Endercreeper01

Why would you rewind time by having negative mass? The only thing that happens is that negative mass repels positive mass and attracts other negative masses. It does not do anything different other then that. It wouldn't be called negative mass. Negative mass would cause this to happen, but there can be other things, too. For example, if an object ran into a spring, the force would towards the object, without requiring negative mass.

Do you mean [latex]f \cdot ds[/latex]? And, is f a vector field? If the above is true, [latex]f \cdot ds[/latex] is just the dot product of the vector field f and an infinitesimal length ds.

Einstein created the equation for gravitational time dilation based on special relativity. Einstein reasoned that because space-time is bent with motion, the same must be with gravitation. The curvature of space-time is described by the metric. The metric is written as [latex]ds^2[/latex], and this describes the length squared of an infinitesimal curve in space-time. When there is no curvature, [latex]ds^2[/latex] takes the form [latex]ds^2=-dt^2 + dx^2 + dy^2 + dz^2[/latex] because of the pythagorean theorem. The above is also written as [latex]ds^2=\sum dx^a dx^b[/latex] When there is curvature, we have to multiply the above by something called the metric tensor, [latex]g_{ab}[/latex]. It is a four by four matrix that corrects the pythagorus in curved space-time. The metric becomes [latex]ds^2=g_{ab} \sum dx^a dx^b[/latex] There are various components of the metric tensor, each corresponding to coordinates. For example, [latex]g_{00}[/latex] is for time, [latex]g_{11}[/latex] is for x, and so on. Now that you understand the metric tensor, you need to know that when you multiply [latex]dt^2[/latex] by [latex]g_{00}[/latex], you get [latex]d\tau ^2[/latex].Therefore, [latex]d\tau = \sqrt{g_{00}} \ dt[/latex] and that is how he found out the equation for time dilation.

It would not work, because motion travels through an object at the speed of sound. It might not move at all on the other end, because energy is lost throughout the object as it is pushed.

Do you want to know the equations that describe how space-time is curved?

We do experiments to see if our hypotheses mach up with reality. But, you have to consider all of the things that would cause something to happen. You won't be able to have a discussion of you don't provide evidence. If you do, you will be able to, but you have not provided any yet.

No, I mean that theories need to have mathematics or else they wouldn't be theories. Your idea doesn't have mathematics, so it is not a theory.

Yes, but that does not mean it will start to spin. The forces on an object in circular motion do not make it spin. The only reason it would rotate is because of the elastic forces when the string is twisted. That creates a torque of [latex]k r^2 \theta[/latex], and it has nothing to do with the circular motion itself. If it doesn't have math, it can't be a theory. Theories need to have mathematics to support them. You have to consider all factors that could cause something, not just one. You can't just do experiments, but also math with it. Without math, your theory doesn't mean much.

Why do you think that objects moving in a circular path would spin? You also need to provide mathematics in your theory.

In the "sweet spot", you would not be traveling in a figure eight for both of them. You would be in a state of equilibrium. There wouldn't be a physical universe in the past. The past is simply everything before the future. There is a way to time travel into the future, but it requires either relativistic speeds, or extreme gravitational fields. In special relativity, proper time is given by [latex]\tau = \int_0^t \sqrt{1-\frac{v^2}{c^2}} \ dt[/latex] At velocities not comparable to [latex]c[/latex], there is almost no time dilation. But, if speeds are close to the speed of light, time will be dilated by a measurable amount. In gravitational fields, [latex]\tau = \int_0^t \sqrt{g_{00}} \ dt[/latex] Or, in the Schwarzschild solution [latex]\tau = \int_0^t \sqrt{1-\frac{2GM}{rc^2}} \ dt[/latex] When [latex]\sqrt{g_{00}}[/latex] is very close to 1, as in weak gravitational fields, time dilation effects are almost nonexistent. However, when gravity is very strong, there will be a noticeable difference between proper and coordinate time.

The total mass is 20,000 kilograms (not including relativistic effects). Because the velocity is 0.1c, [latex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/latex] is 1.00503781526. Solving for the total energy gives us [latex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=1.8065659 \times 10^{21} J[/latex] Working out the kinetic energy, we find that [latex]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2=1.8065659 \times 10^{21} - 20,000c^2=9.05552510888 \times 10^{18} J[/latex] This means that the kinetic energy makes up only 0.5% of the total energy. The energy from the collision is nowhere near the energy of the sun, as that is 1.78771393 × 1047 joules.

Yes, but why would this happen? Why would time need to go backwards?

Can [latex]b(x, y)[/latex] include 0? If so, [latex]b(x,y)[/latex] can be [latex]0xy[/latex], or more generally, [latex]b(x,y)=0f(x,y)[/latex], where [latex]f(x,y)[/latex] is some other function of x an y.

At relativistic speeds, the kinetic energy will play an important factor. In relativity, the energy contained in an object is given by [latex]E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/latex]. The change in energy from being at rest and being in motion is the kinetic energy, i.e kinetic energy is [latex]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2[/latex]. At non-relativistic speeds, kinetic energy makes up a very small amount of the total energy in the object, and so it is only necessary when speeds are relativistic.

To motivate the entire class, teachers need to have a creative approach to teaching. They should also explain the reasons behind things, instead of just stating them. Grades also need to be based on skill, not age. This ensures that each student progresses at their own rate.

If the relationship between mass and volume is linear, [latex]\frac{dM}{dV}[/latex] is just [latex]\frac{M}{V}[/latex]. This is because if [latex]M[/latex] linearly depends on [latex]V[/latex], [latex]M=\rho V[/latex]. [latex]\frac{dM}{dV}[/latex] would get you the same value as [latex]\frac{M}{V}[/latex], as [latex]\frac{dM}{dV}[/latex] is just the limit as [latex]\frac{M}{V}[/latex] approaches 0.

Not really. The main difference between quantum and classical mechanics arises from the fact that the position, time, and momentum have probabilities assigned to them. Replacing [latex]i[/latex] with [latex]e^{\frac{\pi i}{2}}[/latex] would work, but it wouldn't be in it's simplest form. It is similar to replacing [latex]-1[/latex] with [latex]e^{\pi i}[/latex]. The replacement would work, but it wouldn't be in simplest form. It can be useful if you have a wave function that includes an exponent of [latex]e[/latex], as you can add [latex]\frac{\pi i}{2}[/latex] to the exponent and possibly simplify [latex]i \psi[/latex].

Even though the school system is bad in the US, the college system is much better. America holds some of the best colleges, but they have a different way of schooling.

First, you should have good grammar and spelling. To be mature, you need to act logical and make sense. If you want to be more logical, make sure to avoid logical fallacies. In order to be a good science student, you have to take the time to research the sciences.

Do you, or do you not? Spyman's question was very clear.

How is that percentage math? You need equations for it to be math. The equations you posted in the pictures were already known. Show us some equations for your theory.

Yes, but why would there be a negative time zone? Coordinate time always moves into the future, and those equations are the only ways we know of capable to distort time.

That is what I meant. Ideas in physics need to make predictions, and it would be a theory if it makes accurate predictions. By mathematics, I mean that it needs to have mathematical predictions if it is an idea in physics. The big bang has this, as it predicts the CMB, for example. There also needs to be evidence for it. Going back to Rajnish's post, he says the following: There is no evidence of white holes or wormholes, but let's assume there was. He would need a mathematical framework describing the space-time around this white hole or wormhole at the big bang, and also a proof of why this would happen. To make it into a theory, it needs to be tested. If it is untestable, it is not a theory.

Not a nation, since there is only 65000 people (2/3 of which haven't made their first post yet).

The big bang theory is a theory because it has been repeatedly tested, and there is a mathematical proof behind it.