Deriving Special Relativity: Force, Acceleration, and Velocity

[snoopies622]

Wikipedia's 'special relativity' entry gives the following relationship between force, acceleration and velocity:

<br /> \vec f=\gamma m \vec a + \gamma^3 m \frac {\vec v \cdot \vec a}{c^2}\vec v<br />

but without derivation. Could someone tell me how one arrives at this?
Thanks.

 

[JustinLevy]

Sure.

First, a quick non-relativistic primer to make sure we start on the same page.
Newton's second law is F = dp/dt. And p=mv. As long as the mass of an object is constant, this can be rewritten F = dp/dt = m dv/dt = m a. In some introductory classes they only give F=ma for simplicity, but please remember Newton's second law is F = dp/dt.

Okay, now to relativistic dynamics. Since v = dx/dt, and different observers can't agree on measurements of time or distance, it is clear the ratio of them will transform even worse. For convenience, we'd like something that transforms with the lorentz transformations. This can be accomplished if we change dt to d'tau' where tau is the proper time of the object whose velocity we're measuring. Proper time is an invarient and thus all observers will agree upon it and make things easier. So now we have components of a four-vector V = dx / dtau. Of course dx transforms just like x, and as mentioned tau and thus dtau are invariant. It may seem a bit odd that we just redefined velocity as a ratio of a length measured in one frame and a time measure in a different frame, but the simplicity this gives is well worth it.

Okay, now to momentum. Just define momentum P = mV, where mass is the rest mass of the object (and thus an invariant) and V is our four-vector. Since m is invariant and V transforms with the lorentz tranformation, so will P ... thus it is a four-vector as well.

Alright, so now we "update" Newton's second law to use this new definition of momentum. In the non-relativistic limit P = mV -> mv (where v is the ordinary velocity), so this is a reasonable extension (and the standard convention). Also, a quick note that due to time dilation we can also write dtau = (1/gamma) dt, thus the spatial parts of the four vector momentum are gamma m v.

F = (d/dt) gamma m v.

Again, in the non-relativistic limit this reduces to the usual F = (d/dt) mv.

Since gamma = 1/sqrt(1 - v^2/c^2), and v can depend on time, this derivative gets messy, but is merely an exercise at this point. Taking the derivative yields the result you copied from wikipedia.

I hope that helped you see where that came from.

 

[snoopies622]

Got it, thanks!

A quick follow-up, if I may: why don't we use proper time for both derivatives? In other words, why is F = (d/dt) gamma m v instead of (d/dtau) gamma m v?

 

[JustinLevy]

A quick follow-up, if I may: why don't we use proper time for both derivatives? In other words, why is F = (d/dt) gamma m v instead of (d/dtau) gamma m v?

Yes indeed, you can do that to make a force four-vector. This is sometimes referred to as the Minkowski Force.

The force four-vector is also mentioned at the end of the wikipedia section you asked about in your openning post: http://en.wikipedia.org/wiki/Special_relativity#Force

 

[snoopies622]

Is the four-force what is measured by the accelerated observer while the force described in entry #1 what is measured by the inertial observer?

 

[granpa]

Okay, now to relativistic dynamics. Since v = dx/dt, and different observers can't agree on measurements of time or distance,

but velocity is relative. relative to itself everything is stationary. we can only speak of an objects velocity relative to some stationary observer. all observers will agree on how that stationary observer will measure distance and time.

also if the length of the four velocity vector is always c because traveling through space causes it to move slower through time then it seems like the fourth component of four velocity should be the derivative of tau with respect to T (because that gets smaller as you go faster). what am I missing here?

edit:wait, time is imaginary so the fourth component would need to get bigger not smaller. ok. but my first point still seems valid.

edit again:it actually seems to work either way.

 

[matheinste]

Quote:-

----we can only speak of an objects velocity relative to some stationary observer---

And how do we decide who is stationary.

Matheinste.

 

[granpa]

Quote:-

----we can only speak of an objects velocity relative to some stationary observer---

And how do we decide who is stationary.

Matheinste.

ok. I guess I didnt explain myself very well. replace the word 'stationary' with 'nonaccelerating'.it occurs to me that since acceleration isn't relative that it would make sense to define velocity and force in terms of it. someone posted something to the effect that different observers will even agree on the perceived acceleration. I'll have to try to find it.

whats the formula for the amount of acceleration perceived by the accelerated object? (apparently its the rate of change of rapidity with respect to proper time)