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Is there a way to take a moving source already into account when one derives the wave equation for sound waves and derive from that using only math the Doppler effect for moving sources?

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1 Answer 1

If I understand you correctly, you want to desribe sound waves from the viewpoint of a moving observer. To do this, you just have to take the standard wave equation and perform a variable substitution $\mathbf{x}\mapsto \mathbf{y}:=\mathbf{x}-\mathbf{v}\cdot t$.

You start with $$ \partial_t^2 p(\mathbf{x},t) = c^2\nabla^2 p(\mathbf{x},t) = c^2\sum_i\partial_{\!x_i}^2 p(x_1,\ldots,t) $$ The right side transforms as $$ c^2\sum_i\partial_{\!x_i}^2 p(x_1,\ldots,t) = c^2\sum_i\partial_{\!x_i}^2 p(y_1+v_1t,\ldots,t) $$ where we yet have to replace the derivative WRT $x_i$ against the derivative WRT $y_i$, but they are the same: $$ \frac{\partial}{\partial x_i} = \frac{\partial y_i}{\partial x_i}\frac{\partial}{\partial y_i} = (1+0)\cdot\frac{\partial}{\partial y_i}; $$ so $$ c^2\nabla_{\!\!\mathbf{x}}^2p(\mathbf{x},t) = c^2\nabla_{\!\!\mathbf{y}}^2p(\mathbf{y}+\mathbf{v}t,\ t). $$ The left side transforms as $$ \partial_t^2 p(\mathbf{x},t) = \partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot t,\ t). $$

Here, we would like to drop the $+\mathbf{v}t$, but before we can do this we need to take into account the derivative of this term: $$ \partial_t p(\mathbf{y}+\mathbf{v}\cdot t,\ t) = \left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t) $$ now the second derivative: $$ \partial_t\left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} = \left.\bigl(\partial_t\partial_\tau p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} + \left.\bigl(\partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} $$ where $$ \left.\bigl(\partial_t\partial_\tau p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} = \left.\bigl(\partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} $$ and $$ \partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t) = \mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}} \left(\left.\bigl(\partial_t p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot t,\ t)\right) $$ $$ = \left.\bigl(\partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t)\bigr)\right|_{\tau=t} + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}^2\,\mathbf{v}\,p(\mathbf{y}+\mathbf{v}\cdot t,\ t) $$ All together, we get $$ \partial_t^2p(\mathbf{x},t) = \left.\left( \partial_t^2 p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t) + 2 \partial_t\,\mathbf{v}\!\cdot\!\nabla_{\!\!\mathbf{y}}p(\mathbf{y}+\mathbf{v}\cdot \tau,\ t) + \mathbf{v}\cdot\nabla_{\!\!\mathbf{y}}^2\,\mathbf{v}\,p(\mathbf{y}+\mathbf{v}\cdot t,\ t) \right)\right|_{\tau=t}. $$ Now we may drop the $+\mathbf{v}t$ on both sides of the equation, and get $$ \partial_t^2 p(\mathbf{y},t) + 2 \partial_t\,\mathbf{v}\!\cdot\!\nabla p(\mathbf{y},t) + \mathbf{v}\cdot\nabla^2\,\mathbf{v}\,p(\mathbf{y},t) = c^2\nabla^2p(\mathbf{y},t) $$

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Thanks, but I don't want a moving observer but a moving source for example a moving speaker or a siren of an ambulance –  student Jun 23 '11 at 12:46
A moving source is described by exactly the same wave equation as a static source, just with different boundary conditions. So if that was your question, the answer should have been "take the wave equation as it is, and you're done". The problem is just that these boundary conditions make solving the equation quite a bit more complicated. –  leftaroundabout Jun 23 '11 at 13:34
But in the transformed formulation of the wave equation, the moving source is at rest, so you can now solve it like you would normally solve the wave equation for a resting source. If you then want to simulate a resting observer, you just have to take this solution and evaluate it along the path $\mathbf{y}(t)=\mathbf{y}_0+\mathbf{v}t$. –  leftaroundabout Jun 23 '11 at 13:35

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