Tag Info

Hot answers tagged

8

I doubt if your textbook makes it explicit, but the only sacred tenet in here is to respect dimensional homogeneity. One can make no sense of the sum of quantities with different dimensions.


7

From the commentary to the question, the textbook in question appears to be a mathematics textbook rather than a physics textbook. In mathematics, any two elements of a vector space can be added to one another to yield another member of that space. This is one of the requisites of what it means for something to be a "vector" in mathematics. Specifically, a ...


5

There is a definition that $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation is equal to spinor tensor $$ \psi_{a_{1}...a_{m}\dot{b}_{1}...\dot{b}_{n}}, $$ where $\psi_{\dot{b}}$ transforms as complex conjugation of $\psi_{b}$. Why do we assume that $\left( \frac{1}{2}, 0\right)$ and $\left( 0, \frac{1}{2}\right)$ represent spinors? You can think about ...


3

Let's consider this sketch of a man with legs asymmetrically astride: weight : ca.80kg (800N, g = ca. 10) height : 180 length of leg: 1m fore angle 45° rear angle ca. 15° Distribution of weight The rear leg is 73.2 cm and is 19 cm behind the center of mass. Let's assume that the legs are in compression only, (if the person were standing on ice or ...


3

For simplicity let's work in 2D, and take as our axes two unit vectors $\hat{e}_i$ and $\hat{e}_j$. We'll consider some vector $\hat{F}$: In our coordinates we can write the vector as $(F_i, F_j)$, where $F_i$ and $F_j$ are just numbers. The vector $\hat{F}$ is then expressed as the vector sum: $$ \hat{F} = F_i \hat{e}_i + F_j \hat{e}_j $$ I've drawn ...


3

Can we add any two vectors? My book says it is true For example, adding acceleration to velocity (seems impossible). Quite simply, your book meant two vectors of the same type. It's just that simple. As you thought, you can not add two vectors that are "different things"! (If you're just getting started with vectors. Note that indeed vectors have a ...


3

First consider the spatial component of the equation above, you get $$ \frac{d}{d\tau} \left( \gamma(v)\overrightarrow{v} \right) = 0 $$ But $\gamma(v) = \frac{dt}{d\tau}$ so you have $$ \frac{d^2 t}{d\tau^2} v + \gamma(v) \frac{d\, \overrightarrow{v}}{d\tau} = 0 $$ Now since $$ \frac{d \, u}{d\tau} = 0 \implies \frac{du^0}{d\tau} = \frac{d^2 t}{d\tau^2} = 0 ...


3

This is the diagram you need to draw: The red arrow is the net velocity of the boat - the sum of 25 km/h going due North, and a current of 10 m/s at 60 degrees East of South. (not drawn to scale) Just do simple math to determine the size of the two green segments - then compute the angle (heading) from the arctan of their ratio.


3

Yes, there are sometime different conventions for indicating vectors in hand-writing and printing. Yes, overset arrows in handwriting and boldface in printing is one of those conventions. No, it is not the only convention. Yes, you should familiarize yourself with the most common conventions in your sub-discipline. Yes, you should read the section on ...


3

In 3 dimensions, sliding the forces on the line of application until the forces meet does not always work because the two lines of application may be skew. To capture the line of application it is good to work with pairs of forces and torques. (You can transform the line of application of a force by adding a torque acting on the rigid body. This way you can ...


3

Note that: \begin{equation} \begin{aligned} \langle j | A \rangle & = \langle j |\left( \sum\limits_{i} a_i |i \rangle \right) \\& = \langle j| \left(a_1 |1 \rangle + a_2 |2 \rangle + a_3 |3 \rangle + \cdots + a_j |j \rangle + \cdots \right) \\& = a_1 \underbrace{\langle j | 1 \rangle}_{=\delta_{j1} = 0} + a_2 \underbrace{\langle j | 2 ...


2

If you think in terms of vectors and matrices with indices, $a_i$ is a vector and $\delta_{ij}$ is the unit matrix. The summation on the right hand side of your equation represents nothing but a multiplication of the vector $a_i$ with the latter. Since free indices are preserved, the remaining object carries the index $j$. You can write down an arbitrary ...


2

The cross product is defined to be the vector which is perpendicular to both vectors, so for instance the force exerted on a rod moving in a magnetic field is perpendicular to both its velocity and the field, hence is given by their cross product. Now if you work out which vector is perpendicular to both vectors you get the determinant of the two vectors ...


2

The size of the asteroid does not matter: when you jump, the asteroid will move in the opposite direction. Even if you jump up on earth, the earth moves backwards - but only infinitesimally so. Remember that $$F=ma$$ The mass of the earth is $5.97219 × 10^{24} kg$. If your mass is $100 kg$, then the earth's acceleration will be $6 × 10^{22}$ (almost ...


2

This convention is extremely convenient when doing things that physicist often like or need to do, such as computing a flux through a surface. When the area vector is chosen normal to the surface, one can simply take use an dot product to get what you're looking for. In the context of differential forms, this also turns out to be the natural definition, ...


2

Whenever you're confused about how to calculate some quantity, try going back to the definition. Think carefully about what the definition of average acceleration is: $$\vec a_\textrm{avg}\equiv\frac{\vec v_\textrm{final}-\vec v_\textrm{initial}}{\Delta t_\textrm{elapsed}}.$$ Which velocities does this equation depend on? Which velocities does it not ...


1

Use the fact the masses m1 and m3 apply forces along the diagonal. Similarly for the other diagonal. Then convert to the cartesian solution using the two resultants.


1

A square has its diagonals at right angles. So, find the forces along each diag. i.e., m1 and m3 on m5 for one direction and m2 and m4 on m5 for other. You could actually simpy find difference between, "m1 and m3" and "m2 and m4" and the corresponding charges and use them as resultant mass and charge.


1

Recall that when classifying representations of the Lorentz group, we consider \begin{equation} \textbf{N}_{\pm} = \frac{\textbf{J} \pm i\textbf{K}}{2}, \tag{1} \end{equation} where $\textbf{J}$ is the angular momentum (rotation generator) and $\textbf{K}$ is the boost generator. The generators $\textbf{J}$ and $\textbf{K}$ satisfy \begin{equation} ...


1

You have a few basic formulas for solving this kind of stuff: $$\vec{x}_f=\vec{x}_i+\vec{v}_it+\frac{1}{2}\vec{a}t^2,$$ $$\vec{v}_f=\vec{v}_i+at.$$ These are vector formulas, but all you're doing with the vectors is adding/subtracting them. When you add vectors you add the individual components, i.e. $$\vec{a}=(a_1,a_2,a_3)$$ $$\vec{b}=(b_1,b_2,b_3)$$ ...


1

The prime on ∇′ means that you're differentiating with respect to the primed coodinates (x',y',z') treating (x,y,z) as constants. Presumably, r is the magnitude of the vector that points from P' to P-- this is just a simple difference of vectors. Good luck!


1

The key here lies in two observations: If you can jump hard enough to escape the gravitational pull of the asteroid, this will permanently change the momentum of the asteroid; if you just jump and fall back down, nothing of lasting significance happens (conservation of momentum in the system) You want to change the angular momentum of the asteroid's orbit ...


1

To make it easy to see the logic behind, imagine that the first vector is directed to $\vec{i}$ and the second to $\vec{j}$. You get $$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & 0 & 0 \\ 0 & b_j & 0 \end{matrix}\right|$$ and according to Wikipedia the product in this case is $ab\vec{k}$. There are some physical ...


1

Is there any physical significance to this matrix The physical (geometric) relevance to the matrix $$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end{matrix}\right|$$ with regard to the cross product $\vec{a} \times \vec{b}$ is 1: that the three vectors $\vec{i}$, $\vec{j}$, and ...


1

When you translate-and-add, you do get the correct vector. The point of application is something else, not part of the definition of the vector. Additional information that must be supplied. The addition you refer to in your last paragraph is not "something else". It is vector addition. I'm surprised that more people don't have the confusion that you ...


1

In general the base vectors ($\vec{e}_{i}$) are not constant, e.g., in polar coordinates the radial vector do not point in the same direction (unlike the Cartesian base vectors $\hat{i},\hat{j},\hat{k}$. If one takes the j-th base vector $\vec{e}_{j}$ and consider its change if one moves on the direction defined by the i-th vector, mathematically this is ...


1

Other answers being good, i'll try to give a different perspective. What is a vector? As Feynman used to say ("Feynman lectures on physics") not every bunch of numbers (i.e $\left( a_1, a_2,..,a_n\right)$) makes a vector simply because it has $n$ components. Why? Because vectors have a specific relation (or more correctly transformational relation) with the ...


1

In this context, this is just a natural definition that turns out to be useful. For example, the flow rate of a fluid through a plane is the dot product of the fluid velocity with the area vector. [There is a more mathematically sophisticated way of understanding this, which is that the vector area is really a bivector or two-form. In three-dimensional ...


1

While in some contexts, $$ \frac{d\textbf{p}}{dt}dt = d\textbf{p} $$ is correct and is mathematically rigorous, there is a straightforward way to derive impulse as the change in momentum. Consider a function $f(x)$ where $f:\mathbb{R}^m \to \mathbb{R}^n$ and $f \in C^1$. Suppose its derivative is $g = f'(x)$. We can consider integrating $g(x)$ over some ...


1

Obviously I cannot know what Dirac is thinking, but I think it is just that his direction does not correspond exactly to your direction. We "know", just as Dirac does, that quantum states are members of some Hilbert space $\mathcal{H}$. We also know that scalar multiplication should not change the state, so $\lvert \psi \rangle$ and $\lambda \lvert \psi ...



Only top voted, non community-wiki answers of a minimum length are eligible