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1

The origin is assumed to be at rest wrt ground frame. So, your pendulum with lift moves up and the origin stays at its original place only. That explains that factor.


3

The easiest thing for this exercise is to use Levi-Civita symbol for the vector product: $$\vec{a} \times \vec{b} = a_i b_j e_k \varepsilon_{ijk},$$ where I denote by $e_i$ the canonical basis of $\mathbb{R}^3$. Using this notation, we have: $$[L_j,p_i]=[r_k p_l \varepsilon_{klj},p_i]= i \hbar p_l \varepsilon_{ilj}.$$ and $$[L^2,\vec{p}]=e_i[L_j ...


-3

We can consider area as for a small surface and the direction of the area vector is in the direction normal of to that surface. But for a large surface area we cannot consider it as a vector e.g. Earth.


1

While the other answer are all completely correct, I just want to write a more simplified answer. It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$ Adding ...


2

This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i $$ However, this is only true when the relation between the force and the acceleration is linear. Let's take the gravitational force as an example: say you have three bodies and you have already ...


2

$ L = \int \int {\bf F} \cdot d{\bf s}$ is where you should start, where $F$ is the flux in units of Watts/m$^2$. Blackbody flux is given by $\sigma T^4$ and hence an isotropic flux integrated over a sphere $$ L \int^{2\pi}_{0} \int^{\pi}_{0} \sigma T^4 r^2 \sin \theta\, d\theta\,d\phi = 4\pi r^2 \sigma T^4$$


-1

The expression "Total accelration" does not fit if the accelrations have different directions. The vector resultant is actually the "net accelration", or the combined effect of these two accelrations, or equivalently, forces. The vector resultant makes sure that only the effective components are added, and the opposing effects cancel out. Maybe an example ...


4

It makes no sense for a point mass to have 2 accelerations. What you might have done is find accelerations due to 2 forces separately. You can add them as when $m= \text{constant}$, $\vec{F}=\vec{F_1}+\vec{F_2}=m(\vec{a_1}+\vec{a_2})$ When using vectors symbol, its automatically takes care of their directions.


2

There are two kinds of vector multiplications- Dot (Scalar) Product: The dot product of two vectors gives a scalar, that means only the magnitude is left, no direction. Mathematically, it is equal to the product of the magnitude of two vectors times the cosine of the angle between the two. ie. $$\vec{v} \cdot \vec{u}=|\vec{v}||\vec{u}|Cos\theta$$ The ...


0

The co/contra distinction only makes sense when talking about vector fields. Even then the difference only becomes apparent when dealing with curved spaces or at least curvilinear coordinate systems The difference comes from how vectors relate back to the undlying space or manifold on which the fields are defined. Contravariant vectors then are what people ...


0

Thanks for the question. You can derive this equation from the Pythagorean equation which is only for the the case when the two lines a and b are perpendicular to each other. But if they are at an angle then this is the general equation. Now from the picture $$ \begin{align} c^2&=(a+b \cos \alpha)^2 + (b \sin \alpha)^2\\ & = a^2 + 2ab \cos ...


1

We expect a vector to change in a certain way when we change the scale we use to measure distance. Consider the vector $$\vec{x}=(1, 0, 0)\,\mathrm{m}$$ If we change scale and now measure in centimeters this vector becomes $$\vec{x}=(100, 0 ,0)\,\mathrm{cm}$$ Now consider a vector representing a force: $$ \vec{F}=(1,0,0)\,\mathrm{J/m}$$ where I've chosen ...


2

The notion of co- and contravariance depends on context: If you wanted to be as clear as possible, you should actually mention with respect to what the components transform co- or contravariantly. In case of the algebraic dual of finite-dimensional vector spaces, the implied context is a change of basis of the vector space. Then, we can look at how the ...


5

There are two more points that can be made here. Sorry if I repeat someone. In a way you are right that if you have a vector space and its dual there is no intrinsic way to say which space is the original and which is the dual. This is because there is a canonical isomorphism between a vector space and the dual of its dual. In other words if $V$ is a vector ...


2

I) No, it is important to distinguish between covariant and contravariant tensors. OP's link mentions differential geometry. If one has only studied those objects in the context of pseudo-Riemannian manifolds $(M;g)$, which comes equipped with an (invertible) metric $(0,2)$ tensor $g$, then the existence of the musical isomorphism may perhaps unnecessarily ...


0

The cross product does not have a specific direction, but either the left hand rule or right hand rule works, but it has to be applied consistantly. That's the key. So you simply drum in one rule in everywhere. It's kind of like driving to the left or to the right. Either works, but you have to be consistant about it.


0

I will say that the standard definition of vectors and one-forms is not the world's cleanest. A modern definition of vectors would say that a vector space is a mapping from the functions on the space to itself that satisfies the Leibniz rule and is linear (alternately, the vector space is the local linear approximation of the space). Then, the set of ...


22

This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...


1

you have three terms in the equation, which together define the scattered wave. the first two define the amplitude at any given time and place along the vector $\hat{r}$, which points the direction you're interested in. $\hat{r}\times\hat{E}\times\hat{r}$ tells you that the wave is going to be perpendicular to $\hat{r}$ and also in the plane defined by two ...


2

First of all the hats on $\hat r$ and $\hat E_i$ indicate that these are unit vectors. $$(\vec A \times \vec B) \times \vec C \neq \vec A \times (\vec B \times \vec C)$$ See Vector Triple product Which means that $\vec A \times \vec B \times \vec C $ without parenthesis has no meaning. So I shall assume parenthesis around the first pair ie $(\vec A ...


0

Look at the diagram below, Two vectors $B$ (Magnetic field vector ) and $v$ ( velocity vector of a proton ) shown in the diagram are perpendicular to each other. Here, I have considered the simplest case where the two vectors are perpendicular to each other. The cross product of $B$ and $v$ gives us another vector $F$ ( Force vector ) which points in a ...


2

The formula for the force of a particle due to its magnetic field is $F = q \vec v \times \vec B$. The cross product has the property that its result is always perpendicular to both arguments. Its direction is simply a result of how the cross product function is defined and the sign of electric charge (an electron is defined as negative when it should ...


1

The center of mass is not $(0, 0)$. If you take the average of the two positions, you will get $(c, d)$. The angular momentum $\vec L$ is $\vec r \times \vec p$. With mass $m$, this will be: $$ \vec L = m \vec r \times \vec v $$ $\vec v_\pm$ is just $\dot{\vec r}_\pm = \frac{\mathrm d}{\mathrm dt} \vec r_\pm$. So you have to to a time differentiation of ...


1

The dot product is missing. The integral must also be multiplied by a Cosine of the angle between the vectors. $dl$ and $dr$ are the same thing. It's just an infinitesimally small distance on the direction of the field.



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