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### Course: Physics library>Unit 16

Lesson 4: Einstein velocity addition

# Einstein velocity addition formula derivation

Einstein velocity addition formula derivation.

## Want to join the conversation?

• Sal's frame of reference shows S:(x, y). Maybe it should be S:(x, ct)?
(13 votes)
• That would be more consistent, since so far the y axis has always been labelled ct.
(7 votes)
• why does u = delta x / delta t ??
(5 votes)
• u is a reference (designation of) to the 3rd ship which is what he was figuring. The third ship's frame of reference is u (space ship) = delta x/delta t.
(5 votes)
• Won't delta x over delta t equal to velcoity? Delta x over delta t is the formula of time and if we add direction we can get velocity?
(3 votes)
• Delta x over delta t is the velocity, u, of the third person. From Sal's picture, it is the velocity of the light blue object that looks like a boat.
(3 votes)
• In an earlier video x was made equal to ct, and x' was made equal to ct' because they described the "coordinates" of light events. Why now can x be used to describe the position of the third person, and not a light event? Do I have this all wrong?
(2 votes)
• It can be used to describe the position of a third person but x won't be equal to ct because this is only for the velocity of light, the distance x traveled by light will be equal to it's speed multiplied by the time it has been traveling (ct), but for another object it will be different depending on its speed.
(3 votes)
• a particle moves along the x' axis of frame S' with velocity 0.4c. Frame S' moves with velocity 0.6c with respect to frame S . What is the velocity of particle with respect to frame S ?
(2 votes)
• I'm just starting to understand this so I may be wrong (please correct me if that's the case) but since nobody has given an answer in eleven months I'm going to try.

If we wanted to get the velocity of the x particle with respect to frame S' we would use the above formula: (u - v)/(1 - u v / c^2) where u is the velocity of the particle with respect to S and v is the velocity of S' with respect to S.
Now we just need to solve u in the ecuation:
(u - 0.6c)/(1 - 0.6c u / c^2) = 0.4c
Which gives us u = 0.8064 c
(2 votes)
• can x'=gama (x-beta ct) be written as x'= gama (x-beta x) as well, since in previous video, x=ct?
(1 vote)
• x = ct only for the coordinates of objects going at the speed of light.
(2 votes)
• so the lorentz factor is the same thing as gamma?
(1 vote)
• at , wouldn’t the delta x over delta t cancel out with the other delta x over delta t?
(1 vote)
• Why did Sal divide the numerator and the denominator by 1/delta t ?
(1 vote)
• delta x/ delta t is a value we know (u).
We have a delta x in the numerator and denominator and the other delta t's get cancelled out.
(1 vote)
• Would I get the same answer if I just directly divide X' by t'?if I would get the same answer by dividing x' by t', the values of which we derived from Lorentz transformation then why do all this derivation?
(1 vote)

## Video transcript

- [Voiceover] Let's say this is me and I am floating in space. My coordinate system, my frame of reference, we've seen it before, we'll call it the S frame of reference. Any point in spacetime, we give it an x and y coordinates. And let's say that we have my friend. We've involved her in many other videos before. She is traveling with a relative velocity to me of v. So v right over there. Her frame of reference, we call it the S prime frame of reference, and you can denote any event in spacetime by the coordinates x prime and ct prime. Now you could also do it as t prime if you like, but we've been doing it as ct prime so that we have similar units. Now let's say that we have a third character now. This is going to get interesting. Let's say this third character is traveling with a velocity u in my frame of reference. So u is equal to change in x over change in time. If we know all of this information, let's see if we can come up, we can formulate a way if we know what the change in x and change in time is and we know what v is, if we can figure out what is this velocity going to be in the S prime frame of reference, in this purple friend's frame of reference. What we want to figure out is what is, let me do it in that purple color, we wanna figure out what is going to be the change in x prime over change in t prime. If we figure this out then we know what will this velocity look like in the S prime frame of reference. Well let's just go back to the Lorentz Transformations. Let's first think about what our... Let's first think about what change in x prime is going to be. Well our change in x prime is going to be the Lorentz factor times our change in x, change, this in white, times our change in x minus beta times change in ct. Let me close those parentheses. Then we wanna divide that by our change in time. We wanna divide it by our change in time. So change in time or change in t prime I should say. Well let me just go back to the Lorentz Transformation. Let me write it the way that I'm used to writing it. I'm used to writing it as change in ct prime, which is the same thing as c times the change in t prime, is equal to the Lorentz factor times this is going to be change in ct or I could write it as c change in t minus beta times change in x. Once again I could have written this as change in ct or I could write this as c times change in t because the c isn't changing. So we know that all ready. If we wanted to solve for just change in t prime, we could just divide both sides by c. So let's do that. If we divide all of this by c, what do we get? This is a form, we've seen this in other videos that you might recognize and you might see this in some textbooks. Our change in t prime is going to be equal to gamma times, well c times delta t divided by c is just going to be delta t. And then if you take, so minus, beta divided by c. Well let's just remind ourselves that beta is equal to v over c. So beta divided by c is v over c squared. I could write this as minus v over c squared change in x. Change in t prime, let's write that now. That's going to be equal to you have your gamma times change in time in my coordinate system. So not the t prime just change in t minus v over c squared change in x. Immediately there's at least one simplification we can do. We can divide both the numerator and the denominator by gamma. I keep wanting to change... So we can divide the numerator and denominator by gamma that simplifies it a little bit. We can write this as our change in x prime over change in t prime is equal to in the numerator we have change in x and actually let me just write out what beta is. Beta, I'll do it over here, minus v over c that's what beta is And actually let me take the c out. Change in ct is the same thing as c times change in t. So times c delta t. v divided by c times c is just going to be v. Let me write it that way. This part, these cancel out. This is the same thing as change in x minus v change in t. Let me write it that way, simplfy it. Minus v change in t, that's our numerator. And then our denominator is change in t minus v over c squared change in x. It looks like we're getting close here. But we don't have the change in x change in t. We have how much of a change in x we have for a given change in t. So what we could do is we could divide the numerator and the denominator by delta t. We could multiply the numerator by one over delta t. And the denominator by one over delta t which is equivalent to dividing the numerator and the denominator each by delta t. And why am I doing that? Well, if I do that this first term right over here is going to be delta x over delta t which we know. Then we have minus v, cuz delta t divided by delta t is just going to be one. And then our denominator delta t divided by delta t is just one. Then we're gonna have minus v over c squared and then times delta x divided by delta t. This is cool. We've just been able to figure out what our velocity in the primed coordinate system, in the S prime frame of reference is in terms of the relative velocity between the S prime frame of reference and my frame of reference. Delta x delta t the velocity in my frame of reference. And well we know all of this other stuff too, this is just delta x delta t and this is v. Or we can even write it like this. We said delta x over delta t is going to be equal to u. So it's equal to u minus v over one minus, delta x over delta t is just u. So this is just u here. It's u times v or v times u, all of that over c squared. This is a really, really, really useful deriviation which we're going to apply some numbers to in the next video so we can appreciate kinda how fun this is on some level.