Tension, scalar or vector?
Peter R. Oakfield
string. Some knowledgeable people tell me it is a vector; others say
it is a scalar. The implications are important. I think it is a
scalar, because it has no unique direction and I cannot imagine the
meaning of negative tension. Who is right? Could someone please help
me?
Peter
Lubos Motl
the third possible - and a pretty reasonable - answer would probably be
that tension is a *tensor*. ;-) In fact the origins of the words "tensor"
and "tension" are related (tensor is a flexed or tightened muscle). But I
don't want to make your situation even more confusing. Well, all
possibilities - scalar, vector, tensor - can be realized if the notion of
"tension" is made sufficiently precise, but the vector is the worst
choice.
If you consider a (guitar) string, the tension is simply the force exerted
by the string on the guitar; force is the infinitesimal change of the
energy divided by the infinitesimal change of the length, and therefore
its unit is Joule/meter = Newton. Well, the force is a vector, but in this
case it does not make much sense to talk about a vector because the
direction is always aligned with the direction of the string itself.
Once you study the interior of the solid locally, you find out that much
like in liquids, you can define the local "pressure". The difference is
that the pressure of liquids (expressed in Pascals = Newton/meter^2 =
Joule/meter^3) is a scalar - a number that is independent of the direction
i.e. that is isotropic. On the other hand, the "pressure" inside solids
does depend on the direction, and it must be expressed as a tensor (a
matrix).
The tension T of strings in (super)string theory is always a scalar
constant, and the total (relativistic) energy of a stretched string of
length L is simply L.T. In this context - and all physically analogous
contexts of infinitesimally thin strings - the tension is simply a scalar
that tells you how much energy you must spend to change the length of the
string by a unit.
Yes, it is natural to conclude that tension is a scalar or a tensor
because its positive sign is usually determined by general physical
principles. If one acts with a negative force on a guitar string, such a
string can easily twist and shrink, and it does not look like a stretched
string anymore.
Best
Lubos
______________________________________________________________________________
E-mail: lu...@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
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Superstring/M-theory is the language in which God wrote the world.
Igor Khavkine
news:<a7e268ec.03112...@posting.google.com>...
Tension is a scalar. You can always write the force of tension
along a rope as T*t. Where t is the unit vector tangen to the rope.
Since force is a vector and t is a vector, then T, which is the magnitude
of the tension or just "tension", must be a scalar.
Another quick way to see that is to note that tension enters as
a scalar Lagrange multiplier in the action formulatino of the dynamics
of an inextensible rope/string/chain. It enforces the condition of
inextensibility.
The caveat here is that t is not unique, -t works just as well. The
tangent vector is uniquely defined only once a parametrization is
chosen for the curve describing the rope. This is probably related
the fact that geometrically the tangent to a curve is not a vector
but a line! Some times it's also called a "director". Mumble muble...
traceless symmetric tensor representation of the rotation
group ... mumble mumble.
Hmm, negative tension... I don't see a reason why this can't exist,
at least in theory. Consider for example an inextensible rope
positioned perfectly vertically on the ground on one end, and supporting
a weight on the other end.
weight
*
rope |
| ground
--------
If negative tension is allowed, then this system is in equilibrium.
Since the tension is negative, the rope exerts a force of tension
on the ground at one end, and a force of tension that supports the
weight on the other end. This situation can be seen as the limit of
the same configuration of a flexible spring, in the limit as the
spring stiffness goes to infinity.
However, this configuration is obviously unstable to the slightest
disturbance. I would imagine that any configuration of a rope with
negative tension is unstable, as the rope would prefer to bend itself
than be in that situation. So in real life they do not occur. That
would explain why it is hard to imagine.
Hope this helps.
Igor
Rob Woodside
Peter, you'll be happy to learn that tension is neither a scalar, nor
a vector. It is a component of a second rank tensor - The stress-
energy tensor of continuum mechanics, that is the material source for
curvature in Einstein' s equations. A tension or negative pressure
occurs in a stretched liquid just before it cavitates. A tension also
occurs along a field line in electromagnetism.
Charles Francis
<Pine.LNX.4.31.03113...@feynman.harvard.edu>, Lubos
Motl <mo...@feynman.harvard.edu> writes
>Hi Peter,
>
>the third possible - and a pretty reasonable - answer would probably be
>that tension is a *tensor*.
only in the sense that a vector is a special case of a tensor. Tension
is the force exerted by a stretched string, and operates in the
direction of the string. Other forces can be supported in solid
materials, torsion, shearing forces etc and require a tensor to describe
them. IIRC the general tensor describing forces in solid materials is
called stress.
>If you consider a (guitar) string, the tension is simply the force exerted
>by the string on the guitar;
> the force is a vector,
Yes. And that is the answer.
> but in this
>case it does not make much sense to talk about a vector because the
>direction is always aligned with the direction of the string itself.
Of course it is aligned with the string. That is what being a vector is
all about. Any force or any vector can be rotated, and it remains a
vector when it is rotated.
Regards
--
Charles Francis
Charles Francis
In article <a7e268ec.03112...@posting.google.com>, Peter R.
Oakfield <Poak...@msn.com> writes
Its a vector because it is directed along the string. Negative tension
is compression.
Regards
--
Charles Francis
Charles Francis
In article <f1ac2e6e.03113...@posting.google.com>, Igor
Khavkine <k_ig...@lycos.com> writes
>Poak...@msn.com (Peter R. Oakfield) wrote in message
>news:<a7e268ec.03112...@posting.google.com>...
>
>> Hi. I am in a big argument regarding tension, like in a cord or
>> string. Some knowledgeable people tell me it is a vector; others say
>> it is a scalar. The implications are important. I think it is a
>> scalar, because it has no unique direction and I cannot imagine the
>> meaning of negative tension. Who is right? Could someone please help
>> me?
>
>Tension is a scalar. You can always write the force of tension
>along a rope as T*t.
I.e. the tension is a vector.
>Where t is the unit vector tangen to the rope.
>Since force is a vector and t is a vector, then T, which is the magnitude
>of the tension or just "tension", must be a scalar.
I.e. T is the magnitude of tension, a vector.
>Hmm, negative tension... I don't see a reason why this can't exist,
<snip>
>I would imagine that any configuration of a rope with
>negative tension is unstable, as the rope would prefer to bend itself
>than be in that situation. So in real life they do not occur.
A rope does not support compression. All that means is the
correspondence between maths and physics breaks down. Try a rod.
Regards
--
Charles Francis
Charles DH Williams
Charles Francis <cha...@clef.demon.co.uk> wrote:
A 'vector' would change sign under the transformation (x,y,z) -> (-x,-y,-z)
The tension in a string doesn't - it's a tensor, or a pseudo-vector.
So I guess nobody wins the bet.
Charles
Charles Francis
Nice try, but wrong. In any case that would not make it a tensor, which
is a generalisation of a vector with additional components. Tension is
really a set of pairs of equal and opposite forces acting at each point
in the string. In any problem these cancel each other out, leaving you
with only the two equal and opposite forces acting at the ends of the
string.
Because these forces are equal and opposite you only need to specify one
of them to define them both. And since we only need specify one force,
it is a vector, as I said. Under the transformation (x,y,z) ->
(-x,-y,-z) both forces change sign.
The mistake you are making is that you are also swapping which force you
use to specify them, and that cancels the change of signs. But if you
think of the diagram of forces for any mechanics problem then the whole
diagram must be transformed under (x,y,z) -> (-x,-y,-z), not just the
string.
For example in a simple pendulum we are only generally interested in the
tension at the bob. Under the transformation (x,y,z) -> (-x,-y,-z) we
would draw the pendulum upside down, with gravity going upwards, and we
would still be interested in the tension at the bob, which would be
inverted just as expected of a vector.
Regards
--
Charles Francis
Charles DH Williams
Charles Francis <cha...@clef.demon.co.uk> wrote:
> Nice try, but wrong. In any case that would not make it a tensor, which
> is a generalisation of a vector with additional components.
> [snip]
I believe you are mistaken.
1. A tensor is not 'a vector with additional components'. It is an entity
defined by the transformation properties of its component. The things
physicists call 'vectors' (also polar vectors) are contravariant tensors
of tensor rank 1.
2. Tension does not obey the transformation properties associated with
a polar vector.
If you want to split hairs you should pick on my afterthought that it
was a pseudovector (i.e. an axial vector).
But since we know what tension is (it is a component of the stess tensor)
and what it isn't (it is not a vector or a vector component) I can't get
excited about whether it might be a pseudovector.
All the best
Charles (Williams)
Igor Khavkine
news:<WOlb$eJp5uy$Ew...@clef.demon.co.uk>...
> In article <f1ac2e6e.03113...@posting.google.com>, Igor
> Khavkine <k_ig...@lycos.com> writes
> >Poak...@msn.com (Peter R. Oakfield) wrote in message
> >news:<a7e268ec.03112...@posting.google.com>...
> >
> >> Hi. I am in a big argument regarding tension, like in a cord or
> >> string. Some knowledgeable people tell me it is a vector; others say
> >> it is a scalar. The implications are important. I think it is a
> >> scalar, because it has no unique direction and I cannot imagine the
> >> meaning of negative tension. Who is right? Could someone please help
> >> me?
> >
> >Tension is a scalar. You can always write the force of tension
> >along a rope as T*t.
>
> I.e. the tension is a vector.
Note that I said "force of tension". A force is always a vector.
But you must be careful when identifying tension with a force.
At a particular point, tension can be identified with the force
with wich the left piece of the rope acts on the right one, or
vice versa. Since the choice between the two identifications is
arbitrary, it is best avoided.
> >Where t is the unit vector tangen to the rope.
> >Since force is a vector and t is a vector, then T, which is the magnitude
> >of the tension or just "tension", must be a scalar.
>
> I.e. T is the magnitude of tension, a vector.
Well, undoubtably T is a scalar, but you are arguing that when considering
tension, both its direction and "magnitude" must be considered.
That is fine, but a vector is not the only thing endowed with a direction.
Lets rotate a vector and the tension in a rope (effectively rotate the rope
and all the forces acting on it) by a small angle. Both appear to change in
the same way. If you rotate a vector by 180 degrees about an axis
perpendicular to it, you get a different vector, namely the negative
of the original. However, if you do the same with tension, it does not
change. In other words tension can be said to be parallel to some line
(namely the line tangent to the rope at the same point) but it cannot be
said to be in either one or the other direction along that line.
> >Hmm, negative tension... I don't see a reason why this can't exist,
> <snip>
> >I would imagine that any configuration of a rope with
> >negative tension is unstable, as the rope would prefer to bend itself
> >than be in that situation. So in real life they do not occur.
>
> A rope does not support compression. All that means is the
> correspondence between maths and physics breaks down. Try a rod.
Hmm, "a rope does not support compression". Hmm, "any configuration
of a rope with negative tension is unstable". Nope, don't see a
contradiction here.
Igor
Charles Francis
>Charles Francis <cha...@clef.demon.co.uk> wrote in message
>news:<WOlb$eJp5uy$Ew...@clef.demon.co.uk>...
>
>> In article <f1ac2e6e.03113...@posting.google.com>, Igor
>> Khavkine <k_ig...@lycos.com> writes
>> >Poak...@msn.com (Peter R. Oakfield) wrote in message
>> >news:<a7e268ec.03112...@posting.google.com>...
>> >
>> >along a rope as T*t.
>>
>> I.e. the tension is a vector.
>
>Note that I said "force of tension". A force is always a vector.
>But you must be careful when identifying tension with a force.
>At a particular point, tension can be identified with the force
>with wich the left piece of the rope acts on the right one, or
>vice versa. Since the choice between the two identifications is
>arbitrary, it is best avoided.
Both forces are tension. The choice in any given problem is not
arbitrary, but is determined by the body that the string is acting on,
the body whose motion you wish to analyse.
>> >Where t is the unit vector tangen to the rope.
>> >Since force is a vector and t is a vector, then T, which is the magnitude
>> >of the tension or just "tension", must be a scalar.
>>
>> I.e. T is the magnitude of tension, a vector.
>
>Well, undoubtably T is a scalar, but you are arguing that when considering
>tension, both its direction and "magnitude" must be considered.
>That is fine, but a vector is not the only thing endowed with a direction.
>
>Lets rotate a vector and the tension in a rope (effectively rotate the rope
>and all the forces acting on it) by a small angle. Both appear to change in
>the same way. If you rotate a vector by 180 degrees about an axis
>perpendicular to it, you get a different vector, namely the negative
>of the original. However, if you do the same with tension, it does not
>change. In other words tension can be said to be parallel to some line
>(namely the line tangent to the rope at the same point) but it cannot be
>said to be in either one or the other direction along that line.
If you rotate the string you must also rotate the body which the string
is acting on.
>
>> >Hmm, negative tension... I don't see a reason why this can't exist,
>> <snip>
>> >I would imagine that any configuration of a rope with
>> >negative tension is unstable, as the rope would prefer to bend itself
>> >than be in that situation. So in real life they do not occur.
>>
>> A rope does not support compression. All that means is the
>> correspondence between maths and physics breaks down. Try a rod.
>
>Hmm, "a rope does not support compression". Hmm, "any configuration
>of a rope with negative tension is unstable". Nope, don't see a
>contradiction here.
I didn't suggest a contradiction. Only that if you use a rod rather than
a rope you can have a real life situation in which negative tension,
i.e.. compression, does occur.
Regards
--
Charles Francis