As a physics student, I hear this term a lot that this equation is Lorentz Invariant or galilean Invariant e.g Dirac equation is Lorentz Invariant. Even in a non-linear pde class e.g the KdV equation this comes up. How does one check whether an equation is Lorentz Invariant in practice? Could someone please give me an example of these two cases and how to check their in variance?
2 Answers
\begin{equation} x = \gamma x^{\prime} + \gamma v t^{\prime} \qquad t = \gamma t^{\prime} + \gamma \frac{v}{c^2}x^{\prime} \end{equation}
Now we transform the derivatives
\begin{equation} \frac{\partial}{\partial t^{\prime}} = \frac{\partial x}{\partial t^{\prime}} \frac{\partial}{\partial x} + \frac{\partial t}{\partial t^{\prime}} \frac{\partial}{\partial t} = \gamma v \frac{\partial}{\partial x} + \gamma \frac{\partial}{\partial t} \end{equation} \begin{equation} \frac{\partial^2}{\partial t^{\prime^2}} = (\gamma v)^2 \frac{\partial^2}{\partial x^{2}} + 2 \gamma^2 v \frac{\partial^2}{\partial t \partial x} + \gamma^2 \frac{\partial^2}{\partial t^{2}} \end{equation} If you plug this into the 2. Newtonian law, you can observe that this has no longer the required form. Hence the equation is not invariant under LT.
However, if you plug in the derivatives (including those of x) into Maxwell's equations, they will be invariant.
I did not show all this in the most general way, but I hope you get the point.
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$\begingroup$ Thanks a lot this was very helpful $\endgroup$ Commented Nov 4, 2019 at 18:13
The only thing that you have to do is replace all variables, derivatives, ... with your new coordinates and look whether it still has the same form. For example: \begin{equation} m \ddot{x} = F \end{equation} with new coordinates: \begin{equation} x^{\prime} = x - vt; t^{\prime} = t \end{equation} you obtain: \begin{equation} m \ddot{x^{\prime}} = F \end{equation} So the equation is invariant under Galilei transformation.
In general, $F$ can change (friction force) what would make the equation not invariant. But for most cases, the force only depends on relative distances, which makes it invariant as well (here not only the form is invariant but the force itself).
Independent of the transformation, you always just express everything in terms of your new coordinates, and then you've got to look at the equation.
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$\begingroup$ Could you please do it for Lorentz in variance as well? $\endgroup$ Commented Nov 4, 2019 at 17:20