime has been well known for decades and proving it is often assigned as a textbook problem. You should not be persisting in claiming that there are others; you should be looking for where you made a mistake.

 

[stevendaryl]

I'm going to check the last one I posted.
The non vanishing connection coefficients are $$\Gamma^t{}_{rt} = \Gamma^t{}_{tr} = \frac{-m}{2mr - r^2} \\ \Gamma^r{}_{tt} = \frac{m(r - 2m)}{r^3} \\ \Gamma^r{}_{rr} = \frac{m}{2mr - r^2}$$ The Killing equations are $$0 = \nabla_r K_r = \partial_r K_r - \Gamma^t{}_{rr} K_t - \Gamma^r{}_{rr} K_r = 0 \\ 0 = \nabla_t K_t = \partial_t K_t - \Gamma^t{}_{tt} K_t - \Gamma^r{}_{tt} K_r = 0 \\ 0 = \nabla_t K_r + \nabla_r K_t = \partial_t K_r + \partial_r K_t - 2 \Gamma^r{}_{tr} K_r - 2 \Gamma^t{}_{tr} K_t = \frac{-2m}{r^2} - \frac{2m}{2mr -r^2} \bigg(1 - \frac{2m}{r} \bigg) = 0$$

You need three equations:

$\nabla_r K_r = 0$
$\nabla_r K_t + \nabla_t K_r = 0$
$\nabla_t K_t = 0$

 

[kent davidge]

You need three equations:

$\nabla_r K_r = 0$
$\nabla_r K_t + \nabla_t K_r = 0$
$\nabla_t K_t = 0$

There are indeed three equations on my post

The fact that $(1, 0)$ is the only Killing vector field in the $t-r$ plane of Schwarzschild spacetime has been well known for decades and proving it is often assigned as a textbook problem

I've actually seen this through explicit calculation. Is there an other way of knowing it?

 

[PeterDonis]

Is there an other way of knowing it?

Yes, there certainly is. And it is a good idea to consider such a question before even trying an explicit calculation. I think it was John Wheeler who said that you should never try to calculate something unless you already know the answer.

The key is to consider the geometric meaning of a Killing vector field: it is a vector field that generates an isometry. That means that along any integral curve of the KVF, the metric does not change.

This has the following implications if we consider the components of the KVF in a coordinate chart:

(1) The KVF cannot have a component in the direction of any coordinate that the metric depends on (because the metric would have to change along any such component). In the case under discussion, this means a KVF cannot have an $r$ component, since the metric depends on $r$.

(2) The KVF cannot depend on any coordinate that changes along its integral curves (because if it did, you would be able to distinguish points along an integral curve by a geometric invariant--the norm of the KVF--and you can't distinguish points along the integral curves of an isometry by any geometric invariant). In the case under discussion, since we know from #1 just above that the KVF cannot have an $r$ component, it can only have a $t$ component; and therefore it cannot depend on $t$, since $t$ must change along any integral curve.

(3) There cannot be two distinct KVFs that have the same integral curves (because there can only be one isometry that has a given set of integral curves). This means that, once we have found one KVF of the form $(K^t, 0)$, that must be the only one, because any KVF with only a $t$ component will have the same integral curves.

Doing an explicit calculation will show how the above considerations work out in the details of the math.

 

[PeterDonis]

I've actually seen this through explicit calculation.

Do you mean you have just seen this now, or that you had seen it at some time previously before this thread was started?

 

[kent davidge]

Do you mean you have just seen this now, or that you had seen it at some time previously before this thread was started?

that I went to check it after your post #51

 

[PeterDonis]

that I went to check it after your post #51

Ok, good.

 

[kent davidge]

$(1, 0)$ is the only Killing vector field in the $t-r$ plane of Schwarzschild spacetime has been well known for decades and proving it is often assigned as a textbook problem

Is it proven the way you did in post #54?

 

[PeterDonis]

Is it proven the way you did in post #54?

The textbook problems I remember asked you to crank through the detailed math using the Killing equations in component form in Schwarzschild coordinates. My reasoning in post #54 could probably be worked into a proof using coordinate-free math, but I don't remember being asked to do it that way as an actual proof.