# Tag Info

1

I am assuming that zero (rest) mass particles don't interact gravitationally with each other and other particles. That's not a valid assumption in general relativity. Particles with zero invariant mass have energy and momentum and, thus, gravitate. Essentially, in general relativity, the density and flux of energy and momentum are the sources for ...

1

The trajectory of an object, whether massless or not, in a curved spacetime is given by the geodesic equation. $${d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0$$ Solving this is a formidable problem unless there is some helpful symmetry that simplifies the working, but in the particular case of ...

1

No, a massive body is able to bend light around it, which is called gravitational lensing. This has been observed multiple times. EDIT Photons are massless. Otherwise, they would not travel at the maximum speed, which is called speed of light. Keep in mind, that gravitational lensing is not a part of Newtonian mechanics. You need general relativity for ...

0

Almost there. Since this looks like a homework question I'll just give a hint. The force balance equation is due to the buoyancy force which is due to the difference in density inside the ballon vs. outside the ballon. Once you work that out, the weight that you missing will come back into the equation.

0

The air inside the balloon is less dense than the air outside; this difference is what causes the lift for the balloon. When you heat the air in the balloon, it expands until the balloon is full. At this point the balloon is still on the ground since there is not enough lift. You need to heat the air more, which expands the air more and causes some of the ...

1

A better electrical analogy to Newton's second law might be inductance: $$V = L \frac{\mathrm d i}{\mathrm d t}$$ The only reason this looks different is that physics has a name and conventional symbol for the derivative of speed, but electronics does not have a name for the derivative of current. So let's just pretend that the word acceleration does not ...

0

I'm not sure if I'm just misinterpreting your question, but we can most certainly influence the acceleration of an object directly. To do this we must take into consideration what is actually happening as we accelerate; Acceleration, in physics, is the rate at which the velocity of an object changes over time. While it is a sum of the net forces over the ...

1

We should think a bit more carefully what force, mass and acceleration really are: For simplicity, we consider a classical point particle. The force $\vec F$ is something externally applied to the particle, it is a property of its environment. Most often, it is the gradient of a potential, $\vec F = - \nabla V$, but it need not be. In general, it is some ...

5

Your observations are spot on. I usually write Newton's second law this way: $\vec{a} = \vec{F}/m$. This form makes it clear that the law is a relationship between the dynamic variables force and mass, and the kinematic variable, acceleration. $F$ and $m$ describe the situation, $a$ is the result. Cause and effect, if you will. In fact, that's one ...

1

In my view, an expression like V=RI, practically speaking, isn't much different. You have to be concrete with a real-world example. I can connect a resistor across a voltage source, and then I can only alter the voltage or change the resistance value to change the current; I cannot change the current directly in this configuration. Your statement that ...

2

Mark's answer is fantastic and steps through much of the physical reasoning involved wonderfully. But annoyingly, in this particular case the region of interest is pretty much right where both air resistance starts to become nontrivial and important and the details of the velocity of the ball off of the club also has non trivial behavior, so I decided to ...

3

Can someone explain the atomic process, if it even exists, of how this would work to convert energy in to matter, and what form of energy was initially present, and what is required to cause this change? It is not an atomic process, it is an elementary particle process, atoms are made up of elementary particles in a non trivial way. At the level of ...

1

They are actually trying to do so in the lab. They need a very potent laser (I believe they didn't reach the critical power yet). The idea is simple, inside the lase cavity you generate a very powerful electromagnetic field, powerful enough such that the photos have enough energy to transform a virtual electron-positron pair into a real one. See this link ...

0

Your intituition is totally different because ennumerous forces change the situation from the ideal situation predicted by work enrgy theorem, some of them are: Air Drag/Friction, Rotational Friction due to differences in size of tyres, different aerodynamic effects due to different body design.. etc.

0

If the acceleration (presumably measured on the sled) is the same, then the mass of the sled has no effect. It will take 40 times a s much force to accelerate the 40 lb sled. What is more likely to be happening is that at a particularly frequency, then mass of the cap and the spring constant of the legs will be such that you get metal fatigue from the ...

1

You seem to be interested in the concept of dualities. Dualities are incredibly informative in Physics in that every time we've come across one, it's led to unification of the two dual entities. You've already stated the most common one of spacetime. This was of course unified by relativity. You mentioned the Mass-Energy duality. This arises right from ...

3

Disclaimer: while I have a good grasp of GR fundamentals, it is not my area of expertise. The gravitational mass is distinct from the inertial mass as follows. The gravitational mass defines how strongly the body curves space-time, qualitatively it answers the question "how strong is the gravitational force from this object?". The inertial mass defines how ...

5

Are they identifying inertial mass with the ADM mass? There is an old style way of writing $g_{00}$ of a stationary metric, far from the matter source, as $-\left(1 - \frac{1}{2}\phi({\vec x})\right)$, where $\phi$ is the potential function for the metric, and then you can monopole expand this and identify the numerator of the $\frac{1}{r}$ term of the ...

-1

When anything moves at the speed of light, all of our physical models break down. If you were to watch a spaceship speed up to the speed of light, you would see a clock on the ship slow down and come to a complete stop when it hit the speed of light (assuming you could even see it at this point). The ship would also contract so much in the direction of ...

2

It is not the Higgs boson, but rather the Higgs field, that gives mass to the elementary particles, Higgs boson included. In fact, even in Higgsless theories, e.g. such as a technicolor, the W and Z get mass but there is no Higgs boson (although there is a composite Higgs field made of techniquarks). Said this, the Higgs boson has a finite mass because its ...

2

No. The mass of most particles is not a problem. But for the force carriers, i.e. gauge bosons like the gluon or the W and Z bosons, it is a theorem that they must be (naively) massless. But we find that the W and Z bosons act as if they have a mass! The mechanism by which this mass arises is the Higgs mechanism, but we can have masses without it - just not ...

2

Because QED in $D=2$ is a confining theory and as such it develops mass gap. The coulomb potential in $D=2$ is linear with the distance of the charges. It is one of the few exactly solvable confining QFT theories. Perhaps, I should add that by gauge invariance one can always fix $A_x=0$ while for the other component, $A_t$, the equations of motion give ...

2

On a turning roundabout, if you pull your body in to the middle, it will speed up due to conservation of angular momentum. If you lean out, it will slow down, for the same reason. Now, what about if you lie down flat on a large comfy roundabout with your feet at the centre of rotation and your head pointing out. What will happen to the roundabout speed if ...

1

It is not hard to imagine a toy universe in which different fundamental forces propagate at different speeds. However, a necessary consequence of that would be violations of lorentzian symmetry, and the ability to triangulate a preferred rest frame. Although I don't see a theoretical reason why these speeds need be the same (I might be missing something ...

1

By way of analogy, think of what happens when you blow up a balloon and let it go. It spins around, goes this way and that. A balloon rarely goes straight, without spinning. The thrust from a balloon rarely goes through the center of mass. It rotates and translates. Because the thrust vector itself turns with the rotating balloon, the translation is not ...

0

Apparently it's better to ask questions only one-at-a-time on this site, but I understand why you did it like that; these are all related. Now, these are some really interesting questions, hopefully I can help you out (there's some interesting subtleties to consider due to the thrusters). Answer 1 If you fire only B, then it will rotate, and it will ...

0

Assuming the force is in the direction of the velocity, the relativistic form of $F=ma$ is $$F=\gamma^3 m_o a.$$ The term relativistic mass has become outdated recently, but it refers to the quantity $M=\gamma m_o$, where $m_o$ is the rest mass. While not necessary, we can keep this product explicit in the following rearrangement: $$a=\frac{F}{\gamma^2 ... 0 \Sigma F = ma. Therefore, if a net force n is applied to an object, it will accelerate. You are correct that the acceleration will decrease, since m increases, but it will never quite reach zero. 0 Yeah, definitely. One example is an inelastic collision, where both masses will have the same velocity after colliding. In this case, let's say a bullet of mass m and speed v_0 hits a stationary rock of mass M and they stick together and move with a final speed v. Intuitively, you can already tell that their velocity must be in the same direction as ... 0 Consider just a single water drop. During its accelerated motion downward (its decent), does it spend more of its time above the halfway point, below the halfway point, or does it spend an equal amount of time in each? Answering that question may help you. 1 The notion of relativistic mass is an archaic one that most physicists don't use anymore. A particle's mass is a Lorentz invariant, meaning a constant property of the particle, regardless of its energy. Instead, energy and momentum are seen to be non-Lorentz invariant, such that the invariant mass remains, well, invariant:$$E^2-p^2c^2=m^2c^4. Also, the ...

4

No. What happens to an object at high speed has nothing to do with the Higgs field. There are two errors in the argument you've made to come to this conclusion: Energy is not directly equivalent to mass. An object has a certain amount of energy simply by virtue of its mass; or equivalently, an object's mass is the amount of energy it would have in the ...

1

The usual method of "studying" something, is to "take it apart" (break it into its constituent parts). Small "parts" (atoms, neutrons, protons, etc.) are held together by very large forces (nuclear, electromagnetic, etc,). Therefore, high energies are required to break them apart (so we can study them).

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