6 - 32 - What Happens With Perpendicular Velocities- (12-09, Low-Def) (1)

8/10/2019 6 - 32 - What Happens With Perpendicular Velocities- (12-09, Low-Def) (1)

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In a previous video clip we looked atcombining velocities, really addingvelocities is what we were doing when wehad motion in the x direction.So remember we used the LorentzTransformation and you got an equationlike this and what I've done here isadded these x subscripts to the velocityis, use of r remember is the velocity inthe rocket frame, our example was Bob,here, going with some velocity, and Aliceis here, say, get her in in the picture,going with some velocity, v, with respectto Alice's frame of reference, and shotoff in an escape pod in his direction ofmotion, with velocity u sub r compared tohim.And we discovered, then, that thevelocity that Alice would observe.The escape pod to go follow this equationhere.And we did a few examples with that justto check out how it might work.And in particular we saw that even though

the two velocities here, the velocity ofBob with respect to Alice, of hisspaceship, and then plus the velocity ofthe escape pod.If they were both, if they both added upso they were greater than c, just byadding them together, this equation wouldbe such that the result would never begreater than c, in terms of what Aliceobserved, in terms of the velocity of thepod.What we want to do now is do anothercase, where, what happens if Bob shoots

it up maybe straight up, or justperpendicularly off to the side here?And, figure out, what would Alice see inthat situation?And interestingly enough, this will bringus back to the light clock that we aretalking about, recently.So, here's a situation.Bob shoots it up at some velocity u subr, and we'll use a subscript y to in,emphasize this is just in the ydirection, now.You can say it's the z direction but

we'll choose the y direction for, forthat direction.And of course definition of velocity,delta y sub r over delta t sub r.This is for Bob, now, he says, how fardoes it go in the y direction?And what's the elapsed time for how farit's going in that direction?That's going to be the definition ofvelocity.



And so now let's write down a couple ofequations, actually, for u sub ly.That's what we're after, here.What is, what does Alice observe thevelocity in the y direction to be?So U sub ly is going to be delta yl overdelta tl, just by definition of velocity.And this is why I, as a reminder I addedthe y and z components for our LorentzTransformation, remember they don'ttransform at all.They're the same.And so delta y sub l is going to be thesame as delta y sub r.So this is going to be delta y sub r overdelta t sub l.But that does change, of course, by ourequation we used before.Remember we could use deltas in here toindicate change of quantities.So we'll write that as delta y sub r onthe top, and then delta t sub l,remember, using the equation here, isgoing to be gamma times delta t sub rplus v over c squared, delta x of r.

Right?Again, these are the quantities thatthat, that Bob is measuring.And we, we're assuming everything isconstant velocity here.For those of you who've had some physics,you say, well what if it's accelerating?Well we're not dealing with that casehere.And therefore we can define velocity assimply change in, not change indirection, but the direction, elapseddirection if you want, the direction

covered divided by the elapsed time.So, delta y sub r over this quantity,here.Again we'd like to simplify it.We'd like to be able to get some of the usub r rocket frame y components in here.And so let's pull out this delta t sub r,sort of like we did before, and see whathappens here.So we have delta y sub r on the topstill.And, got a gamma here.Gamma's don't cancel this time.

And we're going to pull out a delta t subr from that expression.So we've got delta t sub r, times 1 plusv over c squared, delta x sub r.Over delta r sub r.Okay, so delta r sub r times 1 is thedelta t sub r.Delta t sub r plus this thing gives usthat because the delta t sub r's wouldcancel there, of course.



But then, this is a little strange herebecause, notice that we're, we're doingan equation for, for the y component butwe've got some x components in here aswell.Remember what this was?This is just the velocity of the pod andthe x direction in the rocket frame.Okay.So we could write, in fact just, youknow, to remind ourselves here.U sub r x equals delta x sub r over deltat sub r by definition.Have a good delta there.that's an even worse one, there we go.Okay, now, so, but in our example here,note what happens up for Bob, not whathappens but what is happening for Bobhere.There is no velocity of the pod in the xdirection.Remember Bob's stationary as far as he'sconcerned.He's shooting the pod straight up.

As far as he's concerned, it's straightoff to the side.And therefore, there is no velocity inthe x direction for this example.So, in this example, this equals zero.No velocity of the pod in the x directionas far as Bob is concerned.And so this is nice, cause this wholeturn then here.Is zero for us.So this thing here turns into zero forthis case.And look at what we're left with here.

Delta y sub r divided by delta t sub r,and also divided by gamma.Delta y sub r divided by delta t sub r.That's the velocity in the y directionfor Bob, in the rocket frame.U sub ry.So this simply becomes U sub R in the ydirection divided by gamma.Okay, now we're, we're going to come backto that in a minute, but let's, let's doone other thing here too just to makesure here, because, you know, for Bobit's going just straight up.

Right?But for Alice, Bob is moving that way.So, Alice may, and probably will andactually will, actually see, will observethat the pod is, is going to go up at anangle because you do have the velocity ofmotion here of the spaceship involved.And so let's see if we can get that inhere as well.So, we'll squeeze it in right here.



We just did that.And what we're going to do here make alittle room.So lets do u sub lx.In fact, there's the formula you want touse right here.Don't have to rewrite it.We already wrote it.And so in this case, u sub rx.Remember we just finished talking aboutit.I just erased it.That's zero.Right, for as far as Bob's concerned, novelocity in the x direction.So I get 0 plus v all over 1 plus, whereagain u sub rx is 0 to Bob in thisexample.Therefore this whole thing is 0 so we get1 plus 0.And in fact we did this example.When we were talking about thetransformation equations here for thevelocity.

This just gives us v.So there are two components as we talkabout the velocity here for Alice.One component Is just the velocity v inthe x direction because that's just themotion of the spaceship here, and theother component velocity in the ydirection for Alice is use of our y overgamma so it is, whatever velocity Bobsees there shooting up, then that'sgoing to be divided by, divided by gamma.Note one very interesting thing here.Gamma is in the denominator, so as the

velocity of light gets, not velocity oflight, the velocity of the pod shootingup here gets faster and faster close tothe speed of light.Remember what happens to gamma, gammagets close to infinity.And therefore this whole thing no matterhow big this velocity is.This is like something divided by a very,very, very big number.This goes to zero.So this approaches zero, as v approachesc or gamma.

Another way to say it, gamma approachesinfinity.So that's very interesting, because whatthat's saying is no matter how fast Bobshoots up his escape pod If his relativevelocity is close to the speed of light,Alice sees almost no upward velocity atall.This is Alice's upward velocity in the ydirection that she's observing.



If gamma is very large, if Bob's velocityis very close to the speed of light, thiswhole thing goes to zero, and all shesees is the sideways motion from the xdirection v.And the reason, I mentioned the lightclock a minute.Because remember, we did an example withthe light clock, where of course thelight clock.So here's Bob's clock, moving withvelocity v in that direction.And of course saw from Alice'sperspective the light beam goes up here.And the light beam goes up here and thendown there.So, the light clock has to go a littlefar over there.And we talked about as v gets closer andcloser to the speed of light, rememberthat, put the green in here again, thatthis light beam can never catch up tothat upward mirror because it's, if thisgets close to c, then this light beam has

to travel too far to the upper mirror.And by the time it essentially getsthere, the mirror would be farther away.And in fact, as it gets very, very closeto c, this triangle gets very, veryelongated.And in fact this light beam never gets upto that top mirror and it's frozen.The light clock as we, as Alice wouldobserve it.If Bob could go by at the speed of light,which he can't, but if he gets very, veryclose to it It's almost like the light

clock is frozen.It ticks very very slowly.It's just a time dilation effect.Gamma's very large and therefore Bob'sclock ticks very slow.If you could actually reach the speed oflight.It's frozen.And that's what this mathematics here istelling us another way that we argued forqualitatively earlier.I never put the green in here, but if, ifthis the velocity of light, if we're

going c here, and so, it can get up hereat c, but it never gets up to that topmirror, that's what this is telling us.It's saying, if we can go at the velocityof light, if Bob could shoot.Or if Bob could have a spaceship withvelocity v of light here, then this wouldbe infinite.And Alice's observed velocity for theescape pod, going up no matter what the



velocity of the escape pod was, would be,would be zero in the upper direction.And she'd just see the, the c factor.Going going across here.So tells us another way, it's good tohave these consistency checks here.Because you could say, just to make thepoint in other way here, this light clockis sort of like the escape pod.If the escape pod, instead of being anescape pod, were just a beam of light,going up in that direction.You know, make it squiggly for a beam oflight here.A beam of light going up in thatdirection c.That exactly what the light clock is.For Bob's perspective, it's going up anddown like that.And so that's like shooting light beamdirectly perpendicular to, to Bob.If he's traveling at velocity c.Then what this is saying in this case is,Alice never sees that light beam go up.

She just sees Bob go this way and thelight clock is frozen in that instance.So, an interesting confirmation on someof the other things we've done.This is one of those things sometimes ittakes a while just to sort of mull itover, think about it, work through it,convince yourself of that it actuallymakes sense.Put in some numbers, try things out, sortof get a feel for, for for what's goingon here.

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6 - 32 - What Happens With Perpendicular Velocities- (12-09, Low-Def) (1)