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.