Solutions to Killing's equation in flat spacetime

[LAHLH]

Hi,

Killing equation in flat space is just \partial_{\mu}K_{\nu}+\partial_{\nu}K_{\mu}=0. I've seen in various places the solutions to this written as K^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}+P^{\mu} where P is a constant 4 vector, and \Lambda_{(\mu\nu)}=0 (i.e. symmetric part vanishes and it is antisym on lowered components)

Can anyone show me how to derive this equation? does the symmetry at work here somehow relate to the infinitesmial generators of the Lorentz group that one meets in QFT? or does anyone have any good links/books where I could read more about this?

thanks

 

[arkajad]

The Killing equation comes form rewriting the condition that the Lie derivative of the metric tensor with respect to the vector field K^\mu vanishes. Take the definition of the Lie derivative applied to a covariant rank two tensor, write it down for the constant flat metric, you will get your equation.

 

[LAHLH]

The Killing equation comes form rewriting the condition that the Lie derivative of the metric tensor with respect to the vector field K^\mu vanishes. Take the definition of the Lie derivative applied to a covariant rank two tensor, write it down for the constant flat metric, you will get your equation.

Hi, I think you misunderstand perhaps, what I was asking. I know how to derive Killings equation from Lie derivative etc etc. What I want to know is how to obtain solutions of the form I posted. <br /> K^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}+P^{\mu} <br />

how to see the connection with the generators of Lorentz group and so forth.

 

[arkajad]

Take Taylor expansion

K_\mu(x)=a_\mu+a_{\mu\nu}x^\nu+a_{\mu\nu\sigma}x^\nux^\sigma+...
Apply the Killing conditions. You will get:

a_\mu is arbitrary

a_{\mu\nu} is antisymmetric

All other terms must be zero due to symmetry reasons. a_{\mu\nu\sigma\ldots} must be antisymmetric in two indices and symmetric in \nu\sigma\ldots.

The standard trick works
a_{\mu\nu\sigma}=a_{\mu\sigma\nu}=-a_{\sigma\mu\nu}=a_{\sigma\nu\mu}=a_{\nu\sigma\mu}=a_{\nu\mu\sigma}=-a_{\mu\nu\sigma}

With Grassmann anticommuting variables \theta^\mu instead of commuting x^\mu we would not be done that fast! I do not know about possible non-analytic solutions.

 

[George Jones]

Killing equation in flat space is just \partial_{\mu}K_{\nu}+\partial_{\nu}K_{\mu}=0

Here is another way.

Write this as

K_{\nu , \mu} + K_{\mu , \nu} = 0.

Differentiating again gives

K_{\nu , \mu \alpha} + K_{\mu , \nu \alpha} = 0.

Relabeling indices gives two more equations

K_{\alpha , \nu \mu} + K_{\nu , \alpha \mu} = 0

K_{\mu , \alpha \nu} + K_{\alpha , \mu \nu} = 0.

Taking into account the fact that partial derivatives commute, the last three equations is a set of three linear equations in three unknowns. As a first (of several small) step(s), what is the solution (think linear algebra, not calculus) to this set of linear equations?

 

[LAHLH]

Here is another way.

Write this as

K_{\nu , \mu} + K_{\mu , \nu} = 0.

Differentiating again gives

K_{\nu , \mu \alpha} + K_{\mu , \nu \alpha} = 0.

Relabeling indices gives two more equations

K_{\alpha , \nu \mu} + K_{\nu , \alpha \mu} = 0

K_{\mu , \alpha \nu} + K_{\alpha , \mu \nu} = 0.

Taking into account the fact that partial derivatives commute, the last three equations is a set of three linear equations in three unknowns. As a first (of several small) step(s), what is the solution (think linear algebra, not calculus) to this set of linear equations?

Thanks. How are your second and third equations indepedent to the first? I'm not sure how to solve this system, does it somehow involve writing the partials in a matrix?

 

[George Jones]

It is simpler than you think. Let X = K_{\nu , \mu \alpha} = K_{\nu , \alpha \mu}, where the second equality holds because partials commute. Similary, set Y = K_{\mu , \nu \alpha} and Z =K_{\alpha , \nu \mu}. The system of equations is now

<br /> \begin{equation*}<br /> \begin{split}<br /> X + Y &amp;= 0 \\<br /> Z + X &amp;= 0 \\<br /> Y + Z &amp;= 0,<br /> \end{split}<br /> \end{equation*}<br />

with solution X = Y = Z = 0. Thus, all the K_{\nu , \mu \alpha} are zero.

What do you get when you integrate K_{\nu , \mu \alpha} = 0 once?

 

[arkajad]

Now do the trick:

K_{\alpha,\mu\nu}=K_{\alpha,\nu\mu}

(because partial derivatives commute)

K_{\alpha,\nu\mu}=-K_{\nu,\alpha\mu}

(beacuse of the second equation)

-K_{\nu,\alpha\mu}=-K_{\nu,\mu\alpha}

(because derivatives ...)

-K_{\nu,\mu\alpha}=K_{\mu,\nu\alpha}

(bacause of the first equation)

K_{\mu,\nu\alpha}=K_{\mu,\alpha\nu}

(because derivatives)

K_{\mu,\alpha\nu}=-K_{\alpha,\mu\nu}

(because of the third equation)

Therefore:

K_{\alpha,\mu\nu}=-K_{\alpha,\mu\nu}

therefore

K_{\alpha,\mu\nu}=0

This you should be able to solve.

 

[LAHLH]

Thanks alot.

So we have \partial_{\alpha}\partial_{\mu} K_{\nu}=0 which if I integrate once would give \partial_{\mu} K_{\nu}=M_{\mu{\nu} (where M is just a spacetime constant) and again giving K_{\nu}=M_{\rho\nu}x^{\rho}+P_{\nu} (where again P is just an appropriately indexed constant).

From this approach how does one see that M is antisymmetric? I guess it just follows from K_{\nu,\mu,\alpha}=-K_{\mu,\nu,\alpha} which leads one to M_{\nu\mu}=\partial_{\nu}K_{\mu}=-\partial_{\mu}K_{\nu}=-\left(M_{\mu\nu}\right)

So how does one see that these really are the generators of the Poincare group? I know M is antisymmetric just like the generalized angular momentum operators but...

Also how then would I find the associated Killing vector with say a boost in the x direction, I guess I just use the M that generates this boost to form my K?

 

[arkajad]

Once you have a general solution after all these extra differentiations, you take your general solution and put it back into the starting Killing equation to see what happens, whether it is satisfied or gives some extra conditions.

 

[George Jones]

So how does one see that these really are the generators of the Poincare group?

What relations do the generators of the Poincare group satisfy? Do these Killing vectors satisfy the same relations.

Also how then would I find the associated Killing vector with say a boost in the x direction, I guess I just use the M that generates this boost to form my K?

If you know how to calculate flows (integral curves) of vector fields, it is useful to choose some simple Ms (with P = 0), and to calculate the flows for the specific Killing fields that result from these choices. For example, try M_{01} = -M_{01} = 1 and all other entries zero, and try M_{12} = -M_{21} = 1 and all other entries zero.

 

[George Jones]

If you know how to calculate flows (integral curves) of vector fields, it is useful to choose some simple Ms (with P = 0), and to calculate the flows for the specific Killing fields that result from these choices. For example, try M_{01} = -M_{01} = 1 and all other entries zero, and try M_{12} = -M_{21} = 1 and all other entries zero.

I think that pushing through these exercises will really help. Do you want some help getting started, or do you want to think about it some more yourself?

 

[samalkhaiat]

\partial_{a}f_{b} \ + \ \partial_{b}f_{a} = 0 \ \Rightarrow \ \partial_{c}\left( \partial_{a}f_{b} \ + \ \partial_{b}f_{a} \right) = 0 \ (1)

\partial_{c}f_{b} \ + \ \partial_{b}f_{c} = 0 \ \Rightarrow \ \partial_{a}\left( \partial_{c}f_{b} \ + \ \partial_{b}f_{c} \right) = 0 \ (2)


(1) \ - \ (2) \ \Rightarrow \ \partial_{b} \left( \partial_{c}f_{a} \ - \ \partial_{a}f_{c} \right) = 0 \ \ (3)

Integrate eq(3)

\partial_{c}f_{a} \ - \ \partial_{a}f_{c} = 2 A_{ac} \ (4)

where A_{ac}= - A_{ca} is constant. Write the K-E;

\partial_{c}f_{a} \ + \ \partial_{a}f_{c} = 0 \ \ (5)


(4) \ + \ (5) \ \Rightarrow \partial_{c}f_{a} = A_{ac}

Integrate

f_{a} = A_{ac}x^{c} \ + \ C_{a}

The same method is used in post #1 in

www.physicsforums.com/showthread.php?t=172461

to solve the conformal Killing equation.

Regards

sam

 

[LAHLH]

I think that pushing through these exercises will really help. Do you want some help getting started, or do you want to think about it some more yourself?

I've been unable to give this a go today due to having to do something else all day, but I will give this a go tomorrow. Really appreciate the help.

 

[LAHLH]

What relations do the generators of the Poincare group satisfy? Do these Killing vectors satisfy the same relations.

So the generators of the Poincare group satisfy the algebra:

[X_i,X_j]=i\epsilon_{ijk}X_k
[X_i,Y_j]=i\epsilon_{ijk}Y_k
B_1=\[ \left( \begin{array}{cccc}<br /> \cosh{(\xi)} &amp; -\sinh{(\xi)} &amp; 0 &amp;0 \\<br /> -\sinh(\xi) &amp; \cosh{(\xi)} &amp; 0 &amp;0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp;1 \end{array} \right)\]

So that means the Y_1 generator is given by Y_1=i\frac{d B_1(\xi)}{d\xi}\mid_{\xi=0}

So we have,

Y_1=\[ \left( \begin{array}{cccc}<br /> 0 &amp; -i &amp; 0 &amp;0 \\<br /> -i &amp; 0 &amp; 0 &amp;0 \\<br /> 0 &amp; 0 &amp; 0 &amp;0\\<br /> 0 &amp; 0 &amp; 0 &amp;0 \end{array} \right)\]

(Not antisymmetric as it isn't the form with both indices lowered).In general I believe it turns out that these boost generators are given by (Y_k)^{\nu}_{\mu}=-i\left(\eta_{\mu 0}\eta^{\nu}_{k}-\eta_{\mu k}\eta^{\nu}_{0}\right), so (Y_k)_{\nu\mu}=-i\left(\eta_{\mu 0}\eta_{\nu k}-\eta_{\mu k}\eta_{\nu 0}\right)
and
(Y_1)_{\mu\nu}=\[ \left( \begin{array}{cccc}<br /> 0 &amp; -i &amp; 0 &amp;0 \\<br /> +i &amp; 0 &amp; 0 &amp;0 \\<br /> 0 &amp; 0 &amp; 0 &amp;0\\<br /> 0 &amp; 0 &amp; 0 &amp;0 \end{array} \right)\]

So I believe upto the i factor this is the generator you mentioned when talking about calculating the integral curves. So I'm guessing it gives a killing vector corresponding somehow to an x boost?

If you know how to calculate flows (integral curves) of vector fields, it is useful to choose some simple Ms (with P = 0), and to calculate the flows for the specific Killing fields that result from these choices. For example, try M_{01} = -M_{01} = 1 and all other entries zero, and try M_{12} = -M_{21} = 1 and all other entries zero.

If I use this M (which I think is an x boost gen), then K^{\mu}=M^{\mu}_{\nu} x^{\nu}. Although now the M is not of the form with both indices lowered, so it's no longer symmetric and I think we should have M^{0}_{1}=-1 and M^{1}_{0}=-1. Leading to K^{\mu}=(-1,-1,0,0)=-\partial_t-\partial_x?

As for the integral curves, they would be defined by \frac{dx^{\mu}}{d\lambda}= K^{\mu}

so \frac{dx^{0}}{d\lambda}=-1 and \frac{dx^{1}}{d\lambda}=-1 , and x^2=x^3=const

So t=-\lambda+c_1, x=-\lambda+c_2 , which is effectively t=x+c_3 ?

Am I on the right lines with all this?

 

[George Jones]

If I use this M (which I think is an x boost gen), then K^{\mu}=M^{\mu}_{\nu} x^{\nu}. Although now the M is not of the form with both indices lowered, so it's no longer symmetric and I think we should have M^{0}_{1}=-1 and M^{1}_{0}=-1. Leading to K^{\mu}=(-1,-1,0,0)=-\partial_t-\partial_x?

Not quite.

giving K_{\nu}=M_{\rho\nu}x^{\rho}+P_{\nu} (where again P is just an appropriately indexed constant).

So,

K^{\nu}=M_\rho {}^\nu x^{\rho}
K = K^\nu \partial_\nu = \left( M_\rho {}^\nu x^{\rho} \right) \partial_\nu

As for the integral curves, they would be defined by \frac{dx^{\mu}}{d\lambda}= K^{\mu}

Yes.

 

[LAHLH]

Not quite.



So,

K^{\nu}=M_\rho {}^\nu x^{\rho}
K = K^\nu \partial_\nu = \left( M_\rho {}^\nu x^{\rho} \right) \partial_\nu



Yes.

Oh so, actually K^{\mu}=(-x,-t,0,0) and K=-x\partial_t-t\partial_x ?

Then \frac{dx^{0}}{d\lambda}=-x and \frac{dx^1}{d\lambda}=-t

Leading to t=-x\lambda+c_1 and x=-t\lambda+c_2?

 

[George Jones]

Oh so, actually K^{\mu}=(-x,-t,0,0) and K=-x\partial_t-t\partial_x ?

Yes.

Then \frac{dx^{0}}{d\lambda}=-x and \frac{dx^1}{d\lambda}=-t

Yes.

Leading to t=-x\lambda+c_1 and x=-t\lambda+c_2?

Remember, x and t are functions of lambda, so x and t can't be treated as constants when integrating with respect to lambda. This is a coupled set of ordinary differential equations.

 

[LAHLH]

Oh of course, silly me. So if we differentiate by lambda again, we are led to \frac{d^2 t}{d\lambda^2}=t and \frac{d^2 x}{d\lambda^2}=x

So t(\lambda)=Ae^{\lambda}+Be^{-\lambda} and x(\lambda)=Ce^{\lambda}+De^{-\lambda}

I'm not sure if there are any boundary conditions to fix these further? and get a t(x)? how is this curve then related to an x boost?

 

[George Jones]

So t(\lambda)=Ae^{\lambda}+Be^{-\lambda}

Substitute this into the first de to to relate C and D to A and B.

I'm not sure if there are any boundary conditions to fix these further?

Express the constants in terms of x(0) and t(0).

how is this curve then related to an x boost?

We're getting close, now . It might be more apparent if the exponentials are written in terms of hyperbolic trig.

 

[LAHLH]

OK, so I could have written the solutions as t(\lambda)=A\sinh{(\lambda)}+B\cosh{(\lambda)} and x(\lambda)=C\sinh{(\lambda)}+D\cosh{(\lambda)}

Then subbing the t back into \frac{dt}{d\lambda}=-x leads to,

A\cosh{(\lambda)}+B\sinh{(\lambda)}=-C\sinh{(\lambda)}-D\cosh{(\lambda)}

so

(A+D)\cosh{(\lambda)}+(B+C)\sinh{(\lambda)} =0... (1)

Also t(0)=B and x(0)=D

looking at (1) when lambda=0, gives A=-D

Sim, subbing into the other de leads to,

C\cosh{(\lambda)}+D\sinh{(\lambda)}=-A\sinh{(\lambda)}-B\cosh{(\lambda)}

so

(C+B)\cosh{(\lambda)}+(A+D)\sinh{(\lambda)} =0... (2)

looking at (2) when lambda=0, gives C=-B

So I have now,

t(\lambda)=t(0)\cosh{(\lambda)}-x(0)\sinh{(\lambda)}
and
x(\lambda)=-t(0)\sinh{(\lambda)}+x(0)\cosh{(\lambda)}

which is of course the Lorentz transformation matrix for a boost in the x direction! amazing...

 

[LAHLH]

Thanks a lot for all the help, really appreciate it.

the one thing I'm still not 100% clear on is: in this example we took the generator of the x boosts from the Poincare group, then used this as our M in the killing vector equation for Killing vecs in Minkowski K^{\mu}=M^{\mu}_{\nu}x^{\nu}+P^{\mu}, and I'm still not clear why these M's that we now know are antisymmetric are necessarily the generators. I mean given that all we know about M is that it's antisymmetric is that enough to say it must be a generator of the Lorentz group? (I know that the generators happen to be antisymmetric too).

I'm guessing we have to show that the M's satisfy the commutation relations that the generators of Lorentz satisfy? but I'm not 100% why they should given that they are seemingly just arbitrary 4x4 antisymmetric matrices at this point.

 

[George Jones]

Thanks a lot for all the help, really appreciate it.

the one thing I'm still not 100% clear on is: in this example we took the generator of the x boosts from the Poincare group, then used this as our M in the killing vector equation for Killing vecs in Minkowski K^{\mu}=M^{\mu}_{\nu}x^{\nu}+P^{\mu}, and I'm still not clear why these M's that we now know are antisymmetric are necessarily the generators. I mean given that all we know about M is that it's antisymmetric is that enough to say it must be a generator of the Lorentz group? (I know that the generators happen to be antisymmetric too).

I'm guessing we have to show that the M's satisfy the commutation relations that the generators of Lorentz satisfy? but I'm not 100% why they should given that they are seemingly just arbitrary 4x4 antisymmetric matrices at this point.

Here is what I think happened. We showed that the general solution to Killing's equation for Minkowski spacetime is

<br /> K^{\nu}=M_\rho {}^\nu x^{\rho} + P^{\nu},<br />

where M_{\mu \nu} is an arbitrary antisymmetric matrix. I chose a couple of specific matrices for M (because I knew what would result). The integral curves for the Killing vector field that results from one of these specific matrices are given by

<br /> \left[ \begin{array}{c}<br /> t\left(\lambda\right) \\ x\left(\lambda\right)<br /> \end{array}\right]<br /> = \left[ \begin{array}{cc}<br /> \cosh \lambda &amp; - \sinh \lambda \\<br /> - \sinh \lambda &amp; \cosh \lambda<br /> \end{array} \right]<br /> \left[ \begin{array}{c}<br /> t\left(0\right) \\ x\left(0\right)<br /> \end{array}\right]<br /> = e^{\lambda \left[ \begin{array}{cc}<br /> 0 &amp; -1 \\<br /> - 1 &amp; 0<br /> \end{array} \right]}<br /> \left[ \begin{array}{c}<br /> t\left(0\right) \\ x\left(0\right)<br /> \end{array}\right],<br />

which are orbits of a one-parameter subgroup (of boosts in the x direction) of the Poincare group. The equality

<br /> \left[ \begin{array}{cc}<br /> \cosh \lambda &amp; - \sinh \lambda \\<br /> - \sinh \lambda &amp; \cosh \lambda<br /> \end{array} \right]<br /> = e^{\lambda \left[ \begin{array}{cc}<br /> 0 &amp; -1 \\<br /> - 1 &amp; 0<br /> \end{array} \right]}<br />

can be derived by diagonalizing

<br /> \left[ \begin{array}{cc}<br /> \cosh \lambda &amp; - \sinh \lambda \\<br /> - \sinh \lambda &amp; \cosh \lambda<br /> \end{array} \right].<br />

Following the same procedure for any other antisymmetric matrix M with P = 0 will give other Lorentz transformations. Some of these other transformations will be boosts, some will be rotations, and some will be neither pure boosts nor pure rotations.

I don't think we started with the Poincare group, I think we ended with the Lorentz subgroup of the Poincare group.

I haven't given as much detail as I would like to give, and I don't know if I have answered your questions. I have a deadline of noon tomorrow that I have to meet, and I still have a long way to go, but please add any comments and questions. I'll get back to this thread when I can.

 

[LAHLH]

I think what was troubling me was how the condition of antisymmetry on the M's (that we found when we derived the generic Minkowski Killing solution K^{\nu}=M_\rho {}^\nu x^{\rho} + P^{\nu},) was enough information to say that they can be equated with the generators of the Lorentz group. But after thinking about it a bit longer there are 3 independent boosts and 3 independent rotations, so 6 in total and a generic 4x4 antisym matrix has 6 indep components too, so upto normalization it must be such that if M are an antisymmetric they are the generators of Lorentz. Antisymmetry is strong enough in 4 dimensions.

thanks so much for all the help

 

[arkajad]

Antisymmetry is strong enough in any number of dimensions. The pseudo-orthogonal O(r,s) has n(n-1)/2, n=r+s, generators. This the the dimension of the space of antisymmetric matrices nxn. The pseud-orthogonal group is defined by the condition

S^T\eta S=\eta

where \eta=\mathrm{diag}(1,\ldots 1,-1,\ldots -1)

Differentiating S(t)=\exp(tL) at t=0 we get

L^T\eta+\eta L=0

that is

{L^k}_i \eta_{kj}+\eta_{ik}{L^k}_j=0

which can be written as

L_{ji}=-L_{ij}

where

L_{ij}=\eta_{ik}{L^k}_j

The number of dimensions plays no role in the above.

 

[LAHLH]

OK, yeah, so in any dimension n, the number of independent elements in an indep nxn is n(n-1)/2, and it happens that the number of generators is also equal to this number and the generators are antisymmetric, so upto norm they must be equated.

thanks

 

[BrunoSantos]

How can we compute Killing vectors in Euclidean Space using K?
(I am still a beginner in GR)

 

[Dickfore]

Take two equations:

<br /> \begin{array}{lcl}<br /> K_{\mu; \nu} + K_{\nu; \mu} &amp; = &amp; 0 / \; \partial_{\rho} \\<br /> <br /> K_{\nu; \rho} + K_{\rho; \nu} &amp; = &amp; 0 / \; \partial_{\mu}<br /> \end{array}<br />

The mixed second partial derivatives can have interchanged order (in flat space-time):

<br /> \begin{array}{lcl}<br /> K_{\mu; \rho; \nu} + K_{\nu; \rho; \mu} &amp; = &amp; 0 \\<br /> <br /> K_{\nu; \rho; \mu} + K_{\rho; \mu; \nu} &amp; = &amp; 0<br /> \end{array}<br />

If we subtract the two equations, we get a cancelation of the second term from the first equation and the first term of the second equation. Also, the \partial_{\nu} derivative is common to the remaining two terms. We may write:

<br /> \partial_{\nu} \left(K_{\mu; \rho} - K_{\rho; \mu} \right) = 0<br />

The derivative of the quantity in the parentheses with respect to any coordinate is zero. Therefore, that quantity must be constant. Notice that the quantity is anti-symmetric. We can write:

<br /> K_{\mu; \rho} - K_{\rho; \mu} = 2 \, A_{\mu, \rho}, \; A_{\rho, \mu} = -A_{\mu, \rho}<br />

Combined with one of the original equations (that we had not used until now):

<br /> K_{\mu; \rho} + K_{\rho; \mu} = 0<br />

we have:

<br /> K_{\mu; \nu} = A_{\mu, \nu}, \; A_{\nu, \mu} = -A_{\mu, \nu}<br />

These first order equations are very simple to integrate. The result is:

<br /> K_{\mu} = A_{\mu, \nu} \, x^{\nu} + P_{\mu}<br />