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Lorentz transformation is
t= gamma(t' +(v/c^2)x')
x=gamma(x'+vt')
y=' and
z=z'.

i don't know how to show
ct=ct'cosh(theta) +x'sinh(theta)
x=ct'sinh(theta) +x'cosh(theta)

i'm really stuck to where to start

2007-10-27 04:48:28 · 2 answers · asked by lookincool87 1 in Science & Mathematics Mathematics

2 answers

I see no one has answered this yet.

Have you looked at the page in the link below?

It gives some useful identities and you may be able to work from these to prove the equations given. First of all note the following identities -

cosh(theta) = gamma
tanh(theta) = beta = v/c

Starting with your first equation,

t = gamma(t' + (v/c^2)x') so
ct = c * gamma(t' + x' (v/c^2))
ct = ct' gamma + x' gamma (v/c)
ct = ct' cosh(theta) + x' cosh(theta) tanh(theta)
ct = ct' cosh(theta) + x' sinh(theta)

since sinh = cosh * tanh

I assume the second equation can be similarly transformed.

I hope this helps but I am a bit out of my depth here.

2007-10-27 09:19:24 · answer #1 · answered by tringyokel 6 · 0 0

You won't get anywhere unless you know that
gamma = [1 - {v/c}(^2)](^-1/2).

2007-10-27 19:22:38 · answer #2 · answered by anthony@three-rs.com 3 · 0 0

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