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The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

$$\begin{align} f_1(\vec{r},t) &= \sin{(\vec{k}\cdot\vec{r}+\omega t)} \\ f_2(\vec{r},t) &= \sin{(\vec{q}\cdot\vec{r}-\omega t)} \\ \end{align}$$

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = \pm q_y$ and therefore $\vec{k} = \vec{q}$ or $\vec{k} = \vec{p}$ where $\vec{p}$ corresponds to the minus sign. This $\vec{p}$-vector has the same length as $\vec{q}$ but it makes an angle of $\pi+\theta$ with the positive $x$-axis, if $\vec{q}$ makes an angle of $\theta$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $\omega/|\vec{k}| = c_1 = c = c_2 = \nu/|\vec{q}|$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.

The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

$$\begin{align} f_1(\vec{r},t) &= \sin{(\vec{k}\cdot\vec{r}+\omega t)} \\ f_2(\vec{r},t) &= \sin{(\vec{q}\cdot\vec{r}-\omega t)} \\ \end{align}$$

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = \pm q_y$ and therefore $\vec{k} = \vec{q}$ or $\vec{k} = \vec{p}$ where $\vec{p}$ corresponds to the minus sign. This $\vec{p}$-vector has the same length as $\vec{q}$ but it makes an angle of $\pi-\theta$ with the positive $x$-axis, if $\vec{q}$ makes an angle of $\theta$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $\omega/|\vec{k}| = c_1 = c = c_2 = \nu/|\vec{q}|$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.

The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

$$\begin{align} f_1(\vec{r},t) &= \sin{(\vec{k}\cdot\vec{r}+\omega t)} \\ f_2(\vec{r},t) &= \sin{(\vec{q}\cdot\vec{r}-\omega t)} \\ \end{align}$$

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = q_y$ and therefore $\vec{k} = \vec{q}$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $\omega/|\vec{k}| = c_1 = c = c_2 = \nu/|\vec{q}|$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.

The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

$$\begin{align} f_1(\vec{r},t) &= \sin{(\vec{k}\cdot\vec{r}+\omega t)} \\ f_2(\vec{r},t) &= \sin{(\vec{q}\cdot\vec{r}-\omega t)} \\ \end{align}$$

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = \pm q_y$ and therefore $\vec{k} = \vec{q}$ or $\vec{k} = \vec{p}$ where $\vec{p}$ corresponds to the minus sign. This $\vec{p}$-vector has the same length as $\vec{q}$ but it makes an angle of $\pi+\theta$ with the positive $x$-axis, if $\vec{q}$ makes an angle of $\theta$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $\omega/|\vec{k}| = c_1 = c = c_2 = \nu/|\vec{q}|$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.

The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

$$\begin{align} f_1(\vec{r},t) &= \sin{(\vec{k}\cdot\vec{r}+\omega t)} \\ f_2(\vec{r},t) &= \sin{(\vec{q}\cdot\vec{r}-\omega t)} \\ \end{align}$$

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = q_y$ and therefore $\vec{k} = \vec{q}$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $|\vec{k}|\omega = c_1 = c = c_2 = |\vec{q}|\nu$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.

The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

$$\begin{align} f_1(\vec{r},t) &= \sin{(\vec{k}\cdot\vec{r}+\omega t)} \\ f_2(\vec{r},t) &= \sin{(\vec{q}\cdot\vec{r}-\omega t)} \\ \end{align}$$

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = q_y$ and therefore $\vec{k} = \vec{q}$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $\omega/|\vec{k}| = c_1 = c = c_2 = \nu/|\vec{q}|$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.