🔑:## Step 1: Define the 4-momentum of a photonThe 4-momentum of a photon is given by p^mu = (E/c, mathbf{p}), where E is the energy of the photon, c is the speed of light, and mathbf{p} is the momentum of the photon. For a photon, |mathbf{p}| = E/c.## Step 2: Express the 4-momentum of the two photonsLet's denote the energies of the two photons as E_1 and E_2. Their 4-momenta can be expressed as p_1^mu = (E_1/c, E_1/c, 0, 0) and p_2^mu = (E_2/c, -E_2/c, 0, 0), assuming they are moving along the x-axis in opposite directions.## Step 3: Calculate the total 4-momentumThe total 4-momentum P^mu is the sum of the 4-momenta of the two photons: P^mu = p_1^mu + p_2^mu = ((E_1 + E_2)/c, (E_1 - E_2)/c, 0, 0).## Step 4: Determine the velocity of the center of massThe velocity of the center of mass can be found from the total 4-momentum. The velocity v is related to the energy and momentum by v = pc^2/E, where p is the momentum and E is the energy of the system. For the center of mass, we use the total energy and total momentum.## Step 5: Calculate the velocity of the center of massThe total energy E_{total} = E_1 + E_2 and the total momentum p_{total} = E_1/c - E_2/c. Thus, the velocity v of the center of mass is given by v = frac{p_{total}c^2}{E_{total}} = frac{(E_1 - E_2)c}{E_1 + E_2}.## Step 6: Simplify the expression for velocityThe expression for v simplifies to v = frac{E_1 - E_2}{E_1 + E_2}c. This gives the velocity of the center of mass in terms of the energies of the photons and the speed of light.The final answer is: boxed{frac{E_1 - E_2}{E_1 + E_2}c}

🔑:## Step 1: Understanding the ProblemThe problem asks us to determine the direction in which the electric field does NOT point around a uniformly charged circular disk, using cylindrical coordinates. This involves understanding the symmetry of the disk and how the electric field behaves around it.## Step 2: Symmetry of the DiskA uniformly charged circular disk has rotational symmetry about its central axis (the z-axis in cylindrical coordinates). This means that the electric field at any point around the disk will be the same as at any other point that is rotated about the z-axis by any angle.## Step 3: Electric Field Near the DiskNear the disk, the electric field lines will emanate from the surface of the disk, perpendicular to the surface. This is because the electric field is generated by the charges on the disk, and the direction of the field is determined by the direction of the force that a positive test charge would experience if placed near the disk.## Step 4: Electric Field Far from the DiskFar from the disk, the electric field behaves as if the disk were a point charge located at its center. The field lines will radiate outward from this point, decreasing in intensity with distance. The direction of the field will be radially outward from the center of the disk.## Step 5: Determining the Direction the Electric Field Does Not PointGiven the symmetry of the disk and the behavior of the electric field, the direction in which the electric field does NOT point around the disk is parallel to the surface of the disk, in the direction of the disk's circumference. This is because there is no charge component in this direction (tangential to the disk's surface) that would generate an electric field. The electric field lines will either be perpendicular to the disk (near the disk) or radially outward from the disk's center (far from the disk), but not tangential to the disk's surface.The final answer is: boxed{tangential to the disk's surface}

🔑:## Step 1: Understanding the Basics of Gravitational Force and Relativistic EffectsThe gravitational force of an object is determined by its mass and the distance from the center of the object. According to general relativity, any form of energy contributes to the gravitational field, not just rest mass. When gas is heated, its molecules move faster, which means their kinetic energy increases.## Step 2: Considering the Role of the Stress-Energy TensorIn general relativity, the stress-energy tensor (Tμν) describes the distribution of mass and energy in spacetime. It includes contributions from all forms of energy, including kinetic energy. When the gas is heated, the increase in kinetic energy of the gas molecules contributes to the stress-energy tensor.## Step 3: Relativistic Effects on Gravitational ForceFrom a relativistic perspective, the increased kinetic energy of the heated gas contributes to its inertial mass, which in turn affects the gravitational field. This is because, according to special relativity, the total energy (including kinetic energy) of an object contributes to its relativistic mass, which is the mass that is relevant for gravitational interactions.## Step 4: Applying the Concept to the Sphere of GasFor a sphere filled with gas, heating the gas increases its internal energy, primarily in the form of kinetic energy of the molecules. This increase in energy contributes to the overall stress-energy tensor of the sphere, which in turn affects the gravitational field of the sphere.## Step 5: Conclusion on Gravitational Force IncreaseGiven the principles of general relativity and the role of the stress-energy tensor, the gravitational force of the sphere does indeed increase when the gas inside it is heated, due to the increased kinetic energy of the gas molecules contributing to the sphere's total energy and thus its gravitational field.The final answer is: boxed{Yes}

🔑:## Step 1: Understand the nature of magnetic forces and their relation to velocityMagnetic forces are a result of the interaction between moving charges. For two positively charged objects to attract each other due to magnetic forces, they must be moving relative to each other. The magnetic force between two moving charges is given by the Lorentz force equation, which includes both electric and magnetic components. However, for the attraction to occur due to magnetic forces, we need to consider the conditions under which the magnetic force can overcome the repulsive electric force between like charges.## Step 2: Recall the Lorentz force equationThe Lorentz force equation for a charge (q) moving with velocity (vec{v}) in an electric field (vec{E}) and magnetic field (vec{B}) is given by (vec{F} = q(vec{E} + vec{v} times vec{B})). For two charges moving relative to each other, we consider the magnetic field generated by one charge and its effect on the other.## Step 3: Consider the magnetic field generated by a moving chargeA moving charge generates a magnetic field. The magnitude of the magnetic field (B) at a distance (r) from a charge (q) moving with velocity (v) is given by (B = frac{mu_0}{4pi} frac{qv}{r^2}), where (mu_0) is the magnetic constant (permeability of free space).## Step 4: Determine the condition for magnetic attractionFor two positively charged objects to attract each other, the magnetic force must overcome the electrostatic repulsion. The electrostatic force between two charges (q_1) and (q_2) at distance (r) is given by (F_E = frac{1}{4piepsilon_0} frac{q_1q_2}{r^2}), where (epsilon_0) is the electric constant (permittivity of free space). The magnetic force (F_M) between two moving charges can be attractive if the charges are moving in opposite directions or in the same direction but with a component of motion perpendicular to the line joining them.## Step 5: Derive the condition for minimum velocityTo derive the minimum velocity required for attraction, we equate the magnetic and electric forces. However, since the magnetic force depends on the velocity and the electric force does not, we look for a condition where the magnetic force becomes significant enough to counteract the electric force. The magnetic force on a charge (q) due to a magnetic field (B) is (F_M = qvB). Setting (F_M = F_E) and substituting (B) from the expression for the magnetic field generated by a moving charge, we get (qv cdot frac{mu_0}{4pi} frac{qv}{r^2} = frac{1}{4piepsilon_0} frac{q^2}{r^2}).## Step 6: Solve for velocitySimplifying the equation to solve for (v), we have (v^2 = frac{1}{epsilon_0mu_0}). Recognizing that (frac{1}{sqrt{epsilon_0mu_0}} = c), where (c) is the speed of light, we find (v^2 = c^2), and thus (v = c).## Step 7: Interpret the resultThe minimum velocity required for two positively charged objects to attract each other due to magnetic forces is equal to the speed of light. This result indicates that for magnetic forces to overcome electrostatic repulsion between like charges, the charges must be moving at relativistic speeds. However, it's crucial to note that reaching the speed of light is impossible for objects with mass, according to special relativity. This derivation highlights the theoretical threshold rather than a practical possibility.The final answer is: boxed{c}