Newton's Second Law (relativistic)

[neilparker62]

Does this solution mean anything ?

 

[ChrisVer]

A constant non-zero force is going to give a periodic solution...that will have $v_{max}=c$...

 

[PeterDonis]

Does this solution mean anything ?

It means something, but it has at least two issues:

(1) Your input equation for the force is wrong in general; see here:

http://en.wikipedia.org/wiki/Relativistic_mechanics#Force

Your equation is valid for the special case where the force is always perpendicular to the velocity, i.e., for exactly circular motion. However, that brings up the second point:

(2) You are only solving for one component of the velocity; for circular motion the velocity has two components, not one. The other component is a cosine (with the same argument as the sine you wrote down). You need to tell your differential equation solver that $v$ is a vector in a plane, not a scalar (so $v^2$ is the sum of the squares of both components).

 

[ZetaOfThree]

It is semi-well known that a particle acted upon by a constant force $F$ undergoes "hyperbolic" motion: $x(t)=\sqrt{(\frac{mc^2}{F})^2+(ct)^2}$. I have a feeling that is the problem that you are trying to solve.

 

[neilparker62]

Many thanks for your observations.

Well I thought the solution might be of the form c * e^(jwt) rather than sin(wt). But the frequency looks interesting - very plainly it is ridiculously low for (say) an object in a gravitational field. Even so with small angle approximation, we get the following: v(t) = c sin(Ft/cm) = c . Ft/cm = Ft/m as usual.

You can integrate to obtain x(t) and then you get mc^2/F as the amplitude which you can re-arrange to write:

F * x(t) = mc^2.cos(wt) or maybe F * x(t) = mc^2 * e^(jwt) which is some kind of oscillatory energy equation with a maximum of m * c^2 as might be expected. I am not sure exactly what is oscillating in this instance ?? Once again x(t) = F/(2m) * t^2 falls out if you use a small angle approximation for cos(Ft/cm)

For the frequency w=F/cm not to be ridiculously low you would need to find a situation with a large F/m ratio - I thought maybe Hydrogen electron with F=kq/r^2. Well the Maths and QM at this point is way out of my league but just a thought!

 

[neilparker62]

oops that should be F = k * q^2/r^2

 

[PeterDonis]

I thought the solution might be of the form c * e^(jwt) rather than sin(wt).

You do realize that these are the same thing, right? (I assume that by $j$ you mean the imaginary unit.) $\sin (\omega t) = ( e^{i \omega t} - e^{- i \omega t} ) / 2 i$. Sines and cosines are just different ways of writing exponentials with imaginary arguments. If you treat the full velocity vector instead of just a single component, in polar coordinates it is indeed of the exact form $e^{i \omega t}$.

If you write the correct equation for linear acceleration, rather than circular motion (i.e., acceleration parallel to velocity instead of perpendicular to velocity), you will find that it has solutions that involve real exponentials instead of imaginary ones.

But the frequency looks interesting - very plainly it is ridiculously low for (say) an object in a gravitational field.

The equation you wrote down is not valid in the presence of gravity. Your equation is only valid in special relativity; gravity requires general relativity.

 

[member 141513]

I think that it is a very good expression in linking up the understanding of classical mechanics and relativity (the semi-classical kinematics) .
But Einstein always emphasized that the notation $μ=\frac{m}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ is physically meaningless.
In fact, if it would be more natural to describe a "m" using different reference frame parametrized by $γ=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ to conserve physical laws.(with m not sticked with γ )
Moreover, in relativity, to deal with acceleration, we need to use curved space-time structure to replace the "force" concept .
Likewise , we can physically "prevent" solving this ODE by :
1st : "knowing" that there is an acceleration (the space-time is curved)
2nd : making m to be described invariably (not changing by any v)
3rd : find out how the other reference frames are changing apart from "here" (related by curvature tensor R) and "lets" describe the "m" together with this relationship
4th : "RELATIVITI-ly" solved F=γm v'
I hope this would help.
pliu123123

 

[PeterDonis]

Einstein always emphasized that the notation $μ=\frac{m}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ is physically meaningless.

Reference, please?

it would be more natural to describe a "m" using different reference frame parametrized by $γ=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ to conserve physical laws.(with m not sticked with γ )

First of all, this would not be a different "reference frame"; it would be a different convention for what the symbol $m$ means.

Second, you can't "conserve physical laws" by adopting this definition for $m$; some of the laws still have to change form from their Newtonian versions (I assume what you mean by "conserve physical laws" is "all the laws look exactly the same as their Newtonian versions").

Moreover, in relativity, to deal with acceleration, we need to use curved space-time structure to replace the "force" concept .

We don't do this to deal with acceleration; we do it to deal with gravity--more precisely, with *tidal* gravity.

I'm not sure what the rest of your post means.

 

[member 141513]

First of all, this would not be a different "reference frame"; it would be a different convention for what the symbol $m$ means.

Second, you can't "conserve physical laws" by adopting this definition for $m$; some of the laws still have to change form from their Newtonian versions (I assume what you mean by "conserve physical laws" is "all the laws look exactly the same as their Newtonian versions").

We don't do this to deal with acceleration; we do it to deal with gravity--more precisely, with *tidal* gravity.

Oh, thanks. Do you mean different reference "frames"?
Since physical laws are no "different" versions, I treat accelerations as to equivalence principle.

 

[neilparker62]

@pliu123123 & Peter. All a bit over my head I'm afraid - I simply don't have the requisite understanding of relativistic acceleration. But appreciate your posts all the same.