Gravity works the same under time reversal???

[md65536]

I came across a definition of a white hole as a "time-reversed black hole."

 

I assume that a white hole would only let light out.

 

However, this doesn't make sense to me. If you have a curved geodesic contained within a black hole event horizon, wouldn't light travel along the same path whether it was going forward in time or backward in time?

 

It seems to me that spacetime curvature would determine whether light etc would be confined to a space, or unable to enter that space.

A time-reversed black hole would be gravitationally identical to a black hole.

Perhaps its spin would be reversed or something, but essentially its influence on the universe would be the same???

 

A white hole then would be a "spacetime inversion of a black hole", where the curvature is inverted or negated or something.

 

 

Extrapolating, it seems clear that gravitational attraction is independent of the direction of the arrow of time (it's probably still dependent on the rate of time or the magnitude of the arrow, if that makes sense). Gravitation is a one-way process, regardless of whether or not time is. ???

 

 

 

 

Or would a time-reversal of a black hole allow light to escape along the same paths that let it in?

[ajb]

I am not sure about T-symmetry in general relativity as a whole as we do not usually have a good notion of space and time, rather just space-time.

 

However, one can think about T-symmetry in the context of the path of test particles.

 

The geodesic equation is

 

[math]\frac{d^{2}x^{\lambda}}{dt^{2}} + \Gamma^{\lambda}_{\mu \nu} \frac{dx^{\nu}}{dt} \frac{dx^{\mu}}{dt}[/math].

 

Clearly the above is T-symmetric.

[md65536]

I am not sure about T-symmetry in general relativity as a whole as we do not usually have a good notion of space and time, rather just space-time.

 

However, one can think about T-symmetry in the context of the path of test particles.

 

The geodesic equation is

 

[math]\frac{d^{2}x^{\lambda}}{dt^{2}} + \Gamma^{\lambda}_{\mu \nu} \frac{dx^{\nu}}{dt} \frac{dx^{\mu}}{dt}[/math].

 

Clearly the above is T-symmetric.

Gravitationally accelerated test particles do not move along a single geodesic.

 

If a black hole were time reversed and "became a white hole", yet we at an external perspective were not time reversed, the white hole would continue to gravitationally attract us.

 

What I'm not sure of is whether the particles inside that white hole would follow full reverse paths and appear to be gravitationally repelled from the white hole, or if they would only follow reverse geodesics, and remain gravitationally attracted to the white hole. Only the latter makes sense, because (I assume) the spacetime curvature of the time-reversed black hole would remain the same as before, and that curvature is what dictates the gravitational acceleration.

 

But then again, like you said we don't have a good notion of space and time. I'm mistakenly thinking of curvature only as spatial curvature, so time reversal might imply changes to spacetime curvature that I can't clearly imagine.

Edited May 20, 2011 by md65536

[michel123456]

 

The geodesic equation is

 

[math]\frac{d^{2}x^{\lambda}}{dt^{2}} + \Gamma^{\lambda}_{\mu \nu} \frac{dx^{\nu}}{dt} \frac{dx^{\mu}}{dt}[/math].

 

Clearly the above is T-symmetric.

 

I must be blind. Where is it T symmetric?

[md65536]

I must be blind. Where is it T symmetric?

t only shows up squared, so the equation has the same value whether it is positive or negative. ???

 

It makes sense that a particle traveling along a geodesic would travel in the opposite direction along the same geodesic if time were reversed.

Edited May 20, 2011 by md65536

[ajb]

Gravitationally accelerated test particles do not move along a single geodesic.

 

If we mean by test particle an idealised particle, that is of no finite size then it does follow a geodesic. For finite size objects we can treat them as a collection of points and then one would have to think about geodesic deviation.

 

 

But then again, like you said we don't have a good notion of space and time. I'm mistakenly thinking of curvature only as spatial curvature, so time reversal might imply changes to spacetime curvature that I can't clearly imagine.

 

Of course not knowing what is really meant by time reversal.

[Light Storm]

Time doesn't exist, you certainly can't run it backwards

[ajb]

t only shows up squared, so the equation has the same value whether it is positive or negative. ???

 

It makes sense that a particle traveling along a geodesic would travel in the opposite direction along the same geodesic if time were reversed.

 

Exactly. For massive particles we can interpret the parameter t as the proper time. As you see passing from t -> -t does not affect the geodesic equation.

 

For massless particles we cannot interpret the affine parameter as a proper time. We still have the T-symmetry formally but it does not have such a clear interpretation as for massive particles.

 

Time doesn't exist, you certainly can't run it backwards

 

I must say that this approach is not very useful to us. We have a clear notion of the proper time between two events along a time-like path. That is a path followed by massive particles. If we only have gravity to worry about then the equation of motion is just the geodesic equation and we see that this is T-symmetric as I have pointed out.

[md65536]

Exactly. For massive particles we can interpret the parameter t as the proper time. As you see passing from t -> -t does not affect the geodesic equation.

 

For massless particles we cannot interpret the affine parameter as a proper time. We still have the T-symmetry formally but it does not have such a clear interpretation as for massive particles.

So a geodesic is the same path, passing from t -> -t?

 

I think that implies that spacetime curvature is the same for t as for -t? Or at least, there is no inversion of the curvature that would make time-reversed gravity into a repulsive force.

 

If you send a signal to the moon and time-reverse the process before it gets there, it will reverse course and return to you.

If you drop a neutron off a building and time-reverse the process before it lands, it will continue to fall (I speculate).

 

 

 

 

Admittedly, this depends on some of my own interpretation, which is not accepted science. As you say, there is no clear interpretation.

 

 

Time doesn't exist, you certainly can't run it backwards

Ignoring whether my examples describe practical or possible realities, figuring out the theoretical implications can be useful in figuring out what is possible in the universe and hopefully ultimately how it all works.

 

Large-scale T-symmetrical time reversal is probably impossible (this very thread speculates one of several reasons why), and certainly impossible in any universal way that requires the existence of a fictional concept of "universal time", but it's useful in very small scale interactions such as those described by Feynman diagrams.

Edited May 20, 2011 by md65536

[ajb]

Be careful, t in the geodesic equation is an affine parameter that can, for massive particle be identified with the proper time (using an affine transformation). The definition of the proper time is

 

[math]t = \int_{\gamma} \sqrt{g_{\mu \nu} dx^{\nu}dx^{\mu}}[/math],

 

where the integration is along a path [math]\gamma[/math]. In particular, in general we cannot identify [math]dx^{0}[/math] with [math]dt[/math].

 

So, I am happy with the idea of reversing the proper time, but I am not sure what [math]x^{0} \rightarrow - x^{0}[/math] really means as it is clearly coordinate dependant.

 

Related to this, on Minkowski space-time is CPT-symmetry. It is known that any consistent quantum field theory that is Lorentz invariant must be invariant under inversions of charge, parity, and time simultaneously. I have no idea is there is any significance to similar transformation in general relativity. In particular, the change of coordinates [math]x^{0} \rightarrow - x^{0}[/math] is fine, we can deal with this mathematically as we use tensors. The point is I am not sure it has any deep meaning in general relativity, outside of local Lorentz invariance.

Edited May 21, 2011 by ajb

[michel123456]

Exactly. For massive particles we can interpret the parameter t as the proper time. As you see passing from t -> -t does not affect the geodesic equation.

 

I feel so naive.

Is that what T-symmetric means? that the result is the same for t positive and t negative?

[ajb]

I feel so naive.

Is that what T-symmetric means? that the result is the same for t positive and t negative?

 

Yes in essence. Have a look at the wikipedia article here.

 

What I am confused about is what we mean by T-symmetry in general relativity.

 

The idea if reversing the proper time of a massive particle seems ok to me. Setting [math]x^{0} \rightarrow - x^{0}[/math] seems ok also, but not very interesting (unless I am missing something). The other thing we could examine is the notion of time-orientablity. One can think in terms of future- and past-directed timelike vectors (we have a local arrow of time). I guess we could then think about transforming future-directed timelike vectors into past-directed timelike vector fields. This I expect could be interesting, but I have no idea what is known. Hawking and Ellis [1] is the place to start.

 

References

 

[1]Stephen W. Hawking & G. F. R. Ellis. The Large Scale Structure of Space-Time. Cambridge University Press (March 28, 1975)

[md65536]

Is that what T-symmetric means? that the result is the same for t positive and t negative?

Perhaps "the result is symmetrical" would be more precise than "the same"?

 

I think that to be exactly "the same" with respect to the sign of t, means the process doesn't depend on the direction of the arrow of time, which would be a more specific case of T-symmetry? Eg. constant velocity would be T-symmetrical (reverse time and you reverse the movement), while zero velocity (ie relatively at rest) is not just T-symmetrical, but also "the same" regardless of whether time is going forward or backward.

 

What I am confused about is what we mean by T-symmetry in general relativity.

 

I'm also confused more generally... Is a process T-symmetrical if and only if it is reversible? By reversible I mean able to return exactly to a previous state.

 

 

I haven't been precise with my wording, but with the conjecture that gravity works "the same" under time reversal I mean that it is an attractive force either way, which would mean that it is not T-symmetrical. If gravity is T-symmetrical, then reversing time would mean reversing gravity, and just like playing a movie in reverse, it would appear to be a repulsive force.

 

 

The conjecture is based on other conjectures. I accept that I can't claim anything meaningful or provable at this point.

Edited May 23, 2011 by md65536

[michel123456]

I have a completely different concept of symmetry.

 

To me symmetry acts like a mirror.

 

Take an object A, and his mirrored image B.

 

If the above equation from post#2 is represented by A , under time reversal the result is A (not B).

Which is not what I call symmetry, I'd rather call that invariance (I guess I am wrong somewhere)

Edited May 24, 2011 by michel123456

[ajb]

If the above equation from post#2 is represented by A , under time reversal the result is A (not B).

Which is not what I call symmetry, I'd rather call that invariance (I guess I am wrong somewhere)

 

"Invariance under a transformation is a symmetry".

 

I think by "symmetry" you really mean a "transformation"; a map between objects.

[michel123456]

"Invariance under a transformation is a symmetry".

I have to get used to that.

 

I think by "symmetry" you really mean a "transformation"; a map between objects.

 

yes. Like enantiomorphism or chirality. IIRC we have gone through this already.

Edited May 24, 2011 by michel123456