Tag Info

4

It's because you have to replace the erroneous $\gamma$ by $\gamma^2$ (and similarly $m$ by $m^2$) in the inner product and because $$\gamma^2 m^2 c^2 - \gamma^2 m^2 v^2 = \gamma^2 m^2(c^2-v^2)=\dots$$ and $$\gamma^2 = \frac{1}{1-v^2/c^2} =\frac{c^2}{c^2-v^2}$$ and $c^2-v^2$ from the explicit factor cancels against the denominator of $\gamma^2$, while ...

3

Rotation of a 3-vector We'll find an expression for the rotation of a vector $\mathbf{r}=(x_1,x_2,x_3)$ around an axis with unit vector $\mathbf{n}=(n_1,n_2,n_3)$ through an angle $\theta$, as shown in Figure . The vector $\mathbf{r}$ is analysed in two components $$\mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot \tag{01}$$ ...

2

You have to think about what $\vec j,\mathrm{d}\vec l$ and $\mathrm{d}\vec s$ actually are: $\mathrm{d}\vec l$ points along the flow of the current. So does $\vec j$. So $\mathrm{d}\vec l$ and $\vec j$ are parallel, and indeed $$(\vec j\cdot\mathrm{d}\vec s) \mathrm{d}\vec l = (\mathrm{d}\vec l\cdot\mathrm{d}\vec s)\vec j$$ holds in that case.

2

I use other symbols in order to prevent confusion in the following. Let a point charge $\:q\:$ moving with position vector $\:\boldsymbol{\xi}\left(t\right)\:$ as in above Figure. Then the volume charge density and the charge current density are expressed via Dirac $\:\delta$-function as follows \begin{align} \rho\left(\mathbf{x},t\right) & ...

2

As per, http://en.wikipedia.org/wiki/Four-velocity, we can define four-current density as: $J = \rho_0 U$, where $U$ is the four-velocity. Since it's a scalar times a four-vector, it's another four-vector. $$J = \gamma(v)(\rho_0 c,\rho_0 \vec{v})$$ $$J = (\gamma(v)\rho_0 c,\gamma(v)\rho_0 \vec{v})$$ Now it remains to show that this fits the definition you ...

2

In general it changes although the reason is not exactly because its projections changes. For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change ...

2

Let the velocity at instant be v(vector) = (Vx) i + (Vy) j. i and j denote unit vectors, along x and y axis respectively. dv/dt = a(acceleration) = - gj. dv/dt = [(Vx(final) -Vx(initial))i + (Vy(final) - Vy(initial))j]/dt = - gj. By initial and final I mean Vx and Vy at time t and t + dt. As the resulatant is only along the y axis the X component must be 0 ...

2

Any volume integral of curl $\mathbf A$: $$\int_V \nabla\times \mathbf A \,dV$$ can be calculated also as surface integral of $\mathbf A$: $$\oint_\Sigma d\boldsymbol\Sigma\times\mathbf{A};$$ here $\Sigma$ is boundary of the region and $d\boldsymbol \Sigma = \mathbf nd\Sigma$ is vector whose magnitude is that of area $d \Sigma$ has direction of ...

2

Short and a little inaccurate answer: vector is one-dimensional tensor, matrix is a two-dimensional tensor. More details now: Tensors are multidimensional arrays which have certain properties. Not every multidimensional array is a tensor, check this discussion for more details. There are two types of one-dimensional tensors: vectors and co-vectors. Both ...

1

A charged particle placed in a magnetic field experiences a force that causes it to deflect in the direction of force. Lorentz law is true. It will always be. The best way to apply the direction of the Lorentz force is by using Fleming's left (or right) hand rule. The right hand screw rule is helpful in analyzing the direction of curling of magnetic field ...

1

Yes, the Lorentz force law holds, so whatever rule you're doing with your right hand must be wrong. All of these rules, in the end, come from the right hand cross product rule anyways. There are lots of things you can do with your right hand, though, so I wouldn't be surprised if one of them gave you the right direction.

1

What you have read is only valid in a vacuum. With air resistance the drag is a function of the total velocity, so in reality the deceleration on each axis also depends on the other.

1

The first green part is the Rodrigue's rotation formula. The second green part is a small angle approximation for $\delta \theta$.

1

The question is rather incomplete and confusing. By the way, it is used to consider surfaces as vectors when needed for computing surface integrals, like flux integrals, where the scalar product between a vector field $\vec A$ and a infinitesimal surface $\mathrm d\vec S$ is considered: $\vec A\cdot\mathrm d\vec S$. To this aim, the differential surface is ...

1

In special relativity there are two major assumptions: -the laws of physics are the same in all inertial frames -the speed of light that you observe is always the same, (thus independent of the relative motion between the light source and the observer). From this two assumptions follows the famous Lorentz transformations. In these Lorentz transformations ...

1

The set of transformations that leaves the speed of light unchanged is the Lorentz group. Representation theory enables us to investigate the irreducible representations of the Lorentz group. The lowest-dimensional representations act on scalars four-vectors However, take note that usually we consider representations of the corresponding Lie algebra ...

1

How do we formally define vectors in physics? An excerpt from chapter one, page 12 of "Mathematics of Classical and Quantum Physics" Originally, we introduced a vector as an ordered triple of numbers. The rule for expressing the components of a vector in one coordinate system in terms of its components in another system tells us that if we ...

1

A physical quantity is a vector if it transforms in the same way as a position vector when the coordinate system undergoes a transformation.

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