elfmotat

Here's a way to think about it: Say we place two small objects some distance away from each other so that they are at rest with respect to each other. Now say we graph the position (in one dimension) of these objects vs. time in flat spacetime. What would this graph look like? It would just be two parallel lines, as shown below. But what happens if we draw the same graph on a curved surface (for example a sphere)? The result is shown below: As you can see, the objects start out at rest with respect to each other (the lines start out parallel). But as the objects follow their straightest possible paths through spacetime (i.e. as they follow their respective geodesics) the distance between the objects appears to decrease over time, and the rate at which the objects approach each other increases over time (they are accelerated towards each other). Even though no forces are present and the two objects are simply following the most natural path through curved spacetime, they accelerate toward each other and eventually collide. Now obviously a sphere isn't a realistic spacetime (every geodesic is a CTC), but it demonstrates the idea well.

Also, Newtonian gravity can be represented as time curvature with the following metric: [math]ds^2=-\left ( 1-\frac{2M}{r} \right )dt^2 + dr^2+r^2 d \Omega^2[/math] This is just the Schwarzschild metric without any spacial curvature. So, essentially light is still affected by Newtonian gravity, but it's not affected as much as it is in the "correct" theory of General Relativity.

You, but backwards.

I don't understand the question. Gravity is present whenever there's energy/momentum.

Scale isn't a dimension. The dimension of a space is the minimum number of coordinates you need to specify a point in that space. Spacetime is four-dimensional because you need four numbers (three spacial and one temporal) to describe points in it.

In n-dimensional space, a scalar field (rank (0,0) tensor field) has n0=1 components (i.e. it's just a number). It has the same value no matter what coordinate system you use. An example of a scalar field is the temperature at every point in 3-D space. What you have is a function T(xi). The xi can be any type of coordinates (Cartesian, cylindrical, spherical, some crazy coordinate system). It doesn't matter what you call a point, it has whatever temperature it has. So if yi is the same point in some other type of coordinates, what you find is that T(xi)=T(yi) A vector field (rank (1,0) tensor field) has n1 components. For example, in 3-D space it takes three numbers to make up a vector. The thing that characterizes vectors is that their components transform in such a way that the magnitude of a vector (the scalar product of a vector with itself) is the same in any type of coordinates. This makes sense because, for example, a car traveling at 20 mph is still traveling at 20 mph no matter what type of coordinates you label the road with, what direction you call North, etc. Then there are tensors of higher rank. A (2,0) tensor has n2 components. So a rank (2,0) tensor in 3-D space has 9 components. The components of higher rank tensors transform in such a way that scalar contractions of tensors with other tensors have the same value in any coordinate system. It is this property that makes tensors interesting and useful. Physical quantities need to have to the property of coordinate invariance because they do not depend on how you label points in space, so all physical quantities must be represented by tensors.

A-wal: Trying to explain any physical theory without math is a bad idea. When I was in High School I used to read popsci books that attempted to explain Special and General Relativity, Quantum Mechanics, etc. in a similar fashion to your post. They generally left me more confused than before I had read the books. They also created misconceptions that took me a while to shake when I started learning the theories for real. If you want to learn a theory, you have to learn the math. There's no way around it. The situation is perfectly symmetrical (i.e. both twins see each other's clocks running slowly) until one of the twins turns around. One twin must turn around to get back to his sibling, and in doing so he must accelerate. Only one twin actually feels an acceleration, and that's how the symmetry is broken. The twin that turns around will register less overall time on his clock. You might like this little pdf I wrote last year: http://pdfcast.org/p...al-relativity-2 . It's mostly a compilation of different forum posts I had written addressing people's questions about relativity. I tried to take a historical approach and make everything as clear as possible for someone with very little experience with physics. Most of the math is just mid-High-School-level algebra. I think it should clear up your confusion here.

Nonsensical pseudoscience. It looks like he watched "What the Bleep Do We Know?"

If my coffee is too hot, I wait longer before pouring in the milk because it radiates faster when it's black.

I'm probably well-versed enough to be able to help you, so PM me if you wish. I'll warn you though: I'd be willing to bet a substantial amount that your "theory" is too qualitative/vague to be formulated mathematically. It probably isn't even new. People come to different science boards and make claims similar to yours all the time, and I have never once been impressed by any of them.

The formulas for relativistic momentum and kinetic energy are different from the approximations used in Newtonian mechanics. The difference becomes more and more drastic as v approaches c. Here's a simple way to derive E=pc starting with only the formula for relativistic momentum and a few assumptions (i.e. the Work-Energy Theorem holds valid in relativity, our final equation holds for massless particles). The formula for relativistic momentum can be found by analyzing collisions with the Lorentz transformation, and what you end up with is [math]p=\gamma mv[/math] where [math]\gamma=(1-v^2/c^2)^{-1/2}[/math]. From there you can find the formula for relativistic kinetic energy by using the Work-Energy Theorem, calculating the work done in bringing a mass at rest to a velocity v: [math]E_k=\int Fdx=\int \frac{dpdx}{dt}=\int vdp=pv-\int pdv=\gamma mv^2-m\int \frac{vdv}{\sqrt{1-v^2/c^2}}[/math] Evaluating that integral gives: [math]E_k=\gamma mc^2+\varphi [/math], where [math]\varphi[/math] is some constant of integration. It's easy to solve for [math]\varphi[/math]; all you have to do is take the fact that when v=0 (i.e. the object is at rest) its kinetic energy is taken to be zero, and [math]\gamma=1[/math]. So what you get is: [math]0=mc^2+\varphi~~~ \Rightarrow ~~~\varphi =-mc^2~~~\Rightarrow~~~ E_k=\gamma mc^2-mc^2[/math]. The term "mc2" looks like some intrinsic energy associated to a mass (and is appropriately called "rest energy"). Adding this term to both sides of the kinetic energy formula gives the total energy of a body: rest energy + kinetic energy: [math]E=E_k+mc^2=\gamma mc^2[/math] Now that we have the formulas for the total relativistic energy and the relativistic momentum of a moving body, we can be a bit tricky to find some relationships between them. We can start by solving for [math]\gamma[/math] in both equations: [math]\gamma =\frac{E}{mc^2}=\frac{p}{mv}~~~\Rightarrow ~~~ Ev=pc^2[/math] So there's a neat little formula relating energy and momentum. We can also use this result to get to the equation you're looking for. Start by squaring the energy equation and rearranging: [math]E^2=\frac{(mc^2)^2}{1-v^2/c^2}~~~\Rightarrow ~~~E^2-\frac{(Ev)^2}{c^2}=(mc^2)^2[/math] Now substitute pc2 in for Ev: [math]E^2-\frac{(pc^2)^2}{c^2}=(mc^2)^2~~~\Rightarrow ~~~E^2-(pc)^2=(mc^2)^2[/math] As you're aware, this equation holds for all particles, including massless ones. Setting m=0 gives E=pc.

This is the first time I'm seeing anything like this. From the looks of it, I suppose it might not have a closed form.

If anything, I'd say the opposite. His early work (namely "On the Electrodynamics of Moving Bodies") was inspired by the work of Lorentz and others. His ideas regarding relativistic gravity were much more original.

I think most people would agree that the "present" is infinitesimally short. It's like a point on a number line: it has no duration, but it exists nonetheless. The present is also relative: my present is not the same as someone who's moving relative to me.

Light can be slowed to less than the speed of sound in some materials. This says nothing about "c" itself, because "c" isn't the speed of light. It's still just called "the speed of light" for historical reasons. "c" is defined as 299,792,458 m/s. It just so happens that light travels at this speed in a vacuum, but indeed any massless particle travels at this speed. "c" is better thought of as "the universal speed limit" rather than "the speed of light."

The reason the Einstein Field Equations (and therefore the Cosmological Constant equation you give) have a factor of pi is merely a matter of (historical) convention. When Newton first wrote down his equation for the force of gravity, [math]F=Gm_1m_2/r^2[/math], he included a proportionality constant "G." Instead of breaking this constant up into multiple other constants, he used the logical (at the time anyway) choice of only one constant. Contrast this with Coulomb's Law in electrostatics: [math]F=q_1q_2/4\pi \epsilon_0 r^2[/math]. Instead of using one constant of proportionality, it uses three: 4, pi, and the electrostatic constant (although some introductory textbooks absorb them into a single constant, [math]k=1/4\pi \epsilon_0[/math]). The reason for using three constants instead one is because it simplifies the following Maxwell equation: [math]\nabla \cdot \bold{E}=\rho_q /\epsilon_0[/math]. This is equivalent to Coulomb's Law in simplifying cases, but it's also much more general. The factor of 4*pi appears in Coulombs law when the above Maxwell equation is applied to a spherically symmetric surface (because the surface area of a sphere is 4*pi*radius2). Now look at the analogous equation for gravity: [math]\nabla \cdot \bold{g}=4\pi G\rho_m[/math]. This equation is, again, more general than Newton's force law. The choice of a single constant simplified his force law, but that means an additional factor of 4*pi must be added to the above equation so that when it is applied to a spherically symmetric surface, the 4*pi cancels. When the Einstein Field Equations were first written down (taking the cosmological constant to be zero for simplicity), they looked as follows: [math]R_{\mu \nu }-g_{\mu \nu }R/2=\kappa T_{\mu \nu }[/math]. [math]\kappa[/math] is again some constant of proportionality. Since General Relativity is, in a sense, a generalization of Newtonian gravity, this constant must be chosen so that the EFE's reduce to the gravity equation I wrote above in the low-energy limit. What you find is that [math]\kappa=8\pi G/c^4[/math]. So if Newton had originally chosen to break up his proportionality constant into something like [math]F=Gm_1m_2/\pi r^2[/math], then the EFE's would look like [math]R_{\mu \nu }-g_{\mu \nu }R/2=8G T_{\mu \nu }/c^4[/math]. So, in summary, the pi appears because of an arbitrary choice for the value of G. If G were chosen differently, the factor of pi wouldn't be there.

So are you going to post it or not? If not, what's the point of this thread?

I'm not really sure what you mean when you say you "simplified time dilation." Post what you're talking about.

This just seems like a bunch of nonsense to me.

What do you mean by "in motion?" You can always boost to a reference frame where something is at rest, with the exception of luxons which travel at the speed of light in all reference frames.

He has a point. SR would have been discovered soon with or without Einstein. Einstein just figured it out before everyone else.

It's not worth much. IQ is a terrible reflection of intelligence. For SR, I agree. As for GR, I wouldn't be surprised if something resembling the theory wasn't seen until as late as the 50's if Einstein wasn't around.

That's not a test. You need to white your "theory" mathematically. Once you do that, we can use it to make a numerical prediction. Saying radioactive materials decay "faster" is not a prediction. How much faster? Is the difference measurable?

Physics is an empirical science. Your qualitative description does not make any falsifiable predictions, and is therefore not science.

Your premise that "The Big Bang is not a lazy thing" is ill-defined. How do you figure that? What about the big bang qualifies it as a "non-lazy phenomenon?" The post was also entirely irrelevant to the conversation. In the context of classical mechanics (which, as the section name suggests, is what we're discussing), saying that nature is "lazy" (where I implied in my previous post that "lazy" is by definition doing as little work as possible) is completely accurate. This is the premise of "the least action principle" (which is what the thread is about), and it is an experimentally confirmed principle. If you don't know what the conversation is about then I suggest you refrain from making posts which may make you look silly. In any case, this thread has been sufficiently derailed. **I also see that I've been -rep'd for my first post. I'm not suggesting it was you, but whoever it was is certainly being immature.