Re-Examining Black Holes and the Standard Model

[.Scott]

How can I say this more clearly. I do not believe there is a preferred or "global" reference frame.
When I say that a particular reference frame is "valid", I am not saying that any other reference frame is invalid.

It is correct and constructive to describe events as viewed from any reference frame. After all, in each case, the laws of physics must still appear to be followed.

 

[jartsa]

No, it isn't, because the object isn't "frozen". The light emitted by the object before it reaches the horizon is being redshifted by the hole's gravity. The object itself falls through the horizon; that's why it can't reflect light.

Let's mark two altitudes near event horizon with two buoys. Then we drop down a petawatt pulse laser device, programmed to emit a pulse once every millisecond, proper time. When the device passes a buoy it emits a special pulse.

Now observers far from event horizon can see quite well that:

1: The device seems to spend a long time between the buoys.

2: The device emits few pulses during the time the device spends between the buoys. (No "seems" here)

The device's clock seems to be almost frozen. The device is moving very fast, so the clock should be almost frozen.The point is: The observers saw, with their optical measuring devices with slow response time, a fading continuous laser beam.

 

[.Scott]

If an object is traveling at near the speed of light relative to your inertial reference frame and you measure the amount of gravitational pull from it, it will correspond to a mass much larger than its rest mass.

No, it won't. I realize it seems intuitively like it should, but this is a case where intuition can lead you astray.

To see the issue, consider the scenario as it is seen in the rest frame of the gravitating object. ...

I have no problem in looking at events from different reference frames - and each one will yield a correct view of events.

Let's change the object. Say it is a planet-size bomb and we are in an equatorial orbit. When this bomb explodes, it does so by converting 98% of its rest mass into energy which is directed entirely into pushing the 1% of is mass at each pole (North and South) apart.

So, in an instant, the rest mass is reduced by 98% - but the mass apparent to you in orbit does not change - and, until the poles move a substantial distance, there is little affect on your orbit. That's because in your reference frame, the total mass has not changed.

In fact from any inertial reference frame, the mass has not changed nor has there been a violation of the conservation of momentum - although not everyone will agree on the total mass or on which poll is more massive.

Because there is no preferred reference frame, it would not be accurate to say that, in absolute terms, the two poles are now of equal mass. That would only be true from reference frames that have no polar component in their velocity relative to the center of gravity of the polar masses.

 

[PAllen]

How can I say this more clearly. I do not believe there is a preferred or "global" reference frame.
When I say that a particular reference frame is "valid", I am not saying that any other reference frame is invalid.

It is correct and constructive to describe events as viewed from any reference frame. After all, in each case, the laws of physics must still appear to be followed.

And how can I say this more clearly: invariant facts are true, period, not coordinate dependent. That is what invariant means. There is no such thing as frames AT ALL other than locally, in GR, only coordinates. There is no meaning at all to 'earth frame' for describing global physics, because frames are local. Coordinate choice changes NOTHING invariant (e.g. that a signal sent from an outside observer reaches an infaller inside the horizon, is an invariant fact). That some coordinates don't cover a region of spacetime simply means those coordinates don't describe things outside their coverage - they don't say anything at all about what is for any observer whatsoever. If I draw a polar projection map of the Earth centered on the North pole (which shows only the Northern hemisphere) does that say anything at all about whether the south pole 'exists' for someone at the north pole?

 

[PeterDonis]

1: The device seems to spend a long time between the buoys.

2: The device emits few pulses during the time the device spends between the buoys. (No "seems" here)

The device's clock seems to be almost frozen.

A more compact and precise way of saying all this is: the elapsed time on the distant observer's clock between receiving the light flashes emitted by the device when it passes the two buoys, is much longer than the elapsed time on the device's clock between passing the two buoys, as measured by the number of pulses emitted by the device.

The observers saw, with their optical measuring devices with slow response time, a fading continuous laser beam.

Sure. But in the statement of yours that I responded to, you were talking about reflected light, not light emitted by the device. So you need to add that to the scenario. Here's a way to do that:

The distant observer emits light pulses towards the device; each light pulse has a time stamp, showing the time on the distant observer's clock when the pulse was emitted.

The device has a "reflector" that reflects light pulses from the distant observer, but in such a way as to include a second time stamp, showing the time on the device's clock when the pulse was reflected.

The distant observer then observes the time stamps in the reflected pulses as he receives them. What he will find is that, as the time of reception, on his clock, increases without bound, the two time stamps in the reflected pulses (time of emission and time of reflection) each approach fixed, finite values; they do not increase without bound. These fixed, finite values represent the time, by the distant observer's clock, at which he emits a light pulse that just reaches the device's reflector as it is crossing the horizon; and the time, by the device's own clock, at which the device crosses the horizon (so the light pulse reflected at just this instant stays at the horizon forever). At times beyond these values, the device no longer reflects light that is visible anywhere outside the horizon.

 

[PAllen]

I have no problem in looking at events from different reference frames - and each one will yield a correct view of events.
So, in an instant, the rest mass is reduced by 98% - but the mass apparent to you in orbit does not change - and, until the poles move a substantial distance, there is little affect on your orbit. That's because in your reference frame, the total mass has not changed.

In fact from any inertial reference frame, the mass has not changed nor has there been a violation of the conservation of momentum - although not everyone will agree on the total mass or on which poll is more massive.

This is simply wrong. Rest mass in SR is a synonym for invariant mass. The invariant mass of the exploded Earth has not change at all from the explosion (and will never change if you include all the components). Different flyby observers will attribute different energy to the Earth (exploded or not) but will all agree on its invariant mass (pre or post explosion), and that this does not change (if you include all components of the system).

 

[PeterDonis]

in an instant, the rest mass is reduced by 98% - but the mass apparent to you in orbit does not change

Gravity does not depend on rest mass in GR; it depends on the stress-energy tensor. The "mass" of the planet, as measured by you when you measure your orbital parameters and apply Kepler's Third Law, didn't depend solely on its rest mass even before the explosion; even in the simplest case of a spherically symmetric perfect fluid, the fluid's pressure also contributes to its stress-energy tensor.

Also, how does any of this relate to your original scenario, which I responded to before?

 

[PeterDonis]

I have no problem in looking at events from different reference frames - and each one will yield a correct view of events.

Yes, but if you consider my description of your scenario in the rest frame of the gravitating object, you will see that it requires the description from your rest frame (in which the object is flying past you at near the speed of light) to be different from the one you gave in your previous post: the apparent "mass" of the object as measured by its gravitational effect on you does not increase without bound.

The more general rule, which I mentioned in my post a few minutes ago, is that an object's gravity is not determined by its relativistic mass (or by its rest mass); it's determined by the object's stress-energy tensor.

 

[PAllen]

Yes, but if you consider my description of your scenario in the rest frame of the gravitating object, you will see that it requires the description from your rest frame (in which the object is flying past you at near the speed of light) to be different from the one you gave in your previous post: the apparent "mass" of the object as measured by its gravitational effect on you does not increase without bound.

The more general rule, which I mentioned in my post a few minutes ago, is that an object's gravity is not determined by its relativistic mass (or by its rest mass); it's determined by the object's stress-energy tensor.

Hmm. The oft referenced paper:

https://home.comcast.net/~peter.m.brown/ref/mass_articles/Olson_Guarino_1985.pdf

derives that the active gravitational mass of flyby object on an initially stationary test body is mγ(1+β²). The ultra-relativistic limit is 2γm. This says it does increase without limit.

 

[PAllen]

Hmm. The oft referenced paper:

https://home.comcast.net/~peter.m.brown/ref/mass_articles/Olson_Guarino_1985.pdf

derives that the active gravitational mass of flyby object on an initially stationary test body is mγ(1+β²). The ultra-relativistic limit is 2γm. This says it does increase without limit.

Here is an explanation that squares this result with Peter's argument in #30. I will refer to body (for a massive body) and particle (for test body). I will assume that the speed of flyby is ultra-relativistic. Then, in the frame of the body, you have a deflection that 2 times Newtonian expectation. In the frame of the particle, since the deflection is primarily orthogonal to the relative motion, you have the same amount of deflection occurring in a time interval shorter by a factor of γ than observed in the body frame. Thus, the active gravitational mass (in particle frame), by this definition, of the body is increased by γ over the body frame; thus 2γ times the Newtonian expectation.

 

[.Scott]

...So, in an instant, the rest mass is reduced by 98% - but the mass apparent to you in orbit does not change - and, until the poles move a substantial distance, there is little affect on your orbit. That's because in your reference frame, the total mass has not changed.

In fact from any inertial reference frame, the mass has not changed nor has there been a violation of the conservation of momentum - although not everyone will agree on the total mass or on which poll is more massive.

This is simply wrong. Rest mass in SR is a synonym for invariant mass. The invariant mass of the exploded Earth has not change at all from the explosion (and will never change if you include all the components). Different flyby observers will attribute different energy to the Earth (exploded or not) but will all agree on its invariant mass (pre or post explosion), and that this does not change (if you include all components of the system).

First of all: Oops, you blew up the Earth. I was hoping for some for some other planet-size bomb.

For simplicity, I will change things slightly. Instead of an orbiting observer, I will have the observer at zero velocity relative to the Earth when the explosion occurs.

But using Earth, here are the specifics of the explosion:
Before the Explosion:
Rest Mass of Earth: 6\times10^{24}Kg
Velocity of Earth relative to our observer: 0
After Explosion:
Rest Mass of each remaining piece: 6\times10^{22}Kg
Velocity of each polar fragment relative to our observer: 0.9998c
Total Rest Mass: 12\times10^{22}Kg
Total Mass relative to observer: 6\times10^{24}Kg

Unless you disagree with these calculations, note that the rest mass changed dramatically during the explosion. But the relativistic mass remained constant.

 

[PeterDonis]

In the frame of the particle, since the deflection is primarily orthogonal to the relative motion, you have the same amount of deflection occurring in a time interval shorter by a factor of γ than observed in the body frame.

Yes, I agree with this. To me, it just points out another problem with trying to apply intuitions about "gravitational mass" in relativistic scenarios: the amount of deflection remains finite even though the "active gravitational mass", if it's defined as "deflection per unit proper time", increases without bound.

 

[PAllen]

First of all: Oops, you blew up the Earth. I was hoping for some for some other planet-size bomb.

For simplicity, I will change things slightly. Instead of an orbiting observer, I will have the observer at zero velocity relative to the Earth when the explosion occurs.

But using Earth, here are the specifics of the explosion:
Before the Explosion:
Rest Mass of Earth: 6\times10^{24}Kg
Velocity of Earth relative to our observer: 0
After Explosion:
Rest Mass of each remaining piece: 6\times10^{22}Kg
Velocity of each polar fragment relative to our observer: 0.9998c
Total Rest Mass: 12\times10^{22}Kg
Total Mass relative to observer: 6\times10^{24}Kg

Unless you disagree with these calculations, note that the rest mass changed dramatically during the explosion. But the relativistic mass remained constant.

Of course I disagree because you are making the same mistake again. The invariant mass of a system is not the sum of rest masses of components. It is the norm of the sum of 4-momenta. The sum of rest masses of multiple bodies in relative motion is a quantity of no significance in SR. Since momentum is conserved, the total momentum of all explosion products is zero as it was before (they are going off in all different directions, at magically the same speed, but we can accept this conceit). This means the total 4-momentum is (<total energy>,0,0,0) with time component first. The norm is then <total energy>/c^2. Thus, the invariant mass has not changed (and is invariant).

Since this is a really basic mistake, I am curious if you have ever formally studied SR.

 

[PeterDonis]

the rest mass changed dramatically during the explosion

The rest mass of individual pieces changed, yes. But the rest mass of the total system did not; that was PAllen's point. Rest mass is not additive: a system composed of multiple pieces in relative motion can have a rest mass (a better term is "invariant mass", as PAllen said) that is not the sum of the rest masses of the individual pieces.

 

[PeterDonis]

The OP's question has been answered and he hasn't been back in a while, so this thread is closed.