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### Course: Physics archive > Unit 16

Lesson 4: Einstein velocity addition# Applying Einstein velocity addition

Applying Einstein velocity addition.

## Want to join the conversation?

- At4:15how comes 0.5 times 0.7 = 0.3 instead of 0.35?

Was Sal keeping one significant figure in this calculation?

But if this is the case I don't see why the results of other calculations have had 2 significant figures.(19 votes)- Sal made a mistake. You are correct, it should be 0.35, and so the final answer should be 0.89c, not 0.92c.(17 votes)

- At6:13, Sal mentioned a video game leveraging the effects of special relativity. In 2012 the MIT Game Lab developed just that, called "A Slower Speed of Light": http://gamelab.mit.edu/games/a-slower-speed-of-light/

Has Sal seen this?(25 votes)- The concept of the game is awesome, thanks buddy(4 votes)

- At4:14Sal writes 1 - 0.30, shouldn't it be 1 + 0.30 since velocity v is negative?(1 vote)
- A few seconds later, he changes it to a plus (and so then agrees with you). However, I think the 0.3 should be a 0.35.(34 votes)

- Two bodies moving in opposite direction with velocity v.What is the magnitude of their relative velocity?(0 votes)
- You should use Einstein velocity addition. The relative velocity is: (v+v)/[1+v^2/c^2](3 votes)

- Explored a bit with the numbers using this program I wrote for the addition formula(see link.) In it I ran into the case V = U = 1, and had to write an exception for this case. How would one interpret this situation? In the program, I just left U' undefined, but I'm not certain if this is accurate.

https://www.khanacademy.org/computer-programming/einstein-velocity-addition-exploration/3054910345(1 vote)- You've run into the very reason, mathematically speaking, why things can't go faster than (or even as fast as) light. In the addition formula, you'd have to divide by zero when u=v=1, and in gamma, you'd divide by 0 at exactly the speed of light, or move in to complex space faster than the speed of light(1 vote)

- So I'm a little confused. Nothing can move through spacetime faster than the speed of light, but were talking about relative positions of two thing moving sub light speeds. Nothing says two things can't be approaching each other faster than light, we know galaxies on the edge of the universe are moving apart faster than light. So if this number we calculated is the perceived speed of the second ship as seen from the first ship, but they actually are moving together faster than light, would time dilation and space contraction be what makes appear as they are approaching slower than light? And would it be both, how are they linked?(0 votes)
- In special relativity every inertial observer is at rest in their own reference frame. An observer will never measure a velocity of an object to be moving faster than light. By saying that you have two points moving away from each other faster than the speed of light you are introducing a third frame of reference that is observing each point moving away with a greater than 1/2 the speed of light.

In special relativity there is no one correct set of observations ant the others are wrong, each of the inertial observers measurement is correct from their frame of reference. The need to translate from one frame of reference to another is nothing new, we did this for Newtonian mechanics, the only difference here is that there is a maximum velocity so the translation is a little more complex.(3 votes)

- I was looking forward to using REAL numbers in the scenario set up in the prior video. Now Sal is muddying the waters by changing the direction of the third spaceship, which is now approaching both Sal and Sally. I'm still unclear how to calculate the velocities of vehicle #3 w/r/t Sally when both are flying away from Sal.(1 vote)
- When both ships are flying away from Sal, u=0,5c instead of u=-0,5c. You put the numbers in the equation, and we get (0,5c-0,7c)/(1-(0,5c*0,7c)/c^2) = -0,2c/(1-0,35) = -0,308c. So in Sally's frame of reference, the other ship would be approaching at 0,308 times the speed of light, which is logical because Sally goes faster than the other ship.(1 vote)

- why is Sal's position denoted (x,y) not (x, ct)?(1 vote)
- How to prove that nothing's speed can exceed the speed of light?(1 vote)
- Try plugging v=2c into the transformation equations and see what happens.(0 votes)

## Video transcript

- [Voiceover] Now let's apply the formula we came up with the last video sometimes known as the Einstein
velocity addition formula. And we'll see that it's
a pretty neat thing. So let's say, so this is once
again me floating in space, my frame of reference is just
the s frame of reference, Let's say my friend over here, she's flying away from me, her velocity, let's give it some numbers. Let's say she's flying away from me with a velocity of, let's say, 0.7c, so seven tenths
of the speed of light or 70% of the speed of light. And let's say this third
character right over here, let's say they're flying towards me, so let me change the
direction, the velocity vector. So let me erase that. Gee, that doesn't look like a pure black but it looks fairly
erased, so there you go. And let me give the velocity
vector in the other direction. So the velocity vector, this character is flying towards me at let's say, so u is
going to be equal to, so it's flying towards me x is decreasing with time, let's say it's negative, - 0.5c, so flying towards me at
half the speed of light. And once again, both of these velocities are given in my frame of reference. Now if we were in a Newtonian world or if we used the
Galilean transformations, and this is released as
something on your highway. If I'm on a highway and if
I'm going in one direction at 70 miles an hour, and
someone is coming towards me at 50 miles an hour, it'll look to me like they are coming towards
me at 120 miles per hour. We would essentially
add these two together. And so, if we were in a Newtonian world, this person is flying at 0.7c and this person is
flying this way at 0.5c, you would, in a Newtonian world, say okay from this point of view it will look like this blue character is approaching at 0.7c plus 0.5c or at 1.2 times the speed of light which we know violates
the laws of the universe. And that's where this
formula is really handy. Actually let me even
make the spaceship point in this direction just
so we don't get confused. So the spaceship is going in that direction just like that. And once again, the
magnitude of the velocity coming towards me is
half the speed of light. I put the negative there to show that it's coming towards me, that x is decreasing over time. Well, lucky for us in the
last video, we came up with this Einstein velocity addition
formula, so let's apply it. From this friend's frame of reference, the s-prime frame of reference, the velocity of this character which is change in
x-prime over change in t is going to be this. So what do we have here? So u is the blue velocity
in my frame of reference, So that is -0.5c. v is my friend's frame of reference, the frame of reference that we're trying to get the velocity in. So that's 0.7c, so let me do it in that color. So that is 0.7c. And then, on top of uv over c-squared, so once again, u is - 0.5c. I know I wrote it really small. And then, v is 0.7c. So what is this going to be equal to? Well, the numerator -0.5c minus 0.7c, that's going to be negative. Our numerator, right over here is going to be - 1.2c. And this is the velocity
that you would expect if we were dealing in a Newtonian world. This person says, "Hey, this person looks "like they're coming towards me "at 1.2 times the speed of light." That's what our Galilean
transformations would give us. But luckily, we have all
this business at the bottom that keeps us from
violating the absoluteness of the velocity of light, this notion that nothing can travel faster than the speed of light. Because in our denominator, we're going to get 1 minus 0.5c times 0.7c. Let's see, 0.5 times 0.7 is going to be 0.3. So 0.3, I could say 0.30 if I liked, but I'll just write 0.3c squared. It'll be a negative. I have a negative times a positive. So I could put the negative there but if I'm subtracting a negative, that's just going to be a positive. And I'm just going to divide by c-squared. And so, that cancels out. And so notice, we're going to be dividing - 1.2c by something that's
slightly larger than 1.2. So this is going to be - 1.2c over 1.3. 1.3 And so, lucky for us, this
is going to be less than c, or this absolute value is
going to be less than c. So it's going to be, let's
get a calculator out. So, if we have 1.2 divided by 1.3, this is equal to approximately 0.92. So this is going to be approximately - 0.92c which is cool. It kind of goes with our, let me write it over here. So this delta x prime over delta t prime is -0.92c. So it makes sense that from our
friend's frame of reference, this ship will look like
it's approaching her faster than from my frame of reference. It looks like it's approaching
me in my frame of reference, it looks like it's coming to
me at half the speed of light. In her frame of reference, it looks like it's coming at
0.92 times the speed of light. The negative is just
specifying the direction. But we didn't violate
the fundamental postulate of special relativity,
that nothing can go faster than the speed of light. The speed of light is absolute. So this is pretty cool. I've often even thought about
making a video game somehow where you leverage this, where things are flying in
different relative velocities but special relativity applies.