Lorentz Transformations: Difference between revisions
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\begin{align} | \begin{align} | ||
\Delta t'&=t'_2-t'_1\\ | \Delta t'&=t'_2-t'_1\\ | ||
&=\gamma(t_2-vx/c^2)-\gamma(t_1-vx/c^2) \textnormal{here we are just considering differences in time intervals, so we assume that | &=\gamma(t_2-vx/c^2)-\gamma(t_1-vx/c^2) \textnormal{here we are just considering differences in time intervals, so we assume that x position does not change}\\ | ||
&= | &= | ||
\end{align} | \end{align} |
Revision as of 11:34, 28 November 2022
The Lorentz Transformation is a transformation that allows one to shift between different coordinate systems. Namely, it allows one to transform the cartesian coordinate system of a stationary reference frame to another cartesian coordinate system of a reference frame that is moving with constant velocity [math]\displaystyle{ v }[/math] with respect to the stationary reference frame. It is named after Hendrik Antoon Lorentz, who derived the transformation in 1904 while working on developing the theory of electrodynamics for moving bodies. The transformation is as follows
[math]\displaystyle{ \begin{align}x'&=\gamma(x-vt)\\ y'&=y\\ z'&=z\\ t'&=\gamma(t-\frac{xv}{c^2})\end{align} }[/math]
where [math]\displaystyle{ \gamma=\frac{1}{\sqrt{1-v^2/c^2}} }[/math]. The Lorentz Transformation can be considered a more general formulation of of the ideas of time dilation and length contraction as it gives not just the changes in length and in time when going from one reference frame to another, but rather it gives the full coordinates of objects in space and time when going from one reference frame to another. Because it is a more general formulation, Einstein's equations for time dilation and length contraction can be derived from the Lorentz Transformation (shown below). Going further, we can also use the Lorentz Transformation to relate observed speeds in different reference frames, which is done by taking time derivatives of the Lorentz Transformation (shown below).
Derivation of the Lorentz Transformation
To derive the Lorentz Transformation, we start by considering the movement of light from two difference reference frames. Our unprimed reference frame [math]\displaystyle{ (x,y,z,t) }[/math] will be referred to as the stationary reference frame, and our primed reference frame [math]\displaystyle{ (x',y',z't') }[/math] will be referred to as the moving reference frame. We will define the relationship between the two reference frames such that an observer in the stationary reference frame see another observer in the moving reference frame moving along the positive x-axis with speed [math]\displaystyle{ v }[/math].
Now imagine that the movement of a light wave is observed from both reference frames for some length of time. In the stationary reference frame, we can express the length of this path in two ways as shown below.
[math]\displaystyle{ \begin{align}&l=c(t_2-t_1)\\&l=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\end{align} }[/math]
Now we will rearrange the above expressions for [math]\displaystyle{ l }[/math] to get.
[math]\displaystyle{ \begin{align}&l^2-l^2=0\\&c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2=0\end{align} }[/math]
We can now do the same procedure for the moving reference frame, and we get the expressions below
[math]\displaystyle{ \begin{align}&l'=c(t_2'-t_1')\\&l'=\sqrt{(x_2'-x_1')^2+(y_2'-y_1')^2+(z_2'-z_1')^2}\\&l'^2-l'^2=0\\&c^2(t_2'-t_1')^2-(x_2'-x_1')^2-(y_2'-y_1')^2-(z_2'-z_1')^2=0\end{align} }[/math]
Now we have expressions in both of our reference frames that are equal to zero, so we can set them equal to each other to determine the relationship between our difference coordinate systems.
[math]\displaystyle{ c^2(t_2'-t_1')^2-(x_2'-x_1')^2-(y_2'-y_1')^2-(z_2'-z_1')^2=c^2(t_2-t_1)^2-(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2 }[/math]
Using a clever trick, we can simplify the above equation greatly. We will let the motion of the light wave be purely along the [math]\displaystyle{ x,x' }[/math] axis. Remember that we can set up our coordinate axes however is convenient for us when solving problems, so this trick is not requiring any further assumptions. If the light wave's motion is purely along the [math]\displaystyle{ x,x' }[/math] axis, then we can simplify the above equation to the following.
[math]\displaystyle{ c^2(t_2'-t_1')^2-(x_2'-x_1')^2=c^2(t_2-t_1)^2-(x_2-x_1)^2 }[/math]
We can further reduce our equation by letting the light wave start at the origin of both coordinate systems (i.e. [math]\displaystyle{ t_1=t_1'=0 }[/math] and [math]\displaystyle{ x_1=x_1'=0 }[/math]). Now we can rewrite the above equation as.
[math]\displaystyle{ c^2t'^2-x'^2=c^2t^2-x^2 }[/math]
The general solution solved for the moving reference frame to the above equation is
[math]\displaystyle{ x'=xcosh\alpha+ctsinh\alpha\;\;\;\;\;\;\;\;\;\;\; ct'=xsinh\alpha+ctcosh\alpha }[/math]
where [math]\displaystyle{ \alpha\in\mathbb{R} }[/math] (this is shown to be the general solution to the above equation later on). Our next step in deriving the Lorentz Transformation is to find expressions for [math]\displaystyle{ sinh\alpha }[/math] and [math]\displaystyle{ cosh\alpha }[/math]. This can be done by considering another thought experiment: consider tracking the origin of the stationary reference frame from the perspective of the moving reference frame. This simplifies the above general solution to the following.
[math]\displaystyle{ x'=ctsinh\alpha\;\;\;\;\;ct'=ctcosh\alpha }[/math]
Now we can find expressions for [math]\displaystyle{ sinh\alpha }[/math] and [math]\displaystyle{ cosh\alpha }[/math], which we can then plug back into our general equation for [math]\displaystyle{ x' }[/math] and [math]\displaystyle{ ct' }[/math] to get the Lorentz Transformation. Note: to find expressions for [math]\displaystyle{ sinh\alpha }[/math] and [math]\displaystyle{ cosh\alpha }[/math], we use the identities stated in section 1.1.
[math]\displaystyle{ \frac{x'}{ct'}=\frac{-v}{c}=tanh\alpha\;\;\;\; \Rightarrow \;\;\;\;sinh\alpha=\frac{-v/c}{\sqrt{1-v^2/c^2}} \;\;\;\; cosh\alpha=\frac{1}{\sqrt{1-v^2/c^2}} }[/math]
[math]\displaystyle{ x'=xcosh\alpha+ctsinh\alpha\;\; \Rightarrow \;\; x\frac{1}{\sqrt{1-v^2/c^2}}+ct\frac{-v/c}{\sqrt{1-v^2/c^2}}=\frac{x-vt}{\sqrt{1-v^2/c^2}}=\gamma(x-vt) }[/math]
[math]\displaystyle{ ct'=xsinh\alpha+ctcosh\alpha=x\frac{-v/c}{\sqrt{1-v^2/c^2}}+ct\frac{1}{\sqrt{1-v^2/c^2}}=\frac{ct-xv/c}{\sqrt{1-v^2/c^2}}=c\gamma(t-xv/c^2) }[/math]
[math]\displaystyle{ \Rightarrow \;\;\;\;x'=\gamma(x-vt)\;\;\;\; t'=\gamma(t-xv/c^2) }[/math]
The [math]\displaystyle{ y }[/math] and [math]\displaystyle{ z }[/math] components of the Lorentz Transformation are simply [math]\displaystyle{ y'=y }[/math] and [math]\displaystyle{ z'=z }[/math]. This is because in our derivation for the [math]\displaystyle{ x }[/math] and [math]\displaystyle{ t }[/math] components of the transformation, we had to assume that the motion of our imaginary light wave was purely along the [math]\displaystyle{ x }[/math] axis in the stationary reference frame, and purely along the [math]\displaystyle{ x' }[/math] axis in the moving reference frame. So the [math]\displaystyle{ y,z,y',z' }[/math] coordinates of the light wave do not change over time. Also, since we assumed in our derivation that the origins of our coordinate systems coincide, the initial [math]\displaystyle{ y,y' }[/math] and [math]\displaystyle{ z,z' }[/math] coordinates must agree with each other. Thus, [math]\displaystyle{ y'=y }[/math] and [math]\displaystyle{ z'=z }[/math].
We have now arrived at the full Lorentz Transformation:
[math]\displaystyle{ \begin{align}x'&=\gamma(x-vt)\\ y'&=y\\ z'&=z\\ t'&=\gamma(t-\frac{xv}{c^2})\end{align} }[/math]
where [math]\displaystyle{ \gamma=\frac{1}{\sqrt{1-v^2/c^2}} }[/math].
Hyperbolic Trigonometric Identities
[math]\displaystyle{ \begin{align} &cosh^2\;\alpha-sinh^2\;\alpha=1\\ &sinh\;\alpha=\frac{tanh\;\alpha}{\sqrt{1-tanh^2\;\alpha}}\\ &cosh\;\alpha=\frac{1}{\sqrt{1-tanh^2\;\alpha}} \end{align} }[/math]
Verification of the General Solution
Our general solution to [math]\displaystyle{ c^2t'^2-x'^2=c^2t^2-x^2 }[/math] is [math]\displaystyle{ x'=xcosh\;\alpha+ctsinh\;\alpha\;\;ct'=ctcosh\;\alpha+xsinh\;\alpha }[/math]. We will verify this via direct substitution.
[math]\displaystyle{ ct'^2-x'^2=ct^2-x^2 }[/math]
We will plug in [math]\displaystyle{ x'=xcosh\;\alpha+ctsinh\;\alpha\;\;ct'=ctcosh\;\alpha+xsinh\;\alpha }[/math] to the left hand side of the above equation and show that it is equal to the right hand side.
[math]\displaystyle{ \begin{align} (ctcosh\;\alpha+xsinh\;\alpha)^2-(xcosh\;\alpha+ctsinh\;\alpha)^2&=c^2t^2cosh^2\;\alpha+x^2sinh^2\;\alpha+2ctxcosh\;\alpha\;sing\;\alpha-(x^2cosh^2\;\alpha+c^2t^2sinh^2\;\alpha+2ctxcosh\;\alpha\;sinh\;\alpha)\\ &=(c^2t^2cosh^2\;\alpha-c^2t^2sinh^2\;\alpha)-(x^2cosh^2\;\alpha-x^2sinh^2\;\alpha)+(2ctxcosh\;\alpha\;sing\;\alpha-2ctxcosh\;\alpha\;sing\;\alpha)\\ &=c^2t^2(cosh^2\;\alpha-sinh^2\;\alpha)-x^2(cosh^2\;\alpha-sinh^2\;\alpha)\\ &=c^2t^2-x^2 \end{align} }[/math]
Therefore [math]\displaystyle{ x'=xcosh\;\alpha+ctsinh\;\alpha\;\;ct'=ctcosh\;\alpha+xsinh\;\alpha }[/math] is the general solution to [math]\displaystyle{ c^2t'^2-x'^2=c^2t^2-x^2 }[/math].
A Mathematical Model
What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.
A Computational Model
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Examples
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Simple
Middling
Derive Einstein's equations for time dilation ([math]\displaystyle{ \Delta t'=\gamma \Delta t }[/math]) and length contraction ([math]\displaystyle{ L'=L/\gamma }[/math]) from the Lorentz Transformation.
Solution:
[math]\displaystyle{ \begin{align} \Delta t'&=t'_2-t'_1\\ &=\gamma(t_2-vx/c^2)-\gamma(t_1-vx/c^2) \textnormal{here we are just considering differences in time intervals, so we assume that x position does not change}\\ &= \end{align} }[/math]
Difficult
Connectedness
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History
The Lorentz Transformation was first derived not by Lorentz, but rather by Woldemar Voigt in a paper he published in 1897 (insert link). In this paper, Voigt took an alternative approach to the already derived Doppler Effect in which he postulated that the wave equation should take on the same form in both the stationary and moving reference frames (a postulate for which he gave no reason). This postulate is equivalent to stating that the propagation velocity of the wave must be invariant under transformations between reference frames that are moving with constant velocity relative to one another.
For reasons not known, Hendrik Antoon Lorentz appeared to be inspired by this transformation. He rederived it in his 1904 paper (insert link) in which he investigated the electrodynamics of moving bodies. The transformation was eventually named after Lorentz in 1906 by Jules Henri Poincaré.
See also
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