Flat s-t 4d killing vectors via solving killing equation

[binbagsss]

So I know what these are
4 translation : $\frac{\partial}{\partial_ x^{u}} = \partial_{x^u}$
3 boost: $z\partial_y - y \partial_z$ and similar for $x,z$ and $y,x$
3 rotation: $t\partial_x + x\partial_t$ and similar for $y , z$

however I want to do it by solving Killing equation:

$\nabla_u V^v + \nabla_v V^u =0$
So in flat space, these $\nabla_u$ reduce to partial derivatives $\partial_u$

So Killing equation reduces to : $\partial_u V^v + \partial_v V^u=0$

Without writing things out explicitly, e.g the time translation $\partial x^0 = (1,0,0,0)$ I am confused how to work in index notation. To begin, the translations $\partial_{x^u}$ are covector and not vector, the killing equation works in vector, so rather do I need $\frac{\partial}{\partial x_u}$ instead of $\frac{\partial}{\partial x^u}$, I don't know what this is explictly?

Further I am confused with the indices in the boosts and the rotations, so the translations are given as covectors, which we can raise an index to get a vector but isn't something like:

$z\partial_y - y\partial_z$ a covector multiplied by a vector and so not a vector but a scalar, since $x^u=x,y,z,t$ is a vector but $\partial_x^u$ is a covector.

Anyway once I've cleared these up the HINT is to differentiate Killing equation and then solve the ODE.
Should I do $\partial_u$ or should I choose a different index not already in Killing equation. Does it matter? I don't see how we can convert this PDE itno an ODE since it already has $\partial_u$ and $\partial_v$, if I hit it with $\partial_u$ I get a $\partial^2_u$ but then also the mixed term $\partial^2_uv$

Many thanks

 

[stevendaryl]

So I know what these are
4 translation : $\frac{\partial}{\partial_ x^{u}} = \partial_{x^u}$
3 boost: $z\partial_y - y \partial_z$ and similar for $x,z$ and $y,x$
3 rotation: $t\partial_x + x\partial_t$ and similar for $y , z$

however I want to do it by solving Killing equation:

$\nabla_u V^v + \nabla_v V^u =0$

That's not the Killing equation. The Killing equation is:

$\nabla_u V_v + \nabla_v V_u =0$

(with lowered indices on the vector field $V$). The Killing equation is a relationship between the vector field $V^u$ and the metric tensor, $g_{uv}$. The equation you wrote down doesn't involve the metric tensor at all.

The equation I wrote down doesn't appear to involve the metric tensor, either, but it does, because:

$V_v \equiv g_{av} V^a$

So the simple-looking equation
$\nabla_u V_v + \nabla_v V_u =0$

really means something a little more complicated:

$\nabla_u (g_{av}V^a) + \nabla_v (g_{au} V^a) =0$

$= (\nabla_u g_{av})V^a + g_{av} \nabla_u V^a + (\nabla_v g_{au}) V^a + g_{au} \nabla_v V^a =0$

 

[binbagsss]

That's not the Killing equation. The Killing equation is:
(with lowered indices on the vector field $V$). The Killing equation is a relationship between the vector field $V^u$ and the metric tensor, $g_{uv}$. The equation you wrote down doesn't involve the metric tensor at all.

The equation I wrote down doesn't appear to involve the metric tensor, either, but it does, because:

$V_v \equiv g_{av} V^a$

So the simple-looking equation
$\nabla_u V_v + \nabla_v V_u =0$

really means something a little more complicated:

$\nabla_u (g_{av}V^a) + \nabla_v (g_{au} V^a) =0$

$= (\nabla_u g_{av})V^a + g_{av} \nabla_u V^a + (\nabla_v g_{au}) V^a + g_{au} \nabla_v V^a =0$

hmmm okay, ta.

however, the fundamental theorem of riemann geometry is that you can write the connection, levita cevita tensor in terms of the metric, so therefore via the covariant derivative i do have dependence on the metric. this is consistent with the last equation you've mentioned, meaning something more complicated, because from the fundamental theorem of riemann geometry the fact arises that $\nabla_u g_av = 0$ enabling you to move the metric outside the expression in the bottom line - so you have here chosen to use of the levita cevita tensor as the connection.

However even with this , dependence on the metric via the fundamental theorem, it is a relationship between a covector and not a vector so that part of your point i still see.

 

[binbagsss]

hmmm okay, ta.

however, the fundamental theorem of riemann geometry is that you can write the connection, levita cevita tensor in terms of the metric, so therefore via the covariant derivative i do have dependence on the metric. this is consistent with the last equation you've mentioned, meaning something more complicated, because from the fundamental theorem of riemann geometry the fact arises that $\nabla_u g_av = 0$ enabling you to move the metric outside the expression in the bottom line - so you have here chosen to use of the levita cevita tensor as the connection.

However even with this , dependence on the metric via the fundamental theorem, it is a relationship between a covector and not a vector so that part of your point i still see.

Anyway isn't

$z\partial_y - y\partial_z$ rank $(1,1)$ object whereas we need a $(0,1)$?

 

[stevendaryl]

I'm not sure about what it is that you are doing. Those 10 operations are symmetries of Minkowsky spacetime, but the relevant quantity is a Killing vector field.

A vector field is a vector $V^\mu(\mathcal{P})$ defined at every point in spacetime. If we choose inertial coordinates (so the connection coefficients are all zero), then the Killing equation just becomes:

$\partial_\mu V_\nu + \partial_\nu V_\mu = 0$

It's obvious that any constant vector field will satisfy that equation. But as you suggest, you can come up with other solutions by taking a second derivative:

$\partial^2_\mu V_\nu + \partial_\mu \partial_\nu V_\mu = 0$

We know that $\partial_\mu V_\mu = 0$ (because that's the $\mu - \mu$ Killing equation). So since partials commute, we have:

$\partial^2_\mu V_\nu = 0$

That's a pretty easy differential equation to solve.

 

[PeterDonis]

3 boost: $z\partial_y - y \partial_z$ and similar for $x,z$ and $y,x$

3 rotation: $t\partial_x + x\partial_t$ and similar for $y,z$

You have these backwards; what you are calling boosts are actually rotations, and what you are calling rotations are actually boosts.

 

[PeterDonis]

isnt

$z\partial_y - y\partial_z$ rank $(1,1)$ object

No, it's a vector field--a (1, 0) object--whose components depend on the coordinates.

 

[binbagsss]

You have these backwards; what you are calling boosts are actually rotations, and what you are calling rotations are actually boosts.

apologies, indeeed i do.

 

[binbagsss]

No, it's a vector field--a (1, 0) object--whose components depend on the coordinates.

apologies I was thinking $\frac{\partial}{\partial x^u}$ was a covector and not a vector. So vector multiplied by a vector gives a vector, and if it was vector multiplied by covector it would be rank $(1,1)$ right?

 

[binbagsss]

I'm not sure about what it is that you are doing. Those 10 operations are symmetries of Minkowsky spacetime, but the relevant quantity is a Killing vector field.

A vector field is a vector $V^\mu(\mathcal{P})$ defined at every point in spacetime. If we choose inertial coordinates (so the connection coefficients are all zero), then the Killing equation just becomes:

$\partial_\mu V_\nu + \partial_\nu V_\mu = 0$

It's obvious that any constant vector field will satisfy that equation. But as you suggest, you can come up with other solutions by taking a second derivative:

$\partial^2_\mu V_\nu + \partial_\mu \partial_\nu V_\mu = 0$

We know that $\partial_\mu V_\mu = 0$ (because that's the $\mu - \mu$ Killing equation). So since partials commute, we have:

$\partial^2_\mu V_\nu = 0$

That's a pretty easy differential equation to solve.

Okay thank you, I understand the steps taken to reduce the Killing equation to $\partial^2_\mu V_\nu = 0$
however I'm unsure how to get started, explicitly, the indices are throwing me, I haven't done any integration/ differential equations which involve indices.

To try and see things explicitly I wanted to verify that a rotation, for example, is a solution.
So I have

$V^{a}=x^{3}\partial x^1 - \partial x^1 x^3$, $x^{3}=z, x^1=x, x^0=t ..$ etc
And Killing equation as
$\partial^2_u g_{av} V^a = 0$

I don't know how to even lower the index on $V^{a}$ written as above as it looks to explicit. I have a general form of all rotations that would probably be easier to work with, is this the expected outcome when you solve the differential equation (for rotations)? the general form that covers all rotations , rather than the more explicit form, thanks

 

[stevendaryl]

Okay thank you, I understand the steps taken to reduce the Killing equation to $\partial^2_\mu V_\nu = 0$ however I'm unsure how to get started, explicitly, the indices are throwing me, I haven't done any integration/ differential equations which involve indices.

Well, $V_\nu$ is just a set of 4 (or however many dimensions there are to your space) functions. So this equation is just like the 16 differential equations:

$\frac{\partial^2}{\partial x^2} V_0(x,y,z,t) = 0$
$\frac{\partial^2}{\partial y^2} V_0(x,y,z,t) = 0$
$\frac{\partial^2}{\partial z^2} V_0(x,y,z,t) = 0$
$\frac{\partial^2}{\partial t^2} V_0(x,y,z,t) = 0$
...
(same for $V_1, V_2, V_3$)

$V_0$ is just a function, and the above is just 4 differential equations involving $V_0$. What's the solution for $V_0$?

 

[stevendaryl]

I don't know how to even lower the index on $V^{a}$ written as above as it looks to explicit.

$V_a$ is defined by:

$V_a \equiv \sum_b g_{ab} V^b$

So if your metric is

• $g_{ab} = +1$ when $a = b = 0$
• $g_{ab} = -1$ when $a = b = 1$ or $a = b = 2$ or $a = b = 3$
  
• $g_{ab} = 0$ when $a \neq b$

Then $V_0 = V^0$
$V_i = -V^i$ (for $i=1,2,3$)

 

[PeterDonis]

vector multiplied by a vector gives a vector

No. A vector multiplied by a scalar gives a vector. You can't "multiply" a vector by another vector (at least, not without additional structure on your vector space); but you can form a bivector, which is a rank $(2, 0)$ tensor.

if it was vector multiplied by covector it would be rank $(1,1)$ right?

As above, this is not really "multiplication", but the analogue of a bivector with a vector and a covector gives a rank $(1, 1)$ tensor, yes.

 

[binbagsss]

No. A vector multiplied by a scalar gives a vector. You can't "multiply" a vector by another vector (at least, not without additional structure on your vector space); but you can form a bivector, which is a rank $(2, 0)$ tensor.

So then aren't the rotations in the form $z\partial x - x \partial z$ a vector multiplied by a vector a $(2,0)$ not a $(1,0)$?

[Moderator's note: edited to correct formatting of quote.]

 

[PeterDonis]

So then aren't the rotations in the form $\partial x - x \partial z$ a vector multiplied by a vector

No. The Killing vector $z \partial_x - x \partial_z$ is a vector whose components depend on the coordinates, as I said before. The $z$ and $x$ coefficients are functions of the coordinates, not components of a vector.

 

[binbagsss]

Well, $V_\nu$ is just a set of 4 (or however many dimensions there are to your space) functions. So this equation is just like the 16 differential equations:

$\frac{\partial^2}{\partial x^2} V_0(x,y,z,t) = 0$
$\frac{\partial^2}{\partial y^2} V_0(x,y,z,t) = 0$
$\frac{\partial^2}{\partial z^2} V_0(x,y,z,t) = 0$
$\frac{\partial^2}{\partial t^2} V_0(x,y,z,t) = 0$
...
(same for $V_1, V_2, V_3$)

$V_0$ is just a function, and the above is just 4 differential equations involving $V_0$. What's the solution for $V_0$?

Okay thanks, 16 differential equations allows me to see things clearer.
Solving the equations for $V_0$..

the first one for example, on the first integration I get $c(y,z,t)$, a constant that wrt $x$ depending on $y,z,t$, and then on the second integration I get $x c(y,z,t)$ , similarly for the other three equations.

So I have from each equation:
$V_0 = x c(y,z,t)$
$V_0 = y c(x,z,t)$
$V_0 = z c(y,x,t)$
$V_0 = t c(y,z,x)$

well the constant $c$ shouldn't be the same, but anyway, I now need to solve these simulatenously? or have I done something wrong? no idea how to solve simultaneously? thanks

 

[binbagsss]

No. The Killing vector $z \partial_x - x \partial_z$ is a vector whose components depend on the coordinates, as I said before. The $z$ and $x$ coefficients are functions of the coordinates, not components of a vector.

ah yes apologies thank you, individual components of a vector are scalar

 

[stevendaryl]

Okay thanks, 16 differential equations allows me to see things clearer.
Solving the equations for $V_0$..

the first one for example, on the first integration I get $c(y,z,t)$, a constant that wrt $x$ depending on $y,z,t$, and then on the second integration I get $x c(y,z,t)$ , similarly for the other three equations.

So I have from each equation:
$V_0 = x c(y,z,t)$
$V_0 = y c(x,z,t)$
$V_0 = z c(y,x,t)$
$V_0 = t c(y,z,x)$

well the constant $c$ shouldn't be the same, but anyway, I now need to solve these simulatenously? or have I done something wrong? no idea how to solve simultaneously? thanks

Well, you haven't chosen the most general solution. If

$(\partial_x)^2 V_0 = 0$, then $V_0 = a(y,z,t) + c(y, z, t) x$

Now, plug that answer into the next differential equation:
$(\partial_y)^2 V_0 = 0 \Rightarrow (\partial_y)^2 (a(y,z,t) + c(y, z, t) x) = 0$

So solve that for $a(y,z,t)$ and $c(y,z,t)$. Etc.

We can skip to the answer, unless you want to work it out yourself. Do you?

 

[binbagsss]

Well, you haven't chosen the most general solution. If

$(\partial_x)^2 V_0 = 0$, then $V_0 = a(y,z,t) + c(y, z, t) x$

Now, plug that answer into the next differential equation:
$(\partial_y)^2 V_0 = 0 \Rightarrow (\partial_y)^2 (a(y,z,t) + c(y, z, t) x) = 0$

So solve that for $a(y,z,t)$ and $c(y,z,t)$. Etc.

We can skip to the answer, unless you want to work it out yourself. Do you?

Okay so doing that I get $a''=\frac{-c''}{x}$ where $c' = \partial c / \partial y$, if I now integrate this to get $c$ in terms of $a$ I end up introducing to more constants that now depend on $x,z,t$ instead, so this can't be the right move?

 

[stevendaryl]

Okay so doing that I get $a''=\frac{-c''}{x}$ where $c' = \partial c / \partial y$, if I now integrate this to get $c$ in terms of $a$ I end up introducing to more constants that now depend on $x,z,t$ instead, so this can't be the right move?

I don't understand what you're doing, but let's work it out in 2-D space, and maybe you can see how it would generalize:

If $x, y$ are Cartesian coordinates, then the Killing equation $\nabla_\mu V_\nu + \nabla_\nu V_\mu = 0$
becomes 3 equations:
$\partial_x V_x = 0$
$\partial_y V_y = 0$
$\partial_x V_y + \partial_y V_x = 0$

We take an additional derivative to get an equation for just $V_x$:

$\partial^2_x V_x = 0 \Rightarrow V_0 = a(y) + c(y) x$
$\partial^2_y V_x = 0 \Rightarrow \partial^2_y [a(y) + c(y) x] = 0 \Rightarrow a(y) = b + dy, c(y) = e + fy$

So $V_x = b + dy + e x + f x y$ where $b, d, e, f$ are constants.

Similarly, $V_y = g + hy + i x + j x y$ where $g, h, i, j$ are constants.

Now, plug these back into the original 3 Killing equations, to get:
$\partial_x V_x = 0 \Rightarrow e = f = 0$
$\partial_y V_y = 0 \Rightarrow h = j = 0$

$\partial_x V_y + \partial_y V_x = 0 \Rightarrow i = -d$

So the solution in 2D is: $V_x = b + d y$, $V_y = g - d x$

 

[stevendaryl]

Okay so doing that I get $a''=\frac{-c''}{x}$ where $c' = \partial c / \partial y$, if I now integrate this to get $c$ in terms of $a$ I end up introducing to more constants that now depend on $x,z,t$ instead, so this can't be the right move?

Oh, I see what you're doing.

We have: $\partial_y^2 (a(y,z,t) + c(y,z,t) x) = 0$

The first simplification is to consider the case $x=0$. In that case, you have:

$\partial_y^2 (a(y,z,t)) = 0$

But since $a$ doesn't depend on $x$, if that's true for $x=0$, it's true for all $x$.

The next simplification is that:
$\partial_y^2 (a(y,z,t) + c(y,z,t) x) = \partial_y^2 (a(y,z,t)) + \partial_y^2 (c(y,z,t) x) = 0$. The first term is 0, so this simplifies to:

$\partial_y^2 (c(y,z,t) x) = 0$

But the partial doesn't affect $x$, so we can rewrite this as:
$x \partial_y^2 (c(y,z,t) ) = 0$

which means
$\partial_y^2 (c(y,z,t) ) = 0$

 

[binbagsss]

So $V_x = b + dy + e x + f x y$ where $b, d, e, f$ are constants.

$d,e,f$ here are constants of $x$ though? $d=d(x)$ etc?

$\partial_x V_x = 0 \Rightarrow e = f = 0$

Therefore I do not get $=> e,f=0$ :

$d/dx(e(x)x+f(x)xy) = e'(x)x+e(x)+f'(x)xy+f(x)y=0$...

 

[stevendaryl]

$d,e,f$ here are constants of $x$ though? $d=d(x)$ etc?

No, constants in the sense that they are just numbers. Like 5 or 22.7.

Try for yourself. Let $V_x$ be the following function: $V_x = 12 + 22 y$. Let $V_y = 23.2 - 22 x$. Then calculate $\partial_x V_y + \partial_y V_x$. Is it zero?

 

[stevendaryl]

The point is that

• for one coordinate, the most general function satisfying $\partial_x^2 \phi = 0$ is $\phi = a + bx$.
• for two coordinates, the most general function satisfying $\partial_x^2 \phi = \partial_y^2 \phi = 0$ is $\phi = a + bx + cy + d xy$
• for three coordinates, the most general function satisfying $\partial_x^2 \phi = \partial_y^2 \phi = \partial_z^2 \phi = 0$ is $\phi = a + bx + cy + dx y + ez + f xz + g yz + h xyz$
• for four coordinates, the most general function satisfying $\partial_x^2 \phi = \partial_y^2 \phi = \partial_z^2 \phi = \partial_t^2 \phi = 0$ is $\phi = a + bx + cy + dx y + ez + f xz + g yz + h xyz + i t + j xt + k yt + l zt + m xyt + n xzt + o yzt + p xyzt$
• The pattern is that for $n$ variables, you need $2^n$ constants, so you will use up $2^n + n$ letters of the alphabet. So that's a proof that spacetime must be 4-dimensional, because 5 dimensions would require 37 letter. The Chinese could study 5-dimensional spacetime, but not me.

 

[binbagsss]

• for four coordinates, the most general function satisfying $\partial_x^2 \phi = \partial_y^2 \phi = \partial_z^2 \phi = \partial_t^2 \phi = 0$ is $\phi = a + bx + cy + dx y + ez + f xz + g yz + h xyz + i t + j xt + k yt + l zt + m xyt + n xzt + o yzt + p xyzt$

Okay thanks, so I've worked through that and so $V^0, V^1, V^2, V^3$ all takes this form...
Now from this I am unsure how do to get the three rotations and the three boosts?

thanks

 

[stevendaryl]

Okay thanks, so I've worked through that and so $V^0, V^1, V^2, V^3$ all takes this form...
Now from this I am unsure how do to get the three rotations and the three boosts?

Well, your 10 vector fields are (realize that $\partial_x$ is just a basis vector in the x-direction, so $y \partial_x$ is just a vector with $V^x = y$)

1. Translation in x-direction: $V^x = 1$, $V^y = V^z = V^t = 0$. Then $V_x = -1$ and the other components are zero.
   
2. y-direction: $V_y = -1$ and the other components are zero.
3. z-direction: $V_z = -1$ and the other components are zero.
4. t-direction: $V_t = +1$ and the other components are zero.
5. Rotation in x-y plane: $V^x = y$, $V^y = -x$. So $V_x = -y$, $V_y = x$.
6. Rotation in x-z plane: $V^x = z$, $V^z = -x$. So $V_x = -z$, $V_z = x$.
7. Rotation in y-z plane: $V^y = z$, $V^z = -y$. So $V_y = -z$, $V_z = y$.
   
8. Boost in x direction: $V^x = t$, $V^t = x$. So $V_x = -t$, $V_t = x$.
9. Boost in y direction: $V^y = t$, $V^t = y$. So $V_y = -t$, $V_t = y$.
10. Boost in z direction: $V^z = t$, $V^t = z$. So $V_z = -t$, $V_t = x$.

I think it's clear that for all of these, $\partial_\mu V_\nu + \partial_\nu V_\mu = 0$

 

[binbagsss]

Well, your 10 vector fields are (realize that $\partial_x$ is just a basis vector in the x-direction, so $y \partial_x$ is just a vector with $V^x = y$)

1. Translation in x-direction: $V^x = 1$, $V^y = V^z = V^t = 0$. Then $V_x = -1$ and the other components are zero.
   
2. y-direction: $V_y = -1$ and the other components are zero.
3. z-direction: $V_z = -1$ and the other components are zero.
4. t-direction: $V_t = +1$ and the other components are zero.
5. Rotation in x-y plane: $V^x = y$, $V^y = -x$. So $V_x = -y$, $V_y = x$.
6. Rotation in x-z plane: $V^x = z$, $V^z = -x$. So $V_x = -z$, $V_z = x$.
7. Rotation in y-z plane: $V^y = z$, $V^z = -y$. So $V_y = -z$, $V_z = y$.
   
8. Boost in x direction: $V^x = t$, $V^t = x$. So $V_x = -t$, $V_t = x$.
9. Boost in y direction: $V^y = t$, $V^t = y$. So $V_y = -t$, $V_t = y$.
10. Boost in z direction: $V^z = t$, $V^t = z$. So $V_z = -t$, $V_t = x$.

I think it's clear that for all of these, $\partial_\mu V_\nu + \partial_\nu V_\mu = 0$

aha, but how did that makes use of the general form we just obtained?

it's easy to see that these KVF are of this form, but how do you go the other way and use the form to get them?

 

[stevendaryl]

aha, but how did that makes use of the general form we just obtained?

Well, it's obvious that those are all examples of the general form. The trick is to prove that those are the only ones. That's tedious, but it's just algebra. Do you really want me to do it?

Okay, okay...

The general solution of the second order equation is:

$V_x = A_x + B_{xy} y + B_{xz} z + B_{xt} t + C_{xyz} y z + C_{xyt} y t + C_{xzt} z t + D_{xyzt} y z t$
$V_y = A_y + B_{yx} x + B_{yz} z + B_{yt} t + C_{yxz} x z + C_{yxt} x t + C_{yzt} z t + D_{yxzt} x z t$
$V_z = A_z + B_{zx} x + B_{zy} y + B_{zt} t + C_{zxy} x y + C_{zyt} y t + C_{zxt} x t + D_{zxyt} x y t$
$V_t = A_t + B_{tx} x + B_{ty} y + B_{tz} z + C_{txy} x y + C_{tyz} y z + C_{txz} x z + D_{txyz} x y z$

where the As Bs Cs and Ds are all constants. (I've made sure that $V_x$ does not depend on $x$, etc., because $\partial_x V_x = 0$)

So $V_x$ has 8 constants. Similarly for $V_y, V_z, V_t$. So altogether, there are 32 constants.

But we have the 6 equations:
$\partial_x V_y + \partial_y V_x = 0$
$\partial_x V_z + \partial_z V_x = 0$
$\partial_x V_t + \partial_t V_x = 0$
$\partial_y V_z + \partial_z V_y = 0$
$\partial_y V_t + \partial_t V_y = 0$
$\partial_z V_t + \partial_t V_z = 0$

The first equation gives us the constraint:
$B_{xy} + C_{xyz} z + C_{xyt} t + D_{xyzt} z t + B_{yx} + C_{yxz} z + C_{yxt} t + D_{yxzt} z t = 0$

Setting $z=t=0$ gives us: $B_{xy} + B_{yx} = 0$.
Setting $z=0$ gives us $C_{xyt} + C_{yxt} = 0$.
Setting $t=0$ gives us $C_{xyz} + C_{yxz} = 0$
Then we have, finally: $D_{xyzt} + D_{yxzt} = 0$

Other equations give analogous constraints. So we have the following equations:

1. $B_{xy} + B_{yx} = 0$.
2. $C_{xyt} + C_{yxt} = 0$.
3. $C_{xyz} + C_{yxz} = 0$
4. $D_{xyzt} + D_{yxzt} = 0$
5. $B_{xz} + B_{zx} = 0$.
6. $C_{xzt} + C_{zxt} = 0$.
7. $C_{xyz} + C_{zxy} = 0$
8. $D_{xyzt} + D_{zxyt} = 0$
9. $B_{xt} + B_{tx} = 0$.
10. $C_{xzt} + C_{txz} = 0$.
11. $C_{xyt} + C_{txy} = 0$
12. $D_{xyzt} + D_{txyz} = 0$
13. $B_{yt} + B_{ty} = 0$.
14. $C_{yzt} + C_{tyz} = 0$.
15. $C_{yxt} + C_{txy} = 0$
16. $D_{yxzt} + D_{txyz} = 0$
17. $B_{yz} + B_{zy} = 0$.
18. $C_{yzt} + C_{zyt} = 0$.
19. $C_{yxz} + C_{zxy} = 0$
20. $D_{yxzt} + D_{zxyt} = 0$
21. $B_{tz} + B_{zt} = 0$.
22. $C_{tzy} + C_{zyt} = 0$.
23. $C_{txz} + C_{zxt} = 0$
24. $D_{txzy} + D_{zxyt} = 0$

Note, however, that putting equations 2, 11 and 15 together give:
$C_{xyt} + C_{yxt} = 0 \Rightarrow C_{yxt} = - C_{xyt}$
$C_{xyt} + C_{txy} = 0 \Rightarrow C_{txy} = - C_{xyt}$
$C_{yxt} + C_{txy} = 0 \Rightarrow -C_{xyt} + -C_{xyt} = 0$

So $C_{xyt} = 0$ Similarly for all the C's.

Putting equations 4, 8, 12, 16, 20, 24 together give:
$D_{xyzt} + D_{yxzt} = 0 \Rightarrow D_{yxzt} = - D_{xyzt}$
$D_{xyzt} + D_{zxyt} = 0 \Rightarrow D_{zxyt} = - D_{xyzt}$
$D_{xyzt} + D_{txyz} = 0 \Rightarrow D_{txyz} = - D_{xyzt}$
$D_{yxzt} + D_{zxyt} = 0 \Rightarrow -D_{xyzt} + -D_{xyzt} = 0\Rightarrow D_{xyzt} = D_{yxzt} = D_{txyz} = D_{zxyt} = 0$

So the Ds are all zero, also

The equations involving the Bs give us:

$B_{xy} = - B_{yx}$
$B_{xz} = - B_{zx}$
$B_{xt} = - B_{tx}$
$B_{yz} = - B_{zy}$
$B_{yt} = - B_{ty}$
$B_{zt} = - B_{tz}$

So our V_\mu simplifies considerably:

$V_x = A_x + B_{xy} y - B_{zx} z - B_{tx} t$
$V_y = A_y + B_{yz} z - B_{xy} x - B_{ty} t$
$V_z = A_z + B_{zx} x - B_{yz} y - B_{tz} t$
$V_t = A_t + B_{tx} x - B_{ty} y - B_{tz} z$

So our general Killing vector field is specified by
4 translation constants: $A_x, A_y, A_z, A_t$
3 rotation constants: $B_{xy}, B_{yz}, B_{zx}$
3 boost constants: $B_{tx}, B_{ty}, B_{tz}$.

 

[binbagsss]

Well, it's obvious that those are all examples of the general form. The trick is to prove that those are the only ones. That's tedious, but it's just algebra. Do you really want me to do it?

Okay, okay...

The general solution of the second order equation is:

$V_x = A_x + B_{xy} y + B_{xz} z + B_{xt} t + C_{xyz} y z + C_{xyt} y t + C_{xzt} z t + D_{xyzt} y z t$
$V_y = A_y + B_{yx} x + B_{yz} z + B_{yt} t + C_{yxz} x z + C_{yxt} x t + C_{yzt} z t + D_{yxzt} x z t$
$V_z = A_z + B_{zx} x + B_{zy} y + B_{zt} t + C_{zxy} x y + C_{zyt} y t + C_{zxt} x t + D_{zxyt} x y t$
$V_t = A_t + B_{tx} x + B_{ty} y + B_{tz} z + C_{txy} x y + C_{tyz} y z + C_{txz} x z + D_{txyz} x y z$

where the As Bs Cs and Ds are all constants. (I've made sure that $V_x$ does not depend on $x$, etc., because $\partial_x V_x = 0$)So $V_x$ has 8 constants. Similarly for $V_y, V_z, V_t$. So altogether, there are 32 constants.

But we have the 6 equations:
$\partial_x V_y + \partial_y V_x = 0$
$\partial_x V_z + \partial_z V_x = 0$
$\partial_x V_t + \partial_t V_x = 0$
$\partial_y V_z + \partial_z V_y = 0$
$\partial_y V_t + \partial_t V_y = 0$
$\partial_z V_t + \partial_t V_z = 0$

The first equation gives us the constraint:
$B_{xy} + C_{xyz} z + C_{xyt} t + D_{xyzt} z t + B_{yx} + C_{yxz} z + C_{yxt} t + D_{yxzt} z t = 0$

Setting $z=t=0$ gives us: $B_{xy} + B_{yx} = 0$.
Setting $z=0$ gives us $C_{xyt} + C_{yxt} = 0$.
Setting $t=0$ gives us $C_{xyz} + C_{yxz} = 0$
Then we have, finally: $D_{xyzt} + D_{yxzt} = 0$

Other equations give analogous constraints. So we have the following equations:

1. $B_{xy} + B_{yx} = 0$.
2. $C_{xyt} + C_{yxt} = 0$.
3. $C_{xyz} + C_{yxz} = 0$
4. $D_{xyzt} + D_{yxzt} = 0$
5. $B_{xz} + B_{zx} = 0$.
6. $C_{xzt} + C_{zxt} = 0$.
7. $C_{xyz} + C_{zxy} = 0$
8. $D_{xyzt} + D_{zxyt} = 0$
9. $B_{xt} + B_{tx} = 0$.
10. $C_{xzt} + C_{txz} = 0$.
11. $C_{xyt} + C_{txy} = 0$
12. $D_{xyzt} + D_{txyz} = 0$
13. $B_{yt} + B_{ty} = 0$.
14. $C_{yzt} + C_{tyz} = 0$.
15. $C_{yxt} + C_{txy} = 0$
16. $D_{yxzt} + D_{txyz} = 0$
17. $B_{yz} + B_{zy} = 0$.
18. $C_{yzt} + C_{zyt} = 0$.
19. $C_{yxz} + C_{zxy} = 0$
20. $D_{yxzt} + D_{zxyt} = 0$
21. $B_{tz} + B_{zt} = 0$.
22. $C_{tzy} + C_{zyt} = 0$.
23. $C_{txz} + C_{zxt} = 0$
24. $D_{txzy} + D_{zxyt} = 0$

Note, however, that putting equations 2, 11 and 15 together give:
$C_{xyt} + C_{yxt} = 0 \Rightarrow C_{yxt} = - C_{xyt}$
$C_{xyt} + C_{txy} = 0 \Rightarrow C_{txy} = - C_{xyt}$
$C_{yxt} + C_{txy} = 0 \Rightarrow -C_{xyt} + -C_{xyt} = 0$

So $C_{xyt} = 0$ Similarly for all the C's.

Putting equations 4, 8, 12, 16, 20, 24 together give:
$D_{xyzt} + D_{yxzt} = 0 \Rightarrow D_{yxzt} = - D_{xyzt}$
$D_{xyzt} + D_{zxyt} = 0 \Rightarrow D_{zxyt} = - D_{xyzt}$
$D_{xyzt} + D_{txyz} = 0 \Rightarrow D_{txyz} = - D_{xyzt}$
$D_{yxzt} + D_{zxyt} = 0 \Rightarrow -D_{xyzt} + -D_{xyzt} = 0\Rightarrow D_{xyzt} = D_{yxzt} = D_{txyz} = D_{zxyt} = 0$

So the Ds are all zero, also

The equations involving the Bs give us:

$B_{xy} = - B_{yx}$
$B_{xz} = - B_{zx}$
$B_{xt} = - B_{tx}$
$B_{yz} = - B_{zy}$
$B_{yt} = - B_{ty}$
$B_{zt} = - B_{tz}$

So our V_\mu simplifies considerably:

$V_x = A_x + B_{xy} y - B_{zx} z - B_{tx} t$
$V_y = A_y + B_{yz} z - B_{xy} x - B_{ty} t$
$V_z = A_z + B_{zx} x - B_{yz} y - B_{tz} t$
$V_t = A_t + B_{tx} x - B_{ty} y - B_{tz} z$

So our general Killing vector field is specified by
4 translation constants: $A_x, A_y, A_z, A_t$
3 rotation constants: $B_{xy}, B_{yz}, B_{zx}$
3 boost constants: $B_{tx}, B_{ty}, B_{tz}$.

okay thanks alot,

But there's nothing to say, e.g , all other B' and A's are zero when $B_{xy}$ isn't , so this is probably a really stupid question, from this it looks like one single Killing vector?
We can use that the sum of Killing vectors is a vector, to break it down? but wouldn't you say there's one Killing vector?
thanks
or is there a difference in terminology between Killing vector field and Killing vector , many thanks

 

[stevendaryl]

okay thanks alot,

But there's nothing to say, e.g , all other B' and A's are zero when $B_{xy}$ isn't , so this is probably a really stupid question, from this it looks like one single Killing vector?
We can use that the sum of Killing vectors is a vector, to break it down? but wouldn't you say there's one Killing vector?
thanks
or is there a difference in terminology between Killing vector field and Killing vector , many thanks

The constants A and B are arbitrary, so there are lots of Killing vectors that you get from different choices of those constants. But all the possibilities are linear combinations of 10 basic Killing vectors.

A vector field is just a vector-valued function of location.

 

[vanhees71]

The discussion seems to overcomplicate things. What you want to solve is the Killing equation
$$\partial_{\mu} V_{\nu} + \partial_{\nu} V_{\mu}=0.$$
The trick to find the general solutions is to first take one more derivative of this equation:
$$\partial_{\rho} \partial_{\mu} V_{\nu} + \partial_{\rho} \partial_{\nu} V_{\mu}=0.$$
Now write down also the other two equations obtained from this by simply cyclically permute the indices:
$$\partial_{\mu} \partial_{\nu} V_{\rho} + \partial_{\mu} \partial_{\rho} V_{\nu}=0,\\
\partial_{\nu} \partial_{\rho} V_{\mu} + \partial_{\nu} \partial_{\mu} V_{\rho}=0.$$
Now add the first two and subtract the last equation. Using that partial derivatives commute, this leads to
$$\partial_{\rho} \partial_{\mu} V_{\nu}=0.$$
This means all 2nd deviatives of $V_{\nu}$ vanish, so that it must be a linear equation of $x^{\mu}$:
$$V_{\nu}=T_{\nu} + \omega_{\nu \mu} x^{\mu}.$$
However, the 1st-order equation is more restrictive than the 2nd-order equation. So we must check this result with the first-order equation. Indeed
$$\partial_{\rho} V_{\nu} + \partial_{\nu} V_{\rho}=\omega_{\nu \rho} + \omega_{\rho \nu} \stackrel{!}{=} 0.$$
This implies that
$$\omega_{\mu \nu}=-\omega_{\nu \mu}.$$
Now remember that $V_{\mu}$ describes infinitesimal transformations of the coordinates that are symmetries. It's obvious that the $T_{\nu}$ define spacetime translations and the $\omega_{\mu \nu}$ Lorentz transformations (i.e., boosts and rotations and any combination thereof), and these are the well-known symmetries of Minkowski spacetime.

 

[PeterDonis]

@binbagsss please stop quoting people's entire posts in your responses, it just clutters up the thread. Please quote only the particular things you are directly responding to.