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Very sneaky. Your particle is starting out at 2.67 times the speed of light, so

in order to do this whole thing, we must borrow some math from Dr. Einstein.

-- First of all, if your particle was ever at rest, what was its mass ? Call it M0 kgm.

We have no idea how it ever reached 2.67 times the speed of light, but however that happened, the particle's speed, before we begin to mess with it, is 8 x 108m/s .

The easiest way to approach this whole thing is to tunnel directly to the relativistic

momentum formula, without passing GO or any intermediate stages: Momentum = (rest mass)times(speed)/D kgm-m/sec ' D ' is the dimensionless factor that pops up everywhere in special relativity.

I call it ' D ' because it's so Difficult to type in a straight-text presentation.

' D ' is the square root of [1 minus (speed of the particle/speed of light)2].

Be that as it may, the mass of your particle at its initial speed ... speed1 ... is

M = M0/D1 . (That would be j 0.4045 M0 kgm, which is interesting, but we don't need it

just now.)

Its momentum is P1 = M1 x speed1 = M0/D1 x (speed1) (That would be j 323,615,934 M0kgm-m/sec , which is interesting, but we don't need

it just now.)

Notice that both the particle's mass and its momentum are imaginary. But that's OK.

This whole peculiar exercise is imaginary.

You want to double the momentum by changing the particle's speed.

Seems to me it needs to go down like this: P2 = 2 x P1

M2 x speed2 = 2 x (M1 x speed1)

M0/D2 x (speed2) = 2 x M0/D1 x (speed1) Happily, each side may be divided by M0. The particle's rest mass no longer appears, meaning that it has no effect on the answer.

From here, the rest is just a matter of excruciating algebra. The answer you want

is speed2 . It's right out there in the open on the left side of the equation, but the

square of it is also buried in a square root inside of D2 . The worst part about all of

this is that it's almost impossible to solve for speed2 while typing. So if you'll excuse

me for a few minutes, I'll do it on scratch paper, and I'll be back when I have it in a

form that I can just hand over to you.

===================

Well, I'm back. I worked it 3 times, and got the same answer the last 2 times,

so I'll propose that one as my solution.

The particle's speed must be reduced to 3.386 x 108 m/s , about 42% of its initial

speed, and now only about 13% faster than light, in order to double its momentum.

The mass and momentum both remain imaginary.

It all strikes me just as weird as it does you. But I'm confident that you won't need

this answer to enhance the progress of your research anytime soon, so I decided

that I've done my part, and I'm not to spend any more time on it.

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12y ago

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