General Solution to Killing's Equation in flat s-t

[binbagsss]

Homework Statement

Killing Equation is: $\nabla_u K_v + \nabla_v K_u =0$

In flat s-t this reduces to:

$\partial_u K_v + \partial_v K_u =0$

With a general solution of the form:

$K_u= a_u + b_{uc} K^c$

where $a_u$ and $b_{uv}$ are a constant vector and a constant tensor

QUESTION
Show that $b_{uv}$ is antisymmetric: $b_{uv}=-b{vu}$

Homework Equations

see above

The Attempt at a Solution

[/B]
subbing in the general solution form I have:

$\partial_u (a_v+ b_{va}x^a) + \partial_v (a_u+ b_{ua}x^a)$

From Kiling's equation it is obvious that $\partial_v K_u$ is antisymmetric in $u,v$

Here I need to look at $v$and $a$ in the first term and $u$ and $a$ in the second term, ?where the $a$ is summed over in both terms, which is throwing me of a bit and I'm not sure what to do?

Or if I use the antisymmetry of $u$ and $v$ and substitute in the general form of $K^u$ I have:

$\partial_u(a_v+b_{va}x^a)=-\partial_v(a_u+b_{ua}x^a)$

and now I am not sure what to do
Many thanks

 

[Tio Barnabe]

See, it's far more easy if you consider the problem in the following way.

An infinitesimal Lorentz transformation is of the form $\Lambda_{\alpha \beta} = 1 + \epsilon \ \delta_{\alpha \beta}, |\epsilon| <<1$.
At the same time, a Lorentz transformation must satisfy the Lorentz condition (hopefully you know this) $\Lambda^T \eta^T \Lambda = \eta$.

This condition means that $\Lambda \Lambda^T = 1$. Let's see what we get if we use the above expression for $\Lambda$.

$$\Lambda_{\alpha \beta} \Lambda_{\beta \alpha} = (1 + \epsilon \delta_{\alpha \beta}) (1 + \epsilon \delta_{\beta \alpha}) = 1 + \epsilon (\delta_{\alpha \beta} + \delta_{\beta \alpha}) + \mathcal{O} (\epsilon)^2 \approx 1 + \epsilon (\delta_{\alpha \beta} + \delta_{\beta \alpha}) \stackrel{!}{=} 1$$ The above identity is satisfied only if $\delta_{\alpha \beta} = - \delta_{\beta \alpha}$.

 

[Metmann]

Ku=au+bucKc

Do you maybe mean $K_u = a_u + b_{uc}x^c$?

 

[binbagsss]

Do you maybe mean $K_u = a_u + b_{uc}x^c$?

apologies yes i do

 

[binbagsss]

See, it's far more easy if you consider the problem in the following way.

An infinitesimal Lorentz transformation is of the form $\Lambda_{\alpha \beta} = 1 + \epsilon \ \delta_{\alpha \beta}, |\epsilon| <<1$.
At the same time, a Lorentz transformation must satisfy the Lorentz condition (hopefully you know this) $\Lambda^T \eta^T \Lambda = \eta$.

This condition means that $\Lambda \Lambda^T = 1$. Let's see what we get if we use the above expression for $\Lambda$.

$$\Lambda_{\alpha \beta} \Lambda_{\beta \alpha} = (1 + \epsilon \delta_{\alpha \beta}) (1 + \epsilon \delta_{\beta \alpha}) = 1 + \epsilon (\delta_{\alpha \beta} + \delta_{\beta \alpha}) + \mathcal{O} (\epsilon)^2 \approx 1 + \epsilon (\delta_{\alpha \beta} + \delta_{\beta \alpha}) \stackrel{!}{=} 1$$ The above identity is satisfied only if $\delta_{\alpha \beta} = - \delta_{\beta \alpha}$.

Apologies I didn't say in the OP
but the question is to show this is implied by Killings equation.
thanks

 

[Metmann]

apologies yes i do

But then the answer is trivial. Plugging the general solution into Killing's flat equation yields $$ b_{uc} \partial_v x^c = - b_{vc} \partial_u x^c,$$
which is equivalent to
$$ b_{uc} \delta^c_v = b_{uv} = - b_{vu} = - b_{vc}\delta^c_u $$.

Maybe you missed, that $a$ and $b$ are constant, so $\partial a = 0$ and $\partial ( b x )= b \partial x$.

 

[Tio Barnabe]

the question is to show this is implied by Killings equation

Ok. Then substitute the solution into the Killing Equation. You should get:

$(b_{\nu \sigma} \partial_\mu + b_{\mu \sigma} \partial_\nu) K^\sigma = 0$. This is to hold for any vector $K$,
so we can re express it without the vector $K$ in front: $b_{\nu \sigma} \partial_\mu + b_{\mu \sigma} \partial_\nu = 0$.

Make the possible cyclic permutations for the indices in the above expression. There are two useful cyclic permutations in our case:

1 - swap $\nu$ and $\sigma$;
2 - swap $\sigma$ and $\nu$.

You then obtain

$$b_{\nu \sigma} \partial_\mu + b_{\mu \sigma} \partial_\nu = 0 \\
b_{\sigma \nu} \partial_\mu + b_{\mu \nu} \partial_\sigma = 0 \\
b_{\nu \mu} \partial_\sigma + b_{\sigma \mu} \partial_\nu = 0$$

Now add these three equations together, noticing that 0 + 0 + 0 = 0, and you will get what you are looking for.