Eliminating x in Lorentz Transformation

Smythe · 01-10-2011 06:48PM #1

At the bottom of this page:
http://www.cv.nrao.edu/course/astr534/LorentzTransform.html

It states;
Lorentz factor:

x=(x+vt) and x=(x−vt)

Eliminating x from this pair of equations yields
t=(t+vxc2)

How did they eliminate x?

I tried substituting the stated x value into the equation for x', but didn't get anywhere. [Embedded Image Removed]

ZorbaTehZ · 02-10-2011 02:21AM

Big hint: [latex]\gamma - \frac{1}{\gamma}= \gamma \frac{v^2}{c^2}[/latex]

Smythe · 02-10-2011 12:32PM

ZorbaTehZ wrote: »

Big hint: [latex]\gamma - \frac{1}{\gamma}= \gamma \frac{v^2}{c^2}[/latex]

Using that, I'm able to express gamma in terms of v and c. I'm getting,

gamma^2 = (c^2)/(c^2-v^2)

What equation is the hint which you provided?

ZorbaTehZ · 02-10-2011 04:32PM

The second equation you have in your first post: rearrange it so that you have x=(something). Then stick it into the first equation, and do a bit of manipulation. The hint I gave makes this a lot easier to do.

Smythe · 02-10-2011 06:17PM

ZorbaTehZ wrote: »

The second equation you have in your first post: rearrange it so that you have x=(something). Then stick it into the first equation, and do a bit of manipulation. The hint I gave makes this a lot easier to do.

When I rearrange the 2nd eqn x

=

(x−vt) to equal x I get;

x = x'/y + vt

Then I sub this into the 1st eqn x=

(x

+vt

)
I get x'/y + vt = y(x' + vt')

That's as far as I can get it. I'm not even sure where the c has come from in the answer stated on that website. I relaise it's the SOL, but not sure how to get it into the eqn answer.

* y=gamma

Morbert · 02-10-2011 09:03PM

x = γ(x' + vt') (1)
x' = γ(x - vt) (2)

Equation 1 gives you an expression for x. Just substitute it into equation (2).

x' = γ(x - vt)
= γ( γ(x' + vt') - vt)
= γ^2 x' + γ^2 vt' - γvt

Rearranging to find t

γvt = γ^2 x' + γ^2 vt' - x'
t = γ x'/v + γ t' - x'/vγ
= x'/v(1-γ) + γvt'
= γx'v/c^2 + γt'

= γ(x'v/c^2 + t')

Smythe · 03-10-2011 12:11AM

Thanks Morbert.

I'm not quite sure how you went from;

t = γ x'/v + γ t' - x'/vγ

to

= x'/v(1-γ) + γvt'

Morbert · 04-10-2011 06:24PM

Smythe wrote: »

Thanks Morbert.

I'm not quite sure how you went from;

t = γ x'/v + γ t' - x'/vγ

to

= x'/v(1-γ) + γvt'

Woops that's a typo

t = γ x'/v + γ t' - x'/vγ

to

= x'/v(γ - 1/γ) + γvt'

I then used the identity already mentioned by ZorbaTehZ

γ - 1/γ = γv^2/c^2

Smythe · 04-10-2011 08:53PM

Cheers very much for that.

I understand it now!

Thank you

Eliminating x in Lorentz Transformation

Welcome!

Comments

Welcome!