# Special Relativity handling acceleration

1. Sep 29, 2004

### ostren

If an object were undergoing acceleration in hypothetical flat space, then a distant clock's tick rate (relative to one's own) is given by something like (1+gx)/gamma, where g is the acceleration and x (sign significant) is the significant linear distance to the remote clock. Is this part and parcel of the fact that SR can handle acceleration? Yes, of course it is. And if so, then is the formula above derived solely from the Lorentz transform? Just out of idle curiosity, how is that derivation arrived at?

2. Sep 29, 2004

### jcsd

At any given instant an object can be considered to be the inertial frmae, depending on it's four-velocity at that instant, this refrence frame is called the instaneously co-moving inertial frame. You could approximate by pretending that the objcet has a period of constant velocity, then instaneously accelartes up to another velcoity and so on, the smaller the period the more accuarte the result. Now compare this to the defintion of integration and you should see that to find out exactly what's going on by intergrating our formulas for constant velocity.

3. Sep 29, 2004

### ostren

That's precisely what I suspected, an integration of the Lorentz equation. Thank you!

4. Sep 29, 2004

### jcsd

You can also think of it geometircally. Each inertial frame defines a set of Minkowskian coordinates, an object at rest in this frame has a worldline along the time axis in these coordinates. For an acccelarted refrence frma we can't use Minkowkisan coordinates but we can use curvilinear coordinates with the (curved) wordline of an object in this frame defining the time axis.

5. Sep 29, 2004

### ostren

All that is MUCCH appreciated, thanks. But actually, I wished to see the detailed mathematical derivation of the dt(1+gx)/gamma formula from the Lorentz tranform... just for laughs.

6. Sep 29, 2004

### pervect

Staff Emeritus
The entire subject of accelerated observers is treated in MTW's "Gravitation".

While the answer to your question is left as an exercise, the groundwork is set up for it on pg 173 which is referred to by the exercise.

The coordinate transformation from the coordiante system of an observer in hyperbolic motion (constant proper acceleration of a value g) with local coordinates $$\xi^i$$ to an inertial observer with coordinates xi is:

$$\array{rcl} x^0 &=& (\frac{1}{g}+\xi^1)sinh(g \xi^0)\\x^1 &= & (\frac{1}{g}+\xi^1)cosh(g \xi^0)\\x^2 &=& \xi^2\\x^3 &= & \xi^3\end{array}$$

If you substitute this into the equation for the metric / lorentz interval

$$ds^2 = -(dx^0) ^2 + (dx^1) ^2 + (dx^2) ^2 + (dx^3) ^2$$

you get
$$ds^2 = -(1+g \xi^1)^2(d \xi^0)^2 + (d \xi^1)^2 + (d\xi^2)^2 + (d\xi^3)^2$$

It's reasonably obvious that the coefficient of $$\xi^0$$ is the square of the time dilation factor.

Last edited: Sep 30, 2004
7. Sep 30, 2004

### pmb_phy

People often make this mistake. Special relativity is that theory defined by two postulates. They are

(1) The laws of nature are the same in all inertial frames of reference.
(2) The speed of light has a definite value which is independant of the source.

This theory only addresses what happens in inertial frames. The tensors and are therefore only gaurenteed to be Lorentz tensors. When Einstein later came to update this to arbitrary frames he had to come up with a new postulate - The "Principle of General Covariance" which states that the laws of physics are valid in all frames of reference and in all valid coordinate systems. This does not neccesarily follow from the special theory above just because the laws of physics all hold in inertial frames. It took Einstein to make this leap. Einstein explains all this in his 1916 paper. Pauli also explains this in his relativity text.

SR can handle accelerating objects though.

Pete

8. Sep 30, 2004

### pervect

Staff Emeritus
One other thing I should note is that the equations I presented derive the $$(1+g\xi^1)$$ tick rate of the "stationary" (with respect to the accelerated observer) clock at the fixed coordinate $$\xi^1$$.

In case it isn't obvious, x0 is the time coordinate of the inertial observer, x1 is one of the space coordiantes of the inertial observer (the space coordinate pointing in the direction of motion of the accelerated observer). $$\xi^0$$ is the time coordante of the accelerated observer, and $$\xi^1$$ is the space coordiante of the accelerated observer in the direction of motion.

If the clock were moving in the $$\xi$$ coordinate system, rather than stationary at fixed $$\xi$$, one would get the additional time dilation factor of gamma mentioned.