PDA

View Full Version : What have I imagined? - A math question



Jeff Root
2015-Jan-13, 05:22 AM
I haven't studied algebra very deeply because it tends to
make my brain go into sleep mode...

Just now I thought of something that resembles a trig
function. I can graph it very easily, and describe it in words
almost as easily, but I have no idea how to describe it in
math terms, or what sort of entity it is.

We can optionally begin with a horizontal line on an X-Y
grid, representing a constant value (such as zero) on the Y
axis. This line leads up to the point where the value begins
to change at X=0. It rise in a quarter circle, so that the
curve being drawn reaches the X value equal to the radius
of the circle, which is the maximum X value it ever has.
If the horizontal line was at Y=0, then the value of Y will
also be the radius of the circle. So for example, if the
radius is 8 units, then the arc we draw begins at 0,0 and
ends at 8,8. The curve extends vertically from that point.

Really, really simple.

My first thought was that this describes a function, but as
I understand it, a function can have only a single value of
the dependent variable for each value of the independant
variable, and that isn't the case here.

So what is this thing?

-- Jeff, in Minneapolis

JohnD
2015-Jan-13, 09:54 AM
Try Dr.Math: http://mathforum.org/library/drmath/view/53020.html

There, y=SQR(1-(SQR(-x))^4) gives a semicircle, if that is any help.
There appear to to be methods that can operate an equation within limits, which would give your constant legs outside the curve, but I don't know how to use that notation.

Do you use "PocketCAS" a maths and graphig app on your smart phone? http://pocketcas.com/ I confirmed the above very quickly and easily on that.

John

Grey
2015-Jan-13, 05:24 PM
My first thought was that this describes a function, but as
I understand it, a function can have only a single value of
the dependent variable for each value of the independant
variable, and that isn't the case here.You're correct that a function has to be single valued. What you've described is called a relation (http://en.wikipedia.org/wiki/Binary_relation), although sometimes it's just called a multivalued function (http://en.wikipedia.org/wiki/Multivalued_function), though that's technically an oxymoron. You could parametrize your curve, so that the x and y values would both be functions of a third variable, t.

Jeff Root
2015-Jan-13, 07:36 PM
I thought of this curve while thinking about an effect of a
black hole on the space immediately around it. The X-axis
represents distance, although the way I described the curve
puts the center of the black hole away from the origin.
I put "the last stable orbit", located at 3 R, at the origin.
The initial horizontal line represents the value of the effect
outside the last stable orbit: No effect. Beginning at 3 R
the value of the effect increases in some non-linear way
(probably not like a quarter circle, but it seems plausible)
until at R, the event horizon, the effect becomes infinite.

Does my graph approximately correctly describe the relation
between distance and strength of the force dragging objects
to their doom in a Schwarzschild (non-rotating) black hole?

-- Jeff, in Minneapolis

Colin Robinson
2015-Jan-13, 10:29 PM
I haven't studied algebra very deeply because it tends to
make my brain go into sleep mode...

Just now I thought of something that resembles a trig
function. I can graph it very easily, and describe it in words
almost as easily, but I have no idea how to describe it in
math terms, or what sort of entity it is.

We can optionally begin with a horizontal line on an X-Y
grid, representing a constant value (such as zero) on the Y
axis. This line leads up to the point where the value begins
to change at X=0. It rise in a quarter circle, so that the
curve being drawn reaches the X value equal to the radius
of the circle, which is the maximum X value it ever has.
If the horizontal line was at Y=0, then the value of Y will
also be the radius of the circle. So for example, if the
radius is 8 units, then the arc we draw begins at 0,0 and
ends at 8,8. The curve extends vertically from that point.

Really, really simple.

My first thought was that this describes a function, but as
I understand it, a function can have only a single value of
the dependent variable for each value of the independant
variable, and that isn't the case here.

So what is this thing?

-- Jeff, in Minneapolis

What you've described somewhat resembles the trig function y = tan x in the range from x = 0 (where y = 0) to the asymptote at x = pi / 2 (approaching which, y increases without bound, and the gradient approaches vertical). If you think of trigonometric functions in terms of degrees, then the asymptote is at x = 90 degrees. Although the curve for y = tan x is not an exact arc, and it never even comes close to being a horizontal line.

Colin Robinson
2015-Jan-13, 11:03 PM
Another trigonometric function that resembles yours closer than tangent would be

y = (1 / cos x) - 1 , again in the range from x = 0 to x = pi/2

that gives you the point x = 0, y = 0, and the graph approaches a vertical line as x approaches pi/2. It is a better match than the tangent function, because its curve does approach horizontal as x approaches zero (though doesn't stay horizontal for negative values of x).

Grey
2015-Jan-14, 09:47 AM
I thought of this curve while thinking about an effect of a
black hole on the space immediately around it. The X-axis
represents distance, although the way I described the curve
puts the center of the black hole away from the origin.
I put "the last stable orbit", located at 3 R, at the origin.
The initial horizontal line represents the value of the effect
outside the last stable orbit: No effect. Beginning at 3 R
the value of the effect increases in some non-linear way
(probably not like a quarter circle, but it seems plausible)
until at R, the event horizon, the effect becomes infinite.

Does my graph approximately correctly describe the relation
between distance and strength of the force dragging objects
to their doom in a Schwarzschild (non-rotating) black hole?I'm not quite sure why it has a flat line anywhere, and the sudden discontinuities. I'd expect that a graph of gravitational strength vs. distance for a black hole would look more or less like that for any object when far from the hole (i.e., an inverse square curve), and then as you got closer, you'd start to gradually deviate from that as general relativistic effects become important.

Jeff Root
2015-Jan-14, 10:53 AM
Yes, that is exactly what I would expect, but there apparently
is such a thing as "the last stable orbit" (located at 3 R), so
inside that distance there must be an effect, and outside that
distance there must not be an effect.

Making the graph symmetrical (like an inverse square curve)
was probably just a foolish extrapolation, but it was the most
natural assumption to make. The curve goes out of sight at
the event horizon -- might as well make it mirror the straight
line outside 3 R.

You consider the transitions from straight to curved to be
"discontinuities"? The amount of curvature is discontinuous,
I guess, but the transitions are completely smooth. It isn't
as though there were vertices in the curve.

-- Jeff, in Minneapolis

Grey
2015-Jan-14, 04:32 PM
Yes, that is exactly what I would expect, but there apparently
is such a thing as "the last stable orbit" (located at 3 R), so
inside that distance there must be an effect, and outside that
distance there must not be an effect.

Making the graph symmetrical (like an inverse square curve)
was probably just a foolish extrapolation, but it was the most
natural assumption to make. The curve goes out of sight at
the event horizon -- might as well make it mirror the straight
line outside 3 R.I'm not sure it's that simple. I'd expect that there are deviations from a Newtonian model even before that, but that it isn't enough to prevent stability. But I'll acknowledge that I haven't worked out the math in detail. And even at the event horizon, it probably rises asymptotically, rather than literally going straight up, so then this would be a function mathematically.


You consider the transitions from straight to curved to be
"discontinuities"? The amount of curvature is discontinuous,
I guess, but the transitions are completely smooth. It isn't
as though there were vertices in the curve.Oh! I was thinking you meant to have the semicircle part flipped the other way (so that it makes right angles with each of the straight lines). Like two walls meeting with a piece of quarter-round trim where they join. I was really having a hard time figuring out why you thought it might be like that. I still think it's more complicated, but at least I understand what you mean, now. :)

BigDon
2015-Jan-14, 07:15 PM
Whenever *I* ask "What did I just imagine?" they make me take more meds.

There's some sort of double standard I tell ya!

Jeff Root
2015-Jan-15, 12:51 AM
Grey,

Yeah, my concern about it not being a function resulted from my
completely groundless assumption that the curve approaches the
event horizon the same way I assumed it leaves the last stable
orbit -- as an arc of a circle. It is still an interesting question as
to what such a curve might represent, but it probably doesn't
represent the destabilizing force near a black hole. That would
much more likely be asymptotic as you say.

So that leaves the separate questions of what the curve near a
black hole actually is, what force the curve actually represents,
and why it (apparently) cuts off suddenly at 3 R. That last is
curious, and is why I speculated that the curve might have the
shape of an arc of a circle.


Don,

I'm not much help. After I stopped laughing, my first thought
was what Mel Brooks said about the difference between tragedy
and comedy: Tragedy is when I get a hangnail, comedy is when
you step through a manhole and fall to your death in the sewer.

Definitely a double standard!

-- Jeff, in Minneapolis

Jeff Root
2015-Jan-15, 01:03 AM
In the original post I said: "This line leads up to the point
where the value begins to change at X=0." I interpret that
as implying or suggesting that the change is initially small,
and increases. But someone else reading it wouldn't have
to make that inference, and so could instead think I meant
the line suddenly takes off in a new direction.

Like I said: very easy to draw, *almost* as easy to describe
in words!

-- Jeff, in Minneapolis

Grey
2015-Jan-15, 05:29 PM
Well, if you're mostly curious about how objects behave near a black hole, and what orbits are stable (and why), it's probably best to just look directly at what orbits are possible for the Schwarzschild metric. As usual, Wikipedia (http://en.wikipedia.org/wiki/Schwarzschild_geodesics) has a pretty good, moderately technical discussion that might serve as a good jumping off point.

If you jump down to the section comparing the results to classical physics and discussing circular orbits, you can see that for the right angular momentum, there is actually a curve that looks a little bit like the one you're describing, although this is a graph of potential energy vs. radius, rather than force vs. radius. The text talks about the issues of stability. The short version appears to be that for Newtonian mechanics, you can look at an orbit as a balance between an effective "centrifugal potential" and gravity, but that in general relativity, there's a third term that shows up (that's also attractive), that's proportional to the inverse cube of the distance.* So at large distances, this is significantly smaller than the other terms, and doesn't affect stability, but as you get closer, it gets stronger much more quickly, and eventually overwhelms the other two.

If you look at the graph labeled "Schwarzschild Circular Radii", this gives a really nice explanation for why the last stable orbit is where it is. You can see that for a given angular momentum, there are actually two possible circular orbits. One is stable, out near where the Newtonian orbit would be, where the third term is small enough that the others are dominant. The other is much closer in (between 1.5 and 3 Schwarzschild radii), but these circular orbits are unstable, because here the third term is large enough that it's significant; if there were a slight perturbation inward (for example) the third term will increase enough that the orbiting object will be pulled in (a very common homework problem in a classical mechanics class is to demonstrate that if gravity went like a higher power than the inverse square, like this third term does, circular and elliptical orbit are no longer stable). The last stable orbit ends up being where these two curves meet.


*And that's the potential that's proportional to the inverse cube, so the "force" resulting from that would be proportional to the inverse fourth power of the distance. But remember that in general relativity, gravity isn't a force; it's probably best to stick to talking about the gravitational potential rather than the gravitational force.