elfmotat

Lidal, the solution to any second-order differential equation of position requires two initial conditions: initial position and initial velocity. If you specify those two initial conditions then you get a unique solution. The general solution to [math]GM/r^2 = \ddot{r}[/math] is an ellipse, and the exact form of the ellipse depends on initial conditions.

As I hope has been sufficiently pointed out, names are just names. In relativity it is pretty common to refer to a 3D volume + time as a "four-volume". A 4D spacial volume may also be referred to as a "hypervolume." But just as length is a 1D volume and area is a 2D volume, higher dimensional volumes are most practically referred to as "volumes." Making distinctions between length, area and volume is useful, but when you add additional dimensions the distinction becomes less practical.

What is [math]\phi[/math]? Is it a field or a coordinate? Also, what do you mean by "covariant?" Covariant in the sense that it's a covector, or in the sense that you want it to transform like a tensor? If [math]\phi[/math] is a coordinate then obviously [math]p_\phi[/math] doesn't transform like a tensor because it's just a component of [math]p_\mu[/math]. If [math]\phi[/math] is a field then there's an associated conjugate momentum field [math]\pi (\mathbf{x}) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}(\mathbf{x})}[/math] where [math]\mathcal{L}[/math] is the Lagrangian density of the field.

Lidal, may I ask why you believe elliptical orbits aren't compatible with Newtonian gravity? Because they are.

If you know F(t) and x(t) then you can find F(x), which is what you need to calculate work.

Space isn't "made" of anything. It's not a rubber sheet that decreases in thickness when stretched. You're taking the analogy far too seriously. I'll give you a little technical insight: say the universe is flat and one-dimensional. Now we label each point in the universe with a coordinate x. How do we find the distance between two points x1 and x2? We need to use something called a "metric." It tells us the distance between two infinitely close points. If space isn't expanding then the metric is very simple: [math]ds=dx[/math] where ds is an infinitely small distance and dx is an infinitely small change in the x-coordinate. If we want to find the distance between two points x1 and x2 we just take an integral: [math]s=\int_{x_1}^{x_2}dx=x_2-x_1[/math] Now let's say that the distance between two points isn't always the same, but actually depends on a function of time (i.e. space can be expanding or contracting): [math]ds=f(t)dx[/math] Now the distance between two points is: [math]s=\int_{x_1}^{x_2}f(t)dx=f(t)\left [ x_2 -x_1 \right ][/math] If we let [math]f(t)=kt[/math] for some constant k, then space will expand at a constant rate. If we let [math]f(t)=kt^2[/math], space is expanding with an accelerating rate. [math]f(t)=\frac{k}{t}[/math] would describe a contracting space. The function f(t) which describes our physical universe is determined by the Friedmann Equations (which come from the Einstein Field Equations).

As swansont mentioned at the end of his post, relativistic velocity addition. If I observe you moving at speed v relative to me, and you throw a baseball at speed u relative to you, then I observe the baseball moving at (u+v)/(1+uv/c2) relative to me. For u & v much less than c, this is approximately equal to u+v.

I'm just trying to get a grip on where the OP's confusion is stemming from. If it's coming from a belief in aether theories, I thought it worthwhile to point out that while they can be perfectly consistent with SR and EM, they are incompatible with much of modern physics.

I'm not sure exactly what you're asking. Static charges attract (or repel) each other by Coulomb's Law: [math]F=\frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}[/math] Similarly, masses attract each other by Newton's Law: [math]F=-G \frac{m_1 m_2}{r^2}[/math]

Unless there's a medium which is approximately at rest to the Earth which light travels through at c, dubbed "aether." Of course Michelson-Morley + SR makes this an ad hoc unnecessary assumption that doesn't generalize well to curved spacetimes. Lorentz Ether Theory is perfectly consistent with SR and Maxwell's Equations, but it's incompatible with much of modern physics.

We're just sick of your nonsense, that's all. Do you really think that over a century's worth of physicists are that stupid? Or do you think it's just a little more likely that you're having problems because you don't understand SR? How about you take another look around this thread. You've gotten your answer about a dozen times already. Velocity. Nobody ever said differently.

Unless there's an equal and opposite force to counteract it, then yes.

K40 and Uranium+Thorium radiometric dating of lunar samples.

"Straight lines" are a subtle topic in GR, but essentially the answer is yes.

This is just algebra: [math]\frac{kx}{x}=k~~ \text{if}~x\neq 0[/math] It's undefined if x=0.

The forum is meant for legitimate science questions, not to entertain the delusional fantasies of crackpots. If you want to learn Relativity then go out and buy a textbook. If you have a specific question that comes up in your study, ask here and I'd be more than willing to help. What you're doing now I have very little patience for.

What started off as a good question has now devolved into crank nonsense. If you want to learn SR then go buy a textbook. If you don't, then stop wasting our time.

In the modern formulation of Special Relativity, mass does not increase with velocity. The only mass which is defined is sometimes called "rest mass" or "invariant mass." Mostly, we just call it "mass." It is an invariant quantity, meaning it is the same in all reference frames. It is defined as: [math]m=\frac{1}{c^2}\sqrt{E^2-p^2c^2}[/math] where [math]E[/math] is the total energy of a particle and [math]p[/math] is the magnitude of its momentum. Back when Relativity was a relatively new subject (no pun intended), physicists invented a quantity called "relativistic mass." Relativistic mass does increase with velocity. It is defined as: [math]m_{rel}=\frac{E}{c^2}[/math] This gives us the relationship [math]m_{rel}=\gamma m[/math], where [math]\gamma = (1-v^2/c^2)^{-1/2}[/math] is the Lorentz factor. It turns out that relativistic mass has very little useful application, and hence is no longer used by the vast majority of the physics community. Unfortunately, physics popularizers like Michio Kaku, Brian Greene, etc., tend to write about how "mass increases with velocity" in their books and interviews. I assume this is because they think it makes the concept that "you can't travel faster than light" more intuitive. Even more unfortunately, this tends to confuse the living hell out of people who go on to actually study Relativity. It sure confused the hell out of me when I started learning Relativity, and I see a couple threads a week on various physics sites like this where people are asking about mass increase. So, on to your question: mass doesn't increase. A good way to demonstrate that massive objects obtaining a velocity of [math]c[/math] is impossible is to calculate what its kinetic energy would be. The formula for kinetic energy in SR is: [math]K.E.=(\gamma -1)mc^2[/math] By the work-energy theorem (which still holds in SR), this is equal to the work required to bring a particle at rest to the given velocity. As you can see, as [math]v \rightarrow c[/math], then [math]K.E. \rightarrow \infty[/math]. So you'd need an infinite energy supply to bring a massive particle to [math]c[/math], which is clearly impossible.

This is one of the dumbest, most offensive threads I've ever had the displeasure of reading.

I have one brother and one sister as well. And I've never attempted suicide, nor has the thought ever crossed my mind.

I'm not sure what you're asking. You can have gravitational waves in a vacuum, similar to how electromagnetic waves can exist far away from any charges.

How that paper ended up on arXiv is beyond me.

This looks like journalistic sensationalism to me. I fail to see where the uncertainty principle is violated.

Nevermind, you're right. Fortunately/unfortunately I'm a bit intoxicated at the moment.

Condition 1. is satisfied by setting every element to [math]1/k[/math], but condition 2. is only satisfied if [math]n/k=1[/math], i.e [math]n=k[/math] in which case the condition that [math]n\gg k[/math] is not satisfied.