Lorentz transformation?

Question

Is it correct to say that the Lorentz transformation describes the mass of a particle as a function of velocity? Would the inverse function describe velocity as a function of mass? What would that function be?

Answer 1

Well, I dunno about the Lorentz transformation, but even if your first statement is true, the latter need not be. Just because the fuction f(x)=y does not ALWAYS means the function f-1(y)=x because sometimes f inverse is not a function...for example, take the parabola y=x^2. Its inverse is not a function. So simple mathematics tells us not to extrapolate too much because we may run into errors...

That said I basically don't know.

Answer 2

Yes, you can:

m' = m / √(1 - (v/c)²)

Where:

m' is the relativistic mass, and m the rest mass. I've combined some steps to make it shorter.

m' = m / √(1 - (v/c)²)

√(1 - (v/c)²) = m/m'

1 - (v/c)² = (m/m')²

(v/c)² = 1 - (m/m')²

v = c √(1 - (m/m')²)

You can also do it with time dilation:

t' = t / √(1-v²/c²)

√(1-v²/c²) = t/t'

1 - (v/c)² = (t/t')²

(v/c)² = 1 - (t/t')²

v = c √(1 - (t/t')²)

Edit: Sorry, my mistake. I saw "mass," "particle," and "velocity" and was thinking mass in special relativity.

However, what I gave works. You can calculate the velocity of a particle from it's relativistic mass/energy. Take a look at this cool link:

http://www.fourmilab.ch/documents/OhMyGo...

Edit: If you replace "Lorentz transformation" with "theory of special relativity" and rephrase the question as:

"Is it correct to say that the theory of special relativity describes the mass of a particle as a function of velocity? Would the inverse function describe velocity as a function of mass? What would that function be?"

I think that would work. Einstein said that all of the consequences of special relativity can be derived from examination of the Lorentz transformations.

http://en.wikipedia.org/wiki/Special_rel...

Again, sorry for misunderstanding your question. The word "assume" made an asp out of me, but not you

Edit: In answer to your question:

m' = m / √(1 - (v/c)²)....yes

√(1 - (v/c)²) = m/m' ....ok

1 - (v/c)² = (m/m')² ....I see

(v/c)² = 1 - (m/m')²....huh? <=========

shouldn't this read

0 - (v/c)² = (m/m')² -1

as you are subtracting one from both sides of equation?

After subtracting, I multiplied by -1. It makes it a whole lot easier, and in keeping with the SR equation format.

Answer 3

No it is not. A Lorentz transformation converts coordinates from one coordinate system S=(x,y,z,t) to another S'=(x',y',z',t') that is in relative motion. The inverse of a Lorentz transformation is another Lorentz transformation that converts S' to S. Lorentz transformations have absolutely nothing to do with mass.

EDIT: Ok, *in the theory of relativity,* A Lorentz transformation converts coordinates from one coordinate system S=(x,y,z,t) to another S'=(x',y',z',t') that is in relative motion. The inverse of a Lorentz transformation is another Lorentz transformation that converts S' to S. Lorentz transformations have absolutely nothing to do with mass.