Pete

Some physicists and mathematicians use the term "Cartesian tensor" to describe things that live in a Cartesian space and look a bit like tensors. Some use the term "Cartesian-tensor" to describe these things because Cartesian tensors are not tensors. A Cartesian-tensor has some, but not all, of the characteristics of a tensor.

I disagree. Why do you believe that Cartesian tensors are not tensors?

 

Cartesian tensors are most definitely tensors because they fit the definition of a tensor in the most rigorous sense, i.e. it has all the characteristics of a tensor. What characteristic do you believe it doesn't have?

 

A Cartesian vector can be defined as a geometric quantity whose components transorm in a certain way (i.e. tensorially) under an orthogonal transformation.

 

A Cartsian tensor can be defined in two different, but equivalent, ways. One way a Cartesian tensor can be defined is as a multilinear map from 1-forms and Cartesian vectors into real numbers (i.e. scalars). They can also be defined by how their components transform under an orthogonal transformation.

 

A Lorentz 4-vector is defined as a geometric quantity whose components transorm in a certain way (i.e. tensorially) under a Lorentz transformation.

 

A Lorentz tensor can be defined as a multilinear map from 1-forms and Lorentz 4-vectors into real numbers (i.e. scalars). They can also be defined by how their components transform under a Lorentz transformation.

If we just restrict ourselves to special relativity, we are defining "distance" to be the geometric invariant associated to the Lorentz group. Therefore, I think it is being pedantic to insist on "distance" being strictly positive. Maybe you want to call it pseudo-distance or something?

If one defines "distance" in that way then I wholeheartedly agree that there is absolutely no reason to insist on it being positive. I sure hope I didn't give you the impression that I thought otherwise?

I have never come across the term "Cartesian tensor" before. How are they defined?

You've probably seen/used them but they may not have been called that. A Cartesian tensor is a geometric quantity whose components transform tensorially under an orthogonal transformation. If you have Jackson EM text then he defines them there. I believe that he calls them rotational tensors though. They are also defined in Goldstein. I gave a definition of them in one of my web pages, i.e. in

 

http://www.geocities.com/physics_world/gr_ma/tensors_via_analytic.htm

 

The following are Cartesian tensors

 

1) Stress tensor

2) Moment of Inertia tensor --

http://www.geocities.com/physics_world/mech/inertia_tensor.htm

3) Tidal force tensor -- Moment of inertial tensor

http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm

4) Metric tensor

5) Maxwell's stress tensor --

http://scienceworld.wolfram.com/physics/MaxwellStressTensor.html

6) Stress tensor

 

Pete

The spatial distance, D, between two events is defined as

 

[math]D = \sqrt{x^2 + y^2 + z^2}[/math]

I must say that its a bit irritating not being able to go back and correct a post. In the present case the error here is that D is the distance betwen an event located at the origin of the coordinate system and an event located at the point (x, y, z).

 

Pete

Maybe I should have put distance in quotation marks.

That would be a wise thing to do.

 

I posted that message because I was merely trying to clarify the terminology a bit. The idea of calling the quantity [math]\Delta s[/math] in

 

[math]\Delta s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2[/math]

 

distance has always be a pet peeve of mine. When it is referred to as such it becomes difficult to explain to the layman that it really isn't the distance between the two events, hence the quotations. I believe that some authors of SR texts ractually do refer to it as "distance." It leads people to believe that an object at rest "moves" a finite distance even though it is at rest. It gives rise to phrases like "moving through spacetime" and "the speed of a tardyon through spacetime is always c" which only serves to confuse people in my humble opinion.

 

Such a choice of words Proper distance between two events can't always be defined whereas the actual distance can always be defined. The fact that it is sometimes impossible to define "distance" this way only serves to confuse the layman.

But the fact remains we have time-like, light-like and space-like separation of points in Minkowski space.

While the physicist understands this very well it becomes hard for the layman to understand this idea.

 

I have to think about this some more so please make sure that it hasn't changed during the time you respond to it. Thanks.

 

Best wishes

 

Pete

Thanks Pete.

You're most welcome. Its my pleasure. When I take the time to respond to someone's question I am actually helping myself in the process. In all cases I learn how to provide better responses and in some cases I actually learn something new. This is a result of being forced to look at something from a different perspective.

 

Keep asking these great questions!

 

Pete

Bignose: How do you define a straight line on a curved surface?

 

But what Bignose says is correct and basically what I said. If you use a pseudo-Riemannian metric then you can have positive, negative and zero distances between distinct points.

I disagree. Distance is positive by definition. The magnitude of the fundamental force is correctly called the interval between two points. The term distance only refers to proper distances and even then only certain circmstanes. In relativity a lot of people refer to the square root of the inteval as distance but when this is done it is really a misnomer.

 

Consider special relativity as an example: The first fundamental form is defined as

 

[math]ds^2 = g_{\alpha \beta}d^{\alpha} d^{\beta}[/math]

 

Some physicists call this the metric where g is the metric tesor and [math]g_{\alpha \beta}[/math] are the components of the metric tensor. The term first fundamental form is most often used in differential geometry texts/papers. I usually choose the metric in Minkowski coordinates as

 

g = diag(1, -1, -1, -1)

 

Consider the 4-position 4-vector X = (ct, x, y, z) = (ct, r) where r = (x, y, z). The spatial distance, D, between two events is defined as

 

[math]D = \sqrt{x^2 + y^2 + z^2}[/math]

 

The proper distance is defined only for two events which have a spacelike spacetime seperation. In such case the proper distance, [math]\lambda[/math], is given by

 

[math]\lambda = \sqrt{-s^2}[/math]

 

Pete

With regards to wave-particle duality...plus I'm still quite new to QM i.e beyond anything you'd read in pop sci (so correct me with this). But I'm surprised nobody has mentioned Fourier synthesis (wave packets) and HUP, because these are really the first melding of the two, i.e waves and particles.

Good point. However it has been mentioned in other threads about the uncertainty principle. However the wave particle duality is the physical phenomena for which Fourier describes. The purpose of this thread seemed to me to be about the basic concept of wave particle duality and not the math which is constructed to describe it, which follows from the physical notion.

When you view the wave particle duality from just these principles and math, the distinction does blur somewhat.

I highly disagree. The wave function embeds the wave aspect of the wave-particle duality in that it provides statistical predictions. The particle aspect is embedded in what it predicts, i.e. the distributions of discrete events, e.g. the distribution of precisely localized positions of particles.

 

Pete

If the person wants to ignore the friction, the mass cancels out!

Even when a person does not ignore friction the mass still cancels out.

think about this sentence. It is completely nonsense

On the contrary. This is basic physics (physics 101) and what I said is 100% accurate. One merely has to read a basic physics text to learn this fact.

This is also completely nonsense! What you wrote is a coloumb assumption, but we are talking about sliding.

If I were you I'd restrain from using the term "nonsense". You have used it here to reject the most basic facts of physics so far. What I wrote has absolutely nothing to do with the Coulomb force.

 

The force on a particle is defined as the time rate of change of the particle's momentum, i.e. F = dp/dt where p = mv. If the mass is constant then the force equals F = ma. The force of friction [math]F_f[/math] on a moving body is given ny the relation

 

[math]F_f = \mu_k F_n[/math]

 

where [math]F_n[/math] is the normal force acting on the body. In the case of the gravitational force [math]F_n = w(weight) = mg[/math]. Plug these into the above expression to obtain

 

[math]F = ma = F_f = \mu_k F_n = \mu_k mg[/math]

 

which is identical to what I showed you above and as I also said, the mass cancels out.

 

Since you claim this is wrong then please provide a derivation for explicit expression for the acceleration of a body which is sliding on a rough surface. I am stating that the expression so derived will be independant of the mass of the body. Prove me wrong.

 

That I'm correct can easily be seen from the fact that the force of kinetic friction is dependant only of two materials (the body which is sliding and the material on which it is sliding) which results in [math]\mu_k[/math]. One can see this by considering two identical bodies sliding side by side. Each body will accelerate at the same rate. Now tape/glue them together. The fact that they are glued together will not change the rate at which they accelerate. But now we have a body whose mass has doubled. Since the force of friction is not dependant on the area of the surface of the body which is in contact with the surface on which it is sliding then the acceleration will not change.

 

Pete

Hi,

 

the small guy with 50 kg would win the race!

I disagree.

If we talk about such exercises it is very important to look on the undergound. In an ideal case the mass is completly unimportant.

Not if the person asking the question wishes to ignore things like friction and drag.

The mass fall out and so it don't cares. 100 kg, 150 kg, 2000kg, no matter. But it is not ideal. We have friction and becaus of that the friction force is directed against the motion. If you look now to wintersports, the mass could be useful, because it melts the ice and snow and the friction coefficient becomes better, but on a street is more mass not good.

No. The acceleration of a body is dependant on the coeffiction of kinetic friction but is independant of mass.

[math]\vec{F}_{r} = \mu \vec{F}_{n} = \mu m \vec{g} \cos(\alpha)[/math]

What is [math]\vec{F}_{r}[/math]? I.e. what does the "r" subscript denote? If this is simply the force on the body then once you replace it with ma you will see that the mass drops out yet once again!

This trem ..

What is a "trem"?

..slows the fatter guy down ..

I disagree. Consider a body sliding on a surface under friction for which the coefficient of kinetic friction is [math]\mu_k[/math]. Then

 

[math]F = ma = \mu_k F_n = \mu_k (mg) [/math]

 

Cancel out m to yield

 

[math]a = \mu_k g [/math]

 

Thus the acceleration of a body sliding on a rough surface is a function of [math]\mu_k[/math] but not a function of m.

..and the surface of the street wouldn't melt for a better friction coeffizient.

Huh? What does that mean?

 

Pete

Well, complementarity is the notion that particles must either exhibit particle properties or wave properties, but never simulataneously. The observation that waves can be affected by gravity while still maintaining wave properties (since gravitational force only exists between two objects with mass, a particle property) violates that notion.

Recall "The Feynman Lectures on Physics," by Feynman, Sands and Leighton, Vol -I page 7-11 - Section entitled Gravitation and Relativity

One feature of this new law is quite easy to understand is this: In Einstein relativity theory, anything which has energy has mass -- mass in the sense that it is attracted gravitationaly. Even light, which has energy, has a "mass". When a light beam, which has energy in it, comes past the sun there is attraction on it by the sun.

Light has (passive) gravitational mass and therefore gravity exerts a gravitational force in light. Nothing is violated here.

 

Pete

i was wondering...

how and where did you learn all you know about Physics?

I learned it as both an undergraduate who majored in physics (and math) for 6 years (had to get up to speed due to slacking off in highschool) and 2 years in graduate school. I also spend a great deal of time teaching myself the physics I never learned in college.

 

Pete

Hi Pete,

 

Recall that a black hole with maximal spin (angular momentum) does not have elevated temperature. Electrons have limit (maximal) angular momentum so they don't emit Hawking radiaton.

Sorry but I don't see your point.

 

To simplify this arguement let us not consider an electron but a scalar charge (i.e. a charged particle with zero spin). A black hold emits Hawking radiation by the following mechanism (as I understand it): Pairs of photons are produced (vacuum fluctuations) near the event horizon of the black hole. One photon (the negative energy one) is crosses the event horizon and is captured by the black hole. The other one escapes to infinity. The radiation never comes from the charged particle itself (or the singulartity or anything else inside the event horizon). Even if we were discussing electrons then the process of Hawking radiation has nothing to do with the electron itself emitting radiation.

 

Its not like I believe that electrons are black holes by the way so please don't get the idea that I'm trying to justify the notion as an electron being one (in fact I'd find it hard to believe that they are). I'm just fielding some thoughts, that's all. The reason being is that I find these points hard to explain and thus want to hear opposing ideas about them.

 

Pete

The concept of drag has been introduced and that makes the problem more difficult. Two people can have the exact same drag coefficient and yet have more mass. Anyone who has learned about health and fitness knows that one can loose fat and yet not loose weight since its feasable to work out in a way such that you become stronger in the process and loose fat but gain muscle. It is therefore completely feasible to have two people with the same drag coefficient and yet have different masses.

 

Tell your friend that while the force on a heavier person is greater than a lkight person one mustn't forget that the heavier person will be harder to accelerate (i.e. has more inertia) that the lighter person. As someone mentioned above, these two effects cancel. This is a fact and has been experimentally confirmed. Tell your friend that the laws of physics disagree with him. Does your friend have a reason for his assumption??

 

Consider a body in free fall in a uniform gravitational field (the gravitational field near the surface of the earth can be considered highly uniform if one confines the height to reaonable values (such as a few miles). The gravitational force on his body is given by F = mg where g is the acceleration due to gravity. The force on a body is given by F = dp/dt where p = mv. If the mass, m, is constant then F = ma where a = dv/dt = acceleration. Therefore

 

F = mg = ma

 

The mass cancels to give

 

a = g

 

Therefore the gravitational acceleration is independant of the mass of a body. The drag force is also indepentant of the mass of the body so that a person's weight does not contribute to the time it takes to roll down a hill.

 

If your friend is really interested in this then he can pick up an introductoryu book on physics. However if you'd like I can create a web page to explain these facts. Just let me know.

 

I hope this helped. It probably didn't contribute much because of the very good explainations given above. Let me say Well done! for your efforts here folks.

 

Pete

Insane is right on this one. Blackholes do not appear (for sure semi-classically) stable. In collider experiments they would appear as very short lived particles. This is very different to electrons.

Still, if an electron is truely a point particle then one can calculate a finite Schwarzschild radius for it and its mass would be entirely located within a sphere of this radius thus making it a black hole. Of course there are quantum mechanical considerations to take into account. In what sense do you say that they are short lived particles? I.e. what would they decay into? Suppose a black hole is constructed purely of baryons. Then the baryon number would have to remain constant even though the black hole is radiating. Also while black holes radiate energy by Hawking radiation they are also absorbing energy from the CMBR.

In other words, they get transferred to the particles into which the black hole evapourates. AFAIK it's called Hawking Radiation and even though it's never been demonstrated it's widely accepted as fact.

It should be noted that Hawking radiation is composed entirely of thermal radiation (i.e. electromagnetic radiation, i.e. photons). This means that it can't emit charged particles or any other types of particles. For that reason and the one above (e.g. conservation of baryon number) I can't see how a black hole could evaporate entirely. Hmm! Perhaps its mass would become negative! lol!

 

Pete

They're all wavicles.

 

It was Sir Arthur Eddington who coined that term. Since an electron has wave properties and particle properties it is different than something which has either of them. For this reason a new name was sought and Eddington coined the term wavicle for that reason. I personally don't like that term because it can tend to give the wrong idea of the wave-particle duality. E.g. an electron is a pointlike object and is thus localized, quite unlike a wave. Statistically there is a wave pattern associated with the behaviour of the electron which can't be understood in classical terms and thus can'tbe understood if one thinks of an electron as being a classical particle. Hence the term waveicle. However people tend to take that to mean that an electron is both a particle and a wave simultaneously (i.e. displays both properties at the same time), which is not true.

 

Please note that this is my personal opinion and am not sure how wide spread it is. To be fair I should point out that Feynman wrote the following on page 85 of his book QED. After discussing electrons Feynman writes

It's rather interesting to note that electrons looked like particles at first, then their wavish character was later discovered. On the other hand, apart from Newton making a mistake and thinking that light was "corpuscular," light looked like waves at first, and its characteristics as a particle were discovered later. In fact, both objects behave somewhat like waves and somewhat like particles. In order to save oursleves from inventing new words such as "wavicles", we have chosen to call these objects "particles," but we all know that they obey these rules for drawing and combining arrows that I have explaining. It appears that all the "particles" in Nature - quarks, gluons, neutrinos, and so forth (which will be discussed in the next lecture) - behave in this quantum mechanical way.

However this is not the whole story of course since on page 37 Feynman writes

Quantum electrodynamics "resolves" this wave-particle duality by saying that light is made of particles (as Newton originally thought), but the price of this great achievement of science is a retreat by physics to the position of being able to calculate only the probability that a photon will hit a detector, without offering a good model of how it actually happens.

Well ... nobody ever said that quantum mechanics was easy to understand, right?

 

Pete

Recent experiments have cast doubt upon the Copenhagen interpretation regarding complementarity. An experiment was done that allowed a neutron interferometer to be subject to gravitation, and the neutrons created the interference pattern while simultaneously being affected by the gravitational field, something that should only affect particles.

A photon is often referred to as a particle of light. When that is said it is referring to the particle property of a photon. Since light waves are affected by gravity then I don't see why you find this surprising or why you said it cases done on complementarity. Also, I never heard of the phrase Copenhagen interpretation regarding complementarity. The Copenhagen interpretation refers to the interpretation of the wave function whereas complementarity refers to the uncertainty principle. They are not the same thing.

 

Pete

I disagree, it displays particle like and wave like properties.

Please clarify. To me that is indistinguishable from displays both particle properties and wave properties.

Don't be so pedantic when answering questions like this.

I'm not being pedantic. I'm being precise.

 

Pete

True, that is a better answer but *shrug* in reality our language is just not good enough.

 

Note: I changed This is not precisely correct. to I disagree. The later seems more conversational and less judgemental.

 

Pete

A photon is both a wave and a particle. As would a graviton be.

I disagree. Something like a photon (or electron etc) displays both particle properties and wave properties, but never at the same time. In this sense one can't say that it is both a particle and a wave since stating it implies it displays both properties simultaneously, which it does not.

 

Pete

So, it would be something like this:

 

IF <message body> CONTAINS <email addy> THEN <send to Trash>.

That was the first thing I tried except that there is no slection like that in the program. I'll post the selections that are available later today. Thanks.

 

Pete

"What part of this thread indicates that we are sticking to special relativity?"

 

Nothing indicated that we were STICKING to SR, I was simply responding to the comment earlier stating that the original scenario was "well understood in SR."

 

I was merely using SR scenarios to demonstrate the possibility of FTL within those scenarios.

 

Besides, the possibility of FTL only increases as we attain proximity to General Relativity. The causality violations become more and more easily explained and/or less frequent.

 

PS: I wasn't arguing the point, just throwing out ideas.

I see. Thanks for clarifying. By the way, I wasn't trying to criticize you, just curious as to somethings that you posted, that's all.

 

Pete

Grandfather "paradox" isn't all that neat and it really isn't a paradox at all.

Please prove this assertion

It's just a common misconception due to the failure of one to follow the timeline ...

Please explain what you mean when you use the term timeline. Usually the term timeline refers to a sequence of events. If that is the case your comment failure of one to follow the timeline is not clear.

... rather than the traveller. If you follow the timeline, it's rather simple.

Please clarify. Timeline according to whom? Different worldlines given different timelines, hence the Grandfather paradox.

Sorry to keep bugging: I needed to add clout to what I was saying...special relativity doesn't preclude FTL speeds.

What part of this thread indicates that we are sticking to special relativity?

 

Pete

I use Microsoft Outlook Express to read newsgroup messages. I wish to block a particular poster but he changes his handle everytime so that I can't simply use the Block Sender function and killfile him. Outlook Express has a function called "Rules" or something like that, which allows people to select a criterion which will automatically process messages and, if the user wants, to delete certain messages when they arrive. Using this function would allow me to delete these messages because he always posts the same E-mail address in each and every thread.

 

Question: How do I create a rule which will delete all messages which contain a particular E-mail address?

 

Thanks

 

Pete

Could you please suggest a link for the above?

Carroll's lectures are for the advanced student and not someone who is just learning the mathematics of relativity. Its a great resource otherwise though. But not at the undergraduate level.

 

I myself have been unsatisfied with the current material that is out there. For that reason I decided to start writing my own website for this material. Its mainly to help myself learn the subject and to learn it well enough to teach others. If you'd like to take a look at it then see

 

http://www.geocities.com/physics_world/gr_ma/gr_ma.htm

 

I would love to hear your thoughts on the material and perhaps some suggestions to make it better or perhaps point out any errors that you might find.

 

Applications are found at

http://www.geocities.com/physics_world/sr/sr.htm

http://www.geocities.com/physics_world/gr/gr.htm

 

I also constructed a web page to list all the books I have

http://www.geocities.com/physics_world/ref/books.htm

See the section below which is labeled "Relativity". It lists all the relativity texts I have. I like each one.

 

The following are very mathematical

 

Gravitation, Misner, Thorne and Wheeler, W.H. Freeman & Co., (1973)

Gravitation and Spacetime, Ohanian & Ruffini, WW Norton n& Co., (1994)

Relativity, Thermodynamics and Cosmology, Tolman, Dover, (1987)

General Theory of Relativity, Dirac, Princeton Univ. Press, (1996)

A First Course in General Relativity, Schutz, Cambridge Univ. Press, (1990)

Introducing Einstein’s Relativity, D’Inverno, Oxford Univ. Press, (1992)

Basic Relativity, Mould, Springer Verlag, (1994)

A Short Course in General Relativity, Foster & Nightingale, Springer Verlag, (1994)

Relativity: Special, General and Cosmological, Rindler, Oxford Univ., Press, (2001)

 

Pete

[math]\rho^{i} =T^{0 i}/v^{i}[/math]? Not sure I understand what that means. The indices don't balance and I don't understand the meaning of division here.

T⁰ⁱ is the ith component of momentum density, vⁱ is the ithe component of 3 velocity and [math]\rho^{i} =T^{0 i}/v^{i}[/math] is the ratio of the two.

Anyway, am not going to be about for at lest a week. Sorry for the lack of the near future responses.

Have fun!

 

In the mean time I have to look into this more. It seems that I made an error above. The mass density as given by both Tolman and Rindler is

 

[math]\rho = \gamma^2 (u) (\rho_0 + u^2t_0^{11}/c^4)[/math]

 

which is not the ratio of momentum to speed as I believed. However the term proper mass densit does seem to have a unique meaning and thus is well defined.

 

Pete

[math]\rho[/math] as I have defined it is independent of any coordinates used, but of course depends on the observer via the [math]X[/math] as your rightly say.

 

Thing's don't have to be observer independent, as such, but they should not depend on the coordinates use.

The term observer in SR is normally taken to be synonymous with coordinate system in the sense that different observers correspond to different inertial frames of reference.

The [math]\sqrt{-g}[/math] is needed so that we have a tensor density which (at least formally) can be integrated over the space-time manifold.

Please elaborate. Why can't components be intergrated over spacetime. MTW defines mass-energy density as

 

[math]\rho = T(U, U)[/math]

 

where [math]U[/math] is the observer's 4-velocity. MTW didn't include [math]\sqrt{-g}[/math] in their deifnition.

You can either include it here or in the definition of your integration measure. To me it makes more sense to construct tensor densities. (I am sure you know this.)

Not really. I never payed much attention to tensor densities. Its one of those things that are on my back burner. As I recall they are used to keep integrals invariant under coordinate transformations, right? I've never seen them used in the texts that I learned relativity from, not that they aren't important.

How is the mass-density related to the energy-momentum tensor?

[math]\rho^i = T^{0i}/v^i[/math]

 

Different orientations lead to different directions of momentum for this reason. I.e. momentum is not parallel to velocity in general. Only for closed systems, isolated systems or point (structureless) particles.

[math]T_{\mu \nu} = T(\partial_{\mu}, \partial_{\nu})[/math] depends on the coordinates employed. Again I am sure we both know this.

 

Sure. That's what I meant when I wrote components of a tensor quanity are defined as the scalars which result by inputing unit 4-vectors/1-forms which define the coordinate system.

 

Pete