we know that the action

S=∫ba(−mcds−ecAÎ¼dxÎ¼)

and sinceÎ´S=0 , we can rewrite the above equation as

Î´S=∫ba(−mcÎ´ds−ecÎ´(AÎ¼dxÎ¼))

Î´S=∫ba(−mcdÎ´s−ec(dxÎ¼Î´AÎ¼+AÎ¼Î´(dxÎ¼))

Î´S=∫ba(−mcdÎ´s−ec(dxÎ¼Î´AÎ¼+AÎ¼dÎ´xÎ¼)

usingds=(dxÎ¼dxÎ¼)12 , we can calculate that Î´ds=vÎ¼Î´(dxÎ¼) i.e

\dÎ´s=vÎ¼dÎ´xÎ¼ . We can substitute this term in the previous equation which therefore becomes

Î´S=∫ba(−mcvÎ¼dÎ´xÎ¼−ec(dxÎ¼Î´AÎ¼+AÎ¼dÎ´xÎ¼)

We can rewritevÎ¼dÎ´xÎ¼=d(vÎ¼Î´xÎ¼)−Î´xÎ¼dvÎ¼

\ similarlyAÎ¼dÎ´xÎ¼=d(AÎ¼Î´xÎ¼)−Î´xÎ¼dAÎ¼

substituting the above results in the previous equation leads to

Î´S=∫ba(−mc(d(vÎ¼Î´xÎ¼)−Î´xÎ¼dvÎ¼)−ec(dxÎ¼Î´AÎ¼+d(AÎ¼Î´xÎ¼)−Î´xÎ¼dAÎ¼)

Î´S=∫ba(−mcd(vÎ¼Î´xÎ¼)−ecd(AÎ¼Î´xÎ¼))+∫ba(mcÎ´xÎ¼dvÎ¼)−ec(dxÎ¼Î´AÎ¼−Î´xÎ¼dAÎ¼)

let us rewriteÎ´AÎ¼ as dAÎ¼dxÎ½Î´xÎ½

and similarlydAÎ¼ as dAÎ¼dxÎ½dxÎ½

correcting for the above changes in the equation above gives

Î´S=∫ba(mcÎ´xÎ¼dvÎ¼)−ec(dxÎ¼dAÎ¼dxÎ½Î´xÎ½−Î´xÎ¼dAÎ¼dxÎ½dxÎ½)

Î´S=∫ba(mcÎ´xÎ¼dvÎ¼)−ec(dxÎ½dAÎ½dxÎ¼Î´xÎ¼−Î´xÎ¼dAÎ¼dxÎ½dxÎ½)

Î´S=∫baÎ´xÎ¼((mcdvÎ¼)−ec(dxÎ½dAÎ½dxÎ¼−dAÎ¼dxÎ½dxÎ½))

and sinceÎ´S=0

∫baÎ´xÎ¼((mcdvÎ¼)−ec(dxÎ½dAÎ½dxÎ¼−dAÎ¼dxÎ½dxÎ½))=0

mcdvÎ¼−ec(dxÎ½dAÎ½dxÎ¼−dAÎ¼dxÎ½dxÎ½)=0

mcdvÎ¼=ec(dxÎ½dAÎ½dxÎ¼−dAÎ¼dxÎ½dxÎ½)

mcdvÎ¼=ec(dAÎ½dxÎ¼−dAÎ¼dxÎ½)dxÎ½

dividing the whole equation byds and remembering that dxÎ¼ds=vÎ¼ , we can rewrite the above equation as

mcdvÎ¼ds=ec(dAÎ½dxÎ¼−dAÎ¼dxÎ½)dxÎ½ds

mcdvÎ¼ds=ec(dAÎ½dxÎ¼−dAÎ¼dxÎ½)vÎ½

we now define
FÎ¼Î½=dAÎ½dxÎ¼−dAÎ¼dxÎ½

mcdvÎ¼ds=FÎ¼Î½vÎ½

We can understand from the construction ofFÎ¼Î½ that it is an antisymmetric matrix i.e FÎ½Î¼=−FÎ¼Î½

S=∫ba(−mcds−ecAÎ¼dxÎ¼)

specifically, let’s look atAÎ¼dxÎ¼ . we know that AÎ¼=(Ï•/c,A⃗ ) and that dxÎ¼=(cdt,−dr⃗ )

thereforeAÎ¼dxÎ¼=Ï•dt−A⃗ ⋅dr⃗

S=∫ba(−mcds−ec(Ï•dt−A⃗ ⋅dr⃗ ))

S=∫ba(−mcds−ecÏ•dt+ecA⃗ ⋅dr⃗ ))

we know thatds=cdÏ„ and that dÏ„=dtÎ³ substituting these changes changes the above equation to

S=∫ba(−mc2dtÎ³−ecÏ•dt+ecA⃗ ⋅dr⃗ ))

S=∫ba(−mc2dtÎ³−ecÏ•dt+ecA⃗ ⋅dr⃗ dtdt))

S=∫badt(−mc21Î³−ecÏ•+ecA⃗ ⋅dr⃗ dt))

S=∫badtL

where
L=−mc21Î³−ecÏ•+ecA⃗ ⋅dr⃗ dt

L=−mc21Î³−ecÏ•+ecA⃗ ⋅V⃗

we can get the equations of motion from
∂L∂r−ddt∂L∂r˙=0

∂L∂r=−ec∇Ï•+ec∇⃗ (A⃗ ⋅V⃗ )

∇⃗ (A⃗ ⋅V⃗ )=V⃗ ×(∇⃗ ×A⃗ )−A⃗ (∇⃗ ⋅V⃗ )

sinceA⃗ (∇⃗ ⋅V⃗ )=0 , we can rewrite the earlier equation as

∂L∂r=−ec∇Ï•+ecV⃗ ×(∇⃗ ×A⃗ )

similarly
∂L∂r˙=∂∂r˙(−mc21Î³−ecA⃗ ⋅V⃗ )

∂L∂r˙=∂∂r˙(−mc21−v2c2−−−−−−√−ecA⃗ ⋅V⃗ )

∂L∂r˙=Î³mV⃗ +ecA⃗

going back to
∂L∂r−ddt∂L∂r˙=0

−ec∇Ï•+ecV⃗ ×(∇⃗ ×A⃗ )−ddt(Î³mV⃗ +ecA⃗ )=0

−ec∇Ï•+ecV⃗ ×(∇⃗ ×A⃗ )−dp⃗ dt+ecdA⃗ dt=0

wherep⃗ =Î³mV⃗

dp⃗ dt=−ec∇Ï•+ecdA⃗ dt+ecV⃗ ×(∇⃗ ×A⃗ )

dp⃗ dt=−e(1c∇Ï•+1cdA⃗ dt)+ecV⃗ ×(∇⃗ ×A⃗ )

we’re defining
E⃗ =−1c∇Ï•−1cdA⃗ dt
and
B⃗ =∇⃗ ×A⃗

dp⃗ dt=eE⃗ +ecV⃗ ×B⃗

if we assumeAÎ¼=(Ï•,A⃗ ) ,
E⃗ =−∇Ï•−1cdA⃗ dt

and since

using

\

We can rewrite

\ similarly

substituting the above results in the previous equation leads to

let us rewrite

and similarly

correcting for the above changes in the equation above gives

and since

dividing the whole equation by

we now define

We can understand from the construction of

specifically, let’s look at

therefore

we know that

where

we can get the equations of motion from

since

similarly

going back to

where

we’re defining

if we assume