Derivation of relativistic acceleration and momentum

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Dear all,

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

Many thanks n
happy eqtns
R
 

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  • #2
Hootenanny
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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?
 
  • #3
85
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In relativity, mass is dependent of velocity such that,
[tex]m=\gamma m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}[/tex].
[itex]m_0[/tex] is the mass of the object at rest, [itex]c[/itex] 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:
[tex]p=mv, F=p', a=v', v=x'.[/tex]
Using these, we easily find that,
[tex]p=\frac{m_0v}{\sqrt{1-v^2/c^2}}[/tex]
and
[tex]F=p'=(mv)'=m'v+v'm[/tex].
Now we need to express m' in terms of only v.
[tex]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}}[/tex].
Combining this with the above equation for force,
[tex]F=v^2\frac{m_0v'}{c^2(1-v^2/c^2)^{3/2}}+\frac{m_0v'}{\sqrt{1-v^2/c^2}}[/tex].
Now you can just factor and solve for [itex]a, v'[/itex].
 
  • #4
clem
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You result is for parallel to v. With vectors, there are other terms.
 
  • #5
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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
 
  • #6
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F=ma does work in both SR and GR as long as you are using the 4-vector (tensorial) version.
 
  • #7
clem
Science Advisor
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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.
 
  • #8
601
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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 im waiting for is its dervn
 
  • #9
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Yes, the actual equation is (gamma)ma=(F-F.v/c)v/c as i mentioned in the new thread. What im waiting for is its dervn
Use the fact that
[tex]v\frac{dv}{dt}=\vec{v}\cdot \vec{a}[/tex]
 
  • #10
601
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Use the fact that
[tex]v\frac{dv}{dt}=\vec{v}\cdot \vec{a}[/tex]
do it
 
  • #11
85
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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):
[tex]\vec{F}=m_o \vec{v}\frac{d\gamma}{dt}+m_o\gamma\frac{d\vec{v}}{dt}[/tex]
and using my above post,
[tex]\frac{d\gamma}{dt}=\gamma^3\frac{|\vec{a}||\vec{v}|}{c^2}[/tex],
[tex]\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}[/tex]
So,
[tex]m_o\gamma\vec{a}=\vec{F}-[(m_o\gamma^3\vec{a})\cdot\frac{\vec{v}}{c}]\frac{\vec{v}}{c}[/tex].
According to your author, the following must be true,
[tex]m_o\gamma^3\vec{a}=\vec{F}[/tex].
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.
 
  • #12
clem
Science Advisor
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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.
 
  • #13
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?
 
  • #14
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Dear all,

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

Many thanks n
happy eqtns
R
[tex]a \equiv \frac{d^2x}{dt^2}[/tex]

[tex]p \equiv \frac{\partial L}{\partial v}[/tex]

with [itex]L[/itex] the relativistic Lagrangian and [itex]v[/itex] velocity
 

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