# Geometry of Minkowski space

fcsc
I'm not completely sure were this post must be (math or here).

But i got a question, I want to define velocity, momentum and energy. These looks like a simple task but let me explain the problem.

I'm working in the Minkowski space, and the lorentz transformations (just geometrical one's) and i can only use geometrical arguments.

For example, if we got a lorentz transformation, we could define the velocity of the frame of reference (respect one that is quite, like a lab) as

u_i = -delta_i^4/delta_4^4

where delta^j_i is the (j,i) coeficient of the lorentz transformation.

(Extract from The Geometry of Minkowski spacetime. Naber. Springer-Verlag)

I'm completely sure these work it could be found at wikipedia (some time ago i found it there) but i don't know why i can't find it now.

If somebody could help me I will be very pleased, if you only read this thanks anyway.
(sorry if my english is not good enough)

Mentor
I'm not completely sure were this post must be (math or here).

But i got a question, I want to define velocity, momentum and energy. These looks like a simple task but let me explain the problem.

I'm working in the Minkowski space, and the lorentz transformations (just geometrical one's) and i can only use geometrical arguments.

For example, if we got a lorentz transformation, we could define the velocity of the frame of reference (respect one that is quite, like a lab) as

u_i = -delta_i^4/delta_4^4

where delta^j_i is the (j,i) coeficient of the lorentz transformation.

(Extract from The Geometry of Minkowski spacetime. Naber. Springer-Verlag)

I'm completely sure these work it could be found at wikipedia (some time ago i found it there) but i don't know why i can't find it now.

If somebody could help me I will be very pleased, if you only read this thanks anyway.
(sorry if my english is not good enough)
The four-velocity is easy, it is just the unit tangent to the particle's worldline. To get from the four-velocity to the four-momentum (which contains energy) all you have to do is multiply by the (proper) rest mass. But I don't know any geometric argument for doing that.

I assume you need those properies without reference to coordinate systems.
If P is the relevant four-momentum, and U is an observer's four-velocity, PU is the Energy, P-(PU)U is the momentum.
If V is the particles four-velocity, its ordinary velocity relative to the observer is V/(VU)-U.
I made this up right now, so it may be wrong.

Homework Helper
Welcome to PF!

For example, if we got a lorentz transformation, we could define the velocity of the frame of reference (respect one that is quite, like a lab) as

u_i = -delta_i^4/delta_4^4

where delta^j_i is the (j,i) coeficient of the lorentz transformation.

Hi fcsc ! Welcome to PF! You're asking for a geometrical demonstration that the three-dimensional velocity (u1, u2, u3) is (∆14, ∆24, ∆34)/∆44.

Will you accept a matrix demonstration?

Any observer, at time t on his own clock, has 4-position (0,0,0,t) in his own frame, which is (0,0,0,t) in the other frame, which is t(∆14, ∆24, ∆34, ∆44).

44 is the time dilation, so when the other observer's clocks says t/∆44, the 4-position is t(∆14, ∆24, ∆34, ∆44)/∆44.

So the 3-position is (∆14, ∆24, ∆34)(t/∆44) at time t/∆44.

So the 3-velocity is (∆14, ∆24, ∆34)/∆44 fcsc
Thanks everybody for these quick answer,

tiny-tim, is these demostration the one I'm looking for, would you have an idea in how to find the energy and momentum.

Dale and Ich, you had been helpful too but I'm looking for 3-vectors.

Homework Helper
Thanks everybody for these quick answer,

tiny-tim, is these demostration the one I'm looking for, would you have an idea in how to find the energy and momentum.

Dale and Ich, you had been helpful too but I'm looking for 3-vectors.

Hi fcsc ! The 3-momentum is mass times 3-velocity.

The mass is m times ∆44.

So the 3-momentum is very simple: (p1, p2, p3) = m(∆14, ∆24, ∆34). The 3-energy is more complicated. You see, the 3-momentum is exactly the first three components of the 4-momentum … which is what you'd expect!

But the 3-energy isn't exactly related to the 4-energy … it's only an approximation.

The 4-energy is mc²∆44, which is mc²/√(1 - v²/c²).

And mc²/√(1 - v²/c²) is approximately mc² + 1/2mv².

So the 3-energy, 1/2mv², is approximately the 4-energy minus mc², = mc²(∆44 - 1). I'm completely agree with your point Ich, i know you speak about 3-vectors i don't explain weel myself, my problem is that i don't have U, or PU because these is what i need to define from $$\Delta_{ij}$$.