Derivation of relativistic acceleration and momentum

• phys23
In summary, the author of this summary believes that mainstream theory is wrong and that relativistic mass does not exist. He goes against this by stating that relativistic momentum does not exist, only velocity. He suggests that this is because relativistic mass does not exist and that velocity is all that remains. He also claims that mainstream theory is too linear in its understanding and that this might be why people are confused about it.f

phys23

Dear all,

could anyone please show the full derivation of relativistic acceleration and momentum.

Many thanks n
happy eqtns
R

Dear all,

could anyone please show the full derivation of relativistic acceleration and momentum.

Many thanks n
happy eqtns
R
Welcome to PF,

Have you tried searching the internet?

In relativity, mass is dependent of velocity such that,
$$m=\gamma m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}$$.
$m_0[/tex] is the mass of the object at rest, [itex]c$ is the 299 792 458 m/s.
Most equations still hold true in relativity, the major exception being F=ma.
The following are still true:
$$p=mv, F=p', a=v', v=x'.$$
Using these, we easily find that,
$$p=\frac{m_0v}{\sqrt{1-v^2/c^2}}$$
and
$$F=p'=(mv)'=m'v+v'm$$.
Now we need to express m' in terms of only v.
$$m'=\left(\frac{m_0}{\sqrt{1-v^2/c^2}}\right)'=\frac{-1/2m_0}{(1-v^2/c^2)^{3/2}}(-2v/c^2)(v')=v\frac{m_0v'}{c^2(1-v^2/c^2)^{3/2}}$$.
Combining this with the above equation for force,
$$F=v^2\frac{m_0v'}{c^2(1-v^2/c^2)^{3/2}}+\frac{m_0v'}{\sqrt{1-v^2/c^2}}$$.
Now you can just factor and solve for $a, v'$.

Jan Nebec
You result is for parallel to v. With vectors, there are other terms.

Newton's second law F/m = a Also recall a = dv/dt Also force is related to relativistic momentum by F = dp/dt
Relativistic momentum is defined by p = mv(1 - v^2/c^2)^-.5 You need to use implicit derivation to take the derivative of this with respect to t. Thus you should have dp/dt and dv/dt term. Once you are finished getting the derivative and combining terms you should end up with dv/dt = F(1-v^2/c^2)^3/2 /m

F=ma does work in both SR and GR as long as you are using the 4-vector (tensorial) version.

Newton's second law F/m = a Also recall a = dv/dt Also force is related to relativistic momentum by F = dp/dt
Relativistic momentum is defined by p = mv(1 - v^2/c^2)^-.5 You need to use implicit derivation to take the derivative of this with respect to t. Thus you should have dp/dt and dv/dt term. Once you are finished getting the derivative and combining terms you should end up with dv/dt = F(1-v^2/c^2)^3/2 /m
That result is only valid for a parallel to v.

You result is for parallel to v. With vectors, there are other terms.
Yes, the actual equation is (gamma)ma=(F-F.v/c)v/c as i mentioned in the new thread. What I am waiting for is its dervn

Yes, the actual equation is (gamma)ma=(F-F.v/c)v/c as i mentioned in the new thread. What I am waiting for is its dervn
Use the fact that
$$v\frac{dv}{dt}=\vec{v}\cdot \vec{a}$$

Use the fact that
$$v\frac{dv}{dt}=\vec{v}\cdot \vec{a}$$
do it

do it
If you carry out the same calculations I did in the first post I made in this thread, but use vectors, you get this result (you can do it yourself, it's very easy, esp. since I already did it):
$$\vec{F}=m_o \vec{v}\frac{d\gamma}{dt}+m_o\gamma\frac{d\vec{v}}{dt}$$
and using my above post,
$$\frac{d\gamma}{dt}=\gamma^3\frac{|\vec{a}||\vec{v}|}{c^2}$$,
$$\vec{F}=m_o\gamma^3\vec{v}\frac{|\vec{a}||\vec{v}|}{c^2}+m_o\gamma \vec{a}=m_o\gamma^3\vec{v}\frac{\vec{a}\cdot \vec{v}}{c^2}+m_o\gamma \vec{a}$$
So,
$$m_o\gamma\vec{a}=\vec{F}-[(m_o\gamma^3\vec{a})\cdot\frac{\vec{v}}{c}]\frac{\vec{v}}{c}$$.
According to your author, the following must be true,
$$m_o\gamma^3\vec{a}=\vec{F}$$.
This suggests relativistic mass does not exist and that this author goes against mainstream theory. The only guy I know who goes against this is Levvy. We don't like to trust that guy around here. (Read: http://en.wikipedia.org/wiki/Mass_in_special_relativity#Controversy)

I wouldn't trust this author if I were you, only if you assume relativistic mass does not exist do you get the result you posted.

As I just posted in the other thread, dotting your m\gamma a equation with v
shows the result v.F=m\gamma^3(v.a).
Either you or I are confused about what "mainstream theory" is.

So I've been told that the speed of light remains constant despite length contraction and dime dilation because they each decrease proportionately. So a velocity of 4 meters/2 seconds in a relativistic scenario might see a halving of values of length and time so it may only traverse 2 meters but only 1 second of time elapses. My question is what happens with acceleration if it is measured in distance per time-squared...Shouldn't it decrease proportionately to square root-t? If time dilates, shouldn't time-squared dilate even more? Or is this too linear of an approach?

Dear all,

could anyone please show the full derivation of relativistic acceleration and momentum.

Many thanks n
happy eqtns
R

$$a \equiv \frac{d^2x}{dt^2}$$

$$p \equiv \frac{\partial L}{\partial v}$$

with $L$ the relativistic Lagrangian and $v$ velocity