Recent content by JCMateri

Thanks. I suspected that.

Sorry,but my tex material keeps disappearing when I save.

I don't see how that can be right. Isn't the RHS = sinh(x)+C and its derivative does nor equal \sqrt{1+cosh^2(x)}

I am having trouble with the arc length for hyperbolic sine. Can anyone help? $$L=\int_{0}^{X}\sqrt{1+[\frac{dsinh(x)}{dx}]^2}dx=\int_{0}^{X}\sqrt{1+cosh^2(x)}dx$$ I'm having trouble evaluating the final integral.

I no longer think the last part is correct where I said: It also tells us when ( ##t## in the Rindler frame and ##T## in the stationary frame) the accelerated object catches up (from its starting position at ##(X.T)=(X,0)##) to an object whose world-line is the radial line with starting...

I think I understand it better now. The world-lines in the Minkowski diagram consist of a foliation of hyperbolae indexed by ##x##. The radial lines represent time. Where an hyperbola intersects a radial line corresponding to a given time value (##t## in the Rindler frame) tells us its position...

I'm still working on the above. One thing I notice is that if ##x=1/g## we have ##dx^2/dt^2=1##. Furthermore, we can see from the above that ##X^2-T^2=x^2##. I'll return to this soon.

Let ##S\sim T,X,Y,Z## be stationary coordinates with a source of light at ##(X,Y,Z)=0## and let ##S'\sim t,x,y,z## be the coordinates in an accelerated frame with relative acceleration ##=g## along the ##x## direction for both systems. $$\begin{eqnarray*} T&=&xsinh(gt)\\ X&=&xcosh(gt)...