# Do a comet's tails ever point in the same direction when in an orbit?

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1. Jul 2, 2013

### s3a

1. The problem statement, all variables and given/known data
The problem and its solution are attached as TheProblemAndSolution.jpg.

2. Relevant equations
*The (“instantaneous”) dust tail is opposite the direction of motion.
*The plasma tail is oppositely directed to the star in question (at all times).

3. The attempt at a solution
I know that the (“instantaneous”) dust tail is opposite the direction of motion (so, the (“instantaneous”) dust tail is opposite the direction of the velocity vector of the comet) and that the plasma tail is oppositely directed to the star in question (at all times). I also know that for both tails to be (“instantaneously”) pointing in the same direction (their “instantaneous” part at least – I say this since the dust tail is curved if you don't look at it “instantaneously”), the comet must be moving directly toward the star in question.

I'm posting this thread for a simple question (and related sub-questions): When in (an elliptical) orbit, do the comet's tails ever point in the same direction, if ever? My logic and intuition tell me that this never happens in an (elliptical) orbit and that the only way for the comet's tails to be pointing in the same direction is for the comet to not be in an orbit (and be moving directly toward the star in question).

Is what I'm thinking correct?

If something I said is unclear, tell me and, I will attempt to clarify it.

Any input would be greatly appreciated!

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2. Jul 2, 2013

### Staff: Mentor

Since the plasma tail always points along the radius vector, and the dust tail along the (negative) velocity vector, the only time the tails could coincide would be if the velocity and radius vectors coincide. What would that say about the angular momentum of the orbit?

3. Jul 2, 2013

### s3a

I can't think of a case where the velocity vector coincides with the “radius” (or distance from the center since it's en elliptical orbit) vector. As I asked in my opening post, does that mean that “the only way for the comet's tails to be pointing in the same direction is for the comet to not be in an orbit (and be moving directly toward the star in question)”?

I would say that it varies along the orbit by ω = ν/r though, I don't see the relevance with this and my opening post.

4. Jul 2, 2013

### Staff: Mentor

I'm attempting to provide you with a mathematical argument with which to back up your point

What do you know about the angular momentum of an orbit? Does it really vary?

If not (and I hope you agree that angular momentum is a conserved quantity ), and if the only way the tails can line up is for velocity and radius vectors to be collinear and thus angular momentum = 0, then any orbit with angular momentum cannot have tails which can coincide.

5. Jul 3, 2013

### s3a

Edit:
Make sure to see the edited part further below. I didn't edit what I wrote to emphasize the change in case you're reading what I submitted before an edit (such as when reading an email alert).

Oh, yes, momentum IS conserved (since no external force is applied to the system). :D

I appreciate it. :)

So, to rephrase it myself, for a comet's tails to align, we need the velocity and radius vectors to be collinear so, if they're collinear, that means that momentum must be conserved but, if momentum is conserved, we know that the velocity and radius vectors are always 90 degrees apart therefore, there is a contradiction between our hypothesis (=the vectors being collinear) and our conclusion (=the vectors always being 90 degrees apart) therefore, a comet's tails never align in an orbit (of any kind).

Is the explanation I just gave perfectly correct?

Edit:
Oops, that's for a circular orbit.

Let me change:
"So, to rephrase it myself, for a comet's tails to align, we need the velocity and radius vectors to be collinear so, if they're collinear, that means that momentum must be conserved but, if momentum is conserved, we know that the velocity and radius vectors are always 90 degrees apart therefore, there is a contradiction between our hypothesis (=the vectors being collinear) and our conclusion (=the vectors always being 90 degrees apart) therefore, a comet's tails never align in an orbit (of any kind)."

to

"So, to rephrase it myself, for a comet's tails to align, we need the velocity and radius vectors to be collinear so, if they're collinear, that means that momentum must be conserved but, if momentum is conserved, we know that the velocity and radius vectors are always more than 0 degrees and less than 180 degrees apart therefore, there is a contradiction between our hypothesis (=the vectors being collinear) and our conclusion (=the vectors always being more than 0 degrees and less than 180 degrees apart) therefore, a comet's tails never align in an orbit (of any kind).".

Last edited: Jul 3, 2013
6. Jul 3, 2013

### Staff: Mentor

You have to be a bit careful about stating that something is a consequence of something else when it's really not. Conservation of (angular) momentum doesn't occur because radius and velocity vectors align; one does not follow due to the other. It is simply a fact that angular momentum is a conserved quantity for all orbits (that is, for the classic two-body system isolated from external influences). It is a constant of the motion.

What you can say is, "if they are collinear then angular momentum must be zero for the orbit. Since angular momentum is a conserved quantity, the only comet orbit for which this can happen is one with angular momentum equal to zero. It cannot happen for any orbit with nonzero angular momentum (circular, elliptical, parabolic, hyperbolic)."
Actually, zero is a perfectly good value for angular momentum! The problem is, the "orbit" is then a degenerate orbit --- a straight line which will either escape and never return or plunge into the Sun and disappear for good on its first inbound approach! Only a comet heading straight into the Sun could have tails that point in the same direction, and it would be a one-time only event.

So I suppose that it is possible for that one case of the degenerate orbit. All other orbits have non-zero angular momentum, so they cannot exhibit the effect.

7. Jul 4, 2013

### CWatters

The ion/plasma tail is effected by the suns magnetic field lines. If the comet flew very close to the sun then presumably the tails could point the same way (briefly?) due to fluctuations in the field.

http://www.upi.com/Science_News/201...l-reveals-solar-magnetic-field/3581370612691/

I'm not sure if a dust tail would survive that close though.

Another comet with a wiggly tail..
http://science1.nasa.gov/media/medialibrary/2007/10/01/01oct_encke_resources/movie_short.gif

Last edited by a moderator: Sep 25, 2014
8. Jul 6, 2013

### s3a

Oh, so, basically, I confused angular momentum with the change in angular momentum, right? (Also, thanks for specifying additional, specific information like specifying that the orbit can be circular, elliptical, parabolic or hyperbolic.)

At least in this case, is something that is degenerate something that is in two (or more) states of configuration, for example, the configuration is that the comet is following both a linear path as well as an orbital one (of magnitude 0)?

According to the sources you gave, it seems to me that as long as the comet is close but outside of the Sun's atmosphere, the comet's plasma tail exhibits a behaviour like the one I am describing but, when it reaches deep into the Sun's atmosphere, the behaviour is unknown to humankind. Or, is what I'm saying completely off? If the Sun's magnetic field were perfectly uniform then, my hypothetical/theoretical scenario would have the plasma and dust tails always pointing in the same direction if the comet were headed directly toward the centre (of mass) of the Sun (as opposed to having the plasma tail fluctuate), right?

9. Jul 6, 2013

### Staff: Mentor

Perhaps. You should know that the angular momentum of an isolated system (such as two-body orbiting system) is a constant. It cannot change without external influence.
I don't know what "states of configuration" means in this context; it doesn't appear to me to have a technical meaning here. A degenerate orbit implies that it has lost one or more degrees of freedom, usually due to one or more of the constants of the motion (like angular momentum or total mechanical energy) vanishing; it's a one-dimensional rather than a two-dimensional geometry, so it's "lost one dimension" from the general case of gravitational orbits for the two-body system.
The Sun's magnetic field extends well outside the Sun's atmosphere. Well outside the planetary system, in fact. Of course it grows weaker as distance grows. You would have to compare the strength of the fields at a given location with other influences that are driving the effect. It's possible that random fluctuations in the magnetic field, due to solar activity perhaps, could have an influence, but the problem statement uses the term "usually" in its setup, so I think that you can ignore low probability or unusual circumstances in the background. Magnetic field geometry can be very complex indeed for a fluctuating, rotating source that interacts with lots of moving charged matter (magnetohydrodynamics -- definitely not introductory material!).

10. Jul 6, 2013

### s3a

I recalled that a few posts ago, thanks to you. :)

I made up the term “states of configuration” in an attempt to try and come up with an explanation for the term. ;)

About the rest that you wrote, that makes sense! :D

Okay so, it seems that your answer could be summarized as “the problem assumes an ideal situation” which satisfies me as far as this problem is concerned (though, I do appreciate reading the detail) but, to ask a tangential question, the magnetic field lines in the video did not all seem to be symmetrically coming from the centre of the sphere (that is the Sun) ... could you elaborate on this, please? (I know too little to ask a more specific question.)

11. Jul 6, 2013

### Staff: Mentor

The magnetic field of the Sun, to a crude first approximation, are similar to those produced by a giant bar magnet. So they are concentrated towards the north and south poles of said magnet (which may not be perfectly aligned with the rotation axis of the sun itself). And, they get twisted and entrained by moving charges of the solar mass, solar prominences, and so forth, with all sorts of field line entanglements, breakages, and re-connections occurring. Further, the rotation of the Sun (which varies with latitude) causes the field lines to "wrap" around. It's not a simple situation at all.

12. Jul 6, 2013

### s3a

Thanks, I think I get everything well enough to my satisfaction, now! :D

By the way, what you said also validated something I scribbled which I thought was gibberish!:
L = mνr
∇L(ν,r) = (mr, mν)
mr = 0
mν = 0
m(r + ν) = 0
ν = –r

13. Jul 6, 2013

### Staff: Mentor

If the units don't make sense, neither does the math...

Velocity is not a distance. Distance is not a velocity.

14. Jul 6, 2013

### s3a

Are you sure that it works like that? I thought all that mattered was magnitude and direction (and not units). I ask because, that work seems consistent with what we were doing.

Assuming it really is wrong, what must I do to make it correct?

15. Jul 6, 2013

### Staff: Mentor

You'll have to explain your notation to begin with. What does (mr, mv) signify?

But more importantly, in the final two lines you've apparently added a distance to a velocity and equated a velocity to a distance (assuming that r and v variables have their traditional interpretations).

16. Jul 6, 2013

### s3a

I'm really not sure if I'm following the proper procedure for obtaining the gradient vector but, according to my work, mr is the first component and mv is the second component of the gradient vector.

To get the first component of the gradient vector, I took the partial derivative with v as the variable keeping m and r as constants and, to get the second component of the gradient vector, I took the partial derivative with r as the variable keeping m and v as constants.

I, then, assumed that the gradient's components are each zero since there is no change in momentum since it is conserved and, from this assumption, I get the system of equations:
mr = 0
mν = 0

m(r + ν) = 0

from which I get
ν = –r

Having thought about this again, assuming that I was right about the units not mattering, it still seems strange that the magnitudes of the two have to be the same so, I am, probably, wrong.

Last edited: Jul 6, 2013
17. Jul 6, 2013

### Staff: Mentor

When one says that angular momentum is constant, that means dL/dt = 0.

Also, angular momentum is a vector quantity (well, a pseudo-vector, but let's not split hairs). Angular momentum is then L = (r x mv), where L, r, and v are vectors. You will need to deal with it in vector form (individual components if we're talking Cartesian coordinates). Both r and v are functions of time, r(t) and v(t). They're not simple independent variables since they are intimately interrelated for a given orbit, together defining the trajectory.

∇ is an operator that yields rates of change of a given function with respect to individual variables. If the variables have different units you cannot add the components meaningfully. Usually one would see the terms of the gradient vector being multiplied by small values associated with variable associated with the corresponding differential. The result would be terms all with the same units as the given function (angular momentum in this case).

Any time you have values with unlike units being added, something is wrong.

18. Jul 7, 2013

### s3a

Alright, I have enough information to satisfy my curiosity of the problem posted in the opening post.

Thanks again! :)