# Tag Info

43

To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation. Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" ...

12

Associating a particle with a classical field is what quantization does, practically by definition. Take a classical field, make it an operator, and find the eigenstates of its Hamiltonian. The result is particle states, whatever form the field takes. "Graviton" is just the name we give those hypothetical particles. Even though it's basically just a ...

11

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. Even when at rest in space, the apple still advances in space-time. Here is a visualization of the falling apple in distorted space-time: http://www.youtube.com/watch?v=DdC0QN6f3G4

7

It's tempting to think of gravity as some kind of interaction between the two bodies involved - maybe some form of signal (gravity wave?) sent between the two bodies. If this were the case then you would indeed have to allow for a propagation delay as the signals were sent between the two bodies. However this is not how gravity works. A massive object ...

6

Let's put a more precise description to the other answers, particularly Neil's. First, note that there is a Gauss Law for static gravitational fields, owing to the inverse square nature of the static gravitational attraction. See this answer here and note that the argument it makes uses only the inverse square dependence. (Actually, the Gauss law also holds ...

5

I don't think I can rigorously prove that simulation engines don't need to worry about the (possibly? I don't know if there's a reliable measurement) finite speed of gravity, but I can offer some lines of thought that point in that direction. I'll start with your question 3. Suppose that gravity does have finite speed equal to $c$. Your question seems to be ...

5

As you already suspected, your weight would gradually decrease as you descend into earth. Technically, at the center of the earth you should feel no gravitational force (from earth, I don't know how strong additional sources of gravity affect us, but my guess would be that they are negligible). Once you start "rising" on the other side, gravity would slowly ...

5

If the center of mass is not moving, then because there are no external forces it must remain in place. But if the bodies meet somewhere else, then the place where they meet would then be the center of mass, which is a contradiction since the center of mass must remain where it was at first! Edit: A (hopefully) clearer explanation. Suppose that at $t=0$ ...

4

The height at which the balloon floats is determined by the density of the air outside of the balloon. Once you are high enough in the air that upward force generated by the difference in densities $\rho_{\text{helium}}$ and $\rho_{\text{air}}$ is exactly canceled by the gravitational force on the balloon plus its counterweight, the balloon will remain at ...

4

The way to do problems like this is always to use the Lorentz transformations. Choose some sensible spacetime points in the rest frame $S$ and use the transformations to see what those points look like in the moving frame $S'$. In this case this is what the points look like in $S$: The spacetime points are labelled as $(t, x, y)$ - we'll ignore $z$ since ...

4

No - one of the key differences between electromagnetism and gravity is that the latter only has a single "charge" and everything always attracts. This holds true even in general relativity, but as long as you are only talking about the mass/energy density of material. In GR, pressure itself can cause gravitational effects, and these can indeed look ...

3

Pretty much no. The problem is that you are not (pardon me :-) ) a rigid body, so you're going to feel a certain amount of force from the wind regardless of what sort of weights you're carrying. What can help is walking with your feet farther apart, which gives you a more stable base to work from, and to learn to turn your body sideways to the wind as ...

3

This is a fundamentally pointless question because negative mass doesn't exist (or so we think!), but I'll answer anyway because the answer is so unexpected. Suppose we take our two massive bodies: Then the gravitational force between them is repulsive because: $$F = \frac{G m_1 m_2}{r^2}$$ and $m_1$ and $m_2$ have different signs. But let's ...

3

Please take a look at this link Hole through the earth. It seems that the physics of passing through the earth would not be as it was implied by Total Recall in which gravity switches suddenly. While gravity is normally affected by radius squared (gravity much stronger proportionally as you approach a gravitational body), inside a gravitational body, its ...

3

The atmosphere is more or less at equilibrium with regards to being compressed by gravity, so there is no atmospheric heating caused by gravity (otherwise the atmosphere would gradually get closer and closer to the Earth, which clearly is not the case). Heating of the atmosphere is almost entirely due to radiant energy from the sun, along with terrestrial ...

3

Yes. $c$ is the highest possible speed for light/any information to travel. So for a person 1 light year away, he wouldn't even realise that the object has disappeared, until the light carrying that information has travelled there. You can also think about this another way. Gravitational waves (which researchers today are trying very hard to detect) can ...

3

I'm not sure if this problem was ever solved in classical electrodynamics. However, it is (somewhat) solved in quantum field theory electrodynamics (QED). In QED, self-interaction has noticeable effects on quantities such as the observed mass of a particle. Furthermore, the self-interaction effects create infinities in the theoretical predictions for ...

3

As to the first paragraph, gravity shows up as geodesic deviation; initially parallel geodesics do not remain parallel. Since, for a freely falling particle, the proper acceleration (the reading of an accelerometer attached to the particle) is zero, it is correct to say that a particle whose worldline is a geodesic has no proper acceleration. But it is not ...

3

Re (1): the relativistic extension of velocity is four velocity. Pre relativity we separate the spatial and time coordinates, then define velocity as $dx/dt$ etc. In relativity this no longer makes sense because $x$ and $t$ are both spacetime coordinates and indeed the Lorentz transformations will mix them up. So we define the four velocity $U$ as the ...

2

I suppose my first major question is simply, has this problem been solved yet? After a bit of research I came across the Abraham-Lorentz force which appears to refer exactly to this "problem of self-force". As the article states the formula is entirely in the domain of classical physics and a quick Google search indicates it was derived by Abraham and ...

2

This answer is an complement to Chris White's answer. Fist of there is no explicit equations for the position of an object following a Kepler orbit as a function of time. However, when the initial conditions are known, the path the object will follow can be found, as well as the velocity, acceleration, ect. at any given position. This path can be described ...

2

It seems you've done the hard part already, which is to evolve the object's position as a function of time. And moreover, the simulation seems stable over a number of orbits. (But eventually things start to go wrong; you may want to look at an answer I wrote to What is the correct way of integrating in astronomy simulations?) So my understanding is all you ...

2

The situation is too ill-defined for an answer. The problem there is that in general relativity, you do have general conservation laws that follow from the Einstein field equation. In the asymptotically flat case, you have conservation of a global ADM mass, and in all cases there is a local covariant conservation law that requires the stress-energy to be ...

2

Have a look at Why do same/opposite electric charges repel/attract each other, respectively? for an explanation of the attraction between charges (I'm using the word charge in the general sense). It looks long and intimidating at first glance but persist because it isn't really. For there to be a gravitational repulsion there would have to be negative mass ...

1

In most cases, it doesn't really make sense to talk about a lowered effective mass caused by sitting in a gravitational potential well, since the equivalence principle says that locally the spacetime looks flat, and hence it looks like the gravitational field vanishes. However, in certain special cases, there is a sensible notion of energy that is ...

1

Whenever it seems like two water levels should be equal but aren't, either there is a physical restriction preventing flow (like a dam keeping upstream waters higher than downstream, or surface tension causing meniscuses or capillary action), or there is energy being expended to put water back upstream as fast as gravity is pulling water downstream. In a ...

1

You are quite correct that the position of the tectonic plates does affect Earth's gravity. The most accurate measurements of this were made by NASA's GRACE satellites. The results are generally shown as a geoid, that is a deformed globe showing where gravity is high and where it's low: Note the scale is exaggerated to highlight the differences. The scale ...

1

The rear of the bus normally overhangs the back wheel further than the front of the bus overhangs the front wheel - especially on traditional yellow school buses. When the bus goes over a bump it pivots around the other set of wheels, the distance you move is the ratio of the distance between the wheels and the distance from your back seat to the front ...

1

I suggest you to take a look at Appendix E of Wald's General relativity book. There he derives all boundary terms which appear in the variation of Hilbert's action. There are only 3 terms coming from this variation. Two of them give Einstein's equation. The surface term comes from the other term, $g^{ab}\delta R_{ab}$, which is a total derivative, as ...

1

You are correct - the force is constant in all four cases. Since each of the situations describes a "uniform spherical shell of matter," you can assume that the mass is concentrated at the center of that shell, as per the shell theorem cited. If you've learned Gauss's Law for electric fields, it can be applied to this problem. Gravitational force, following ...

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