🔑:## Step 1: Define the four-vector and the Lorentz transformationThe four-vector for a particle is given by p^mu = (E, p_x, p_y, p_z), where E is the energy and p_i are the components of the momentum. The Lorentz transformation is defined as p^mu = Lambda^mu_{,,nu} p^nu, where Lambda^mu_{,,nu} is the Lorentz transformation matrix.## Step 2: Determine the Lorentz transformation matrix for a boost in the negative x-directionFor a boost in the negative x-direction with velocity v, the Lorentz transformation matrix is given by:[Lambda^mu_{,,nu} =begin{pmatrix}gamma & -gamma beta & 0 & 0 -gamma beta & gamma & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1end{pmatrix}]where gamma = frac{1}{sqrt{1 - beta^2}} and beta = frac{v}{c}, with c being the speed of light.## Step 3: Express gamma and beta in terms of E and pFor a particle, gamma = frac{E}{m} and beta = frac{p}{E}, where m is the rest mass of the particle and p is the magnitude of the momentum. However, in the context of the Lorentz transformation matrix, beta is directly related to the velocity of the boost, not the particle's momentum. The relation of beta to the particle's momentum comes from the definition of beta in the context of relativistic kinematics, where beta = frac{v}{c} and v is the velocity of the particle. In the transformation matrix, beta refers to the relative velocity between the two frames.## Step 4: Relate the matrix to the particle's rest frameIn the particle's rest frame, p^mu = (m, 0, 0, 0), where m is the rest mass of the particle. Applying the Lorentz transformation to this four-vector gives the four-momentum in the boosted frame: p'^mu = Lambda^mu_{,,nu} p^nu = (mgamma, -mgammabeta, 0, 0), which corresponds to E = mgamma and p_x = -mgammabeta in the boosted frame.## Step 5: Express the transformation matrix in terms of E and pSubstituting gamma = frac{E}{m} and beta = frac{p}{E} into the Lorentz transformation matrix, we get:[Lambda^mu_{,,nu} =begin{pmatrix}frac{E}{m} & -frac{E}{m}frac{p}{E} & 0 & 0 -frac{E}{m}frac{p}{E} & frac{E}{m} & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1end{pmatrix}=begin{pmatrix}frac{E}{m} & -frac{p}{m} & 0 & 0 -frac{p}{m} & frac{E}{m} & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1end{pmatrix}]The final answer is: boxed{begin{pmatrix}frac{E}{m} & -frac{p}{m} & 0 & 0 -frac{p}{m} & frac{E}{m} & 0 & 0 0 & 0 & 1 & 0 0 & 0 & 0 & 1end{pmatrix}}

🔑:## Step 1: Understanding Black Body RadiationBlack body radiation is a theoretical concept where an object at a certain temperature emits radiation across a wide range of wavelengths. The distribution of this radiation is described by Planck's law, which shows that the peak wavelength of the radiation shifts with temperature.## Step 2: Applying Wein's Displacement LawWein's displacement law states that the peak wavelength (λ_max) of the radiation emitted by a black body is inversely proportional to its temperature (T) in Kelvin: λ_max = b / T, where b is a constant. This means that as the temperature of the object increases, the peak wavelength of the emitted radiation decreases.## Step 3: Calculating Peak Wavelength at Room Temperature (20°C)First, convert the temperature to Kelvin: T1 = 20°C + 273.15 = 293.15 K. Using Wein's displacement law, we can find the peak wavelength at this temperature, but since the exact value of b (Wein's displacement constant, approximately 2.8977685(51) × 10^(-3) m·K) is not provided, we'll focus on the conceptual change rather than calculating exact wavelengths.## Step 4: Calculating Peak Wavelength at Elevated Temperature (50°C)Convert the elevated temperature to Kelvin: T2 = 50°C + 273.15 = 323.15 K. According to Wein's displacement law, the peak wavelength at this higher temperature will be shorter than at 293.15 K because λ_max is inversely proportional to T.## Step 5: Comparing the Distribution of RadiationAs the cell phone's temperature increases from 20°C to 50°C, the peak of the radiation distribution shifts to shorter wavelengths. This means the cell phone emits more radiation in the shorter wavelength (higher energy) part of the spectrum at 50°C than at 20°C. The overall radiation intensity also increases with temperature, as described by the Stefan-Boltzmann law, but the key point here is the shift in the peak wavelength due to the temperature increase.## Step 6: Relevance of Wein's Displacement LawWein's displacement law is crucial for understanding how the temperature of an object affects the radiation it emits. In this scenario, it predicts that as the cell phone heats up, the radiation it emits will peak at a shorter wavelength, indicating a shift towards more energetic radiation.The final answer is: boxed{Wein's displacement law shows that as the cell phone's temperature increases from 20°C to 50°C, the peak wavelength of the emitted radiation decreases, indicating a shift towards shorter wavelengths and higher energy radiation.}

🔑:## Step 1: Understand the phi^4 Theory ContextThe phi^4 theory is a quantum field theory involving a scalar field phi with a quartic interaction term. The Lagrangian density for this theory is given by mathcal{L} = frac{1}{2}(partial_mu phi)^2 - frac{1}{2}m^2phi^2 - frac{lambda}{4!}phi^4, where m is the mass of the scalar particle and lambda is the coupling constant.## Step 2: Identify the One-Loop Diagram for 4 Outgoing ParticlesThe one-loop diagram for a process involving 4 outgoing particles in phi^4 theory can be visualized as a loop with four external lines, each representing an outgoing particle. This diagram represents the process where two virtual particles interact to produce four real particles.## Step 3: Calculate the Number of Inequivalent PermutationsTo calculate the number of inequivalent permutations of the external legs, we need to consider the symmetry of the diagram. For a diagram with 4 external legs, if all legs were distinguishable and the diagram had no symmetry, there would be 4! ways to arrange them. However, in phi^4 theory, the particles are identical bosons, and the diagram has a symmetry due to the possibility of relabeling the vertices.## Step 4: Account for Symmetry and Avoid OvercountingGiven the symmetry of the interaction (all particles being identical) and considering the one-loop diagram, we recognize that the permutations of the external legs that result in the same diagram (due to rotational symmetry or reflection) should not be counted multiple times. For the one-loop diagram with 4 external legs, there's essentially one unique way to connect the legs to the loop, considering the particles are indistinguishable.## Step 5: Consider Vertex RelabelingIn the context of phi^4 theory, vertex relabeling does not change the diagram's topology or the physical process it represents. Since the particles are identical and the vertices are indistinguishable in terms of the particles they connect, we do not need to further adjust our count for vertex relabeling in this specific step of calculating permutations of external legs.The final answer is: boxed{1}

🔑:Based on the symptoms described, the most likely diagnosis for this individual is Transient Global Amnesia (TGA). TGA is a rare and temporary condition characterized by a sudden and transient episode of memory loss, typically lasting from a few hours to a few days.The typical characteristics of TGA include:1. Sudden onset: The memory loss begins suddenly, often without warning, and can be triggered by physical or emotional stress, medical procedures, or other factors.2. Short-term memory loss: The individual is unable to form new memories, making it difficult to recall recent events, conversations, or tasks.3. Difficulty accessing older memories: The individual may also experience difficulty recalling older memories, although this is typically less severe than the short-term memory loss.4. Preserved self-identity and autobiographical memory: Despite the memory loss, the individual typically retains knowledge of their self-identity, including their name, age, occupation, and relationships with close family members.5. Intact perceptual skills: The individual's perceptual skills, such as vision, hearing, and spatial awareness, remain intact.6. Preserved complex learned behavior: The individual can still perform complex tasks, such as playing a musical instrument or driving a car, which suggests that their procedural memory is intact.7. Mental alertness and lucidity: The individual appears mentally alert and lucid, with no signs of confusion, disorientation, or altered consciousness.8. No other neurological symptoms: TGA is typically not accompanied by other neurological symptoms, such as seizures, weakness, or sensory deficits.Other key features of TGA include:* The episode is usually self-limiting, with memory function returning to normal within 24 hours.* The individual may experience a "fugue state," where they repeatedly ask the same questions or exhibit repetitive behavior due to their inability to form new memories.* TGA is more common in middle-aged and older adults, with a peak incidence in the 50-70 age range.* The exact cause of TGA is still unclear, but it is thought to be related to temporary disruptions in blood flow to the brain, particularly in the hippocampus and temporal lobe.It's essential to note that TGA is a rare condition, and other causes of memory loss, such as stroke, traumatic brain injury, or neurodegenerative diseases, should be ruled out through thorough medical evaluation and diagnostic testing.