Gradient - Engineering Mathematics Video Lecture

oh I know
- happy holiday does mean ah ha ha ha ha
teach me
you
you
you
good morning last time I had introduced
a new quantity namely the electrostatic
potential by showing that a particular
integral of the electric field depended
only on n points let us review if you
have a charge and it's sitting at the
origin charge Q and if I took another
charge small Q and took it on some path
from a position or not or not is this
vector to another position R and you
went however you liked then the electric
then you can write an equation it says
go from R naught to our along this path
C so usually you do not write the
endpoints we write C and you work out
what this integral is what this integral
means is I choose a lot of points along
this curve and
I draw little arrows and on each arrow I
find the local electric field and I
compute e dot DL one then add T dot DL
two etcetera and I take that whole sum
and then I let dl 1 dl 2 dl 3 etcetera
become very small in other words I put
more and more points on this curve
uniformly so that these arrows become
very small when I do that I am
essentially having an integral and that
is the integral that is meant here
integral R naught 2 R of e dot DL now in
general this integral would depend on
how I went from are not to R however we
saw last time that if you work this
integral out for a charge placed at the
origin you find that this integral
becomes the charge Q over 4 PI epsilon
nought times integral R naught to R of
all divided by mod R cubed dot DL but
this direction R at any point is
pointing outwards this DL points in some
general direction but DL dot r only
cares about that part of DL that is
along the radial direction so you can
take this entire integral and replace it
with an integral that is only in the
radial direction so you find the radial
part of are not call it ro not radial
part of our call it row and then you are
really only now concerned with the
radial direction D Rho and R over R
cubed 0 so
this becomes a scalar integral and when
you work it out you find that this is
nothing but the integral Rho not to Rho
zero over Rho squared which is equal to
minus one over Rho between the limits
which is 1 over Rho naught minus 1 over
so here Rho is nothing but the radial
coordinate of our and what you are
seeing is the electric field dotted with
D l's integrated from a point R naught
to a point R along some path could be
any path is equal to Q divided by 4 PI
epsilon naught times 1 over Rho naught
minus 1 over Rho where Rho naught and
Rho other are coordinates of the
beginning and the ending point it does
not depend on how you got there it only
depends on where you started from
and where you ended up I could have gone
this way or I could have gone this way
or I could have gone this way it does
not matter all that matters is the
beginning of the final value of the
radial coordinate
then this is the situation then we can
call this a function of the final point
alone and it is called the electrostatic
potential which is a function of
position in fact in this case it is a
function only the radial component of
that position okay
so this is what we call the
electrostatic potential but we have
derived it only for a charge kept at the
origin
supposing the charge was not at origin
supposing it was somewhere else
supposing my charge is sitting here so
the distance from the origin to the
charge is a distance I will call R Prime
it is very standard in electromagnetics
that when I use a symbol R I mean
observer and then we use a symbol R
Prime
I mean source so this Q is generating
the electric field it is the source of
the electric field so I use R Prime now
I have got some curve going from R
naught to earth this this equation won't
work because this equation assume the
charge was at origin however what is
this origin the universe does not really
have knowledge of an origin universe
does not say this lab is the center of
the universe he does not say the center
of the Milky Way is the center of the
universe if we put origin here it's it's
in our minds universe does not define it
so we might as well put an origin at the
charge
now if you put the origin at the charge
we know the answer we know that the
electric field are not 2 R of P dot DL
is equal to some files which is a
function of R and not of how we got from
are not to R so how can we take this
information and use it in general well
the answer is quite easy what we do is
we say let us consider a new variable
called U which is equal to R minus R
Prime
now as we move on this curve what is
changing is r so I will do a change of
variables D U is d R - 0 because R prime
remained constant so D u equals D R so
what that what is that saying that is
saying that if I take a point on this
curve earlier I have this was our
now I have from the new origin I have
you and they are different vectors but
if I now go to the next point that is r
plus d r that corresponds to u Plus D U
and when we changed coordinates like
this R Plus D R minus R that is the D R
is nothing but u Plus D u minus u so it
says D U and D are are the same thing
even though I change coordinates so my
coordinates are now very different but
the change in U and the change in R are
the same and it is only the change that
I concern myself about when I do an
integral it is DL that comes not L
itself so let us see where this
transformation of coordinates takes us
so we start with the integral R naught 2
R 1 I am changing this final point to R
1 because I used R for the intermediate
point as well so R naught R 1 electric
field dot DL that is why we start now I
put I will change my coordinates I will
go to this coordinate system u equals R
minus R naught so that is like working
in this coordinate because this is you
and this is U plus D
so when I do that the integral becomes u
naught u 1 the electric field as a
function of U dot G U and I already know
that D U and DL are the same thing so
that is why I am able to write this u
naught is nothing but R not in the you
coordinate system it is it is this
vector starting from this origin to R
naught and u 1 is this vector starting
from the new origin to R 1 but in this
new coordinate system my charge Q is
sitting at the origin which means I know
how to write down the electric field we
can write it down if integral u naught u
1 Q over 4 PI epsilon naught u over u
cubed dot d u since the electric field
in the you coordinate system dot T u
along the curve but this is the problem
we have already done we have already
done the problem of integral of e dot DL
when the charge is at the origin and the
answer we got for that was it is equal
to Q over 4 PI epsilon naught integral I
replace the vector by just the distance
the radial distance from the origin so
without the line underneath such as a
radial coordinate you know tu 1
D u over u square + the integral became
as we saw last time Q over 4 PI epsilon
not 1 over u naught minus 1 over u wasp
but what is u naught u naught is the
magnitude of the vector u naught is the
radial length so that is the magnitude
but u naught itself is defined by this
equation so it is equal to magnitude R
minus R naught similarly u 1 is equal to
magnitude R minus R 1 sorry I think I
made a mistake there equal to R naught
minus this this is R minus R Trent R
naught minus R Prime and this is R 1
minus R prime so I substitute these in
here and what I get is Q over 4 PI
epsilon naught 1 over magnitude R naught
minus R prime minus 1 over magnitude R 1
minus after
so once again even though the charge is
not at the origin this integral turned
out to be a function only of the initial
position the final position and of
course where the charges it did not
depend on how we went from are not to r1
that got taken care of by evaluation of
this integral and the integral depended
only on how where we started and where
we ended it didn't depend on the route
and usually we take this initial point
such that it does not contribute at all
for instance in 3d we will choose this
to be infinity in which case we get
integral R naught which is infinity to R
1 now I am going to just call it R P dot
DL is equal to minus Q over 4 PI epsilon
naught 1 over R - R
what this means is that we can again
define a function a scalar function
which is derived from the electric field
so we will define potential Phi of R
which is equal to minus integral
infinity to R of e dot d R
however we come from infinity to R this
answer will be unchanged it only depends
on the fact that we came to this point
so this is a generalization it means
that regardless of where the charges
whether it was of the origin or not at
the origin my potential function is the
same and now I have one more
generalization I can put many charges Q
1 Q 2 Q 3 Q n and I can use this same
technique and I can say integral from
infinity this is the symbol for infinity
and H that is turn horizontal to R a dot
DL this electric field is due to all
these charges so I can write this as
infinity to R summation over say K
equals 1 to infinity
you sub K over 4 pie epsilon-not
well let me call this R 1 after R minus
R prime K divided by R minus R prime K
cubed dot DL now each of these terms is
nothing but this integral that we have
already done so the answer must
therefore be just the sum of all these
answers so it becomes equal to some K
equals one to n Q sub K over four PI
epsilon naught 1 over R minus R K right
and this is now BR 1 the limit R 1 so I
have now an expression for potential is
actually the negative of this is what I
want so minus minus minus that became
minus so this is plus Phi of R 1 is
equal to so this electrostatic potential
is the sum of electrostatic potentials
due to a charge Q 1 Q 2 Q 3 up to QN
now this potential is completely
fictitious at least as far as the
electric field is concerned we know
there is a force therefore we can say we
believe there is a force per unit
Coulomb but as far as the electric
potential is concerned there is no way
of measuring the electric potential
however it is an extremely useful idea
because you see the electric field has
three quantities in it so there is a
electric field along X and electric
field along Y and electric field along Z
so it is a complicated field it has
three components in it but the electric
potential Phi has only one component Y
itself so it is a much simpler concept
than the electric field so I have gone
down from having three pieces of
information at every point to having one
piece of information at every point
there is more to it than that if this is
all I was doing for example supposing a
plane is flying over ground okay so the
plane has a location and this R would
have x y and z information x and y talk
about where horizontally it is z it is
how much higher than ground now I could
ask what is the altitude of the plane
that maps from this 3d value to just Z
so that also is like going to a very
simple idea so you could think of
finding altitude finding height
which takes a three dimensional vector
and gives me back an answer it is one
kilometer above the ground and finding
potential I go from electric field which
is e ex ey ez to a single value Phi so
it looks like I am doing the same sort
of thing but there is something very
surprising if I told you the height of
the plane you could not tell me where
the plane is you can only tell me the
height of the plane you cannot tell me
that it is 100 kilometers east of Madras
but if you are trying to if you give if
you give me the electric potential it so
turns out I can tell you what all these
three components are it is the
information that is inside Phi is enough
to reconstruct the entire electric
vector this is an extremely unusual
thing in fact what it means is that the
electric field is a very simple field it
is not really a vector field at all that
in some sense it is so simple its entire
information is captured in a scalar
field now let us show this supposing we
have found potential so we know that for
a point source Phi is equal to Q over 4
PI epsilon 0 1 over magnitude R minus R
prime this is what the potential is
potential itself is a function of R and
the point source Q
is at a point our time all right now I
want to go from knowing this potential
to knowing this P and I am claiming that
unlike in the case where I am giving the
height and asked to find the plane and I
cannot do it in the case where I am
given the electric potential I can find
the entire electric field let us see how
we do it
first things first if I have got an
electric potential Phi of our this is a
scalar field it means there are numbers
associated with every position this
electric potential came from an integral
now I am sure you know that when we
integrate things we get smooth functions
in other words integral of any function
is continuous so what it means is Phi is
smooth now what does that mean it means
that Phi is like temperature Phi is like
the height of a mountain
I can draw contours of five so I can if
I work only in x and y I can draw
contours this is X this is y and I say
that Phi is 4 volts 5 volts 6 volts and
so on and so forth so I have because it
is a smooth function I have the ability
to draw surfaces on which potential is
constant now what is the advantage of
this
supposing I look at this part alone I am
going to blow it up
the 5-volt surface looks like this the
6-volt surface looks like this they're
both slightly curved so I have drawn
them slightly curved now I am sitting
here and I move in some direction and
after I move some distance call it DL I
go from Phi of 5 volts to Phi of 6 volts
however we have a separate result we
know that Phi of R with H 5 volts or 6
volts is equal to integral R naught to R
of electric field dot DL and it says it
does not care how you got there if you
start at R naught and reach our however
you reach it is Phi of R so now let us
see how that works let us say that my
point or not is here to calculate Phi at
this point and find that it is 5 volts I
have done an integral and come here to
find that Phi at this point 6 volts I
did another integer and came there and
said the difference in potential is 1
volt but this curves c1 and this curve C
2 are completely arbitrary I can choose
any curve I like for c1 and c2 so
specifically when I when I choose C 2 I
do not choose this one I am going to
choose a curve that looks like this I am
going to retrace c1 itself go along c1
come to five volts and then I will move
along DL itself and reach them so what
does that mean it means I did integral e
dot DL in order to do this integral I am
going to do it as integral from are not
to are to this point P dot DL plus this
piece a of r dot here what have I done I
can calculate this potential any way I
like
I can go on any curve I like but I
choose chose to go to that curve which
touches our before it goes to our plus D
our this part of the curve is this and
this additional part is this and you
know that if you have an integral A to B
of some f of X DX you can always write a
to C f of X DX plus integral C to B f of
X DX so that is what I have done I have
broken it up into two pieces
the second piece is very simple it is
just e dot DL but this first piece I
know the answer already this first piece
is nothing but Phi at all I am sorry
here this should be Phi of R plus DF
so I have an expression which says Phi
at this point R Plus D DL is an integral
R naught R + dl of e dot DL that is by
definition I have broken it into two
parts one part which goes to R and E dot
DL and this part is nothing but Phi of R
itself so if we write it out now
we have an equation that says phi of r
plus dl is equal to phi of r plus e dot
d h of course it assumes that DL is very
small so that second integral is e dot
DL I can take this Phi of R to the other
side and I can write it as Phi of R + dl
minus Phi of R is equal to P dot HTM
this is the Delta Phi now if we divided
both sides by DL then in a certain sense
we get e I mean you look at this
equation it is sort of saying I am going
to put this within quotes D Phi T L is
equal to minus T well there is a minus
sign coming somewhere if the whole thing
has - right so Phi is minus of this so
there is a minus sign so it is minus e
okay
except we do not know how to define this
quantity we do not know how to say
derivative of a function with respect to
a vector a very funny kind of the
derivative we understand things like DF
DX DF dy but what does it mean to say DF
g el let us look at this diagram again
at 6 volts I have 5 volts
I started at a point R I went to R + DL
and I am saying that this Delta Phi is
equal to the electric field at r dot da
now there is another way of calculating
Delta Phi supposing these are straight
lines and they are for very small
distances they are straight lines so if
they are straight lines can draw this
diagram as Phi equals 5 volts pi equals
6 volts this is my DL this is our I can
draw a normal between these two planes
and I can take the angle between them
that is theta if this distance is D
then it is clearly true that D is equal
to D L cos theta because this is the
diagonal this is the base so D over the
diagonal is cos theta so D is equal to D
L cos theta so in a certain sense when I
look at Delta Phi I am saying move along
any direction you like till you go a
distance along the base which is equal
to G because if I've entered a different
kita event along an angle theta - I
would have to go a different distance DL
but I would have to go that distance DL
such that my base was equal to D so I
can write the same relation as Delta Phi
is equal to D Phi D L times so D Phi D n
times DL cos theta
meaning go in this direction this is the
n direction find go in this direction
how far you go in this direction this
length cos theta you would have gone
there for a distance D the answer is you
have gone a distance so you changed Phi
by Delta Phi
now if you look at these two this
equation you have one magnitude you have
a second magnitude and you have cos
theta and we know a vector operation
that introduces cos theta we can call
this D Phi DN along the normal direction
dot here
all right
I want to repeat the steps I made so
that you're with me first of all we have
a reference somewhere or not when I
calculate potential at any point I am
doing integral R naught to R of e dot DL
with a minus sign I have not kept the
minus signs properly
so I integrate from are not to R and I
integrate from R naught R plus DF they
give me different answers for Phi and
the difference in the answers is what I
call Delta Phi now I can calculate this
Delta Phi in a clever way by saying the
Phi at this point is equal to
calculating this integral e dot DL up to
this R and then integrating along deal
that gave me that the change in Delta
Phi is only this piece a dot DL with a
minus sign okay minus sign here so I
have Delta Phi is equal to minus e dot
Da r dot DL and I can calculate the same
Delta Phi in a different way namely I
can say that if I move in any direction
I move a distance DL
it is equivalent as far as the change in
Phi is concerned with moving in the
normal direction a distance DL cos theta
that is the normal distance I move is
the sloping distance I move times cos
theta the change in potential is nothing
but the change in potential as I move in
this shortest distr Direction times this
projected distance which can be written
as a vector operation that is I am
constructing a special vector a vector
whose direction is the direction of the
normal whose magnitude is the derivative
along this direction now look at these
two equations both are equal to the same
thing so I can write minus e dot DL is
equal to D Phi DN n hat dot DL
which means I can write that the
electric field is equal to minus D Phi D
n along MN now the textbook gives the
algebra so you do not have to really
understand the algebra that much but
what is important here is that from the
electric field I did an integral I got a
scalar quantity I can do that for many
fields as I told you if I if you tell me
the location of a plane I can tell you
its height that is like extracting a
scalar field out of a vector field but
given the height I cannot locate the
plane given the height I only have the
height I do not have the X&Y coordinates
but this is a case where given the
scalar field Phi I have now gone back
and found the vector field E so it says
the electric field is particularly
simple it is so simple that given the
electric potential I can calculate back
the electric field this is one of the
reasons why the electric potential is so
useful because by working with a single
scalar quantity
I have found out a vector quantity by by
studying a function of with one
component I have actually solved for
three components and therefore there is
a great saving in labor the best way of
showing that it is a useful idea is to
work out a problem
couple of lectures ago I derived what
the dipole field the field due to a
dipole is I did that because I wanted to
show that the Coulomb field does not
fall fast enough if you remember I was
looking at a problem that the Coulomb
field was 10 to the 36 times 36 orders
of magnitude times fast
stronger than the gravitational field so
one of the explanations that is given is
there are equal numbers of positive and
negative charges so I said let us assume
these charges are separated by a
distance of 1 angstrom and let us work
out what the field would be and at that
time I used Coulomb's law and calculated
the field let us do the same thing with
the electrostatic potential so I have a
charge capital Q I have another charge
minus Q this is the Z direction
and some distance far away is an
observer so I am putting my origin here
and this is a position or so
this distance between the charges is a
distance D and I have this is R - R 1
this is our - R 2 as you worked out the
electric field at R is Q over 4 PI
epsilon naught times R - R 1 divided by
magnitude R minus R 1 cubed minus R - R
2 divided by magnitude R - R 2 cubed it
is a vector quantity it depends on R
depends on R 1 and it depends on R 2 and
at that time I didn't want to do all the
algebra so I only worked out roughly
what the R dependence of this vector
field is going to be now I can work out
what the electric potential of this is
we have already worked it out Phi of R
equals Q over 4 PI epsilon naught 1 over
magnitude R - R 1 minus 1 over that is
the electric field depended on the
direction and the magnitude of each of
these lines the electric potential
depends only on the distance it does not
care about the direction is an enormous
simplification why is it in
simplification
supposing I wanted to find out what this
field is for are much much greater than
D so I am talking about a point which is
far away then what is going to happen
these two lines let me redraw this
because it
not drawn very well these two lines are
nearly parallel by that I mean this
angle Delta Theta is nearly zero so
these lines are nearly parallel which
means that if I draw an arc with this
line as my as my radius actually it will
look like this and this it is
essentially a straight line how much is
this distance well
this position are has an angle with
respect to the horizontal theta that
same angle shows up here and so this
distance becomes D sine theta if you
choose your angle the other way which is
what the book does actually you will get
D cos theta so now let us see if we can
work out what Phi should be I am going
to take common denominator so Phi of R
equals Q over 4 PI epsilon naught times
small R minus R 2 minus mod R minus R 1
divided by the product of the 2 R minus
R 1 R minus R 2 so it is the difference
in distances between R and these two
charges divided by the product of the
distances now each of these distances is
more or less equal to net distance R so
the denominator can be simplified so
what you get is Phi of R is equal to Q
over 4 PI epsilon naught times the
difference in distances we worked out is
d sine theta divided by the denominator
is nothing but R square is an
approximation to approximations are
involved one is
that if I put a right angle line here it
is equivalent to saying it is a arc this
is a very good approximation the second
approximation is that the product of
these two distances is approximately R
square so both of these are very good
approximations for charges that are far
away now why do I call this an advantage
I mean I have calculated Phi but how
does it help me well now all I have to
do is to go from Phi to e so I need to
do electric field is equal to minus D
Phi D n n hat I can plot out this Phi I
can find the what are called
equipotentials of Phi and if I navigate
along the normals to those potentials I
get the electric field out of it now
there is a standard mathematical
machinery to carry out this action let
us try and see how we can do it if you
if you look at what we did earlier we
said Delta Phi is equal to
d phi d n n hat dot dl so in a certain
sense what this is saying can be written
out this piece I keep D Phi D n n hat
dot DX along X plus dy along y plus DZ
along Z so this quantity has an X
component it has a Y component and a Z
component so it is equal to D Phi D n
along X DX plus D Phi D n along Y dy
Plus D Phi G n along Z research the
component along X the component along y
the component along Z but we also know
that this same quantity is equal to
minus e dot DL so the same equation can
also be written as minus e X DX plus
minus ey dy plus minus e Z DZ which
means that e X is nothing but the X
component of this normal derivative e Y
is nothing but the Y component of this
normal derivative and easy it is nothing
but the Z component of this normal
derivative
now if you know your partial derivatives
you know that if you have any function f
of X Y Z then you know that DF is equal
to del F del X DX plus del F del y Gy
plus del F del Z DZ the standard result
in mathematics and here we have D Phi is
equal to D Phi D n along X DX Plus D Phi
DN along y dy plus D Phi D and along Z
DZ so if you just look at these two it
becomes obvious how to calculate this D
Phi DN so so I have three equations let
me write them all here this D Phi is
equal to this it is also equal to del
Phi del X DX plus del Phi del Y dy plus
del Phi del Z DZ it is also equal to
minus E X DX plus minus ey dy plus minus
ez DZ and they are all equal to each
other for any DX any D Y and any DZ
which means these coefficients are
saying the same thing these coefficients
are saying the same thing and these
coefficients are saying the same thing
so I can write down px is equal to minus
del Phi del x ey equals minus del Phi
del Y
and easy equals minus del v del Z
this procedure of extracting three
values out of one function is given a
name it is called the gradient and it is
also given a symbol so we write the
electric field is equal to minus
inverted triangle Phi and when we write
this we mean this we mean that the
inverted triangle Phi is a vector whose
X component is del Phi del X whose Y
component is del Phi del Y whose Z
component is del Phi del Z so minus of
that puts in a minus sign for each of
these now this is in Cartesian
coordinates what happens if we do not
stay in Cartesian coordinates up to now
we have worked only with XYZ but it
turns out that there are many many
interesting uses of cylindrical and
spherical coordinates so let us look at
a case where you want to go away let us
say we have spherical polar coordinates
spherical polar coordinates as I
remember talking about little while back
you have the radial direction you have a
theta direction that is the angle with
the z axis and you have
a cone that is traced out by this vector
if you rotate it around the z axis that
point if you take this point and project
it onto the XY plane you get a vector in
the XY plane this is X this is y there
is an angle here this angle is called
Phi so you have Z is equal to R cos Phi
cos theta sorry that is it is the this
height square root of x squared plus y
squared is equal to R sine theta that is
this length
the x-coordinate itself would become R
sine theta times cos Phi cos Phi because
this length is our sine is R sine theta
and you have to multiply that R sine
theta to get by cos Phi to get the
distance along X similarly Y is equal to
R sine theta sine Phi is an
exceptionally useful coordinate system
we should know all about it
but in this particular case what I am
interested in finding out is if I wrote
the potential Phi as a function of R
theta and Phi what can I get if I want
to get the gradient well the answer is
that I want to know what Delta Phi is
Delta Phi is del Phi del R Delta R plus
del Phi del theta Delta Theta plus del
Phi del
well I should use a different symbol
Delta first so this is nothing but if I
have a function of three variables the
change in the function is given by this
expression but I I also want to know
what is DL DL is Delta R along the R
Direction plus R Delta Theta along the
theta direction so along the R direction
I just go but along the theta direction
if I increase in theta by an amount
Delta theta this distance is R Delta
Theta and then it is equal to R sine
theta Delta Phi along the Phi direction
I put a twiddle over the files to
separate it out from the electric
potential where is this R sine theta
come from well the projection on to the
XY plane is R sine theta and if I change
Phi this distance is given by R sine
theta Delta Phi now if you look I want
to say Delta Phi is equal to minus e dot
DL so I have to read off this is the DL
part this is the Delta Phi so what is
what are the components of e so minus e
R is del Phi del R minus e theta is
equal to 1 over R del Phi del theta and
minus e Phi is equal to 1 over R sine
theta del Phi del
third angle I will come back to this the
next class but this will allow us to
calculate the electric field out of a
dipole thank you