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0

The momentum is given by $P^\mu = (E,\vec{P})$, and the operator acting as $[:P^\mu:,a^\dagger_k] = k^\mu a^\dagger_k$. Energy is an eigenvalue of the Hamiltonian, which is given by $H = \int d^3x~T^{00}$ and $\vec{P}$ is given by your relation above, which from the definition of the stress energy tensor also equals $$P^j = \int d^3x~T^{0j}$$ You can ...


0

Bullet A would need twice the (de)acceleration than bullet B in order to have twice the force of impact. I though of 2 scenarios that would explain the results. If the target is suspended, then bullet A starts the target swinging. If the target is moving away when bullet B strikes, it takes more time to transfer the momentum and the recorded force is less. ...


-1

Many things that you have said is right but something is ... . Force: Force is the agent that changes or tends to change the rest or uniform motion of the body. It accelerates the body as a result of which the velocity and momentum changes. Time must flow to make something change. Nothing can change at an instant. So, force must act on the body for a ...


0

"Isn't always a force actually an impulse?" Force and impulse have different units--impulse has units of momentum, mass * distance/time (mass times velocity), while force has units of mass * distance/time^2 (mass times acceleration). If only a single force is acting on an object for a given time period, then multiplying the force by the time (or integrating ...


1

It's hard to think of a physical system involving a force that acted for zero time. However I think it's useful to consider a collision, perhaps between two billiard balls. When the balls collide they change momentum. We know that the change of momentum is just the impulse, and we know that the impulse is given by: $$ J = \int F(t)\,dt $$ where I've used ...


1

A force is not an impulse, it's a force. A force can exist without producing any work, stresses in materials are typically generated by forces applied to the same solid that oppose themselves and therefore do not produce any work. Some forces require contact, some forces don't (infinite range, decreasing with distance). As long as you're under the influence ...


-2

The force of gravity acting between the Earth and the Moon will be in effect for billions of years, until one of the bodies is destroyed. For the purpose of many experiments, this can be considered to be indefinitely.


0

The pattern does not continue. In fact it doesn't always hold as written. In particular: $\sum_i m_i = \sum_j m_j$ only if rest mass is conserved, which it is not when, say, you have particles and antiparticles being created by and annihilating into photons. $\sum_i m_i \vec{v}_i = \sum_j m_j \vec{v}_j$ only if all momentum in the system is of the form ...


0

Case A) Moving ball hits stationary man: $F_{1A}=\frac{m_{1}(v_{1A}-u_{1A})}{t}=-\frac{m_{2}v_{2A}}{t}$ Case B) Moving man hits stationary ball: $F_{2B}=\frac{m_{2}(v_{2B}-u_{2B})}{t}=-\frac{m_{1}v_{1B}}{t}$ You point out that $u_{1A}=u_{2B}$, let's just call it $u$. I don't see what is wrong with that. That forces in both cases must be equal just means ...


1

The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean. When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) ...


1

First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$. Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ ...


0

Momentum is a vector quantity. For a particle with mass, the momentum equals mass times velocity, and velocity is a vector quantity while mass is a scalar quantity. A scalar multiplied by a vector is a vector. A moving body would be a particle with a mass. If the body moves through space, relative to an observer, it will have a velocity, momentum, and ...


0

The net force on raindrop plus wagon is zero. Consider a single rain drop. Let the momentum in the direction of travel of the combined wagon/raindrop system be p. Now p = p_wagon_before + p_raindrop_before where p_wagon_before & p_raindrop_before are the momentum of the wagon and the raindrop before the drop hits the wagon. We have then: ...


0

This is the opposite of the rocket problem. In a rocket, acceleration occurs becaue mass is thrown out the back end. F = d/dt(mv) = m.dv/dt +v.dm/dt. If F= 0, then m.(-dv/dt) = v.dm/dt <-- note the negative term with the acceration. In a "typical" problem mass does not tend to change significantly, but in the rocket this mass term is highly ...


6

I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like \begin{equation} ig \sum_i p_i \ , \end{equation} where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum ...


0

But if the velocity of the wagon changes, the net force can't be zero, right? Only true if the mass is constant (it's not if a wagon is filling up with water). If mass and velocity both change, you can't say anything about the force. Record the experiment and play the video backwards. You will see a wagon moving backwards. The wagon is spraying water ...


0

Note that in your suggested motion after the collision, momentum would not be conserved. The Law of Conservation of Momentum is kind of a big deal in Physics, and especially useful when analyzing collisions. It states that for a closed system, momentum is conserved. It's also sometimes restated that for a closed system experiencing a collision, the momentum ...


0

Because in your example the action is not in the same direction than the velocity. The action and reaction are normal to the wall of the table, so the billiard ball only feels a reaction force (and thus a change in its velocity) in that direction. The component of the velocity of the ball parallel to the wall is not affected.


1

The force from the ball on the wall is exactly equal and opposite to the force of the wall on the ball. Both forces however are perpendicular to the wall (and must be assuming the wall is frictionless) and not necessarily perpendicular to the ball's initial direction of motion. Being perpendicular to the wall the force on the ball has absolutely no effect on ...


1

I think we are supposed to assume that the buoyancy force of the balloon is equal to the weight force of the balloon, ladder, and climber. If this is the case, the system is in equilibrium with its environment, with no net forces to the environment. You could look at it as a center of mass problem. We assume the ladder has negligible mass. When the climber ...


45

To deal with this type of problem, you must be careful to define exactly what system you are dealing with, and then not change that system part way through the problem. This definition allows you to be very clear about whether the "system" has any external forces acting, and thus whether the momentum of the system is constant or not. In this case, you seem ...


32

When the raindrops hit the wagon's surface, they aren't moving relative to the tracks. Friction is required to accelerate the raindrops to the wagon's speed. By Newton's third law, there must therefore be a reaction force on the wagon surface by the raindrops.


9

The mass of the cart is changing! This is the variable-mass system, which says, $$ F_{ext}+v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ where $v_{rel}$ is the relative velocity of the escaping/entering mass. In your case, there are no external forces so, $$ v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ So the change of velocity comes from the change in the mass.


1

The same question in the OP has been asked and replied in this question (already linked in one comment here). Probably, if you had made the example of the glass, your post would not have been misunderstood. I have just observed a simple fact there: if a glass can take, support only 100N if we exert a greater force, it will break and offer no support, the ...


2

The are defined just once you have fixed an inertial reference frame and a Cartesian orthonormal coordinate system co-moving with it. Changing inertial reference frame, the new operators are related with the initial ones by means of a strongly continuous projective-unitary representation of (connected) Galileo group $G \ni g \mapsto U_g$, $$P'_k = U_g P_k ...


0

I do not see any other possibility than doing the integral along the path. Do your telemetry data include the horizontal position? because if does you can calculate an approximation to the integral. Of course, there is no guarantee that the error will be larger than the effect you want to measure. You should have to do some pre-tests (using situations that ...


1

Short answer: No. Momentum-energy conservation arises by dint of Noether's theorem, which says that if a system's Lagrangian is invariant with respect to a continuous transformation, there is one conserved quantity, called the "Noether Charge" for each such transformation (technically: for each linearly independent tangent vector in the Lie algebra of the ...


2

In as far as I know, the universe is actually gaining energy. And as far as we can see, the universe is expanding as a product of pressure, in the direction of the difference between opposing pressures, blah blah.... So would momentum be conserved under this rule? I suppose, if the exact same amount of energy were being "destroyed" as "created" within the ...


0

When you act $\phi(t,\vec{x})$ on the vacuum, you get a particle at $(\vec{x},t)$, that is $|\vec{x},t>=\phi(t,\vec{x})|0>=\int \frac{d^3 k}{(2\pi)^3}e^{-ikx}a^\dagger _k|0>$. It's a Fourier transformation. And it means that if a particle wants to be in $(\vec{x},t)$ then it should have all kinds of momenta, which is actually the uncertainty ...


1

Your scenario of ball-girl-rail is not completely clear to me - but I can tell you this: both Newton's laws, and conservation of energy, are quite generally applicable. When the girl pushes against a light object (like the ball), then that object will accelerate away from her. If she maintains a certain force for a certain time, she will impart the same ...


1

Note that force is just the momentum per unit of time: $F=\frac{dP}{dt}$ I.e. the force is the speed of momentum transmission. When two bodies interact, they will exchange momentum. The quantity of momentum that will be transmitted from one body will be the same quantity, received by another. This is depicted by following picture which is the third ...


3

To answer the question in your title - no, the kinetic energy of an electron is a function of its velocity, same as for any other particle. The charge of an electron is always $1.6\cdot 10^{-19}C$, and so the electron will pick up $1.6\cdot 10^{-19}J$ of energy for every Volt of accelerating potential. The key to answering the question (part d in the ...


0

0.16 aJ is the energy an electron acquires by going through 1 volt. If it goes through 500,000 volts, it gets an energy almost equal to its mass.


2

Conservation of momentum works here like everywhere else. When A (with mass $m_A$) throws a ball with mass $m_b$ with velocity $v$, then $v_A=-v_b\frac{m_b}{m_A}$ so that after the ball is thrown, the net momentum is zero; note that the ball will not be moving towards $B$ at velocity $v$ but instead at $v-v_A$ since $A$ started moving backwards... When B ...


1

Would A move backwards at the same speed? no.. Momentum will be conserved,not speed and since A has greater mass than the ball so his speed will be less.. if A throws the ball with speed 'u1',then his speed(in the backward direction) will be m1u1/M1 where m1 is the mass of the ball and M1 is the mass of the person A Now, what would happen when B catches the ...


2

No. Kinetic energy depends on how much energy you give to an electron. $Volt = Work$ $done$ / $ unit$ $ test$ $ charge$. $1.6 * 10 ^ -19 J$ is the amount of work done to accelerate an electron of charge $1.6 *10^-19$ to $1$ $volt$ potential.


2

A lot of the answers get distracted by the "feeling pain" part of the question. So let me start by focusing on that - simplifying a very complex psycho-physiological issue to a simple physical statement (this is a physics forum - we leave the other stuff for other sites): During the impact of two bodies, there will be an exchange of momentum. The force ...


0

yes, you assumption is correct, for an isolated system, conervation of linear momentum is equivalent to the velocity of the center of mass being constant. The term with variable mass from another answer is incorrect. You can only have variable mass in a non-isolated sysrtem.


1

The linear momentum of a system is given by $\vec{p} = m \vec{v}$. If you differentiate this with respect to time in an inertial frame, you have: $$ {d\vec{p} \over dt} = m {d\vec{v} \over dt} + {dm\over dt} \vec{v} $$ If $\vec{v}$ is constant with time, this becomes $$ {d\vec{p} \over dt} = {dm\over dt} \vec{v} $$ Which means, for $d\vec{p}/dt$ to be ...


8

Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.


4

No mater the inertial referential: The pain is linked to the energy dissipated by the change of speed of the 2 objects in any inertial referential. Remarque 1: a part of the energy can be absorbed by the ball by deformation or heating. So a soft ball would make less pain. Remarque 2: The pain depends on what is static behind you and what is static behind ...


29

Look at it this way: Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball ...


-1

Yes you will.. what really matters here is the relative momentum !


2

Pavel Petrman gave a different perspective, but since impact remains same pain should remain same, given that properties of the situation remain same. To try take a ball in your hand and run towards a weighing scale and hit it and in next case exchange ball and weighing scale and repeat the procedure. Take a note of readings on weighing scale.


13

Yes, unfortunately. Because of the equivalence of inertial reference frames, the the physical laws are the same in both reference frames. However, another possibility, which is non abelian, is that instead of feeling the same amount of pain, you could be feeling the opposite amount of pleasure. It depends if pain (X) are fermions or bosons, that is, if ...


0

Whether or not your idea produces thrust depends on at least one of the following statements being true: The laser doesn't travel with the vehicle. The photon eventually leaves the mirror structure. Forget the structures holding the mirrors and consider each component as the photon interacts with it. In all cases, I assume each emission and reflection ...


1

Unfortunately you got answers that are wrong in many aspects. 1) I read all the previous answers concerning the 3rd law and I have seen that it is definitely not universal The third law is universal in classical mechanics. You stop to talk about forces in quantum field theory, but that is another issue. 2)Conservation of momentum is a universal ...


2

The laws of physics are discovered through a mixture of heuristics, modelling and inference. In case of momentum, the story goes like this: It is possible to 'transfer motion' from one body to another. However, experiment shows that it is not velocity that is conserved during such transfers, but another 'quantity of motion'. We give that quantity the name ...


7

If classical mechanics were valid at cosmological levels, the answer would be yes. But general relativity is what describes the dynamics at this larger scale, and it is not generically possible in GR (in an arbitrary spacetime) to define conservation of momentum or conservation of mass-energy.


1

This is simply a Fourier transform. The wave equation, Hamiltonian, and Green's functions are all simpler in momentum space than in position space. This is because, for the free, theory, the different momenta decouple, and you can create a particle with momentum $k$ with $a^\dagger_k \left|0\right\rangle$, and that state will evolve in a nice way. This is ...



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