Sections | |||
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Inertial Reference Frames | 14 mins | 0 completed | Learn |

Special Vs. Galilean Relativity | 17 mins | 0 completed | Learn |

Consequences of Relativity | 53 mins | 0 completed | Learn |

Lorentz Transformations | 46 mins | 0 completed | Learn |

Concept #1: Lorentz Transformations of Position

**Transcript**

Hey guys, in this video we're going to start talking about Lorentz transformations when we first introduced relativity we talked about the addition of velocities or the addition of speeds which was the way in Galilean relativity to take a measurement in one reference frame and believe it to an inertial reference frame now in special relativity we can't do something as simple as addition to speeds as addition of speeds but we can still move measurements from one inertial frame to another using more complicated Lorentz transformations. Alright let's get to it alright so as I said in Galilean relativity the process of moving one measurement into another relatively simple the position in one reference frame is easily relatable to the position in another reference frame based on how quickly that reference frame is moving relative to the first so in X if you make a measurement sorry in S if you make a measurement of X. and S. prime is moving at a velocity or speed U relative to S then X prime is just going to be X plus or minus U times T it depends on the relative direction OK but it's basically just the initial position in S. plus how far the frame the moves in your time measurement T`.

It's much much more complicated in special relativity because this thing right here time sort of messes it up for everybody because you can't just say that the time measured in one frame is equal to the time measured in another frame you have to consider time dilation and it makes the complication calculation a lot more complicated and that's what I said here the time duration is not the same between two inertial frames. So Lorentz transformations are what are going to allow us to relate these measurements in different frames taking into account phenomena like time dilation and length contraction. Galilean relative Galilean transformations like this one right here will always work if the speed is low enough and that's something that we're going to see in the example below but they are not going to work as you get closer and closer and closer to the speed of light and once again just like we've been saying the entire time we've been talking about relativity this only works if you are considering inertial frames it does not work if you're talking about accelerated reference frames like non inertial frames, let me minimize myself here so to use Lorentz transformations we need to have the two things first. First of all we need two inertial frames here we're saying S is the rest frame just like we've been using. In all of our other videos the convention we've been using so far and S prime is the moving frame right and they are inertial so the velocity U is constant now what we're choosing is that their origin's are aligned at the initial time so this is at some time T.

The coordinate systems are going to look at this but originally here's Y prime here's Z prime here's X prime this is at T prime equals T equals 0 we are choosing to align their origins. Now what we need to do is we need to choose a particular orientation of our frames such that the boost which is what we call this relative velocity I've probably used that word before but the boost is along some axis. Commonly we are all we are going to choose the X direction if you look at the images here the axis along which the boost lies is the X axis there is not going to be any relative velocity for the Y axis or for the Z axis there are not going to be any relative velocity so like we saw when talking about length contraction there is not going to be any length contraction along those axis length contraction will only apply along the X direction because that's the direction of the boost so now we can actually write down our transformations our coordinate transformations, now first of all because there's no boost in the Y or the Z direction that means that there's no length contraction along those directions so there's no change in those positions that Y prime equals Y Z prime equals Z right if this is S and this is the S prime and this is X and this is X prime where this is some speed U you can see that if I were to measure something here.

That is going to have a completely different measurement from the origin in S prime because as time goes on X prime is moving away sorry the origin of X prime is moving away and there's also length contraction due to relativistic effects but if I were to look at U versus U prime and let's say that this event occurred here where this was Unot we can see that because there's no relative velocity between the Y directions that Ynot prime is going to be the same length as Ynot there's no relative velocity there's no length contraction there's nothing and the same sort of thought applies to the Z direction as well so only in the X direction are we going to get length contraction and for time we're also going to have time dilation so we need to take that into account as well the equation is gamma where this is the same Lorentz factor that we've used times time minus U, X. over C squared where U remember is the boost that is the boost velocity and then X prime is going to be gamma again right this is special pretty much every equation is going to have a gamma in it a Lorents factor and this is going to be X minus U, T so interestingly enough if we look at our X prime equation it looks exactly like what we had for Galilean relativity just with a factor of gamma in front of it because that gamma when you multiply it out you're going to get length contraction from the first term and then you're going to get time dilation with the second term and that's basically what happens the time equation is little bit more complicated.

The first term and the time equation is clearly just time dilation but the second term there's no easy way to just look at it like oh yeah that second term is obvious so I'm not going to talk about that just accept the second term as fact it has to be there and there's a mathematical reason why there's just not a good sort of intuitive reason why so let's take a look at this problem we have two frames S and S prime which have their origins aligned at this time, that's one of those requirements and we have a boost of 580 Kilometers per second in the X direction remember that's the second requirement and by convention these are typically going to have boosts in the X direction so we can always use these above Lorentz transformations what is the position and time in S prime. So first of all the position in S prime is going to be gamma the position in S U times the time by the way this time this position this time this position those are always measured in S just look at them they don't have primes so they are not measurements in S prime, so before we can do anything we do need to know gamma and remember gamma is 1 over the square root of 1 minus U squared over C squared so this is going to be 1 over the square root of 1 minus 580 kilometers per second is the same as 580 times 10 to the 3 meters per second speed of light 3 times 10 to the 8 meters per second squared and this whole thing's going to be 1.00000187 so if you look basically one that's going to be very very very little effects due to relativity because your not really going that fast. So 1.00000187.

X is going to be in units of meters just because our speed is in kilometers per second so it's easiest to put everything in meters you could represent this as 10 centimeters but then this would be 580 times 10 to the 3 to get it to meters times 10 to the 2 to get it to centimeters per second and then everything gets all messed up like it would just be easier if we stuck with meters for now so 0.1 meters minus the speed of 580 times 10 to the 3 meters per second times 5 seconds is how long this occurs obviously this number is going to be way way way bigger than 10 centimeters so this is going to be -2900005 meters , notice that U Times T let me just write this in red just because it's easier notice that you times T if you plug it this in for U this in for T is actually just 2900005

You can see that the length contraction is 5 meters and we know that the length contraction should be very very small because we're not actually going that fast and now time is going to be gamma time measured in S frame the rest frame, minus U X over C squared gamma once again 1.00000187 time measured is 5 seconds minus the speed was 580 kilometers per second times right so times 10 to 3 meters per second the position at 0.1 meters you need to get this in the same units. So you want it all to be meters and then this is going to be 3 times 10 to the 8 meters per second squared this number right here is going to be really really really really small it's 580 times 10 to the 2 because that 0.1 point one drops X1 by 1 right 580 times 10 to the 2 divided by 9 times 10 to the 16 so that number is basically 0.

So what you're going to get is you're going to gamma times the time interval and that's just length that's just time dilation right so this ends up being a 5.00000935. So you can see that the time dilation is all the way out here at 10 to the -6 seconds 9 times 10 to the -6 seconds is the time dilation very very small as we expected because gamma is very very near 1 because we're going pretty slowly so this is the basics of the Lorentz transformation what you are essential doing is time dilation and length contraction just at once and it's often times much much easier to relate these two because you don't need to worry about what's the proper frame what's the non proper frame you know S you know S prime you know the boost of S prime you just plug in the numbers into equations and you're done. Alright guys that wraps up this discussion on Lorentz transformations we're going to follow this up with some practice problems. I'll see you guys in another video.Thanks so much for watching.

Practice: In a lab frame, S, an object crosses a distance of 15 m in 10 s. In an initially aligned frame S', moving at 1000 km/s in the x-direction relative to S, how far a distance does the object have to travel, and in what time does it travel the distance?

Concept #2: Lorentz Transformations of Velocity

**Transcript**

Hey guys, so far we've talked about Lorentz transformations for individual positions so if an object is at position X at time T in S where will it be in S prime what position what time now we want to talk about how we use Lorentz transformations for velocity and these are much much more commonly used then just pure Lorentz transformations for position, so this is what you're much more likely to be tested on when you actually see questions about special relativity alright let's get to it so remember for Galilean relativity it's easy to find a speed in some moving frame relative to some rest frame or vice versa you just have simple addition of velocities if a car is moving at 10 miles an hour and throw a ball out the window at 20 miles an hour a guy on the street sees the ball moving at 30 miles an hour addition of velocities is very simple it's much much more complicated when talking about a relativistic problem because of things like length contraction and time dilation. Remember the two things that we need for a Lorentz transformation we need to have two frames with their axes and origins aligned at time T equals 0 this actually isn't that important for velocity this was incredibly important for position but as long as if this is true right Lorentz transformations will work for position and velocity.

The second one is definitely true we need to choose a direction for the boost and typically we're always going to choose that to be in the X direction so these equations are going to be written assuming that there is a boost in the X direction alright now, what is the X component of the velocity going to be in S prime in the moving frame well it's going to be the X component in the rest frame S minus the speed of the boost divided by 1 minus the X component of the velocity times U divided by C squared. This is exactly the Lorentz transformation of velocity along the X direction what about the Y in the Z. direction remember that positions don't transform unless the boost is going in those direction there's no length contraction I could just as easily have written this as delta Y, delta Z, so now these are specifically distances specifically lengths because there's no length contraction you might want to say that there's not going to be any change in the velocity but bear in mind that a velocity is not just the distance right velocity is going to be that distance over time and while the distances are going to remain the same the time intervals will not remain the same right there is definitely going to be time dilation and because of time dilation you are going to change the denominator it's very easy to see that you should have some sort of gamma term in the denominator in order to transform these it's not going to be as simple as that but you could see that it should have something to do with that it's. Now the velocity in the Y direction in the prime frame is going to be the velocity in the Y direction in the unprimed frame divided by gamma times this same term. It's very very important to recognize that that velocity term in the denominator is in the direction of the boost it's V.X. it's not V.Y and we get basically the same equation for Z we're going to get again a gamma in the denominator 1 minus V X U over C squared and once again we have that X component of velocity in the denominator there we do not have the Z component those denominator terms are incredibly important because of the time dilation.

Let's do a problem right here let me minimise myself to get out of the way. So a space ship is passing the Earth at 0.5 C. From an observer on the ship right so this is in S prime a missile is fired forword at 0.1 C. According to an observer on Earth how fast is the missile moving this is going to be our VX that we want to find so we are going from X Sorry hit my knee against the table we are going from VX prime to VX. So this is from S prime to S so this is sort of like the opposite transformation of what we've been doing but don't worry doing the inverse transformation is really really easy and it uses the same equations they're just going to be changed in one respect. So if this is S and this is S prime moving forward as speed U this is absolutely equivalent to S moving backwards at a speed U and S prime being stationary. These two are absolutely equivalent. So we can say that the transformation from S prime to S is going to look the exact same as the transformation from S to S prime. There's just going to be one major difference, the one major difference is we're normally dealing with U, now we're dealing with negative U. The direction is opposite because the direction's opposite we're gonna take off a negative sign so this was negative now it's going to be positive. This term was negative but the U is going to become negative so this is going to be positive and this is exactly the Lorentz transformation we're going to use to go from S prime to S. Now we were told the missile which is our thing that's moving is moving at 0.1 C and the frame, the frame is going to be the ship in this case, is moving at 0.5 C, this guy is U. Divided by 1 plus 0.2 C, 0.5 C over C squared. This is why it's really nice to use everything in terms of C because those are going to cancel and if you plug this into your calculator you will get 0.57 C. Now this answer makes perfect sense. If we were just going to add the velocities we would have gotten not 0.57 C but 0.6 C.

The problem with just adding the velocity is that as they get higher and higher and higher you will eventually cross the speed of light. If the ship fired the missile forward at 0.6 the speed of light relative to its captain it's going at 0.6, the ship's going at 0.5, if you just add those two numbers that's 1.1 times the speed of light that violates special relativity that cannot be it. So it has to be the addition but slightly less and how slightly less it is depends on how fast you're going right. The faster and faster and faster you're going the more and more and more less you're actually going to be then just the addition of what you would expect it to be. Another Lorentz transformation problem, a spaceship passes the Earth at 0.7 times the speed of light from the observer on the ship a missile's fired laterally. So here is the observer on Earth, here is the ship moving at a speed of 0.7 C when it fires a little missile laterally at 0.2 C relative to the ship. This 0.2 C is V prime but now this isn't actually VX prime, assuming that the booster's in the X direction I'm going to assume that this is the X direction and this I'll call the Y direction actually so this is VY prime. This guy is still U in the X direction and what we're looking for is how fast is the missile, we're looking for VY prime in the S frame. So remember what's really important well first of all let's do the same thing to our Lorentz transformation equation for the Y component as we just did for the X component. This was originally VX times U over C squared but remember when going backwards from S prime to S we're going to take U make it -U so instead of this being a negative here it's now a positive. Right away the thing that you have to remember, sorry, this is also times gamma. 1 plus VX prime U over C squared. Right off the bat the thing that you have to remember that you cannot say is that the velocity in a perpendicular direction to the boost is going to be the same because it's not because of time dilation it will be different. Let's find what gamma pm is really quickly. Let me minimise myself for this. Gamma is going to be one over the square root of one minus U squared over C squared. The speed of the frame we're told is 0.7 C so it's going to be 0.7 squared and, sorry, I wrote the wrong number in my calculator in my calculator I used 0.2 times the speed of light it's not 0.2 because the frame is not moving at 0.2 the speed of light the frame is at 0.7 the speed of light so this is actually 1.4. We are not going to use 0.2 times the speed of light because that is the speed of the missile not the speed of the frame. Now what about this term right here? Well that is definitely zero right the missile has no component in the X direction. Imagine if it was fired at an angle. Well then it would have a component in the X direction and a component in the Y direction but that's not the case here it's only fired in the Y direction so this term is absolutely zero and all we're going to get is the speed in the Y direction which we're told was 0.2 times the speed of light divided by gamma which we just calculated to be 1.4 and that is going to be 0.14 times the speed of light. So the speed that you measured is definitely reduced because the time interval that you measure in the non-proper frame which is on Earth the time that you measure is going to be longer than the time that you can measure in the proper frame on the ship and because that distance in the Y direction is the same but you measure a longer time you have a lowered velocity. So velocities absolutely do change in directions perpendicular to the boost but it's actually a pretty easy change. Okay guys, thanks so much for watching and I'll see you guys in another video hopefully shortly.

Practice: In a particle accelerator, a neutron is traveling at a speed of 0.7 c, as measured by you in a laboratory. This neutron decays (becoming a proton), ejecting an electron. If you measure the electron’s speed to be 0.5 c, traveling in the same direction as the neutron, what was the relative speed between the electron and neutron when the neutron decayed? Was the electron ejected forward or backwards relative to the neutron’s motion, as “seen” by the neutron?

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A spacecraft passes an Earth-bound osberver at a speed of 0.5 c. If the ship then fires a missile at 0.4 c relative to the captain of the ship, what speed does the Earth-bound observer measure the missile at?

A distant galaxy is moving away from the Milky Way at a speed of 0.8 c. If the galaxy ejects some material at a velocity of 0.9 c towards the Earth, measured by astronomers, how fast is the material moving relative to the distant galaxy?

In a rest frame, a mass m1 moves with a speed of 0.2c towards a mass m2, at rest. In the center-of-mass frame, both mass move towards each other with the same speed. What must the speed of the center-of-mass frame be, relative to the rest frame, in order to achieve this?

A high-speed train is traveling at a constant velocity along a straight-line track from Capital City to Shelbyville. According to an observer at rest on the ground, the clocks at the railroad stations in Capital City and Shelbyville both strike noon at the same time. According to a passenger on the train, when the Capital City clock strikes noon, what time is it in Shelbyville?A. noonB. before noonC. acer noonD. The answer depends on the speed of the train.E. The answer depends on the speed of the train and the distance between Capital City and Shelbyville.

Santiago stands on the ground as Miriam flies directly toward him in her spaceship at 0.5c. She fires a small rocket directly toward Santiago that flies at a speed of 0.8c relative to her spaceship. According to Santiago, the speed of the rocket isA. 1.3cB. faster than c but slower than 1.3cC. cD. faster than 0.8c but slower than cE. 0.8c

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