Meir Achuz

E's eq. is irrelevant to T. T depends only on E and p. The problem (usually insoluble) posed by E's eq. is finding g. Writing E's eq. does not change the fact that the trace of T, if diagonalized is E^2-p^2. The short answer to your original question is "Yes."

Revprez, The trace is E^2-p^2 so the sign of E is not involved.

Positive definite for a matrix means that if the matrix is diagonalized the trace will be positive. This trace is just the length squared of the E-p four-vector, so I think my previous answer applies. By the way, I see you used Latex. I know latex, but how do you get this forum to typset it? Thanks.

revprez: I assume your question refers to an energy-momentum four-vector with E>p (with c=1). The velocity, v=p/E, will always be less than 1, implying a lengthened geodesic. By "indefinite energy momentum state", I assume you mean one with E<p. This would have v>1, implying a tachyon. Some speculative theories do postulate tachyons, but they are not in the "standard model", and have never been observed.

You could ask the same quesiton in NR physics: How many three-vectors are there?

A rotating object cannot be treated in SR. It is a difficult (still unsolved) problem iin GR.

No. The emitted photons exert a very small repusive force on the Earth. This "radiation pressure" is not usually considered in calculating the orbit of the Earth.

The wave function of a photon is the EM four-potential, A^\mu.

If a spaceship travels at v=.9c toward a star 5.4 LY away, it wll reach it in 6 years. (5.4/.9)

My answer is no.

Your three questions remind me of the story of the Emporer's cloths. Because you are only 15 (but certainly thoughtful), you have asked the three most difficult questions for practioners of those theories. Your first question is to the point, and has also been asked by Nobel laureates. This is the first time since Galileo that an untestable theory has been called science. I will leave your other two questions for cosmologistgs.

"The energy and mass of an object are closely related, but not equivalent. To within factors of $c$, the energy is the time-like component of the momentum four-vector, while the mass is the invariant length of the four-vector. The energy in the rest system equals $mc^2$, but while in motion, the energy varies with velocity".

yourdadonapogos: Your algebra is correct, but your interpretation of f\lambda as a particle's velocity is wrong. f\lambda is the wave velocity (which is greater than c) of the particle's wave function. The velocity at which we would interpret the particle to be moving is the group velocity of the wave packet. This is given by dE/dp, which equals pc^2/E which is less than c.

There are a large number of four vectors in SR. You list some of them, but also some that are not four vectors. 1. The velocity four-vector is (g,g v), where g is the SR gamma and v is the usual three-vector velocity we would measure as dx/dt. 2. Acceleration is not a four vector in SR. 3. Force (and acceleration) are not useful variables in SR. A four-vector force, the so called "Minkowski force" can be defined by F=(g dW/dt,g dp/dt} where W is the energy and p the three-vector momentum of a particle. 4. The energy-momentum four-vector is (W,p). 5. The displacement four-vector is (t,x,y,z). 6. The four-potential is usually written as (phi,A), where phi was the NR scalar potential and A the NR vector potential. A is no longer referred to as "magnetic" because it also affects the electric field. 7. The current four-vector is (rho,j). 8. There is a four-divergence, which is a four-vector. 9. Taking the four-divergence of tensors lead to other four-vectors. 10. ... and so on.

An elliptic integral is an integral. Converting one integral to another integral is another integral, and not an answer.

In special relativity, if the frequency in a moving spacehip is f, the frequency measured on Earth is given by: f'=\gamma f(1+v cosA/c), where v is the ships speed, c the speed of light, \gamma=1/sqrt(1-v^2/c^2), and A is the angle between the photon direction and the observation point on earth, as measure in the space ship. This equation is different than the acoustic Doppler effect, but is usually called the "Relativistic Doppler shift".

Look in any UG mechanics book that derives conservation of momentum from NII, and follow the steps backward. In a grad text (like Goldstein), conservation of momentum is derived from the Lagrangian. Then NIII is shown to follow from that.

I am puzzled by your question. As Johnny5's first step shows (The constant C should not be there.), the answer you give is wrong.

NIII follows from conservation of linear momentum, which is more basic than simple action and reaction. Newton never incorporated magnetism into his mechanics. The magnetic interaction between two moving charged particles does not satisfy N III. If the changing momentum of the electromagnetic fields of the two particles is incuded, then overall momentum is conserved. Energy has nothing to do with NIII. I do not understand the wording of your original question, and the quote seems to be archaic.

The relative velocity, V, of the two particles (each with velocity v) will be given by the formula V=2v/(1+v^2/c^2), which is still less than c. The energy involved in their collision depends on their energy, which is not directly related to their relative velocity V, but is given by E=2mc^2/sqrt(1-v^2/c^2).

The simple nonrelativistic rule for addition of velocities is wrong, because it disagrees with experiment. It disagrees with conclusions from the Lorentz transformation, which has had numerous experimental verifications. More directly, if a fast moving elementary particle emits a photon in the forward direction, the photon velocity is still measured as c. Relativistic velocity addition has also been verified in decays like pi-->mu+nu.

Most of the errors in this thread arise because some people have implicitly assumed, as obvious, the simple nonrelativistic rule for addition of velocities, but it is wrong.