Relativistic addition question

[rajeshmarndi]

Observer A measures B moving east at .8c and measures C moving west at .8c . Then wouldn't the impact of the two will be a result of relative motion greater than c in frame A. I understand B and C frame will not see any object speed greater than c. So the impact in frame A and frame B will not be a result of speed greater than c. But how would frame A explain in his frame?

Thanks.

 

[sweet springs]

From the addition rule of speed,
$\frac{u+v}{1+\frac{u}{c}\frac{v}{c}}=1.6c/1.64=0.976c$.

In IFR of A, speeds are,
B:0.8c C:-0.8c

In IFR of B
A:-0.8c C:-0.976c

In IFR of C
A:0.8c B:0.976c

 

[Ibix]

Observer A measures B moving east at .8c and measures C moving west at .8c . Then wouldn't the impact of the two will be a result of relative motion greater than c in frame A. I understand B and C frame will not see any object speed greater than c.

There are two separate concepts here. First, there is the "separation rate". In the frame of A, the distance between B and C does indeed grow at 1.6c. This is fine. Neither object is exceeding c.

The other concept is the "relative velocity" which is the velocity of B as measured by C. You can't naively add velocities to get this, because C will be measuring distances and times using (according to A, anyway) length contracted rulers and time dilated clocks which are synchronised differently due to the relativity of simultaneity. Once you account for that you will find that the velocity, u', of B as measured by C is$$u'=\frac{u-v}{1-uv/c^2}$$ where u is the velocity of B and v is the velocity of C, both as measured by A. Note I'm using a different sign convention to @sweet springs - all my velocities are positive if they are in the +x direction, whereas sweet springs' version has u and v with the same sign if they are in opposite directions.

In Newtonian physics separation rate and relative velocity are the same concept, but not in relativity. Note that if u and v are a lot less than c then the denominator is very close to one, and your naive Newtonian calculation is very close to the relativistic prediction. You might like to try u and v being ±60mph to see how small the differences are at every day speeds.

 

[rajeshmarndi]

There are two separate concepts here. First, there is the "separation rate". In the frame of A, the distance between B and C does indeed grow at 1.6c. This is fine. Neither object is exceeding c.

Does this mean in frame A, C will be approaching B at 1.6c and B will be approaching C at 1.6c but none individual speed can exceed c in frame A.

Even if C approaches B at speed greater than c in frame A, but in its frame B(rest), C approaching B will never exceed c according to Lorentz transformation. Same is for frame C.

 

[Ibix]

Does this mean in frame A, C will be approaching B at 1.6c and B will be approaching C at 1.6c but none individual speed can exceed c in frame A.

I'd say both B and C are approaching their collision point at 0.8c from oppositrme sides. I'm being pedantic, but there is a very clear distinction between the concept of "changing distance between two objects" and "velocity". And it's important to use language that keeps that in mind, especially when you're learning.

Even if C approaches B at speed greater than c in frame A, but in its frame B(rest), C approaching B will never exceed c according to Lorentz transformation. Same is for frame C.

As above, I'd avoid talking about things approaching each other faster than light. Talk about their velocities in one frame or other.

 

[rajeshmarndi]

So the impact between B and C would be observed more in frame A because of the closing speed greater than c. The impact will also be observed differently by B and C if there relative velocity were different.

 

[Ibix]

So the impact between B and C would be observed more in frame A because of the closing speed greater than c. The impact will also be observed differently by B and C if there relative velocity were different.

What do you mean by "observed more"? That the energy released by the impact would be higher? No. It can't be - a frame of reference is just a choice of coordinates. It can't change how much damage a crash does.

 

[sweet springs]

So the impact between B and C would be observed more in frame A because of the closing speed greater than c.

As Ibix mentioned as

but there is a very clear distinction between the concept of "changing distance between two objects" and "velocity".

your "closing speed greater than c" is not velocity that relativity deals with. Say light sources are 300,000km away from here one right side and the other left side and emit light simultaneously. They will meet 1 second later here. The distance between the lights 600,000km / traveling time 1 second = 600,000 km/s = 2c > c. It's OK. It is not velocity. Velocity of an object is defined as the rate of change of its position with respect to a frame of reference. Relativity says such velocity of an object has an upper limit of c.

As my first post said NO velocity of A, B,C in any A,B,C frame exceed c.

 

[Imager]

It can't change how much damage a crash does.

I'm not following how is the same force of impact between the frames is explained. Assuming objects A and B have the same mass, the velocity of impact from C’s frame is at 1.6c, but from A or B frame it is less than c. I understand the force should be the same, I just don't see how?

 

[Ibix]

Assuming objects A and B have the same mass, the velocity of impact from C’s frame is at 1.6c,

I think you are confusing your objects. A sees B and C doing 0.8c in opposite directions. B sees A doing 0.8c and C doing 0.976c. But the key point is that no one is doing 1.6c.

You can't work out the force because we don't know anything about the elasticity of the materials. You can certainly work out the energy and momentum change in the collision. If you use the relativistic formulae you'll find that the change is the same in either frame.

The force will, in general, be different between frames, by the way. Although the frames agree on the total momentum change they don't agree on time, so the rate of change of momentum isn't the same.

 

[Nugatory]

'm not following how is the same force of impact between the frames is explained. Assuming objects A and B have the same mass, the velocity of impact from C’s frame is at 1.6c, but from A or B frame it is less than c. I understand the force should be the same, I just don't see how?

Assume for simplicity that B and C have the same mass $m$ and that the collision is completely inelastic so that after the collision the tangled lump formed by smashing B and C together will be at rest in their center-of-momentum frame (which is also the frame in which A is at rest). The damage done by the impact is determined by the amount of kinetic energy that was spent crushing and deforming the two objects into that one lump; this will be the difference between the kinetic energy before the collision and the kinetic energy after the collision.

Do the calculation in the frame in which A is at rest and the initial kinetic energy will be $2m(\gamma(.8)-1)c^2$ for two masses moving at .8c and the final kinetic energy will be zero for one lump with mass $2m$ at rest.

Do the calculation in a frame in which B or C are at rest before the collision and the initial kinetic energy will be $m(\gamma(.976)-1)$ for one mass moving at .976c and the final kinetic energy will be $2m(\gamma(.8)-1)$ for one lump with mass $2m$ moving at speed .8c.

 

[stevendaryl]

I'm not following how is the same force of impact between the frames is explained. Assuming objects A and B have the same mass, the velocity of impact from C’s frame is at 1.6c, but from A or B frame it is less than c. I understand the force should be the same, I just don't see how?

Well, what causes damage is the magnitude of the proper acceleration in a collision. The proper acceleration is the acceleration as measured in a frame in which the object is momentarily at rest.

 

[SiennaTheGr8]

The force will, in general, be different between frames, by the way. Although the frames agree on the total momentum change they don't agree on time, so the rate of change of momentum isn't the same.

Wait, isn't it the other way around?

The "total momentum change" would be $\Delta \mathbf{p} = m \Delta ( \gamma \mathbf{v} )$. Isn't its magnitude frame-dependent even in the rectilinear case under discussion? Differences in rapidity $\Delta \phi$ are invariant in the rectilinear case (under Lorentz boosts along the relevant axis), but differences in celerity $\Delta (\gamma v) = \Delta \sinh{\phi}$ aren't (since, e.g., $\sinh{3} - \sinh{2} \neq \sinh{2} - \sinh{1}$).

However, I do believe that the three-force $\mathbf{f} = \dot{\mathbf{p}}$ has an invariant magnitude in the constant-mass rectilinear case (again, under boosts along the relevant axis):

$\dot{p} = m \dfrac{d}{dt} \sinh{\phi} = m \cosh{\phi} \, \dfrac{d \phi}{d t} = m \dfrac{d \phi}{d \tau}$.

(Or maybe I've misunderstood you!)

 

[Mister T]

Even if C approaches B at speed greater than c in frame A, [...]

C approaches B at a speed of about $0.976\ c$. This is not altered in any way by the fact that you have inserted A between them. Or by the fact that you used observations made by A to reach this conclusion.

 

[Ibix]

(Or maybe I've misunderstood you!)

I was simply applying conservation of momentum and conservation of energy to the whole system. All frames must agree that momentum is conserved and all frames must agree about the difference in total energy (kinetic plus rest) before and after, since that could be extracted to do work.

I agree that frames may not agree on the change of momentum of individual masses. Regarding forces, there is a non-trivial force transformation equation (see http://www.sciencebits.com/Transformation-Forces-Relativity), which was what I based my claim on. Can't immediately see a flaw in your reasoning, though. I'll have a think when it's not the end of a too-long Friday.

 

[SiennaTheGr8]

... all frames must agree about the difference in total energy (kinetic plus rest) before and after, since that could be extracted to do work.

Hm. I don't think this is so. For a fixed-mass system, $\Delta E = \Delta E_k = E_0 \Delta \gamma$, and $\Delta \gamma$ isn't invariant. Even in the rectilinear / standard-boost case, we have invariant $\Delta \phi$, but not invariant $\Delta \cosh{\phi} = \Delta \gamma$. (This is obviously similar to what I wrote above about three-momentum and celerity.)

Regarding forces, there is a non-trivial force transformation equation (see http://www.sciencebits.com/Transformation-Forces-Relativity), which was what I based my claim on. Can't immediately see a flaw in your reasoning, though. I'll have a think when it's not the end of a too-long Friday.

The equation $F^\prime_x = F_x$ at the very bottom of your link corresponds to what I was getting at.

 

[SiennaTheGr8]

In general, the component of 3-force that's parallel to a particle's 3-velocity is equal to the corresponding component of the particle's proper force (proper force being the 3-force as measured in the particle's instantaneous rest frame). This component is therefore invariant under a boost along the axis of the particle's motion (which may vary from moment to moment).

A special case is when this component of 3-force is the only non-zero component (i.e., $\mathbf{f} \parallel \mathbf{v}$). Then the 3-force magnitude itself is Lorentz-invariant, and not just under a boost in a special direction.

I think.

 

[Ibix]

Sigh. @SiennaTheGr8 is correct. The initial four momentum and the final four momentum and (hence) the difference are invariant. Neither the energy nor the three momentum, being components of the four momentum, are invariant. Although each is independently conserved.

The force is, in general, different between frames. But in the case where everything is colinear it isn't.

I did not have a good day on Friday.

 

[SiennaTheGr8]

I did not have a good day on Friday.

Sorry to hear that!

One thing I still haven't cleared up for myself is whether in the colinear case the 3-force (and for that matter the change in rapidity) is invariant only under a colinear boost, or whether it's invariant under any boost.

 

[stevendaryl]

Sorry to hear that!

One thing I still haven't cleared up for myself is whether in the colinear case the 3-force (and for that matter the change in rapidity) is invariant only under a colinear boost, or whether it's invariant under any boost.

I'm not sure what you mean by that.

If everything is colinear, then we can do everything using just 2D spacetime. Let $x$ be a coordinate for distances measured in the direction of motion. Then we have:

• Position: $X$ where $X^0 = t, X^1 = x$
• 4-velocity $V = \frac{dX}{d\tau}$ where $V^0 = \frac{dX^0}{d\tau}$ and $V^1 = \frac{dX^1}{d\tau}$. In terms of 3-velocity $v$, $V^0 = \gamma$ and $V^1 = \gamma v$
• 4-acceleration $A = \frac{dV}{d\tau}$ where $A^0 = \frac{dV^0}{d\tau}$ and $A^1 = \frac{dV^1}{d\tau}$

Since $V \cdot V \equiv c^2 (V^0)^2 - (V^1)^2 = c^2$, we can immediately relate $A$ and $V$:

$\frac{d}{d\tau} (V \cdot V) = 0 \Rightarrow A \cdot V = 0$

This implies that $c^2 A^0 V^0 = A^1 V^1$. So in terms of 3-velocity, this means $A^0 = \frac{v}{c^2} A^1$

Under a Lorentz transformation,

$(A^1)' = \gamma_u (A^1 - u A^0) = \gamma_u (1 - \frac{uv}{c^2}) A^1$
$(A^0)' = \gamma_u (A^0 - \frac{u}{c^2} A^1) = \gamma_u (\frac{v - u}{c^2}) A^1$

Since 3-force is just $m A^1$, it isn't invariant.

 

[DrGreg]

Since 3-force is just $m A^1$

But it isn't. The 3-force is $dp/dt$ whereas $mA^1 = dp/d\tau$ (for constant $m$).

 

[stevendaryl]

But it isn't. The 3-force is $dp/dt$ whereas $mA^1 = dp/d\tau$ (for constant $m$).

For some reason, I mistakenly remembered someone defining 3-force to be the spatial components of the 4-force.

Okay, then the 3-force $f$ would be related to the spatial component of the 4-force, $F^1$ through:

$F^1 = \gamma f$

But that's still not invariant under Lorentz transformations, either.

 

[SiennaTheGr8]

For some reason, I mistakenly remembered someone defining 3-force to be the spatial components of the 4-force.

Okay, then the 3-force $f$ would be related to the spatial component of the 4-force, $F^1$ through:

$F^1 = \gamma f$

But that's still not invariant under Lorentz transformations, either.

Your $f$ is invariant in the colinear case, at least under boosts along the colinear axis (I'm not sure about arbitrary boosts, but maybe?). See my post #13.

 

[stevendaryl]

Ah! That's a new fact for me. In the case of 1 spatial dimension (or any number, as long as everything is collinear), the transformation for the spatial component of 4-acceleration is:

$(A^1)' = \gamma_u (A^1 - u A^0) = \gamma_u A^1 (1-\frac{uv}{c^2})$

The transformation for the time component of 4-velocity is:

$(V^0)' = \gamma_u (V^0 - \frac{u}{c^2} V^1) = \gamma_u (\gamma -\frac{u}{c^2} \gamma v) = \gamma \gamma_u (1-\frac{uv}{c^2}) =\gamma_u V^0 (1-\frac{uv}{c^2})$

So dividing them gives: $\frac{(A^1)'}{(V^0)'} = \frac{A^1}{V^0}$

That's interesting. Is there some direct way to understand why that works out?

 

[SiennaTheGr8]

That's interesting. Is there some direct way to understand why that works out?

I guess it's just canceling Lorentz factors: one upstairs from the celerity-derivative, one downstairs from time dilation, leaving only invariants (though again, I'm struggling over whether the rapidity differential is invariant only under a boost along the colinear axis—perhaps the relevant rapidity-component differential is invariant, period?).

 

[Ibix]

That's interesting. Is there some direct way to understand why that works out?

Four acceleration and four velocity are orthogonal and the modulus of the four velocity is one:$$\begin{eqnarray}
0&=&A^0V^0-A^1V^1\\
1&=&V^0V^0-V^1V^1\end{eqnarray}$$Eliminating $V^1$ yields $$\left(A^1\right)^2=\left(V^0\right)^2\left(\left(A^1\right)^2-\left(A^0\right)^2\right)$$which tells us that your ratio is $i$ times the modulus of the four-acceleration, and therefore manifestly invariant. It also tells me that the four-acceleration is spacelike, which I didn't know. Or that I've made a sign error, which is always possible.

 

[DrGreg]

the four-acceleration is spacelike

That's always true because it's orthogonal to the timelike 4-velocity. The reason why it's orthogonal is because the 4-velocity has constant length -- differentiate $\mathbf{V} \cdot \mathbf{V}= c^2$ with respect to proper time.

 

[Ibix]

That's always true because it's orthogonal to the timelike 4-velocity. The reason why it's orthogonal is because the 4-velocity has constant length -- differentiate $\mathbf{V} \cdot \mathbf{V}= c^2$ with respect to proper time.

Indeed - I can always pick V to be the time-like basis vector, and then non-null vectors orthogonal to it are clearly candidates for a spacelike basis vector. I just hadn't put two and two together.

 

[SiennaTheGr8]

In case anyone's interested, I returned to the question of "longitudinal" invariance with respect to three-force (via rapidity).

First: under a Lorentz boost, changes (or differences) in the component of rapidity that's parallel to the boost are invariant. For example, if the boost is along the $x$-axis, then $\Delta \phi_x = \Delta (\tanh^{-1} v_x)$ is invariant ($c=1$).

Now, as for three-force, we have the following vector equation ($c=1$, $\dot{m} = 0$):

$\vec f = \dfrac{m}{\cosh{\phi}} \, \dfrac{d}{d \tau} \left( \sinh{\phi} \, \hat{v} \right) = m \left( \dfrac{d \phi}{d \tau} \hat{v} + \tanh{\phi} \dfrac{d \hat{v}}{d \tau} \right).$

If I'm not mistaken, the component vector of $\vec f$ that's parallel to the velocity $\vec v$ would then be:

$\vec f_{\parallel \vec v} = m \left[ \dfrac{d \phi_{\parallel \vec v}}{d \tau} \hat{v} + \tanh{\phi_{\parallel \vec v}} \left( \dfrac{d \hat{v}}{d \tau} \right)_{\parallel \vec v} \right].$

(Did I do that right? I'm unsure about the $\phi_{\parallel \vec v}$'s.)

But the component vector $( d \hat{v} / d \tau )_{\parallel \vec v}$ is necessarily a zero vector, since the derivative of a vector of fixed magnitude (like $\hat{v}$) must be perpendicular to the differentiated vector. Thus:

$\vec f_{\parallel \vec v} = m \dfrac{d \phi_{\parallel \vec v}}{d \tau} \hat{v}$

$f_{\parallel \vec v} = m \dfrac{d \phi_{\parallel \vec v}}{d \tau}$.

The $m$ and $d \tau$ are "fully" invariant, and the $d \phi_{\parallel \vec v}$ is "longitudinally" invariant (i.e., invariant under a boost in the $\pm \hat{v}$ direction, as discussed above). Therefore the component of three-force that's parallel to the three-velocity is likewise "longitudinally" invariant. Of course, in the special case that this is the only non-zero component of three-force, the three-force magnitude itself is invariant under a longitudinal boost:

$f = m \dfrac{d \phi}{d \tau}$.

 

[SiennaTheGr8]

If I'm not mistaken, the component vector of $\vec f$ that's parallel to the velocity $\vec v$ would then be:

$\vec f_{\parallel \vec v} = m \left[ \dfrac{d \phi_{\parallel \vec v}}{d \tau} \hat{v} + \tanh{\phi_{\parallel \vec v}} \left( \dfrac{d \hat{v}}{d \tau} \right)_{\parallel \vec v} \right].$

(Did I do that right? I'm unsure about the $\phi_{\parallel \vec v}$'s.)

Well, certainly $\tanh{\phi_{\parallel \vec v}}$ was redundant, since that's just $\tanh{\phi} = v$ (duh!). But what about that infinitesimal $d \phi_{\parallel \vec v}$? Seems right to me, but again I'm not quite sure.

 

[SiennaTheGr8]

And now I'm thinking I was overthinking this, and that it should just be $\phi$ every time:

$\vec f = \dfrac{m}{\cosh{\phi}} \, \dfrac{d}{d \tau} \left( \sinh{\phi} \, \hat{v} \right) = m \left( \dfrac{d \phi}{d \tau} \hat{v} + v \dfrac{d \hat{v}}{d \tau} \right).$

$\vec f_{\parallel \vec v} = m \left[ \left( \dfrac{d \phi}{d \tau} \hat{v} \right)_{\parallel \vec v} + \left( v \dfrac{d \hat{v}}{d \tau} \right)_{\parallel \vec v} \right]$

$\vec f_{\parallel \vec v} = m \left( \dfrac{d \phi}{d \tau} \hat{v} + \vec 0 \right),$

so:

$f_{\parallel \vec v} = m \dfrac{d \phi}{d \tau},$

invariant under a boost in the $\pm \hat{v}$ direction (though in general $\hat{v}$ may rotate as the force is applied).

If that's right, then I guess I was getting myself confused earlier by the fact that $\Delta \phi$ isn't in the $\pm \hat{v}$ direction. That felt off (still kind of does).

 

[SiennaTheGr8]

If that's right, then I guess I was getting myself confused earlier by the fact that $\Delta \phi$ isn't in the $\pm \hat{v}$ direction. That felt off (still kind of does).

And now I'm very confident that it's right. The key point is that $\vec v$ is (at least) momentarily parallel to the boost axis, and so the infinitesimal $d \phi$ is indeed invariant under a collinear boost. Nothing feels off anymore, and I was wrong to bring the component $\phi_{\parallel \vec v}$ into this at all.

(Sorry if I've derailed the thread. That's my last word on this unless someone responds.)