Special Relativity and Velocity

[thequantumcat]

I can't seem to wrap my head around it: if an object is moving at speed v in frame S, and its observed to move at speed v' in S', what is the relationship?

 

[pervect]

You also need to define the speed of frame S' relative to frame S. Then (assuming you are using special relativity), you use the relativistic velocity addition formula, https://en.wikipedia.org/w/index.php?title=Velocity-addition_formula&oldid=96826716.

The case with 1 spatial dimension is simpler than the case with all three.

The relativistic velocity addition formula is different than just summing the velocities. Are you interested in just the formula, how it was derived, or some motivation as to why relativistic velocity addition is not just adding the velocities?

 

[thequantumcat]

Yes this is SR. I am confused about the actual formula. In this case, what is v, u' and u to plug into the velocity addition formula. Moreover, what exactly is the frame at "rest"? The full question is actually:

 

[PeroK]

This is homework, so you need to give it your best shot. Are you stuck on all of these questions?

 

[PAllen]

There appears to be some missing explanation of the problem context (that you didn’t quote?). Specifically, it appears you are to assume S‘ is moving at v in the x direction per S.

 

[thequantumcat]

This is homework, so you need to give it your best shot. Are you stuck on all of these questions?

I am, and no. I think if I can grasp the set up, then I'll be able to show the gamma's, and momentum and energy equations are fine.

 

[jambaugh]

Here's how I've learned to think about it. No matter what form of relativity we use, adding velocities is a matter of adding the transformation parameters we use to parameterize velocity transformations.

A velocity of an object becomes the slope (for horizontal t-axis) of it's world-line graph in space-time coordinates.

If we are using Newtonian physics (Gallilean relativity) then time is unchanged for every frame and we add velocities by adding the distance traveled an a unit of time which means adding slopes. Thus Gallilean relativity transformations are shears (imagine drawing a line on the side of a deck of cards and then fanning the deck slightly to form a parallelogram instead of a rectangle when viewed from the side.)

If you instead unified space and time in an euclidean space-time you would rather transform velocity frames by rotating. The result would be that the velocity is the tangent of the angle of the object's world-line.
Your velocity addition formula would be the slope addition formula:
v_{\alpha+\beta}/c = \tan(\alpha + \beta) =\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=\frac{v_\alpha + v_\beta}{1-v_\alpha v_\beta/c^2}

But space-time is pseudo-euclidean. We pseudo-rotate with hyperbolic trigonometric functions and a velocity as a slope is then the hyperbolic tangent of a pseudo-angle (boost parameter). Then adding velocities is adding pseudo-angles and we get the velocity addition formula:

v_{\alpha+\beta}/c = \tanh(\alpha + \beta) =\frac{\tanh(\alpha)+\tanh(\beta)}{1+\tan(\alpha)\tan(\beta)}=\frac{v_\alpha + v_\beta}{1 + v_\alpha v_\beta/c^2}

If we lived in an euclidean space-time then a constantly accelerating object would eventually loop around and travel back in time tracing out a circle in space-time as it's velocity (slope) passes 90degrees (infinite velocity) an onward.

Since we live in a pseudo-euclidean space-time constant acceleration of an object causes it to trace out a hyperbola with the limiting velocity:
v/c \to \lim_{\beta\to \infty} \tanh(\beta)=1
It also negates time travel which, I believe, is a good thing.

What we experience as Newtonian physics is the small angle limits. As \beta\to 0, \tanh(\beta)\approx \beta \approx \tan(\beta). And so v_{\alpha+\beta} \approx v_\alpha + v_\beta.

 

[thequantumcat]

There appears to be some missing explanation of the problem context (that you didn’t quote?). Specifically, it appears you are to assume S‘ is moving at v in the x direction per S.

So then is this correct:

S is at rest--inertial frame.
u is speed of object in that frame
S' is "moving observer" and its speed is v
u' is the speed of object in S' (i.e speed as seen by this frame)

In that case, the velocity formula would be:

u = (u' - v)/(1-(u'v/c^2)) ??
I am confused by this part really. Which quantity goes where?
I always get confused...

 

[PeroK]

So then is this correct:

S is at rest--inertial frame.
u is speed of object in that frame
S' is "moving observer" and its speed is v
u' is the speed of object in S' (i.e speed as seen by this frame)

In that case, the velocity formula would be:

u = (u' - v)/(1-(u'v/c^2)) ??
I am confused by this part really. Which quantity goes where?
I always get confused...

In the non-relativistic limit we have $u = u' + v$. Note that these are all velocities, not speeds.

 

[Sagittarius A-Star]

Note that these are all velocities, not speeds.

What is the difference between "speed" and "velocity"? I ask, because I am a non-native English speaker, and dictionaries translate both to the same set of words:
https://dict.leo.org/englisch-deutsch/velocity
https://dict.leo.org/englisch-deutsch/speed

 

[Ibix]

What is the difference between "speed" and "velocity"?

Velocity means the vector. Speed is its magnitude. I seem to recall reading on here that this convention was established by one textbook in the 1920s (or something like that), so this may not apply to older texts.

 

[etotheipi]

Also, what do they mean by 'massless energy' and 'massless momentum'?

 

[Ibix]

Also, what do they mean by 'massless energy' and 'massless momentum'?

What does who mean by that?

 

[etotheipi]

What does who mean by that?

In the problem in post #3, it reads "these masses interact to produce masses $\tilde{m}_i$ moving at velocities $\tilde{\mathbf{u}}_i$ with massless energy $\Delta E$ and massless momentum $\Delta \mathbf{P}$". Seems a bit odd to me... I don't quite know what that means...

 

[PeroK]

Also, what do they mean by 'massless energy' and 'massless momentum'?

I've never heard the term before, but I assume they mean energy/momentum taken away by massless particles.

 

[Ibix]

In the problem in post #3

...which is an image, which is why "find in page" only found your post. Sorry.

I don't quite know what that means...

Me neither. Energy or momentum per unit mass, perhaps? Or kinetic energy $(\gamma-1)mc^2$, although what massless momentum would be in that case I don't know.

 

[Ibix]

I've never heard the term before, but I assume they mean energy/momentum taken away by massless particles.

That makes more sense. Something like a nuclear fission process where you get energy carried away as radiation.

 

[vanhees71]

Once more, the confusion seems to be due to the use of old-fashioned ideas which were long abandoned in the physics community but still being used in popular-science writings and (which is a crime) even in some textbooks. One such things are the various relativistic masses. We had a long thread about it recently, which unfortunately ended up in the insistence of some posters to introduce even more nonsensical notions like norms which can get imaginary. I'll stick to the modern covariant formulation.

Masses are real scalar quantities. To emphasize that this one and only useful definition of mass is used, one sometimes call it invariant mass.

Another quantity which usually makes trouble is the "three velocity". This is somewhat harder to eliminate, because sometimes it's convenient to have it in the context of various density, where it can be used to build covariant densities from it, e.g., in any frame $(c \rho(t,\vec{x}), \vec{v}(t,\vec{x}) \rho(t,\vec{x}))$, where $\rho$ is a charge density (as measured in the arbitrary frame) and $\vec{v}$ is the three-velocity-field of the charges (described in a continuum-mechanical manner, which is the natural description in any field theory). is a four-vector field. We come back to the fact that this is really a four-vector field in a moment.

Now consider first point charges in special relativity. You can describe its motion covariantly by a world line, $x^{\mu}(\lambda)$, where $\lambda$ is a scalar parameter. Now the three-velocity is defined by $\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t$, which makes it a complicated non-covariant object, because it uses the arbitrary components with respect to the chosen Minkowski basis, to which the components $x^{\mu}$ refer.

For massive particles it's much more convenient to choose the proper time of the particle, which is invariant as the world-line parameter. It's defined by
$$c^2 \mathrm{d} \tau^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}$$
and a scalar. Then the socalled four-velocity
$$U^{\mu}=c u^{\mu} = \mathrm{d}_{\tau} x^{\mu}$$
are four-vector components, which transform under Lorentz transformations as any four-vector. It's easy to see that the relation between the spatial part of this "proper velocity" and the "three-velocity" is given by
$$\vec{v}=\mathrm{d}_t \vec{x} = \frac{\mathrm{d} \tau}{\mathrm{d} t} \mathrm{d}_{\tau} \vec{x}=\sqrt{1-\vec{v}^2/c^2} \vec{U} \; \Rightarrow \; \vec{U}=\gamma \vec{v}, \quad \gamma=1/\sqrt{1-\vec{v}^2/c^2}.$$
You can also write
$$(U^{\mu})=\gamma (c,\vec{v}).$$
With this the velocity-addition law becomes easy to derive. Just first consider the proper velocity. Say in some reference frame $\Sigma'$ the proper velocity of a particle is $U^{\prime '}=(c,v',0,0,0) \gamma_{v'}$ and now let this frame move with a three-velocity $\vec{w}$ in $x^1$-direction wrt. a reference frame $\Sigma$. Since $U^{\mu}$ is a four-vector it transforms as any four-vector under the appropriate (rotation-free) Lorentz transformation,
$$U^0=\gamma_w (U^{\prime 0}+w/c U^{\prime 2}), \quad U^1 = \gamma_w (U^{\prime 0}+w/c U^{\prime 0}, \quad U^2=U^{\prime 2}, \quad U^{3}=U^{\prime 3}.$$
Now the three-velocity in $\Sigma$ is given by
$$\vec{v} = c \frac{\vec{U}}{U^0}.$$
Using the above Lorentz transformation result this gives
$$v^1=c \frac{U^1}{U^0}=\frac{ \gamma_w(c U^{\prime 1}+ w U^{\prime 0})}{\gamma_w (U^{\prime 0}+w/c U^{\prime 1})}.$$
Now dividing both numerator and denominator by $U^{\prime 0}$ yields
$$v^1=\frac{v'+w}{1+ v w/c^2}.$$
Of course in this case $v^2=v^3=0$.

Now let's briefly come back to the above claim that $(j^{\mu})=(c \rho,c \vec{v} \rho)$ is a four vector. Writing it in terms of the four-velocity vield $(U^{\mu}) = \gamma (c,\vec{v})$ leads to
$$j^{\mu}=\frac{\rho}{\gamma} U^{\mu}=\rho_0 U^{\mu}.$$
it's clear that $\rho_0=\rho/\gamma$ is the charge density as measured in the instantaneous local rest frame of the charge-fluid cell. The charge is a scalar, but the volume of the fluid cell is Lorentz contracted in the frame, where this cell is moving, i.e.,
$$\mathrm{d} V=\mathrm{d} V_0/\gamma \; \Rightarrow \; \rho=\mathrm{d} Q/\mathrm{d} V = \mathrm{d} Q_0/V=\mathrm{d} Q_0/\mathrm{d} V^0 \gamma=\rho_0 \gamma.$$
One can also see this directly, because the four-current in the local rest frame is
$$(j^{\prime \mu})=(c \rho_0,0,0,0) \; \Rightarrow \; (j^{\mu})=\gamma \rho_0 (c,\vec{v})=\rho(c,\vec{v}).$$

 

[DrStupid]

Once more, the confusion seems to be due to the use of old-fashioned ideas which were long abandoned in the physics community but still being used in popular-science writings and (which is a crime) even in some textbooks. One such things are the various relativistic masses.

Where do you see relativistic masses?

 

[etotheipi]

Please may we not have yet another digression about relativistic mass

 

[vanhees71]

In #1, or how do you interpret "massless energy". But you are right, let's not digress again on nonsensical fight about old-fashioned ideas.

 

[Vanadium 50]

Where do you see relativistic masses?

A useful tool for derailing threads.

 

[DrStupid]

In #1, or how do you interpret "massless energy".

In #1 there is no mass mentioned at all. The mysterious "massless energy" first appeared in #3 and I interpret it as PeroK did in #15 - as the energy of something that has no mass (e.g. photons). I agree that the question is unclear and confusing, but not due to old-fashioned ideas. There is information missing (e.g. the meaning of $\Delta E$ and $\Delta P$) that might be given outside the citation above.

 

[DrStupid]

A useful tool for derailing threads.

I wouldn't call that "useful".

 

[vanhees71]

Sorry if I misunderstood the OP. It's indeed not in #1 but in #3. If you agree with all this confusion, could you then explain to me, what "massless energy" and "massless momentum" is? I interpreted this as if this might be a language stemming from usual wrong concepts related to mass.

 

[DrStupid]

If you agree with all this confusion, could you then explain to me, what "massless energy" and "massless momentum" is? I interpreted this as if this might be a language stemming from usual wrong concepts related to mass.

Maybe it's just me, but "massless energy" sounds like energy without mass. As the concept of mass you are referring to is equivalent to energy it would mean energy without energy. That makes no sense.

However, only @thequantumcat can tell us what the terms actually mean.

 

[vanhees71]

Sigh! I knew that it is about the oldfashioned concept of mass :-((. Mass is invariant mass (a scalar) and nothing else, and energy is energy (the temporal component of the energy-momentum four-vector). I agree that #3 doesn't make sense in the sense that it is not clear what the writer wanted to say.

 

[DrStupid]

Sigh! I knew that it is about the oldfashioned concept of mass :-((. Mass is invariant mass (a scalar) and nothing else, and energy is energy (the temporal component of the energy-momentum four-vector).

I don't know what you are talking about. In this thread I don't see any indications that "mass" and "energy" have been used for something different than invariant mass and energy - until you mentioned "relativistic masses" in #18.

 

[vanhees71]

Ok, I give up. If you can't specify what's the meaning of these strange ideas in #3, I don't know what you are talking about.

 

[PeterDonis]

As post #3 shows, this thread is about a homework problem. Homework problem questions belong in the homework forum, with the homework template filled out properly. @thequantumcat, please start a new thread in the homework forum if you have questions about a homework problem.

This thread is closed.