Geometry of Signals and its Relation to Time Shift

[latentcorpse]

On page 7 of the attached document, it says that because "the above geometry does not depend on t", the trajectory of the first signal must be the same as the trajecotry of the second, just shifted by $\Delta t$.

I don't understand:
a) how it doesn't depend on t? there's a dt in the formula? or are we treating dt as different from t?

b) if it doesn't depend on t - why does this mean they follow the same path just shifted by $\Delta t$?

Thanks!

 

[Tangent87]

When he says it doesn't depend on t he means the metric doesn't explicitly depend on t, so the geodesic equations will be time invariant and hence the trajectories of the two signals will be the same, the only difference being the second signal was sent out at a time delta't' later than the first. Hope this helps.

 

[latentcorpse]

When he says it doesn't depend on t he means the metric doesn't explicitly depend on t, so the geodesic equations will be time invariant and hence the trajectories of the two signals will be the same, the only difference being the second signal was sent out at a time delta't' later than the first. Hope this helps.

Why couldn't we apply a similar argument in section 2.5 then?

 

[Tangent87]

You can and in fact he refers back to equation (13) if you read to the end of section 2.6. I think the reason he uses the actual trajectory equations in section 2.5 is just because it's easy to write them down in that case and also for completeness.

 

[latentcorpse]

You can and in fact he refers back to equation (13) if you read to the end of section 2.6. I think the reason he uses the actual trajectory equations in section 2.5 is just because it's easy to write them down in that case and also for completeness.

i don't follow. it seems to me like he does two different things. in section 2.6, he shifts them both by delta t but in section 2.5 he shifts one by delta ta and the other by delta tb.

 

[latentcorpse]

In the attached notes I have quite a few problems with one of the proofs. In the proof of the proposition on p15,

a) he says to note that $\nu(0)=0$. why is this?

b) he goes from
$\{ \frac{d}{dt} [ \alpha ( x^\mu ( \lambda(t)) - x^\mu(p)) + \beta (x^\mu(\kapa(t))-x^\mu(p)) + x^\mu(p)] \}_{t=0} = [ \alpha ( \frac{d x^\mu ( \lambda (t))}{dt})_{t=0} + \beta ( \frac{dx^\mu ( \kappa ( t))}{dt} )_{t=0}]$
I really don't understand how these two lines are equal at all!
And also how can we change the $\phi$'s to $x^\mu$'s in going from eqn 25 to the defn of $Z_p(f)$?

c) where does eqn 27 come from? isn't $( \frac{\partial}{\partial x^\mu})_p (f) = \frac{\partial f}{\partial x^\mu})_p$
is it something like if we compose the numerator with $\phi^{-1}$ then we have to cancel that out by composing the $p$ with $\phi$ to give the $\phi(p)$? I don't really get why this is allowed though?

d)Where does eqn 29 come from?

Thanks a lot for any help. I really need to get my head round all this vector business over the holidays!

 

[arkajad]

a) it is assumed that $\lambda(0)=\kappa(0)=p$ then

$$\nu(0)=\phi^{-1}[\alpha(\phi(p)-\phi(p))+\beta(\phi(p)-\phi(p))+\phi(p)]=\phi^{-1}(\phi(p))=p$$

b) When you differentiate you skip constant terms, those that do not depend on t. You use formula (24) with $\lambda$ replaced by $\nu$ and $X_p$ replaced by $Z_p$

c) Replace first the dummy index $\mu$ in (24) by $\alpha$. The notice that for the curve given by (26) we have

$$dx^\alpha (\lambda_\mu(t))/dt=\delta^\alpha_\mu$$

You will probably able to figure it out why.

 

[arestes]

Hi, well, this might be off-topic, but... those notes from Harvey Reall are nice. Where can I get the rest? thanks

 

[latentcorpse]

that is all of them i believe.

 

[latentcorpse]

When he says it doesn't depend on t he means the metric doesn't explicitly depend on t, so the geodesic equations will be time invariant and hence the trajectories of the two signals will be the same, the only difference being the second signal was sent out at a time delta't' later than the first. Hope this helps.

Hi. I was hoping you could help me with a quick question about these notes. On page 82, about half way down he says that "a free particle incident from large r will spiral into $r=2M$.

Why does it spiral into $r=2M$ and not $r=0$? The only significant thing I know about $r=2M$ at this point is that the components of the Schwarzschild metric diverge at this point. Has this got anything to do with it?

Thanks a lot!

 

[latentcorpse]

a) it is assumed that $\lambda(0)=\kappa(0)=p$ then

$$\nu(0)=\phi^{-1}[\alpha(\phi(p)-\phi(p))+\beta(\phi(p)-\phi(p))+\phi(p)]=\phi^{-1}(\phi(p))=p$$

b) When you differentiate you skip constant terms, those that do not depend on t. You use formula (24) with $\lambda$ replaced by $\nu$ and $X_p$ replaced by $Z_p$

c) Replace first the dummy index $\mu$ in (24) by $\alpha$. The notice that for the curve given by (26) we have

$$dx^\alpha (\lambda_\mu(t))/dt=\delta^\alpha_\mu$$

You will probably able to figure it out why.

Hi. I was hoping you could help me with a quick question about these notes. On page 82, about half way down he says that "a free particle incident from large r will spiral into $r=2M$.

Why does it spiral into $r=2M$ and not $r=0$? The only significant thing I know about $r=2M$ at this point is that the components of the Schwarzschild metric diverge at this point. Has this got anything to do with it?

Thanks a lot!

 

[Tangent87]

Hi. I was hoping you could help me with a quick question about these notes. On page 82, about half way down he says that "a free particle incident from large r will spiral into $r=2M$.

Why does it spiral into $r=2M$ and not $r=0$? The only significant thing I know about $r=2M$ at this point is that the components of the Schwarzschild metric diverge at this point. Has this got anything to do with it?

Thanks a lot!

Well I know that r=2M is only a *coordinate* singularity, it's not a physical singularity like r=0 is, but as soon as a particle or light ray crosses over r=2M it will spiral into r=0 because the gravity is so strong, there is no escape.

 

[arkajad]

You may like to check http://en.wikipedia.org/wiki/Eddington-Finkelstein_coordinates" .

 

[latentcorpse]

You may like to check http://en.wikipedia.org/wiki/Eddington-Finkelstein_coordinates" .

Thanks but he says under (260) that t diverges logarithmically as $r \rightarrow 2M$. Why is that?

Also, if we have a white hole (the time reversal of a black hole) and we say that any timelike curve starting in the region $r < 2M$ when working in outgoing Eddington Finklestein coordinates must cross the $r=2M$ surface within finite proper time, then surely by time reversal any timelike curve in the $r > 2M$ region when working in ingoing Eddington Finklestein coordinates must cross the $r=2M$ surface within finite proper time - does this mean that we are all doomed to be swallowed by a black hole?

And finally, do you have any ideas about how to derive eqn (304) or eqn (305)?

Thank you!

 

[Tangent87]

Thanks but he says under (260) that t diverges logarithmically as $r \rightarrow 2M$. Why is that?

Also, if we have a white hole (the time reversal of a black hole) and we say that any timelike curve starting in the region $r < 2M$ when working in outgoing Eddington Finklestein coordinates must cross the $r=2M$ surface within finite proper time, then surely by time reversal any timelike curve in the $r > 2M$ region when working in ingoing Eddington Finklestein coordinates must cross the $r=2M$ surface within finite proper time - does this mean that we are all doomed to be swallowed by a black hole?

And finally, do you have any ideas about how to derive eqn (304) or eqn (305)?

Thank you!

We can show that for a radially infalling timelike geodesic that $$\frac{dr}{d\tau}=\sqrt{E^2-(1-2M/r)}$$ (we can derive this from the schwarzschild metric). You can then take E=1 for convenience and integrate to get an equation for r in terms of proper time, then you just set r=2M and you'll get a finite value for proper time. So yes if we're on radially infalling timelike geodesics we're all doomed to be swallowed by the black hole.

P.S. You don't need Eddington Finklestein coordinates to get that equation.

 

[latentcorpse]

We can show that for a radially infalling timelike geodesic that $$\frac{dr}{d\tau}=\sqrt{E^2-(1-2M/r)}$$ (we can derive this from the schwarzschild metric). You can then take E=1 for convenience and integrate to get an equation for r in terms of proper time, then you just set r=2M and you'll get a finite value for proper time. So yes if we're on radially infalling timelike geodesics we're all doomed to be swallowed by the black hole.

P.S. You don't need Eddington Finklestein coordinates to get that equation.

When I derived that equation, I did not make any use of the infalling/outfalling property so I see that equation as just describing a radial timelike geodesic. Is that correct?
And then we interpret it as
a) we will hit the singularity in finite proper time in infalling case
b) we will cross r=2M surface in finite proper time in outfalling case (not sure if this is correct interpretation here though - what do you think?)

 

[Tangent87]

When I derived that equation, I did not make any use of the infalling/outfalling property so I see that equation as just describing a radial timelike geodesic. Is that correct?
And then we interpret it as
a) we will hit the singularity in finite proper time in infalling case
b) we will cross r=2M surface in finite proper time in outfalling case (not sure if this is correct interpretation here though - what do you think?)

Truth be told I'm not exactly sure what outfalling even means, are we saying that a outfalling is the same as outgoing? I.e. it's moving away from the singularity (opposite direction of the infalling particle)? Or is it still going towards the singularity just in a different way/curve?

Also do you have any idea of how to show that a distant observer would see the horizon
crossing only after an infinite time, I've seen loads of notes that state that a light ray at r=2M will stay there or that it takes infinite time for a far away observer to see something cross the horizon but I'm not sure what equation we need to show the maths for this.