Explaining Curved Space-Time to a Friend

[pervect]

Things don't physically travel along space-like geodesics, of course. However, both space-like and time-like geodesics are described by a function which assigns a position (vector) to an affine parameter, lambda. (i.e. we have functions x(lambda) and t(lambda) in the 2-d example I was talking about).

Reading back over the thread, I'm not quite sure why you thought I implied objects could physically travel along a space-like geodesic, so I'm not sure if this remark clears up the miscommunication.

On a related note:

The fact that a^2(t)*dx/dlambda is a constant has a physical interpretationa as a consered momentum when the geodesic is time-like, by setting lambda=tau. This exact interpretation is not strictly available when the geodesic is space-like, yet mathematically the quantity remains conserved. The straightforwards but tedious way of showing this is to write down the geodesic equations from the metric and the Christoffel symbols. This confirms the fact that this quantity is conserved, as one of the geodesic equations reduces to

d/dlambda (a^2(t)*dx/dlambda)=0.

 

[Berislav]

Reading back over the thread, I'm not quite sure why you thought I implied objects could physically travel along a space-like geodesic, so I'm not sure if this remark clears up the miscommunication.

In one of your previous posts:

It appears to me that if you follow an actual "straight line" in space (i.e you follow a space-like geodesic),

You showed in your equations that \frac{d^2t}{d \tau^2} must be negative, which implies that after a sufficient amount of proper time objects following geodesics will go backward in coordinate time in a "flat" universe.
This (very interesting result) holds for every time-like geodesic trajectory, as you calculated them for \tau.

All in all, I'm not sure why you mentioned space-like geodesics.

 

[pervect]

In one of your previous posts:


You showed in your equations that \frac{d^2t}{d \tau^2} must be negative, which implies that after a sufficient amount of proper time objects following geodesics will go backward in coordinate time in a "flat" universe.
This (very interesting result) holds for every time-like geodesic trajectory, as you calculated them for \tau.

Note that tau is formally equivalent to lambda, don't let my (perhaps unfortunate) choice of notation confuse you on the physical significance of the equatiions.

Note that the result you cite follows not for every geodesic, but rather for every geodesic in which dt/dtau is zero. But dt/dtau = 0 implies a space-like geodesic whenever dx/dtau > 0. Having both dt/dtau = 0 and dx/dtau = 0 does not give any interesting solutions (the solution is a point, in this case, rather than a curve).

We know that dt/dtau = 0 and dx/dtau > 0 imples a spacelike geodesic because

a^2(t) (dx/dtau)^2 - dt^2 > 0, which is a spacelike interval, not a timelike one.

 

[Berislav]

Note that the result you cite follows not for every geodesic, but rather for every geodesic in which dt/dtau is zero

But,

t'' + a da/dt (x')^2 = 0

implies that t'' will be negative as a and da/dt are positive, and (x')^2 is positive even for a time-like geodesic.

Sorry if I'm being a bother.

 

[pervect]

No problemo...

The actual solution for the timelike geodesics may help illustrate what's going on

A convenient form for the time-like geodesics for the metric

(which is again ds^2 = a(t)^2 dx^2 - dt^2 for the 2d case)

x := k + (1/(2*H))*(ln(1+H*C/lambda));
t := sqrt(lambda^2+H*C*lambda);

this particular form for the geodesic is convenient becase t=0 when lambda=0

k and C are abitrary constants, H is the constant in a(t) = HT (not to be confused with the Hubble constant).

Both of the following statements are true for 0 < lambda < infinity

t'' < 0, t' > 1

and the limit as lambda-> infinity is t' = 1, t''=0

A sample plot is attached for k=0, C=2. Since the axes are not labelled, I should explain that it's a plot of the cosmological coordinate x which runs vertically versus the cosmological time t which running horizontally, for lambda (equivalent to proper time tau) varying between 0 and 5.

You can see that becase t' >= 1, the cosmological time coordinate t always increases when lambda (which in this case IS equal to the proper time tau) increases.

 

[Berislav]

I see. I naively assumed that t' will become negative for some large value of tau.

 

[Chronos]

Mathematically, it can never be negative. Although it could be imaginary. Imaginary time is, however, hard to conceptualize.

 

[Berislav]

Mathematically, it can never be negative. Although it could be imaginary. Imaginary time is, however, hard to conceptualize.

I said that I assumed that t', (i.e,\frac{dt}{d \tau} ) would eventually become negative even for a time-like geodesic. I never though that t itself would become negative, I thought that it would go to 0.

 

[pervect]

I got myself confused a while ago on a very similar issue. One thing I want to add in addition to the the specific solution I quoted for a specific expansion function a(t) is that we know in general that t' could not pass through zero for a timelike geodesic. This follows from the requirement of a timelike geodesic that

a^2(t) dx^2 - dt^2 < 0

after dividing both sides by d lambda^2 we get

a^2(t) (x')^2 - (t')^2 < 0

t' can never be zero and make the interval timelike, the best we could do is to make it lightlike (and that would be a boring, pointlike solution).

 

[Flatland]

Like what infinitetime said, how can the universe be finite and flat? That would mean the universe has an edge. What's beyond the edge? That would also mean the universe has a center which violates relativity because there is now an absolute reference frame. The only way the conditions exists where there is no absolute reference frame is either
1. the universe is infinite or 2. the universe is closed.