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49

You've calculated the speed of a remote-triggered gun after it fires the bullet, true. However, there's actually nothing about space in your calculation, as @ACuriousMind noted. In theory, a gun fired on Earth could fly off just as fast, at least for a second. What you should use is not $m_\mathrm{gun}$ but $m_\mathrm{gun} + m_\mathrm{person}$. The gun never ...


20

For most guns, you can roughly hold them in place while fired. That is, the repulsion will not only "hit" the gun's mass but the astronaut's mass too, not allowing the gun to gain such high speed. With your numbers this leaves at most $$ v \approx 0.11~\text{ms}^{-1} = 0.38~\text{kmh}^{-1} $$ for an astronaut + spacesuit + gun with $m=225~\text{kg}$, if no ...


11

Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ...


7

As the other answers have stated, the primary oversight in the original question is the mass of the astronaut/cosmonaut holding the firearm. However, your original number for the mass of the projectile is off by an order of magnitude. Therefore, the original calculation - as well as some of the other samples provided afterward - are all still an order of ...


7

Mass, or more correctly, rest mass is not conserved in special relativity. Particles are able to be created and annihilated in special relativity, for instance, an electron and a positron can interact to produce two photons: $$e^++e^-\rightarrow 2\gamma $$ Here mass is clearly not conserved, because both the electron and positron are massive but photons are ...


6

I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from ...


4

We (physicists) believe the reason is this: known symmetries and conservation laws. For example, the mutual annihilation of a proton and positron would remove $2\,e$ charge units from the Universe. This violates the conservation of charge principle, which can be seen to arise from the application of Noether's theorem to the global gauge symmetry of the ...


4

No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ...


4

The rest mass of the system is conserved, it's just that the rest mass of the system isn't the sum of the masses of the parts. The rest mass of a system is just the length of the total energy-momentum vector. And that vector is conserved, so the length is conserved. The sum of the rest masses of the parts is not conserved. But that simply isn't the rest ...


4

Any energy principle is not being violated since the speed of the photon is never less than $c$ and hence the momentum is unchanging (in the classical sense). Why light travels slower than $c$ in a medium is because of the photons being absorbed and reradiated by atoms in the material. In a sense you can make the analogy of light traveling a longer path in ...


3

For the rest mass we have $$m^2=p^2=p^{\mu}p_{\mu}$$ where $p^{\mu}=(E,\vec{p})$. It is Lorentz invariant, which means the rest mass of the particle is always the same no matter in which frame the observer is in. While for relativistic mass, it's simply equivalent to the total relativistic energy, which is always conserved. Note the difference between ...


3

I suppose you read this passage in the famous Feynman Lectures. I am fairly certain that what Feynman is referring to (and what you are looking for) is a proof that an electrostatic field is conservative. There are a number of equivalent ways of stating that a vector field is conservative, each of which can be taken as a definition. Let $\vec{F}(x)$ be a ...


3

Here is my comment in more details For any system or single elementary particle mass $M$ is defined as $$ M = \sqrt{E^2 - \textbf{P}^2} $$ where $E$ is total energy and $\textbf{P}$ is total momentum. For an elementary particle (like electron) mass is always conserved. For a system $M$ is conserved as long as $EdE - \textbf{P}d\textbf{P} = 0$, in ...


3

Disclaimer: Let us here avoid the discussion of how to assign a stress-energy-momentum (SEM) pseudo-tensor $t^{\mu\nu}$ to the gravitational field. The word pseudo here refers to the fact that $t^{\mu\nu}$ is not a tensor wrt. general coordinate transformations; only a rigid subgroup thereof. In other words, the pseudo-tensor ...


3

To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ...


3

Let $\xi^\alpha$ be a Killing vector of a metric $g_{\mu\nu}$, i.e. it satisfies $$ \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = g_{\mu\alpha} \partial_\nu \xi^\alpha + g_{\nu\alpha} \partial_\mu \xi^\alpha + \xi^\alpha \partial_\alpha g_{\mu\nu} $$ Then the quantity $$ Q = \xi^\alpha u_\alpha $$ is conserved along any geodesic. To see this, we can compute $$ ...


3

Yes. The so called rest mass $m_0$ is the magnitude of energy-momentum four vector and is always conserved. It is not just conserved but is also invariant since it is the magnitude of a four vector. Perhaps you are confused between rest mass and relativistic mass. The relativistic mass is the total Energy divided by $c^2$. $m\equiv \gamma m_0 $ This ...


3

One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ...


3

I will be assuming that the system has some angular momentum about the center of mass initially. If the system has no angular momentum then both the stars would accelerate towards each other and end up colliding. The problem is a two body problem. In such cases, both the stars would revolve around the center of mass. If the distance between two stars is ...


3

What's the difference between where you are and space? Atmosphere (ie air pressure), temperature, gravity, radiation. Think about how each of those affects what happens to someone after they pull the trigger of a gun, on earth. Atmosphere: has no significant impact on the effects of recoil. It will help slow the bullet down but that's not relevant to ...


2

Let's dive right into an example -- let's say you are simulating a fluid. First, you need to pick your reference frame. Are you going to simulate a fixed domain in space and have your fluid move through it, meaning you have a grid and at each point on the grid you store and solve for the fluid properties (Eulerian frame)? Or will you track each discrete ...


2

I'll make an example, to make things clear. Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$. This is a holonomic constraint, since $$ | \vec{r}_1 - \vec{r}_2 | = d $$ with the particle-positions $\vec{r}_1$ and $\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus ...


2

the speed of both should be according to the conservation of energy If both balls have the same speed after the collision, the collision is inelastic, i.e., kinetic energy is not conserved. If the balls are identical, then conservation of momentum requires that $$\mathbf v'_1 + \mathbf v'_2 = \mathbf v_1 + \mathbf v_2 = 150 \mathrm{\frac{m}{s}}$$ If ...


2

I think the source of your confusion is the misnomer "conservation of parity". Because a conserved charge is a result of the invariance of the s-matrix under a continuous symmetry (best understood in the lagrangain formulation with Noether's theorem). Parity is a discrete symmetry, and there for does not have any corresponding conserved charge (in the ...


2

Consider a control volume $\Sigma(t)$ of a fluid with density $\rho(\mathbf x,t)$. The mass inside $\Sigma(t)$ is clearly given by $$M(t):=\int_{\Sigma(t)}\rho(\mathbf x,t)\text d^3\mathbf x.$$ The way $\Sigma(t)$ is defined is that its mass content doesn't change with time, that is, a control volume is representing the time evolution of a certain amount of ...


2

The quantum version of Noether's theorem are the Ward-Takahashi identities. Classical conservation laws only hold "inside expectation values", so you should not expect a classically conserved operator to have no time evolution quantumly. Additionally, switching to the Hamiltonian formalism breaks manifest Lorentz invariance, so you should also not expect ...


2

When you compute the final velocity of the parcel you have forgotten that it's no longer traveling at 37 degrees - that was the angle at the end of the chute. While it drops, the horizontal component of velocity doesn't change - it is still $3.4\cdot \cos 37° = 2.71 m/s$. With that, you should be able to solve this.


2

When we say something is conserved or that there is a conservation law for a given thing, we mean that the quantity of it does not change. You neither lose nor gain any of that thing. More specifically, conservation can come in two flavours. Something can be globally conserved. This means that the total amount of that something in the universe does not ...


2

When you say If something goes outside, then it will decrease inside! what you assume is exactly a conservation law. It may seem trivial, but it is not necessarily. Consider the population of a city, for example. At one point in time, you measure how many people are within the city borders; let's call this number $N_0$. Then, you observe all city ...


2

For two objects to remain in a stable circular orbit, the force acting on them must be equal to the centripetal force corresponding to their rotation. $$F=\frac{mv^2}{r}$$ or in terms of angular velocity $$F=m\omega^2r$$ where $r$ is the radius of orbit in this case. As the gravitational force acting on the two stars is the same. $F$ is equal in both ...



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