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Space-Time Symmetry Properties of Space • Three Dimensionality: If we consider any arbitrary point in space then maximum three perpendicular lines can be drawn from this point. These mutually perpendicular lines are called three axes of the coordinate system. Therefore position of an arbitrary point in space can be defined using three coordinates (x, y, z). So space is three dimensional. • Flatness: Space is Flat. This means (i) In a right angle triangle, (Hypotenuse)2 = (Base)2 + (Normal)2 (ii) The sum of three angles of a triangle is equal to π radians. (iii) The shortest distance between two points in the space is a straight line. For most of Classical Mechanics Problems, the space is assumed to be completely flat. • Isotropy: Isotropy of space means uniformity of direction. All the directions of space are equally preferred. E.g.: if we have a source of light in space, then it will send light rays in all the directions with equal speed. Properties of Space • Homogeneity: Homogeneity of space means space is alike everywhere. If an experiment is performed in space at one place then the identical experiment performed anywhere in space will give the identical results. Also If F=ma is valid in one coordinate system then it will also be valid in another coordinate system. • Space Reflection: Space reflection means transformation of coordinates under which coordinates change sign. For example if r xi yj zk in right handed coordinate system then in left handed coordinate system xi yj zk However laws of physics (e.g. F=ma) remains invariant. Properties of Time • One Dimensionality Time is specified by only one coordinate ‘t’. Hence time is one dimensional. • Homogeneity (i) The laws of physics does not change with time. For example: If we perform an experiment today and repeat it after one month, the result of the experiment will be the same. (ii) interval of time are not affected by origin of time. One minute of today will be exactly equal to one minute of yesterday or tomorrow. • Isotropy The laws of physics remains invariant by changing t to –t. Conclusion • The properties of space do not change with time. • Time interval has same value for all times i.e., time flows uniformly. • Free space is homogenous. • Free space is isotropic. • Free space has property of reflection. These symmetry properties of space & time are called space-time invariance principle. According to these, laws of nature are same at all points in space & for all times. Taylor’s Theorem If r1 & r2 are two variables , having infinitesi mal change dr1 & dr2 respective ly, then according to Taylor' s theorem in vector form f( r1 dr1,r2 dr2 ) f(r1,r2 ) dr1.1 f dr2 . 2 f U ˆ U ˆ U ˆ where 1 f i j k x1 y1 z1 U ˆ U ˆ U ˆ & 2 f i j k x2 y2 z 2 Frame of Reference This train platform serves as the “rest” frame for Observers A and B, but it moves with a speed v towards Observer C, who stands on the roadside. A B C v Homogeneity of Space & Newton’s Third law of motion • Consider two particles A & B interacting with each other. Let r1 & r2 be the position vectors of the particles in coordinate system S. U ( r The P.E. of interaction of two particles will be 1 , r2 ) • Consider another coordinate system S’ which is displaced infinitesimally through a displacement dr. The position vectors of the particles in S’ will be r1 dr & r2 dr Then P.E. of interaction of two particles will be U (r1 dr , r2 dr ) Z’ S’ Z S m1 dr A r1 O Y r2 X B m2 X’ Y’ O’ Homogeneity of Space & Newton’s Third law of motion • According to Principle of Homogeneity U(r1,r2 ) U( r1 dr ,r2 dr ) ...(1) From Taylor' s theorem in vector form f( r1 dr ,r2 dr ) f(r1,r2 ) dr .1 f dr . 2 f U ˆ U ˆ U ˆ where 1 f i j k x1 y1 z1 U ˆ U ˆ U ˆ & 2 f i j k x2 y2 z 2 eq.(1) U(r1,r2 ) U(r1,r2 ) dr .1U dr . 2U dr .1U . 2U 0 dr .(1U 2U ) 0 since dr 0 (1U 2U ) 0 ...(2) but for conservati ve forces F U F12 1U & F21 2U eq. (2) becomes F12 F21 0 F12 F21 ...(3) which is Newton' s Third Law. Homogeneity of Space & Law of Conservation of Linear Momentum • If m1 & m2 be the masses of the particles A & B which are moving with velocities v1 & v2 respectively at any time then according to Newton’s second law of motion dv1 m1a1 F12 or m1 F12 ...(4) dt dv2 & m2 a2 F21 or m2 F21 ...(5) dt adding (4) & (5) dv1 dv2 m1 m2 F12 F21 dt dt according to Newton' s third law ( eq. (3) ) F12 F21 dv1 dv2 d (m1v1 ) d (m2 v2 ) m1 m2 0 0 dt dt dt dt d (m1v1 m2 v2 ) 0 dt (m1v1 m2 v2 ) constant which is law of conservati on of linear momentum. Rotational Invariance of Space & Law of Conservation of Angular Momentum • Consider two particles A & B interacting with each other. Let r & r be the position vectors of the particles in coordinate system S. Distance between two particles will be given by r1 r2 1 2 • The P.E. of interaction between the particles depends upon the scalar distance between two particles i.e., U (r1 , r2 ) U ( r1 r2 ) Z’ Z A S S’ r1 r2 O X r2 r1 B Y’ Y X’ Fig. 1 Suppose the system is rotated through a small angle d then the position vectors will get changed to r1 dr1 & r2 dr2 but according to isotropy of space, the distance between two particles will remain same. Continued… • Therefore P.E. U (r1 , r2 ) before rotation is equal to P.E.U (r1 dr1, r2 dr2 ) after rotation U (r1 , r2 ) U (r1 dr1 , r2 dr2 ) ...(1) • Applying Taylor’s theorem U (r1 dr1 , r2 dr2 ) U(r1,r2 ) dr1.1U dr2 . 2U ...(2) • Substituting the value to eq. (1), we get U (r1 , r2 ) U(r1,r2 ) dr1.1U dr2 . 2U dr1.1U dr2 . 2U 0 ...(3) Continued… • We know F U F12 1U & F21 2U • Hence equation (3) becomes dr1.F12 dr2 .F21 0 dr1.F12 dr2 .F21 0 ...(4) • We know, when coordinate system is through a vector angle d then, r d dr ...(5) Z X where dr is an arc which subtends a vector angle d at the origin. r1 d dr1 ...(6) r2 d dr2 ...(7) rotated r dr dr d r O Y from Newton' s Second law dp1 dp2 F12 & F21 ...(8) dt dt Substituti ng eq. 6,7 & 8 in eq. (4) dp1 dp 2 (r1 d). (r2 d). 0 dt dt Interchang ing Dot & Cross product dp1 dp 2 d.r1 d.r2 0 dt dt dp1 dp 2 d. r1 r2 0 dt dt since d 0 dp1 dp2 r1 r2 0 ....(9) dt dt d (r1 p1 ) dr1 dp1 Now p1 r1 dt dt dt dp1 v1 mv1 r1 but v1 v1 0 dt d (r1 p1 ) dp1 r1 ...(10) dt dt d (r2 p2 ) dp2 Similarly r2 ...(11) dt dt hence eq.(9) becomes d (r1 p1 ) d (r2 p2 ) 0 dt dt d (r1 p1 ) (r2 p2 ) 0 dt but angular momentum is given by Lrp d L1 L2 0 L1 L2 constant dt Implicit & Explicit Time Dependence • Suppose we are looking at the movement of a classical particle. The relevant variables here are position x(t) and momentum p(t). • For example, angular momentum L⃗=x⃗×p⃗ . Since x and p depend on the time, L also depends on time, but in this case it does so only because x and p depend on time. We have basically a function L=L(x,p) which then becomes L(x(t),p(t)). This is because in the definition of L , the time does not play a role. Therefore, we say that this quantity has only an implicit time dependence. In particular, ∂L/∂t=0 . • If, however, derived quantity f is defined such that the time occurs explicitly in the definition, for example a=∂v/∂t then a=a(v,t) there is direct dependence (explicit time dependence) on time & ∂a/∂t will not be equal to zero. Homogeneity of flow of Time & Law of Conservation of Energy We know, force is related to P.E. by F U ...(1) r We know gravitational force between two masses m1 & m2 is given by Gm1m2 F r 2 r It is clear from the expression that force does not depend on time explicitly (directly). Therefore P.E. will also not depend on time explicitly. U 0 ....(2) t Continued… • Also K.E. will also not depend on time explicitly but depends on time implicitly K.E. T 1 2 mv 2 T 0 ....(3) t • Total energy of system E=T+U where E is function of r & t. i.e. E=E(r,t) Continued… as E E (r , t ) E E dE dr dt r t (T U ) (T U ) dr dt r t U T U T dr dr dt dt r t r t using eq. (2) & (3), U T dE dr dr r r dE T U dr ...(4) dt r r dt U but F ....(5) r T 1 2 m v 2 & mv r r 2 2 r m v v 2v mv 2 r r T r v m r t r T v m. r t T ma ....(6) r substituti ng eq. (5) & (6) in eq. (4) dE dr ma F dt dt but F ma dE 0 dt or E constant Frames of Reference Inertial & Non-Inertial Frames • Inertial Reference Frame: Any frame in which Newton’s Laws are valid! Any reference frame moving with uniform motion (nonaccelerated) with respect to an “absolute” frame “fixed” with respect to the stars. • Non-Inertial Reference Frame: Any frame in which Newton’s Laws are not valid! Any reference frame moving with non-uniform motion (accelerated) with respect to an “absolute” frame “fixed” with respect to the stars. • The most common example of a non-inertial frame = Earth’s surface! • We usually assume Earth’s surface is inertial, when it is not! – A coord system fixed on the Earth is accelerating (Earth’s rotation + orbital motion) & is thus non-inertial! – For many problems, this is not important. For some, we cannot ignore it! RELATIVITY Problems with Classical Physics Classical mechanics are valid at low speeds c But are invalid at speeds close to the speed of light Relativity a special case of the general theory of relativity for measurements in reference frames moving at constant velocity. predicts how measurements in one inertial frame appear in another inertial frame. How they move wrt to each other. Reference Frames The problems described will be done using reference frames which are just a set of space time coordinates describing a measurement. eg. r r x, y, z, t z t y x Galilean-Newtonian Relativity According to the principle of Newtonian Relativity, the laws of mechanics are the same in all inertial frames of reference. i.e. someone in a lab and observed by someone running. Galilean-Newtonian Relativity Galilean Transformations Galilean Transformations We will consider effect of uniform motion on different quantities & laws of physics. We will establish a relationship between the space & time coordinates in two inertial frames of reference. The basic relations were obtained by Galileo & are known as Galilean Transformation Equations. allow us to determine how an event in one inertial frame will look in another inertial frame. assume that time is absolute. Galilean Transformations When second frame is moving relative to first along positive direction of X-axis. In S an event is described by (x,y,z,t). How does it look in S′? z z′ S v S’ vt x′ t P(x,y,z.t) (x’,y’z’.t’) t′ x O O′ x′ Y Y’ Galilean Transformations We will assume that time of occurrence is same in both the frames t = t′ From the diagram, x x' vt x' x vt And there is no relative motion in Y & Z-directions z v z′ vt x′ y' y t z' z t′ O Y O′ Y’ x x′ • So the Galilean transformations are x' x ut y' y z' z t' t • Inverse Galilean transformations are x x'ut y y' z z' t t' Galilean Transformations for Velocity Velocities can also be transformed. Using the previous equations, dx' d ( x vt) dx v' x v dt dt dt dy ' dy v' y vy dt dt dz ' dz v' z vz dt dt v' x v x v addition law for velocities Galilean Transformations for Acceleration dv' x d (vx v) dvx dv a' x dt dt dt dt dv since v is constant 0 dt dvx a' x ax dt & a' y a y , a' z a z Hence accelerati on remains invariant Galilean Transformations for Force • The mass is a constant quantity in Newtonian Mechanics and does not depend on state of rest or motion of the object. • According to Newton’s second law of motion F ma • w.r.t frame of reference S & S’ F ma (in S) F ' ma ' (in S') • But acceleration is invariant under Galilean Transformations a a' ma ma ' or F F ' Galilean Transformations Transforming Lengths Galilean Transformations How do lengths transforms transform under a Galilean transform? Note: to measure a length two points must be marked simultaneously. Galilean Transformations S S′ XA v XB Consider the truck moving to the right with a velocity v. Two observers, one in S and the other S′ measure the length of the truck. Galilean Transformations S v S′ XA XB In the S frame, an observer measures the length = XB-XA In the S′ frame, an observer measures the length = X′B-X′A Galilean Transformations S XA v S′ XB We know x’=x-vt each point is transformed as follows: x 'B xB vtB x ' A xA vt A Galilean Transformations x' B xB ut B S U S′ 0 x' A x A ut A Therefore we find that 0 XA XB x' B x' A xB x A u(t B t A ) if we measure poeition simulatenously then tB t A x' B x' A xB x A Hence for a Galilean transform, lengths are invariant for inertial reference frames. Important consequence of a Galilean Transformations All the laws of mechanics are invariant under a Galilean transform. Galilean Transformations When second frame is moving relative to first along a straight line along any direction In S an event is described by (x,y,z,t). How does it look in S′? Z’ P(x,y,z.t) (x’,y’z’.t’) S’ Z r' r S Y’ vt O Y X’ X O’ • S & S’ are two frames of references. • S’ is moving w.r.t. S with velocity ‘v’ • The coordinates of point P are (x,y,z,t) & (x’,y’,z’,t’) relative to S & S’ respectively. • Time is measured from the instant at which the two origins O & O’ coincide • Then after time t, O’ is separated from origin O by displacement vt • Then OO ' vt & let OO ' r , O ' P r ' • Applying the triangle law of vectors in triangle OO’P, we get OP OO ' O ' P r r ' vt or r ' r vt ...(1) • Now r xi yj zk r ' x 'i y ' j z 'k & v vx i v y j vz k • Therefore eq. 1 becomes ( x ' i y ' j z ' k ) ( xi yj zk ) (vx i v y j vz k )t comparing coefficients of i , j & k , we get x ' x vx t y ' y vyt z ' z vz t Invariance of Law of Conservation of Linear momentum • Under Galilean Transformations, velocity is not invariant, therefore linear momentum is also not invariant. • We have to check that whether law of conservation of linear momentum holds or not • According to whether law of conservation of linear momentum “in the absence of external force, the total momentum of the system remains unchanged or conserved” • Consider two particles m1u1 m2u2 m1v1 m2v2 u1 u1' v u2 u 2' v v1 v1' v v2 v 2' v m1 (u1' v) m2 (u 2' v) m1 (v1' v) m2 (v 2' v) m1u1' m2u 2' m1v1' m2 v 2' Invariance of Law of Conservation of Kinetic Energy • In S 1 1 1 1 m1u12 m2u22 m1v12 m2 v22 2 2 2 2 1 1 1 1 m1 (u1.u1 ) m2 (u2 .u2 ) m1 (v1.v1 ) m2 (v2 .v2 ) 2 2 2 2 1 1 1 1 m1 (u1' v).(u1' v) m2 (u 2' v).(u 2' v) m1 (v1' v).(v1' v) m2 (v 2' v).(v 2' v ) 2 2 2 2 u1 u1' v u2 u 2' v v1 v1' v v2 v 2' v 1 1 1 1 m1u1'2 m2u2'2 m1v1'2 m2v2'2 2 2 2 2 Are all the laws of Physics invariant in all inertial reference frames? Problems with Newtonian- Galilean Transformation Are all the laws of Physics invariant in all inertial reference frames? For example, are the laws of electricity and magnetism the same? Problems with Newtonian- Galilean Transformation Are all the laws of Physics invariant in all inertial reference frames? For example, are the laws of electricity and magnetism the same? For this to be true Maxwell's equations must be invariant. Problems with Newtonian- Galilean Transformation Problems with Newtonian- Galilean Transformation From electromagnetism we know that, c2 1 0 0 Since and constants then the speed of light are 0 0 is constant .c 3 108 m / s Problems with Newtonian- Galilean Transformation From electromagnetism we know that, c2 1 0 0 Since and constants then the speed of light are 0 0 is constant .c 3 108 m / s However from the addition law for velocities cv Problems with Newtonian- Galilean Transformation Therefore we have a contradiction! Either the additive law for velocities and hence absolute time is wrong Or the laws of electricity and magnetism are not invariant in all frames. Fictitious Forces • Some problems (like rigid body rotation!): Using an inertial frame is difficult or complex. Sometimes its easier to use a non-inertial frame! • “Fictitious Forces”: If we are careful, we can the treat dynamics of particles in non-inertial frames. – Start in inertial frame, use Newton’s Laws, & make the coordinate transformation to a non-inertial frame. – Suppose, in doing this, we insist that our equations look like Newton’s Laws (look like they are in an inertial frame). Coordinate transformation introduces terms on the “ma” side of F = mainertial. If we want eqtns in the non-inertial frame to look like Newton’s Laws, these terms are moved to the “F” side & we get: “F” = manoninertial. where “F” = F + terms from coord transformation “Fictitious Forces” !