# 1N Force and going near the speed of light?

Gold Member
Lets suppose we are in a space and theres no external force that affects our system,which our system is simply one object which we can think it is a box and it has 1 kg mass and there is us.

The object ,lets call it A,Its initally rest relative to us so it can be our inertial referance frame So We pushed the A and we applied a force to the A ( Lets call it 1N).It started to accelerate.(I am not sure at this point A will be still our inertial referance frame or not).

In this case we are observer and we will think we are stationary so the A will start to accelerate with constant acceleration ## \vec F=m \vec a## (The motion is only in one direction like +x) so, ##F=1N## and ##m=1kg## so ##a=1\frac m s^2##

We know that theres no external force affecting our system.So after some time the objects speed will increase.And after a period we will use relativity theories to calculate A 's speed.The thing is I just wanted know In this circumtances , With ##1N## force object can reach ##0.9999999....c## if its given enough time ?

If you also give an answer to the question that I wrote in bold I'll happy,

Thank you

You say .... "We pushed the A and we applied a force to the A " well that makes it accelerate away from you , so how are you going to continue to apply a force ??

But if you could find a way of applying a 1N force to a 1kg mass ... then yes , in theory you could get it to 0.9999999 c

Merlin3189
Homework Helper
Gold Member
I don't think A can be an inertial frame of reference, as it is accelerating.

Since you say, we think we are stationary observer, then you are defining yourself, the observer as the frame of reference. Since you are applying the force to A, then A is applying an equal and opposite force to you (Newton's 3rd law) and you are also accelerating. So you also are not an inertial reference and neither can be any other frame in which you are stationary.

Arman777
Gold Member
You say .... "We pushed the A and we applied a force to the A " well that makes it accelerate away from you , so how are you going to continue to apply a force ??

But if you could find a way of applying a 1N force to a 1kg mass ... then yes , in theory you could get it to 0.9999999 c
Well It gained some acceleration right.Theres no external force so it will keep its state of motion.It will accelerate with constant ##a##.I dont need to apply force constantly.
Thats my idea

Merlin3189
Homework Helper
Gold Member
... it will keep its state of motion.It will accelerate with constant ##a##.I dont need to apply force constantly.
Thats my idea
Keeping it's state of motion, means no acceleration: constant velocity.
If you want acceleration, a force needs to be applied. ##a=\frac{F}{m}## if the force is 0, so is the acceleration.

ZapperZ
Staff Emeritus
Well It gained some acceleration right.Theres no external force so it will keep its state of motion.It will accelerate with constant ##a##.I dont need to apply force constantly.
Thats my idea

Your "idea" is wrong. When there's no longer an applied force, it will not accelerate (Newton's First Law).

Zz.

Arman777
Gold Member
Ok I got it thanks.I also understand the referance frame case thanks for that too.It will accelerate and gain some velocity (lets call it v) but when I cut force there will be no acceleration so it will move with a constant velocity v.
For referance frame.A cannot be an inertial referance frame cause its accelerates.We need constant velocity object to choose an inertial referance frame.

russ_watters
Mentor
But if you could find a way of applying a 1N force to a 1kg mass ... then yes , in theory you could get it to 0.9999999 c

...but you cannot get to 99.999...C

Gold Member
...but you cannot get to 99.999...C

İs it some kinda of joke .I didnt understand

A.T.
...but you cannot get to 99.999...C
Not even to 0.999... c

russ_watters
Gold Member
sure we need huge energy a lot of energy.Even for an atom we need huge amount of energy for a 1 kg object we need a lot

russ_watters
Mentor
Not even to 0.999... c
Oops, lol.

russ_watters
Mentor
İs it some kinda of joke .I didnt understand
Badly played joke. In math language, 0.999...=1 and since you can't go C, you can't go 0.999...C either.

mfb
Mentor
Accelerating 1 kg to 99.99% the speed of light needs 6*1018 J, about the world energy consumption of a week.
Accelerating a proton to 99.99% the speed of light needs a large accelerator. The SPS and the LHC, Tevatron (and its preaccelerator), HERA and RHIC can/could do that. The energy per proton is tiny - 0.01 microjoule - but getting that energy per proton is not easy.

Accelerating a proton (well, actually 6*1014 protons) to 99.9999990% the speed of light (1.06 microjoule per proton, 600 MJ in total) is done at the LHC, it is the current record for accelerators.

Arman777 and russ_watters
Accelerating 1 kg to 99.99% the speed of light needs 6*1018 J, about the world energy consumption of a week.
Accelerating a proton to 99.99% the speed of light needs a large accelerator. The SPS and the LHC, Tevatron (and its preaccelerator), HERA and RHIC can/could do that. The energy per proton is tiny - 0.01 microjoule - but getting that energy per proton is not easy.

Accelerating a proton (well, actually 6*1014 protons) to 99.9999990% the speed of light (1.06 microjoule per proton, 600 MJ in total) is done at the LHC, it is the current record for accelerators.

You're figures are way off mfb ... at .9999c mass is 10 tonnes needs 4.5 * 10 ^20 J...

but at 0.9999999 c the original 1kg mass would be so much greater .....trillions of tonnes I would guess...

So are you saying the 1kg mass at 0.9999999c would not be "so much greater in mass"?
Are you saying the equation in post 15 is not valid?
If so , what is the mass at 0.9999999c?

1kg. Read again the article that I gave you.

1kg. Read again the article that I gave you.

Perhaps you can explain it to me ....I don't understand it.

ZapperZ
Staff Emeritus
Perhaps you can explain it to me ....I don't understand it.

It looks like we are back on this issue again.

There's nothing wrong with the equation. It is the "concept" of "relativistic mass" that is the issue. Please note that even the "originator" of this idea of relativistic mass, Albert Einstein, stopped using it after he realized that this might be problematic. Read this post as well as that Insight article:

Zz.

It looks like we are back on this issue again.

There's nothing wrong with the equation. It is the "concept" of "relativistic mass" that is the issue. Please note that even the "originator" of this idea of relativistic mass, Albert Einstein, stopped using it after he realized that this might be problematic. Read this post as well as that Insight article:

Zz.

So if there's nothing wrong with the equation , the mass would be 10 tonnes at 0.9999c , and trillions of tonnes at 0.9999999c , is that right?

ZapperZ
Staff Emeritus
So if there's nothing wrong with the equation , the mass would be 10 tonnes at 0.9999c , and trillions of tonnes at 0.9999999c , is that right?

There's nothing wrong with A LOT of equations. For example, in the simple kinetic energy equation

KE = 1/2 mv2

I can also write it as (m/2)v2. Someone may wish to interpret it as having only HALF of the mass having the kinetic energy, while the other half doesn't! Can't you see how this makes it sound absurd?

The same applies to relativistic physics. You are extracting a part of a larger, more general form, and then trying to "interpret" what it means. Simply look at how that form is derived, and you'll see that what is "primary" here (as used in one of the papers I cited) is "relativistic MOMENTUM" and "relativistic ENERGY". The "mass" can and should only be defined as the inertial or invariant mass.

Please note that in high energy physics, where a lot of particles are zooming around at relativistic speeds, if you look in the Particle Data book on their masses, they NEVER cite masses at specific speeds.

Zz.

My understanding is this has all been verified in particle accelerators up to 0.9999999c and beyond ?
There's nothing wrong with A LOT of equations. For example, in the simple kinetic energy equation

KE = 1/2 mv2

I can also write it as (m/2)v2. Someone may wish to interpret it as having only HALF of the mass having the kinetic energy, while the other half doesn't! Can't you see how this makes it sound absurd?

Honestly I can't ... this ke formula is very clear and works .....

Please tell me how I find the the mass of 1kg rest mass object at 0.9999c

Gold Member
My understanding is this has all been verified in particle accelerators up to 0.9999999c and beyond ?

Honestly I can't ... this ke formula is very clear and works .....

Please tell me how I find the the mass of 1kg rest mass object at 0.9999c

mfb
Mentor
Please tell me how I find the the mass of 1kg rest mass object at 0.9999c
It is 1 kg. Because "mass" means "invariant mass"="rest mass".
You can use the concept of a relativistic (velocity-dependent) mass, but no one does that any more.

Your factor of 1000 is wrong.

$$\gamma_{99.99\%c} = \frac 1 {\sqrt {1-0.9999^2} }= 70.71$$

##E=\gamma_{99.99\%c} 1kg~c^2 = 6.36 \cdot 10^{18} J##

Can we come back to the original topic?