What do you think of the following transformation?
The tangent vector of a time-like curve in Minkowski space is (1,{v^1}). Now, we construct a new tangent vector w, so that
{w^0} = \frac{{\sqrt {1 - {v^1}{v^1}} }}{{{v^1}\sqrt {1 + \frac{1}{{{v^1}{v^1}}}} }}
{w^1} = \frac{{\sqrt {1 -...
What I've been trying to do is to take the tangent vectors of a curve and transform each of them so that they point in the same direction but so that their magnitude is equal to their proper time. My thinking is that this would leave the shape of the curve mostly unchanged but its length would...
Is there a way to map time-like curves in Minkowski space to curves in a Euclidean space such that the length of the curve in the Euclidean space is equal to the proper time of the curve in Minkowski space?
The energy of a particle is defined as the "time-component" of the 4-momentum, vector. In non-geometrized units, that component of the 4-momentum is E/c. The -(E^2)/(c^2) term comes from multiplying out the 4-momentum with itself: p*p = -p0*p0 + p1*p1 + p2*p2 + p3*p3. Of course, p0 = E/c, so p*p...
Ask this question: Why can't they both be right? They are running two totally different experiments, and there is no reason at all a priori to think their measurements should be exactly the same. In fact, if spacetime had no geometry, their time measurements would be random (and meaningless)...
That depends on the velocity of whatever the car is driving on. In that case you use the velocity-addition formula. If, however, the car is out in empty space, then its speedometer would have to measure its speed as 0, not 1/2 c.
The Einstein tensor, which describes the curvature of spacetime, is a rank 2 tensor, so the stress-energy tensor that appears in the field equations (and describes the distribution of mass and energy throughout spacetime) must also be a rank-2 tensor. It just so happens that the "time-space"...
Agreeing with the above poster, but to clear up another misconception, time dilation isn't due to the warping of spacetime but rather to the Lorentzian character of flat spacetime.
The only way to fail for sure is to never even try. Don't deify Einstein. At the end of the day he was human like everyone else and you can definitely understand GR if you put in the effort.
The worst thing you can do is convince yourself that you can't do something. It's perfectly possible for you to master the math necessary for GR regardless of your age so long as you have the necessary materials and determination to succeed.
EDIT: I realize how incredibly lame this sounds, but...
Your equations are wrong. For S':
\Delta t = \gamma \left( {\Delta t' - v\Delta x'} \right)
While for S:
\Delta t' = \gamma \left( {\Delta t - v\Delta x} \right)
Where the primed coordinates are the measurements in the primed frame and the unprimed coordinates are measurements in the...