Gaining Kinetic Energy Through Time: Explained

[nemosum]

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? So wouldn't something moving through time, since it's just another dimension according to Einstein, gain kinetic energy for that too? And, if so, isn't it possible that on object's potential energy is just kinetic energy for moving through time? Especially since as time slows down for an object its potential energy becomes less. Which means that it's kinetic energy for poving through time is lessening. Do I make sense? Yeah, I was just wondering about that.

-nemosum

 

[metrictensor]

very interesting question about traveling through time affecting KE. I'll have to think about it.

 

[JesseM]

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? So wouldn't something moving through time, since it's just another dimension according to Einstein, gain kinetic energy for that too? And, if so, isn't it possible that on object's potential energy is just kinetic energy for moving through time? Especially since as time slows down for an object its potential energy becomes less. Which means that it's kinetic energy for poving through time is lessening. Do I make sense? Yeah, I was just wondering about that.
-nemosum

The problem is, what do you mean by "moving through time"? Kinetic energy is based on velocity, and velocity is (change in position)/(change in time), so it seems like kinetic energy already takes time into account as well as space. In relativity, the technical definition of kinetic energy is (\gamma - 1)mc^2, where m is the object's rest mass and \gamma is the following function of velocity: 1/\sqrt{1 - v^2/c^2}

Also, what do you mean when you say "as time slows down for an object its potential energy becomes less"? Where'd you hear that from?

 

[jtbell]

When you move through space 3-dimensionally you gain Kinetic energy, right?

Not when you're moving at constant speed.

 

[Pengwuino]

Also add to this that time is not a spatial dimension so I would assume kinetic energy doesn't apply here. I mean exactly how would it come into play mathematically?

 

[nemosum]

First of all, Something I should have stated before was that I'm only 16 and I just barely got accepted to a college and am just barely starting my major in Math and Physics, I haven' even taken Calculus yet, so my question was a truly innocent one.
Second, the only things I know about Relativity Theory came out of a 20 year-old book, that merely gave the theoretical part of the theory, and not the math. I don't know anything about the mathematics involved. So it was a truly innocent question indeed.
It was in that book I referred to before that it stated that as you get closer to a center of gravity that time slows down, AND on object's potential energy becomes less. It also said that as you get closer to the speed of light time slows down, AND an object's potential energy becomes less. I think there might have been a couple of other cases too, but the point is that I just thought it was a little interesting that every time time slows down on objects potential energy becomes less (or, at least that is what it seemed like to me).
And jtbell said that you don't gain KE when you're moving at a constant speed? But wouodn't you still have more KE than if you were just standing still right?

 

[Jimmy Snyder]

When you consider time as a separate coordinate and space as three other separate coordinates, you are thinking in a frame dependent way. For instance when you speak of a particle that is at rest in space and moving in time, you are describing that particle as it would look to an observer in a particular frame.

Energy and momentum stand in a similar relation to each other as time and space. When considered individually, they are frame dependent concepts. In the particular frame described above, the energy is rest mass and the momentum is zero.

From that point of view, the energy of a particle that is moving through time alone is its rest mass.

 

[El Hombre Invisible]

First of all, Something I should have stated before was that I'm only 16 and I just barely got accepted to a college and am just barely starting my major in Math and Physics, I haven' even taken Calculus yet, so my question was a truly innocent one.
Second, the only things I know about Relativity Theory came out of a 20 year-old book, that merely gave the theoretical part of the theory, and not the math. I don't know anything about the mathematics involved. So it was a truly innocent question indeed.
It was in that book I referred to before that it stated that as you get closer to a center of gravity that time slows down, AND on object's potential energy becomes less. It also said that as you get closer to the speed of light time slows down, AND an object's potential energy becomes less. I think there might have been a couple of other cases too, but the point is that I just thought it was a little interesting that every time time slows down on objects potential energy becomes less (or, at least that is what it seemed like to me).
And jtbell said that you don't gain KE when you're moving at a constant speed? But wouodn't you still have more KE than if you were just standing still right?

Moving at constant speed and standing still are the same thing. That's the most basic concept of relativity you need to grasp. You may choose your x, y and z axes such that you are at rest, but I may choose mine such that you are moving at constant non-zero speed.

Also, your speed does not effect your potential energy - your potential energy function does, and that is a function of distance. It may well be that the author you refer to meant that because you are accelerated towards the speed of light, whatever your potential energy was (say, gravitational energy due to an initial orbit around the Earth), it will decrease because you will be moving away from it. However, it is not your speed that is causing the decrease in potential energy. Two objects orbiting the Earth at a height h with different speeds (one closer to c) will have the same gravitational potential energy.

The short answer to your original question is that things don't gain energy just because they are moving forward in time, just as they don't simply by moving forward (or backward) in any of the spatial dimensions. That would be a violation of the conservation of energy, since everything would be gaining energy.

Good to question all these things, though, and it's great you have a keen interest in pondering the nature of such things.

 

[El Hombre Invisible]

Having said that, though, I guess you could argue a link between how fast time passes for a fast-moving body and how much kinetic energy it has. If you supply energy /\E to a body moving at, say, 0.8c, and you didn't know about relativity, you would expect its kinetic energy to increase by /\E and time to pass the same as before. In reality, time would seem to pass more slowly than before, and its kinetic energy would increase by considerably less than /\E.

It's a tenuous link, but it's a link!

But Jimmy's statement is more realistic. The energy you have by virtue of moving through time alone is your rest mass energy.

 

[El Hombre Invisible]

From that point of view, the energy of a particle that is moving through time alone is its rest mass.

No, thinking about it: what about massless particles?

 

[Jimmy Snyder]

No, thinking about it: what about massless particles?

There is no frame in which they are at rest.

 

[nemosum]

Moving at constant speed and standing still are the same thing. That's the most basic concept of relativity you need to grasp. You may choose your x, y and z axes such that you are at rest, but I may choose mine such that you are moving at constant non-zero speed.

Sorry, you're right, I forgot that part.

 

[nemosum]

But Jimmy's statement is more realistic. The energy you have by virtue of moving through time alone is your rest mass energy.

I think that's what I meant. You have rest mass energy for moving throught time, even if you're standing motionless in reference to the up, down, left and right. So, is rest mass energy different than potential energy?

 

[Garth]

Welcome to these Forums nemosum!

Keep asking those questions, that is how we learn. If you keep an open questioning mind and learn to "stand on the shoulders of giants" you will become a great physicist! Keep working at it.

Garth

 

[El Hombre Invisible]

There is no frame in which they are at rest.

Precisely, and yet they move through time. So the energy a particle has by which it moves through time alone cannot be it's rest mass energy, since not all particles that move through time have rest mass.

 

[El Hombre Invisible]

I think that's what I meant. You have rest mass energy for moving throught time, even if you're standing motionless in reference to the up, down, left and right. So, is rest mass energy different than potential energy?

Well, I don't think you can say a particle moves through time by virtue of its rest mass or vice versa for the reasons above.

Yes, rest mass is different from potential energy. You can describe a massive particle's total energy as the total of its mass energy, its kinetic energy and its potential energy.

For a massless particle, such as a photon, it's just 'energy'. Unless we describe gluons as having potential energy due interactions with quarks and other gluons? Anybody know? I've never heard of such a thing.

 

[Jimmy Snyder]

Precisely, and yet they move through time. So the energy a particle has by which it moves through time alone cannot be it's rest mass energy, since not all particles that move through time have rest mass.

For the photon, no time passes.

 

[El Hombre Invisible]

For the photon, no time passes.

True, but to us it does travel through time. So traveling through time cannot depend on rest mass (or vice versa) in normal frames. Relativistic mass, maybe.

 

[masudr]

Energy is the time component of the energy momentum 4-vector.

 

[Jimmy Snyder]

For the photon, no time passes.

I'm sorry I wrote that, although I do think it is pertinent. I wrote it because I lost track of my own train of thought. When I wrote my explanation, I wrote the words "In that sense". I meant that the explanation was limited to the sense to which it refered, namely for particles in rest frames. When you asked about massless particles, I responded "There is no frame in which they are at rest." What I meant by that is that the case you are talking about is not covered under "In that sense". You can take that to mean that you're case is not a counter-example since it doesn't match the sense, or you can take it to mean that you are right, there are other senses.

 

[El Hombre Invisible]

I'm sorry I wrote that, although I do think it is pertinent. I wrote it because I lost track of my own train of thought. When I wrote my explanation, I wrote the words "In that sense". I meant that the explanation was limited to the sense to which it refered, namely for particles in rest frames. When you asked about massless particles, I responded "There is no frame in which they are at rest." What I meant by that is that the case you are talking about is not covered under "In that sense". You can take that to mean that you're case is not a counter-example since it doesn't match the sense, or you can take it to mean that you are right, there are other senses.

You are right - nemosum did explicitly end his question that you replied to with the case of a body at rest. However this is a special frame of reference, so I was extending it to all inertial frames. I stopped considering the rest frames when this opened up to photons. There were some thought progressions I kept to myself. My fault entirely. The thing is that when you take relativistic mass into account, the more you have the slower time passes for you, be it in someone else's rest frame (as in SR) or your own (as in GR, where the relativistic mass is also your rest mass).

Going back to the OP, I don't know GR. I cannot say whether or not the increase in relativistic mass in SR that corresponds to a decrease in the rate by which time passes for it compared to its rest frame looks much like the increase in the gravitational field in GR that corresponds also to a decrease in the rate by which time passes for it compared to negligible gravitational field.

But, yes, the similarities look interesting. Relativistic mass seems to be inertial in time as it is in space. This still has nothing to do with potential energy so far as I can fathom.

 

[masudr]

Energy is the time component of the energy momentum 4-vector.

Sorry I wrote more, but it seemed to have been swallowed up by my browser before I clicked submit.

What I meant to say was, that the energy-momentum 4-vector is defined as

p^\mu = m\frac{dx^\mu}{d\tau}

where \tau is proper time and x_\mu = (-ct, x, y, z) (I've written it as a row vector, hence the minus sign on the time component). 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.

 

[nemosum]

K, assuming that rest mass energy is a result of moving through time. Then perhaps you could look at it like a kind of scale, where, on the one hand you have KE for moving 3-dimensionally, and on the other you have you rest mass energy for moving though time alone. If you accelerate, then the scales will tip, and you will have more KE and less rest mass energy as a result of moving more slowly through time. Hipothetically you could say that as soon as you get going fast enough (perhaps c) then time would stop, and the whole of you energy would be kinetic energy. Basically turning yourself into a big photon. This is just an idea I'm throwing out there.

And I didn't think photons and massless particles moved through time anyway. Am I wrong?

 

[El Hombre Invisible]

Sorry I wrote more, but it seemed to have been swallowed up by my browser before I clicked submit.
What I meant to say was, that the energy-momentum 4-vector is defined as
p^\mu = m\frac{dx^\mu}{d\tau}
where \tau is proper time and x_\mu = (-ct, x, y, z) (I've written it as a row vector, hence the minus sign on the time component). 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.

When you say "proper time", do you mean time in the observer's rest frame? I assume, then, that t is dilated time.

I'm not entirely sure how that answers the question. Could you elaborate?

 

[El Hombre Invisible]

K, assuming that rest mass energy is a result of moving through time. Then perhaps you could look at it like a kind of scale, where, on the one hand you have KE for moving 3-dimensionally, and on the other you have you rest mass energy for moving though time alone.

If you choose a frame such that the body is at rest, it will have no KE. If you choose a frame such that it is moving through space, it does have KE. To travel through time in your own frame, you have to have mass. In any other frame, you have relativistic mass and travel through time more slowly, that is: you travel through less time than an observer in his frame.

If you accelerate, then the scales will tip, and you will have more KE and less rest mass energy as a result of moving more slowly through time. Hipothetically you could say that as soon as you get going fast enough (perhaps c) then time would stop, and the whole of you energy would be kinetic energy. Basically turning yourself into a big photon. This is just an idea I'm throwing out there.

No, I think you've gone astray somewhere. Acceleration does not lower your rest mass. If the acceleration is positive, it increases your relativistic mass when viewed from your original frame. Your rest mass is unchanging.

And I didn't think photons and massless particles moved through time anyway. Am I wrong?

In normal frames, it takes some time between emission and absorption. To that extent, photons travel through time.

 

[my_wan]

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

Actually this hypothesis can be invalidated with a fairly straight forward thought experiment. Consider the physical situation I described https://www.physicsforums.com/showthread.php?t=96927". Place particle A inside this sphere. Place particle B far removed from any gravitational fields but comoving with particle A such that no kinetic energy can result from any relative motion between the two. So here we have a clear situation were B is traveling through time faster than A even though no relative motion and no kinetic energy is definable. There does exist a potential energy well between the two particles though.

 

[nemosum]

In normal frames, it takes some time between emission and absorption. To that extent, photons travel through time.

Well, I'm not exactly sure what you mean by "normal" frames. But, just because the photon exists doesn't mean that it has to be moving through time, at least not at a normal rate. Couldn't you look at it as if the photon's clock has stopped. I mean, when you send a particle through an accelerator it still travels around even though it's clock is much slower than when it's at rest. So a couldn't a photon still exist and travel even of it's clock has stopped? It would still take time (or that's what it would seem like to us at any rate) to travel from place to place, but to the photon there would be no time at all.

And another question, relativistic mass is the KE gained from traveling at high speed? And that's why nothing can go the speed of light. Because it would take an infinite amount of energy to push all that relativistic mass around. Am I right?

 

[Mortimer]

Couldn't you look at it as if the photon's clock has stopped. I mean, when you send a particle through an accelerator it still travels around even though it's clock is much slower than when it's at rest.

Yes. The photon does not move in it's own proper time \tau, which is the time recorderd by a clock moving along with the photon (if that would be possible) and defined as \tau=t \sqrt{1-v^2/c^2} (obviously zero for any particle moving at lightspeed). The value of it's speed however is calculated by stationary observers using their proper time, which is not the same as the proper time of the photon. The proper time for a stationary observer is equal to its normal (coordinate) time t.

 

[nemosum]

No, I think you've gone astray somewhere. Acceleration does not lower your rest mass. If the acceleration is positive, it increases your relativistic mass when viewed from your original frame. Your rest mass is unchanging.

But if your rest mass is constant then it would not be the result of moving through time. Because as you accelerate time slows down, therefore, if rest-energy is the result/cause of moving through time then it could not be constant. Hmm...

 

[jtbell]

what do you mean by "moving through time"?

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.

 

[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.