# Lorentz Transformation of Velocity 4 Vector Help

1. Apr 6, 2014

### cooev769

So i am working on a question, which is beyond my knowledge of Lorentz transformations and some help is greatly appreciated.

I have a 4 velocity, u=γ(v vector,c) and its transformation properties are the standard lorentz boost. I don't quite understand how I am supposed to do this given that there will be a transformation for each component, and I don't know whether for each component given x1*=γ(x1-Vt). Is the V in the direction of the component and if so isn't it equal to x1 anyway because x1 in this case is a velocity?

Obtain the lorentz transformation for each component and then deduce the velocity addition formula for v vector. You will end up with products of different γs not entirely sure why either.

Any help on this would be greatly appreciated. I have already done some assuming the velocity to be only in the first component direction which i know is wrong but i have never actually been taught the general boost yet they chuck it in the questions.

Thanks.

2. Apr 6, 2014

3. Apr 6, 2014

### cooev769

Thanks for the help bt we are only given

$x' = \gamma (x - V t)$
$y' = y$
$z' = z$
$t' = \gamma (t - \frac{Vx}{c^2})$

Cheers

4. Apr 7, 2014

### Fredrik

Staff Emeritus
Let O and O' be two inertial observers. Let S and S' be the inertial coordinate systems they use. The Lorentz transformation from S to S' is given by the formulas you posted only when the velocity of O' in the coordinate system S is (v,0,0). This v has nothing to do with the velocity that you're supposed to transform.

5. Apr 7, 2014

### cooev769

Yeah i figured that but when I transform each component do I turn V into the velocity of the inertial frame in the given direction?

6. Apr 7, 2014

### Fredrik

Staff Emeritus
I'm not sure I understand exactly what you mean, but I'm pretty sure the answer is no.

Note that the Lorentz transformation from S to S' is just one function that takes (t,x,y,z) to (t',x',y',z'). This function is determined up to a rotation by the velocity of O' in S. If this velocity is in the direction of one of the spatial axes of S, then it's extremely convenient to let the other two spatial axes of S' coincide with the corresponding two spatial axes of S. If we do this, then the Lorentz transformation is completely determined by the velocity of O' in S. (So a different velocity would give us a different Lorentz transformation).

What you're suggesting sounds like this: First apply a Lorentz transformation to (t,something,something,something), then another Lorentz transformation to (something,x,something,something), then another to (something,something,y,something), and finally another to (something,something,something,z). You would be applying four different transformations (at least three of them different from the one you were given) to four different vectors (all of them different from the one you were given).

7. Apr 7, 2014

### cooev769

Yeah this question may just be really poorly worded. Given they asked for the standard configuration boost, I'm assuming they mean the movement is along one of the axis.

8. Apr 7, 2014

### Fredrik

Staff Emeritus
You should post the full problem statement. I just noticed that you're supposed to derive the velocity addition law. This makes it pretty clear (to me) that you're dealing with a 4-velocity that corresponds to a velocity in the x direction, and will have to use a Lorentz boost with a velocity in the x direction.

Since this is a textbook-style problem, we will have to treat it as homework. We will give you hints, but we won't solve the problem for you.

9. Apr 7, 2014

### cooev769

That's cool I don't really have a problem doing an standard Lorentz boost but this is worded ambiguously at least to me that a nudge in the right direction would be helpful.

Q. Since the four velocity u=gamma(v vector, c) is a four vector, it's transformation properties are simple. Write down the standard Lorenz boost for all four components of u. Use these to deduce the velocity addition formula for v vector. Hint you will get products of different gamma's, don't try to algebraically simplify them, instead loom for ways to eliminate them.

Cheers.

10. Apr 7, 2014

### Fredrik

Staff Emeritus
OK, I see. Your first task is to do a Lorentz transformation to find the components of u in another coordinate system. Then you need to use the result to calculate the velocity of the particle in that coordinate system. (How do you obtain a velocity from a four-velocity?)

11. Apr 7, 2014

### cooev769

So is the Lorentz transformation for a four velocity the same as a position vector?

12. Apr 7, 2014

### cooev769

Or do I define u related to the derivative of a given x and transform the x?

13. Apr 7, 2014

### Fredrik

Staff Emeritus
It's the same as for a position vector. Note by the way that since a four-velocity is a tangent vector to the world line, and we're dealing with a world line through the origin, the four-velocity is equal to the coordinate 4-tuple of a point on the world line. So in a way, you are transforming the coordinates of an event.

14. Apr 7, 2014

### cooev769

Okay and given the general transformation is

X1' = gamma(x1 - Vx4)

Do I just use x1 as the value for V? Because that would be the v in the direction of v1.

15. Apr 7, 2014

### Fredrik

Staff Emeritus
That's how you calculate one of the components, yes.

I don't fully understand what you're asking, but you don't use anything as the value for V. You're doing the calculation for an arbitrary V.

16. Apr 7, 2014

### cooev769

Well we just started relativity and the only transformation we have been given is the following.

X1' = gamma (x1 - vx4)
X2' = x2
X3' = x3
X4' = gamma (x4 - vx1)

And we are giving x1 and so on for a velocity four vector. So if it is done as I think and as you suggested you just transform each component. It doesn't seem natural that you just chuck in

V1' = gamma (v1 - v x4)

And so on. I'm assuming this is wrong but this is the extent of my knowledge on Lorentz transformations as we are in our first week and I have a horrible lecturer and the book is quite convoluted. What do I use for v in the transformation if this is the way it is to be transformed?

17. Apr 7, 2014

### Fredrik

Staff Emeritus
As I said earlier, v is just the velocity of O' in S. (O and O' are inertial observers; S and S' are their inertial coordinate systems). The four-velocity u is the components in S of a tangent vector to the world line of some object that's moving independently of O and O'. There is no relationship between u and v.

What you wrote as V1' = gamma (v1 - v x4) should be u1' = gamma (u1 - v u4), but I assume that the x was just a typo. Don't use the letter v for two unrelated things (the velocity of O' in S, and the four-velocity of some object in S).

I agree that it doesn't seem natural, but it's the right way to transform components of four-velocity. You may want to try to prove that when you're done with this problem.

If you don't like your book, I would recommend that you find a copy of "A first course in general relativity" by Schutz. The book starts with an excellent introduction to special relativity.

18. Apr 7, 2014

### cooev769

Thanks very much for your patience. That has definitely clarified things. I'll give the book a go.

19. Apr 7, 2014

### cooev769

Is this in the right direction?

$u1 = \gamma (u1 - v1 t)$

And then do this for each u up to 3 what about 4 usually that's just

$t4 = \gamma (t - v1 x1/ c)$

20. Apr 8, 2014

### Fredrik

Staff Emeritus
No, it's not. Why is there a t in there?

I think I may have caused some of the confusion by not noticing that you said "u=gamma(v vector, c)". The "v" I've been talking about has nothing to do with this v. My v is the V that appears in the first version of the Lorentz transformation that you posted. And I haven't even been explaining it accurately enough. I said that it's the velocity of O' in S, but that's only accurate in 1+1 dimensions, where velocities are numbers rather than vectors. In 3+1 dimensions, the velocity of O' in S is a vector. If we know that the y and z components of this vector are zero, then we can write it as (w,0,0), where w is a real number. (I'm using the symbol w now, since the problem is using both v and u for other things). This w is what you denoted by V (uppercase) earlier. Note that it's not the velocity of O' in S, it's just the x component of that velocity.

If we do a Lorentz boost of u in the x direction, we have $u'_1=\gamma(u_1-wu_4)$ and so on.

The only thing the problem says about u is that it's a four-vector, and that we should use the following notation:
$$u=(u_1,u_2,u_3,u_4)=\gamma\left(\frac{u_1}{\gamma},\frac{u_2}{\gamma},\frac{u_3}{\gamma}, \frac{u_4}{\gamma}\right)=\gamma(v_1,v_2,v_3,c)=\gamma\left(\vec v,c\right).$$ It doesn't say that $\vec v$ is in the x direction. It also doesn't say that the Lorentz boost we should do is in the x direction (i.e. that the velocity of O' in S is in the x direction). I'm just assuming that everything is in the x direction, because the problem is unreasonably hard if it's not. If everything is in the x direction, and we write $\vec v$ as (v,0,0), the velocity addition law that you're supposed to find is (in units such that c=1):
$$w\oplus v=\frac{w+v}{1+wv}.$$ If we don't make this assumption, the velocity addition law looks like this instead:
$$\vec w\oplus\vec v=\frac{1}{1+\vec w\cdot\vec v}\bigg(\vec w+\vec v+\frac{\gamma_{\vec w}}{1+\gamma_{\vec w}}\vec w\times(\vec w\times\vec v)\bigg).$$ Since this one is unreasonably hard to prove in the early parts of an introductory course, I have to assume that you should prove the first one.