Return to Answer

So would the real part just be ��(��,��)=��(��)?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \mathbf{E}(\mathbf{r}) \sin(\omega t)$

If you have not taken the real part yet, $$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} = j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) = \omega \mathbf{E}(\mathbf{r})[-\sin(\omega t) + j \cos(\omega t)]$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.

• In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.

So would the real part just be $\mathbf{e}(\mathbf{r}, t) = \mathbf{E}(\mathbf{r})$?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \mathbf{E}(\mathbf{r}) \sin(\omega t)$

If you have not taken the real part yet, $$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} = j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) = \omega \mathbf{E}(\mathbf{r})[-\sin(\omega t) + j \cos(\omega t)]$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.

• In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.

So would the real part just be 𝐞(𝐫,𝑡)=𝐄(𝐫)?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \ \mathbf{E}(\mathbf{r}) \sin(\omega t)$

If you have not taken the real part yet, $$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} =-j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) =- \omega \ \mathbf{E}(\mathbf{r}) \{\sin(\omega t) + j \cos(\omega t)\}$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.

• In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.

So would the real part just be ��(��,��)=��(��)?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \mathbf{E}(\mathbf{r}) \sin(\omega t)$

If you have not taken the real part yet, $$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} = j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) = \omega \mathbf{E}(\mathbf{r})[-\sin(\omega t) + j \cos(\omega t)]$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.

• In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.

So would the real part just be 𝐞(𝐫,𝑡)=𝐄(𝐫)?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \ \mathbf{E}(\mathbf{r}) \sin(\omega t)$

So would the real part just be 𝐞(𝐫,𝑡)=𝐄(𝐫)?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \ \mathbf{E}(\mathbf{r}) \sin(\omega t)$

If you have not taken the real part yet, $$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} =-j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) =- \omega \ \mathbf{E}(\mathbf{r}) \{\sin(\omega t) + j \cos(\omega t)\}$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.

• In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.