If V',A' must continue to satisfy the Lorentz gauge, then the extra "lambda" field must satisfy the wave equation (D'Alembertian = 0). The time & space derivatives acting on lambda in the transformation have the character of the energy & momentum operators of QM, such that perhaps you could think of the extra lambda field as some sort of "virtual photon field" whose energy & momentum represent a voltage & vector potential?
Using different gauge transformations will give the SAME E and B fields, but DIFFERENT V and A fields! Remember, there are an infinite number of V and A fields that will give you the same E and B fields. Trivially, you can choose whatever 0 point you want for the V field and always obtain the same E field. You can do something similar for the A field. Using different gauge transformations allows us to choose the one that will make the problem we are solving easier to solve.
Thanks for the presentation. So lamda is the 'guage'? If so are there better lamda choices for different situations? Examples would help. (Minor critique - choosing V to symbolize the scalar potential is a bit off-putting - first thing that comes to mind for V would be that it would be the vector potential , I have seen texts that use phi.)
The Gauge Transformation is a small set of assumptions on how the four fields relate to each other. You need to make some set of assumptions, and the different sets of assumptions make different problems easier or harder to solve. There is no universally useful or "best" gauge transformation.
I don't think you make sense there at the end, where you say beta = -time derivative of lambda (plus some t-dep term k(t) ). The previous line says that the GRAD of (beta+timedev of lambda) = 0; which does NOT imply 0=(beta+timedev of lambda) (plus some time dependent k(t)). I think you're misreading the book. The book actually says "The term in parentheses is therefore independent of position (it could however depend on time); call it k(t)". This means 'call the term in parentheses k(t)' NOT 'call some time-dependent term k(t)'
1:14 the curl of a divergence is not zero but undefined since you cannot take curl of a scalar quantity.
This is such a nice detailed video. We need more like these!
If V',A' must continue to satisfy the Lorentz gauge, then the extra "lambda" field must satisfy the wave equation (D'Alembertian = 0). The time & space derivatives acting on lambda in the transformation have the character of the energy & momentum operators of QM, such that perhaps you could think of the extra lambda field as some sort of "virtual photon field" whose energy & momentum represent a voltage & vector potential?
Thanks for your explanation! With it, the result seems kind of simple now :)
Whether By using guage transformation we can obtain the different scalar and vector field where E and B are the same??
Using different gauge transformations will give the SAME E and B fields, but DIFFERENT V and A fields! Remember, there are an infinite number of V and A fields that will give you the same E and B fields. Trivially, you can choose whatever 0 point you want for the V field and always obtain the same E field. You can do something similar for the A field.
Using different gauge transformations allows us to choose the one that will make the problem we are solving easier to solve.
Thanks for the presentation. So lamda is the 'guage'? If so are there better lamda choices for different situations? Examples would help. (Minor critique - choosing V to symbolize the scalar potential is a bit off-putting - first thing that comes to mind for V would be that it would be the vector potential , I have seen texts that use phi.)
Why time derivative of A is zero??
The time derivative of A appears on both sides of the equal sign, so it cancels itself.
one Question..why we need to use gauge transformation? why is it importatnt? Doesn't using this make mathematics more complex.
The Gauge Transformation is a small set of assumptions on how the four fields relate to each other. You need to make some set of assumptions, and the different sets of assumptions make different problems easier or harder to solve. There is no universally useful or "best" gauge transformation.
U will get it when u get into field theory
What is beta here?? ,as my understanding beta has potential fields.
Thank you sir
Thank you sir.
thanks
I don't think you make sense there at the end, where you say beta = -time derivative of lambda (plus some t-dep term k(t) ). The previous line says that the GRAD of (beta+timedev of lambda) = 0; which does NOT imply 0=(beta+timedev of lambda) (plus some time dependent k(t)). I think you're misreading the book. The book actually says "The term in parentheses is therefore independent of position (it could however depend on time); call it k(t)". This means 'call the term in parentheses k(t)' NOT 'call some time-dependent term k(t)'
5:35 omg bad sign, how can you get - in front without changing minus in paranthesis ()..omg
ya should be +d/dt(lambda) inside the bracket
Good 😎