Transcript for:
Galilean Transformations

so let's continue our discussion on Galilean transformations so once again a Galilean transformation is basically a mathematical way of transforming the quantities that represent objects and events between two different inertial reference frames now in the previous lecture we were able to show that the equations that describe the transformation of the position of an object between two different reference frames in which one frame is stationary and the second frame is moving with a velocity V in the positive direction along the x axis is given by these three equations so basically if we know the position the exact position of our object within the second reference frame that is moving with a velocity V we can use that to actually find the equivalent position of our object within the first frame that is stationary so let's suppose that frame two has coordinates X prime Y Prime and Z prime and frame one has coordinates x y&z so if we know X prime Y Prime and Z prime we can use these three equations to calculate what XY and Z is so X is given by X prime plus V times T where V is simply the velocity of the second reference frame and T is the time that has elapsed so remember we assume that initially before the movement began the two frames essentially were found at the same exact location the origin of those two frames exactly coincided and then when time began progressing frame number two began moving in the positive direction along the x axis so there was no movement along the Z and along the y axis so velocity just like position can be transfer form between two different inertial reference frames so now let's find the Galilean velocity transformation equations the transformation equations that will give us the transformation of our velocity between two different reference frames so once again suppose that two frames frame F and frame F prime begin at the same position at T equals zero seconds and as time progresses reference frame F prime moved to the right along the x axis with the velocity V while frame F is stationary so to see exactly what's taken place let's look at the following two diagrams so in this diagram we have time equals zero seconds and at that point these two frames these two coordinate planes exactly coincide so the origin of frame 1 lies on the origin of frame 2 so that means X 1 and y or X 1 and X 2 will lie on top of one another y 1 and y 2 will lie on top of one another and Z 1 and Z 2 will also lie on one another now after some time passes let's suppose at time T 2 what happens is the frame given by F is stationary it stays where it initially began while frame F Prime the second frame essentially is moving to the right along the x axis with a constant velocity given by V so this entire frame is moving to the right with a velocity V now suppose at the time of t 2 a particle that is given by point P so this is point P it's a particle and it has a velocity vector in frame given by F prime so the velocity vector thin the frame f prime is given by this equation so W prime is equal to W prime along the x axis w prime along the y axis and W prime along the z axis so the velocity vector of this particle within the second frame f prime is given by this equation now what exactly is the definition of velocity well instantaneous velocity is the rate of change of displacement with respect to time so that means we have the following three equations so W prime along the x axis is equal to the derivative of X Prime with respect to time now this is equal to so W Prime in the y axis is equal to dy prime DT and finally W prime Z is equal to DZ prime divided by DT these will become important in just a moment so now we want to actually use the Galilean transformation equations for position that we saw in the previous diagram so we want to use these three equations so remember we want to determine what the velocity vector is for this particle in reference frame one given by F so that means we want to find what these three quantities are so W XW Y and W Z so the velocity of the particle within frame F with respect to the x axis is equal to DX divided by DT now if we go back to these equations what exactly is X with respect to X prime well X is equal to X prime plus V multiplied by T so we replace X with X prime plus V times T the same thing can be done for these so W Y is equal to dy / DT by the definition of instantaneous velocity and y is equal to Y Prime and Z is equal to Z Prime so W Y is equal to this and W Z is equal to this now let's look at the following equation we can distribute this derivative and we get the following result DX prime DT + DV T divided by DT so notice that the T will cancel and we're simply left with V so W X the velocity of this particle within frame 1 given by F along the x axis is equal to DX prime DT plus V so now we have three different equations so these three different equations basically give us the velocity of our particle P within frame F so equation one equation two an equation three and these equations are known as the Galilean velocity transformation equations