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

DiracPool
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Is the energy gained by a particle moving at a high velocity a product of acceleration or can it arise in an object moving at constant velocity? I mean...OK, thought experiment. I haved a 100 GeV spherical mass moving through space relative to the Earth at 10 m/s. I now accelerate that mass to a velocity, v', relative to the Earth whereby now the traveling mass has an energy of 200 GeV. I now stop accelerating the mass. I'm assuming now that the mass will continue on indefinitely traveling at v' with an energy of 200 GeV relative to the Earth. Is that correct?

I guess the larger question I'm asking is, is it only the acceleration of the mass that imparts energy to that object? Such that the actually velocity per se has nothing to do whatsoever with imparting energy to an object? In other words, accleration gives energy/mass to an object and deceleration removes energy/mass? While the actual velocity per se of the object has nothing to do with it?
 
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DiracPool said:
In other words, accleration gives energy/mass to an object and deceleration removes energy/mass? While the actual velocity per se of the object has nothing to do with it?

i don't know whether i am fully right or wrong,but if other posters can see anything wrong,they can point out. Well,making mistakes is not mistakes!

By looking at the velocity we can understand how much energy it has.

And we provide energy with the help of force.

And we know that force causes acceleration.

So energy is provided by acceleration.

To achieve higher velocity,we need to put more force,so more energy.

Earlier kinetic energy was given by the equation ## \frac{1}{2}mv^2 ##

Now because of the effects of relativity,it has been replaced by the equation:

## \frac{mc^2}{\sqrt {1-\frac{v^2}{c^2}}} ##
 
if v=0, you can see that ## E=mc^2 ##

So we find from the equation that mass in rest is same as energy and energy increases when something moves,so we can also say that mass increases when something moves(not sure!)
edit: from the above total energy calculation equation,you can find that when an object is made to move at slower velocity,you will get rest mass plus kinetic energy of the object. For slower velocity that kinetic energy part is same.for bigger velocity,kinetic energy will not be same as kinetic energy found out from special relativity. Infact,it is a bigger number. We see from the above conclusion that objects resistance to acceleration has increased(inertia). So mass is increased. But you should not confuse with the mass that calculates the total amount of matter. Total amount of matter is still same but inertia has increased.. This one is called relativistic mass.
 
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and i have also heard that there are two types of masses.
For example,light bends under the influece of gravity(general theory of relativity). Source of gravity is energy. Light is energy.so light bends. Since energy is mass, light has this type of mass.. But don't consider the mass that you think.light has no mass,the way you think,because there are two types of masses...
One of them is relativistic mass and the other is rest mass..
 
i think you can answer your questions yourself from my replies!
 
DiracPool said:
I guess the larger question I'm asking is, is it only the acceleration of the mass that imparts energy to that object? Such that the actually velocity per se has nothing to do whatsoever with imparting energy to an object? In other words, accleration gives energy/mass to an object and deceleration removes energy/mass? While the actual velocity per se of the object has nothing to do with it?

I'm not sure exactly what you mean.

We can say that energy is a function of velocity - relativistically

E = m/sqrt(1-v^2/c^2). (m here is rest mass.)

Acceleration is the rate of change of velocity with time, if the acceleration is zero the energy is constant.

To impart velocity to an object and make it accelerate requires some force. This force does work in accelerating an object to a higher velocity / higher energy. Work = force * distance, and power = rate of work = force * velocity.
 
DiracPool said:
Is the energy gained by a particle moving at a high velocity a product of acceleration or can it arise in an object moving at constant velocity?
How does an object come to be moving at constant velocity if it doesn't accelerate to that velocity?
I now stop accelerating the mass. I'm assuming now that the mass will continue on indefinitely traveling at v' with an energy of 200 GeV relative to the Earth. Is that correct?
Yes.
I guess the larger question I'm asking is, is it only the acceleration of the mass that imparts energy to that object? Such that the actually velocity per se has nothing to do whatsoever with imparting energy to an object? In other words, accleration gives energy/mass to an object and deceleration removes energy/mass? While the actual velocity per se of the object has nothing to do with it?
You cannot slice apart two sides of the same coin.
 
russ_watters said:
You cannot slice apart two sides of the same coin.

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

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?
 
DiracPool said:
I guess the larger question I'm asking is, is it only the acceleration of the mass that imparts energy to that object? Such that the actually velocity per se has nothing to do whatsoever with imparting energy to an object? In other words, accleration gives energy/mass to an object and deceleration removes energy/mass? While the actual velocity per se of the object has nothing to do with it?
No, this is exactly backwards. The speed is the only thing that matters, the energy is a function of speed only, not acceleration and not velocity (except insofar as they affect speed in a given frame). If an object accelerated in one frame then the object's speed and energy increased in that frame. But there exists another frame where the object decelerated and its speed and energy decreased, and there exists another frame where the object changed direction and its speed and energy remained constant.
 
  • #10
DaleSpam said:
No, this is exactly backwards. The speed is the only thing that matters, the energy is a function of speed only, not acceleration and not velocity (except insofar as they affect speed in a given frame). If an object accelerated in one frame then the object's speed and energy increased in that frame. But there exists another frame where the object decelerated and its speed and energy decreased, and there exists another frame where the object changed direction and its speed and energy remained constant.

just curious. Can you tell me an example?
 
  • #11
ash64449 said:
just curious. Can you tell me an example?
Every example. Let's use low velocities so that relativistic effects can be ignored. Let an object start at rest in frame A and accelerate to a velocity of 10 m/s to the right. Let frame B be the frame moving 10 m/s to the right wrt frame A. Let frame C be the frame moving 5 m/s to the right wrt frame A.

In frame A the speed increases from 0 m/s to 10 m/s. In frame B the speed decreases from 10 m/s to 0 m/s. In frame C the speed starts at 5 m/s and ends at 5 m/s.
 
  • #12
DaleSpam said:
Every example. Let's use low velocities so that relativistic effects can be ignored. Let an object start at rest in frame A and accelerate to a velocity of 10 m/s to the right. Let frame B be the frame moving 10 m/s to the right wrt frame A. Let frame C be the frame moving 5 m/s to the right wrt frame A.

In frame A the speed increases from 0 m/s to 10 m/s. In frame B the speed decreases from 10 m/s to 0 m/s. In frame C the speed starts at 5 m/s and ends at 5 m/s.

so isn't same amount of force applied in different frame and ended up with different velocity?
 
  • #13
ash64449 said:
so isn't same amount of force applied in different frame and ended up with different velocity?
Yes, the same force, the same acceleration, the same change in velocity, but different changes in speed and therefore different changes in KE.
 
  • #14
DiracPool said:
I'm not sure what you mean by that last statement?:confused:

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?
yeah, it is all relative. But this is general relativity you are talking about now. If you want to use the reference frame of the spaceship, where the spaceship is accelerating with respect to the earth, then these two reference frames are not 'inertial reference frames' with respect to each other. Therefore, you're going to get gravitational fields appear in one reference frame where there was none in the other reference frame.
 
  • #15
BruceW said:
yeah, it is all relative. But this is general relativity you are talking about now. If you want to use the reference frame of the spaceship, where the spaceship is accelerating with respect to the earth, then these two reference frames are not 'inertial reference frames' with respect to each other. Therefore, you're going to get gravitational fields appear in one reference frame where there was none in the other reference frame.
No, no, no! Special relativity alone is perfectly adequate to handle acceleration, and acceleration is not the same as gravity.
 
  • #16
special relativity is not enough to handle using reference frames with relative acceleration. It is only able to handle inertial reference frames. And in DiracPool's post, he is talking about what happens according to the reference frame of an accelerating rocket. This is possible in GR, but the metric will be weird (which can be interpreted as a kind of 'gravity' where there was none before).
 
  • #17
BruceW said:
special relativity is not enough to handle using reference frames with relative acceleration. It is only able to handle inertial reference frames.
This is not correct. See: http://www.edu-observatory.org/physics-faq/Relativity/SR/acceleration.html

"It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames. It is claimed that general relativity is required because special relativity only applies to inertial frames. This is not true."

What defines GR is the Einstein field equations, i.e. curved spacetime or tidal gravity. As long as your spacetime is flat then you can use SR, otherwise you need to use GR and the EFE.
 
  • #18
I see. so if we have a situation where we have:
- reference frame [1]
- reference frame [2] (which is non-inertial with respect to reference frame [1] )
- the coordinates of an object according to reference frame [1]
- a mapping from coordinates in reference frame [1] to reference frame [2]

Then we can calculate the coordinates of the object with respect to reference frame [2]. Note, this doesn't require flat spacetime at all. (If I have interpreted it right). This is pretty interesting. I will keep it in mind in the future, thanks.
 
  • #19
Yes, that is all correct with one very minor exception.
BruceW said:
- reference frame [2] (which is non-inertial [STRIKE]with respect to reference frame [1][/STRIKE] )
(strikeout added)

A reference frame is inertial or non-inertial on its own. Inertial-ness is not relative nor is it defined with respect to some other reference frame. So the part of the quote that I struck out is not necessary and doesn't really make sense.
 
  • #20
It depends on how you define inertial reference frames. I'm guessing the definition you are using is something like: An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed. But I really do not like this definition.
In wiki's page http://en.wikipedia.org/wiki/Inertial_frame_of_reference they talk a bit about the problems associated with this kind of definition (e.g. we must now assume that we can ascertain whether a particle is subject to forces or not). I would prefer a definition something like A set of inertial reference frames can be related to each other by Lorentz transforms or something like that, so that reference frames are only inertial 'with respect to' other frames, and not inertial in their own right.
 
  • #21
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.
 
  • #22
DaleSpam said:
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.

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?
 
  • #23
DaleSpam said:
This is not correct. See: http://www.edu-observatory.org/physics-faq/Relativity/SR/acceleration.html

"It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames. It is claimed that general relativity is required because special relativity only applies to inertial frames. This is not true."

What defines GR is the Einstein field equations, i.e. curved spacetime or tidal gravity. As long as your spacetime is flat then you can use SR, otherwise you need to use GR and the EFE.

Lorentz transform connect inertial reference frames.

And Lorentz transform is based on special theory of relativity.

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)
 
  • #24
ash64449 said:
Lorentz transform connect inertial reference frames.

And Lorentz transform is based on special theory of relativity.

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)
If you write ##\beta=a\tau## ( where a is proper acceleration and τ is proper time ) in the LT, the transformation works as usual between the frames, and leaves the metric unchanged.

The domain of special relativity is Minkowski space-time, which is flat and so does not include gravity - but certainly can handle proper acceleration. See for instance the 'relativistic rocket' and Rindler coordinates.
 
  • #25
Mentz114 said:
If you write ##\beta=a\tau## ( where a is proper acceleration and τ is proper time ) in the LT, the transformation works as usual between the frames, and leaves the metric unchanged.

The domain of special relativity is Minkowski space-time, which is flat and so does not include gravity - but certainly can handle proper acceleration. See for instance the 'relativistic rocket' and Rindler coordinates.

how do you calculate the proper time? So that i can understand what beta is?(i know beta is v/c)
 
  • #26
Mentz114 said:
If you write ##\beta=a\tau## ( where a is proper acceleration and τ is proper time ) in the LT, the transformation works as usual between the frames, and leaves the metric unchanged.

The domain of special relativity is Minkowski space-time, which is flat and so does not include gravity - but certainly can handle proper acceleration. See for instance the 'relativistic rocket' and Rindler coordinates.

then there must be something wrong in my understanding of equivalence principle.
Isn't principle mean proper acceleration of object in flat space-time same as object in rest in which space-time is curved?
 
  • #27
ash64449 said:
then there must be something wrong in my understanding of equivalence principle.
Isn't principle mean proper acceleration of object in flat space-time same as object in rest in which space-time is curved?
That is not relevant. We are dealing with the flat spacetime case here.

how do you calculate the proper time? So that i can understand what beta is?(i know beta is v/c)

##d\tau/dt=\sqrt{1-\beta^2}##, so if you know the range of t this can be integrated.
 
  • #28
Mentz114 said:
That is not relevant. We are dealing with the flat spacetime case here.



##d\tau/dt=\sqrt{1-\beta^2}##, so if you know the range of t this can be integrated.

but i cannot understand what ' v'? There are two 'v's over here. Initial velocity and final velocity.
Is it possible for you to elaborate and take an example to explain it to me?
 
  • #29
Mentz114 said:
That is not relevant. We are dealing with the flat spacetime case here.
you must elaborate this. Because acceleration in flat space-time is same as object in curved space-time(object in rest). So if SR cannot wrk in curved spac-time,then it cannot wrk with acceleration because they are exactly same.
 
  • #30
ash64449 said:
but i cannot understand what ' v'? There are two 'v's over here. Initial velocity and final velocity.
Is it possible for you to elaborate and take an example to explain it to me?
If you want to find the proper time to go from vi to vf, convert the velocities to ti and tf and integrate over that range.

[edit]It looks as if this cannot be done. But we can find out how the velocity has changed in the interval tf-ti in the coordinates of the initial inertial frame.
 
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  • #31
ash64449 said:
you must elaborate this. Because acceleration in flat space-time is same as object in curved space-time(object in rest). So if SR cannot wrk in curved spac-time,then it cannot wrk with acceleration because they are exactly same.
Nonsense. They are not exactly the same. One is in curved spacetime, one in flat spacetime.

I'm not going to argue about whether SR can handle acceleration. I know it can. If you bothered to look up the two examples I gave, and also Born coordinates and the Langevin frame you'd have to something to think about, instead of acting on recent and ill-formed judgements.
 
  • #32
Mentz114 said:
Nonsense. They are not exactly the same. One is in curved spacetime, one in flat spacetime.

I'm not going to argue about whether SR can handle acceleration. I know it can. If you bothered to look up the two examples I gave, and also Born coordinates and the Langevin frame you'd have to something to think about, instead of acting on recent and ill-formed judgements.

ok... If you think like that,then it's ok.and you see,i wasn't argueing with you whether SR can handle acceleration. I just asked doubt and to elaborate.
 
  • #33
Mentz114 said:
Nonsense. They are not exactly the same. One is in curved spacetime, one in flat spacetime.

I'm not going to argue about whether SR can handle acceleration. I know it can. If you bothered to look up the two examples I gave, and also Born coordinates and the Langevin frame you'd have to something to think about, instead of acting on recent and ill-formed judgements.

and can this be transformed?
Knowing the coordinates in accelerating frame and then tranforming them to inertial refernce frame(opposite of what i asked)
 
  • #34
Mentz114 said:
Nonsense. They are not exactly the same. One is in curved spacetime, one in flat spacetime.

i didn't say curved space-time same as flat space-time. I said acceleration can be said to be same with curved space-time.
 
  • #35
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
 
  • #36
ash64449 said:
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 ?
 
  • #37
Mentz114 said:
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:biggrin:

Mentz114, can you address my post #8?
 
  • #38
Mentz114 said:
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.
 
  • #39
oh.. Mentz.. This is not off-topic.OP is also talking about acceleration. So it isn't off-topic
 
  • #40
DiracPool said:
I'm not sure what you mean by that last statement?:confused:

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.
 
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  • #41
ash64449 said:
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}##
 
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  • #42
Mentz114 said:
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..
 
  • #43
ash64449 said:
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##.
 
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  • #44
Mentz114 said:
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!)
 
  • #45
ash64449 said:
Now answer what to do with v?? (please,i am stuck here!)

I've edited my previous post.
 
  • #46
Mentz114 said:
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.:-p
 
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  • #47
Mentz114 said:
I've edited my previous post.

sorry,i was in mobile..
Ok..so in place of v, we need to put a*proper time?
 
  • #48
mentz,i will start a new discussion.. I will try to understand more from there..
 
  • #49
DiracPool said:
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.:-p
I don't know anything about tee-shirts so I'll skip this one.
 
  • #50
ash64449 said:
sorry,i was in mobile..
Ok..so in place of v, we need to put a*proper time?
Yes ! You got it.
 
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