Does Velocity or Acceleration Determine the Energy of a Moving Object?

[Mentz114]

and i have lots of doubts and trouble in transforming acceleration into SR's lotentz transformation. I am not able to express them. So i asked you to help me with an example. So can you provide an example? It would be more clear to me.
This reply is to mentz114

This is getting seriously off-topic. What exactly do you want an example of ?

 

[DiracPool]

This is getting seriously off-topic. What exactly do you want an example of ?

Well, let's get it back on-topic then. OP here

Mentz114, can you address my post #8?

 

[ash64449]

This is getting seriously off-topic. What exactly do you want an example of ?

i want an example of transforming the coordinates of a moving object in accelerating frame when the coordinates of moving object in a inertial reference frame is known.
Note that you have not provided an example but only half-explantion which i didn't understand. So take some values and explain the above.

 

[ash64449]

oh.. Mentz.. This is not off-topic.OP is also talking about acceleration. So it isn't off-topic

 

[Mentz114]

I'm not sure what you mean by that last statement?

Ok, here's a related question that may "root out" a larger picture here. Take the exact same scenario I used in the original post, only replace the Earth as the "stationary" frame relative to the traveling sphere, with me in a rocket ship. The first phase of the thought experiment is identical to the above, the sphere is accelerated until its energy is 200 GeV, and then it starts coasting.

Ok, now in the second phase, I accelerate myself in my rocket ship in the opposite direction to the moving sphere until the moving sphere is traveling at a velocity, say v'', relative to me whereby its energy is now 400 GeV.

In this instance the sphere's velocity relative to me and its energy have increased again, but the sphere has not been given any energy. It has not been accelerated. So here is a case that an object that is moving at a constant velocity gains energy/mass. I know I painted myself into a corner on this one and the response is going to be "it's all relative", i.e., in principle the sphere was actually accelerated "relative to me" when I took off in the opposite direction.

That answer is unsettling, though. Perhaps someone can unsettle me? Or is my unsettledness unfounded?

This is post#8.

Surely someone has pointed out that kinetic energy depends on relative velocity ? Although one party in the scenario accelerated, only the instantaneous relative velocity determines the energy.

So the fact that something gained or lost energy in some frame but was not accelerated is not strange.

 

[Mentz114]

i want an example of transforming the coordinates of a moving object in accelerating frame when the coordinates of moving object in a inertial reference frame is known.
Note that you have not provided an example but only half-explantion which i didn't understand. So take some values and explain the above.

The LT does that. If you use $a\tau$ as the relative velocity. One thing that is apparent is that in the original rest-frame coordinates, the relative velocity never reaches c. Because

$\frac{dx}{dt} = \frac{d\tau}{dt} \frac{dx}{d\tau}$

 

[ash64449]

The LT does that. If you use $a\tau$ as the relative velocity.

ok,then.. In one of the LT,there is x'=(x-vt)alpha. So what 'v' there? Please provide an example.. It would be very helpful to me..

 

[Mentz114]

ok,then.. In one of the LT,there is x'=(x-vt)alpha. So what 'v' there? Please provide an example.. It would be very helpful to me..

That is not the LT. The LT is

$t'=\gamma t + \beta\gamma x$
$x'=\gamma x + \beta\gamma t$
$\beta=v_x/c$

I always work in units where c=1 so that space and time have the same units.

[Edit]I misread your post - I missed the 'alpha'. Try to express your maths better.

To get an accelerating frame, v is $a\tau$.

 

[ash64449]

That is not the LT. The LT is

$t'=\gamma t + \beta\gamma x$
$x'=\gamma x + \beta\gamma t$
$\beta=v/c$

I always work in units where c=1 so that space and time have the same units.

u mean Lorentz Transformation.. Right?
Well,these are transformations found in theory of relativity written by Albert einstein:
x'=(x-vt)/alpha
y'=y
z'=z
t'=(t-(v/c^2)*x)/alpha
where alpha is sqrt of 1-(v^2/c^2)
i earlier made mistakes.. Sorry.. But this one is from a book..
Now answer what to do with v?? (please,i am stuck here!)

 

[Mentz114]

Now answer what to do with v?? (please,i am stuck here!)

I've edited my previous post.

 

[DiracPool]

So the fact that something gained energy but was not accelerated is not strange.

You don't think that's strange? Really? Let's extrapolate my thought experiment to the accelerated spreading of galaxies in the universe. Aren't these galaxies gaining incredible amounts of mass-energy, due to that acceleration? The further out ones gaining even more than the closer ones. Why aren't we seeing these galaxies implode inward on themselves from our reference frame due to the tremendous increase in gravity they've incurred due to their acceleration? Is it because "space" is actually expanding, so these relativistic effects don't hold as they do in my #8 example? Well, what about galaxies and/or stars that may be accelerating away from us due to an effect not caused by dark energy?

Or, is the whole gaining of energy due to relative velocity just a perceptual phenomenon? Just for fun, I accelerate away from you because I think it's funny how heavy you get when I do so. So now, after accelerating away from you continuously for a few light-hours, I cease that acceleration and begin to coast at a constant velocity relative to you as I arrive at my favorite department store at the edge of the galaxy. I buy you a shirt that says, "My friend took a trip across the galaxy and all he got me was this lousy t-shirt!" However, as I'm gauging the size I should purchase, I take a look back at your massiveness and say to the cashier, yes, I'll take a Double XL. You, however, are waving frantically at me and holding up a sign that says "No, I'm just a large." And I say, "Not from my reference frame, buddy!" So, when I get back home you get the XXL. But you never did actually ever gain any weight, did you?

Seems almost absurd, doesn't it? But that's the way it is. Did I get any of that wrong?

By the way, the store has a NO RETURN policy due to relativistic travelers misrepresenting their times of purchases. Sorry. It's a cotton t-shirt, though, so maybe it'll shrink.

 

[ash64449]

I've edited my previous post.

sorry,i was in mobile..
Ok..so in place of v, we need to put a*proper time?

 

[ash64449]

mentz,i will start a new discussion.. I will try to understand more from there..

 

[Mentz114]

You don't think that's strange? Really? Let's extrapolate my thought experiment to the accelerated spreading of galaxies in the universe. Aren't these galaxies gaining incredible amounts of mass-energy, due to that acceleration? The further out ones gaining even more than the closer ones. Why aren't we seeing these galaxies implode inward on themselves from our reference frame due to the tremendous increase in gravity they've incurred due to their acceleration? Is it because "space" is actually expanding, so these relativistic effects don't hold as they do in my #8 example? Well, what about galaxies and/or stars that may be accelerating away from us due to an effect not caused by dark energy?

I thought we were talking about SR. To model the expanding cosmos we need GR. This changes everything. The FLRW or Einstein-DeSitter models do expand/collapse/oscillate depending on the energy density, but not in the naive SR way.

Or, is the whole gaining of energy due to relative velocity just a perceptual phenomenon?

If two objects collide, the damage done depends on their relative velocities in a spectacular way. Two cars colliding when each going 30mph in the ground frame is much the same as if one was stationary and hit by one going at 60mph. It is very real.

Just for fun, I accelerate away from you because I think it's funny how heavy you get when I do so. So now, after accelerating away from you continuously for a few light-hours, I cease that acceleration and begin to coast at a constant velocity relative to you as I arrive at my favorite department store at the edge of the galaxy. I buy you a shirt that says, "My friend took a trip across the galaxy and all he got me was this lousy t-shirt!" However, as I'm gauging the size I should purchase, I take a look back at you and say, yes, I'll take a Double XL. You, however, are waving frantically at me and holding up a sign that says "No, I'm just a large." And I say, "Not from my reference frame, buddy!" So, when I get back home you get the XXL.

Seems almost absurd, doesn't it? But that's the way it is. Did I get any of that wrong?

By the way, the store has a NO RETURN policy due to relativistic travelers misrepresenting their times of purchases. Sorry. It's a cotton t-shirt, though, so maybe it'll shrink.

I don't know anything about tee-shirts so I'll skip this one.

 

[Mentz114]

sorry,i was in mobile..
Ok..so in place of v, we need to put a*proper time?

Yes ! You got it.

 

[DrGreg]

DiracPool, like many things in relativity, energy is relative to the observer. So it's not an absolute property of an object, it's a relationship between object and observer. Different inertial observers measure different energies for the same object. And if one observer is initially inertial, then accelerates, then becomes inertial again, the energy of some other object will have changed relative to the observer, even though nothing happened to the object.

Conservation of energy isn't always true in all circumstances. In special relativity (no gravity, no expanding universe) it's always true relative to a single inertial frame.

To get conservation of energy to work in a non-inertial frame you need to introduce the concept of "potential energy". We already have this concept for Newtonian gravity, and a similar idea works for non-inertial coordinate systems provided they are "stationary", a technical term that means, roughly speaking, that the coordinate system doesn't change over time in some sense.

In non-stationary coordinate systems there need not be any conservation of energy at all (other than locally).

 

[DiracPool]

I thought we were talking about SR. To model the expanding cosmos we need GR. This changes everything. The FLRW or Einstein-DeSitter models do expand/collapse/oscillate depending on the energy density, but not in the naive SR way.

OK, so maybe that was a bad example.

If two objects collide, the damage done depends on their relative velocities in a spectacular way. Two cars colliding when each going 30mph in the ground frame is much the same as if one was stationary and hit by one going at 60mph. It is very real.

Who said anything about objects colliding? If I accelerate away from you at a velocity that begins to approach c, you get heavier and heavier, and/or your energy increases (if you want to keep that your inertial rest mass stays constant) to a point that approaches infinity. How do you stay alive? In your example with the car, everyone dies because that energy DOES have real effects. But is has no real effects on you when I accelerate away from you at a velocity appraoching c...according to you nothing has changed.

I don't know anything about tee-shirts so I'll skip this one.

Fair enough, I'll just give it to my brother, he's been bugging me for one anyway.

 

[BruceW]

Mathematically, an inertial reference frame is one where the metric is the Minkowski metric. Experimentally, an inertial reference frame is one where all accelerometers read the same as the second time derivative of their position. Neither condition requires reference to any other reference frame.

eh... So you're saying if we have a Minkowski spacetime, and choose a coordinate system with a metric that is diag(-1,1,1,1) and zero off-diagonal entries, then this defines an inertial reference frame? This is a pretty nice definition. Although I think it just gets rid of the question "how do we know the particle has no forces acting on it" and instead introduces the question "how do we know we have a Minkowski spacetime". But if we assume (for a particular physics problem), that we are in a Minkowski spacetime, then yes, this does give a good definition. I'll try to remember it :)

 

[DiracPool]

DiracPool, like many things in relativity, energy is relative to the observer. So it's not an absolute property of an object, it's a relationship between object and observer. Different inertial observers measure different energies for the same object. And if one observer is initially inertial, then accelerates, then becomes inertial again, the energy of some other object will have changed relative to the observer, even though nothing happened to the object.

Thanks DrGreg, I'm going through Hartle right now, hopefully I will become more comfortable with these notions as I progress

 

[Dale]

Don't you need an independent inertial reference frame, A, to determine the position that B is accelerating relative to? How else would you calculate the second time derivative?

You only need one reference frame (inertial or not) and a set of accelerometers. You then calculate the 2nd derivative of each accelerometer's position wrt the one reference frame. If the reading on every accelerometer equals the 2nd derivative of its position then the frame is inertial. There is no need for a second frame, everything is calculated relative to one.

 

[Dale]

If Acceleration can be handled by special theory of relativity,What is the equation that helps to determine the position of an Object(or coordinates of object) with respect to an accelerating frame when the coordinates of the object in an inertial reference frame is known?(answer should be on the context of SR)

The coordinate transform. Here is a prototypical example: https://en.wikipedia.org/wiki/Rindler_coordinates

 

[Dale]

eh... So you're saying if we have a Minkowski spacetime, and choose a coordinate system with a metric that is diag(-1,1,1,1) and zero off-diagonal entries, then this defines an inertial reference frame?

Yes. Although I haven't seen anyone define it this way so I would call it a "test" rather than a definition. I.e. if a coordinate system has this property then it is an inertial frame.

This is a pretty nice definition. Although I think it just gets rid of the question "how do we know the particle has no forces acting on it" and instead introduces the question "how do we know we have a Minkowski spacetime".

Since you have the metric you can also determine that quite easily. Simply use the metric to calculate the Riemann curvature tensor. If it is 0 then the spacetime is Minkowski.

 

[PeterDonis]

it just gets rid of the question "how do we know the particle has no forces acting on it" and instead introduces the question "how do we know we have a Minkowski spacetime".

These are two different questions that are independent of each other. By that I mean that all four possibilities can be realized:

(1) A particle can have no forces acting on it in flat Minkowski spacetime.

(2) A particle can have forces acting on it in flat Minkowski spacetime.

(3) A particle can have no forces acting on it in curved, non-Minkowski spacetime.

(4) A particle can have forces acting on it in curved, non-Minkowski spacetime.

So you can't substitute one question for the other.

DaleSpam explained how to tell whether spacetime is Minkowski--compute the Riemann curvature tensor. To tell whether a particle has forces acting on it, you compute the path curvature of its worldline. (Physically, this corresponds to attaching an accelerometer to the particle and seeing whether it reads zero or not. A nonzero reading means forces are acting on the particle.)

 

[WannabeNewton]

eh... So you're saying if we have a Minkowski spacetime, and choose a coordinate system with a metric that is diag(-1,1,1,1) and zero off-diagonal entries, then this defines an inertial reference frame?

There's a distinction in GR between a global inertial reference frame and a local inertial reference frame. Global inertial frames exist in Minkowski space-time and in these frames you can set up coordinates in which the metric tensor takes the form diag(-1,1,1,1) everywhere, and in which the Christoffel symbols vanish everywhere (i.e. you have an identically flat connection), in space-time. On the other hand, in an arbitrary curved space-time, you cannot do this globally but what you can always do is find a neighborhood of an event in which you can set up geodesic normal coordinates by applying exp to the tangent space at that event. In these coordinates, the metric reduces to the Minkowski metric at that event, and similarly the Christoffel symbols vanish identicallty at that event (again we can always do this because we are using the levi-civita connection). However, I'm not fond of attaching the notion of a coordinate system to the notion of an inertial frame as a definition but rather as a logical consequence of the usual, more physical SR definition of an inertial frame.

 

[BruceW]

The coordinate transform. Here is a prototypical example: https://en.wikipedia.org/wiki/Rindler_coordinates

Ah, this is gold. thanks. for doing relativistic rocket problems, I have always just used the inertial reference frame, and compared things to that. But this gives some good information on some of the mathematical properties of the rocket's frame. (for constant proper acceleration, which is the most common relativistic rocket problem).

edit: ah, actually maybe the most common relativistic rocket problem is one where the rocket loses mass as propellant...But this is still a nice example of accelerating reference frames in SR. I was really unaware of such things until now. (I never took a class in GR, and my class in SR was pretty basic).

 

[BruceW]

Experimentally, an inertial reference frame is one where all accelerometers read the same as the second time derivative of their position. Neither condition requires reference to any other reference frame.

Going back to this, an accelerometer works because its internal mechanism is not subject to any forces from outside (apart from the force transmitted by contact by whoever might be swinging it around). So by assuming we can make reliable accelerometers, we are automatically assuming that we can tell that the internal mechanism has no external forces acting on it. This seems to me to be the same as assuming that we can tell if a particle has no forces acting on it. Well, the internal mechanism could be a beam of light. But in this case, we must be able to say whether the internal mechanism is being acted on by an external field or not.

This is a bit similar to the question "can we make reliable clocks?". And we do assume this in relativity, so maybe I am just being pedantic when it is really not necessary to be. I am trying to think of a counter-example, but I can't think of any. And judging from other posts, it seems that it is common to assume that we can make reliable accelerometers?

These are two different questions that are independent of each other.

Ah yeah, sorry about that. I didn't really explain very well. I meant that if we answer the question "are we in a Minkowski spacetime, and using metric diag(-1,1,1,1)?" then this answers the question "how do we know a particle has no forces acting on it", because we can just look and see if the particle moves in a straight line or not.

There's a distinction in GR between a global inertial reference frame and a local inertial reference frame. Global inertial frames exist in Minkowski space-time and in these frames you can set up coordinates in which the metric tensor takes the form diag(-1,1,1,1) everywhere...

Yeah, I was just thinking of the global inertial frame. I have no problem with the local inertial reference frame, because it is clear to me that you can define this at wherever you are, so It is not necessary to define local inertial frame as being 'relative' to another reference frame.

 

[PeterDonis]

if we answer the question "are we in a Minkowski spacetime, and using metric diag(-1,1,1,1)?" then this answers the question "how do we know a particle has no forces acting on it", because we can just look and see if the particle moves in a straight line or not.

True, but it's worth noting that this only works if the answer to the first question is yes; it doesn't if the answer is no. The method I gave, computing the path curvature of the particle (or, physically, attaching an accelerometer to the particle), works regardless of whether spacetime is Minkowski or not and regardless of what coordinates we use to express the metric.

 

[Dale]

This is a bit similar to the question "can we make reliable clocks?". And we do assume this in relativity, so maybe I am just being pedantic when it is really not necessary to be. I am trying to think of a counter-example, but I can't think of any. And judging from other posts, it seems that it is common to assume that we can make reliable accelerometers?

Yes. Recall that this is for the experimental definition of an inertial frame. The problem you mention here is not specific to accelerometers but rather it is a general problem with ANY experiment using any type of measurement device to measure any quantity. You always need to evaluate the reliability of your measurement devices, so this is just "standard fare" for experimental work.

 

[BruceW]

The definition I was using (when this thread began) was that two reference frames need to be moving at constant relative velocity to be inertial 'with respect to each other'. So my experimental definition required that we could make reliable devices that could measure this.

I was a bit apprehensive at the 'reliable accelerometers' definition, which people seem to be more familiar with. But really I guess it is not much more radical than my old definition was. Thanks for helping explain to me, everyone :) Also, am I right in thinking that in a Minkowski spacetime there is only one unique set of global inertial reference frames? (and therefore, it is obvious that we don't need to specify what they are inertial 'relative to', because there is only one set which they can be inertial 'relative to', so we might as well drop the words 'relative to' in this case).

Also, sorry for wandering away from the topic of the thread a bit.