GR: find covariantly constant vector on a given curve

[binbagsss]

Homework Statement

I am stuck on finding $W^u$

Homework Equations

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I have computed the christoffel symbols via comparing the Euler-Lagrange equations to the form expected from geodesic equation.

geodesic equation: $\ddot{x^a}+\Gamma^a_{bc}\dot{x^b}\dot{x^c}=0$

covariantly constant equation: $V^a \nabla_a W^b = V^a (\partial_a W^b) + V^a \Gamma^b_{ac} W^c= 0$ [1] where $V^a$ is the tangent vector to the geodesic.

I have computed the christoffel symbols as:

$\Gamma^{x}_{tt}=\frac{-1}{2x^2}$ and $\Gamma^{t}_{tx}=\frac{-1}{2x}$

The Attempt at a Solution

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From the information given $x^u=(t,1) \implies V^u=(1,0)=\delta^u_t$

Therrefore [1] non-zero equations reduces to:

$\nabla_t W^b = (\partial_t W^b) + V^t \Gamma^b_{tc} W^c= 0$

Using the christoffel symbols non-zerro equations further reduce to:

$\partial_t W^t - \frac{1}{2x}W^x=0$

and $(\partial_t W^x) -\frac{W^t}{x^2}= 0$

MY QUESTION:

so it is at this point that I am stuck. the only way I can see to proceed is to differentiate either one of the equations again wrt $t$ to get a second-order equation and then substitute in the other equation. However to then solve completely we would need 2 boundary conditions, but are only given one.

Many thanks in advance

 

[Orodruin]

No, you have two boundary conditions. If you are rewriting two first order differential equations as one second order one, you can use the original differential equations to obtain an initial condition on the derivative of the variable you are keeping.

 

[binbagsss]

No, you have two boundary conditions. If you are rewriting two first order differential equations as one second order one, you can use the original differential equations to obtain an initial condition on the derivative of the variable you are keeping.

mmmm okay, so for the 2nd PDE I obtain: $\partial_t^2W^t-\frac{1}{2x^3}W^t=0$

If I plug in the boundary condition into the first ODE I have:

$\partial_t W^t=0$ at $(0,1)=(t,x)$

and

$\partial_t W^x-1=0$ at $(0,1)=(t,x)$

BUT... to get these I have only fed in an $x$ value and not a $t$ value? so, really, do I not have that the above two hold for $(t,1)$ unless it's one-to-one such that $x=1 \iff t=0$?

 

[binbagsss]

mmmm okay, so for the 2nd PDE I obtain: $\partial_t^2W^t-\frac{1}{2x^3}W^t=0$

If I plug in the boundary condition into the first ODE I have:

$\partial_t W^t=0$ at $(0,1)=(t,x)$

and

$\partial_t W^x-1=0$ at $(0,1)=(t,x)$

BUT... to get these I have only fed in an $x$ value and not a $t$ value? so, really, do I not have that the above two hold for $(t,1)$ unless it's one-to-one such that $x=1 \iff t=0$?

no?

(anybody out there, does anyone care, cause baby I'm not f-f-f-fooolin')

 

[binbagsss]

No, you have two boundary conditions. If you are rewriting two first order differential equations as one second order one, you can use the original differential equations to obtain an initial condition on the derivative of the variable you are keeping.

please is there a reason you didnt reply to my reply, I am still stuck, many thanks for your time.

 

[stevendaryl]

You have the second-order equation:

$(\partial_t)^2 W^t - \frac{1}{2x^3} W^t = 0$

What's the solution to that? Note, since $x$ doesn't vary along the path, you just have a function of $t$ and you can use ordinary derivatives instead of partial derivatives, treating $x$ as constant. So letting $W^t = f(t)$, this becomes:

$\frac{d^2 f}{dt^2} - k f = 0$

for some positive constant $k$

What's the solution(s) to that equation?

 

[Orodruin]

To be honest, the best way of solving this is to not solve a differential equation at all. Instead, you can use well known properties of parallel transport. Note that $x = 1$ is a geodesic when parametrised properly since the geodesic equations are given by
$$
L = \dot t^2 /x - \dot x^2, \quad \frac{d}{ds} \frac{\partial L}{\partial \dot t} = 2\frac{d(\dot t /x)}{ds} = 0,
\quad \frac{d}{ds}\frac{\partial L}{\partial \dot x} - \frac{\partial L}{\partial x} = - \ddot x + \frac{\dot t^2}{x^2} = 0.
$$
For $x = 1$ you would have the appropriate normalisation when $\dot t$ is constant. So indeed $t = s$, $x = 1$ is an affinely parametrised geodesic.

The question becomes, what can you say about the field that is parallel transported along $x = 1$ based on the properties of parallel transport? (Assuming you are using a metric compatible connection, but I took it as implicit that you are using the Levi-Civita connection.)

 

[binbagsss]

You have the second-order equation:

$(\partial_t)^2 W^t - \frac{1}{2x^3} W^t = 0$

What's the solution to that? Note, since $x$ doesn't vary along the path, you just have a function of $t$ and you can use ordinary derivatives instead of partial derivatives, treating $x$ as constant. So letting $W^t = f(t)$, this becomes:

$\frac{d^2 f}{dt^2} - k f = 0$

for some positive constant $k$What's the solution(s) to that equation?

Ok so the solution is $f=Ae^{\sqrt(k)t} + Be^{-\sqrt(k)t}$, $A, B$ some constants.

Regarding $W_t$ being a function of only $t$ , this is the same for $x$ right?
This makes sense for when the vector we are after is specifically on the curve $x=1$, however my interpretation of the question is to find $W_t$ and $W_x$ in terms of both $(x,y)$ such that the dependence of $x$ satisfies the intial data etc and so must be constant on that curve, however general expressions that reduce to that?

 

[stevendaryl]

Ok so the solution is $f=Ae^{\sqrt(k)t} + Be^{-\sqrt(k)t}$, $A, B$ some constants.

Regarding $W_t$ being a function of only $t$ , this is the same for $x$ right?
This makes sense for when the vector we are after is specifically on the curve $x=1$, however my interpretation of the question is to find $W_t$ and $W_x$ in terms of both $(x,y)$ such that the dependence of $x$ satisfies the intial data etc and so must be constant on that curve, however general expressions that reduce to that?

There is no $y$, since they're talking about a 2-D spacetime. So, the "constants" $A$ and $B$ are actually constants with respect to $t$. They are still functions of $x$. So you have:

$W^t = A(x) e^{\sqrt{k} t} + B(x) e^{-\sqrt{k} t}$ where $k = \frac{1}{2x^3}$

At $t=0, x=1$, $W^t = 1$

You have a similar constraint for $W^x$, which is related to $W^t$, as you derived in the first post.

 

[stevendaryl]

There is no $y$, since they're talking about a 2-D spacetime. So, the "constants" $A$ and $B$ are actually constants with respect to $t$. They are still functions of $x$. So you have:

$W^t = A(x) e^{\sqrt{k} t} + B(x) e^{-\sqrt{k} t}$ where $k = \frac{1}{2x^3}$

At $t=0, x=1$, $W^t = 1$

You have a similar constraint for $W^x$, which is related to $W^t$, as you derived in the first post.

It seems to me that the vector field $W^\mu$ is not uniquely determined anywhere except on the line $x=1$.

 

[binbagsss]

$W^t = A(x) e^{\sqrt{k} t} + B(x) e^{-\sqrt{k} t}$ where $k = \frac{1}{2x^3}$

But also we haven't found general $A(x)$ and $B(x)$ for the above, but instead $A(x=1)$ and $B(x=1)$?

 

[stevendaryl]

But also we haven't found general $A(x)$ and $B(x)$ for the above, but instead $A(x=1)$ and $B(x=1)$?

Right, but it seems to me that we can choose them arbitrarily in the case $x \neq 1$. The problem statement doesn't pin them down.