Johnny5

Yes they are the same thing. Can you state Nernst's theorem?

Suppose that we have two enormous spheres, and let each of them be a rigid body. Let one be the size of the earth, but far more dense, and let the other be the size of the moon, and also much more dense. Now, let them be touching, the smaller one on top of the bigger one. Would gravity cause the smaller one to start rolling around the bigger sphere, or would it just sit on top?

Case where one of them is standing on the surface of a planet. In such case, the centers of mass of each object are at rest with respect to one another. So how in the world could you specify which one is accelerating? In other words, define a frame, whose origin is located where the center of mass of the two objects is, and let the center of mass of each object be at rest in this frame. This completely specifies what I will henceforth refer to as the CM frame. In this frame, neither of the centers of mass of the two objects is moving, so that the speed v, of either of these two objects in this frame is equal to zero. Acceleration in a frame is defined as: [math] \vec a = \frac{d\vec v}{dt} [/math] Where [math] v [/math] is velocity. The magnitude of the velocity vector is the speed of the thing in question, in whatever frame you have defined it's speed in. Since the speed of the center of mass of either object in the CM frame is zero, neither is accelerating in that frame.

You are going to have to help me out, since I have no idea where to begin. Well actually that's not true, but I want to see how you start off.

You know what, lets do something which will be constructive for me. In the standard Big Bang model, if space didn't exist at the first moment in time, then matter didn't exist at the first moment in time. Then later, space existed, and matter was where exactly? Thank you Martin

Do you mean Nernst's theorem?

See, now I have to decipher this. The goal was to understand a particular definition of rest frame. I confess I can't make sense out of it, can you?

In the case of a mechanical clock, any of the parts. In the case of a digital clock, the electronics. In the case of an atomic clock, in the atoms.

I don't know what that means.

What I was doing the other day didn't go anywhere. In fact, I shouldn't have approached the problem the way I did. You gave part of the answer, but I shouldn't have tried to go from the answer you gave to the question. I happen to have my own arsenal for solving integrals. Let me have a look at this guy... [math]F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}[/math].

I know that simple is not equivalent to correct, but it's a beginning.

Swanson answered you correctly. A lot of times, physicists use the word 'force' when what they really mean is "magnitude of force." I am as guilty of this as the next fella. But you can always clear things up. Force is a vector quantity, not a scalar quantity, but I am not aware of any special name for "magnitude of force." So in Newton's third law you have two things interacting. Each exerts a force on the other, during the interaction. Let [math] \vec F_{12} [/math] denote the force that object one exerts on object two. Let [math] \vec F_{21} [/math] denote the force that object two exerts on object one. Here is Newton's third law in mathematical form: [math] \vec F_{12} + \vec F_{21} = 0 [/math] From here, you are one mathematical step away from this: [math] \vec F_{12} = - \vec F_{21} [/math] The minus sign tells you that the forces have antiparallel directions. Since the directions are along the same line of action, you can divide both sides of the equation above by the direction, to obtain a vector independent statement, which is this: [math] | \vec F_{12} | = | \vec F_{21} | [/math] This states that the magnitude of the forces are equal. Regards

I am with you on this syntax, that is exactly what I think is happening. The internal parts are being affected, and so the interior functioning of the clock is being altered, so there must necessarily be a difference in clock readings. But not because time was altered, because the CLOCK was altered. Yes, I am with you, if that is in fact what you really think.

Not recently.

I'm not trying to say that, but that would follow from this line of thought. I would say this though for sure: You cannot push on a vacuum, and accelerate yourself. I guess we could do it mathematically too, using Newton's second law. You are about to push the vacuum. Your inertial mass is M, and any acceleration you recieve is a. There is a buoy a few meters off to your left, with a blinking light. And no forces are acting on the buoy, and it's at rest relative to you. Here's what you are going to do, you are going to try to kick space, and make yourself move away from the buoy. Let the inertial mass of the space which you kick really equal zero. The force you exert on it, will equal the force it exerts on you, by Newton's third law. The force on it, will be equal to its mass, times the acceleration you give it. Since its inertial mass is zero, the net force you exert on it, is equal to zero. Therefore you have this: [math] 0 = M a [/math] Where M is your inertial mass, and a is any acceleration you recieve (relative to the buoy) by kicking the vacuum as hard as you can. Your inertial mass is nonzero, so you can divide both sides of the equation above by M, and you get this: [math] 0 = a [/math] So you cannot push off the vacuum, it is nothing. I wasn't trying to say that, but in a sense it follows from what I did say.

The m in the formula above' date=' should be called "inertial mass." By [i']rest mass[/i], I presume you mean the inertial mass of an object measured in any reference frame in which it's center of inertia is at rest.

I think I understand you, and yes. But just to be sure, can you explain your statement more? Intuitively, if the gravitational mass of the earth increased, you would slowly be crushed. If you take your scale to the moon with you, the scale will measure your weight on the surface of the moon, and you will discover that its less than on earth. So the scale is measuring the force of earth's gravity upon you, and when you take it to the moon, it's measuring the force of the moon's gravity on you. And we call what we feel "our weight."

We have to go slowly. You have some object. You weigh it on a scale here on earth, and the reading is W. W is a force, called "the weight of the object on the surface of the earth." <--- That's what the scale measured. So the scale has measured a force. Now, suppose that you release that object, at a height h above the surface of the earth. The empirical fact, studied in great detail by the great physicist Galileo Galilei, is that the object does not hover at a height h above the surface of the earth, but instead freely falls towards the surface of the earth, moving perpetually faster and faster, i.e. it is accelerating. The rate of change of speed of the object, in the rest frame of the earth, is called "the acceleration due to earth's gravity," and is almost always represented by the letter g. Let v denote the speed of the center of inertia of a falling body, in the rest frame of the earth, at any moment in time after the object is released. Using the differential calculus we have: [math] g = \frac{dv}{dt} [/math] Let us define the gravitational mass of the object being weighed as follows: [math] M_g = W/g [/math] The RHS of the equation above, is composed purely of measured quantities, and the equation itself is a definition. We can perform a repeatable experiment... we can keep dropping the object, and measure g. Galileo did this. We use rulers at rest in the gravitational rest frame of the earth to measure distance, and clocks at rest in the gravitational rest frame of the earth to measure "amounts of time." Using the 'meter' as our unit of length, and 'second' as our unit of time, the measurable value of g is: [math] g = 9.8 \frac{m}{s^2} [/math] Obviously, the exact value of g depends on where you are on the earth, it's slightly less at the equator, than at the poles, but think of 9.8 as the average value, or let it be exactly what you measured where you are at. Now, you can drop objects of different weights, and discover, as did Galileo, that they all accelerate at g, independently of their weights. So let's look at the tentative definition of gravitational mass. [math] M_g = W/g [/math] There would be a problem with the formula, if we ever found a place on the surface of the earth where g=0, because that is the division by zero error of algebra. Now, we all know that astronauts in space float weightless. We can alleviate the problem if we write this instead: [math] W = M_g g [/math] Now, we can think of W as a variable, which reaches zero once an object is in circular orbit about the center of inertia of the earth. So you can now measure your weight while you are orbiting the earth, using your trusty scale. You put it below your feet, and you stand straight up, and your scale reads zero, you are weightless, in other words, W=0. Now, there are three ways that W could equal zero in the formula above. Either your gravitational mass Mg is equal to zero, or the acceleration of earths gravity g, where you are at is equal to zero, or both are equal to zero. Ok, your weight is going to be zero once you are in space, regardless of how you got there. So let there be a ladder that you can climb, embedded firmly into the surface of the earth, that extends all the way into space. This is probably how people imagined getting into space, thousands of years ago, since they didn't have an idea of rocketry, and it would in fact work, if your ladder material were strong enough. So you would just climb up your space-ladder, straight into space. Now, the higher you climbed, the lower would be what we are calling your weight (that which is measured by your trusty scale). Once you got far enough away from the center of inertia of the earth, your weight would finally be zero, so that if you let go of the ladder, you wouldn't free-fall to earth at all. The point is, for the statement [math] W = M_g g [/math] to be true, something on the RHS must also go to zero, as you climb higher and higher up the ladder. Now comes the question of geosynchronous orbit. A geosynchronous orbit, is an orbit in which the satellite hovers above the same position above the earth, and never falls. Suppose you climb all the way to geosynchronous orbit, and then let go of the space-ladder. If you are looking down at the surface of the earth, the base of the ladder will remain at rest in your rest frame. Additionally, the ladder will also stay at rest in your frame (this will be true as long as your orbit is truly geosynchronous). Now, think of the ladder as a ruler. Each pair of rungs is one unit apart, in other words, the distance between any two rungs, as measured in the gravitational rest frame of the earth, is exactly one meter. The point is, that as you climb up that ladder, the center of inertia of the earth is moving in your rest frame, and vice versa. In the rest frame of the earth, your center of inertia is moving. The absolute fact of the matter is this, as you climb up that ladder, it is not the case that the center of inertia of earth, and your center of inertia are at rest with respect to one another. That's only true at any moment when you aren't climbing, like for example when you are standing upon the surface of the earth, and weighing yourself. Here is the thing that's bugging me. If you climbed all the way to geostationary orbit, then your acceleration towards the earth, in either your rest frame, or the earth's rest frame would be equal to zero. Yet, Newton showed that centripetal acceleration is given by [math] a_c = \frac{v^2}{R} [/math] v is your "tangential speed". R is the distance between your center of inertia, and the earth's center of inertia, which was steadily increasing as you made your way up the space-ladder, all the way into space. The direction of your centripetal acceleration is towards the earth, yet if you are in geo-stationary orbit, and you let go of the ladder, you will hover over the same spot above the earth, long after you let go. This is why a satellite dish can be pointed to just one spot in the sky, and not have to be adjusted. The satellite is hovering above one spot over the surface of the earth, so the homeowners satellite dish doesn't have to have an expensive tracking system. Such satellites are geostationary. Now, if it is your centripetal acceleration that is decreasing as you climb the ladder, then when you finally reach a position where you have geostationary orbit, then at that height R above the center of inertia of the earth, your tangential velocity v, must be equal to zero (or your radius is infinite... which is impossible). So, let us presume that the acceleration due to earths gravity, is in fact centripetal acceleration, and see what happens. Thus, at the surface of the earth we have this: [math] a_c = g = 9.8 = \frac{v_t^2}{R} [/math] In the formula above, R would be the radius of the earth. But there is a problem. While you are standing on the surface of the earth, your tangential speed v, is zero. A contradiction is reached (g=0 and not (g=0)). The error has occurred, because tangential speed was not clearly defined. The way out is simple. Define your tangential speed at the surface of the earth as follows: [math] v_t \equiv \sqrt{gR}[/math] Where g is the local acceleration due to gravity, and R is the center to center distance, i.e. the distance between earth's center of inertia, and the falling body's center of inertia. Since both quantities on the RHS are non-zero, it follows that the LHS must also be nonzero. Therefore, your tangential speed at the surface of the earth, is not defined as the relative speed between your center of inertia, and a fixed point on the surface of the earth. We, any of us can know, that in the gravitational rest frame of the earth the stars orbit, to a good approximation, once every 24 hours. This fact is empirically verifiable. Tomorrow night at exactly 8:00 PM look straight up in the sky. Hopefully you can see the stars in your area, if not this won't work, and you will have to go to a place on earth where you can see the stars. Presuming you can see the stars from your position on earth, wait exactly 24 hours. The next night, again at 8:00 PM, look straight up into the sky, and again you should see the same star configuration. The big dipper should be at the same location in the sky, etc. The correct inference is that the earth isn't spinning wildly in space. The correct inference is that the earth is spinning, with a constant period, which we can denote by Te. So by experiment we have: [math] T_e = 24 hours [/math] Now, because we measured the acceleration of gravity in units of seconds, it would be nice to also have the period of earth in seconds as well. There are sixty seconds per minute, and sixty minutes per hour, therefore: [math] 24 h = 24 h \frac{60m}{h} \frac{60s}{m} = 86,400 seconds[/math] Now, we don't know the distance from the earth to the stars. And it isn't easy to figure out the radius of the earth either, even though it appeared in the formula above, as R. But lets proceed as if we know R. Using Euclidean geometry, the circumference of the earth is given by: [math] C_e = 2\pi R [/math] Certainly, a distance divided by a quantity of time, has units of speed. Let us divide the circumference of the earth, by Te, to obtain a quantity which has units of speed. [math] V = \frac{2\pi R}{T_e } [/math] Let us at least temporarily postulate that the speed above, is our tangential speed v, when we are weighing ourself on the earth. We then have this: [math] v_t = V= \frac{2\pi R}{T_e } = \sqrt{gR}[/math] Therefore: [math] \frac{2\pi R}{T_e } = \sqrt{gR}[/math] Let us square both sides of the statement above to obtain: [math] \frac{(2\pi)^2 R^2}{(T_e)^2 } = gR [/math] Dividing both sides by R we have: [math] \frac{(2\pi)^2 R}{(T_e)^2 } = g [/math] All the quantities on the LHS can be measured, and the quantity on the right can be measured, so we can check to see if the assumption is true. A quick check on google tells us that the Radius of the earth is 6, 378.1 kilometers. [math] \frac{(2\pi)^2 (6378100)}{(86,400)^2 } = .0337 [/math] The RHS is not equal to 9.8 m/s^2, in which case a false statement was used someplace. It must have been from equating V with vt.

Let's do it this way. [math] M_g = 1 [/math]