Lorentz Transformations - Online Tutor, Practice Problems & Exam Prep

Lorentz transformations are essential in special relativity for relating measurements between inertial frames, accounting for time dilation and length contraction. Unlike Galilean transformations, which simply add velocities, Lorentz transformations involve more complex equations. The x-component of velocity transforms as $v^{\text{'}}=(v-u)/(1-vu/c^2)$. Understanding these transformations is crucial for analyzing relativistic effects in high-speed scenarios.

Lorentz Transformations of Position

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 move it to an inertial reference frame. Now in special relativity, we can't do something as simple 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.

So like I said, in Galilean relativity, the process of moving one measurement into another is 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. For example, if you make a measurement of x in s, and s' is moving at a velocity or speed u relative to s, then x' is going to be x plus or minus u times t. It depends on the relative direction. But it's basically just the initial position in s plus how far the frame moves in your time measurement t. Now, it's much more complicated in special relativity because this thing, 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. It makes the calculations a lot more complicated.

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 transformations like this one will always work if the speed is low enough, but they are not going to work as you get closer and closer to the speed of light. 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.

To use Lorentz transformations, we need two things. 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, and s' is the moving frame. They are inertial so the velocity, u, is constant. We are choosing that their origins are aligned at the initial time t0, so coordinate systems are going to look like this at t' 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, is along some axis. Commonly, 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 axes or the z axes, so like we saw when talking about length contraction, there's not going to be any length contraction along those axes. 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. Because there's no boost in the y or z direction, that means there's no length contraction along those directions, so y' equals y, z' equals z. The equations for time dilation and length contraction become: t ' = γ ( t - u x c 2 ) and x ' = γ ( x - u t )

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.

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?

• A. 0.99998889 s
• B. 9.9998889 s
• C. 1.0000056 s
• D. 10.000056 s

Lorentz Transformations of Velocity

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 than just pure Lorentz transformations for position. Okay? 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 you 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. Okay? Addition of velocity is very simple. It's much much more complicated when talking about a relativistic problem because of things like length contraction and time dilation. Okay? Remember the two things that we need for a Lorentz transformation. We need to have 2 frames, okay, with their axes and origins aligned at time t equals 0. This actually isn't that important for velocity. Okay? This was incredibly important for position but as long as if this is true, right, Lorentz transformations will work for position and velocity. Okay? The second one is definitely true. We need to choose a direction for the boost, and typically you're always going to choose that to be in the x direction. Okay? So these equations are gonna 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 s prime in the moving frame? Okay? 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. Okay? This is exactly the Lorentz transformation of velocity along the x direction. What about the y and the z direction? Remember that positions don't transform unless the boost is going in those directions. 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 gonna be any change in the velocity. But bear in mind that a velocity is not just a distance. Right? Velocity is going to be that distance over time. Right? 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. Okay? Now the velocity in the y direction in the primed frame is going to be the velocity in the y direction in the unprimed frame divided by gamma times this same term. Okay? It's very very important to recognize that that velocity term in the denominator is in the direction of the boost. It's vx it's not vy. Okay? And we get basically the same equation for z. We're gonna get a gamma in the denominator, 1 minus vxu over c squared. And once again, we have that x component of velocity in the denominator there. We do not have the z components. Okay? Those denominator terms are incredibly important because of time dilation. Okay? Let's do a problem right here. Let me minimize myself to get out of the way. Okay. So a spaceship is passing the earth at point 5 c. From an observer on the ship, right, so this is in s prime, a missile is fired forward at 0.1c. According to an observer on earth, how fast is the missile moving? This is going to be our v x that we wanna find. So we are going from, 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 at speed u, this is absolutely equivalent to s moving backwards at a speed u, and s prime being stationary. Right? These 2 absolutely equivalent. So we can say that the transformation from s prime to s is going to look the exact same oops. That's not c, that's a u, as the transformation from s to s prime. I don't want it to be negative. There's just going to be one major difference. The one major difference is we're no longer dealing with u, now we're dealing with negative u. Right? The direction is opposite. Because the direction is opposite we're gonna pick up 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 are going to use to go from s prime to s. Okay? Now, we were told that the missile, which is our thing that's moving, is moving at 0.1c. And the frame, right, the frame is going to be the ship in this case, is moving at 0 point 5 c. Right? This guy is you divided by 1+ 0.1c.5c overc squared. This is why it's really nice to use everything in terms of c because those are gonna cancel. And if you plug this into your calculator, you will get 0 point 57 c. Okay? Now this answer makes perfect sense. If we were just going to add the velocities, we would have gotten not 0.57c but 0.6c. 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. Right? If the ship fired the missile forward at 0.6 the speed of light relative to its captain, 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 than, just the addition what you would expect it to be. Okay? Another Lorentz transformation problem. A spaceship passes the earth that points 7 times the speed of light. From observer on the ship, a missile is fired laterally. K. So here is the observer on earth. Here's the ship moving at a speed of 0.7 c when it fires a little missile laterally at point 2c, relative to the ship. Okay this point 2c is v prime but now this isn't actually the x prime, right? Assuming that the boost is in the x direction, I'm going to assume that this is the x direction and that this I'll call the y direction actually. So this is v y prime. This guy is still u, right, in the x direction, and what we are looking for is how fast is the missile. We're looking for v y non prime, right, 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 words from s prime to s, we're gonna take u and make it negative u. So instead of this being a negative here, it's now a positive. Right away the thing that you have to remember oh sorry, this is also times gamma, 1+vxprimeuovercsquared. 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. Okay? Let's find what gamma is really quickly. Let me minimize myself for this. Gamma is gonna be 1 over the square root of 1 minus u squared over c squared. The speed of the frame, right, we were told was point 7 c, so it's gonna be 0.7 squared. Okay? 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 moving at 0.7 the speed of light. So this is actually 1.4. Okay. 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. Okay? Now what about this term right here? Well, that is definitely 0. 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 0, and all we're going to get is the speed in the y direction, right, 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 0.14 times the speed of light. Okay? 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 gonna be longer than the time that you would 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. Okay? So velocities acts absolutely do change in directions perpendicular to the boost, but it's actually a pretty easy change. Alright? Okay, guys. Thanks so much for watching, and I'll see you guys in another video hopefully shortly.

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?

• A. -0.55c; Forward
• B. -0.31c; Forward
• C. -0.31c; Backward
• D. -0.55c; Backward

Here’s what students ask on this topic:

What are Lorentz transformations and why are they important in special relativity?

Lorentz transformations are mathematical equations that relate the space and time coordinates of two inertial frames moving at a constant velocity relative to each other. They are crucial in special relativity because they account for the effects of time dilation and length contraction, which occur at high velocities close to the speed of light. Unlike Galilean transformations, which simply add velocities, Lorentz transformations ensure that the speed of light remains constant in all inertial frames, preserving the principles of special relativity. The transformations are given by:

$\frac{x\text{'}}{=\gamma (x-ut)}$

$\frac{t\text{'}}{=\gamma (t-\frac{ux}{c²})}$

where γ is the Lorentz factor, defined as:

$\gamma =\frac{1}{\sqrt{1-\frac{u²}{c²}}}$

How do Lorentz transformations differ from Galilean transformations?

Galilean transformations are used in classical mechanics to relate the coordinates of two inertial frames moving at a constant velocity relative to each other. They assume that time is absolute and the same for all observers, and they simply add velocities. The Galilean transformation for position is:

$x\text{'}=x-ut$

In contrast, Lorentz transformations are used in special relativity and take into account the constancy of the speed of light and the effects of time dilation and length contraction. They modify both space and time coordinates and are given by:

$\frac{x\text{'}}{=\gamma (x-ut)}$

$\frac{t\text{'}}{=\gamma (t-\frac{ux}{c²})}$

where γ is the Lorentz factor:

$\gamma =\frac{1}{\sqrt{1-\frac{u²}{c²}}}$

What is the Lorentz factor and how is it used in Lorentz transformations?

The Lorentz factor, denoted as γ (gamma), is a crucial component in Lorentz transformations. It accounts for the relativistic effects of time dilation and length contraction. The Lorentz factor is defined as:

$\gamma =\frac{1}{\sqrt{1-\frac{u²}{c²}}}$

where u is the relative velocity between the two inertial frames, and c is the speed of light. The Lorentz factor increases as the relative velocity approaches the speed of light, leading to more significant relativistic effects. In Lorentz transformations, γ is used to modify both space and time coordinates:

$\frac{x\text{'}}{=\gamma (x-ut)}$

$\frac{t\text{'}}{=\gamma (t-\frac{ux}{c²})}$

This ensures that the speed of light remains constant in all inertial frames, preserving the principles of special relativity.

How do Lorentz transformations affect time and length measurements?

Lorentz transformations significantly impact time and length measurements due to the effects of time dilation and length contraction. Time dilation means that a clock moving relative to an observer will tick slower compared to a clock at rest with respect to that observer. The time transformation is given by:

$\frac{t\text{'}}{=\gamma (t-\frac{ux}{c²})}$

Length contraction means that an object moving relative to an observer will appear shorter along the direction of motion. The length transformation is given by:

$\frac{x\text{'}}{=\gamma (x-ut)}$

where γ is the Lorentz factor:

$\gamma =\frac{1}{\sqrt{1-\frac{u²}{c²}}}$

These transformations ensure that the speed of light remains constant in all inertial frames, preserving the principles of special relativity.

How do you apply Lorentz transformations to velocity?

Applying Lorentz transformations to velocity involves more complex equations than for position. For the x-component of velocity, the transformation is:

$\frac{v\text{'}x}{=\frac{vx-u}{1-\frac{vxu}{c²}}}$

For the y and z components, the transformations are:

$\frac{v\text{'}y}{\frac{vy}{\gamma (1-\frac{vxu}{c²})}}$

$\frac{v\text{'}z}{\frac{vz}{\gamma (1-\frac{vxu}{c²})}}$

where γ is the Lorentz factor:

$\gamma =\frac{1}{\sqrt{1-\frac{u²}{c²}}}$

These transformations account for the relativistic effects of time dilation and ensure that the speed of light remains constant in all inertial frames.