New answers tagged

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Schwartz uses negative signature, so $(-i\gamma^2)^2=1$.


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The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious. The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change. If the volume does change, then either the pressure must change or the temperature must change. In the "...


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In some cases there are obvious reasons. The particle $J/\psi$ was named by Samuel Ting and Burton Richter, separately. You can check the first letter of the Chinese ideogram for the name of Samuel C.C. Ting A Lax pair is a pair of time-dependent matrices or operators that satisfy a corresponding differential equation, called the Lax equation. They ...


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So you don't confuse $G$ with grams I expect.


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It's just a matter of notation, you can use whatever symbol you want to denote them. Historic reasons are also taken into account I guess.


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The physics and chemistry examples you gave describe two different processes. In the physics example, the gas is subjected to an isothermal reversible (quasi-static) expansion, where the gas pressure and the external pressure decrease very gradually. For this case, using the ideal gas law to determine both the gas pressure and the external pressure are ...


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Both versions of the first law are valid. In engineering the version is Δ𝑈=𝑄−𝑊. The Chemistry version is Δ𝑈=𝑄+𝑊. In physics it varies. Both versions give the same magnitude of work done. The difference is the sign of the work is assigned a positive value in the version if the work is done by the system (expansion), and the sign is negative for the ...


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Both the equations are right,the second equation is just a special case of the first equation(when pressure is a constant, we note $\frac{RT}{V}$ is also constant and it comes outside the Integral sign which leaves us with the equation -A result from the ideal gas equation $P(V_2-V_1)$). Hope this helps


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Your first set of $\sigma_a$ satisfies the standard angular momentum algebra, $$[\sigma_a, \sigma_b] = 2 i \varepsilon_{a b c}\,\sigma_c .$$ Your second set (name it differently!!), $\Sigma_a: ~~ \{\Sigma_1=\sigma_2, \Sigma_2=\sigma_1, \Sigma_3=\sigma_3\}$, satisfies an algebra with the opposite structure constants, $$[\Sigma_a, \Sigma_b] = -2 i \...


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Under the convention that $1 \leftrightarrow x, 2 \leftrightarrow y, 3 \leftrightarrow z$, the ordering we have ties to the actions performed about $x, y, z$ axes. In the context of Pauli's work, $σ_k$ represents the observable corresponding to spin along the $k^{\text{th}}$ coordinate axis in three-dimensional Euclidean space $\mathbb{R}^3$. Also, the ...


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We say that angular momentum is perpendicular to the plane of rotation (in a simple case like a rotating disk) only because Josiah Willard Gibbs and Oliver Heaviside popularized vector algebra (including a vector product using the right-hand rule) and vector calculus in the early 1900’s. There are more modern formalisms — not generally taught in high school ...


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1) If the body is moving in a plane then it's position and velocity/momentum vectors are in that plane too. This is easy to verify. 2)Angular momentum is a cross product of position and momentum 3)A cross product is always perpendicular to the plane in which the vectors to be crossed are. 4) We conclude that the angular momentum is perpendicular to the ...


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The set of points unaffected by rotation forms a line, the axis of rotation, which is perpendicular to the plane of rotation. One can take this axis as defining the direction of angular momentum. However, AM, is a pseudovector . It is in reality an antisymmetric tensor and such objects have 3 components in 3D. Coincidentally this can be treated as something ...


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As with all mathematical conventions in physics, the reason why we represent something a certain way is because it is useful. Angular velocity is a pseudovector, so its direction is defined by the axis about which an object rotates, which makes the angular velocity vector normal to the plane of rotation. Angular momentum is defined by $$\vec L = I\vec\omega$...


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It depends on the field. There is an ISO standard to use a from latin annus, but in astronomy and cosmology yr is far more normal. I think the same is true in geology and paleontology. The metric symbols should be capitals, M not m (which stands for milli-, $1/1000$). b is deprecated for billions, use G. ya is often used for years ago. I use Myr, Mya, Gyr, ...


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It's Ma for million years and Ga for billion years.


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Another approach would be - adapted to the case of symmetric matrices in the OP but also valid for general symmetric arrays as well as in many other contexts - to consider the space $S^2_n$ of $n\times n$ symmetric matrices, viewed as the coordinate space $S^2_n=\mathbb R^{n(n+1)/2}$ with standard coordinates $$(s^{ij})_{i\le j}=(s^{11},s^{12},...,s^{1n},s^{...


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There is really very little standardization of this kind of thing. Various textbooks have various conventions about all kinds of things related to phasors and the complex number representation of sine waves. Some say that phasors rotate, others consider them as static objects. Some say that phasors are complex numbers, others avoid mentioning complex numbers ...


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A phasor is a rotating vector. If, at time t it makes angle $(\omega t + \phi)$ with the x axis, its projection (for a phasor of unit magnitude) on to x axis is, by definition, $\cos(\omega t + \phi)$, and its projection on to y axis is $\sin(\omega t + \phi)$. Which of these we choose to represent a sinusoidal displacement, voltage or whatever is pretty ...


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Consistent notation is needed for clear communication, but there is no one "correct" notation for anything. If you can remember what things mean in your own work, you can do whatever you want. If you are writing something that others must read too, like a graded homework assignment, it is important to define any notation. That being said if you clearly ...


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Sorry for answering my own question, i just noticed a really simple solution for my question. The answer is not about the physics itself, but absence of mind , The current indeed decides the direction of magnetic field I still have a doubt on the conclusion that whether the Area vector convention is strictly necessary for the thing which prof. Lewin ...


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With the right hand thumb in the direction of positive current, the fingers wrap around in the direction of the magnetic field. The field comes out of the north (seeking) pole of a magnet or current carrying solenoid. (When working with a vector (cross) product, hold your right hand so that you can curl your fingers from the first named vector toward the ...


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In a winding, use a right hand grip rule (thumb stuck out). Conventional current flows in the direction of the fingers, then magnetic field within the winding is in the direction of the thumb. The rule also works for magnetic field around a current. Thumb in the direction of current, fingers curl in the direction the field goes. To test whether a pole is ...


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When considering Lorentz force, right hand rule is for negative charges, so electrons. Left hand rule is for conventional current(notice that current is always just pointing in the opposite direction to velocity in the right hand side, which is like the derivation of conventional current that positive charges going one way is the same as negative charges ...


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The simple form of Faraday's law is $\mathcal E = - \dfrac{d\phi}{dt}$ where $\mathcal E$ is the induced emf in a loop and $\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux though the loop. In the left-hand diagram the magnetic flux is $BA$ and it is increasing so the induced current, $I_{\rm induced}$, must try and produce a downward ...


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It is not a convention.It just means that induced emf opposes the magnetic flux passing thriugh the closed loop. Emf can be negative because it is defined as the amount of work done by an external mechanism on a unit charge to displace it from one position to another.


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Emf can't be negative. Actually the negative sign in Faraday's law show that the emf is induced in such a way that it opposes the cause of it. Suppose you have a close loop of a conducting wire and you place it in a region where magnetic field is increasing perpendicular to the plane of the wire. So according to the Faraday's law, the induced emf or ...


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I know I'm very late to this question, but I thought I would add some maths to explain it for anyone that's interested. As every other answer the sign of $g$ depends on which coordinate system you are in. Normally, we choose the coordinate system to have it's $x, y$ axes parallel the ground and the $z$ axis perpendicular to those and pointing away from the ...


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Well smaller stars can transfer there heat quicker to their surface and this means the temperature gradient can be greater than the adiabatic gradient. So the forces of temperature and heat transfer takes over and convection happens. This convection process actually prolongs the life of the red giant. Red giant is going to burn same amount of hydrogen our ...


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With your definition of θ, The Biot-Savart formula reduces to (km)I/R times the integral of sin(θ)(dθ) which integrates to the -cos(θ) and evaluates to (1+1) at the limits 0 to 180 degrees.


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If the text says calculate the potential energy, it means that calculate the potential energy difference from the choice of you reference point. Taking the following example where I choose $y_0$ to be my reference point. $$\int_{U(y_0)}^{U(y)}dU=\int_{y_0}^{y}mgdy$$ $$U(y)-U(y_0)=mg(y-y_0)$$ Now, if I choose $U(y_0)$ to be $0$, its the matter of convenience,...


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Angular momentum is given as $\textbf{L}=\textbf{r}\times\textbf{p}$. Thus, we obtain that torque, defined as $\frac{d\textbf{L}}{dt}$ is $\textbf{r}\times\frac{d\textbf{p}}{dt}=\textbf{r}\times\textbf{F}$. As you know, the $\frac{d\textbf{r}}{dt}\times \textbf{p}$ factor vanishes because it is the cross product of two parallel vectors. Now, the reason as ...


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The formula for torque is ultimately set by an arbitrary choice in the way that we define angular velocity. We have chosen to define angular velocity according to the right-hand rule - in other words, we have arbitrarily chosen that counterclockwise motion corresponds to an angular velocity vector pointing upward. Defining angular velocity this way means ...


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The order of indices is indeed important, but we can define tensors with inverted indices if we wish: for instance, the tensor $A_{ij}$ is in general different from $A_{ji}$, but if you are given $A_{ij}$ you can define $B_{ij} = A_{ji}$. This is somewhat like defining the transpose of a matrix: it is a different matrix, but still a valid one. So, the ...


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All of the answers about vectors and matrices seem technically correct, but I am not sure that that is what is being asked here. If I understand the question correctly, the short answer is "it is entirely arbitrary". However, there is a convention. If you only have one direction then it is conventionally labeled "the x axis". It is conventionally drawn ...


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Kilowatt-hours without time information are useless "1 kW⋅h" does not give any information about how much time was needed in order to transform this amount of energy. It could have been either 1000W during an hour or ~114mW during a year. Similarly, my household could also consume 12 gigawatt-hours worth of electricity. It would just need 4000 years to do ...


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I think it would help to read up on the definition of a vector. Velocity, Acceleration, Force, and many other quantities in Physics are described by vectors. From a geometrical point of view most texts begin with the idea of a directed line segment. Just as we have the idea of a line segment from points P to Q, where only the length has meaning, we ...


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The 3D Cartesian Axes are usually defined as Z 'upwards' as positive Z direction, Y 'right' as positive Y direction and X 'out-of-the-page (towards you)' as the positive X direction. 2D obviously the Y and Z axes swap, there is no Z axis, and the X axis 'becomes the 3D Y axis' in effect. Each axis is independent from each other so have their own positive ...


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In general velocity is a vector. In 1D motion we can get away with saying the velocity is positive or negative by associating each sign with a direction. However, once you move in more dimensions you can't say the velocity is positive or negative. You just have the vector (in Cartesian coordinates) $$\mathbf v=v_x\,\hat x+v_y\,\hat y+v_z\,\hat z$$ Of ...


3

The way you phrase the question is slightly ambiguous, it doesn't double count units per time. One is per time, the other is $\times$ time. And therefore cancels to get joules. However just using joules, you get no idea on the time period that amount of energy was used, just the amount. Using $kW\cdot h$ gives you a reference for how long you used an amount ...


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There is historical reason behind using kWh as well as matter of general convenience. Many people around the world pay their electricity bills using the "per unit of electricity" concept that energy provider companies have defined as 1kWh. Generally, the supplier has a base charge/connection charge for a certain number of electricity units (say "$b$"). Any ...


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Practically speaking, maybe you want to know how much energy it takes to keep your refrigerator running for one hour. On the back of a fridge you can find the power in watts. If you measure energy in units of $kW\cdot h$, it's easier to figure out how long you can keep it running. You can call it the "natural unit for a thrifty man".


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The agent is always forcing the unit mass with a continuously changing Force, $\vec{F}$(x) ... = $\frac{GM}{x^2}\hat{x}$ By your force definition, the agent is not the attractive gravitational force but is something which is restricting the motion to constant velocity because the mass M is pulling in the $-\hat{x}$ direction with a force equal in magnitude ...


1

You made a mathematical error in trying to prove the result. It arises in many scenarios. To give you an insight into your mistake I would like to tell you the correct method of integration in physics. Remember that we always consider an element $dx$ at a distance $x$ from origin in the direction of $x$. What you did was physically correct but while ...


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Your calculation looks like you want to calculate the work done by gravitation. In that case, you're missing a sign. Newton's law of universal gravitation as a vector equation is actually ($m=1$ in your example): $$\vec F(\vec x) =- G {M m \over {\vert \vec x \vert}^2} {\hat{x}}$$ The missing sign perpetuates through the whole calculation, leading to the ...


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By definition of electric potential energy $$dU=-dW_{electric}$$ In general potential energy is defined as $$dU=-dW_{int,cons}$$ The above equation says that the change in potential energy of a system is defined as the negative of work done by the internal conservative forces of the system In fact, $$dV=\frac{-dW_{elect}}{q}$$ If you solve the above ...


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A unit positive charge will have a force on it due to the electric field in the direction of the electric field. In moving in that direction from position $A$ to position $B$ the electric field does work on the positive charge $\displaystyle \int ^B_A E(x)\, dx$ and this is a positive quantity. By definition the difference in potential in going from ...


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I hope you know about the components of forces... if not read this part : Any vector quantity can be separated as sum of two different vectors. That means a force applied along North East direction can be seen as two forces, of same strength one along East and the other along North acting together. (If one of them was stronger, the net force would not be ...


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It will be hard not to use any calculus to derive the equation, by I will try my best to give you reasoning why is it so. Work is defined as force component in line with displacement direction. That's where the $cos\theta$ term is coming from. It may be easier to see from this picture why is the case: Our direction of displacement denoted by $\Delta\vec s$...


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If you apply a force on a body at θ to its displacement, magnitude of velocity(speed) only changes when you apply a acceleration parallel to its displacement vector ( v = ds/dt) so, here acceleration according to your equation is a cosθ. When you put this acceleration to your equation and multiply both side by mass gives you the same result as work energy ...


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