# I Derivative of Lorentz factor and four-acceleration

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1. Apr 14, 2017

### Frank Castle

As far as I understand it, the Lorentz factor $\gamma(\mathbf{v})$ is constant when one transforms between two inertial reference frames, since the relative velocity $\mathbf{v}$ between them is constant.
However, I'm slightly confused when one considers four acceleration. What is the physical reasoning for why the derivative of the Lorentz factor is non-zero (i.e. $\dot{\gamma}(\mathbf{v})\neq 0$) in this case? Is it simply that, if an object is accelerating, then the relative velocity between it and an inertial frame is no longer constant. Accordingly, the Lorentz factor becomes time-dependent.

Is the point that the reference frame of the object that is accelerating is non-inertial, and so relative to an inertial frame the Lorentz factor will become time-dependent (and so $\dot{\gamma}(\mathbf{v})\neq 0$) since the relative velocity between the objects frame and the inertial frame is constantly changing?! Furthermore, if we subsequently transform between two inertial frames observing the same accelerating object, would it be correct to say that the Lorentz factor relating these two frames is constant?!

2. Apr 14, 2017

### PeroK

3. Apr 14, 2017

### Frank Castle

Not so much doubts, just want to check that I'm understanding the reasoning correctly?!

4. Apr 14, 2017

### PeroK

$\gamma = \frac{1}{\sqrt{1-v^2}}$

Therefore: $\gamma$ is constant (over time) if and only if $v$ is constant (over time).

5. Apr 14, 2017

### Frank Castle

I've been trying to reason (to myself) how one can define four-velocity for an accelerating object? Is it simply defined as the velocity as measured in an inertial reference frame at a given instant in time, where the relative velocity between the inertial frame and the accelerating object (such that $\gamma$ is constant).

6. Apr 14, 2017

### PeroK

The four-velocity is a function of the three-velocity, hence of time. Explicitly:

$u_x = \gamma v_x = \frac{v_x}{\sqrt{1-v^2}}$

Mathematically, that defines the components of the four velocity as functions of the three velocity. You could measure the components of the three velocity and then compute the four velocity, as above.

Or, you could, equivalently, define:

$u_x = \frac{dx}{d\tau}$ (where $\tau$ is the proper time of the particle)

You could, theorectically, measure the four-velocity by measuring the change in displacement of the particle in your frame and using the proper time on a clock attached to the particle.

If the particle is accelerating this is no different from the classical case. You don't have a direct measurement of an instantaneous velocity. It relies on calculus to define velocity - or any continuously changing quantity.

7. Apr 14, 2017

### PeroK

PS: alternatively, you could think about acceleration as a sequence of discreet changes in velocity with periods of constant velocity in between. You don't have to do this, however, if you have calculus up your sleeve.

That said, it is useful sometimes to think about acceleration this way.

8. Apr 14, 2017

### vanhees71

Four-velocity is defined for any motion of a particle, not only uniform ones with constant velocity (SRT would be a poor theory if it couldn't describe the mechanics of point particles moving under the influence of forces). It is defined by
$$u^{\mu} = \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}$$
with the proper time defined by (for massive particles)
$$\mathrm{d} \tau^2 = \eta{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}>0$$
and thus you have
$$u^{\mu} u_{\mu}=1.$$
The relation to the non-covariant three-velocity in an inertial frame is
$$\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \frac{\mathrm{d} \tau}{\mathrm{d} t}=\frac{1}{u^0} \vec{u}.$$
Thus you have
$$u^0 \vec{v}=\vec{u} \; \Rightarrow \; (u^0)^2 \vec{v}^2=\vec{u}^2=(u^0)^2-1 \; \Rightarrow \; u^0=\frac{1}{\sqrt{1-\vec{v}^2}}=\gamma.$$
Proper acceleration is defined as
$$a^{\mu}=\frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d}^2}{\mathrm{d} \tau^2} x^{\mu}.$$

9. Apr 14, 2017

### pervect

Staff Emeritus
I'm not sure when you write $\dot{\gamma}(\mathbf{v})$ whether the dot refers to differentation by coordinate time t, or proper time $\tau$.

Either way, the way to find it is to use the chain rule from calculus.

Since you are talking about four-acceleration, lets assume you've parameterized everyting in terms of proper time $\tau$, which has the added benefit that $\tau$ is independent of the frame of reference by the way it's defined.

Then we have some function $v{\tau}$ that gives the velocity as a function of proper time, and we want to find
$$\frac{d}{d\tau} \left( \gamma(\tau) \right) = \frac{d\gamma}{dv} \frac{dv} {d\tau} = \frac{d}{dv} \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2} }} \right) \cdot \frac{dv}{d\tau}$$

10. Apr 14, 2017

### Frank Castle

Is any of what I wrote in my OP correct at all then? (The dot is meant to be a derivative with respect to proper time).
If a particle is accelerating then its frame is non-inertial and so relative to an inertial observer, the Lorentz factor between the two frames will be time-dependent (since the relative velocity between the two frames is not constant).

Would it correct to say that essentially a particles trajectory is parametrised by its proper time, $x^{\mu}(\tau)$. The particle can either be moving at a constant velocity relative to an inertial observer, or accelerating. In the former case, the particles frame is also inertial and so the Lorentz factor between the particles frame and the inertial observers frame is constant. In the latter case, the particles frame is non-inertial (since it is accelerating) and hence the Lorentz factor between the two frames is no longer constant, but changes in time, i.e. $\frac{d\gamma}{d\tau}\neq 0$. In both cases, the particles velocity $u^{\mu}=\frac{dx^{\mu}}{d\tau}$ is defined, however, in the former case $a^{\mu}=\frac{du^{\mu}}{d\tau}=\frac{d^{2}x^{\mu}}{d\tau^{2}}=0$, whereas in the latter case $a^{\mu}\neq 0$.

Last edited: Apr 14, 2017
11. Apr 14, 2017

### PeroK

How can there be any doubt about that?

You can parameterise a trajectory any way you like, but coordinate time and proper time of the particle are the two obvious candidates.

I'm struggling to see where you have any serious doubts.

12. Apr 14, 2017

### Frank Castle

You're right, I think I'm overthinking things now. I shouldn't doubt myself so much!

13. Apr 14, 2017

### Staff: Mentor

It is defined as the unit tangent vector to the worldline. The same for accelerating and non accelerating worldlines

14. Apr 15, 2017

### Frank Castle

Thanks for clarifying. For some silly reason I was confusing myself over the definition when relating the coordinate time expression to the proper time expression: $$u^{\mu}=\frac{du^{\mu}}{d\tau}=\gamma(\mathbf{v})\frac{du^{\mu}}{dt}$$ and the fact that, for an accelerating particle, $\gamma(\mathbf{v})$ will become time-dependent (such that $\frac{d\gamma(\mathbf{v})}{d\tau}\neq 0$), and $\frac{d^{2}u^{\mu}}{dt^{2}}\neq 0$.
Apologies for getting stuck over a silly point like this!

15. Jan 9, 2018

### Jacques Smeets

Assume a general reference frame (Observer O) along a worldline L and 4-velocity u (not necessarily inertial) and a particle P along a worldline L' and with 4-velocity u' (again not necessarily inertial). Let us assume that at instant t (=O's proper time) the observer is at "spacetime event" O(t) and the particle is (as perceived by O) at "spacetime event" M(t). Let dt be the proper time elapsed according to O and Ot(t) --> O(t+dt) and Mt) --> M(t+dt). Let now dt' be the proper time elapsed according to the particle. The Lorentz factor Γ is then defined by dt=Γdt'.
Using the fact that the vector u(t+dt) and the vector OM(t+dt) are orthogonal it can be simply shown that
Γ = (u. u')/(1+(a.OM)),
where (*.*) is the inner product w.r.t. the metric g, a is the 4-acceleration of the Observer and OM is the vector connecting O(t) and M(t).
Note that this shows that the Lorentz factor in general is not-symmetrical (unless a=0 or OM=0).