bob012345's latest activity

That’s like saying I’m a billionaire but I just happen to have zero billion dollars at the moment.

Perhaps, but we are dealing with a gamma ray and a radio wave in the same reference frame. Consider what happens if both interact with a...

I wonder if that is just an approximation? I would think that in principle, the energy difference between a gamma Ray and a radio wave...

Well something needs to have mass in order to curve spacetime.

So what effective mass is used when starlight is bent by the sun? Isn’t it just ##m=\frac{E}{c^2}##?

Alternative to @Sagittarius A-Star's rapidity method, you can use ##v/c=\frac{at/c}{\sqrt{1+a^2t^2/c^2}}##. Using years and light years...

I do not know what typo you are referring to? Perhaps it’s the way I specified the termination term. I meant if the ##n^{th}## stage has...

@Baluncore, I think that evaluating the ladder from left to right, instead of right to left, can be better (for ##\vert b \vert<1##). A...

Take the limit of ##\frac{R_n R}{R_n+R}## as ##R_n→0## which gives zero. So if one shorts the first stage the minimum resistance is 1 unit.

In post #17, I gave an accelerator that works well on an alternating series like the golden ratio. The accelerator takes three...

If one takes the Golden Ladder equation ##R^2 -R -1=0## and instead of solving it exactly, write it as ##R = \sqrt{1+R}## then start...

Yes @Baluncore, that is what I found, and I put in the figure above. In fact, the continued fraction that computes the equivalent...

I like this problem. It overturns every assumption we would normally make.

I was wrong to say changing the tail doesn’t change the ladder but like I said, it literally changes the ladder to be a different...

Ok, but then changing the tail to low values is changing the problem significantly. It no longer the ‘tail’ but a different ladder.