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Despite what some of the other answers are mentioning, the following equation you have is correct $$\vec{r} \cdot d \vec{r} = r dr$$ You can check this by noting $$\vec{r} = x {\hat i} + y {\hat j} + z {\hat k} \implies d \vec{r} = d x {\hat i} + d y {\hat j} + d z {\hat k}$$ Then $$\vec{r} \cdot d \vec{r} = x dx + y dy + z dz$$ Further note $$r = ... 3 You can actually infer the difference of the approaches just by looking at their names. A vector has direction and an energy (a scalar quantity) does not. Therefore, when you are trying to figure out scalar quantities such as distance and speed, you may find energy method more advantageous; when you look for velocity, acceleration, you have to use vector ... 3 Well, a good example is thinking in term of components. In several areas of physics, the math gets more intuitive when you think in terms of components of the vectors. So, instead of writing the vector \mathbf r for the position of a particle, you write x^i as the i-th component of a vector. The i in the top is to indicate a contravariant vector, ... 2 Physicists tend to be a bit casual about sign conventions when it seems to be obvious. So let's attempt to be completely rigourous. The key step is getting the flight time t since the range is just v\cos\theta\, t. We do this using the SUVAT equation:$$ v = u + at $$We'll use the usual conventions that up and right are positive, so v_y and v_x ... 2 I've tried to do it the following way, but I don't know if I can use the product rule as such when matrices and vectors are involved. \begin{equation*} U'=\frac{dX'}{d\tau}=\frac{d}{d\tau}\Lambda{X}=\frac{d\Lambda}{d\tau}X+\Lambda\frac{dX}{d\tau}=\Lambda\frac{dX}{d\tau}=\Lambda{U} \end{equation*} since the Lorentz transformation matrix ... 2 The sign of a scalar depends upon the scale with which it is measured. In the case of temperature, a Fahrenheit scale arbitrarily says 32 degrees is the freezing point of water, and zero degrees is a mark on a scale of numbers. Any temperature less than zero has a negative sign. The Celsius scale says zero is the freezing point of water, so temperatures ... 2 Torque \vec{\tau}(\vec{r}) is an vector multiplication of radius-vector \vec{r} on applied force vector \vec{F}, i.e. \vec{\tau}=[\vec{r}\times\vec{F}] Here the radius-vector (or position vector) \vec{r} is the vector from the point where the torque is defined to the point where the force is applied (see image). On the picture you shown ... 2 What I have done is to add the 569\;\mathrm{N} vector with μ_kn and finding the horizontal component of that resultant vector. Something's wrong here. The resultant force vector should not have any other components than the horizontal one - otherwise the object should be moving vertically also, which it doesn't. In the sentence here it seems that ... 1 The mistake is in the step where you go from$$\vec{r}.\frac{d\vec{r}}{dt} = r\frac{dr}{dt}$$to$$\vec{r}.d\vec{r} = r.dr$$The integrand is time dependant (and involves a dot product as well) and hence the result is non-trivial. You simply cannot "cancel" off the times as you have done. 1 Instead of r\, dr \cos \theta = r\, dr, that line should read r \, ||{d\vec{r}}|| \cos \theta = r\, dr. Since ||d\vec{r}|| \neq dr, the argument does not follow. If you are not sure why ||d\vec{r}|| \neq dr, ask yourself whether ||\frac{d\vec{r}}{dt}|| = \frac{dr}{dt}. 1 Start by rewriting the scalar product as a covariant-contravariant contraction, like so:$$ {\bf u}\cdot{\bf v} = g_{ij}u^iv^j = (g_{ij}u^i)v^j = u_jv^j $$Now transform the components with your S and T matrices,$$ u_jv^j = \left( S_j^a {\bar u}_a \right) \left( T^j_b {\bar v}^b \right) = (S_j^a T^j_b) {\bar u}_a {\bar v}^b = \delta^a_b {\bar u}_a {\bar ...

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Think of a vector as having direction in space (north, south, east, west). Scalars may or may not be capable of having negative values. It just depends on the nature of the quantity. The statement that negative values for scalars are just convention is rather misleading. Some "conventions" just naturally make a whole lot of sense, and changing them would ...

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The modern notions that separate "scalars" and "vectors" goes as follows: Scalars are elements of fields. Examples of fields include the rational numbers, the real numbers, and the complex numbers. Scalars can be added and multiplied and divided. Vectors are spaces over fields. That is, lists of elements of fields. Velocity vectors, for example, are ...

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There are two different questions that are unrelated to each other. The first one is how the 4-velocity is defined. By definition, velocities are tangent flows on a differential manifold, therefore derivatives must be taken with respect to the parameter you are using to describe the flow with. In the context of special relativity such parameter is the ...

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Your proof is entirely correct. Relativity doesn't have to be difficult :) To be clear, the steps in your proof are simply the definition of $U'$, the definition of $X'$, the product rule, the constancy of $\Lambda$, and the definition of $U$, respectively. There is nothing wrong with any of these steps. As for why $\tau$ and not $t$: $\tau$ is defined ...

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I think this can be best understood by an example. The distance between two cities is a scalar quantity, say 500 km. If you want to know how much gas you are going to burn when you drive from one city to other, the distance is the thing you want to know. However position of a city with respect to another one (or with respect to any point in space) is a ...

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