V=dx/dt (relative velocity between two inertial observers)

[etotheipi]

There's also a nice way to find the magnitude ##V## of the relative velocity between two particles of masses ##m_1## and ##m_2## and 4-momenta ##p_1^a## and ##p_2^a##. If we work in the rest frame of ##m_1##, then ##p_1^a = (m_1, \vec{0})## and ##p_2^a = (\frac{m_2}{\sqrt{1-V^2}}, \vec{p}_2')## the product of the two 4-momenta can be written\begin{align*}
(p_1)_a (p_2)^a &= \frac{m_1 m_2}{\sqrt{1-V^2}} \implies V = \sqrt{1- \frac{(m_1m_2)^2}{[(p_1)_a (p_2)^a]^2}}

\end{align*}Let's now switch to any arbitrary frame of reference, in which ##p_1^a = (E_1, \vec{p}_1)## and ##p_2^a = (E_2, \vec{p}_2)##. We can write ##(p_1)_a (p_2)^a## as\begin{align*}

(p_1)_a (p_2)^a &= E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2 \\ \\

&= \frac{m_1 m_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} - \frac{m_1 m_2 \vec{v}_1 \cdot \vec{v}_2}{\sqrt{1-v_1^2}\sqrt{1-v_2^2}} \\ \\

&= \frac{m_1m_2 (1-\vec{v}_1 \cdot \vec{v}_2)}{\sqrt{(1-v_1^2)(1-v_2^2)}}

\end{align*}Hence, we can determine ##V## in terms of ##\vec{v}_1## and ##\vec{v}_2##,\begin{align*}

V = \sqrt{1- \frac{(1-v_1^2)(1-v_2^2)}{(1-\vec{v}_1 \cdot \vec{v}_2)^2}} = \frac{\sqrt{v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 - \left( v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2 \right) }}{1-\vec{v}_1 \cdot \vec{v}_2}

\end{align*}However, ##v_1^2 -2 \vec{v}_1 \cdot \vec{v}_2 + v_2^2 = (\vec{v}_1 - \vec{v}_2)^2##, and from vector analysis we also know that\begin{align*}
(\vec{v}_1 \times \vec{v}_2)^2 = (\vec{v}_1 \times \vec{v}_2) \cdot (\vec{v}_1 \times \vec{v}_2) &= (\vec{v}_1 \cdot \vec{v}_1)(\vec{v}_2 \cdot \vec{v}_2) - (\vec{v}_1 \cdot \vec{v}_2)(\vec{v}_2 \cdot \vec{v}_1) \\

&= v_1^2 v_2^2 - [\vec{v}_1 \cdot \vec{v}_2]^2

\end{align*}which results in\begin{align*}
V = \frac{\sqrt{(\vec{v}_1 - \vec{v}_2)^2 - (\vec{v}_1 \times \vec{v}_2)^2 }}{1-\vec{v}_1 \cdot \vec{v}_2}

\end{align*}which is, of course, symmetric in the two particles!

 

[vanhees71]

The magnitude ("relative speed") is symmetric, the vectors are not, except for ##\vec{v}_1 \parallel \vec{v}_2##.

 

[Kairos]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

 

[Sagittarius A-Star]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

As you can see at the example of the moving light clock, the angle between the vectors is in general differently in the 2 frames. This effect is called "aberration".

If the angle is for example 90° in the receivers frame, then there is a Doppler redshift by ##1/\gamma##.
But if the angle is 90° in the senders frame, then there is a Doppler blueshift by ##\gamma##.

So, that in each frame the moving clock ticks slower than the clock at rest does not lead to a contradiction.

 

[PeroK]

As you can see at the example of the moving light clock, the angle between the vectors is in general differently in the 2 frames. This effect is called "aberration".

If the angle is for example 90° in the receivers frame, then there is a Doppler redshift by ##1/\gamma##.
But if the angle is 90° in the senders frame, then there is a Doppler blueshift by ##\gamma##.

So, that in each frame the moving clock ticks slower than the clock at rest does not lead to a contradiction.

That's not the question. The question is: if relative motion is not symmetric, then can we see differences in the Doppler shift?

 

[Sagittarius A-Star]

That's not the question. The question is: if relative motion is not symmetric, then can we see differences in the Doppler shift?

Sorry, I did not read the complete thread :-(

 

[PeroK]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

If we imagine that object ##B## has coordinates ##(t, x_0 + v_xt, y_0 + v_yt, z_0 + v_zt)## in object ##A##'s rest frame. The components ##(v_x, v_y, v_z)## define a three-vector, which we can take to be the new x-axis. Likewise we can take the new y-axis so that object ##B## has coordinates ##(t, x_0 + vt, y_0, 0)##, where we take ##v > 0## and ##y_0 > 0##..

Now object ##B## can do the same and choose axes so that ##A## has coordinates ##(t', v't', y'_0)##, where again we take ##v' > 0, y'_0 > 0##.

The question is: do we necessarily have ##v = v', x_0 = x'_0, y_0 = y'_0 ##?

My argument is that, by the isotropy and homogeneity of spacetime, we must. We must have complete symmetry of these coordinates. Relative motion must be physically completely symmetric.

Moreover, we could take the point of closest approach to define ##t = 0## and ##t' = 0## in each case.

Note that I've changed strategy to make the displacements and velocities the same, rather than negatives of each other, which further emphasises the symmetry.

By this argument, there must also be completely symmetric Doppler shift.

 

[vanhees71]

The effect of velocity on time dilation is orientation-independent but the Doppler effect depends on orientation, so when the vectors are not parallel, can the two inertial observers perceive a different Doppler effect?

Concerning the Doppler effect, I've just written a short note on the general case for waves with arbitrary phase velocity (##<c##, i.e., in the region of "normal dispersion"):

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

 

[vanhees71]

That's not the question. The question is: if relative motion is not symmetric, then can we see differences in the Doppler shift?

Yes, as was nicely explained in #39, the angles between the wave vector and the velocity of the source and/or observer are frame dependent. If you have in addition a medium, where the waves propagate in (e.g., a fluid for sound waves or a dielectric for electromagnetic waves), then everything also depends on the velocity of the medium too. You find the kinematics of the Doppler effect in all generality here:

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

 

[Kairos]

Thank you, I already knows that and #39, but that's not the point:

Is the "non-reciprocity" of relative velocity (introduced above in this thread) compatible with the reciprocity of the Doppler effect. If not, is the admitted reciprocity of the Doppler effect in SR also matter of debate.

 

[vanhees71]

What do you mean with "reciprocity"? There is no matter of debate concerning the Doppler effect in SR (nor in GR). For em. waves in vacuo you find the complete explanation of the Doppler effect already in Einstein's paper of 1905!

 

[PeroK]

Thank you, I already knows that and #39, but that's not the point:

Is the "non-reciprocity" of relative velocity (introduced above in this thread) compatible with the reciprocity of the Doppler effect. If not, is the admitted reciprocity of the Doppler effect in SR also matter of debate.

The apparent non-reciprocity of relative velocity is a coordinate effect, resulting by considering the velocities in a wider context, with a necessary convention for spatial coordinate axes.

Post #42 shows that if we consider only the two objects in relative motion, then we can choose coordinates so that the relative motion is seen to be physically completely symmetric. Using those coordinates, we can see that any effects, such as the Doppler effect,must be symmetric.

Note that if either A or B uses a different coordinate system than that proposed in post #42, then the complete physical symmetry of the situation may not be immediately apparent. In that case, it may take yet more mathematical bravura to show the complete physical symmetry of relative inertial motion.

 

[Kairos]

Many thanks! It is a coordinate effect but not a physical effect! I was destabilized by post #9:

There are some subtleties to the subject of the reciprocity of the relative velocity. It does not hold true in special relativity! But we ought to put discussion of that on hold until you have a firmer grasp of the underlying machinery.

 

[etotheipi]

Sigh! This quote refers to the fact that the relative velocities each measures for the other are elements different local rest spaces (see the image in #30), and so when considered as vectors in ##\mathbf{R}^4## they obviously cannot be parallel not least because although the intersection ##E_{u_1} \cap E_{u_2}## of those rest spaces is still non-trivial, the relative velocities do not lie within that intersection.

Even the three dimensional relative velocities are not simply the negations of each other except in the special case where the three velocities of the two particles in some frame are parallel as showed in #35.

 

[PeroK]

Sigh! This quote refers to the fact that the relative velocities each measures for the other are elements different local rest spaces (see the image in #30), and so when considered as vectors in ##\mathbf{R}^4## they obviously cannot be parallel not least because although the intersection ##E_{u_1} \cap E_{u_2}## of those rest spaces is still non-trivial, the relative velocities do not lie within that intersection.

Even the three dimensional relative velocities are not simply the negations of each other except in the special case where the three velocities of the two particles in some frame are parallel as showed in #35.

The relative velocity (and indeed everything else about the relative motion) is completely symmetric, as it must be. However ##A## describes the motion of ##B##, ##B## must be able to describe the motion of ##A## identically.

The relative motion of ##B## with respect to ##A## must be physically indistinguishable from the relative motion of ##A## with respect to ##B##.

This was the original point at issue.

 

[etotheipi]

The relative velocity (and indeed everything else about the relative motion) is completely symmetric, as it must be. However ##A## describes the motion of ##B##, ##B## must be able to describe the motion of ##A## identically.

This is not possible in the relativistic framework, again because the relative velocities are elements of different vector spaces. The actual relationship I derived in #28! You must use projection operators.

 

[vanhees71]

The apparent non-reciprocity of relative velocity is a coordinate effect, resulting by considering the velocities in a wider context, with a necessary convention for spatial coordinate axes.

Post #42 shows that if we consider only the two objects in relative motion, then we can choose coordinates so that the relative motion is seen to be physically completely symmetric. Using those coordinates, we can see that any effects, such as the Doppler effect,must be symmetric.

Note that if either A or B uses a different coordinate system than that proposed in post #42, then the complete physical symmetry of the situation may not be immediately apparent. In that case, it may take yet more mathematical bravura to show the complete physical symmetry of relative inertial motion.

It's important to keep in mind that this is a special case, i.e., the velocities of the two points are collinear (then they are collinear in any inertial frame!). The additional rotation between the two relative velocities in the general case must be kept in mind. It's due to the fact that the composition of two rotation-free Lorentz boosts in non-collinear directions is not again a rotation-free Lorentz boost but a boost followed by a rotation (or a rotation followed by a boost). That's related to the Thomas precession.

It's also related to the fact that the "addition of velocities" is not a commutative operation. That's also discussed on the Wikipedia page:

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[PeroK]

It's important to keep in mind that this is a special case, i.e., the velocities of the two points are collinear (then they are collinear in any inertial frame!).

You must be talking about something different from me. If we take an inertial reference frame - e.g. the rest frame of an inertial observer ##A##, then there is only the three-velocity of object ##B## under discussion here. There is no concept of ##B##'s velocity being "collinear" with ##A## or anything else. ##A## has zero velocity in this reference frame.

Likewise, if we change to the rest frame of ##B##, there is only ##A##'s velocity under discussion. ##B## has zero velocity in that frame, so again there is no concept of collinearity.

The motion of ##B## relative to ##A## must be physically identical to the motion of ##A## relative to ##B##.

This should be an elementary observation.

 

[etotheipi]

What is means is that, take an arbitrary reference system in which the two particles possesses three-velocities ##\vec{v}_{\mathcal{A}}## and ##\vec{v}_{\mathcal{B}}##, then the relative three-velocity of each particle with respect to the other satisfies ##\vec{V}_{\mathcal{AB}} = - \vec{V}_{\mathcal{BA}}## iff ##\vec{v}_{\mathcal{A}} \parallel \vec{v}_{\mathcal{B}}##. Otherwise you also need to take into account a rotation (cf. 35), although of course ##||\vec{V}_{\mathcal{AB}}|| = ||\vec{V}_{\mathcal{BA}}||## is still always satisfied (cf. 36).

 

[vanhees71]

You must be talking about something different from me. If we take an inertial reference frame - e.g. the rest frame of an inertial observer ##A##, then there is only the three-velocity of object ##B## under discussion here. There is no concept of ##B##'s velocity being "collinear" with ##A## or anything else. ##A## has zero velocity in this reference frame.

Likewise, if we change to the rest frame of ##B##, there is only ##A##'s velocity under discussion. ##B## has zero velocity in that frame, so again there is no concept of collinearity.

The motion of ##B## relative to ##A## must be physically identical to the motion of ##A## relative to ##B##.

This should be an elementary observation.

I'm talking about the general case that ##A## moves with some velocity ##\vec{v}_1## and ##B## with some velocity ##\vec{v}_2## wrt. an inertial frame of reference. By definition the relative velocity of particle 2 wrt. particle 1, ##\vec{v}_{\text{rel}2}##, is the velocity of particle 2 in the inertial frame, where particle 1 is at rest. To calculate this you need a Lorentz boost with ##\vec{v}_1## (as detailed in one of my postings above).

Correspondingly the relative velocity of particle 1 wrt. particle 2, ##\vec{v}_{\text{rel}1}## is the velocity of particle 1 in the rest frame of particle 2, to which you get by the Lorentz boost with ##\vec{v}_2##. As the above explicit calculation unambigously shows in general ##\vec{v}_{\text{rel}2} \neq -\vec{v}_{\text{rel}1}##, and the explanation is that rotation-free Lorentz boosts in different directions don't form a group but their composition is not a rotation-free boost.

That's obviously if ##u_1## and ##u_2## are the four-velocity vectors of particle 1 and 2, respectively in the original frame you have
$$u_{\text{rel}2}=\hat{\Lambda}(\vec{v}_1) u_{2}, \quad u_{\text{rel} 1}=\hat{\Lambda}(\vec{v}_2) u_1.$$
Obviously this implies that
$$u_{\text{rel}2}=\hat{\Lambda}(\vec{v}_1) \hat{\Lambda}(-\vec{v}_2) u_{\text{rel} 1},$$
where I have used that ##\hat{\Lambda}^{-1}(\vec{v}_2)=\hat{\Lambda}(-\vec{v}_2)##. The additional term with the double-cross product derived above is thus due to the fact that the decomposition of the above two rotation free Lorentz boosts only lead to a rotation free Lorentz boost if and only if ##\vec{v}_1 \parallel \vec{v}_2## and then and only then the additional cross-product terms cancel. These additional cross-product terms arise because the composition of two rotation free Lorentz boost in non-collinear directions leads to a Lorentz transformation which is composed of some rotation-free Lorentz boost followed by a rotation (or a rotation followed by a rotation-free Lorentz boost). These rotation(s) are called "Wigner rotations".

That the relative speeds are the same comes from the much simpler calculation that
$$\gamma_{\text{rel}1}=\gamma_{\text{rel}2}=u_1 \cdot u_2,$$
which is symmetric under exchange of particles 1 and 2. For details, see

https://en.wikipedia.org/wiki/Wigner_rotation

 

[PeroK]

I'll leave it to the OP to ask any further questions if he has them.

 

[Kairos]

Many thanks. I thought that in special relativity in absence of any reference to a fixed medium, the reciprocity of a relative velocity between two inertial objects A and B interchangeable is just natural. As a theoretical observer, you can change your point of view as you wish, invert it, rotate it, change your origin, assign the velocity entirely to one of them, or split it into v1 and v2 using velocity composition rules, and so on. This gives you some excellent mathematical exercises, but in no way question the intrinsic reciprocity of the relative velocity between A vs B and B vs A. I have the impression that the claim:

There are some subtleties to the subject of the reciprocity of the relative velocity. It does not hold true in special relativity! But we ought to put discussion of that on hold until you have a firmer grasp of the underlying machinery.

is not rigorous and can mislead interested non-specialist readers, like me, asking for help on this forum. Please correct me if I am wrong!

 

[vanhees71]

I think the calculation in #35 is not so complicated. You can do it knowing matrix multiplication and some vector algebra. It is thus a bit surprising, how much confusion exists about it even among practitioners of relativity theory. Where is something unclear with this calculation?

 

[PeroK]

Many thanks. I thought that in special relativity in absence of any reference to a fixed medium, the reciprocity of a relative velocity between two inertial objects A and B interchangeable is just natural.

You can, of course, consider the Centre of Momentum frame for two objects of nominally the same mass. In that frame, the three-momenta and three-velocities are reciprocal.

 

[vanhees71]

Of course, because in the CM system you have ##\vec{p}_1=-\vec{p}_2## by definition (no matter whether the particles have the same or different masses; mass is always invariant mass!), and there ##\vec{p}_1 \parallel \vec{p}_2## in this specific frame. The same holds true for the "lab frame", where by definition ##\vec{p}_1=0##.

 

[PeroK]

Of course, because in the CM system you have ##\vec{p}_1=-\vec{p}_2## by definition (no matter whether the particles have the same or different masses; mass is always invariant mass!), and there ##\vec{p}_1 \parallel \vec{p}_2## in this specific frame. The same holds true for the "lab frame", where by definition ##\vec{p}_1=0##.

If I am not wrong, this is precisely what the OP believes that you have been denying all through this thread!

 

[Ibix]

I thought I'd have a shot at some Minkowski diagrams, which may be helpful. I'm restricting motion to one dimension, but this is actually completely general for intersecting inertial worldlines, since in either one's rest frame we can simply define the x direction as "##\pm## the direction in which the other guy is moving". So these diagrams are all drawn in the hyperplane @etotheipi shaded in #30.

So we have two observers, red and blue. Here's a Minkowski diagram in red's frame - red's worldline is straight up the page and blue's is slanted because they're doing 0.6c.

Red can drop a perpendicular to the x-axis and draw blue's three-velocity, the heavy dotted blue line. This is what red calls "blue's velocity through space", so it must be a vector in the spatial direction. Blue's four-velocity is, of course, parallel to blue's worldline.

(Apologies that there's no arrowhead on the three-velocity - I didn't add the capability to draw arrowheads when I wrote my Minkowski diagram software.)

Now we can do the same thing in blue's frame:

Aside from the colours, this is an exact mirror image of the first diagram. This is the "reciprocity" between the red and blue's three-velocities that we've been talking about.

But the complexity comes in if we draw what red called "blue's velocity through space" on the diagram above:

I've included red's axes and the dropped perpendicular and three-velocity from the first diagram. And here we can see the problem with comparing three-velocities in different frames - three velocities in the red frame lie along the red spatial axis and three-velocities in the blue frame lie along the blue spatial axis. You can't compare them because they lie in different spatial planes. You can "promote" them to four-vectors (which I have kind of implicitly done by drawing them both on one diagram here), and then you can compare them, but they aren't parallel and boosting the blue dotted line to be parallel to the blue axis would not yield blue's three velocity in this frame (a problem @vanhees71 alluded to in #35).

So, the "reciprocity" of the three-velocities each observer assigns to the other in their respective rest frames just means that we can draw the first two Minkowski diagrams and they are mirror-symmetric. It follows from the symmetry of the situation - it must be true unless there's some difference between the red and blue observers, and there isn't. However, the three velocities don't lie in the same space and can't be meaningfully compared as vectors - you are, of course, free to compare their magnitudes.

I hope that's helpful, anyway.

 

[robphy]

It might be helpful to repeat this for Euclidean case, again to emphasize that (in this geometric view) the Galilean geometry is the special (atypical) case.

 

[Ibix]

It might be helpful to repeat this for Euclidean case, again to emphasize that (in this geometric view) the Galilean geometry is the special (atypical) case.

You mean, two intersecting lines shown on two rotated maps with dropped perpendiculars? I wonder if I can just replace the Lorentz transforms in my code...

 

[Ibix]

So that was simple: add a flag saying to use Minkowski or Euclidean geometry and change the transform function to look at the flag and use the appropriate coordinate transform. I'm almost tempted to change the Lorentz transform code to use cosh and sinh instead of ##v## and ##\gamma## just for the elegance, but it ain't broke so I'm going to resist the urge to fix it.

Here are the three graphs using Euclidean geometry. In this case, "horizontal lines" are the frame-dependent phenomenon and you can't compare "the x component of the direction vector of the line" from the two frames without doing something like promoting the line to a two-vector (but it still can't be interpreted as "the horizontal component of the line" in any frame except the one we started in). Note that, apart from the axis labels, the first two diagrams appear identical to the first two diagrams in my previous post #62.

I'm wondering if I can add a third option that does Galilean transforms now. Would that be called Cartan geometry?

 

[Ibix]

Well, whatever it's called, I went ahead and did it. So here are the equivalent Newtonian displacement-time graphs, if Newtonian physics were remotely valid at 0.6c. Note how, once again, the first two diagrams are identical to the first two diagrams in my recent posts. It's only the third diagram where things differ - and in this case the three-velocities do share a plane and can be directly compared in a meaningful way.

Note the messy horizontal axis labelling in the third diagram - this is because the ##x## and ##x'## axes coincide in this version.

 

[robphy]

I'm wondering if I can add a third option that does Galilean transforms now. Would that be called Cartan geometry?

I would call it Galilean geometry.
(Newton-Cartan would also be okay but maybe more appropriate... if considering constant curvature space[times].)

Check out my Spacetime diagrammer
for a variable parameter “E”
https://www.desmos.com/calculator/kv8szi3ic8

 

[Ibix]

Check out my Spacetime diagrammer
for a variable parameter “E”
https://www.desmos.com/calculator/kv8szi3ic8

That ##E## slider is neat. I couldn't see documentation (I found lots for the Desmos tool, but not for your application), but I think that the ##E## parameter works like this: $$
\left(\begin{array}{c}x'\\t'\end{array}\right)=\frac 1{\sqrt{1-Ev^2}}\left(\begin{array}{cc}1&-Ev\\-v&1\end{array}\right)\left(\begin{array}{c}x\\t\end{array}\right)$$So ##E=1## gives you the Lorentz transforms and ##E=-1## gives you Euclidean rotations. Is there a name for the "intermediate" geometries when ##-1<E<1##?

 

[robphy]

That ##E## slider is neat. I couldn't see documentation (I found lots for the Desmos tool, but not for your application), but I think that the ##E## parameter works like this: $$
\left(\begin{array}{c}x'\\t'\end{array}\right)=\frac 1{\sqrt{1-Ev^2}}\left(\begin{array}{cc}1&-Ev\\-v&1\end{array}\right)\left(\begin{array}{c}x\\t\end{array}\right)$$So ##E=1## gives you the Lorentz transforms and ##E=-1## gives you Euclidean rotations. Is there a name for the "intermediate" geometries when ##-1<E<1##?

Everything in desmos is a definition of a slider, point, or function or an equation... not necessarily presented in any order... they are simultaneous equations. One just has to open everything up and track everything down, unfortunately.

There are really three cases, +1, 0, -1.
The fractional values are really just scaling factors (like varying the maximum signal speed) that interpolate the three cases.

 

[Ibix]

There are really three cases, +1, 0, -1.

Ah! In that case I have my matrix transposed, and it should be$$
\left(\begin{array}{c}x'\\t'\end{array}\right)=\frac 1{\sqrt{1-Ev^2}}\left(\begin{array}{cc}1&-v\\-Ev&1\end{array}\right)\left(\begin{array}{c}x\\t\end{array}\right)$$Then ##E=1## corresponds to Lorentz, ##E=0## corresponds to Galileo, and ##E=-1## corresponds to Euclid. Neat.