Orodruin

I think he is explaining it quite well. When you accelerate, this is actually equivalent to adding a gravitational component in the other direction (as seen from the car). The definition of "down" in the car system would therefore still be down but slightly back. Now, two things to note: When the car is moving at constant velocity, i.e., not accelerating, then the flame will point in the "normal" up direction. When the car is breaking, the flame points to the back - the acceleration is directed in the other way.

So I believe I have identified your misconception about Newtonian mechanics. Momentum is always conserved. What is not conserved in an inelastic collision is the total kinetic energy. Elastic collision of two objects with mass m (one dimension): Before the collision: p1 = m v1 E1 = m v1^2 / 2 p2 = 0m = 0 E2 = 0 P = p1 + p2 = m v1 E = E1 + E2 = m v1^2/2 After collision: p1 = 0m = 0 E1 = 0 p2 = m v1 E2 = m v1^2/2 P = m v1 E = m v1^2/2 Now for the totally inelastic collision: Before: p1 = m v1 E1 = m v1^2/2 p2 = 0 E2 = 0 P = m v1 E = m v1^2/2 After: Equal velocity of both objects gives a velocity v. Momentum is conserved (as a result of Newton's third law), i.e. P = m v1 = (m+m)v = 2m v This gives v = v1/2 and thus E =(2m) v^2/2 = m v1^2 / 4 < m v1^2 /2 Thus, the total kinetic energy after the collision is smaller but the total momentum is still the same. Do not confuse conservation of momentum with conservation of kinetic energy.

This depends on the equation of state of the matter that dominates the Universe. The equation of state is typically given on the form [math]p = w \rho[/math] where [math]\rho [/math] is the energy density of the Universe and p the pressure. For non-relativistic matter w is 0 and for radiation 1/3. However, for dark energy or a cosmological constant w = -1 or even less (in the case of more exotic dark energy contributions). If this dominates, the energy density of the Universe can be constant or even increase.

Then in some sense your definition is circular or dependent on the definition of what "is". Cogito ergo sum? This debate is then completely philosophical in nature and has nothing to do with science.

The point is that your definition of what is real and not seems a bit arbitrary and mainly philosophical in nature. In addition, your philosophical view of things directly contradicts how they are described in the prevailing physical models. In the end, all physical models and concepts are just descriptions and make no claim whatsoever to describe something "real" - only to provide a good description of what has been observed and make predictions about what will be observed.

In hindsight I would think that learning the Lagrange and Hamilton formulations of classical mechanics would be most useful when learning quantum mechanics. Not because it is absolutely necessary, but rather because things might seem a bit less taken out of the blue. When it comes to mathematics, I would consider linear algebra to be an absolute necessity. Unfortunately, being a theoretical physicist is not only having fun, even if there is lots of it (although this may just be the part of me that corrected 140 exams a week back talking ...). Overall, I would take it rather than a "normal" job any day.

Well simply put the Universe did not contain enough matter and radiation for it to contract immediately. Instead the recent BICEP2 results seem to be pointing to the early Universe undergoing an inflationary phase, in some sense similar to the one we currently seem to be about to enter. It should be said that, although we have a pretty good picture of the formation of the Universe after very early times, at some point we would typically run into problems of our currently established theories not being applicable. It should also be said that there are also some things we do not yet understand about the standard cosmology as well, such as how the dark matter was created and why there is an overabundance of matter over antimatter.

Since you continue to ignore the facts, let me tell you exactly where your video with the bouncing balls goes wrong: You assume that only the moving object is affected by the frictional force. According to Newton's third law, when one body exerts a force on another, the other body exerts the same but opposite force on the first one. This is not the situation in your animation and as I said in my previous post, you are forgetting about one of the forces. The moving object is subjected to a frictional force from the stationary one, but the stationary one is subject to a frictional force from the moving one (and thus should start moving, which is not the case in your simulation). To answer your last sentence, you need to think a bit more about it. The reason they do not bounce to the same height is that the rubber ball transfers more of its momentum to the Earth because the collision is elastic. Now, the momentum transfer in both cases is so small that you will not notice the change in the Earth's velocity, but if you were able to measure it, the Earth's velocity change would be larger in the elastic case.

Since when does physics depend on the symbols used to represent it? The representation of the same quantity may be dependent on the field or even the specific paper you read. Potential energy will typically be denoted U, V, or EP and who knows what else may be used by people. The question was for the potential energy of the system, I chose to represent it using a V, which is something I am free to do as long as it is clear from context (answering the OP).

A non-damped car oscillates for longer because it has less damping and therefore it exchanges momentum back and forth with the Earth for quite some time before settling. And there *is* something wrong with your simulation. Since it violates conservation of momentum it cannot be built upon newtonian mechanics. Most likely you have forgotten that the friction force acts on *both* bodies but with opposite direction.

Momentum is still conserved in cars with shock absorbers and cars in general. It is also conserved when you jump. There is a force between the car and the ground, resulting in a change of momentum of both but their combined momentum is still the same. So why dont you notice the Earth's change in velocity? Well ... It is pretty heavy ...

Sorry, but if you accept that distance is real due to it being measurable using atoms, you also have to accept time to be "real" in the same sense. Just as length can be defined as what you measure with a ruler, time can be defined as what you measure with a clock, or if you want to go to the atomic level by the number of periods of a given oscillatory system. It even turns out that the concepts of time and space are intimately tied together within relativity. Now as a physicist I do not like to talk about things being real or not because that would imply a philosophical argument about what real is (much like this one). All I do care about is whether or not a given theory fits with my observations My favorite example here is the wave-particle duality in QM: is the electron (for example) a wave or a little billiard ball? The answer to this should be that it is neither - it is an electron and an electron happens to have some properties reminiscent of wave properties and some reminiscent of billiard balls. Both of these are more or less pronounced in different setups, some of which show signs of both and typically this is what seems to confuse people...

I did not bother with this before, but it is also wrong. You are always assuming that s = vt and that Fv is constant during the acceleration. Since the mass is accelerating, this is definitely wrong. Instead we have [math]W =\int Fv dt[/math] [math]\frac{dW}{dt} = Fv[/math] [math]\frac{dW}{dt} =v \frac{dp}{dt}[/math] [math]v^{-1} \frac{dW}{dt} = \frac{dp}{dt}[/math] Now by the chain rule [math]v^{-1} \frac{dW}{dt} = \frac{dt}{ds} \frac{dW}{dt} = \frac{dW}{ds} = F = \frac{dp}{dt}[/math] So you end up right where you started. Again you are just ignoring the facts. The ship must gain momentum if it is to be accelerated. By conservation of momentum, this ends up somewhere else. Unless you break conservation of momentum, the ship will not accelerate. You do not need simulations to end up with this conclusion. You have agreed on the second point and I sincerely hope that you realize the first is also true. So, unless you can show us how point 3 is not true (which you just agreed that it is), point 4 is moot and nobody will lift a finger to investigate it further.

I could have been more clear. When I just say "potential" I mean potential energy. To drop the second sum sign by lazyness when intention is clear (and I did indicate in the text that summation was over particle pairs) is standard in my field and several others I am aware of so this is in my backbone, as is dropping additional integral signs. When it comes to 1/2 vs i<j I find the latter more illustrative.

She asked for the potential energy of the system. Not the potential of a test charge. The potential energy of the system I can only interpret in one way. I agree that your formula with the sum multiplied by the test charge charge is the potential energy added by a test charge in the origin. Again, what is shown is the electric potential (measured, e.g., in Volts). What was asked for was the electrical potential energy (measured in, e.g., Joules) of the system. See http://en.wikipedia.org/wiki/Electric_potential_energy

It *is* superposition. Every charge will feel the electric potential of all the others, which is easy to check. I am 100% positive on this one, Although it strikes mr now that you may be thinking of the potential of a test charge introduced into the system (although lacking the charge of the test charge). The potential energy of the entire system is given by the expression I wrote down. In fact, it is easy to derive the full expression from your expression. Start with one charge, potential zero. For every charge you add, the potential energy is changed by an amount given by your formula (corrected with the charge of the new particle) with the sum going to j-1 when you introduce the jth charge. Hence the sum over i<j.

All charges in the system will interact with each other. The correct expression would therefore be [math]V = \frac{1}{4\pi \varepsilon} \sum\limits_{i < j} \frac{q_i q_j}{r_{ij}}[/math] where [math]r_{ij}[/math] is the distance between charge i and charge j. The sum is over all pairs of charges.

You do not need to do a simulation or do any consideration of energies to see that your overall thrust is zero. For continuous "propulsion" your accelerated mass would have to have the same velocity every time it arrives at the accelerating zone. If this is not true, it will just mean that your device wobbles more and more every time the mass passes. So, assume that this velocity is v0 and that the accelerated velocity is v1. This gives us the following cycle: During the acceleration, the mass has gained momentum m(v1-v0) which has been transferred to the ship. The ship has thus received the momentum m(v0-v1). In the turn, the mass has velocity v1. Since it is turning around completely and direction of momentum matters, the mass gains momentum -2m v1, since it is going from velocity v1 to -v1 and therefore from momentum m v1 to -m v1. This momentum has been transferred to the ship which now has momentum m(v0-v1)+2m v1 = m(v0+v1). During the friction phase, the mass slows down to v0 again simply because this is the only place where it can slow down and it needs to be at this speed when it comes back to the acceleration phase. However, it is moving in the negative direction so its linear momentum goes from -m v1 to - m v0, meaning that the momentum change is -m(v0 - v1). Due to conservation of momentum, the ship receives the opposite momentum and now has momentum m(v0+v1)+m(v0-v1) = 2m v0. During the second turn, the mass momentum changes from -m v0 to m v0 and the momentum change is thus 2m v0. The ship receives the opposite momentum change and its total momentum is therefore 2m v0 - 2m v0 = 0. Conclusion: No net momentum is gained by the ship throughout one cycle of operation of your device. No amount of simulation can change this and if you want to argue otherwise you must find exactly where my argumentation is wrong. Since my argumentation is solely based on the conservation of momentum, finding something wrong with it is equivalent to arguing that momentum is not conserved. You may rewrite integrals as much as you like, but momentum conservation will remain and give you exactly the argumentation as presented above.

This depends on the equation of state for the dark energy (typically parametrized by w, the ratio between its pressure and energy density). If the dark energy is simply a cosmological constant the equation of state will be w = -1 and we would simply have a constant Hubble parameter. However, we do not know the equation of state for the dark energy and there are models which could accommodate w < -1, leading to a Big Rip where the Hubble parameter increases and eventually rips the Universe apart. With today's experimental observations of the acceleration of the expansion, we do not know whether or not w < -1, w = -1, or w > -1.

Since you have only one v I am here going to assume you are only considering the energy of the ship. Yes, the energy and momentum of the ship do change during the friction phase. However, the total energy and total momentum of the ship+ball system does not change assuming we also take into account heat energy in the ship that is being generated due to the friction. Whenever the ball is transferring momentum to the ship, the ball loses the very same momentum. This happens both in the friction area, the area where you accelerate the ball, and the curves (while the absolute value of linear momentum does not change here, the direction does - the vectorial difference is transferred to the ship). As long as the force accelerating the ball is not external to the entire contraption, the total momentum of ball+ship will not change. It is very possible that the momentum transferred during the friction segment is not of the same magnitude as that transferred during the acceleration segment. However, this difference will be exactly the same, but with opposite sign, as the difference of the momentum transferred in the two different curves. The end result will simply be generating heat in the device due to the friction.

Light bulbs emit photons because they are hot, not the other way around. The photon emission cools the bulb by carrying away energy. However, during normal operations the same amount of energy is added by electric work. Once the electric current stops the bulb (the glow thread to be precise) quickly cools down to a temperature where it is no longer emitting visible light.

What you suggest sounds very similar to the actual topological definition of continuity, for which it is not even necessary to have a metric space where epsilons and deltas have a meaning. As a student I admit to have been in awe of the beauty of topology as it allowed much more elegant proofs of many theorems. However, it is relatively abstract and I am not sure it is for everyone.

Let us assume that we do not know that the containers contain clips. Then it turns out when we do measurements the mass content of the containers come in multiples of 2.5 g. What would we conclude about the container contents? Compare to the Millikan experiment. We measure the charge of the oil drops and find that they come in multiples of 1.6e-19 C. What do we conclude about the oil drop charges?

It can and has been done. It is called a heat engine and can convert differences in temperature to mechanical or other forms of work.

Just to be complete: Separation speed *is* limeted to 2c as long as we stay in special relativity so that space itself cannot expand.