Tag Info

New answers tagged

1

Is this the correct way to find the derivative of kinetic energy? $$ K=\frac{1}{2}m v^2 \\ $$ So: $$ \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right) $$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$ And finally $$ \frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right) $$ So simplifying: $$ ...


3

Here is the procedure: $KE = 0.5mv^2$ $\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$ So the question becomes,how do we find the derivative of $v^2$ with respect to time? One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom) Therefore, $\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ...


2

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.


0

The velocity is always, by definition, the derivative of the position $\textbf{r}(t)$ with respect to the time and likewise for the acceleration $$ \textbf{v}(t)=\frac{d}{dt}\textbf{r}(t),\qquad \textbf{a}(t)=\frac{d}{dt}\textbf{v}(t)=\frac{d^2}{dt^2}\textbf{r}(t). $$ The position is, in turn, always a function of the time, although often not explicitly ...


0

Acceleration is defined as the derivative of velocity with respect to $t$: $$a=\frac{dv}{dt}$$ It is the instantaneous change of velocity. Just like velocity is defined as the instantaneous change of position $r$: $$v=\frac{dr}{dt}$$ If you agree that: $$a=-\frac{GM}{r^2}$$ then it is a simple thing to exchange $a$ with its definition $dv/dt$.


0

Because by definition acceleration is change in velocity over time. The fact that acceleration is a function of radius changes nothing. Basically what the equation says is that acceleration as a function of radius is equal to some formula. We then know from the definition of acceleration that it is change in velocity over time. If this helps clear it up: ...


1

Divergence at a point measures the flux flowing out from an infinitesimal volume around that point. Suppose you have a volume $v$ in vector field $\mathbf{h}$. Divide it into two parts: $$v = v_1 +v _2$$. Now the flux out of volume $v$ is given by : $\int \mathbf{h}\cdot da_v = \int \mathbf{h} \cdot da_{v_1} + \int \mathbf{h}\cdot da_{v_2}$. Start dividing ...


1

Think about it one more time. If $\vec{F}$ has continuous partial derivatives, then $$\vec\nabla\cdot\vec{F}=\sum_i \frac{\partial F_i}{\partial x_i}$$ is also continuous. If a function is continuous, it's approximately constant on sufficiently small volumes: that's pretty much the definition of continuity! So your original understanding was just fine. ...


0

Indeed as was commented before usually in physics these derivatives are 'derived' by postulating how the object transforms. Obviously there are more rigorous definitions, as is usually the case. In this particular case you could try to find the structure which gives you these 'covariant derivatives' in the first place. In books they are often referred to ...


0

They're just applying the chain rule. $$\frac{d}{dx}f^{n}(x)=nf^{n-1}(x)\frac{df}{dx} \Rightarrow \frac{d}{dt}D^{3}(t)=3D^{2}(t)\frac{dD}{dt}$$


2

This is more of a math question, but it's just a trigonometric identity. $$\cos\Big(\frac{\pi}{2}-x\Big)=\sin(x)\\$$ Or $$\cos\Big(x+\frac{\pi}{2}\Big)=-\sin(x)$$


0

A physical derivation of that result can be provided in the framework of general relativity using the equivalence principle and the principle of general covariance. The principle of general covariance tells us that if a physical reult is valid in a frame of reference, then it must be valid in every other frame of reference. On the other hand, the equivalence ...


9

This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here. We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions ...


3

As one way, consider the parallel transport of two vectors $p^\mu$ and $q^\nu$ along the same curve $C$ such that the same angle between the two, $$ \cos\phi=\frac{p^\alpha q_\alpha}{\sqrt{p^\beta q_\beta}\sqrt{p^\gamma q_\gamma}} $$ is preserved. Parallel transport tells us that for a general vector $f^\alpha$, $$ ...


2

First of all, derivative of $1.5t-9.75t^3$ is $1.5-29.25t^2$, you missed first part. Secondly, consider ideal pendulum. It's speed is zero at extreme points, however it never stops for ever. it depends on the problem and additional conditions (like, object hitting ground, so speed=0 is good enough condition for finding time of impact)


4

Nice question. One way to think about it is that given a metric $g$, the statement $\mathcal L_Xg = 0$ says something about the metric, whereas $\nabla_Xg = 0$ says something about the connection. Now what $\mathcal L_Xg = 0$ says, is that the flow of $X$, where defined, is an isometry for the metric, while $\nabla_Xg = 0$ says that $\nabla$ transports a ...



Top 50 recent answers are included