sieger35

Hey, I was wondering if anyone here could give me a little help with this problem.
I'm having trouble starting the proof. Thanks!

Statement:

Show [Prove] that |x| is not an affine function of x.


If I'm not mistaken, an affine function is a linear form + a constant term.

Any help would be greatly appreciated!

kebo

if the definition of an affine function is a linear function, then show that the first derivitive for |x| is constant (of couse this assume you know calculus and can use it in your proof.)
kevin

Hain

Kebo, but I do know calculus and I know that d(|x|)/dx is not constant.

d(|x|)/dx = { -1 , x<0 ; undefined , x=0 ; 1 , 0<x }

If you need to see how that is done, just break |x| into three functions: y(x) = { -x , x<0 ; 0 , x=0 ; x , 0<x } . The derivative of y(0) can't occur because of the conflicting limits as you approach 0- and 0+ (You are in calculus right?).

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kebo

that's absolutely correct. But according to sieger35's definition of an affine function the derivative must be constant.
kevin
edit:
or at least it must be constant over the range of values in question

sieger35

kebo,

Since the derivative of |x| is not constant, doesn't that show that |x| is not an affine function of x, by the definition I provided? That is, if I'm following this discusion correctly.
Thanks.

Hain

Yeah. It's not an affine function. And for the years of math that I have taken, I have never heard that term used to describe a linear linear function. I have never even heard it before now :laughs:

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kebo

I agree it is not unless you define the function to exsist only for a given input say for x>=0 in which case you are talking about f(x) = x rather than f(x) = |x|.

Hain

I need help with a math problem. This is a hypothetical situaton:
Particle A and Particle B have a mass m and an attractive force acting on them proportional to the inverse square law. The force pulling them together is F=-K/(r^2) (negative sign is a formality to make sure you know that the force is pulling the objects closer).

So we set up our equations before we differentiate. However this system encounters Lorenz dialations of mass (oh great).

so m, c (c is the spesky speed of light), and K are constants. [This K will allow for the set up of using .5r (that's to the origin) because both particles will move exactly the same]

So remember, acceleration is the second derivative of position: d^2 r /dt^2 = a
SOLVE: r(t) = ?
v(t) = r`(t) = ?
a(t) = r``(t) = ?
All those should be in terms of m, c, K and initial radius.

So there is some differential calc needed here and I went... Fuck Waffles! (-George Carlin).

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Yakk

m*a/(1-(v/c)^2)^.5 = -K/r^2

m/K * a * r^2 = -(1-(v/c)^2)^(1/2)

Let c = 1 by choice of units (ie, time seconds and distance in light-seconds -- always a good idea when working in relativity!)

r`` * r^2
---------------- = (-K)/m
(1-(r`^2))^(1/2)

This looks like a non-linear second order differential equation. I think you are fucked? My knowledge of this area of math is very sparse.

By 'fucked' I mean I don't know of any non-numeric means of modelling this. Numeric models exist, of course, but they look more like computer programs than equations.

I'm wondering -- wouldn't there be time and space dialations/contractions as well, which are missing from your equation? This could either simplify or complicate the situation. I'd have to hit up my relativity-knowing friend.

Second, initial conditions will matter. It wouldn't be that hard to solve the position/accelleration/velocity if the two objects where moving in a perfect orbit around each other, or so I'd guess...

I also wonder if conservation of mass/energy might be of help here... From your perspective, as you see someone falling towards you, they are losing potential 'gravitic' energy and gaining kinetic energy. Isn't this a net zero change in their energy, and hence a net zero change in their mass?

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Hain

MSIE ERROR... goddamnit...

Hain

Ok, I guess I'll let the c = 1 thing slide even though I forsee problems doing so... :shrugs:The time dialations aren't significant to the particles. The only contraction necessary from an outside observer is the mass dialation which is represented by m over gamma. Like I said, they start from rest at a distance d apart. I don't think so. You contradicted yourself. The net energy is not zero because relative to one object, the other is moving rather fast. But still, 2 x KE of one will be the energy of the system at that instant. Don't we just love infinities???

Yakk

The energy of the entire system is constant -- there is no energy coming in or going out of the system.

Doesn't that mean that the mass of the entire system is constant?

Net energy change.

As something falls down a gravity well, does it gain mass?

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Hain

Slightly yes it does. As your velocity increases (really only close to that of light) so does your mass. That is why there is the term c / sqrt( c^2 - v^2 ): that times the mass will tell you relativistic mass. Einstein said God doesn't play dice with the universe... then it's apparent that God has a sense of humor.

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Yakk

I know adding energy causes added mass.

But a closed system should not have a mass change. Converting gravitational potential energy into kinetic energy shouldn't cause a mass change.

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Hain

Yes it will. Or else these obects would be able to break light speed. Adding energy comes from where? A force acting on something over distance. Same thing is occuring.

Hain

Anyways new weirdness of math.

I was bored out of my skull in calculus the other day and began experimenting with polar graphs and finding areas. Easy enough--generate infinitely small triangles and take the sum of them, Area = 1/2 * r^2 * dTheta. I got to this by saying that the sum of the area of arcs will be the area total. The area of a triangle is 1/2 r(theta) r(theta + change in theta) sin(change in theta)
So: The sum from a to b of the product of 1/2, r(theta), r(theta + change in theta), and sin of (change in theta) is area. Further you can take the limit of that expression for when the change in theta becomes zero.
I saw that the limit made the two radii approach each other, and that sin x as it approaches 0, approaches x.
So: the integral of the product of 1/2 and the square of r(theta) of dTheta from a to b was actually the area of polar curve! My teacher was amazed that I figured that out, kinda sad aint it for any calc nerds out there.

All in a day's work. Then I began working on polar arc lengths. Anyone that knows the method, you convert to parametric.
Let: x = r(theta) times cos(theta) and y = r(theta) times sin(theta).
To find a length you integrate the squareroot of the sum of the squares of r(theta) and r`(theta) by differential change in theta from a to b... that's simplified (maybe someone else wants to send a letter to TFP for a pretty print feature for math equations since copying from word doesn't work in here???). Well I wanted to do the same method.

Well law of Cosines: a = sqrt(b^2 + c^2 -2bc cos A)

By my logic, as the angle between approaches 0 then the two sides, b and c, become the radius taken at theta. Then that can be simplified to the integral of absolute value of r times root 2 times the squareroot of 1 minus cosine dTheta. I stumped even my teacher trying to find a different method from this point. Anyone want to take a stab at where I might have gone with this if I knew a little more about limits?

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Yakk

Could you get those objects to break light speed, using gravity, without them already being inside a black hole?

My knowledge of relativity is .. sketchy.

It just seems wrong that if you took two objects, and placed them in orbit around each other, the mass of the system could fluxuate without any outside energy input...

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Hain

This system has potential energy. The mass changes because of the relative velocity of the objects to the rest of the universe.

And no you cannot break the speed of light even with gravity--this is one of those safety features that prevents such things from occuring in this universe. You'd have to punch a hole through the universe first--that's exactly what would occur at impact of these two objects. For a brief instant the fabric of space and time will warp almost infinitely and then ripple back when the objects collide, scatting their mass from impact.

I'm still curious as to the math of this situation occuring as predicted. I am seeing my physics prof. some times soon with this problem. If he doesn't know then se la vi.

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Yakk

Energy is mass. Why do you consider potential energy not to be energy?

The conversion of potential energy into kinetic energy shouldn't cause a change in mass, as far as I know. I haven't seen a convincing arguement stating otherwise.

Sigh. I said:
Things already inside a black hole don't happen -- no outside observer can witness something actually getting inside a black hole, because it blue-shifts out of perception, and hence existance. As such, if the only way to 'break light speed using gravity' was 'already being inside a black hole', this means 'you cannot break the speed of light with gravity'.

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Hain

I didn't say potential energy isn't energy. And what I wanted to show is that the potential energy would increase as well as kinetic energy as they moved closer because of that inverse square relation. I don't truly care about impact, I want to see what it looks like going towards impact.

And a change in kinetic energy does increase relative mass. What else is the Lorentz contraction equation used for then?

Here is a site on Lorentz Transformations. And it says that:That is why my situation includes the relative mass ratio in it.

And my comment about "being inside the black hole" was not such for I was referencing a worm hole which wont be occurring.

And something that just occured to me that you may have been saying that I didn't realize is: That as the objects move closer and increase speeds, the mass increases. How does the gravitational force start to play into this? Then the K being used has to transform as well.... I say that I am still asking my Prof. tomorrow.

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Yakk

I thought the mass increase was due to kinetic energy buildup in the objects.

As you add kinetic energy to an object, it gains mass. The amount of velocity a given amount of energy represents is described by the Lorentz equation and E=MC^2.

In the case of a closed system, the if you get the kinetic energy from other energy in the system (gravitational potential energy), the energy total doesn't change, and hence the mass.

ie:
M_moving = M_0 * (1/sqrt(1-(v/c)^2))
M_kinetic = M_moving - M_0 (ie, the additional mass added by moving)
M_kinetic = M_0 * (1 - sqrt(1-(v/c)^2)) / sqrt(1-(v/c)^2)
E = MC^2
E_kinetic = M_kinetic * C^2
E_kinetic = C^2 * M_0 * (c - sqrt(c^2-v^2))/sqrt(c^2-v^2)
let C=1 lightsecond/second
E_kinetic = M_0 * (1 - sqrt(1-v^2))/sqrt(1-v^2)

IIRC the above equation, with v very small, approaches E_kinetic = 1/2 M_0 v^2 (unless I made some math errors!)

What is the Taylor series expansion of M_0 * (1 - sqrt(1-v^2))/sqrt(1-v^2)? If my math was right, it the lowest order power should be 1/2 M_0 v^2. With v << 1, higher order powers disappear compared to v^2.

I might be wrong, but that is how I thought of it.

Interestingly, for one object, a change in mass should have no change on the accelleration nor the velocity. However a change in mass in the other body would increase accelleration. If two bodies falling towards each other gained mass, they would accellerate faster, not slower, because of it. It wouldn't defend you against breaking light speed at all.

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

fckm

err, shouldn't potential energy decrease?

U=-GM/r

as r decreases, U decreases.

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vinaur

Yes it does. Way to conserve energy, Augi.