An invariant while crossing the horizon

[Tomas Vencl]

TL;DR

Is the fact that the infalling observer crossed the horizon at the time when the black hole had mass m_1 an invariant?

While reading the neighboring thread I came up with the following question: consider an evaporating black hole and an observer falling into it. The observer crosses the horizon at a finite proper time \tau_1; at the moment of horizon crossing the black hole has mass m_1 from the infalling observer’s point of view (I understand that the horizon is a global property and the infalling observer may not notice that they have just crossed it; however, retrospectively—e.g. by tracking the beacon of the infalling observer—one can infer which signals continue forever). Is the fact that the infalling observer crossed the horizon at the time when the black hole had mass m_1 an invariant?

 

[PeterDonis]

While reading the neighboring thread

What thread? Please give a link.

The observer crosses the horizon at a finite proper time \tau_1

Yes, that's fine, although the specific value of $\tau_1$ has no meaning off of that observer's worldline.

at the moment of horizon crossing the black hole has mass m_1

At the event of the horizon crossing, meaning at the specific point on the infalling observer's worldline where it crosses the horizon, one can assign a "mass" to the hole, yes--it would be the parameter $m$ that appears in the metric. Note that for an evaporating hole, $m$ is not constant, but it does have a definite value at each event in the spacetime.

However, note that this value of $m$ only applies at that particular event.

Is the fact that the infalling observer crossed the horizon at the time when the black hole had mass m_1 an invariant?

Only in a very restricted sense: that at that particular event in spacetime, as above, the hole has that mass.

But when you use the phrase "at the time", you might be assuming that we can assign some kind of "time" to the hole having that mass that applies more broadly than just that event, by saying that other events happened "at the same time". We can't, at least not in any invariant sense. Any such thing would be a choice of simultaneity convention, which is not invariant. The only invariant way to know what the mass is at any particular event in spacetime is to look at the value of the parameter $m$ in the metric at that event.

 

[Tomas Vencl]

……The only invariant way to know what the mass is at any particular event in spacetime is to look at the value of the parameter $m$ in the metric at that event.

Yes, that exactly is the point of the question.

 

[Nugatory]

consider an evaporating black hole and an observer falling into it. The observer crosses the horizon at a finite proper time \tau_1; at the moment of horizon crossing the black hole has mass m_1 from the infalling observer’s point of view....
Is the fact that the infalling observer crossed the horizon at the time when the black hole had mass m_1 an invariant?

This is tricky even without introducing the complexities of an evaporating black hole. "Crosses the horizon at a finite proper time $\tau_1$" doesn't make sense unless and until we specify the point on the infaller's worldline where the proper time has been arbitrarily considered to be zero.

But even after we allow for that... "the back hole has mass $m_1$ from the infalling observer's point of view" is an invariant. They somehow measured/calculated the mass and came up with an answer. We all agree about what they measured/calculated (we could even transform to the coordinates they used and repeat the calculation using the values of the measurements they made to get the same answer) which makes it an invariant.

 

[martinbn]

I understand that the horizon is a global property

In a way so is mass. It is not an easy concept.
https://en.wikipedia.org/wiki/Mass_in_general_relativity

 

[Tomas Vencl]

Yes, thank you all. The infalling observer’s proper time (\tau_1) is ultimately not important. What matters is that all observers agree that at the moment the infalling observer crosses the horizon, the black hole had mass m_1 from his point of view. We take this mass to be computed by the infalling observer from the circumference (or area) of the horizon. In other words, all observers agree that the infalling observer crossed the horizon at a specific r_1 (r_1= circumference / 2\pi).

 

[PeroK]

Yes, thank you all. The infalling observer’s proper time (\tau_1) is ultimately not important. What matters is that all observers agree that at the moment the infalling observer crosses the horizon, the black hole had mass m_1 from his point of view. We take this mass to be computed by the infalling observer from the circumference (or area) of the horizon. In other words, all observers agree that the infalling observer crossed the horizon at a specific r_1 (r_1= circumference / 2\pi).

Every measurement is itself an invariant. The observer could record the result and promulgate that round the universe (subject to certain limitations). In any case, no one can disagree about what was recorded.

 

[PeterDonis]

the black hole had mass m_1

Yes.

from his point of view

No. The value of the metric parameter $m$ at that event is an invariant. It doesn't depend on any "point of view".

We take this mass to be computed by the infalling observer from the circumference (or area) of the horizon.

No, we can't, because that is not a local measurement. The observer has to locally measure the value of the parameter $m$ in the metric.

all observers agree that the infalling observer crossed the horizon at a specific r_1

The areal radius $r$ is not a local measurement, since the horizon area is not. It's a global parameter.