- #1

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

Many thanks n

happy eqtns

R

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- Thread starter phys23
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- #1

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

Many thanks n

happy eqtns

R

- #2

Hootenanny

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Welcome to PF,

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

Many thanks n

happy eqtns

R

Have you tried searching the internet?

- #3

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[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

Meir Achuz

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

- #5

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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

Meir Achuz

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That result is only valid for a parallel to v.

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

- #8

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

- #9

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Use the fact thatYes, 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

[tex]v\frac{dv}{dt}=\vec{v}\cdot \vec{a}[/tex]

- #10

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do itUse the fact that

[tex]v\frac{dv}{dt}=\vec{v}\cdot \vec{a}[/tex]

- #11

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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):do 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

Meir Achuz

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shows the result v.F=m\gamma^3(v.a).

Either you or I are confused about what "mainstream theory" is.

- #13

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- #14

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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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