Vector field equality Curl Proof of Moving Magnet & Conductor Problem

[PeterDonis]

I believe its the other way round

The quote you gave in post #5 makes it clear that the assumption $\vec{E}^\prime = \vec{v} \times \vec{B}$ is used to show that the curls are equal, not the other way around.

Can you give a link to the previous thread? As I noted, your quote in post #5 does not include a link.

 

[tade]

ok searching for it

The quote you gave in post #5 makes it clear that the assumption $\vec{E}^\prime = \vec{v} \times \vec{B}$ is used to show that the curls are equal, not the other way around.

it doesn't appear to start with the assumption that $\vec{E}^\prime = \vec{v} \times \vec{B}$
that would be just two (or one) steps.

instead, it starts with the Maxwell-Faraday equation.
then, expressing the rate of change of the B-field in terms of the uniform velocity field, and finally re-arranging that to give the curl of $\vec{v} \times \vec{B}$

 

[PeterDonis]

it doesn't appear to start with the assumption

The equality of the curls is only stated in one place: right after the assumption that $\vec{E}^\prime = \vec{v} \times \vec{B}$. And it is explicitly stated that the equality of the curls is obtained from that assumption: the quote says "with $\vec{E}^\prime = \vec{v} \times \vec{B}$ you would have [equality of the curls]".

Either give a link to the previous thread or this thread will be closed, since you seem unable to state clearly what it is you are concerned about or even to correctly read things you have quoted yourself.

 

[tade]

The equality of the curls is only stated in one place: right after the assumption that $\vec{E}^\prime = \vec{v} \times \vec{B}$. And it is explicitly stated that the equality of the curls is obtained from that assumption: the quote says "with $\vec{E}^\prime = \vec{v} \times \vec{B}$ you would have [equality of the curls]".

Either give a link to the previous thread or this thread will be closed, since you seem unable to state clearly what it is you are concerned about or even to correctly read things you have quoted yourself.

But we don't need that assumption to show the equality of curls.

Just need to follow these steps, all featured in #5:

instead, it starts with the Maxwell-Faraday equation.
then, expressing the rate of change of the B-field in terms of the uniform velocity field, and finally re-arranging that to give the curl of $\vec{v} \times \vec{B}$

the last step which features the curl of E shows the re-arranging.
And we don't need to assume that $\vec{E}^\prime = \vec{v} \times \vec{B}$ in order to do that re-arranging.

 

[PeterDonis]

the last step which features the curl of E shows the re-arranging.

Yes, after the assumption $\vec{E}^\prime = \vec{v} \times \vec{B}$ has been introduced to derive the equality of curls.

we don't need to assume that $\vec{E}^\prime = \vec{v} \times \vec{B}$ in order to do that re-arranging.

You do to get the equality of curls that then gets rearranged. See above.

Since you continue to make incorrect statements about something you quoted yourself, and are either unable or unwilling to clearly state your issue or to provide a link to the previous thread you are quoting from so we can see the context, this thread is closed.