Special relativity - frame of reference

[m4r35n357]

Length contraction must be accounted for when transforming densities between frames.

I don't think there are that many SR "newbies" who are looking to transform densities ;) As I said, that other stuff can all be done after the LT and spacetime interval are assimilated, as these enable the student to check their own answers and understanding.

 

[Orodruin]

I would strongly disagree about TD and LC. We are witness pretty much daily here to casualties of that approach. You end up with people who don't know what (me or you) is dilated, think that clocks really change their speed, and try to do SR and GR problems with Newtonian equations and γγ\gamma.

This depends a lot on how it is presented and in what context. I agree that, if it is done improperly, it will certainly lead to issues. With such an approach it is necessary that the relativity of simultaneity is treated properly in connection to it. The fact remains that you can derive the Lorentz transformations using these arguments and that you need to delve a bit deeper to understand why length contraction and time dilation works both ways.

That is the motivation behind my approach above. It gives the right answers (eg. the twin , is nothing more than three spacetime intervals) most simply, which enables the "tricky" stuff like TD, LC to be postponed (perhaps indefinitely, as they are rarely used to calculate anything but muons!) until students have the mathematics to see what they really mean (again, not much IMO), and to get the right answers for themselves.

I think omitting it altogether is to do the students a disservice. You can bet your hat on at least 90% of the students having read about it before and those students will likely either feel cheated you did not treat it or maintain their misconceptions because you did not (or both).

Length contraction must be accounted for when transforming densities between frames.

No, it must not. Take energy density for example, what needs to be considered is its proper interpretation as a component of the energy-momentum tensor. For charge density you must properly take into account its interpretation as the time-component of the 4-current density.

 

[m4r35n357]

I think omitting it altogether is to do the students a disservice. You can bet your hat on at least 90% of the students having read about it before and those students will likely either feel cheated you did not treat it or maintain their misconceptions because you did not (or both).

I said perhaps ;) You could make it a learning point at the end of a course to explain how various "pop science" scenarios really work, maybe.

 

[SiennaTheGr8]

No, it must not. Take energy density for example, what needs to be considered is its proper interpretation as a component of the energy-momentum tensor. For charge density you must properly take into account its interpretation as the time-component of the 4-current density.

I don't quite follow. Doesn't transforming a volume-density necessarily involve transforming a volume?

(If this is too off-topic, another thread would be fine with me.)

 

[Orodruin]

I don't quite follow. Doesn't transforming a volume-density necessarily involve transforming a volume?

(If this is too off-topic, another thread would be fine with me.)

This opens a whole new can of worms (such as the volume being a 3-form and that volume in the new frame not being a simultaneous volume) and is probably better suited for a new thread. Long story short is that a density is essentially the time component of a current. This is true also in classical mechanics.

 

[vanhees71]

To make the connection to 3-volume elements in this connection, you have to properly define them as hyper-surface elements and the proper "hypersurface-element normal", and all mysteries wrt. densities and the corresponding integrated quantities (charges) are gone, but I'd also say that's not a topic for the introductory SRT lecture (which at my university is taught in the 1st semester at the end of the theory 1 course lecture about classical mechanics, which is of course about point particles and no continuum mechanics or field theories yet; this comes only in the theory 3 course lecture about classical electrodynamics.

 

[David Lewis]

…which enables the "tricky" stuff like TD, LC to be postponed (perhaps indefinitely, as they are rarely used to calculate anything but muons!)

There will be some students don't plan to use their knowledge of Relativity Theory to do anything useful except pass tests. They're in it for personal enrichment and because they want to be well educated. Now if you are training professional scientists then the focus is on practical application.

 

[Dale]

There will be some students don't plan to use their knowledge of Relativity Theory to do anything useful except pass tests. They're in it for personal enrichment and because they want to be well educated.

In this case I think “well educated” is to be achieved the same way for both students that plan to use it later and for students who will not pursue it further.

 

[vanhees71]

There will be some students don't plan to use their knowledge of Relativity Theory to do anything useful except pass tests. They're in it for personal enrichment and because they want to be well educated. Now if you are training professional scientists then the focus is on practical application.

A good test should lead to let them fail... I now, I'm a bad guy, but any physicist who has not understood special relativity to some extent should not be able to earn a degree in physics. That's in the interest of everybody who wants to get a physicist since then the high reputation of physics degrees worldwide is also guaranteed in the future!

 

[pervect]

I don't quite follow. Doesn't transforming a volume-density necessarily involve transforming a volume?

(If this is too off-topic, another thread would be fine with me.)

More if you start a new thread, but I suggest drawing a space-time diagram of the 1-space + 1-time case. Draw the 1-d volume element for a stationary observer on the 2d (1+1) space-time diagram. The "volume" element here is just a line segment because there's only one spatial dimension in this case. Then draw the 1-d volume element for a moving observer, and note that the volume element of the stationary and moving observer are not the same set of points. It's not just length contraction, one must include the relativity of simultaneity. The 2-space + 1 time can be visuzlized with a bit more effort, above that and I think visualization is too difficult and would suggest an abstract approach.

If the diagrams would be helpful and you do start another thread, I could draw them, but I sincerely believe eople have better luck with diagrams they draw themselves.

 

[vanhees71]

Easier than drawing a diagram is to know that a density (e.g., electric charge per volume of a relativistic plasma) is always accompagnied by the correspondin current density. Both together build a Lorentz-vector field,
$$j^{\mu}(x)=(c \rho(x),\vec{j}(x)).$$
A "three-volume" is in fact a 3D space-like hypersurface with elements with a Lorentz-normal vector $\mathrm{d}^3 \sigma_{\mu}$. A space-like hypersurface is defined to have time-like hypersurface-element vectors only. The charge in this hypersurface is then invariant
$$Q_{\sigma}=\int_{\sigma} \mathrm{d}^3 \sigma_{\mu} \frac{1}{c} j^{\mu}.$$
The "naive" total charge
$$Q_{\text{tot,naive}}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(t,\vec{x})$$
is a Lorentz scalar only if it is conserved, i.e., if the continuity equation
$$\partial_{\mu} j^{\mu}=0$$
holds.

For a thorough discussion of this, see Jackson, Classical Electrodynamics.

 

[Orodruin]

Easier than drawing a diagram is to know that a density (e.g., electric charge per volume of a relativistic plasma) is always accompagnied by the correspondin current density. Both together build a Lorentz-vector field,
$$j^{\mu}(x)=(c \rho(x),\vec{j}(x)).$$
A "three-volume" is in fact a 3D space-like hypersurface with elements with a Lorentz-normal vector $\mathrm{d}^3 \sigma_{\mu}$. A space-like hypersurface is defined to have time-like hypersurface-element vectors only. The charge in this hypersurface is then invariant
$$Q_{\sigma}=\int_{\sigma} \mathrm{d}^3 \sigma_{\mu} \frac{1}{c} j^{\mu}.$$
The "naive" total charge
$$Q_{\text{tot,naive}}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(t,\vec{x})$$
is a Lorentz scalar only if it is conserved, i.e., if the continuity equation
$$\partial_{\mu} j^{\mu}=0$$
holds.

For a thorough discussion of this, see Jackson, Classical Electrodynamics.

I want to stress again that this is not just the case in relativity, but also in classical mechanics - although it is perhaps not as evident there, since the time-component of 4-vectors are invariant under Galilei transformations. The current is still a spacetime 4-vector, but the transformation rule is instead based on the Galilei transformation
$$
\rho' = \rho, \quad \vec j' = \vec j - \vec v \rho.
$$
The current integrations over different spacetime hypersurfaces works exactly the same as in relativity, it is just that their relations between their interpretation in different frames are different due to there being an absolute simultaneity concept and the volume element is therefore the same between frames (as it should, the density does not transform!) and it is much more important how the spatial parts transform. The continuity equation can still be written on the form $\partial_\mu j^\mu = \kappa$, where $\kappa$ is the source.

The charge in classical mechanics is a scalar because simultaneity is well defined.

 

[SiennaTheGr8]

To be clear, I'm aware that charge density is the temporal component of the four-current-density.

What I was thinking of (but not expressing well) is the relation $V = V_0 / \gamma$ between a volume $V$ and the corresponding proper (rest-frame) volume $V_0$ (because of length contraction along the axis of observers' relative motion). This is relevant to the relation between a charge density and the corresponding proper charge density.

 

[Orodruin]

To be clear, I'm aware that charge density is the temporal component of the four-current-density.

What I was thinking of (but not expressing well) is the relation $V = V_0 / \gamma$ between a volume $V$ and the corresponding proper (rest-frame) volume $V_0$ (because of length contraction along the axis of observers' relative motion). This is relevant to the relation between a charge density and the corresponding proper charge density.

This is not due to length contraction. It has to do with the Lorentz transformation of the 4-current density.
$$
\rho' = \gamma(\rho - v \vec j).
$$
Letting $\vec j = 0$ defines the zero-current frame and therefore
$$
\rho' = \gamma \rho_0.
$$
To be clear, the only reason your heuristic works is because the current is conserved and there is no current in the frame where your charges are at rest. This has the same pitfalls as length contraction, in that what defines the length of an object in a frame where it is moving are not simultaneous events in the frame where it is moving.

 

[SiennaTheGr8]

I understand that the Lorentz transformation of the 4-current density leads to the same result, and I understand that "length" is a concept that must be handled carefully because of the relativity of simultaneity.

What I don't understand is how the length-contraction argument is "only" a heuristic. Have I been misled by (or simply misunderstood, more likely) several textbooks?

 

[Grinkle]

length-contraction argument is "only" a heuristic

To me calling it a heuristic and jumping straight to a calculation for it means that the more complete way of thinking about it requires an understanding of relativity of simultaneity.

There is a bit of judgement involved imo when one calls something a heuristic.

 

[pervect]

To be clear, I'm aware that charge density is the temporal component of the four-current-density.

What I was thinking of (but not expressing well) is the relation $V = V_0 / \gamma$ between a volume $V$ and the corresponding proper (rest-frame) volume $V_0$ (because of length contraction along the axis of observers' relative motion). This is relevant to the relation between a charge density and the corresponding proper charge density.

The volume element $V_0$ associated with the rest frame of an object is a different entity than the volume element V in a different frame, a frame in which the object is moving, due to the relativity of simultaneity. Writing that $V = V_0 / \gamma$ treats them as if they were the same object, but they're really not the same object in a geometrical sense. Hence the cautions.

The most elementary way I'm aware of showing this is to draw a space-time diagram. One colors in the set of points represented by the volume V in some moving frame, and contrasts it to the set of points representing the volume $V_0$ in some stationary frame. (Moving and stationary are of course regarded as a matter of convention). Then one shows that the two sets of points are different sets.

The fundamental issue here is the relativity of simultaneity, and the space-time diagram is the most elementary tool that I'm aware of that illustrates the issues associated with the relativity of simultaneity.

Of course it's difficult to draw 4-d diagrams, so I advocate reduce the number of dimensions to get the insight. I suggest starting with a 2d diagram, and then drawing (or just imagining) a 3-d diagram. The volume element in the 1+1 diagram is a line segment, in the 2+1 diagram it's a segment of a plane.

 

[vanhees71]

No, it's not only a heuristic, but in a way it specializes to very special "foliations" of spacetime, namely hypersurfaces of constant coordinate times in inertial reference frames.

The most general case is however much easier formulated in terms of covariant objects, i.e., the four-vector current and, for the corresponding integrated quantities via the covariant integration over space-like 3D hypersurfaces, which lead to scalar quantities, no matter whether you integrate a conserved or some non-conserved current.

The trouble with the integrated quantities is that they are not local quantities like the fields. That lead to very confusing problems before the full theory of relativity has been discovered and even for some years after 1905. The breakthrough was Minkowski's mathematical analysis of SRT spacetime, leading to Minkowski vectors, tensors, and all that.

The most (in)famous classic problem within SRT is the definition of (integrated) electromagnetic field energy at the presents of charge-current distributions. When integrating the energy-momentum-stress tensor naively over an arbitrary constant-coordinate-time hypersurface of an arbitrary intertial frame you get a contradiction with the total-energy-momentum relation ("on-shell condition"), but that's of course because you integrate over a non-conserved current. Only the total energy, i.e., the electromagnetic an mechanical energy and momentum are of course conserved, and you can get these quantities by naive integrating over these special hypersurfaces, and then energy and momentum build together a proper four-vector as it must be.

See Jackson, Classical Electrodynamics for a very careful treatment of these problems.

 

[Alan McIntire]

I get the following from a book by Paul Davies.

Assume Alpha Centari is exactly 4 light years away, and one twin is
traveling there at 4/5 speed of light. (Using a 3,4,5 triangle I avoid
complicated messy fraction in my computations.

Traveling at 4/5 the speed of light, from the point of view of the
stay at home twin, the trip will take 10 years, 5 years there, 5 years
back.

Time for the traveller T' = T( sqrt( 1- (v^2/c^2))) = 3/5 T
Likewise, the distance for the traveler, D' = 3/5 D

The traveler on the spaceship sees himself traveling a distance of
4*3/5 = 2 2/5 light years in a time of 3 years, and likewise the 2 2/5
light years back
in a time of 3 years, so the traveler will see the trip as lasting 6
years.

Suppose the twins have super telescopes and can see each other throughout
the trip.
As long as they are traveling apart, the twins will see each other as
aging at 1/3 speed. As long as they are traveling towards each other,
the twins will see each other as aging at triple speed.

The difference is, the traveling twin will see the stay at home twin
as aging at 1/3 speed for the 3 years to Alpha Centauri,
and at triple speed for the 3 year trip back to Earth for a total of
3* 1/3 + 3*3= 1 + 9= 10 years.
The stay at home twin will see the travel age at 1/3 speed for 9
years, the 5 years it takes the traveler to get to Alpha Centauri,
plus the 4 years it takes the light to get back to earth. Since the
total trip will take 10 years, the stay at home twin will see the
traveler age at triple speed during the one year he observes the
traveler coming back to earth. The Earth observer sees the traveler
age at 1/3 speed for 9 years, and at triple speed for 1 year, for a total of
1/3*9 + 3*1 =3+3=6 years.
Both observers see each other aging at the same slow rate while moving
apart, they see each other aging at the same fast rate while moving
together. The difference lies in one observer deliberately changes
the relative motion of his rocket from moving away from Earth to
moving towards earth, and the other observer remaining passive, and
not seeing the change until the light from the
traveler reaches earth. If the Earth could be accelerated like a
rocket ship, and the earthbound observer decided to change his frame
so the rocket appeared to be moving towards him at 4/5 lightspeed
rather that away at 4/5 lightspeed, while the rocket remained in
motion past Alpha Centauri, then it would have been the Earth twin who
appeared to age less.
Of course you could have some intermediate situation where BOTH
observers decide to change their relative motion before they see the
other observer change his motion.

Say A and B are flying apart at 12/13 the speed of light. As long as
they are receeding from each other, they will see each other as moving
at ((1 - 12/13)/(1+ 12/13))^(1/2) = 1/5 normal speed. When they
approach each other, they will
see each other as moving at 5 times normal speed.((1 + 12/13)/(1 -
12/13)^(1/2).
If they are separated by 10 light years when A changes direction, A
will see B approaching immediately, and see B aging at 5 times speed
immediately. B, who hasn't done anything to change relative motion,
will see A continue to recede at 1/5 speed, and won't see A approach
at 5 times normal speed until 10 years have passed, the 10 years it
takes light from A to reach B. Whoevher changes direction, A or B,
will IMMEDIATELY see the other change direction, and start aging more
rapidly than normal. The other party will not see the change until
the light reaches him or her. It doesn't matter who fires their
rocket or feels acceleration, but if they're ever going to meet again,
one of them must do so.
With special relativity, two observers traveling apart will see each
other appear to age at a rate slower than normal, each will see the
other age at the same slower rate. Two observers approaching will see
each other appear to age at a rate faster than normal, the same fast
rate. The DIFFERENCE is, if they do nothing, they will continue to
travel on different paths, never meeting again, so no one could tell
who is "REALLY" aging faster. One of the two will have to fire his
rocket and change direction. This is the observer who will age less.