- #1
electricspit
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Hello, I'm going through Landau and Lifshitz "The Classical Theory of Fields" this summer with a friend and in section 4 I've come to a bit of a math problem.
Assume you have an inertial frame [itex]K'[/itex] moving at speed [itex]V[/itex] relative to an inertial frame [itex]K[/itex] in the [itex]x[/itex]-direction. In order for invariant intervals we require:
[itex]
(ct)^2 - x^2 = (ct')^2 - (x')^2
[/itex]
For this to be true:
[itex]
x = x'\cosh{\Psi}+ct'\sinh{\Psi}
[/itex]
[itex]
ct=x'\sinh{\Psi}+ct'\cosh{\Psi}
[/itex]
Where [itex]\Psi[/itex] is the angle of rotation in the [itex]xt[/itex] plane. Which makes sense. Now if we just look at the origin of the [itex]K'[/itex] frame moving these can be reduced to:
[itex]
x=ct'\sinh{\Psi}
[/itex]
[itex]
ct=ct'\cosh{\Psi}
[/itex]
and dividing the equations yields:
[itex]
\tanh{\Psi}=\frac{x}{ct}
[/itex]
But the speed [itex]V[/itex] is given by [itex]V=\frac{x}{t}[/itex] so:
[itex]
\tanh{\Psi}=\frac{V}{c}
[/itex]
The next part is what is a bit confusing. If this were regular trigonometry to find both [itex]\sin{\Psi}[/itex] and [itex]\cos{\Psi}[/itex] would just require the construction of a triangle. In this case however, they end up with very, very similar results except where the hypoteneuse would be they end up with something like this:
[itex]
\sinh{\Psi}=\frac{\frac{V}{c}}{\sqrt{1-({\frac{V}{c}})^2}}
[/itex]
[itex]
\cosh{\Psi}=\frac{1}{\sqrt{1-({\frac{V}{c}})^2}}
[/itex]
If anyone can point me in the right direction of deriving these relationships either algebraically or using geometry that would be extremely helpful!
Assume you have an inertial frame [itex]K'[/itex] moving at speed [itex]V[/itex] relative to an inertial frame [itex]K[/itex] in the [itex]x[/itex]-direction. In order for invariant intervals we require:
[itex]
(ct)^2 - x^2 = (ct')^2 - (x')^2
[/itex]
For this to be true:
[itex]
x = x'\cosh{\Psi}+ct'\sinh{\Psi}
[/itex]
[itex]
ct=x'\sinh{\Psi}+ct'\cosh{\Psi}
[/itex]
Where [itex]\Psi[/itex] is the angle of rotation in the [itex]xt[/itex] plane. Which makes sense. Now if we just look at the origin of the [itex]K'[/itex] frame moving these can be reduced to:
[itex]
x=ct'\sinh{\Psi}
[/itex]
[itex]
ct=ct'\cosh{\Psi}
[/itex]
and dividing the equations yields:
[itex]
\tanh{\Psi}=\frac{x}{ct}
[/itex]
But the speed [itex]V[/itex] is given by [itex]V=\frac{x}{t}[/itex] so:
[itex]
\tanh{\Psi}=\frac{V}{c}
[/itex]
The next part is what is a bit confusing. If this were regular trigonometry to find both [itex]\sin{\Psi}[/itex] and [itex]\cos{\Psi}[/itex] would just require the construction of a triangle. In this case however, they end up with very, very similar results except where the hypoteneuse would be they end up with something like this:
[itex]
\sinh{\Psi}=\frac{\frac{V}{c}}{\sqrt{1-({\frac{V}{c}})^2}}
[/itex]
[itex]
\cosh{\Psi}=\frac{1}{\sqrt{1-({\frac{V}{c}})^2}}
[/itex]
If anyone can point me in the right direction of deriving these relationships either algebraically or using geometry that would be extremely helpful!
Last edited: