Johnny5

No, you are confused about E=Mc^2 in the first place. Let's see your derivation of the formula, if you don't have one, I can provide you with one. Regards

This was terrific, thank you very much. I couldn't remember how to do it. There was some sort of procedure to start you off in the general case, which resembles the middle part of your proof. I'm still trying to recall it, I think it involved synthetic division. Hmm I'm sure it will come to me. Thanks again

Why do you say that the perpendicular component does no work? A proton is flying through a linear accelerator, something is applying a force to it, to cause it to linearly accelerate. Now, someone way down the tube is waiting for it to pass them. As the proton flies by, they "punch" the proton, in a direction perpendicular to its original motion. In that scenario, work is done by the "punch" force. So I am not following you. I do better with equations.

I have a question. What standard things do you think the final 'model' of physics should explain. Right now there are thousands of experiments discovering the tiniest minutia, but no mathematical theory is going to be able to explain one billion bits of minute experimental data. I can sort of start this off. 1. magnetic moment of proton 2. magnetic moment of neutron 3. magnetic moment of electron 4. g factors 5. Why the nuclear force range is what it is. 6. Fine structure constant. 7. Formula for gravity 8. Show how the fundamental constants of nature are related. 9. Equation of motion for particles 10. Explain masses of elementary particles 11. Explain superconductivity etc You get the idea. Here is the question again... What things must the final supertheory of physics explain?

Does anyone know how to use the finite discrete difference calculus to find a formula for the following finite sum? [math] \sum_{n=1}^{n=m} n^2 [/math] Thank you

From the perspective of a photon at rest, the speed of other things would not be constant, nor would they all be c, but to be sure, large things would be zipping by. (I am using the Galilean transformations by the way) Regards

Exactly why do you get a given force causing different accelerations in different directions? I understand what you are saying, I just want to know why you are saying it. Suppose that: [math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math] I don't see any vector symbols there, so could you explain yourself more? Regards

To continue a bit, we have: [math] \frac{dc}{dt} = \frac{d}{dt}(\omega R) = R\frac{d\omega}{dt} + \omega \frac{dR}{dt} [/math] Now, invoke the postulate that the speed of light relative to the emitter is a fundamental constant of nature. Hence dc/dt=0 so that we have: [math] 0 = R\frac{d\omega}{dt} + \omega \frac{dR}{dt} [/math] From which it follows that: [math] R\frac{d\omega}{dt} = - \omega \frac{dR}{dt} [/math] Has anyone seen this before?

My question was, does the difference calculus take this into account. A - B is totally different from B - A In order to define the difference operator [math] \Delta [/math] the values of the physical quantity which changed are necessarily values at different moments in time. So a convention must be used, in order to specify which is the latter and which is the earlier value. So I am thinking that there is already a convention in place, for the unidirectionality of the "flow of time". To make this totally clear, let Q denote an arbitary physical quantity. The difference in Q is defined as: [math] \Delta Q = Q2 - Q1 [/math] So if I don't choose a convention right now, a student will come along and ask, which came first the Q2 or the Q1. Am I right in saying this? Regards

Here was your model for space expansion: T: R3xR-->R3xR, T(x,y,z,t)=((x+|a|Dt,y,z,t+Dt) You have the x axis stretching. Use this model of space expansion to analyze a rigid bar for me, if you don't mind. At some moment in time, let one end of a steel bar have x coordinate 0, and let the other end of the bar have x coordinate 3. So the bar is three meters long presently, at moment in time t=0. Then exactly one moment in time later, let space have expanded in that model. Tell me where the center of inertia is now, and where the ends of the bar are now. One more thing... at moment in time t=0, the temperature of the bar is absolute zero, the density of the bar is uniform, and the bar is at rest in the frame. You can introduce any other physical concepts you need. Regards

Yes, that's what is being said. That the points in space have a 1-1 mapping to the points in the coordinate system.

In order to mathematically define the "spatial location of an object" one needs to have already chosen some reference frame in which to define that location. The location of the object is to be taken as the location of the objects center of inertia. The spatial separation of two objects would then be measured by a rigid ruler (ruler at absolute zero kelvin) at rest in the measurement frame. Now the location of a center of inertia is just a point in the reference frame. So to say that two objects got further apart we do not mean that two points in the reference frame moved. Instead we mean that there was relative motion of material, and the center of inertia of at least one of the objects, now has a new location in the original frame. Or to put this another way. Take a rigid steel bar, and locate the center of inertia. Now bend the bar into a V shape. The center of inertia no longer lies inside the material, it lies exterior to the material. You did not bend space in order to change the coordinates of the center of inertia... no sir... you bent the bar. Regards

Its a conclusion, not an assumption. Set up a three dimensional rectangular coordinate system whose origin is the center of mass of the universe. Let the center of mass of the universe be permanently at rest in this frame. Make this frame a minimum energy frame. That means you don't want the axes spinning wildly relative to the fixed stars. Mathematically this means we want Galileo's law of inertia to be true in this coordinate system. Ok all that being done, consider two locations in space A,B relative to the center of mass of the universe, which is at point C. I say that the distance between A and B cannot change. It is a temporal constant. For suppose not. Let it be the case that A can move relative to B. Let a force act upon A, to give it an impulse in the direction of B, with just the right final speed so that at some point in the future, A coincides with B. At that moment in time, the point A is spatially equivalent to the point B. But the points are different by stipulation. Therefore, point A cannot move through the coordinate system. Regards (If the argument seems corny, that's not my fault. It constantly boils down to not (2=1), because if at any moment in time the point A moved it would be on top of and in (coinciding) with some other distinct point D in the frame. And in a rigid frame such as this, the Pythagorean theorem is a true statement)

I don't want this post to end here. Suppose that before a photon is emitted, it is whirling in a circle in the rest frame of the emitter. Just like Newton's pail. Then we let it go, and it flies off in a straight line. There were parameters to the motion of the photon before emission. The photon had a period T, and a circumference [math] 2 \pi R [/math] Its tangential speed would be: [math] v_t = \frac{2 \pi R}{T} [/math] f = frequency = 1/T, hence [math] v_t = 2 \pi f R [/math] and [math] \omega = 2 \pi f [/math] Which gets us here: [math] v_t = \omega R [/math] Which is correct for circular motion. Setting the tangential speed vt equal to the speed of light relative to the emitter gives: [math] c = \omega R [/math] Now I guess use the galilean transformations from here. Has anyone seen this before? Regards

That's cute, but in the example I provided, the argument goes something like this... 2 points = 1 point Therefore, 2=1. And 2 does not equal 1, hence.... Regards

You say that the differential is always positive, do you mean df, or do you mean dx? The following is basic calculus: [math] y = f(x) = x^{2} + 3x - 2 [/math] The picture is of a parabola. And we can discover its vertex. Taking the first derivative we have: [math] dy/dx = \frac{df}{dx} = 2x + 3 [/math] Setting that equal to zero gives: [math] 0 = 2x + 3 [/math] So if dy/dx=0 then x=-3/2 dy/dx is the slope of the tangent line as a function of x. Setting dy/dx=0 is the criterion for the tangent to the curve to be a horizontal line. This is the place where the function assumes a local max, or local min. The second derivative is given by: [math] d^2y/dx^2 = 2 [/math] Positive so concave up. So when x=-3/2, the function has a local minimum. [math] y = (-3/2)^{2} + 3(-3/2) - 2 = -4 [/math] So the vertex of the parabola is at (x,y)=(-3/2,-4). Picking up from here, what do you mean that dx is necessarily positive? Regards

No, one cannot equal two, this has nothing to do with assumptions. My conclusions about the vacuum are reached by using what I might as well call spatio-temporal logic. Logicians are still working out the details of temporal logic, physicists have actually worked out spatio-temporal logic. In a way, that's what reasoning about motion leads to. Regards

You are right, that reason doesn't logically necessitate that position. You are absolutely right. Ok I guess I will have to show you my real argument, though I don't want it ridiculed because I like it. Suppose that space could be created. Therefore, points in a coordinate system could move relative to one another. If a point in a coordinate system could move in a coordinate system, then two points in the coordinate system would simultaneously be located at one point in the coordinate system, whence 1=2. Therefore, locations in the vaccum cannot be created. That was my secret argument for maintaining my position. Regards

Ok, help me out here. You are saying the relativistic mass formula isn't true? What is the problem you described? Regards