Pete

and, is kinetic energy not energy?

Yes. Most certainly.

i believe if you put a value close to c in the equation for kinetic energy, then you will get a large energy.(pretty close to half the energy of the object at rest)[math]E_{kinetic}=\frac{1}{2}mV^2[/math]

The correct expression for the kinetic energy, K, of a particle with proper mass m₀ is

 

[math]K =(\gamma - 1)m_0 c^2[/math]

 

is it just me or would that added energy increase gravity?

Its not just you. As I mentioned above, the increase in energy (regardless of the type of energy) will result in an increase in the strength of the gravitational field. I.e. increase in energy - > increase in active gravitational mass.

 

Hence the widely used term mass-energy.

 

Pete

By "physics university" I mean one of the universities which is quite well known for physics. E.g. MIT, Cal Tech, etc.

 

When people say that physicists don't use inertial mass anymore they claim that they use energy instead since "its the same thing". I don't think I gave an example of what that's not true in general. If it was always true that E = m² then one might want to say that this is a definition of inertial mass. But since the correct definition of inertial mass is m p/v then one can defined m as m = E/c² if and only if this expression is true in all possible/concievable cases. If one can give 1 single counter example then such an assertion must be abandoned.

 

A simple example is that of a rod at rest in the inertial frame S'. Let the rod lay on the x' axis and have a force exerted on each end such that the total force is zero. Then the rod does not accelerate in any frame. In this case the realtion E = mc² is invalid. For those of you who understand the math/physics see a related example at

 

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

 

I'll make a more explicit example soon and post it.

 

Pete

WHAT ABOUT ENERGY????????????????????????????

Since mass and energy are related then yes. An increase in the energy of a body will increase the mass of the body and hence increase the strength of the gravitational field.

actually, the effects of acceleration and gravity are indistinguishable. that is the basis for general relativity.

To be precise - The Principle of Equivalence states that a uniform gravitational field is equivalent to a uniformly accelerating frame of reference. This does not mean that all gravitational fields can be replaced by an accelerating frame. When the gravitational field is not uniform (or in the language of general relativity - when the spacetime is curved) then the principle of equivalence applies locally (restrict observations to regions of spacetime where the spacetime curvature, aka tidal forces can be neglected).

Rotation doesn't create gravity. Acceleration and gravity are indistinguishable, according to GR, but not synonymous.

If one is at rest in a rotating frame of reference then according to Eintein there is a gravitational field present. As Einstein wrote in The Foundation of the General Theory of Relativity, Albert Einstein (1916), Annalen der Physik 49

It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates.

 

In Einstein's general relativity' date='

the effects of acceleration and the gravity is the same.[/quote']So long as one understands that the Principle of Equivalence is a local principle if the gravitational field is not uniform. For example; Is it possible for me to determine if I'm in a gravitational field which is not uniform by observing a point charge? The answer is yes. If I'm on Earth and thus in a non-uniform gravitational field and I decided to weigh the charge then the weight of the charge will depend not only on the gravitational acceleration at the location of the weight scale but the weight will also depend on the strength of the spacetime curvature. The reason for this is that the electric field is not local to a charge so this experiment is not a local one. The electric field sort of "reaches out" to feel the tidal gradients in the gravitational field which then effects the weight measurement. The weight would be different if one is in a uniformly accelerating frame in flat spacetime when the gravitational acceleration at the location of the scale has the same value as the previous experiment.

 

whilst this may seem impossible due to lack of gravity (and the fact such a formation is not natural) it is possible due to centrifugal there would appear to be gravity (although not actually).

Newton would say no and Einstein would say yes.

 

Centrifugal force doesn't exist though right? Well it is not a force, we just call inertia of an object having centripedal acceleration...I think

According to Newton it is an inertial force. According to Einstein its a gravitational force. Einstein postulates of GR imply that inertial forces are identical in nature to gravitational forces. He then concluded that since gravitational forces are "real" then so too must coriolis forces and centrifugal forces be real.

 

you do get more gravity for moving, but it is not anywhere near noticeable. kinetic energy is energy(duh!), and mass AND ENERGY both contribute to gravity.

If by this you mean that an increase in energy produces an increase in active gravitational mass then yes, that is correct.

 

they spin the space station around an axis. this is called centrifugal force.

If you're in a rotating frame of reference and there is a particle moving in your frame then it will accelerate. The force related to this acceleration is called centrifugal force. It points radially away from the axis of rotation. There is another force acting on the particle as well - the Coriolis force which is a velocity dependant force (much like the magnetic force). Each of these are inertial forces. If the particles are also acted on by the wall then the it experiences other kinds of forces. But the gravitational forces here are the centrifugal force and the Coriolis force.

 

Mind you - If you're at rest in an inertial frame and someone started spinning a space station around you then no gravitational forces will appear. The reason being that spinning the space station does not mean you've changed your own frame of reference. For gravitational forces to be produced one must change their frame of reference too. When you start to rotate with the space station then, according to general relativity, a gravitational field will be produced.

 

Pmb

Typing a message in which every sixth word is some variant of, or akin to, "relativistists". Warning: I'm all caprice, in December!

I see. I this particular case I found it hard not to due to the nature of the subject.

 

Pete

That's some impressive relativityping!

What is relativityping?

So, I understand that in a gravitational field, time slows down the deeper one goes into the field, but what difference does it make whether one is falling or on the ground? If I understand GR and SR correctly, then I'd assume when one is falling, you simply sum up the time dialation due to the process of falling (SR) and due to how deep one is at any point during the fall (GR). That is, when one is falling, time slows down because of the principles of SR. But in addition to that, when one falls, one ends up at deeper and deeper points in the gravitational field, and at each point, time slows down because of the principles of GR. When one is on the ground, SR would say time dialation = 0, so only GR must be taken into account. So am I right in assuming the net time dialation is the sum of the SR and GR?

There are two time dilation effects in relativity. One is gravitational time dilation and there is time dilation due to a motion. The specifics of the problem and question will determine the exact response that you're looking for. I.e. who is measuring what etc.

 

Let consider a specific example ("time orthogonal spacetime"/no frame dragging). In such a field time intervals dt (time measured by 'coordinate' observer - precise meaning depends on exact situation) are related to proper time intervals, dT (time read by clock which is in g-field and moving), by

 

dt/dT = 1/sqrt(1 + 2*Phi/c² - v²/ c²)

 

where v = speed of particle and Phi = gravitational potential.

 

Consider, for example, a person who is in free-fall in a uniform g-field. Then to that free-fall observer no gravitational field exists anywhere (Phi = 0 everywhere in region where field is uniform) so there is no gravitational time dilation to speak of. There is only time dilation due to motion. If, however, the gravitational field is not uniform then the gravitational field does not vanish everywhere for a free-fall observer and in general there will be gravitational time dilation.

 

In case I made an error please see the Global Positioning System (GPS) example in Taylor and Wheeler's book at

http://www.eftaylor.com/pub/projecta.pdf

 

Pete

Teasing:

I'm sorry' date=' your question falls beyond the boundaries of GR and I would be unable to answer it in that context. Please post in a more apropriate forum.[/quote']

At least it can be said that the question was answered.

 

For those interested in what is actually used you can go to http://www.physicsforums.com and talk to ZapperZ. He is an accelerator physicists at Argonne National Lab who says he uses the so-called relativistic mass. I was wondering how people came to conclude the researches, while they might write about a concept in their text that they don't use it in theor own research. ZapperZ responds

And for whatever it is worth, I used to teach about relativistic mass, and I currently USE relativistic mass for the dynamics in the particle accelerator that I work with. So the assumption that scientists who teach relativistic mass do not use it in their research can be safely thrown out of the window.

 

But then again Steve Carlip (GRist at UC Davis) also claims 'nobody' uses it. sigh!

 

I'll have to wait until I can get to a decent research library and see for myself.

 

Pete

and so... rather that say "Thats Incorrect" wouldn`t it be much nicer to say, yes, and there`s This Also )

When it would apply, sure.

 

I see no reason for someone would take offense at someone saying that something was incorrect. Why would anyone find that offensive???

 

Pete

Pmb. although your answers are Great too! and wouldn`t be out of place if Swansont or Jakiri asked them' date=' it`s a little OTT for a basic question [/quote']I tried to answer a basic question with a basic answer. The first basic question I responded to was

But how can light be influenced by gravity in this way if light, being composed of photons, is massless? Doesn't it require mass to be "heavy"?

And the basic answer I gave was

Light does have mass (aka inertial mass aka relativistic mass). It has zero proper mass (aka rest mass) and that's what you were thinking about.

The second basic question I responded to was

can someone explain the exact difference between:

 

inertial mass' date='

relativistic mass &

rest mass.

[/quote']I responded with the basis answer

The inertial mass of a body is defined as the quantity m such that the quantity mv is conserved in particle collisions as observed from an inertial frame of reference. If you're discussing relativity then some people like to use the term relativistic mass for this same term.

The rest was in response to swan's comments. His responses were about invariants, scalars etc. so II responded to that.

 

Its my experience that people who make the assertions swan did seem to be particle physicists/nuclear/atomic scientists. But such groups have a limited use of relativity, since they rarely do anything which requires speaking of anything other than proper mass. For that reason such groups use only the term "mass" rather than "proper mass" or rest mass and are unaware of what GRists and cosmologists use. I've looked into this one particular subject point in more detail over the years and that research brought me to applications not found in either particle physics, general relativity or cosmology, the main users of relativity.

 

One might have otherwise thought it strange to hear that E = mc² is not always correct. I know that most people would find that fact strange to hear. Yet its true. Its all described in the relativity literature if one knows where to look and it can't be found in particle/GR/cosmology texts etc.

 

I see no point in leaving out interesting points I know of. One can simply feel free to ingore them if they wish.

 

Let me give you an example of certain points I'm refering to; The expression F = ma was not an expression given by Newton. It was given by Euler. For this reason F = ma has been come to be know as Euler's definition of mass. This definition can be used to define what can be called Mach's definition of mass. However Newton's was F = dp/dt. This is the definition used in relativity (the topic of this forum). This yields a definition of mass which has come to be called Weyl's definition of mass. This is the definition I stated above and the one people in relativity use.

 

If a basic question is asked then I'll give a basic response. If that response is objected to or contradicted and I believe that objection/contradiction is incorrect then I'll state so. I see that as the purpose of a discussion forum.

 

Thanks

 

Pmb

By the way, some of what I've relayed has come from speaking to research physicists. For example, on this point I once asked a very well-known research physicist if "light has mass" to see his response. He told me that sometimes its useful/helpful to think in those terms. He thinks this stongly enough to teach it to his students in upper level courses at a major "physics university."

 

So let us not neglect how a research physicists thinks and only focus by what appears in his published papers only (not that he leaves it out but I haven't scanned Physical Review D for example for such usage).

 

Pete

But I don't see how you can contend that you are presenting the viewpoint of the "relativity community" with a list of college intro courses' date=' beginner's textbooks and the above list of journal articles - the American Journal of Physics is not a research journal, it is a teaching journal.

[/quote']And I don't see how you can contend that you're presenting the viewpoint of the "relativity community" by saying that they don't use something which is widely used in the physics literature. I've given you examples from that literature. And these examples are not all from intro college courses. And because the journals I spoke of are teaching journals it can't be taken to mean that they don't reflect the relativity community. In fact it accurately reflects what some relativistists in the relativity people want to relay to students entering the relativity community. I suppose I can go to the research journals but I don't see the point.

 

You say that I've presented a list of college intro courses. That is correct. But that is not the only thing I presented. Many of the texts are quite advanced textbooks. For example if you followed the link to the examples in relativity that I posted they you'll see the following texts listed

 

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

Cosmological Physics, John A. Peacock, Cambridge University Press, (1999) [Cosmology text used for MITs graduate cosmology course]

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

 

These are advanced special/general relativity texts. I neglected to add

 

General Relativity, Robert W. Wald, University of Chicago Press. This is the GR text which is considered to be the most advanced GR text that is in publication. Note from page 72

However, Posson's equation tells us that

 

del²Phi = 4*pu*rho

 

where rho is the mass (i.e. energy) density of matter ...

One can't call rho mass and also refer to it as energy without it meaning that the mass is relativistic mass.

 

Recall a proof used by MTW on patge 141

Calculate in a specific Lorentz frame. Consider first the momentum density (components T^j0) and the energyt flux (components T 0j). They must be equal because energy = mass ("E = Mc² = M")

 

T^j0 = (energy flux)

 

= (energy density)x(mean velocity of energy flow)j

 

= (mass density)x(mean velocity of mass flow)j

 

= (momentum density) = T^0j

The mass used here is what you call "relativistic "mass." You also neglected to mention that the links I gave are from particle accelerator laboratories, not from intro courses.

All you've shown, really, is that some people use it as a teaching tool, or to explain it to the public (e.g the www).

And you've shown .... what? Why would you think that those research physicists who teach relativity are teaching something that they don't use or expect their students to use? And why would you think that its merely a teaching tool? Simply because its being taught? That could be said about anything that appears in any text. Simply put - a teacher would not teach a student something that they thought to be a bad idea or something that is invalid/meaningless. Also, Jammer's book is on mass from the philosophical point of view. That means it is a book designed to get to the very heart of this topic in physics.

 

You said that the relativity community doesn't use it anymore. It now appears that you mean something different. What is it you mean please? Do you mean that they teach it but don't use it in certain journals?

 

I'd ask you to demonstrate that it doesn't appear in the research literature but that, of course, is an impossible task since one can't prove a negative. Its very important to note that what appears in research journals is quite watered down. By this I mean that you never get to see what thought processes lead research physicists to the end product. The end product rarely, if ever, shows this process, only the end results of the thinking process, not what led the research scientist to his conclusions. It doesn't show what insights he used or where they came from etc.

 

My problem is that I'm disabled at the moment and only gave access to the Am. J. Phys. I'll get to other journals some day and pick through them to see what's there. But whatever it is its quite limited. Research physicists restrict their research to areas like general relativity, cosmology and particle physics. Those areas don't use all the aspects of relativity. I sincelerly doubt that you'd be able to find a journal which discusses the mass of a capacitor. Yet its a legitimate topic. Particle Physicists only deal with objects which can be treated as particles so their use is limited. Plus it is their concern not to study dynamics (where this term is meaningful). They study the inherent properties of particles, such as lifetime, charge, "mass" etc. If you ever see a particle physicist refer to the "lifetime" of a neutron they'll most likely be speaking of the proper lifetime. However proper lifetime and lifetime are two different things which two different meanings and values measured in the lab. But all particle physicists know that the lifetime of a particle is relative whereas the proper lifetime isn't. I see no reason for them to use terms which are inconvenient to them. I expect them to simplify terms. I.e. its simpler to say 'lifetime" so long as everyone unders what it means - proper lifetime. Same with mass. Its simpler to say "mass" rather than "proper mass".

 

Pmb

Whilst swan's explanation may be technically incorrect, it still illustrates the correct definitions of the terms; I don't think the answers are geared towards physicists.

I don't believe he was "technically" incorrect per se. I only meant to say that he was incorrect as far as how everyone in the relativity community defines the term mass.

 

There are two ways in which people use the term "mass" in relativity. One is as I've described and the other, i.e. proper mass, is proportional to the square of the magnitude of the particle's 4-momentum. However this is a vauge term since 4-momentum can't always be defined for an object. 4-momentum can only be defined for an isloated object or a pointlike object (no spatial dimensions). But some people use the term mass to refer to proper mass (which is what swan calls rest mass).

 

However it is my opionion that the magnitude of some 4-vectors are best refered to as proper quantities. Thus if

 

dX = (cdt, dx, dy, dz)

 

is a spacelike spacetime displacement then c^2dT^2 = dX*dX where dT is referred to as the proper time interval between the two events. If dX is a spacelike spacetime displacement, then c^2ds^2 = dX*dX where ds is proper distance.

 

Likewise

 

P = (mc, p)

 

is the 4-momentum of a particle whose mass is m and 3-momentum is p.

 

m₀²c² = P*P

 

where m₀ is the particle's proper mass.

 

The terms proper mass and rest mass do not refer to the same thing. Rest mass denotes the value of proper mass when the velocity is zero. However m(0) does not equal m₀ unless dt/dT = 1.

 

For example; In a wide variety of circumstances (i.e. time-orthogonal spacetimes, where g_0k = 0 -- no spacetime dragging) dt/dT is a function of both the particle's speed and the gravitational potential of the particle. dt/dT = 1 only when v = 0 and the gravitational potential also is zero (more on this can be found in Moller's text on general relativity).

 

Pmb

Inertial mass is the mass in F=ma.

That is incorrect. That is not the correct definition of inertial mass. That is a relationship between force and inertial mass in those cases when the mass is constant. Force is not defined as F = ma. Its defined as F = dp/dt.

 

As far as anybody can measure, this is the same as the gravitational mass in F=GMm/r²

The qauntity F/a does not have the same value as the quantity that might be considered to be "gravitational charge" as is the quantity m in in F=GMm/r². The expression for gravitational force in relativity is not F=GMm/r². Its the m in

 

[tex]G_k = m\Gamma^{\alpha}_{k\beta} v_{\alpha}v^{\beta}[/tex]

 

..relativistic mass is the mass from E = mc² if you use total energy.

That is incorrect. The m in that expression is an equality under certain circumstances. Not all circumstances however. There are cases when E = mc² is invalid. Note: E = mc² is only total energy when the potential energy V is zero.

so what is a photon's relativistic or intertial mass?

For a photon the inertial mass can be expressed in various ways. If the magnitude of the photon's momentum is p then its inertial mass is given by m = p/c. If the photon's frequency is f then since E = hf the inertial mass is m = E/c² = hf/c².

 

In a sense they are undefined ...

That is incorrect. The inertial mass of a photon is a very well defined quantity.

..- science doesn't use the terms anymore.

That is also incorrect. The concept is widely used. For example;

 

Textual Examples ---

Relativity: Special' date=' General and Cosmological,[/i'] Rindler, Oxford Univ., Press, (2001)

Cosmological Physics, John A. Peacock, Cambridge University Press, (1999)

Understanding Relativity: A Simplified Approach to Einstein's Theories, Leo Sartori, University of California Press, (1996)

Basic Relativity, Richard A. Mould, Springer Verlag, (1994)

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

Gravitation from the Ground Up, Bernard F. Schutz, Cambridge University Press, (2003)

 

Journal Examples

Apparatus to measure relativistic mass increase, John W. Luetzelschwab, Am. J. Phys. 71(9), 878, Sept. (2003).

Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995).

In defense of relativistic mass, T. R. Sandin, Am. J. Phys. 59(11) 1032 (1991).

A simple relativistic paradox about electrostatic energy, Wolfgang Rindler and Jack Denur, Am. J. Phys. 56(9), Sept. (1988).

An elementary development of mass-energy equivalence, Daniel J. Steck, Frank Rioux, Am. J. Phys. 51(5), May (1983).

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

 

Observed Relativistic Mass Increase for 0.3 eV Electron, G. Gabrielse and H. Dehmelt, Bull., Am. Phys. Soc. 25, 1149 (1980).

 

Online Examples

 

http://web.uniud.it/cird/girepseminar2003/abstracts/pdf/mulaj.pdf

http://physics.syr.edu/courses/PHY106/Slides/PPT/Lec16-Special-Relativity_2.pdf

http://physics.syr.edu/courses/PHY106/Slides/PPT/Lec17-Special-Relativity_2.pdf

http://www.physics.mq.edu.au/~jcresser/Phys378/LectureNotes/SpecialRelativityNotes.pdf

http://astro.wsu.edu/allen/courses/astr150/Einstein.pdf

http://www.phys.virginia.edu/classes/109N/lectures/mass_increase.html

http://www.eas.asu.edu/~holbert/eee460/Relativity.pdf

http://www.ucls.uchicago.edu/Academics/depts/science/physics/Relativity.pdf

http://www.dur.ac.uk/Physics/students/physics_specialrelativity.html

http://www.ph.rhul.ac.uk/course_materials/PH154/Relativistic%2520mass%2520and%2520dynamics.pdf

http://www.lima.ohio-state.edu/physics/113sp03/113Lectures/113rela2.pdf

http://www.lima.ohio-state.edu/physics/113sp03/113Lectures/113rela2.pdf

http://www.physics.fsu.edu/users/ng/courses/phy2054c/hw/Ch26/ch26.pdf

 

http://www.astro.washington.edu/tmurphy/phys110/faqs/AB05.05.html

But the most honest answer to your question is yes--light has mass.

 

Particle Accelerator Labs

 

http://humanresources.web.cern.ch/humanresources/external/training/tech/special/AXEL2003/AXEL-2003_L02_24Feb03pm.pdf

http://www.neutron.anl.gov/hyper-physics/inertia.html

http://aether.lbl.gov/www/classes/p139/animation/sr.html

http://www.fpm.wisc.edu/safety/Radiation/2000%2520Manual/chapter12.pdf

 

 

mass is the invariant mass (aka rest mass)..

That assertion is dependant on how one defines "mass" in the first place.

 

.. - that's the only term you can really discuss, because it's reference-frame independent.

That is incorrect. One can always speak of frame dependant quantities in relativity because those are the only quantities which are measureable. Its not quite accurate to say that relativistic mass is not invariant unless you're quite specific about what you mean by that statement. It can be very tricky. For example: If the 4-momentum of a particle is P and the 4-velocity of an observer is U then the relativistic mass as measured by that observer is proportional to the scalar product of P and U[/b]_obs, i.e.

 

m_obs = P*U_obs

 

In case you think this sort of thing doesn't appear in the relativity literature then I recommend reading

 

Energy Conservation as the Basis of Relativistic Mechanics II, J. Ehlers, W. Rindler, R. Penrose, Am. J. Phys. 33, 995-997 (1965). From page 996

For convenience, we introduce instead of u a new variable, namely the Lorentz factor

 

[tex]\gamma = \gamma(u) = (1-u^2/c^2)^{-1/2}[/tex]

 

and we temporarily write E(u,S) = E[[itex]\gamma[/itex]] when only one scalar state is under discussion. We recall that the 4-velocity associated with the 3-velocity u is given by U = [itex]\gamma[/itex](u)(u,c); hence the 4-velocity of an inertial observer relative to his own rest frame is V = (0,c). Thus

 

[itex]\gamma[/itex] = V*U/c²

 

i.e., the Lorentz factor of a particle relative to any inertial observer is given by the scalar product of the two corresponding 4-velocities divided by c². And this product, being invariant, can be calculated in any inertial frame.

 

It is good to keep in mind that the components of a vector or tensor are defined in terms of the scalar product of the tensor or vector with a basis vector and as such is a scalar.

 

..for the people that still refer to relativistic mass, it's E/c2,

For people who do that they run the risk of making an error in those cases when p/v does not equal E/c². There are cases where the relationship p/v = E/c² does not hold and therefore E = mc² does not hold. A simple example would be the mass of a gas in a box.

 

...but it's not a useful quantity since it's not invariant. So it's not generally used.

Its both useful and as such still used.

 

inertial mass is even more problematic, because if you look at it from point of view of resistance to acceleration (i.e. m=F/a), ..

 

Since that expression is invalid this arguement is flawed. Mass is the proportionality quantity between velocity and momentum. It therefore represents the resistance to changes in momentum, not velocity.

 

 

More on this subject can be found listed in the references here

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

 

and in the journal articles listed here

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

 

Pete

can someone explain the exact difference between:

 

inertial mass' date='

relativistic mass &

rest mass.[/quote']The inertial mass of a body is defined as the quantity m such that the quantity mv is conserved in particle collisions as observed from an inertial frame of reference. If you're discussing relativity then some people like to use the term relativistic mass for this same term. I don't like it since it makes people think that its somehow different than inertial mass. If the body can travel at speeds less than the speed of light then it turns out that the mass is a function of speed, i.e. m = m(v). The quantity m₀ = m(0) is called "rest mass" since v = 0, i.e. the body is at rest.

 

Pete

Thx for the answer Pmb.

 

However' date=' I either didn´t understand your answer or there has been a misunderstanding because your answer actually backed up what I said: That it´s not possible to tell space from time in spacetime.

I´ll also assume a flat spacetime:...

[/quote']Perhaps I was unclear as to why you can tell the difference between space and spacetime but its quite true that you can. You do it all day long. I can move my car from x = 0 to x = 1 and I can then move my car from x = 1 back to x = 0. However, while its true that I can move my car from t = 0 to t = 1 it is not true that I can move my car from t = 1 back to t = 0. (x,y,z) detnotes a point in "space" while "t" denotes a momentum in "time." Spacetime is the collection of points whose coordinates are (t,x,y,z). It is true that mathematically space and time can in some ways be treated on the same footing mathematically they cannot be treated on the same footing physically.

 

- >>"You measure "space" with a ruler and "time" with a clock."

Yes. And the time I measure is the length of my path through spacetime.

 

Why do you think that's true and what is it you mean by "length"?

 

Normally in GR the term "length" as it pertains to spacetime is not defined the same way as it is in Euclidean geometry. typically the term is used to denote the "interval" between two points. For instance, the spacetime interval between two points on a worldline of a photon in flat spacetime in an inertial frame is zero, even if the actual photon has traveled light years. The spatial displacement equals t = x/c and the temporal displacement equals x = ct, the spacetime interval [itex]\Delta s[/itex] is zero. Althought the spatial and temporal coordinates are proportional in this case they are not identical.

 

- Your elevator example is a good example for my claim that you cannot tell if a trajectory is curved or not:

I hope we can agree that the path of the light is completely independent of what the elevator does. As you said the beam will be a line for an observer in the elevator if it isn´t accellerating but will appear curved if it accellerates. So since both reference systems are equivalent there´s no way telling if the trajectory is curved or not.

Why do you think that there are "special" coordinate systems in which light is "really" moving on a straight spatial trajectory? Why true in Newtonian mechanics it is false in GR.

 

Same as above: A judgement like "this path is curved" can only be made within a coordinate system and is dependant on it.

That's correct. And that's why I was explaining to you. I.e. that whether the spatial trajectory of a beam of light is curved or not depends on the frame of reference.

 

 

Frame-dependant statements, however, are quite useless.

Then the question "Is the spatial trajectory is curved" is quite useless. But I can still speak of it. Its not true that frame-dependant statements are useless. The observer lives in a particular frame and the observer himself is a geometric object. Therefore statements like "spatial path is curved" has a geometric meaning independant of a coordinate system and depends only on observer and phenomena.

 

That´s the point in the whole tensor stuff (so you can tell if a trajectory is a geodesic or not because that´s a tensorial statement).

The whole point of tensor stuff is that relativity demands that the laws of physics not depend on a particular coordinate system. It does not mean that observations are coordinate independant.

 

Pmb

Partly agreed. I rather see a straight line as a special case of a geodesic for a flat space. Well, I can live with your statement but I think it cries for new confusion (keep in mind that most people here don´t know GR).

People should keep examples of geodesics in mind so as not to be confused. A few examples go a long way. Here are a few which are illuminating

 

(1) geodesics on an infinite flat plane

(2) geodesics on a sphere

(3) geodesics on a cylinder

(4) geodesics on a cone

 

For #1 the geodesics are straight lines.

For #2 the geodesics are great circles.

For #3 the geodesics are spirals. See Figure 4 at

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

For #4 the geodesics are hard to picture. See figure 2 at

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

 

Is it really possible to tell space from time in spacetime in general?

Absolutely. Its quite important to distinguish the two. You measure "space" with a ruler and "time" with a clock. As Einstein himself said in Science in Feb 17, 1921 issue page 783

..it follows that, in respect to its role in the equations of physics, though not with regard to its physical significance, time is equivalent to space coordinates.

Consider the inside of an elevator in flat spacetime in an inertial frame of reference. Let there be a rectangular set of spatial axes, i.e. an xy coordinate system whose axes are mutually perpendicular. A beam of light will move on a spatial curve described by x = ay + b. This is a straightline in space and is thus also a geodesic in space. The worldline is a line in spacetime for which x = ct (when x and "ct" axes are perpendicular). This too is a straight line, a straightline in spacetime - a geodesic. However if the elevator is uniformly accelerating then the curve the light will move on a spatial curve which is not a straight line. The curve in spacetime is also not a staight line. However it is a trajectory in spacetime is still a geodesic.

 

Pete

I'm confused about something. I've been hearing that photons are massless, yet it is predicted that blackholes exist. The appearant contradiction here is that blackholes are predicted to exist because light is subject to gravity in the same way matter is. That is, light curves toward the source of gravity. They say that blackholes are objects with so much gravity that they do not let light escape (i.e. Light curves so much that it always ends up travelling directly towards the object's center of gravity). But how can light be influenced by gravity in this way if light, being composed of photons, is massless? Doesn't it require mass to be "heavy"?

 

Light does have mass (aka inertial mass aka relativistic mass). It has zero proper mass (aka rest mass) and that's what you were thinking about. As such it is effected by gravity. One simply has to turn to The Feynman Lectures 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.

That´s absolutely correct. Those "straight lines" are called Geodesics which should sound familiar to you. In fact, not only light but all particles move on Geodesics.

That's not quite right. While it is true that light moves on geodescics is not quite right to say that a geodesic is a "straight line". Its more appropriate to say that its the straightest possible line. In this case it is a geodesic, not in space, but in spacetime. Light does not travel on a straight line is space. It travels on a curved line in space.

 

Pete