Gaining Kinetic Energy Through Time: Explained

[Jimmy Snyder]

How do you measure or calculate the rate at which something is "moving through time?"

It was my inability to adequately answer this question to myself that motivated the first response I made in this thread. Time is a frame dependent concept. The only frame I can think of where "moving through time" might have meaning is the rest frame. Because in that frame time is proper time and proper time is frame independent. We have that in this frame (and in no other) an object "moves through time" at a speed of 1 second per second. I realize that this is a lame definition of "moving through time" I just thought it might address the inner meaning of nemosum's question. Apparently it did not.

 

[Mortimer]

The notion of "speed in time" was discussed in a couple of older threads already (I don't recall which ones exactly). My favorite definition of 'speed in the time dimension' if the one that can be found in Brian Greene's The Elegant Universe. You can rewrite the Minkowski 4-velocity components:
c^2=(dct/d\tau)^2-(dx/d\tau)^2-(dy/d\tau)^2-(dz/d\tau)^2
into the alternative form:
c^2=(cd\tau/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
which gives 4-velocity components for an Euclidean space-time geometry, while maintaining the invariant scalar value c for this 4-velocity.
Here the factor cd\tau/dt represents the temporal speed component which has value \sqrt{c^2-v^2}.
In essence this promotes \tau to the actual fourth dimension, while t becomes a parameter or a fifth dimension.

 

[robphy]

The notion of "speed in time" was discussed in a couple of older threads already (I don't recall which ones exactly). My favorite definition of 'speed in the time dimension' if the one that can be found in Brian Greene's The Elegant Universe. You can rewrite the Minkowski 4-velocity components:
c^2=(dct/d\tau)^2-(dx/d\tau)^2-(dy/d\tau)^2-(dz/d\tau)^2
into the alternative form:
c^2=(cd\tau/dt)^2+(dx/dt)^2+(dy/dt)^2+(dz/dt)^2
which gives 4-velocity components for an Euclidean space-time geometry, while maintaining the invariant scalar value c for this 4-velocity.
Here the factor cd\tau/dt represents the temporal speed component which has value \sqrt{c^2-v^2}.
In essence this promotes \tau to the actual fourth dimension, while t becomes a parameter or a fifth dimension.

From http://www.rfjvanlinden171.freeler.nl/dimensionshtml/index.html ,
which is linked from your homepage, appears this sentence in the abstract:

The velocity addition formula shows a deviation from the standard one; an analysis and justification is given for that.

So, it seems to me that this alternative velocity addition formula [justified or not] suggests that this "Euclidean Relativity" is not "the Theory of Relativity". In addition, I note the following statement from the first sticky in this forum:

This forum is meant as a place to discuss the Theory of Relativity and is for the benefit of those who wish to learn about or expand their understanding of said theory. It is not meant as a soapbox for those who wish to argue Relativity's validity, or advertise their own personal theories. All future posts of this nature shall either be deleted or moved by the discretion of the Mentors.

 

[nemosum]

Whoa robphy! I'm the new kid on the block, and I'm not even qualified to post a theory of my own considering that I know hardly anything about physics. I was simply wondering if STR left the implications of what I thought it did, so you don't have to bite my head off!
As to measuring movement through time, I agree with jimmysnider when he said that in a frame of rest you move through time at a rate of 1 sec./sec. Therefore, isn't it possible to measure the rate at which an object in a different frame moves through time by taking a certain fraction of that? Like I said before, I'm still in pre-calculus, but it seems to me that this wouldn't have to be that complicated. But you guys are the experts here so...

 

[robphy]

Whoa robphy! I'm the new kid on the block, and I'm not even qualified to post a theory of my own considering that I know hardly anything about physics. I was simply wondering if STR left the implications of what I thought it did, so you don't have to bite my head off!
As to measuring movement through time, I agree with jimmysnider when he said that in a frame of rest you move through time at a rate of 1 sec./sec. Therefore, isn't it possible to measure the rate at which an object in a different frame moves through time by taking a certain fraction of that? Like I said before, I'm still in pre-calculus, but it seems to me that this wouldn't have to be that complicated. But you guys are the experts here so...

nemosum,
relax... carefully look over what I have quoted and my comments. I was not commenting on your comments or your question, which seems like a reasonable question from someone trying to understand this subject.

 

[El Hombre Invisible]

I may be dense, but I haven't seen an answer to JesseM's question yet, so I'd like to bring it back into view.
How do you measure or calculate the rate at which something is "moving through time?" Slinging words around doesn't do much for me, but if you can give me a specific numeric example of what you're talking about, I might be able to make some sense out of those words.

That's easy enough. Get an old-style egg-timer and set it going. As the grains of sand slip to the bottom, you are moving through time.

I think in this case, though, we're talking about the differences between the rate by which time passes for an object in its own rest frame compared to that by which it passes in a frame in which it is traveling close to c.

In this case, if you define the time it takes for the egg timer to finish as 1, and the time it takes for the traveller's egg-timer to finish as t, then the traveller "moves through time" 1/t times as "fast" as the observer in his rest frame.

 

[jtbell]

I think in this case, though, we're talking about the differences between the rate by which time passes for an object in its own rest frame compared to that by which it passes in a frame in which it is traveling close to c.

In that case, it's just the time-dilation factor \gamma, or maybe 1/\gamma depending on which way you're looking at it.

To me, the problem with terminology like "moving through time" is that (to me) it seems to imply that there is some kind of absolute time that an object's "motion though time" can be reckoned relative to. But there is no such thing! These different rates of "motion through time" for an object are always observed by different observers, and two different observers in different reference frames observe different rates. The object itself (or an observer riding along with it) always observes its own "motion through time" to be at the same rate.

The problem with using words to discuss stuff like this is that different people tend to associate different things with the same words. Also, if different peole use different sets of words to describe the same things, it gets confusing. I consider terms like "motion through time" to be more like metaphors than terms for practical discussion. I'm pretty sure you won't find it in any real textbook, as opposed to hand-waving popularizations of relativity.

 

[eon_rider]

Still a good way of putting it.

Just my opinion.

The problem with using words to discuss stuff like this is that different people tend to associate different things with the same words. Also, if different people use different sets of words to describe the same things, it gets confusing. I consider terms like "motion through time" to be more like metaphors than terms for practical discussion. I'm pretty sure you won't find it in any real textbook, as opposed to hand-waving popularisation of relativity.

Where as this may be very true.

"moving through time" is that (to me) it seems to imply that there is some kind of absolute time that an object's "motion though time" can be reckoned relative to. But there is no such thing! These different rates of "motion through time" for an object are always observed by different observers, and two different observers in different reference frames observe different rates. The object itself (or an observer riding along with it) always observes its own "motion through time" to be at the same rate.

This was very understandable and very well put.

Perhaps there is a time (excuse the pun) for talking about things using words and concepts regardless of the possibility of confusion, and a time for just doing the maths. Not sure.

But I found your comments about "motion through time" very helpful and made total sense, to me.

Thanks,

Eon.

 

[nemosum]

To me, the problem with terminology like "moving through time" is that (to me) it seems to imply that there is some kind of absolute time that an object's "motion though time" can be reckoned relative to. But there is no such thing! These different rates of "motion through time" for an object are always observed by different observers, and two different observers in different reference frames observe different rates. The object itself (or an observer riding along with it) always observes its own "motion through time" to be at the same rate.

Alright then, picture to astronaughts sitting motionless in an area of pretty much flat space (or however you say space where there is little or no gravitational effect). Each astronaught's time would be excatly the same as the others, therefore there would be no confusion about who had the right time, and they're motionless, hovering in flat space, meaning there is no time dilation due to either motion or gravity. Couldn't we reasonably assume their time to be "True" time? A time by which all others may be measured?
Of course, when it comes to the measuring there would naturally be confusion as to who is moving in what frame, and it would be hard to know what to measure. BUT, there is always truth. I mean, someone is the one moving, and someone is not. No matter who says what. Some things just are no matter who is watching.

 

[Mortimer]

BUT, there is always truth. I mean, someone is the one moving, and someone is not. No matter who says what. Some things just are no matter who is watching.

I'm afraid this is where you go wrong. Your reasoning assumes a preferred reference frame. One of SRT's basic statements is that there is no such preferred frame. In any inertial (i.e. non-accelerating) frame the laws of physics are identical to any other inertial frame. There exists only relative motion between two astronauts A and B.
If they move relative to each other, astronaut A thinks B is moving and astronaut B thinks A is moving. Astronaut A observes B's clock to slow down but astronaut B observes A's clock to slow down. Both are correct.
If you think this is weird, compare e.g. two observers standing a few kilometers apart on the curved surface of the Earth. Both think they are standing 'on top' of the Earth and the other is 'lower' or partly behind the horizon. Although the physics of this example is of course not the same, it reflects the basic idea.

 

[nemosum]

Yeah, you're right, I kind of regreted what I posted after I posted it, realizing my big mistake. I get the symultaniety. But if things all depend on who's looking at them, then the whole universe is in chaos. In that case, you could basically move planets just by looking at them as you rush past in your spaceship. It's just hard to make sense of. The only book I read "Einstein's Univers" by Nigel Calder explains symultaniety in I think one chapter, but then he hardly ever referres to it again, making it a little hard to understand.
And another question,if Astonaught A thinks B is moving, and visa-versa then when they both come together and stop why don't both of them see each other's clock as being slow. If A was really the one moving, then his clock should be the appearently slower one, but A would see B's clock as being slower. So what happens when A stops? Does A suddenly see B's clock jump ahead?

 

[nemosum]

I have another question. In that book the guys says that as you approach a center of gravity a little bit of your rest-energy is shed. He makes reference to a certain Roger Penrose (I think) who designed a hypothetical machine which would launch a bucket full of garbage towards a rotating black whole. As the garbage approached, it woud emit large amounts of its rest-energy, which the bucket would absorb in the form of kinetic energy. The bucket would then swing around the black hole (it didn't reach the inner horizen), and come back to the machine, which would turn it's kinetic energy into power by means of a turbine. At least that's how I think it was supposed to go. The problem is that many people have been telling me that being closer to a center of gravity doesn't make your rest-energy less. What am I missing?

 

[Mortimer]

And another question,if Astonaught A thinks B is moving, and visa-versa then when they both come together and stop why don't both of them see each other's clock as being slow. If A was really the one moving, then his clock should be the appearently slower one, but A would see B's clock as being slower. So what happens when A stops? Does A suddenly see B's clock jump ahead?

Whenever they get together, one or both of the astronauts has to de- or accelerate. During the acceleration, frames are no longer inertial (a requirement for the application of simple SRT calculations). http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#gap"of John Baez gives an idea w of what happens with the clock readings of the astronauts. Accelerations can indeed imply a seeming 'jump' in the clock readings.
This is essentially a "twin paradox" situation about which a zillion threads have already run, so we might better not turn this one into it.

 

[Intuitive]

Doesn't the relaxation of an Atoms vibration (XYZ) tend to slow it's relativity of time to an outside observer, Isn't this true also in cryo Physics? If this is the case.

How does near absolute zero temperatures effect Radio active Materials?

Does Radioactive decay of a Radio Active Isotope slow down as the material approaches absolute zero temperature?

How does Sub-Atomic Relativity differ from Atom Relativity?

Can Magnetic Relaxation techniques lead to a technology of Suspended animation like Cryo-Physics?

Some Curious questions.

 

[pmb_phy]

Alright, I'm new at this, so I'll try to explain my question as best I can, and I hope it's not a stupid one. When you move through space 3-dimensionally you gain Kinetic energy, right?

Hi nemosum - Welcome to physicsforums!

Forgive me for entering the discussion late. Other things have taken a higher priority and this is an interesting thread.

The kinetic energy of a body changes if and only work is being done on the body. So the kinetic energy may or may not change.

So wouldn't something moving through time, since it's just another dimension according to Einstein, gain kinetic energy for that too?

The phrase "moving through time" brings motion to mind and when the term "motion" is used it has a very specific meaning. Also note that Both energy and kinetic energy are defined as they are and I don't see what it'd mean for a body which has no work being done on it having its kinetic energy changed simply because time has changed. It'd mean a violation of the law of conservation of energy.

Pete

 

[pmb_phy]

This 4-vector captures everything to do with motion through spacetime, and kinetic energy is the 0th component, and the other 3 components are just the ordinary momenta.

That is only true for a point particle. In general one needs the energy-momentum tensor to describe everything about an object. Its a shame that this is under-emphasized nowadays.

Pete

 

[nemosum]

Mr. Pete, I stated my question wrong in the first place. What I really was asking was if it was possible that an object's rest-energy was a result of "moving" through time. I'm realizing more and more that I shouldn't have used the word "moving". I think of it more like a speedometer, were, if say you were traveling at 99% of the speed of light, you "timeometer" would be very close to 0. And if you aren't traveling at all, then it's at whatever the normal setting can be defined as, or 1 sec./sec.
So you're not measuring "movement" as much as you are the rate at which you...hmm...still working on how to define this.

 

[pmb_phy]

Mr. Pete, I stated my question wrong in the first place.

Hence my appology for jumping in late. That happens often. Its too difficult for me to sit here and read the entire thread right now (back pain due to surgery etc) so I did my best and took a look through a survey of representative posts. I guess I failed at getting the gist of the conversation. Sorry.

What I really was asking was if it was possible that an object's rest-energy was a result of "moving" through time.

I don't see how. A photon has zero rest energy and it "moves through time." That its proper time is zero makes little difference to me as I see it.

I'm realizing more and more that I shouldn't have used the word "moving".

Yeah. That happens a lot. It seems to be a common error so don't worry about it.

I think of it more like a speedometer, were, if say you were traveling at 99% of the speed of light, you "timeometer" would be very close to 0.

You do understand, don't you, that regarless of your "speed" (which is relative so something which you have left unstated and wherein lies the problem) that if you look at your wristwatch (time-o-meter) it will never run any different. Its workings/rate will always be measured by you to be the same - no matter what your speed is relative to something else.

Pete

 

[masudr]

That is only true for a point particle. In general one needs the energy-momentum tensor to describe everything about an object. Its a shame that this is under-emphasized nowadays.
Pete

The 4-vector is a tensor, (but i imagine you're referring to the symmetric (0,2) rank tensor that is commonly called the stress-energy tensor.

It does simplify matters to talk about point particles, and I felt restricting the discussion to that as opposed to general systems (esp. fields) would suit the discussion better.

I don't believe this is under-emphasized "nowadays", although I wouldn't really know; I haven't been in the relativistic learning scene for very long. In fact it is clearly emphasised when discussing the field equations for the metric; this tensor is a clear feature of the equation describing (what is commonly called) "mass-energy" (but of course it describes more than that) of the system.

 

[Intuitive]

Relativity of a Watch changes if it is in the local of a Powerfully but (randomly) synched Electromagnetic Pulse, because the Watch will absorb a random pulse and the pulse will change the kenetics of the watch, This is also true of both analog watches and Electronic watches.

The time difference is proportional to the randomness of the pulse.

These are possible factors in Relativity Experiments using Watches so there can be an error factor involved.

 

[pmb_phy]

The 4-vector is a tensor, (but i imagine you're referring to the symmetric (0,2) rank tensor that is commonly called the stress-energy tensor.

Yup. Sorry. Thanks for the clarification.

It does simplify matters to talk about point particles, ..

So long as the domain of applicability is noted as it must be with all definitions. Its rarely, if ever, noted that a second rank tensor is required in general. I know only of Einstein, Rindler and Tolman to make this statement.

In fact it is clearly emphasised when discussing the field equations for the metric; this tensor is a clear feature of the equation describing (what is commonly called) "mass-energy" (but of course it describes more than that) of the system.

What does the metric tensor have to do with it? I was referring to special relativity only and not GR.

To make sure we're on the same page please see the example in SR that I worked out.

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

Thanks

Pete

 

[jtbell]

But if things all depend on who's looking at them, then the whole universe is in chaos. In that case, you could basically move planets just by looking at them as you rush past in your spaceship.

But this is true in Newtonian mechanics too. It wasn't new with Einstein.

 

[nemosum]

Quote:
Originally Posted by nemosum
But if things all depend on who's looking at them, then the whole universe is in chaos. In that case, you could basically move planets just by looking at them as you rush past in your spaceship.

But this is true in Newtonian mechanics too. It wasn't new with Einstein.

But the universe ISN'T in chaos! Thing DON'T just fly around because someone looked at them!

 

[daniel_i_l]

These are possible factors in Relativity Experiments using Watches so there can be an error factor involved.

If during motion with constant velocity different watches can give different results, than that could be a way that an observer could tell if he is moveing or not, by comparing two watches? In motion with constant velocity all kinds of watches must give the same time.

 

[daniel_i_l]

But the universe ISN'T in chaos! Thing DON'T just fly around because someone looked at them!

It is true that an observer moving at a constant speed past a star is jusified to say that he is at rest and the star is moving past him. But that doesn't mean that the star is flying around the universe, relative to you, the whole universe is moving past you but it is at rest relative to itself (two stars that are moveing relative to you are at rest relative to each other) so there is no chaos, but all views are correct.

 

[masudr]

But the universe ISN'T in chaos! Thing DON'T just fly around because someone looked at them!

No one ever said that. There are (at least we think so, and have found most of these) strict rules as to how the universe behaves, and very few of them permit things "flying around just because someone looked at them" (in certain limits anyway, esp. in the relativistic limit).

 

[masudr]

I was referring to special relativity only and not GR.

Even in SR, for example, classical electrodynamic field theory, we are taught that we require the stress energy tensor to describe the 4-Lagrangian. Similarly, we are taught that Noether's theorem comes out of the fact that the 4-divergence of the stress energy tensor is 0. So I feel that there is sufficient emphasis on the fact that the stress energy tensor is required to describe a system.

Furthermore, in your web page, you are describing a box. And E=mc^2 only applies to particles and in it's own rest frame. Particles in general are constrained by E^2 = p^2 + m^2 (where I have set c=1).

 

[pmb_phy]

So I feel that there is sufficient emphasis on the fact that the stress energy tensor is required to describe a system.

You'd think so. But the problem is that people always ignore that fact. I see it on a constant basis in my travels in SR (last 15 years).

Furthermore, in your web page, you are describing a box. And E=mc^2 only applies to particles and in it's own rest frame. Particles in general are constrained by E^2 = p^2 + m^2 (where I have set c=1).

That is incorrect. What "E = mc²" means and in what frame it holds will depend critically on how "m" is defined. It seems to me that you're using the convention wherein "m" means "rest mass." I do not follow that convention (For many many reasons which I've considered over the last 15 years. They are listed in a paper I wrote on this subject. Its 80 pages long and pretty boring to read. )

In your convention the equation should read E₀ = mc². I've modified my web page to clarify terms. Have you ever read Rindler's Intro to SR text in your travels?

I also have no idea where you see E = mc² in that page! It does not apply to the mass here so it is not used. I mention energy only to show that matter (where I use the term "matter" as Einstein defined it) is in the box.

I have updated my web page to clarify these things.

Pete

 

[nemosum]

I have another question. In the book "Einstein's Universe" Nigel Calder says that as you approach a center of gravity a little bit of your rest-energy is shed. He makes reference to a certain Roger Penrose (I think) who designed a hypothetical machine which would launch a bucket full of garbage towards a rotating black whole. As the garbage approached, it woud emit large amounts of its rest-energy, which the bucket would absorb in the form of kinetic energy. The bucket would then swing around the black hole (it didn't reach the inner horizen), and come back to the machine, which would turn it's kinetic energy into power by means of a turbine. At least that's how I think it was supposed to go. The problem is that many people have been telling me that being closer to a center of gravity doesn't make your rest-energy less. What am I missing?

 

[jtbell]

An individual elementary particle (such as an electron) can't "shed rest-energy", as far as I know. That would change the rest mass, which is a fundamental property of an elementary particle. I can think of three possibilities:

1. Maybe you misunderstood what Calder wrote.

2. Maybe Calder misunderstood what Penrose wrote.

3. Maybe Penrose was referring to complex objects, i.e. bound systems of elementary particles. The rest mass or rest energy of a complex object does not equal the sum of the rest mass or rest energy of its component particles. It also includes the potential energy that binds the system together, and the kinetic energy of the individual particles relative to the center of mass of the system (for example thermal energy of the atoms in a brick).