Velocity addition ain't what it used to be

[yre]

In special relativity, the velocity-addition for colinear velocities u and v is given by (u+v)/(1+uv/c^2), but for non-colinear velocities the formula is more complicated and is not symmetric in u and v. But since the reciprocal velocity of v is -v then the reciprocal of (u+v) should be -(u+v) but also starting from the other end the reciprocal should also be (-v + -u) however -(u+v) is not equal to (-v + -u) according to the addition formula. So what is going on?

The resolution of the velocity composition paradox as described at http://en.wikipedia.org/wiki/Velocity-addition_formula makes it clear that Einstein's original non-commutative velocity addition, that Einstein said only obeys the parallelogram law to first approximation, is not actually a velocity composition law - it is a boost parameter composition law which only partially describes boost composition.

Ungar has been writing about this since 1988, but given that it is well known that a boost followed by a boost results in a boost plus a rotation then I'm guessing that people familiar with boost composition must have known the fact that the original addition formula doesn't give the actual composite velocity, except when the velocities are colinear. So can anyone point to a reference prior to Ungar's work that explicitly states this fact ?

 

[Fredrik]

I don't have a reference for you, but I can outline a way to prove the addition law. I'll use the notation $u\oplus v$ for the velocity associated with a Lorentz transformation that's the product of two boosts with velocities u and v. This velocity can also be interpreted as the velocity of the object that has velocity v in the frame that has velocity u.

A 3+1-dimensional Lorentz transformation can be written as

$$\Lambda=\gamma\begin{pmatrix}1 & -v^T\beta \\ -v & \beta\end{pmatrix}$$

where v is a 3×1 matrix representing the velocity difference, $$\gamma=1/\sqrt{1-v^2}$$, and $\beta$ is a 3×3 matrix that's orthogonal when v=0.

For a pure boost $\beta$ is not I, it is

$$\beta = \gamma^{-1} I + (1 - \gamma^{-1})\frac{vv^T}{v^Tv}$$

It acts as the identity on v, but as $\gamma^{-1} I$ on the orthogonal complement.

Now if we just multiply two boost matrices together, we see that

$$u\oplus v=\frac{u+\beta_uv}{1+u^Tv}$$

and it isn't too hard to show that this agrees with the formula in Ungar's book, which (except for factors of c and minor differences in the notation) is

$$\vec u\oplus\vec v=\frac{1}{1+\vec u\cdot\vec v}\bigg(\vec u+\vec v+\frac{\gamma_{\vec u}}{1+\gamma_{\vec u}}\vec u\times(\vec u\times\vec v)\bigg)$$

I found it easier to start with Ungar's formula and show that it can be rewritten as my formula.

 

[yre]

But Fredrick, the point is that it is Ungar's contention that this formula does not give the velocity addition. Instead he posits an alternative addition formula which is symmetric in u and v as given in http://www.phil-inst.hu/~szekely/PIRT_Bp_2/Papers/Ungar_09_ft.pdf"

 

[bcrowell]

Ungar has been writing about this since 1988, but given that it is well known that a boost followed by a boost results in a boost plus a rotation then I'm guessing that people familiar with boost composition must have known the fact that the original addition formula doesn't give the actual composite velocity, except when the velocities are colinear. So can anyone point to a reference prior to Ungar's work that explicitly states this fact ?

L. H. Thomas, "Motion of the spinning electron", Nature 117 (1926) 514.

Ungar seems like a crank, based on the link you posted.

But Fredrick, the point is that it is Ungar's contention that this formula does not give the velocity addition. Instead he posits an alternative addition formula which is symmetric in u and v as given in http://www.phil-inst.hu/~szekely/PIRT_Bp_2/Papers/Ungar_09_ft.pdf"

There is the question of whether he is arguing about physics or just about what words he likes to use to describe the physics. In the paper you linked to, he makes some vague references to experimental evidence. Actually all the experimental evidence confirms what Thomas figured out in 1926.

 

[yre]

Since no one seems to be understanding what I am asking, let me ask it in another way:

The Lorentz transformation B(u) B(v) where B(u) and B(v) are the boosts of velocities u and v, is the same as B(u+v)T(u,v) where T(u,v) is the Thomas rotation arising from B(u) B(v) and B(u+v) is the boost of velocity u+v where u+v means the full non-colinear velocity addition formula.

Do you agree with that ?

If so then what is the resulting velocity after applying the transformation B(u+v)T(u,v) ?

It's not u+v. Is it ?

Applying B(u+v)T(u,v) to where you are at A to get to C, means rotating your frame by T(u,v) to get to A' and then applying the boost B(u+v) of velocity u+v. This means that u+v does not give the velocity of C relative to your original unrotated frame at A - it gives the velocity relative to the frame A' that is rotated relative to A.

So the actual relative velocity of C relative to your unrotated frame at A is not given by u+v.

 

[bcrowell]

Since no one seems to be understanding what I am asking, let me ask it in another way:

The Lorentz transformation B(u) B(v) where B(u) and B(v) are the boosts of velocities u and v, is the same as B(u+v)T(u,v) where T(u,v) is the Thomas rotation arising from B(u) B(v) and B(u+v) is the boost of velocity u+v where u+v means the full non-colinear velocity addition formula.

Do you agree with that ?

Yes.

If so then what is the resulting velocity after applying the transformation B(u+v)T(u,v) ?

It's not u+v. Is it ?

No. It's not u+v even in the collinear case. In the collinear case, it's (u+v)/(1+uv/c^2). For the relevant equations in the noncollinear case, see http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.5

 

[yre]

Yes.


No. It's not u+v even in the collinear case. In the collinear case, it's (u+v)/(1+uv/c^2). For the relevant equations in the noncollinear case, see http://www.lightandmatter.com/html_books/genrel/ch02/ch02.html#Section2.5

You are still not getting what I wrote. I said "u+v means the full non-colinear velocity addition formula."

I should have used a different symbol.

 

[bcrowell]

You are still not getting what I wrote. I said "u+v means the full non-colinear velocity addition formula."

I should have used a different symbol.

Then aren't you defining something and then asking whether it equals what you defined it as...?

 

[yre]

Let me repeat my last reply but with a different symbol so as not to confuse it with ordinary addition.

Let's use @ for the velocity addition formula.

The Lorentz transformation B(u) B(v) where B(u) and B(v) are the boosts of velocities u and v, is the same as B(u@v)T(u,v) where T(u,v) is the Thomas rotation arising from B(u) B(v) and B(u@v) is the boost of velocity u@v.

Do you agree with that ?

If so then what is the resulting velocity after applying the transformation B(u@v)T(u,v) ?

It's not u@v. Is it ?

Applying B(u@v)T(u,v) to where you are at A to get to C, means rotating your frame by T(u,v) to get to A' and then applying the boost B(u@v) of velocity u@v. This means that u@v does not give the velocity of C relative to your original unrotated frame at A - it gives the velocity relative to the frame A' that is rotated relative to A.

So the actual relative velocity of C relative to your unrotated frame at A is not given by u@v.

 

[yre]

Then aren't you defining something and then asking whether it equals what you defined it as...?

The point is that the velocity addition formula u@v does not give the relative velocity of C relative to A. It gives the parameter of the boost of C relative to A.

Therefore it is not a velocity composition formula in the sense that you'd normally expect something called a velocity-addition formula to be.

 

[DrDu]

Well, you start with a toaster at rest in A and end up with a rotated toaster with velocity u@v in C. I'd say u@v is as close as you can get to a combined velocity.

 

[bcrowell]

The point is that the velocity addition formula u@v does not give the relative velocity of C relative to A. It gives the parameter of the boost of C relative to A.

OK, so let's use v_CA for the velocity of C relative to A, and u@v for the parameter of the boost of C relative to A. I don't see how v_CA and u@v could differ in magnitude. I could see how they could differ in direction, depending on the specific definition of u@v. Depending on whether you define u@v in terms of a boost done before the Thomas rotation or after it, I suppose you could get the direction of u@v differing by the angle of the Thomas rotation.

 

[yre]

OK, so let's use v_CA for the velocity of C relative to A, and u@v for the parameter of the boost of C relative to A. I don't see how v_CA and u@v could differ in magnitude. I could see how they could differ in direction, depending on the specific definition of u@v. Depending on whether you define u@v in terms of a boost done before the Thomas rotation or after it, I suppose you could get the direction of u@v differing by the angle of the Thomas rotation.

Exactly. Same magnitude. Different direction.

 

[yre]

Well, you start with a toaster at rest in A and end up with a rotated toaster with velocity u@v in C. I'd say u@v is as close as you can get to a combined velocity.

Not quite. You get a velocity with the same magnitude as u@v, but u@v doesn't give the correct direction of C relative to A.

 

[yre]

I suppose you could get the direction of u@v differing by the angle of the Thomas rotation.

According to Ungar the velocity v_CA with the correct direction is given by v_CA=u@T(u,-v)v

 

[yre]

Exactly. Same magnitude. Different direction.

Actually not so exactly. u@v and v@u have the same magnitude as each other but I don't know about the magnitude of Ungar's formula v_CA=u@T(u,-v)v.

I don't know how he got that formula.

 

[DrDu]

There's a similar situation already in ordinary non-relativistic setting:
Consider the rotations in 3-dimensions. I call the rotations around the z-axis "true rotations" and the rotations around an axis in the x-y plane a "boost". As long as the rotation angles are <<2 pi, the combination of two boosts will again look like a boost, but when the angles become substantial, then there is an additional rotation around z. Furthermore, the sign of the rotation around z depends on the order of the two "boosts".

 

[bcrowell]

According to Ungar the velocity v_CA with the correct direction is given by v_CA=u@T(u,-v)v

The problem is that Ungar seems to lack the ability to make definite statements about the connection between theory and experiment. If "correct" means in better agreement with experiment, then he hasn't pointed out what the experiment is. If "correct" means he just likes his way of saying it better, then it's just a different formalism for expressing the same thing.

 

[Fredrik]

But Fredrick, the point is that it is Ungar's contention that this formula does not give the velocity addition. Instead he posits an alternative addition formula which is symmetric in u and v as given in http://www.phil-inst.hu/~szekely/PIRT_Bp_2/Papers/Ungar_09_ft.pdf"

Where does he say that? The article is 35 pages long.

The Lorentz transformation B(u) B(v) where B(u) and B(v) are the boosts of velocities u and v, is the same as B(u+v)T(u,v) where T(u,v) is the Thomas rotation arising from B(u) B(v) and B(u+v) is the boost of velocity u+v where u+v means the full non-colinear velocity addition formula.

Do you agree with that ?

If so then what is the resulting velocity after applying the transformation B(u+v)T(u,v) ?

It's not u+v. Is it ?

We seem to agree that B(u)B(v) is a Lorentz transformation, and that it's not a boost. But there's always a velocity associated with a Lorentz transformation, and its components are $-\Lambda_{i0}/\Lambda_{00}$ (regardless of whether $\Lambda$ is a boost or not), where $\Lambda_{\mu\nu}$ denotes row $\mu$, column $\nu$ of the matrix $\Lambda$. That's the velocity I calculated. (I called it $u\oplus v$).

If I understand you correctly, you're saying that if we rewrite B(u)B(v) as a boost times a rotation, the velocity of the boost wouldn't be $u\oplus v$. My question is, why do you consider the velocity of that boost to be "the actual relative velocity"? I would say that $u\oplus v$ is the actual relative velocity, because that's the velocity of the Lorentz transformation.

Edit: This is a better way to motivate it: $1/(u\oplus v)$ is the slope of the line that the 0 axis (of the coordinate system in which the object is stationary) is mapped to. So $u\oplus v$ is the velocity associated with the world line of the object, and I think that's what we should be calling the "relative velocity".

Edit 2: I changed my mind. There is no "actual relative velocity". See post #28 below. When I made this edit, I also corrected $\Lambda_{0i}$ to $\Lambda_{i0}$ above.

 

[Passionflower]

Ungar seems like a crank, based on the link you posted.

Really? How did you conclude that?

 

[yre]

Where does he say that? The article is 35 pages long.

Page 9 eq(22) gives the formula for "coaddition" u#v in terms of @

Page 10 eq(29) gives u#v in symmetric form.

Page 13 says "Einstein velocity coaddition ... does give rise to an exact "velocity gyroparallelogram" in hyperbolic geometry."

Page 23 poses the question "whether, in the limit of negligible force, relativistic velocities add (1) according to Einstein’s 1905 velocity addition law (2), or
(2) according to the Einstein gyroparallelogram addition law (54).", eq.(54) being w=u#v

and continues: "We will find that owing to the validity of wellknown relativistic stellar aberration formulas, (i) Einsteinian, relativistic velocities are gyrovectors that add according to the gyroparallelogram addition law (54), Fig. 3, which is commutative,"

Page 27 describes NASA's gyroscope experiment and says this corroborates the relativistic stellar aberration formulas and hence also validates the gyroparallelogram addition law of Einsteinian velocities a.k.a the coaddition formula.

Other eqs:

Page 3. eq(2) gives the formula for u@v

Page 5. eq(12) gives the formula for Thomas rotation (gyr[u,v]) in terms of @

Page 9. eq(22) gives the formula for "coaddition" u#v in terms of @

Page 10. eq(29) gives u#v in symmetric form.
(probably by combining eqs 12 and 22 and using computer algebra)

 

[yre]

My question is, why do you consider the velocity of that boost to be "the actual relative velocity"?

We should be consistent about what is considered to be the "actual relative velocity" of things relative to A, which means fixing the coordinate frame of A, not rotating it.

If B is boosted from A by B(u) and we say the relative velocity of B from A is u, then if the boost of C from A is B(w) then the relative velocity of C from A is w.

We need to compare like with like. The existence of B doesn't physically affect C. Even if B didn't exist, there would be a boost from A to C and that would be the relative velocity.

Earlier in the thread we used B(u)B(v) although in this setup we should be talking about B(v)B(u).

 

[yre]

If I understand you correctly, you're saying that if we rewrite B(u)B(v) as a boost times a rotation, the velocity of the boost wouldn't be $u\oplus v$.

We can rewrite B(u)B(v) as B(u@v)T(u,v).

I would say that $u\oplus v$ is the actual relative velocity, because that's the velocity of the Lorentz transformation.

but we could also write B(u)B(v) as T(u,v)B(v@u).

so B(u)B(v)=T(u,v)B(v@u)=B(u@v)T(u,v).

u@v is not equal to v@u so there isn't a unique velocity of the Lorentz transformation.

 

[JesseM]

The point is that the velocity addition formula u@v does not give the relative velocity of C relative to A. It gives the parameter of the boost of C relative to A.

"Velocity" ordinarily just means (change in position)/(change in time) in some inertial frame. Do you disagree that if A is moving inertially in the x-direction of frame B with (change in position)/(change in time) = v in that frame, and B is moving in the x'-direction of frame C with (change in position)/(change in time) in frame C, then in frame C the (change in position)/(change in time) for A is equal to (u + v)/(1 + uv/c^2)? This is very easy to check by picking two arbitrary events on the worldline of A which have (difference in x)/difference in t) = v in frame B, then applying the Lorentz transformation to find the (difference in x')/difference in t') for the same events in frame C.

 

[yre]

"Velocity" ordinarily just means (change in position)/(change in time) in some inertial frame. Do you disagree that if A is moving inertially in the x-direction of frame B with (change in position)/(change in time) = v in that frame, and B is moving in the x'-direction of frame C with (change in position)/(change in time) in frame C, then in frame C the (change in position)/(change in time) for A is equal to (u + v)/(1 + uv/c^2)? This is very easy to check by picking two arbitrary events on the worldline of A which have (difference in x)/difference in t) = v in frame B, then applying the Lorentz transformation to find the (difference in x')/difference in t') for the same events in frame C.

You mean A,B and C are all traveling in the same direction, not necessarily on the same straight line, but all parallel to each other then do we get (u + v)/(1 + uv/c^2)?

I think so, I should have been using the phrase parallel rather than colinear. and in this case there is no rotation and so the result of B(u)B(v) is B(u@v) with no rotation, and in the parallel case u@v=v@u as the formula is symmetric.

"Velocity" ordinarily just means (change in position)/(change in time) in some inertial frame.

I'm not sure why you made the point of saying "ordinarily".

The change (which is a 3-dimensional directional vector) must be relative to some coordinate frame. Two coordinate frames both set up at A, call them A, and A' could have their origin at the same point but different orientations in which case the velocity (change in position ) of C using the coordinate system of A would be different to the velocity of C using the coordinate system of A'.

 

[JesseM]

If A,B and C are all traveling in the same straight line then we get (u + v)/(1 + uv/c^2).

So you have no problem with the traditional velocity addition formula in this scenario, and your comment "the velocity addition formula u@v does not give the relative velocity of C relative to A. It gives the parameter of the boost of C relative to A" was not meant to apply to this scenario? If so, then if we continue to define velocity in terms of (change in position)/(change in time), would you disagree that in a situation where motion is not restricted to the x-axes (and perhaps the x-axes of B and C aren't even aligned), the more general form of the velocity addition formula still correctly gives the velocity of A frame C if we know the velocity of A in frame B? Again this could be checked by actually picking the coordinates in frame B of two events on A's worldline, then using the appropriate form of the Lorentz transformation to find the coordinates of these same events in frame C.

 

[JesseM]

You mean A,B and C are all traveling in the same direction, not necessarily on the same straight line, but all parallel to each other then do we get (u + v)/(1 + uv/c^2)?

B and C are frames, not objects, so I don't understand what it means to distinguish between "on the same straight line" and "parallel". A is moving along the x-axis of B and along the x'-axis of C, and the x-axis of B is parallel to the x'-axis of C as per the traditional simple form of the Lorentz transform (where we assume each frame's spatial origin is moving along the x-axis of the other frame, and the two spatial origins coincided at a time coordinate of 0 in each frame)

 

[Fredrik]

Even if B didn't exist, there would be a boost from A to C and that would be the relative velocity.

After some thought, I now understand that I was wrong to say that u@v is "the relative velocity". But I don't agree with the quote above.

Instead, I say that the notion of "relative velocity" is inappropriate in 3+1 dimensions. (In what follows, A and C are objects, not frames). In 1+1 dimensions, "the velocity of particle C relative to particle A" is defined as "the velocity of C in the unique positively oriented inertial frame associated with A's motion by the standard synchronization procedure". The reason why it makes sense to talk about relative velocity in 1+1 dimensions is that the coordinate system is uniquely determined. In 3+1 dimensions, the standard synchronization procedure applied to the world line of a non-accelerating massive particle, only determine a positively oriented inertial frame up to a proper rotation.

Since none of these frames can be preferred over the others, I say that the concept of relative velocity is ambiguous, and that we therefore have to be explicit. We should simply specify which positively oriented inertial frame we associate with A's motion, and then talk about C's velocity in that frame instead of "C's velocity relative to A".

(Note that the question of what frame to associate with C's motion doesn't arise, since we are only interested in the velocity of C's world line in A's coordinates).

there isn't a unique velocity of the Lorentz transformation.

Every Lorentz transformation $\Lambda$ has a unique velocity, and it's equal to $-\Lambda_{i0}/\Lambda_{00}$. It's determined by the slope of the line it maps the 0 axis to.

 

[PAllen]

After some thought, I now understand that I was wrong to say that u@v is "the relative velocity". But I don't agree with the quote above.

Instead, I say that the notion of "relative velocity" is inappropriate in 3+1 dimensions. (In what follows, A and C are objects, not frames). In 1+1 dimensions, "the velocity of particle C relative to particle A" is defined as "the velocity of C in the unique positively oriented inertial frame associated with A's motion by the standard synchronization procedure". The reason why it makes sense to talk about relative velocity in 1+1 dimensions is that the coordinate system is uniquely determined. In 3+1 dimensions, the standard synchronization procedure applied to the world line of a non-accelerating massive particle, only determine a positively oriented inertial frame up to a proper rotation.

Since none of these frames can be preferred over the others, I say that the concept of relative velocity is ambiguous, and that we therefore have to be explicit. We should simply specify which positively oriented inertial frame we associate with A's motion, and then talk about C's velocity in that frame instead of "C's velocity relative to A".

(

How about the idea that for talking about velocity of C relative to A, there is a preferred subset of these coordinate systems: ones in which C's velocity is in the direction of a basis vector (e.g. all +x, all +y, or all +z). Then, presumably, for any of these, C's velocity will be the same. It seems reasonable to me to call this the speed of C relative to A. Thus we can give a unique meaning to relative speed, but not relative velocity (?).

In any of the 'A' frames describe by Fredrik, would it be true that that the norm of C's velocity would be this same speed?

 

[yre]

B and C are frames, not objects

I'm taking A, B and C to be moving objects at rest w.r.t. inertial frames whose origin is at A, B and C.

 

[JesseM]

I'm taking A, B and C to be moving objects at rest w.r.t. inertial frames whose origin is at A, B and C.

Well, the speed (magnitude of velocity vector) of A in C's frame given a known velocity in B's frame should be the same regardless of where you place the origins of B and C's frames. Are you specifically concerned with the actual coordinate direction of the velocity vector apart from just the speed? If you don't assume that the origins of B and C coincide at times t=0 and t'=0 in those frames, then to get the actual vector you'll need to use some appropriate form of the velocity addition formula that's derived from the Poincaré transformation rather than the Lorentz transformation...I'm not sure what the formula would be but I'm sure you could find it somewhere.

Can you state more clearly what your objection to the velocity addition formula is? There are obviously going to be different versions of the formula depending on how you define the coordinate systems (where their x-axes are parallel, whether the origins coincide at a time of zero in each frame), are you objecting that sometimes people use the wrong version in a given scenario? Or are you arguing that even when people use a version that's tailored to the pair of coordinate systems being used (like using the (u + v)/(1 + uv/c^2) version in a scenario where B and C's x-axes are parallel and their spatial origins coincided at t=t'=0, and A is moving along the x-axis) there's still some problem with it?

 

[yre]

Even if B didn't exist, there would be a boost from A to C and that would be the relative velocity.

After some thought, I now understand that I was wrong to say that u@v is "the relative velocity". But I don't agree with the quote above.

Ok, yes, I was thinking C would be moving away from A in a straight line from A's perspective in which case I'd say there would be a boost from object A to object C.

However in 3+1 dimensions C could also be moving across A's line of sight, so to speak, which wouldn't be a boost.

 

[yre]

Are you specifically concerned with the actual coordinate direction of the velocity vector apart from just the speed?

Yes. The full 3D velocty addition formula is about directed velocities.

If you don't assume that the origins of B and C coincide at times t=0 and t'=0 in those frames, then to get the actual vector you'll need to use some appropriate form of the velocity addition formula that's derived from the Poincaré transformation rather than the Lorentz transformation

This thread has gotten side-tracked into talking about translations and parallel movements, but what this discussion is about is the composition of two boosts, i.e. Lorentz transformations, i.e. the origin of B and C did coincide at some time and the origin of A and B did coincide at some time.

Can you state more clearly what your objection to the velocity addition formula is? There are obviously going to be different versions of the formula depending on how you define the coordinate systems (where their x-axes are parallel, whether the origins coincide at a time of zero in each frame), are you objecting that sometimes people use the wrong version in a given scenario? Or are you arguing that even when people use a version that's tailored to the pair of coordinate systems being used (like using the (u + v)/(1 + uv/c^2) version in a scenario where B and C's x-axes are parallel and their spatial origins coincided at t=t'=0, and A is moving along the x-axis) there's still some problem with it?

The 3D-velocity formula u@v reduces to (u + v)/(1 + uv/c^2) in the 1-dimensional case.

Even if BC is not in the same line as AB they are all still in a plane, so it reduces to 2-dimensions. The full 3-dimensions is only needed when composing three boosts which might not all be in a plane.

Anyway u@v gives the parameter of the boost composition B(u)B(v)=B(u@v)gyr[u,v].

In the 1-dimensional case B(u@v)gyr[u,v] reduces to B(u@v), and so boost-parameter and composite-velocity mean the same thing in 1-dimension.

Ungar says that for non-colinear boosts in 3-dimensions, boost-parameter and composite-velocity do not mean the same thing.

 

[yre]

This discussion has made me think I am misinterpreting what Ungar says about coaddition. It's not that the magnitude |u#v| gives the speed of C which I think is given by |u@v|=|v@u|. It's that neither u@v or v@u give the direction.

The role of u#v is that when comparing the trigonometric formulae of Newtonian dynamics with relativistic dynamics, u#v appears in expressions in an analogous way to ordinary vector addition u+v and in this sense relativistic velocities behave more like u#v than u@v. In which case u#v isn't a departure from the laws of special relativity, u#v is just a quantity which can be used to compare Newtonian dynamics with relativistic dynamics.