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The natural choice is actually $\mathcal{L}=\text{something}$, the reason being is that the $\sqrt{-g}$ term is naturally paired with the volume form $d^4x$. Even before considering curved spacetime, consider non Cartesian coordinates. For example spherical coordinates $$dt\,dx\,dy\,dz = r^2 \sin\theta\ dt\,dr\,d\theta\,d\phi$$ Where does that term $r^2 ... 1 If I take a mass$m$and apply a force$F$(greater than$mg$) to it for a distance$h$upwards then I will do work of: $$W = Fh \tag{1}$$ The force$F$has to be greater than the force due to gravity,$mg, or the object won't move upwards, so let's write the force I apply to the mass as: $$F = mg + F'$$ then equation (1) becomes: \begin{align} W ... 1 You might want to change your question title to "Work done by gravity," because that is what is implied by the variables mgh. Of course, you can add a greater force than that of gravity, which would cause whatever object to which the force is applied to accelerate (since the forces are not in balance). No matter how much force you apply, the force of ... 1 It depends on where you are standing. A convex surface is one that bulges out towards the person who is talking about it. Since we do not live inside glass, everyone knows what a bi-convex glass lens is - one that bulges out on both sides as seen by someone who lives in air. Note that while it bulges out towards the incident ray arriving from the air, the ... 0 How would you define a convex surface? I'd call a set \mathcal S (of five or more elements) "convex" if for each element A \in \mathcal S any two circumspheres of A and any triples of distinct elements of \mathcal S other than A have the property that the center of the smaller circumsphere is inside the larger circumsphere; in other words: ... 1 In your example #1, you need to remember that mass is conserved. The stream of exhaust left behind by the rocket grows in mass by exactly the rate the rocket consumes fuel. In your example #2, there's often a desire to make constants positive. If that constant is actually negative in your interpretation of the problem, you need to add a minus sign. In your ... 5 It's a matter of history. When George Stoney developed Stoney units in 1881, or when Robert Millikan performed the oil drop experiment in 1909, it wasn't yet known that it was possible for anything to have a charge smaller in magnitude than the charge of an electron. By the time the quark model was proposed, in 1964, the use of the "elementary charge" being ... 4 If the particle moves from the point x to x+dx, and assume dx\gt 0 for simplicity, then its potential energy increases by dU = \frac{dU}{dx}dx $$Well, it increases if dU is positive and decreases if dU is negative. So far I have only used the definition of the derivative – pure mathematics. However, the total energy is conserved. The sum of ... 1 The Lorentz force on a charge in an electromagnetic field is$$F=q(E+v \times B) \ \ .$$For an electron between the cylinders, q is negative, and E is defined as pointing outward, so the electron will experience a force radially inward. But due to the unfortunate sign convention used for currents, electrons flowing inward means that the conventional ... 2 Yes - electrons flow from the negative to the positive, so in the opposite direction to the conventional direction of the electric field (which points from positive to negative). So if the E field points outwards, the electrons flow from the outer to the inner cylinder. The direction does not affect the answer (the calculation of the flow) though - at least ... 2 If you say that the acceleration of gravity is towards the ground and positive, then you must have distance increasing in that direction as well - so top of building is zero, and ground is 50. In that case$$y(t) = y(0) + v_0t + \frac12gt^2\\ 50 = 0 + 4.9 t^2$Or you say that the vertical direction is "up is positive"; then the acceleration of gravity is ... 2 You want$X_i=-50$. The ground is zero, down is positive, so the top of the building is at$-50$. There's no universal convention. You're stuck figuring it out from scratch each time. Fortunately once you do it several times you'll get the hang of it. 0 Yes, the reason for this is that in maths$\nabla$is often used as the vector differential opertator. This is a vector. When typesetting the convention to denote a vector is bold text e.g.$\bf{x}\$. However for handwriting you can't really write bold font so other conventions are needed. Common ones are putting an arrow over as in your example or ...