Question on the energy needed to go at a relativistic speed

[John SpaceY]

TL;DR Summary

If an object goes very fast, and if we want to make it go faster, do we need more and more energy to accelerate it or less and less?

Hello,

I consider to be in a relativistic area, where an object is moving very fast, seen from our Earth, at a speed v where v is less than the speed of light c.
I have considered the following equations (relativity equations) :

T = β * t

M = β * m

L = l / β

β = 1 / (√(1 – v2/c2) )

v = (l / T) = (L / t)

M, T and L are parameters seen from our Earth.
These parameters become m, t and l in a Spacecraft reference for example.

If I consider that an object is moving at a speed v seen from the Earth
and has a mass m at rest, when we go in the relativistic area (where v is near to c so when Beta becomes higher than 1), the object mass becomes M.
And this mass M is increasing when v is increasing. When v tends to c, Beta tends towards infinity and so M tends to infinity.
Current physics consider that if we want to continue to increase v (and being always less than c), as M becomes infinite, an infinite force should be needed and so the energy needed to continue to increase v should be higher and higher. And the more v is near of c, the more energy is needed in order to continue to increase v, because M is increasing.

I have another way of thinking ant it is on this point that I have my question:
For me, in the relativistic area, M tends towards infinity when v tends to c, but not only M is changing !
If we consider that we want to increase v when v is closed to c, on a distance L and during a time dT, an energy W should be needed :

W = M . Gamma . L

with Gamma = dv / dT

and so. W = M . dv/dT . L

I consider all the parameters seen from our Earth and so it is M, L and T and not m, l and t
When v tends to c these parameters are changing, according to the above equations :

M is changing like Beta : see equation (2)

dv tends towards 0 because v tends to c and as c is the maxi speed, dv tends to 0, and dT is changing like Beta (see equation (1))

and so dv/dT tends towards 0 when v tends towards c, because Beta tends to infinity.

And L is changing like 1/Beta : see equation (3). And this will cancel the evolution of M…

And if I consider W, for me this energy with change as

Beta x zero x 1/ Beta when v tends towards c

And so W will tend towards 0 when v tends towards c.

I come here to the reverse conclusion : when v tends towards c, less and less energy is needed to continue to increase v.
I would like to know if my last way of thinking is the good one. And if not where is the mistake ?
Because the current physics is considering that more and more energy will be needed to continue to increase v when v tends to c…

But is my reasoning correct?
If an object goes very fast, and if we want to make it go faster, do we need more and more energy to accelerate it or less and less?
If the object is in a vacuum we consider its mass as zero and therefore to accelerate it there is not a need for a lot of energy: is that correct ?
On the other hand, if the object is on our Earth, what is the answer?
does it take a lot of energy or not a lot to continue increasing speed?
(for example in a particle accelerator at CERN or ...)

I thank you very much in advance for your answers

Best regards

John

 

[PeroK]

I come here to the reverse conclusion : when v tends towards c, less and less energy is needed to continue to increase v.
I would like to know if my last way of thinking is the good one. And if not where is the mistake ?

You have come to the wrong conclusion. I can't follow your working at all.
$$E = \frac{mc^2}{\sqrt{1- v^2/c^2}}$$
Clearly, therefore, $E \rightarrow +\infty$ as $v \rightarrow c$.

 

[Ibix]

First off, "mass" in modern physics means "rest mass". If you intend what you say to be interpreted as relativistic mass then you need to say "relativistic mass". Second, the term you are calling $\beta$ is usually given the symbol $\gamma$.

The correct result is that the kinetic energy of an object of (rest) mass $m$ moving at speed $v$ is $(\gamma-1)mc^2$, where $\gamma=1/\sqrt{1-v^2/c^2}$ (edit: note that PeroK has included the rest energy of the mass, so his formula is equal to mine plus $mc^2$).

Your maths is very difficult to follow because you don't really explain where you are getting your formulae from. One thing you appear to be doing incorrectly is regarding $L$ as a speed dependent quantity. You should be using your own rulers which are at rest - their length is unaffected by the speed of the accelerating object.

 

[John SpaceY]

Thank you for your comments

Yes I agree that kinetic energy tends towards infinity when v tends towards c
But if you want to increase a high speed v by dv, if you do an infinite addition of energy that are decreasing to get an infinite addition of dv when v tends to c, the final kinetic energy will tend also towards infinity.
In the same way then if you do an infinite addition of energy that are increasing to get an infinite addition of dv when v tends to c.
What I would like to know is if you want to increase v by a quantity dv, do you have to add more and more energy or less and and less energy to the moving object, when v is tending to c ?

For the equations I have used the normal physics equation to accelerate a mass M
F = M . Gamma (Gamma is the acceleration and is égal to dv / dt)
The energy needed to move M on a distance x is legal to : W = F . x
And so the energy needed is W = M . dv/dt . x
And after this I have taken the relativistic values of M, t and x in the Earth reference
M, T and L for the mass, the time and the distance

Thank you in advance for your answers
John

 

[Ibix]

What I would like to know is if you want to increase v by a quantity dv, do you have to add more and more energy or less and and less energy to the moving object, when v is tending to c ?

From the formula in my last post, $$\frac{dE}{dv}=m\gamma^3v$$which means that for a fixed small velocity increase $\delta v$ the energy required to make that increase, which is $(dE/dv)\delta v$, increases with $v$.

 

[Mister T]

And the more v is near of c, the more energy is needed in order to continue to increase v, because M is increasing.

The only reason $M$ is increasing is because of the way it was defined. It always takes more energy to increase the speed of an object. What we find is that each successive increase in speed requires a greater amount of energy, and this progresses in such a way that no matter how much energy we transfer to the object, the speed never reaches $c$. This is best explained as a feature of the geometry of spacetime, not an increase in some property of the object.

I have another way of thinking ant it is on this point that I have my question: [...]

and so. W = M . dv/dT . L

This can't be right, the quantity on the right hand side of your equation doesn't have units of energy.
Edit: I see that it does indeed have units of energy, but I can't follow the logic of how you derived it.

If the object is in a vacuum we consider its mass as zero

No, we don't.

On the other hand, if the object is on our Earth, what is the answer?
does it take a lot of energy or not a lot to continue increasing speed?
(for example in a particle accelerator at CERN or ...)

The inside of the beam tube in particle accelerators like those at CERN is a vacuum.

All you have to do is look at the sizes of the thousands of particle accelerators that people have built. The reason the biggest ones make the particles go the fastest is because it takes more energy to make particles go faster.

 

[Ibix]

This can't be right, the quantity on the right hand side of your equation doesn't have units of energy.

The units are OK, aren't they? At least if you interpret it as $M.(dv/dT).L$.

The problem (I think) is that what he's trying to do (I think) is a relativistic form of $\int\vec F.d\vec x$, but he's fixed the distance the force moves instead of fixing the velocity change, and hasn't constrained his acceleration profile at all. So the equation isn't really saying anything. (And I haven't worked through whether he's using the correct definition of force.)

Edit: actually, given that there's no integral in @John SpaceY's formula, that means that the force has been implicitlyconstrained. But it's not clear to me that it's been correctly constrained.

 

[Ibix]

If the object is in a vacuum we consider its mass as zero

I missed that last time. As @Mister T says, no we do not. What does vacuum have to do with mass?

 

[PeroK]

Thank you in advance for your answers
John

Let's take an example, for a particle of mass $m$. First, we'll start with $v_1= 0.9c$ and accelerate it to $v_2 = 0.91c$. That's an increase of $0.01c$. We have:
$$E_1 = \gamma mc^2 = 2.29mc^2, \ \ E_2 = 2.41mc^2, \ \ \Delta E = 0.12mc^2$$

Now, we'll start with $v_1 = 0.95c$ and accelerate it to $v_2 = 0.96c$. That's also an increase of $0.01c$. Now, we have:
$$E_1 = 3.20mc^2, \ \ E_2 = 3.57mc^2, \ \ \Delta E = 0.37mc^2$$
That's three times the additional energy requirement for the same acceleration.

In general, if you have a simple mathematical equation and then some elaborate mathematics that contradicts it, then the elaborate mathematics is wrong.

 

[Janus]

If the object is in a vacuum we consider its mass as zero and therefore to accelerate it there is not a need for a lot of energy: is that correct ?
On the other hand, if the object is on our Earth, what is the answer?John

Being in a vacuum or on the Earth has no bearing on an object's mass. A 1kg object on the surface of the Earth has the same mass as that same object floating in deep space. It is only their weight that is different. Your confusion probably comes from the fact that we tend to equate mass and weight in normal everyday parlance.
But this is just because we are used to everything on Earth being subject to the same gravity.
If you were standing on the surface of the Moon, for example, you would weigh about 1/6 of what you do on the Earth, but your mass would still be the same.

 

[Ibix]

Being in a vacuum or on the Earth has no bearing on an object's mass. A 1kg object on the surface of the Earth has the same mass as that same object floating in deep space. It is only their weight that is different.

Just to add to this, @John SpaceY (because I don't think Janus said so explicitly), the vacuum is irrelevant. The difference in weight is due to different gravity and states of motion. You can be weightless in air (e.g. people inside a space station) and have weight in vacuum (e.g. on the moon).

 

[John SpaceY]

Kinetic energy of a moving particle is

W = ( ϒ - 1) m c2

The normal way of thinking in physics is the following :

When we inject some energy inside the particle its speed will increase and all the energy injected will be transformed in kinetic energy.

W injected is known and so with the above equation of W, we can calculate ϒ and so we will calculate v.
But if I take the following equation :

W = M . dv/dT . L

And if we inject an energy W1 to the particle we can also calculate the new value of v for the particle.

Initial particle speed v is supposed to be known (data measured)

We inject W1 (known) on a distance L (known : because on Earth reference and so measurable) and during a time T (known because also measurable on Earth).

Initial speed v is known and so M is known (M = ϒ . m and m is known (m is the mass at rest) )

The equation W1 = M . dv/dT . L gives dv because all the other parameters are known.

v + dv gives the new value of v and now we can calculate the new ϒ

And after we can calculate the new kinetic energy of the particle with the following equation : with the new ϒ

W2 = ( ϒ - 1) m c2


Is it possible that W2 could be higher than W1, thanks to a relativistic amplification ? in the same way that relativistic equations will increase M when v tends to c ?

Or W2 has to be in any case equal to W1 ?

 

[Ibix]

But if I take the following equation :

W = M . dv/dT . L

This isn't correct, even in Newtonian physics, if the acceleration is not constant. The correct expression is $$W=\int \frac{d\vec p}{dt}.d\vec l$$

If you insist on taking this approach, you need to write down an expression for $dp/dt$. It is not $M dv/dt$ in relativity. Then you need to do the integral, probably by writing $dl=v\ dt$ and defining an acceleration profile. The answer will be $(\gamma_f-\gamma_i)mc^2$, where the two $\gamma$s are associated with the final and initial speeds.

 

[John SpaceY]

Thanks for this answer
I have understood my mistake !

And so if I will calculate W2 with your equations could yo answer to my question below ? :

Is it possible that W2 could be higher than W1, thanks to a relativistic amplification ? in the same way that relativistic equations will increase M when v tends to c ?

Or W2 has to be in any case equal to W1 ?

 

[Ibix]

I have no idea what you mean by "relativistic amplification". Conservation of energy applies in relativity, so if you accelerate a body then its kinetic energy gain is equal to the work done on it, if that's what you are asking.

 

[John SpaceY]

What I don't understand is the following :

The particle energy in the particle frame is mc2
The particle energy in the Earth frame is the same mc2 but we have to add the kinetic energy of the particle because it is moving seen from the Earth : this kinetic energy is (ϒ - 1)mc2
So the total energy seen from the Earth of the particle is ϒmc2
So for me the energy is the particle frame mc2 is not the same than in the Earth frame ϒmc2
Seen from the Earth there is a relativistic amplification which is ϒ : there is not conservation of energy (or I don't understand the way of conservation : for me mc2 is not equal to ϒmc2)
this amplification is around 1 when we are not in a relativistic speed for the particle

My question is that if there is an amplification for the particle energy, is it possible that when you inject some energy inside the particle, this energy will increase the speed and when you calculate the particle energy in the Earth frame the result ϒmc2 will be higher than expected ?
Or in other words, could the speed be higher than expected ?

 

[Ibix]

Energy is not frame invariant. This is also the case in Newtonian physics - if you sit in a moving car your kinetic energy in your rest frame is zero, but it's definitely not zero as measured by a pedestrian.

Could the speed be higher than expected? Not if you calculated your expected speed correctly. Anything else would imply that relativity was internally inconsistent. It isn't. It may or may not be an accurate description of the world (we are not aware of any inaccuracies so far), but it is completely consistent with itself.

 

[PeroK]

So for me the energy is the particle frame mc2 is not the same than in the Earth frame ϒmc2
Seen from the Earth there is a relativistic amplification which is ϒ : there is not conservation of energy (or I don't understand the way of conservation : for me mc2 is not equal to ϒmc2)
this amplification is around 1 when we are not in a relativistic speed for the particle

"Conservation" of any quantity means it doesn't change over time. "Invariance" of a quantity means it is the same in all (inertial) reference frames.

Energy of a closed system is conserved. The energy of your particle in this case is not conserved, as there is some external source of energy accelerating the particle.

Energy is not invariant and never has been. Kinetic energy always varies between frames moving with respect to each other, and always has done! For example, you might study a collision of snooker balls in the reference frame of the table. The balls have low kinetic energy. But, in the reference frame of the solar system, the snooker table is orbiting the sun and rotating with the Earth. The kinetic energy of the balls is much greater in that reference frame. Kinetic energy is never going to be a generally invariant quantity.

However, in both the table frame and the solar system frame energy is conserved (assuming an elastic collision).

 

[John SpaceY]

Thanks
I think I have understood
John

 

[John SpaceY]

I have other questions

Today I have understood that the maximum energy the CERN particle accelerator can transfer to a particle is 10-6 Joules. And so this energy will be transformed into speed for the particle.
Supposing that we find a way to increase more the quantity of energy transfer, this will increase the speed of the particle.
the speed will become 0,999...999.c
and the more energy injected the more the number of 9 will increase

Is it possible, after a certain number of 9 that because the kinetic energy will be very high, that the moving particle could create new particles ? in this case the moving particle speed and energy will be reduced
The kinetic energy will be transformed in Mass
I have read that in the CERN particle accelerator they see Neutrinos when they see a lot of other particles : could these particles have been created by the Neutrinos when they have reached a high speed ?
if not where are the new particles coming from ?

Is it possible also that the high speed neutrinos could disappear ?
I explain : in the neutrinos particle reference time is going smaller and smaller because of the relativistic equation T = ϒ . t
(it is like the mass M that is increasing M = ϒ . m)
And at the limit, time t in the Neutrinos reference will tend to 0 when v tends to c
Is it possible that in the neutrinos reference, as time is decreasing, when arriving at a very high speed, time decrease and becomes negative ?
In this case all the particle kinetic energy could have been used to do this change and the neutrinos speed will be zero : but we couldn't see it because it will be in another space-time ?
For exemple I have read also that Higgs Bosons destroyed themselves after 10-22s
is it possible that they have not been destroyed but they have changed in the negative space time because they have reached a too high speed ? they have disappeared and not been destroyed in other particles.
Maybe they have also created other particles but could it be possible that they go in a negative time ?

If it is possible for high speed particles to go in a negative space time how this could be explained in a scientific way ? a physics demonstration for example
Do you have heard about negative space time ?

Thanks for your comments
John

 

[Nugatory]

Is it possible, after a certain number of 9 that because the kinetic energy will be very high, that the moving particle could create new particles ?

No.
A particle moving very fast relative to CERN is the same situation as the particle sitting still while CERN is moving very fast in the opposite direction. If there's no reason for the new particles to be created using one picture, there's no reason for new particles to be created using the other picture.

It would be a different story if our particle interacted with something else, like maybe a stray air molecule that happens to be in its way. Then we could say that the particle was at rest when the fast-moving molecule slammed into it, or we could say that the fast-moving particle slammed intoa n air molecule that was at rest - but either way, we have a high-speed collision that could create new particles.

 

[PeroK]

I have other questions

Today I have understood that the maximum energy the CERN particle accelerator can transfer to a particle is 10-6 Joules. And so this energy will be transformed into speed for the particle.
Supposing that we find a way to increase more the quantity of energy transfer, this will increase the speed of the particle.
the speed will become 0,999...999.c
and the more energy injected the more the number of 9 will increase

Is it possible, after a certain number of 9 that because the kinetic energy will be very high, that the moving particle could create new particles ? in this case the moving particle speed and energy will be reduced
The kinetic energy will be transformed in Mass
I have read that in the CERN particle accelerator they see Neutrinos when they see a lot of other particles : could these particles have been created by the Neutrinos when they have reached a high speed ?
if not where are the new particles coming from ?

Is it possible also that the high speed neutrinos could disappear ?
I explain : in the neutrinos particle reference time is going smaller and smaller because of the relativistic equation T = ϒ . t
(it is like the mass M that is increasing M = ϒ . m)
And at the limit, time t in the Neutrinos reference will tend to 0 when v tends to c
Is it possible that in the neutrinos reference, as time is decreasing, when arriving at a very high speed, time decrease and becomes negative ?
In this case all the particle kinetic energy could have been used to do this change and the neutrinos speed will be zero : but we couldn't see it because it will be in another space-time ?
For exemple I have read also that Higgs Bosons destroyed themselves after 10-22s
is it possible that they have not been destroyed but they have changed in the negative space time because they have reached a too high speed ? they have disappeared and not been destroyed in other particles.
Maybe they have also created other particles but could it be possible that they go in a negative time ?

If it is possible for high speed particles to go in a negative space time how this could be explained in a scientific way ? a physics demonstration for example
Do you have heard about negative space time ?

Thanks for your comments
John

Not this again! We went through all this on your thread in January:

https://www.physicsforums.com/threa...bject-goes-near-of-the-speed-of-light.982767/

 

[Vanadium 50]

Maybe Relativity has changed since January.

 

[John SpaceY]

OK
Sorry to having disturbing you
I have understood now my mistakes
Best regards
John

 

[Mister T]

Seen from the Earth there is a relativistic amplification which is ϒ : there is not conservation of energy (or I don't understand the way of conservation : for me mc2 is not equal to ϒmc2)
this amplification is around 1 when we are not in a relativistic speed for the particle

There is no definitive separation between relativistic and non-relativistic speeds. That nomenclature simply means that if the speed is low enough and the accuracy required is low enough, we can use the Newtonian approximation. For example, GPS satellites move at speeds that are a tiny fraction of $c$ yet because of the accuracy required GPS engineers cannot use the Newtonian approximation.

And the fact that $\gamma \approx 1$ at low speeds doesn't necessarily mean the kinetic energy can be ignored. Sure, it's small compared the the rest energy $mc^2$, but the relevant factor is $\gamma-1$ and that tells us that the kinetic energy is approximately $\frac{1}{2}mv^2$ when $\gamma \approx 1$.

 

[PAllen]

Worth noting: at the precision currently achievable at NIST, speeds which can be used to verify relativistic corrections to Newtonian physics are several meters per second. So a good sprint is a ‘relativistic speed’.

 

[MikeeMiracle]

John, sometimes you just have to admit that you just don't understand enough of the underlying physics to understand whether any of your questions and perceptions are valid or not. The people on this site are very knowledgeable and when they say your are wrong you need to accept that and try and understand why and listen to them.

Arguing and repeating the same thing just because it makes sense to you but no one else is not how you learn science.

 

[Vanadium 50]

Electron drift velocities are quite low (mm/s range) and yet we can see their magnetic fields.

 

[John SpaceY]

There is something that is disturbing be in the current Relativistic Theory :
If we find a way to make a spacecraft that is able to go at a speed near to the speed of light c (not done today), and if we want to go to the nearest galaxy the travel will need minimum 2,55 million of years (seen from our Earth). It is because the Andromede Galaxy is located at a distance which is 2,5 millions of light years from our Sun.
The good point with the current Relativistic Theory is that, seen from inside the spacecraft , the travel could take only one hour, if the speed v is closed to the speed of light c. And so people inside the spacecraft will not have aged.
If these people will make the return travel, they will see inside the spacecraft that this travel will need also one hour (or less, just fonction af the spacecraft speed v). But the time seen from our Earth will be also 2,5 millions of years (and this is a minimum because v will be less than c).
And so seen from our Earth, this travel, done at a very high speed, will need, in the best case 5,1million of years. And here is my disturbing point : at the speed where things are going for our Earth, the possibility to have no Earth at our return is high ! we will never see our Earth is the same situation than it was when we have left. And so long travels into space, when we will find a way to do very high speed spacecraft , will be only one way travel. These kind of travels will have no return.

The reason why I have asked all my questions in January and now in March is because I would like to find a theory that could solve this issue.
But I have understood, according to all your answers, that my theory can not be explained by current physics and so it has to stay at a fictional scientific essay level.

I explain:
My Theory was based on the observation that when the spacecraft speed is increasing at a speed v very closed to c, as the mass M, seen from the Earth (the relativistic mass) is increasing and tends towards infinity when v tends to c. And so the spacecraft energy will tend to infinity.
Inside the spacecraft time is decreasing and will tend to 0 when v tends towards c.
I have imagined that if we continue to inject energy in the spacecraft speed : for example a pulse of energy, this would have the following effect : the inside time of the spacecraft will continu to decrease, reach 0 and after will continue to decrease (by continuity) and will become negative. All the spacecraft energy will be used to do this time change and the new speed would be 0 in the new space time where t is negative.
And so the speed c will be be exceeded.
I have imagined another explanation when the spacecraft energy will become to high : it was the creation of new particles. These new particles would have decreased the spacecraft energy and its speed.
But you have explained me that it is not possible.
And so I stay with my theory of negative time.
But you have also explained me that it is not possible.
If we continue to inject energy inside the spacecraft , its speed will increase more and more
(we will add more 9 to the following speed v : v = 0,999...99999.c)
And that is all what will happen : the spacecraft energy seen from the Earth will continu to increase with no other effect : no creation of particles, no negative time and no explosion (except if something will be in contact with the spacecraft ).
So my theory cannot be explained by current physics.

But if someone could find an explanation it will be nice.
I just continu my theory :
if we give an pulse of energy after the travel an goes into the negative time the situation will be the following :
Seen from the Earth the first travel will need 2,55 millions of years and 1 hour inside the spacecraft
when airing to the Andromede galaxy, we give the pulse of energy and the time is now negative.
We return to our Earth, the time is negative : and the spacecraft is also going at a speed v near of c.
Seen from the Earth the time of the return travel will be - 2,55 millions of years : this will compensate te firs travel.
And inside the spacecraft the time will be - 1 hour and this will also compensate the time for the first travel.
When airing on our Earth, we need to give again a pulse of energy and the time will now become positive and the spacecraft will stop at a zero speed.
And so the complete travel time will be zero : seen from our Erat and inside the spacecraft .
And here is the possibility to make a long travel into space, with people inside the spacecraft and on the Earth who will not have aged.
All the people will have the same age before and after the travel : people on our Earth and inside the spacecraft .
And this will be the same four all travels distance : the return time will compensate the first travel time.

But I understand that I cannot prove this with the current physics and I need some help if somebody has other ideas or some physics concrete explanation.
Now all my theory is only science fiction.
But I wanted to explain that I am just searching a way to solve the today problem of people on Earth that will aged a lot when other people inside a spacecraft will travel (and will not have aged).
I understand the current relativistic theory but I don't like the consequences for people on Earth and also for people inside the spacecraft : when they will do a space travel they will know that they will lose everything they have experienced on Earth !
I have a feeling that there should be a solution for this : and that a space travel cannot be only a non return travel.

Thanks in advance to understand my questions and thanks for your help if you have some ideas or some remarks.

Best regards
John

 

[vanhees71]

The funny thing as that it wouldn't help to travel with humans in space close to the speed of light for a long time since then cosmic microwave background radiation will be hard $\gamma$ radiation due to the Doppler effect. In other words, you'll be dead from the radiation damage before getting very far.

 

[russ_watters]

I understand the current relativistic theory but I don't like the consequences for people on Earth and also for people inside the spacecraft ...

The universe does not care if you like the way it works and we don't have the power to change the laws of the universe to something we like better. The origional question was adequately answered, so this thread is locked.