Archive for the ‘gravity’ Category

Black Holes and Those Pesky Event Horizons

October 8, 2017

In Leonard Susskind’s book, The Black Hole War, page 240, he states, “To a freely falling observer, the horizon appears to be absolutely empty space. Those falling observers detect nothing special at the horizon…” In Amanda Gefter’s book, she points out that the distant observer sees the event horizon, while the falling observer detects no event horizon at all. Of course, she took a lot of her ideas from Susskind. In the meantime, Hawking treats the event horizon as a fixed boundary where virtual particles can split apart (Hawking radiation).


I think none of these is right. The idea between the “escape velocity being faster than the speed of light” is relative to the delta between the gravitation potential of the observer and the potential at the event horizon. From an infinite distance, we observe an event horizon at a certain radius. Should the event horizon suddenly disappear if we are in an inertial frame starting our fall into the black hole? Starting at what distance? A thousand miles? A light year?

The more likely result is that the event horizon moves inward as you approach it. You are in a deeper gravity well as you approach the black hole, thus the difference between your local gravity potential and that of the event horizon, to maintain a high enough value for the escape velocity to equal the speed of light, requires that the event horizon continuously move away from you (toward the singularity) as you move toward the singularity. You never quite catch up with it. There’s a Wikipedia article that says this explicitly, but then, it’s a Wikipedia reference (Event Horizon). Sometimes they’re wrong, but usually they’re dead-on.

An interesting consequence of this is that if you maintain a certain orbit near the event horizon, and your version of the event horizon is closer to the singularity than that of a more distant observer, then a photon just outside your observed event horizon could reach you just fine, even though it cannot reach the more distant observer. Having received that photon, you could transmit the data from it outward, (boosting the frequency) as the distance from your gravity well to the distant observer requires an escape velocity somewhat less than the speed of light. Is this a loophole?

Why, then, do we think that a photon below the event horizon (for the observer at infinity) can’t escape the confines of the black hole? Is it only because it would be red-shifted to a zero frequency? Or is that false?

Escape velocity is merely a calculation of the velocity required to go from one gravity potential to another. If you are already in a gravitational well (like the outer edge of the Milky-Way galaxy) with an escape velocity of 300km/s, this has no effect on the escape velocity from Earth (11km/s), or the velocity needed to orbit Earth (7.5km/s). Likewise, consider a photon trapped just beyond the event horizon as viewed from an observer at infinity. To the guy in orbit around the black hole, the difference in potential is much smaller, and his relative event horizon is closer to the singularity. Won’t he see that photon? Can’t he receive it from the domain outside his apparent event horizon, but inside the event horizon of the observer-at-infinity? And then capture the photon and retransmit it?

So, even though a photon by itself can’t escape the event horizon of the observer-at-infinity, an intermediate process (natural or human) could conceivably pass a photon up through overlapping light cones, even though the light-cones at either end don’t overlap. This might eliminate the question of whether information can escape a black hole or not. The infalling observer can see what’s happening beyond the outer event horizon, and pass the information on, since his own event horizon is even closer to the event horizon.


A Variant Geometry for Spacetime

September 18, 2017

There are a lot of odd characteristics of existing spacetime physics, creating a lot of questions that are difficult or impossible to answer, such as, “Why is the speed of light approximately 300,000 km/s?” or “Why can’t you go faster than the speed of light?” or “Is there such a thing as a tachyon?” Or, “If you can’t go faster than the speed of light, how is it possible to age only 4 months due to time dilation while traveling 4 light years?” I hope to offer an alternate geometry to provide some reasonable answers to these questions.

Let’s start with some fundamental concepts about photons. It’s generally believed that photons are their own antiparticles, and also that the speed of light is the ultimate speed past which nothing can travel. Also, in a photon’s frame of reference, the distance from source to destination appears to be zero, and it takes zero time to travel that distance. This led me to speculate that light might, in fact, travel at an infinite speed, and that somewhere out there, there is a geometry in 4D space-time where that makes sense, where, when we try to measure it, we see light ambling along at a tedious 300,000 km/s. It would also explain why you can’t travel faster than the speed of light; the speed is, in fact, infinite. It’s really hard to go faster than that. The difficulty lies in finding that geometry. My second supposition regarding photons is that they are always emitted perpendicular to the path of travel (through time) of the originating particle. In a standard space-time diagram, assuming a velocity of c, this leads to the light cone diagram. In the new geometry, assuming an infinite photon speed, the picture is a little different, but still leads to the well-known equations we are used to.


Looking at Figure 1, the vertical line A represents the source of photons a and b, which travel instantaneously to the observer on line B. Line B observes the two events a and b separated by time ct2, and from B’s perspective, the object has moved away a distance x, which equals a-b. In A’s proper time, the time between the two events is merely c time t, and the distance is zero, so the interval is s=ct1. From B’s proper time, the duration is ct2 and the distance A has traveled away from him is x, so the measured interval between the two events is s=√(ct22-x2), which should be familiar to everyone.


In Figure 2, we see what happens as B gets closer to the X axis. But it still produces the common formula for the interval. What the diagram does not explain is why the speed of light appears to be roughly 300,000 km/s.

However, what Figure 1 can do is allow us to derive the standard time dilation formula:


How do we get there?

Note the velocity of B away from A is


 So x2=(vt2)2

From before, we had s=ct1 in A’s reference frame and s=√(ct22-x2) from B’s perspective. Set the two equations equal, and we get


Substituting for x2 we get


 Divide it all by c2 and pull the t2 out of the two terms on the right gives us


Or, ∆t1=∆t2•√(1-v2/c2), which should be familiar to a lot of you out there. It’s the standard formula for time dilation due to relative velocity.

So Figure 1 works out that if the line is at a 45 degree angle, then v=c/√2, which shouldn’t be a surprise. And as v gets closer and closer to c, then the graph gets flatter and flatter. But this graph is based on the idea that c is infinite. Why is it that when we measure it, it’s always 300,000 km/s? Clearly, if you set up an experiment that bounces light back and forth between mountain tops, and c=∞, then your test should show that light moves instantaneously. Not 300,000 km/s. And yet, we always measure c at the same boring 300,000 km/s (yeah, I know this isn’t exact value-it’s 299792.458 km/s. Get over it-this is easier to type).

So what is it that makes an infinitely fast photon measure as a finite number in all frames of reference? Is it the expansion of spacetime? Is it the curvature of spacetime due to gravity affecting the value of c that we measure? Is it that this theory is just wrong? I don’t know yet – this would obviously have to be resolved before this model made any sense, and may or may not appear in a later entry. Any speculations about this are welcome.


A Fool’s Physics

September 18, 2017

I’ve read a lot of physics in my life and have a lot more to read, a lot more to learn. It’s hard to read any general physics text without stumbling across some interesting tidbit that makes me sit back and ponder how that tidbit fits in with my mental model of the universe. Some things make sense, some don’t. When I heard of the Unruh effect, I was dumbfounded (to my understanding, this is the emergence of energy out of a vacuum relative to an accelerating object). When I learned that photons are their own antiparticles, I was confused. When I realized that the time component in the spacetime interval produces a hyperbolic curve in the formula, it was an enlightenment years in the coming. When I read that antiparticles are just regular particles going backward in time (Feynman, I think), that, too, messed with my mental models of the universe. To say nothing of dark matter, the accelerating expanding universe, and so on. So, I try to organize all this hodge-podge of apparently related information into a single model that makes sense. As most physicists will tell you, it is an insurmountable task. But, I am not a physicist, really. I have a Masters in Astronautical Engineering, and as Sheldon Cooper would tell you, I’m really just an engineer. I create mental models that make sense to me, but may not have any practical use or truth in the larger sense of things. But we all have to start somewhere. I’ll keep reading, and revise the incorrect bits as I go along.

In this log of ramblings, I’ll offer up a bunch of foolish ideas on physical reality. I’m a big fan of determinism, so be forewarned. I also think of time as an actual, physical dimension. If you happen to join me on this warped (!) journey of speculation, I’d love for you to tear my arguments apart, tell me what’s wrong, and perhaps help shape the speculations into something that makes a coherent sense of reality, or assure their demise.

Background Gravity – Time Dilation in a Flat Field

August 11, 2017

I got into an argument with a physics buddy not that long ago (a year, maybe), about gravity. We have an intermittent arrangement where we go drink beer and talk about physics at two or three pubs in San Luis Obispo. Usually, the physics becomes a little less coherent as the evening wears on.

One of the discussions centered on whether there is a “background field” of gravity or not, or whether it’s even sensible to discuss such a thing since, in an infinite field of equally distributed mass (or gas, or 1 atom per cubic light year, whatever), all the forces around you seem to cancel out. The mass of the universe to your left is equal in size to the mass of the universe on your right; you feel a net acceleration of zero. I argued that even though the field was “flat”, there was still a field there. He argued that a field implied a gradient; there is always a force.

We did not come to a satisfactory conclusion. It might merely have been the fact that we were defining the same terms in different ways in our heads. I’m not sure. I thought my argument was rock-solid.

So, here is my side of it.

Some of you are probably familiar with Newton’s Shell Theorem. It’s in his Principia Mathematica, and if I remember right, he solved it without using calculus. Basically, what it says is this; if you are inside a spherically symmetric shell of mass, then you feel no gravity pulling you any direction. It’s a bit non-intuitive. Let’s say the Earth is hollow, and the entire planet’s mass has been compressed into a thin spherical shell a few centimeters (or meters, it doesn’t matter) thick. If you are floating around in your Nike Space Suit inside this shell, you will not be pulled toward the center, or the inner surface of the shell, or anywhere else inside the shell. Wherever you are put, you will remain.

Personally, I think this is one of the coolest theorems ever.

It’s also true that if you are outside the surface of a spherically symmetric planet, then it doesn’t matter how dense it is, at a given radius you will feel the pull of a certain amount of gravity. If you are in orbit above the Earth, and the Earth suddenly becomes a black hole of the exact same mass, you will remain in orbit, totally unaffected by that change. That’s pretty cool, too. Given that the gravitational force is based on F=GMm/r2, this should be kind of obvious. Neither your mass (m) or the mass of the Earth (M) has changed, your orbital radius is the same, and G is the gravitational constant. Ergo, the density of the object you are orbital at a radius “r” from the center of the object is irrelevant.

So, that was me drifting from the actual subject. Shell Theorem—let’s get back to that.

As you might know, the clock of anyone in a gravity field runs slower than that of a clock outside of that gravity field. This is called gravitation time dilation (and is equal to ∆t’ = ∆t √(1-2MG/rc2) for a non-rotating sphere). A person on Earth actually ages slower than a person in deep space, according to relativity. This was verified with clocks flying around the Earth in the Hafele-Keating experiment. Before you ask, yes, they took into account Earth’s rotational speed, the speed of the airplane (in both directions relative to Earth’s rotation) with regard to Special Relativity’s time dilation due to velocity. It was a nice experiment.

Let’s say we’re using a hollow Earth from Newton’s shell theorem. As you get closer to the Earth, you are in a deeper gravity well, and the outside observer sees your clock slow down. There’s a small hole in the planet, and you pass into the planet, where everything is pulling you in opposite directions equally, so you seem to feel no force. And yet, your time dilation effect does not suffer a discontinuity, jumping suddenly to that of the outside observer. You are in a denser gravity field, but a flat gravity field. [to the physics majors out there, for god’s sake, if my terminology sucks, please correct me]. Your time dilation will be just the same as if you were standing on the surface of the planet.

So now you have a flat gravitational field (no “force” pulling you in one direction, that is, all forces pulling you equally in all directions). And yet, even in this apparent lack of gravity, where you can’t actually tell that you’re in a gravity well, your time runs slower than the time of someone far from the planet.

I extended this argument to the rest of the universe. If mass was distributed equally around you, even though you felt no force one way or the other, there would still be a background gravitational field. Gravitational time dilation implies a gradient; for time dilation to be relevant, you need someone in a weaker gravitational field measuring your time. However, both the measured and the measurer can be in locally flat gravitational fields.

Does a flat gravitational field curve spacetime by itself? Or is it only the gradient between two different gravitational fields that curves spacetime? My general opinion is that you don’t need the gradient for the curvature of spacetime. If you have an infinite universe with equally distributed mass, then from some arbitrary center, it will appear to curve spacetime until it closes the dimensions of universe into closed loops, like the inside surface of an event horizon (though other arbitrary centers will have different, yet overlapping event horizons – a subject I will touch upon another day).

How would you test the curvature of space inside a shell? I’m not entirely sure. I think the universe we have is a good test case, however.