Lorentz transformations formulas

[blueberrynerd]

I'm slowly trying to understand sp relativity. I admit I got lost in the last thread I posted . But thanks to all who replied!

I have a question about the Lorentz transformations formulas. This is more of a mathematical question about how the formulas are derived.


If you have the two formulas,

x'= γ( x- vt) and x= γ(x' + vt')

which represent the x components for two reference frames S and S', and where γ is the Lorentz factor,

and you combine them to solve for t:

x'= γ[γ(x' + vt') - vt]

how do you arrive at the formula

t= γ(t' + vx'/c^2) ?

I know that you simply solve for t from the other formula, but I really cannot figure out how. Sorry, I realize this is more of a math-related problem, but I'm wondering if anybody can give me some tips?

 

[tiny-tim]

hi blueberrynerd!

(try using the X² icon just above the Reply box )

x'= γ[γ(x' + vt') - vt]

how do you arrive at the formula

t= γ(t' + vx'/c^2) ?

x'= γ[γ(x' + vt') - vt]

γvt = (γ² - 1)x' + γvt'

t = (γ² - 1)x'/γv + t'

(the https://www.physicsforums.com/library.php?do=view_item&itemid=19" is often easier to understand if you use the rapidity, α, defined by tanhα = v

then coshα = 1/√(1 - v²), sinhα = v/√(1 - v²), cosh²α - sinh²α = 1 )​

 

[Dale]

You also have to use the formula:
\gamma=\frac{1}{\sqrt{1-v^2/c^2}}

 

[grav-universe]

x'= γ[γ(x' + vt') - vt]

γvt = (γ² - 1)x' + γvt'

t = (γ² - 1)x'/γv + t'

Oops, just a small mistake there. It should be

x'= γ[γ(x' + vt') - vt]

γvt = (γ² - 1)x' + γ^2 vt'

t = (γ² - 1)x'/γv + yt'

Using y = 1 / sqrt(1 - (v/c)^2), it can then be reduced further to

t = (y - 1 / y) x' / v + y t'

t / y = (1 - 1 / y^2) x' / v + t'

t / y = (1 - (1 - (v/c)^2)) x' / v + t'

t / y = x' v / c^2 + t'

t = y (t' + x' v / c^2)

 

[GrayGhost]

Blueberrynerd,

This may be the easiest way. You start with this ...

. x = γ(x' + vt')

and since x = ct, then ...

. x = γ(x' + vt')
. ct = γ(x' + vt')

and t' = x'/c, so ...

. ct = γ(x' + vt')
. ct = γ(x' + v(x'/c)

and x' = ct', so ...

. ct = γ(x' + vx'/c)
. ct = γ(ct' + vx'/c)

dividing thru by c ...

. t = γ(t' + vx'/c²))

GrayGhost

 

[DrGreg]

...and since x = ct, then ...

Er, why should that be? The Lorentz transform applies to all values of x and t, not just the subset you consider.

 

[blueberrynerd]

Oops, just a small mistake there. It should be

x'= γ[γ(x' + vt') - vt]

γvt = (γ² - 1)x' + γ^2 vt'

t = (γ² - 1)x'/γv + yt'

Using y = 1 / sqrt(1 - (v/c)^2), it can then be reduced further to

t = (y - 1 / y) x' / v + y t'

t / y = (1 - 1 / y^2) x' / v + t'

t / y = (1 - (1 - (v/c)^2)) x' / v + t'

t / y = x' v / c^2 + t'

t = y (t' + x' v / c^2)

This reply is almost a month delayed, but THANK YOU! You helped a lot!