Derivation of relativistic acceleration and momentum

[phys23]

Dear all,

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

Many thanks n
happy eqtns
R

 

[Hootenanny]

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?

 

[gamesguru]

In relativity, mass is dependent of velocity such that,
$$m=\gamma m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}$$.
[itex]m_0[/tex] is the mass of the object at rest, $c$ 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 $a, v'$.

 

[Meir Achuz]

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

 

[genesis1]

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

 

[Cusp]

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

 

[Meir Achuz]

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.

 

[vin300]

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

 

[gamesguru]

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
$$v\frac{dv}{dt}=\vec{v}\cdot \vec{a}$$

 

[vin300]

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

do it

 

[gamesguru]

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.

 

[Meir Achuz]

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.

 

[LorentzWasWro]

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?

 

[juanrga]

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