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Topic Review (Newest First)
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| Dec 13th, 2005 09:58 PM | ||
| Sethomas | If you're relating force to friction, then the force has to be an improper integral. But I don't see what the point of an equation would be if it just relates a value to some mysterious force. | |
| Dec 10th, 2005 01:43 PM | ||
| camacazio |
You seem to still be having trouble with the idea that if m*a=n*g*m, then m cancels and all we care about is accelleration a. it doesn't matter if m is a million times larger or a thousand times smaller, it's the same accelleration. I use friction for the original idea of modeling a time delay caused by it in expansion, but if it's really causing you that much trouble then let's just use some accelleration a. We're not dealing with force, we're dealing with linear one dimensional uniform accelleration. To reiterate the equation of velocity for that, it's v(t)=v0-a*t. Mass doesn't come up. You can replace each ng I used with a, and everything's fine for rates of expansion. Changing mass is not a consequence of lorentz contraction. That is incorrect. The only thing that is a consequence of lorentz contraction is lorentz contraction. Lorentz contraction is a consequence of changing time intervals between events used to measure a length. The same rest mass is squished into less volume, so one could argue that density is increased, but this says nothing about mass. These are things that do not matter with regards to mass. This relavistic mass you keep bringing up is a consequence of kinetic energy. Something moving near the speed of light has it's rest mass, E=mc^2, added to it's kinetic energy. The total energy is gamma*mc^2. Momentum, equally important, relativistically is gamma*mv. The m in those cases is the same rest mass used with equations involving a factor of gamma. Rest mass is a poor term because the gamma does not come from mass, it comes from the proper time. "Relatavistic mass" has a factor of gamma not because of mass but because of velocity. Velocity is defined as dx/d(tau), tau being the proper time. Tau=t/gamma. When you make the change of variables, you have gamma*dx/dt. That's the very simple way the equation for momentum appears. |
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| Dec 10th, 2005 01:36 AM | ||
| Sethomas | Mass is variable when anything approaches light speed. It's a consequence of Lorentz contraction. | |
| Dec 9th, 2005 11:32 PM | ||
| camacazio | In case it wasn't clear, what I was saying is that it would take millions of atoms before the change in mass matters at all, so I don't care if I pick up a few with my space ship. The change in mass is unimportant. If I have a 1kg block, and add 1,000,000 iron atoms, i'm adding 1kg + 9.345*10^-20kg. I feel confident that I can leave it at 1kg. | |
| Dec 9th, 2005 01:13 PM | ||
| camacazio | Yeah supa and that's an important thing to keep in mind in any physics. If the values are negligable, it's best to ignore them (the transfer of mass from atoms during friction is almost nothing). That is why so many equations lasted so long without going into question. Newtonian mechanics are perfect for the world on our scale. Quantum mechanics and relativistic physics were fields of science that began discovering ways in which the old newtonian mechanics were flawed under deep scrutinization--the super fast and the super small. Frictional force as it is has not been researched much. At the quantum level, one would have to factor in ideas including some such as sethomas has brought up, but it's not an issue of math. It's an issue of research. Or statistical mechanics, but I've had only a little class time on that field outside of the kinetic theory of gasses. As far as my equation goes, I'd have to assume that the surface doesn't get entirely melted by the energy transfer of the rocket ship. | |
| Dec 9th, 2005 01:05 PM | ||
| Sethomas | But mass is a variable here, so you can't just represent it as "m". | |
| Dec 9th, 2005 01:01 PM | ||
| Supafly345 |
Re: Relativistic Frictional Force Quote:
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| Dec 9th, 2005 03:47 AM | ||
| Sethomas | Well, if you're going that way, then your entire superstructure is flawed. If the ground is exerting force on the spaceship, then the tangent molecules (the only ones relevant) become part of the spaceship's frame of reference, so relativistic effects go out the window. So, deriving an equation is pointless, as you'd have to treat the frictional surface as a fluid for a realistic representation of how the tangential surface becomes unified with the spaceship while the ground below it doesn't. I doubt you're that smart. I for one wouldn't want that for an assignment. | |
| Dec 9th, 2005 03:32 AM | ||
| camacazio | The net force on the moving object is friction. I defined it as exactly that when I said F=ma=ngm. There is no other force on the system. After rereading your post a couple times, I think what you're trying to say is that I'm not modeling some sort of thrusting power of this object moving near the speed of light. I don't care about how it got to this speed. I don't even care why it's on earth, or how it found a run way long enough to slide along for any decent amount of time. All I care about is that it's coasting, and there is a linear force slowing it down--that being what is causing an expansion. This same equation I have come up with could be easily modified for when it accellerates to a certain speed, instead of a negative accelleration for slowing it down. In the end, as I stated above, all I want is a linear accelleration such that I can equate it to ma and also use it in a simple linear velocity equation. Friction in the macroscopic approximation gives an excellent and simple form for this, and that's why I use it. Typically special relativity avoids accelleration because it ventures into general relativity, so most things in special relativity ignore how and why something is moving near the speed of light. As long as I don't jump into the reference frame of this object that I'm modeling expansion of, I should be in the clear. | |
| Dec 8th, 2005 09:08 PM | ||
| Sethomas | Umm, no. mg(nu) only refers to the force of friction. Saying that it cancels out would be equivalent to saying that critical force against friction equals an indeterminate speed. It doesn't. You have force responsible for overcoming friction, and you have force responsible for acceleration. They're totally separate. | |
| Dec 8th, 2005 08:48 PM | ||
| camacazio | F=ma=mng, so a=ng. M cancels when making my equations. | |
| Dec 8th, 2005 08:35 PM | ||
| Sethomas | Umm, with force being a constant as it would be on a space ship, m=F/a, so you can't possibly ignore it and have an equation that means anything. | |
| Dec 8th, 2005 08:11 PM | ||
| camacazio |
Well as far as my equations go I didn't bother to go very in depth with the derivations. My equations start from simply the equation for length contraction (Proper length divided by gamma), and using an equation for linear accelleration for my v term in gamma. As for mass, that is not an issue in this case. I'm only dealing with length contraction, not with any sort of relativistic momentum-energy of any system. As a matter of convention, relativistic mass is a term I prefer to avoid. In my most recent special relativity class, our professor made a point of saying how some use relativistic mass, some don't. I like avoiding the term because all equations with mass in relativity involve the rest mass, a factor of gamma, and another factor. Rest mass stays the same for a particle. It's gamma times the other factor that is changing relativistically with this mindset. What changes from different values of energy and momentum is the mass of the system, but that is still different from relativistic mass. It's all the same math in the end, just how I prefer to think about it. If I had a better way to type out my equations I would do just that. I'll try anyway though. When something is travelling at a velocity near the speed of light with respect to our reference frame, we will see it contract in the direction of motion. The equation for this is L/gamma, where L is the proper length. Proper length is the length of the object at rest. Gamma = 1/sqrt(1-v^2/c^2). This is always a number great than one, and at velocities much below .1 times the speed of light comes out to nearly 1--for this equation, that gives proper length. For my v, I used (in this case, and with my previous letter defintions) v(t)=v-ngt, ng being the coefficient of friction. Realistically something going near the speed of light would take forever to slow down reasonably by friction, but it gave some simple equations for a case when something would slow down. I could use simply a for some constant of accelleration. What's important is the accelleration is linear to avoid a third derivative of anything. So once this constant accelleration is applied opposing the motion, such as friction, the object slows down at a constant rate. Thus, it will expand at some rate. That's what my first derivative represents. It really has no use that I can think of though. |
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| Dec 8th, 2005 07:32 PM | ||
| Sethomas | You're missing a m0 factor in there. Also, I don't see where you factor in the relativistic increase in mass. I don't want to look up the equations to derive it myself, since I'm not that bored. | |
| Dec 8th, 2005 06:53 PM | ||
| CaptainBubba |
I couldn't follow your equation's progression because you didn't explain anything except your variable declarations and your final derivatives. This seems like interestin stuff for someone into physics. I was just mostly confused because I havent taken physics in years and seeing equations just appear upsets me as a rule.  for example I do not know what any of the original formulas were which you derived. That would be helpful to someone reading this so that they could see where you started and how you progressed. There is no shame in showing too much work. |
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| Dec 8th, 2005 02:51 PM | ||
| camacazio |
You wouldn't by any chance still have that test, would you? Well I started out with this chunk of physics hypothesis but it degenerated into mathematics. There really isn't any use for rates of expansion or compression of objects accellerating near the speed of light anyway, unless someone can come up with one. |
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| Dec 8th, 2005 01:17 PM | ||
| AChimp |
I used to know a lot of math, and I was really good at it. I haven't used it in several years, and I've forgotten it all now.
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| Dec 8th, 2005 12:50 PM | ||
| Sethomas |
On my first collegiate physics test, I had to derive an equation for the rate at which at elephant pushes a tub of water up an incline plane with its trunk while decreasing its mass by drinking from it. I think the applied force was variable. I got a 12 out of 30, which was one of the best scores in the class. Having a string theorist for a professor has its ups and downs. |
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| Dec 8th, 2005 12:34 PM | ||
| camacazio | I don't like math that uses numbers. | |
| Dec 7th, 2005 11:41 AM | ||
| kahljorn |
You know, I haven't been in highschool for something like 5 years. I had to take an assessment test for mathematics last night and I did so horribly. I used to be able to do math without a calculator, even most of the advanced shit in trig and geometry, but I couldn't even remember how to do some of the basic shit. No big deal, though, I just need to find some kind of Math remembering thing on the internet and use that. I couldn't even remember how to subtract decimals, though .At least I'm sure I passed all the other parts of it, although having to write an essay in 20 minutes where they expect you to rough draft, proof read then write a final draft is kind of ridiculous. |
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| Dec 7th, 2005 08:14 AM | ||
| ziggytrix |
math forum, asshole. 
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| Dec 7th, 2005 03:10 AM | ||
| camacazio |
I screwed up my chain rule. These derivatives of Length with respect to time needed to be redone. My function for the accelleration of expansion is too messy to type up. For the rate of expansion at a given time, however, is not so difficult. (L/c)*(vng-(ng)^2*t)/[1-((v-ngt)/c)^2] The second derivative with respect to time gave me a nasty function. It also continues to not tend towards 0. This function, however, does. It's units are in m^2/s^2 though, which is a huge set-back. I should have units of m/s, since it should be a Length devided by a Time. |
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| Dec 6th, 2005 01:33 AM | ||
| camacazio |
After some simple frictional mechanics work, and some fancy derivative work with relativistic values, I got a value for the rate of change of expansion for the rocket ship on the frictional surface: v = initial velocity n=coefficient of static friction g=gravitational accelleration on earth=9.8m/s^2 L=proper length of the rocket ship sliding on the high friction surface in meters c=speed of light in a vacuum t=time in seconds accelleration of expansion = (-1/2)(ng)^2*L/(1-[(v-ngt)/c]^2)^(3/2) However, that accelleration seems faulty. All my other values seem correct, but this one doesn't look so good. It has no initial value to reduce from, so this implies that the accelleration will go to (-1/2)(ng)^2*L as t goes to infinite. It should, of course, go to 0 as t goes to infinite, because the ship will reach it's proper length and then stop having any change in length. |
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| Dec 5th, 2005 11:31 PM | ||
| camacazio |
Relativistic Frictional Force I had some free time between classes today and I started thinking about the force of friction. The macroscopic approximation for it is simply a constant related to the type of materials in contact and proportional to the normal force of the object. Say we had--and I'll make this an idealized situation--a rocket ship with a perfectly flat surface on it's underside. It is travelling near the speed of light (for completeness let's say .8c) when it starts to travel along a frictional surface on Earth. Once it begins sliding, the force of friction will cause an accelleration slowing down the rocket ship. At first the simplified equation for friction between the rocket ship and the surface is all good and fine aside from some minor relativistic changes to force. However, this rocket ship as it slows down will lose it's length contraction gained from it's speed. As it slows more, it will expand more. Our simplistic equation for force of friction does not rely on surface area, so that is not a problem. The equation is, though, an approximation to simplify an assortment of small jagged edges along any real surface. As the space ship slows and expands in its direction of motion, the jagged edges on its underside will spread out--effectively reducing the force of friction. In the reference frame of the rocket ship, the road was lorentz contracted and is now what is expanding. Both the road and rocket ship reference frames should agree in a reduction of force of friction due to an observed change in the nature of the surfaces in contact. To further complicate things, the expansion of the rocket ship will do this: as it's edges lengthen along the ground, it's lengthening is pushed against by the road. As it expands, the friction is opposing it's expansion. The time dialation that will occur between the front and back of the space ship with this force will effectively cause the front and back to expand at diffierent rates, though this part I'm not too sure about. If it's expanding with it's decrease in velocity, then every discreet change in velocity will give a discreet change in length. This discreet change in length across an integral for the change in rocket ship velocity across the surface could give a useful term for the accelleration with which each end expands, and use that in discreet chunks of mass for finding what discreet chunk of opposition to expansion each small expansion has. |
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