# Tag Info

0

Step back and ask how you know whether a Lagrangian, $L=L(Q_i,\dot Q_i),$ is correct. At the classical level the only answer is whether it gives the correct equations of motion as the Euler-Lagrange equations: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot Q_i}\right)= \frac{\partial L}{\partial Q_i}.$$ Which is unchanged if you replace ...

1

Conventions do not change physics. If they would, we would not call them conventions. When studying Lagrangian mechanics, you may have noticed that you can multiply a lagrangian by any constant, and receive the same dynamics. Thus, we often (Or always) choose the constant such that the term $(\partial_0\phi)^2$ appears with a positive sign. (And often with ...

2

No, the Lagrangian density is different: $$\mathcal{L} = \pm \frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi.$$ The Hamiltonian density is actually the same in both conventions. However, this has no physical meaning. The choice of the signature is purely conventional.

0

you can also take negative charges to define potential of a point.there would be electric field due to a -ve charge Q also and when you bring a point charge q(having -ve charge) near Q then you have to do work against the field.(because same charge repel each other )

1

Other people have answered your questions about the signs; I want to address your comment, "why does this sound like the normal force?" By Newton's third law, pairs of forces always come in equal and opposite pairs. If the ground exerts an upward normal force on you, you exert a downward normal force on the ground. If sliding on the ground exerts a ...

1

What everyone said, but also: the way the question is phrased, the model it uses, is a 1-dimensional model. Imagine a point travelling along the real number line/the x-axis on a cartesian plane. The model in the problem assumes there is no vertical movement, or 'side to side', but only 'pure' 1-D movement along a straight horizontal line. There is only that, ...

1

It's all a matter of how you've chosen your coordinate system. Think of how the velocities were calculated. They essentially looked at two points in time and the two corresponding points in space. If your positive direction is to the right, and the guy has advanced to the right, then its a positive number for velocity. Force carries the same logic here. The ...

0

The force/ acceleration of the skater comes out as negative because: the force or acceleration is in the opposite direction of motion. You can also think of it like this, frictional force always tries to oppose relative motion. In this case the skater is moving forward w.r.t the ice and the ice is moving backwards w.r.t the skater; so the ice exerts a ...

14

It's very common to get mixed up about signs. The only recommendation I can give is to establish a clear sign convention and stick carefully to it. To show what I mean let's consider your skater: I'm going to use the convention that positive is to the right and negative is to the left. remember that quantities like velocity and acceleration are vectors, ...

2

When the skater is on the ice, friction stops him/her in 3.52 seconds as you said. The molecules in the skates rub against the molecules in the ice, and the ice molecules absorb some of the skate molecules's energy, slowing the skater down. The reason the force is negative is because the friction is acting in the opposite direction of the skater's motion. ...

1

Both the zeroes are "real" to answer the question. There is absolutely no problem with multiple points in a space having the same electric potential. Note that we have ASSUMED the potential to be zero at infinity and based upon that assumption we have found out the potential to be again zero at the equidistant point between two opposite charges. So there ...

3

If you the rate of change of mass is $\dot{m} = \mathrm{d}m/\mathrm{d}t$ and we're talking about passing into the future direction ($\mathrm{d}t > 0$), then $\mathrm{d}m$ is necessarily negative. Thus, $m\mapsto m+\mathrm{d}m$ represents a decrease of the rocket mass. The opposite convention would require that $\dot{m} = -\mathrm{d}m/\mathrm{d}t$ works ...

0

The sign is simply a convention: a convention is being used which identifies motion away from the large, gravitational body (against gravity) as having a positive sign. Thus motion in the direction of gravity is negative. This convention is useful because when we perform work on an object by raising it against against gravity, we increase its gravitational ...

0

There is a subtle difference between displacement and position. Position is a vector measured relative to a defined origin and is often designated by the vector symbol $\vec{r}$. In a Cartesian coordinate system, $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},$$ where $x$, $y$, and $z$ are the individual coordinate values, positive or negative, relative to the ...

0

Suppose I throw a ball from my chest-height straight up into the air: it travels upwards for a time $\tau$ before it reaches its maximum, then it falls down for a time $\tau$ too. We know that it achieves the height $h = \frac 1 2 ~g~\tau^2$ above my chest-height in this time where $g \approx \text{9.81 m/s}^2$, so if we take my chest-height as the value ...

1

Let's make this concrete by using a $\text{spin-}\frac 1 2$ state in the $z$ direction, where we get to use the Pauli matrices, usually written as\sigma_x = \left[\begin{array}{cc}0&1\\1&0\end{array}\right]; ~~~\sigma_y = \left[\begin{array}{cc}0&-i\\i&0\end{array}\right]; ~~~\sigma_z = ...

1

Displacement is the vector version of distance, so it has a magnitude (the distance) and a direction. If the motion is only in one dimension, as in free fall, then the direction manifests only as positive and negative, or up and down but you are free to define whether up is positive or negative (and similarly for down) as long as you are consistent within ...

Top 50 recent answers are included