Understanding the Relativistic Velocity Addition Formula

[Kutuzov]

Homework Statement

In this thread the author performs the following calculation under "method 1":
v+dv=\frac{v+dv'}{1+vdv'}=v+(1-v^2)dv'\implies dv'=\frac{dv}{1-v^2}

He's set c=1 so the second expression is the relativistic velocity addition formula. What I don't understand is how he gets the third expression from the second one.

Homework Equations

Velocity addition formula
u=\frac{v+u'}{1+\frac{v u'}{c^2}}=\left[ c=1 \right]=\frac{v+u'}{1+vu'}

The Attempt at a Solution

When I solve for dv' between the first and second expression, I get
dv'=\frac{dv}{1-v^2-vdv}
Instead of what he gets
\frac{dv}{1-v^2}
Perhaps vdv is negligible or something like that? I can't figure it out.

 

[Mute]

Homework Statement

In this thread the author performs the following calculation under "method 1":
v+dv=\frac{v+dv'}{1+vdv'}=v+(1-v^2)dv'\implies dv'=\frac{dv}{1-v^2}

He's set c=1 so the second expression is the relativistic velocity addition formula. What I don't understand is how he gets the third expression from the second one.

Homework Equations

Velocity addition formula
u=\frac{v+u'}{1+\frac{v u'}{c^2}}=\left[ c=1 \right]=\frac{v+u'}{1+vu'}

The Attempt at a Solution

When I solve for dv' between the first and second expression, I get
dv'=\frac{dv}{1-v^2-vdv}
Instead of what he gets
\frac{dv}{1-v^2}
Perhaps vdv is negligible or something like that? I can't figure it out.

Yes, the $dv$'s are infinitesimal, and so terms of order $(dv')^2$ or $(dv)^2$ are neglected. In the original derivation, the author used the binomial expansion on the denominator and dropped any terms $(dv')^2$ or higher.

If you use the binomial expansion on your derivation, you will similar find that you get the desired result plus terms of order $(dv)^2$ or higher, which you neglect. (You are not just neglecting $v dv$; it's only because there is already a $dv$ in the numerator that the $v dv$ term ends up vanishing.)

 

[Kutuzov]

I need an example of said binomial expansion to see what it's about. I'm sure I will recognize it straight away, it's just what I think is binomial expansion is stuff like (a+b)^3=a^3+3a^2b+3ab^2+b^3. In this case (1+vdv') the exponent is one, so an expansion like this would leave the denominator unchanged...

 

[Mute]

I need an example of said binomial expansion to see what it's about. I'm sure I will recognize it straight away, it's just what I think is binomial expansion is stuff like (a+b)^3=a^3+3a^2b+3ab^2+b^3. In this case (1+vdv') the exponent is one, so an expansion like this would leave the denominator unchanged...

You want to expand $(1-v^2 - vdv)^{-1}$. Actually, it's better if you pull out a factor of $1-v^2$, so that you have

$$\frac{dv}{1-v^2}\left(1-\frac{vdv}{1-v^2}\right)^{-1}.$$

You want to expand the factor $\left(1-\frac{vdv}{1-v^2}\right)^{-1}$ now.

 

[Kutuzov]

Thank you. Works perfectly! I call that kind of expansion Taylor series expansion, and only say binomial expansion if I use the binomial theorem. But it might be the same thing.

 

[Mute]

Thank you. Works perfectly! I call that kind of expansion Taylor series expansion, and only say binomial expansion if I use the binomial theorem. But it might be the same thing.

That particular Taylor series is called the Binomial series, hence why I called it a binomial expansion. :) It reduces to the binomial theorem in the case of positive integer exponent, as well.

 

[diazona]

Thank you. Works perfectly! I call that kind of expansion Taylor series expansion, and only say binomial expansion if I use the binomial theorem. But it might be the same thing.

Like Mute said, binomial expansion is a special case of Taylor expansion, so you could say they're the same thing in that sense.

It's standard practice to say "binomial expansion" to mean any expansion of this sort, even if it doesn't use the binomial theorem for nonnegative integer exponents.