How do we find the null geodesic for a 2D spacetime with a positive constant?

[quasar987]

We consider a 2 dimensional spacetime with coordinates (t,y) and metric

$$g=dt\otimes dt - (t^2+a^2)^2dy\otimes dy$$

where a is a positive constant.

In previous subquestions, I have calculated the Christoffel symbols for the metric-compatible connexion and now I am asked to find the null geodesic (for a photon) starting at t=0, y=pi/a and which is directed towards y=0.

The geodesic equations write

$$\frac{d^2t}{dp^2}+\Gamma_{yy}^t\left(\frac{dy}{dp}\right)^2=0$$

$$\frac{d^2y}{dp^2}+2\Gamma_{yt}^y\frac{dy}{dp}\frac{dt}{dp}^2=0$$

I didn't know what a null geodesic was so I looked it up wiki and it says it's a geodesic whose tangent vector whose norm is 0.

But this is crazy because the length of a curve is $L=\int_{\gamma}||\dot{\gamma}(p)||dp$, so a null tangeant vector means null lenght, means it's just a point. :grumpy: :grumpy: :grumpy:

 

[nrqed]

We consider a 2 dimensional spacetime with coordinates (t,y) and metric

$$g=dt\otimes dt - (t^2+a^2)^2dy\otimes dy$$

where a is a positive constant.

I assume you meand $ds^2$, not g (what you gave is not the metric itself)

In previous subquestions, I have calculated the Christoffel symbols for the metric-compatible connexion and now I am asked to find the null geodesic (for a photon) starting at t=0, y=pi/a and which is directed towards y=0.

The geodesic equations write

$$\frac{d^2t}{dp^2}+\Gamma_{yy}^t\left(\frac{dy}{dp}\right)^2=0$$

$$\frac{d^2y}{dp^2}+2\Gamma_{yt}^y\frac{dy}{dp}\frac{dt}{dp}^2=0$$

I didn't know what a null geodesic was so I looked it up wiki and it says it's a geodesic whose tangent vector whose norm is 0.

But this is crazy because the length of a curve is $L=\int_{\gamma}||\dot{\gamma}(p)||dp$, so a null tangeant vector means null lenght, means it's just a point. :grumpy: :grumpy: :grumpy:

The metric in GR is not positive definite so a vector may have a zero "length" and still be non zero.

For a null geodesic, just set $ds^2 =0$ and then you get a differential equation of y as a function of t. Solve to find y(t) and use conditions given to fix the function.

 

[quasar987]

So I don't even need those geodesic equations. Ok, thanks nrqed, you're saving me!

 

[shadi_s10]

can anyone help me with this null geodesic?
in d'inverno it is written:
when two metrics are conformal,the null geodesic of one metric coincides with the null geodesic of the other. how can we prove it?
I know the equation of a null geodesic but how can these two coincide?
is it because they are somehow proportional?

 

[Mentz114]

If two metrics g, h are conformal then
$$g_{ab}=\lambda^2 h_{ab}$$

The equation for the null geodesic is
$$c^2d\tau^2=0$$

If you expand the above in terms of g and h the constant $\lambda^2$ cancels.

 

[shadi_s10]

If two metrics g, h are conformal then
$$g_{ab}=\lambda^2 h_{ab}$$

The equation for the null geodesic is
$$c^2d\tau^2=0$$

If you expand the above in terms of g and h the constant $\lambda^2$ cancels.

if we take:
c^2 dt^2=0
Then we have :
g_ab (dx^a)/du (dx^b)/du=0
and ofcourse it leads to:
h_ab (dx^a)/du (dx^b)/du=0

does it mean coincide?
i thought it should be something more clear.
this means h and g both are with one equation
but does it really mean a coincide?!

 

[shadi_s10]

if we take:
c^2 dt^2=0
Then we have :
g_ab (dx^a)/du (dx^b)/du=0
and ofcourse it leads to:
h_ab (dx^a)/du (dx^b)/du=0

does it mean coincide?
i thought it should be something more clear.
this means h and g both are with one equation
but does it really mean a coincide?!

by the way thanks for helping me!

 

[Mentz114]

What I mean is this. Using g the null geodesic satisfies

$$c^2d\tau^2=c^2g_{00}dt^2 - \sum_{ij}g_{ij}(dx^i)^2(dx^j)^2=0$$

which is the same equation as the one we get using h

$$c^2\lambda^2h_{00}dt^2 - \lambda^2 \sum_{ij}h_{ij}(dx^i)^2(dx^j)^2=0$$

because you can cancel $\lambda^2$ from the second.

 

[shadi_s10]

Thanks for helping!

 

[shadi_s10]

Is it right with the timelike geodesics too?
I mean if with g it's timelike, can it be still time like with h too?
as landa^2 is positive and can vanish.

 

[Altabeh]

Is it right with the timelike geodesics too?
I mean if with g it's timelike, can it be still time like with h too?
as landa^2 is positive and can vanish.

It makes no difference since you are just setting the equation to be an inequality so you CAN still cancel landa^2. For a better understanding look again at the new setup:

$$\lambda^2(c^2h_{00}dt^2 - \sum_{ij}h_{ij}(dx^i)^2(dx^j)^2) > 0$$

Since landa^2>0, so it can readily be eliminated, leaving the sign of inequality unchanged. The same thing also is true in the case of spacelike geodesics.

AB

 

[shadi_s10]

Thanks a lot for helping!

 

[shadi_s10]

I have another problem!
can you tell me how can I use the variation method for geodesics for a spacelike one?
What I mean is that in d'inverno the variational method and the euler lagrange method is done for a timelike one.
for a space like I know that we should write ds=-dσ
but I think this is not used as we just use the lagrangian and it is not involved with ds!
I am so confused !

 

[Altabeh]

I have another problem!
can you tell me how can I use the variation method for geodesics for a spacelike one?
What I mean is that in d'inverno the variational method and the euler lagrange method is done for a timelike one.
for a space like I know that we should write ds=-dσ
but I think this is not used as we just use the lagrangian and it is not involved with ds!
I am so confused !

Please specify what σ is here. Even if I take it the affine parameter of a curve, I don't think d'inverno shows us a very strange equation other than that of geodesic for the curve. One important thing is that we usually do not need the specification of geodesics when wanting to get their equation which in terms of christoffel symbols is written similarly for all types as was shown earlier at this thread.

Maybe I'm not getting your idea correctly, so a little bit more explanation would lead us to what you want.

 

[shadi_s10]

o my god! I typed it wrong!
this is it:

for the spacelike geodesics we have: -ds^2=dσ^2
(I mean as ds^2 is negative we have to introduce a positive thing as we say L=√(g_ab (x) (x^a ) ̇(x^b ) ̇ ) and we can not have something negative under the square root)

and in chapter 7 (section 7.6) it is written that for a timelike geodesics if we use the variation method we finally get: (x^a ) ̈+Γ_bc^a (x^b ) ̇(x^c ) ̇=((d^2s/du^2)/(ds/du) ̇)(x^a ) ̇
What I want to know is that how would it be if we used the spacelike geodesic?
would it be different?

By the way thank you for being so patient and helping me.

 

[Altabeh]

o my god! I typed it wrong!
this is it:

for the spacelike geodesics we have: -ds^2=dσ^2
(I mean as ds^2 is negative we have to introduce a positive thing as we say L=√(g_ab (x) (x^a ) ̇(x^b ) ̇ ) and we can not have something negative under the square root)

and in chapter 7 (section 7.6) it is written that for a timelike geodesics if we use the variation method we finally get: (x^a ) ̈+Γ_bc^a (x^b ) ̇(x^c ) ̇=((d^2s/du^2)/(ds/du) ̇)(x^a ) ̇
What I want to know is that how would it be if we used the spacelike geodesic?
would it be different?

By the way thank you for being so patient and helping me.

I think I now know what you are trying to find out.

Since the sign of ds^2 does not change during integration, you can simply replace the integrad [(ds/du)^2]^.5 with (ds/du)^2. So the Lagrangian can be squared as for you to avoid any possible misleading. :)

AB

 

[shadi_s10]

And by using it, I'm assuming the euler-lagrange equation will be the same as the timelike one.
Is it true?

 

[Altabeh]

And by using it, I'm assuming the euler-lagrange equation will be the same as the timelike one.
Is it true?

As I said earlier, yes! Because the equation of geodesic is in general unique for all kinds of geodesics. It means that the equation d'inverno led to for timelike geodesics is the same as the one you are going to get for spacelike geodesics, too!

AB

 

[shadi_s10]

Thank you so much!
you really helped me