D Alembert Wave Equation is not Gallean Invariant but Why y=Asin(wt-kx) is Gallean Invariant?

Asked Jul 10 '23 at 16:00

Active Jul 10 '23 at 17:20

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I just watched this video from MIT 8.04 Quantum Physics I by Barton Zwiebach, explaining Galilean transformation of y=Asin(wt-kx).

I have a confusion, are ordinary waves Galilean invariant or not? Because d'Alembert's wave equation is not Galilean invariant. So, this is a contradiction between the two. What is wrong?

edited Jul 10 '23 at 17:20

asked Jul 10 '23 at 16:00

Dibyajit Bhattacharyya

4

Does this answer your question? Isn't D'Alembert's wave equation enough to see that Galilean transformations are wrong? – Mauricio Jul 10 '23 at 16:19
No my actual question is why y=Asin(wt-kx) is Gallean Invariant but DAlemberts Wave Equation is not Invariant? – Dibyajit Bhattacharyya Jul 10 '23 at 17:24

D Alembert Wave Equation is not Gallean Invariant but Why y=Asin(wt-kx) is Gallean Invariant?

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