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Simplify this equation.

I'm going out of my mind here but I can't seem to simplify this equation.


∂2Ex∂t2=1μμ0ε0∂2Ex∂z2 \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}

What do you get when substituting Ex(z,t)=Ex(z)e−jωtE_x(z, t) = E_x(z)e^{-j\omega t} into both sides?

EDIT: 1μμ0ε0 \frac{1}{\mu \mu_0 \varepsilon_0} is a constant.

Would I be right in saying on the left side you would get −ω2e−jωtE(z)-\omega^2 e^{-j\omega t} E(z)?
(edited 9 years ago)
Original post by Dr Alcoholic
I'm going out of my mind here but I can't seem to simplify this equation.


∂2Ex∂t2=1μμ0ε0∂2Ex∂z2 \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}

What do you get when substituting Ex(z,t)=Ex(z)e−jωtE_x(z, t) = E_x(z)e^{-j\omega t} into both sides?

Let me just check with Stephen Hawkins :wink:
Original post by Old_Simon
Let me just check with Stephen Hawkins :wink:


Who?
Original post by Dr Alcoholic
Who?

Hawking. typo don't stress.
Original post by Old_Simon
Hawking. typo don't stress.


Maybe you can ask him to give me a new brain while you're at it.
(edited 9 years ago)
Reply 5
Avatar for Namch
Namch
4
That question is a joke.
I don't mean to be rude, but this is stressing me out so could you only leave a comment if you know how to do this please.
Calculate the second partial derivatives of your given substitution. Substitute it in.
Reply 8
Avatar for Reety
Reety
12
I had a go, didn't really get it, but here is what i came out with:

Unparseable latex formula:

\[(z,t)]





*I tried my very best, so don't be mad if i didn't help :s-smilie:

All the best,
Reety.
(edited 9 years ago)
Original post by Reety
I had a go, didn't really get it, but here is what i came out with:

Unparseable latex formula:

\[(z,t)]





*I tried my very best, so don't be mad if i didn't help :s-smilie:

All the best,
Reety.



TBH it's further than I've got.
Original post by BlueSam3
Calculate the second partial derivatives of your given substitution. Substitute it in.




Does that imply that Ex(t)=VpEx(z) E_x(t) = V_p E_x(z), where Vp V_p is the phase speed?
What is this for? I'm sensing a mechanics module?
Original post by RhymeAsylumForever
What is this for? I'm sensing a mechanics module?


No it's to do with transmissions. I'm not a maths student, it's engineering but this is a maths based problem hence I'm asking for help from maths people.
Original post by Dr Alcoholic
No it's to do with transmissions. I'm not a maths student, it's engineering but this is a maths based problem hence I'm asking for help from maths people.


Oh ok, looked a bit like further maths with physics.
Reply 14
Avatar for Reety
Reety
12
I had another crack at it, here is what i got:

Unparseable latex formula:

\[(z,t)]



Hope this helped :smile:
Original post by Dr Alcoholic
Does that imply that Ex(t)=VpEx(z) E_x(t) = V_p E_x(z), where Vp V_p is the phase speed?


The image is broken, so I can't tell you, sorry.
Original post by Dr Alcoholic
I'm going out of my mind here but I can't seem to simplify this equation.


∂2Ex∂t2=1μμ0ε0∂2Ex∂z2 \frac{\partial^2 E_x}{\partial t^2} = \frac{1}{\mu \mu_0 \varepsilon_0} \frac{\partial^2 E_x}{\partial z^2}

What do you get when substituting Ex(z,t)=Ex(z)e−jωtE_x(z, t) = E_x(z)e^{-j\omega t} into both sides?

EDIT: 1μμ0ε0 \frac{1}{\mu \mu_0 \varepsilon_0} is a constant.

Would I be right in saying on the left side you would get −ω2e−jωtE(z)-\omega^2 e^{-j\omega t} E(z)?

If this hasn't been sorted yet, yes the LHS there is correct.

The RHS is even easier - notice that the "e−jwte^{-jwt}" factor in Ex(z,t)=E(z)e−jwtE_x(z,t) = E(z) e^{-jwt} is a constant w.r.t. zz. So when you differentiate w.r.t. z twice for the RHS, the e−jwte^{-jwt} can be pulled out of the derivative, leaving only the E(z) factor to be operated upon.

This should give you a second order linear differential equation for E(z)E(z), which you should be able to solve.

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