studiot

Again I didn't say that the man on the roof has no potential energy. What I said was that applying your statement means he does not. So let us re-examine your statement.

Yes the OP starts from a misconception and several have tried to explain this in different ways. A point has no length, of any value.

Thank you. But the question is about what happens after annihilation. Perhaps I made the scenario too dramatic. If I just said that the charges are moved further apart, no one would bat an eyelid about the signal that ensues. But the question of what happens to the energy in the field, before the signal arrives, still remains unanswered.

What bearing does this have on the question posed in the OP?

The ancient Greeks thought long and hard about this question and they came up with a notion that, when put in modern terms boils down to the idea of dimension. Euclid definition 1 A point is that which has no part Euclid definition 2 A line is a breadthless length Euclid definition 3 The extremities of a line are points (John made this point) Euclid definition 15 A circle is a plane figure contained by one line such that all the lines falling upon it from one point are equal. A modern translation of the above is that you require zero dimensions for a point, one dimension for a line and two dimensions for a circle. Extending that you require three dimensions for a sphere. The corresponding figure for dimensions greater than three is called a ball or an n-ball, and sometimes a ball is used in three dimensions.

Thank you MigL, you have encapsulated the question I posed exactly. +1 Indeed if you only annihilate one of the charges, this still must change the field and the question of what happens to the energy in the field (if there is any) before the 'change' signal has arrived from the destroyed charge.

I don't think it's as simple as that. When you are standing on the roof you are subject to a reaction force from the roof, of opposite direction to gravity. Applying your logic above, you would be subject to a negative potential equal and opposite to the gravitational potential., making your net potential zero. Potential and potential energy are not necessarily the same thing. Just as the reaction force and the gravitational field are not the same.

OK let us examine what you said in post19 more closely. I have underlined the bit which does not make sense. Any body which is stationary under the opposing action of two (or more) forces neither gains nor looses energy. Now you said that both airfoils (the free one and the constrained one) gain energy. Again I have underlined the part which is at variance with basic mechanics, for the same reason. It is quite simple. If the body does not move its energy does not alter, apart from any heating or cooling which is not determinate. This exemplifies the difficulty that you can get into with an energy analysis, rather than a force analysis, or mixing the two. It can be done but you have to be careful.

Because if it is truly STATIONARY it moves neither up nor down, nor at all.

I have no idea what you are studying or at what level. Geophysics is a college or university subject, not a school subject. Here is an extract from Kearney & Brooks An Introduction to Geophysical Exploration. We can discuss this list if you like, but only if you are prepared to answer questions, but as you can see, they have a different list of possibilities for your application. You have not answered my question about cost.

Well I've wasted enough time on this one sided conversation. If you have difficulty with English tell the next person who tries to help you before she starts.

If the airfoil is truly STATIONARY what form of energy do you think it gains? It obviously can't be kinetic or potential energy. True it may gain some heat due to friction, but it may loose heat by convective cooling.

I don't see you putting much in. Here is another hint, and the last one without some posted effort from you. How does the resolution of a seismic survey vary with detector spacing?

How would you conduct a resistivity survey to show the ditch?

Well homework help is meant to help you do some work. You have listed 5 methods but have not thought about any of them, so i will start you off. Will Radionucleide methods show anything without some digging? If you can dig, will you not find the trench anyway? So what do you think about each method? Is cost important?

What ditch?

Given the shape of your 'foil' I suggest you look at standard wind loading codes for buildings, as it is like a pitched roof. This will give you both the uplift and drag forces on your shape. I'm sorry I don't know these for Texas, only England. https://www.google.co.uk/search?hl=en-GB&source=hp&q=wind+load+calculator&gbv=2&oq=Wind+l&gs_l=heirloom-hp.1.0.0l7j0i10j0l2.1531.3953.0.7468.6.6.0.0.0.0.110.594.4j2.6.0.msedr...0...1ac.1.34.heirloom-hp..0.6.594.sz0Y8pfMcgA

You will have to ask MigL exactly what he meant. A body may, on occasion, extract temporary energy from the air, but all non powered 'heavier than air' bodies eventually fall back to the ground. The path of a glider is basically a longer or shorter fall to ground. The fall may be extended by temporary 'borrowing' energy from some fortuituous local pressure difference and/or or air current. A leaf may be blown up in the wind, but it returns the energy to the fluid by displacement when it falls back to earth. For sustained, heavier than air, flight the power comes from the engines which develop thrust in some form. You should be careful and not think in horizontal and vertical terms because the forces may well not be horizontal/vertical. There are three main forces involved in a heavier than air craft. Weight, which always acts vertically. Thrust which always acts in the direction of motion (the craft may be climbing or descending) The force exerted by the fluid on the craft. This force is usually resolved into two components The drag which acts in the same line as the thrust but in the opposite direction The lift which is perpendicular to the drag, in the opposite sense to the weight. In level flight only the lift is vertical and the drag is horizontal. After resolution into lift and drag there four forces acting. This the usual description. However the four forces do not, in general, meet at a point. This means that there is also a residual moment to consider. The moment is countered by the air force on the subsidiary planes, particularly the tailplane. So there are five forces to consider. Finally the weight is actually distributed and measures to pump fule and other fluids about are used to maintain trim. As to your calculation, I have no idea from your sketchy figures and this site is not a do your calculations for you site.

Ultimately, yes. But in pulling the plates apart your pull is separating the charges against the electric field, thereby doing electrical work.

Does it? When the aircraft has passed by is the fluid any different from before the aircraft arrived? In what way does it have less energy?

Aminoxyl I should check the stochiometry of your correction equation in post#5. Although you did not start the thread, you seem interested, did you understand my entropy comment?

Have you considered the entropy changes between the reactants and the products?

Not more plants but bigger plants!

Two recent event would offer a powerful counterargument. In France multiple ordinary members of the public, including the your paramilitary police (who after all are really drawn from ordinary members) Some terrorists died. In Belgium No ordinary members suffered. Some terrorists died. I am not sorry to see the back of any of those terrorists.

No you need observed reality. All our theories, without exception, are incomplete models that do not explain everything. However you make a good point that perhaps classical field theory is inadequate to handle the situation. Can this happen outside the relativistic horizon?