EP 222: Classical Mechanics - Lecture 12 Dipan K. Ghosh Indian - PowerPoint PPT Presentation

EP 222: Classical Mechanics - Lecture 12 Dipan K. Ghosh Indian Institute of Technology Bombay dipan.ghosh@gmail.com Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 12 August 17, 2014

Review of Lecture 11: Concept of Symmetry We discussed symmetry of physical laws under space and time translation as well as under rotation. homogeneity of space, expressed by invariance of the Lagrangian under translation leads to conservation of linear momentum. We also saw that isotropy of space leads to symmetry of physical system under rotation. This leads to conservation of angular momentum. Homogeneity of time implies that the Lagrangian of the system does not have explicit time dependence. This leads to the conservation of the total energy of the system. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 1 / 15

Gauge Transformation We discussed Galilean invariance and found that the coordinate transformation corresponding to Galilean transformation is equivalent to Lagrangian Gauge transformation : L → L + dF dt Electromagnetic potentials satisfy gauge transformations A ′ = � � A + ∇ F ( � r , t ) ϕ ′ = ϕ − ∂ F ∂ t These transformations leave Maxwell’s equations unchanged. For instance, consider Faraday’s Law ∇ × E + ∂ B ∂ t = 0 � � E + ∂� A Substitute � B = ∇ × � � A , = ⇒ ∇ × = 0 ∂ t Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 2 / 15

Gauge Transformation � � E + ∂� E + ∂� A A � ⇒ � ∇ × = 0 = ∂ t = −∇ ϕ ∂ t � � ∇ ϕ + ∂ϕ � E = − ∂ t � � �� ∂� � ϕ − ∂ F � ∂ t + ∂ ( ∇ F ) A = ∇ + ∂ t ∂ t Thus A ′ = � � A + ∇ F ( � r , t ) ϕ ′ = ϕ − ∂ F ∂ t Electromagnetic forces can be included in Lagrangian by defining a v · � velocity dependent potential e ϕ − e � A L = 1 2 mv 2 − e ϕ + e � v · � A Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 3 / 15

Gauge symmetry Suppose we add a term e dF ( � r , t ) to the Lagrangian. dt The new Lagrangian becomes L = 1 A + e dF ( � r , t ) 2 mv 2 − e ϕ + e � v · � dt = 1 � � ϕ − ∂ F 2 mv 2 − e � � ( e � + � v · A + e ∇ ϕ ) ∂ t We have used dF dt = ∂ F ∂ x v x + ∂ F ∂ y v y + ∂ F ∂ z v z + ∂ F v + ∂ F ∂ t = ∇ F · � ∂ t This gives L = 1 2 mv 2 − e ϕ ′ + e � v · � A ′ Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 4 / 15

Coordinate Transformation vrs. Symmetry Operations Under coordinate transformation q → q ′ , L → L ′ ( q , q ′ , t ) Suppose L = 1 q 2 − 1 2 kq 2 . Change q → q ′ = q + a 2 m ˙ L ′ = 1 q ′ 2 − 1 2( q ′ − a ) 2 2 m ˙ L ′ ( q ′ , ˙ q ′ , t ) = L ( q ( q ′ , ˙ q ′ , t ) , ˙ q ( q ′ , ˙ q ′ , t ) , t ) The functional dependence of L ′ on the primes coordinates is not the same as the functional dependence of L on the unprimed coordinates Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 5 / 15

Active and Passive Transformation Relabelling done in a passive manner in which we do not change the system but change the coordinate system itself In active transformation, we leave the coordinate system unchanged but move the entire system relative to the original coordinates so that a point which coordinates q has coordinate q ′ Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 6 / 15

Symmetry Operation Translation and rotations are examples of continuous transformation. In each case, there is a parameter – a for translation, θ for rotation. Discrete Transformation : Reflection x → x ′ = x − a Let the parameter of continuous transformation be s . q → q ′ = Q ( s , t ) Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 7 / 15

Continuous Transformation Suppose we demand L ′ ( q ′ , ˙ q ′ , t ) = L ( q ′ , ˙ q ′ , t ) An example could be a situation like L ′ = 1 q ′ 2 − 1 2 kq ′ 2 2 m ˙ This could happen by a reflection transformation q ′ = − q . Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 8 / 15

Continuous Transformation An example of continuous transformation : A spring mass system with spring force being radially inward z , t ) = 1 z 2 ) − 1 x 2 + ˙ y 2 + ˙ 2 k ( x 2 + y 2 + z 2 ) L ( x , y , z , ˙ x , ˙ y , ˙ 2 m (˙ Consider a coordinate transformation x ′ = x cos θ − y sin θ, y ′ = x sin θ + y cos θ, z ′ = z Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 9 / 15

Test of Symmetry Operation z ′ , t ) = 1 x ′ cos θ + ˙ y ′ sin θ ) 2 L ′ ( x ′ , y ′ , z ′ , ˙ x ′ , ˙ y ′ , ˙ � 2 m (˙ x ′ sin θ + ˙ y ′ cos θ ) 2 + ˙ z ′ 2 ) � + ( − ˙ = 1 ( x ′ cos θ + y ′ sin θ ) 2 � 2 k + ( − x ′ sin θ + y ′ cos θ ) 2 + z ′ 2 ) � = 1 z ′ 2 ) − 1 x ′ 2 + ˙ y ′ 2 + ˙ 2 k ( x ′ 2 + y ′ 2 + z ′ 2 ) 2 m (˙ = L ( x ′ , y ′ , z ′ ) However, a translation by an amount a would change the potential energy term to 1 ( x ′ + a ) 2 + y ′ 2 + z ′ 2 � � , which is not a symmetry operation for 2 k this system. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 10 / 15

Test of Symmetry Operation An alternate prescription is to check if L ( x ′ , y ′ , z ′ ) = L ( x , y , z ) L ( x ′ , y ′ , z ′ ) = 1 z 2 ) − 1 x 2 + ˙ y 2 + ˙ 2 k ( x 2 + y 2 + z 2 ) 2(˙ Now L ( x ′ , y ′ , z ′ ) is given by L ( x ′ , y ′ , z ′ ) = 1 y sin θ ) 2 + (˙ y cos θ ) 2 + ˙ z 2 � � 2 m (˙ x cos θ − ˙ x sin θ + ˙ − 1 ( x cos θ − y sin θ ) 2 + ( x sin θ + y cos θ ) 2 + z 2 � � 2 k = 1 z 2 ) − 1 x 2 + ˙ y 2 + ˙ 2 k ( x 2 + y 2 + z 2 ) 2 m (˙ = L ( x , y , z ) Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 11 / 15

Noether’s Theorem Whenever a physical system exhibits a continuous symmetry, there is a conserved quantity associated with it. In field theory, the name“Noether’s Charge” is given to this conserved quantity. Suppose, there exist a parameter s which can be continuously varied. (e.g. θ for rotation, a for translation, while reflection does not have any associated parameter.) Let Q ( s , t ) be a symmetry operation such that the Lagrangian does not depend on it. d Q ( s , t ) , t ) = d ds L [ Q ( s , t ) , ˙ ds L [ q ( t ) , ˙ q ( t ) , t ] = 0 What it means is that though Q ( s , t ) has explicit dependence on s , the Lagrangian does not. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 12 / 15

Noether’s Theorem Since s does not have t dependence, we have, ˙ ∂ L ds + ∂ L dQ Q s = 0 ∂ ˙ ∂ Q Q Use Euler Lagrange equation on the first term � ∂ L � dQ � ∂ L d ˙ d ds + ∂ L Q d dQ � ds = 0 = ⇒ = 0 ∂ ˙ ∂ ˙ ∂ ˙ dt dt ds Q Q Q Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 13 / 15

Noether’s Theorem Canonical momentum corresponding to Q is p = ∂ L ∂ ˙ Q Λ = pdQ ds | s =0 = constant If the Lagrangian possesses a set of n continuous symmetry operations, there can be n conserved quantities, If the Lagrangian is a function of k generalized coordinates and velocities, the conserved Noether charges are k dQ i � Λ j ( { q k } , { ˙ q k } , t ) = p i | s j =0 for s j = 1 , 2 , . . . n ds j i =1 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 14 / 15

Cyclic Coordinates A coordinates is said to be cyclic (or ignorable) if L does not explicitly depend on it. The canonical momentum corresponding to a cyclic coordinate is conserved, since, � ∂ L ∂ L d � ∂ q = 0 = ⇒ = 0 ∂ ˙ dt Q This follows trivially from Noether’s theorem. If we consider a continuous translation q ( t ) → q ( t ) + s ; q ( t ) → ˙ ˙ q δ L = L ( q + s , ˙ q , t ) − L ( q , ˙ q , t ) ≡ 0 since L does not depend on the coordinate q . Dipan Ghosh (I.I.T. Bombay) Class. Mech. -12 August 20, 2014 15 / 15