# SRT two observers will agree on their relative velocity

1. Jun 6, 2009

Ok I am confused. As far as I understand it, in SRT two observers will agree on their relative velocity.

How do they calculate it?

$$v = \frac{\mathrm{d}x}{\mathrm{d}t}$$

In my rest frame. But in the others rest frame:

$$v' = \frac{\mathrm{d}x'}{\mathrm{d}t'}$$

but now for me the length is contracted and the time delated:

$$\mathrm{d}t' = \frac{\mathrm{d}t}{\gamma}$$
$$\mathrm{d}x' = \gamma \mathrm{d}x$$
$$\arrow v' = \gamma^2 \frac{\mathrm{d}x}{ \mathrm{d}t}$$

So we disagree on our differential speed, this must be wrong. What did I miss?

2. Jun 6, 2009

### George Jones

Staff Emeritus
Re: Confusion

Time dilation and Lorentz contraction don't tell the whole story. The worldline of the unprimed observer is the t-axis, i.e., the line $x = 0$. One of the Lorentz transformation expressions is (with $c = 1$)

$$\Delta x = \gamma \left ( \Delta x' + v \Delta t'} \right).$$

So, $x = 0$ gives

$$v = -\frac{\Delta x'}{\Delta t'},$$

which is what one would expect, since, conventionally, the unprimed observer moves along the primed observer's negative x-axis.

3. Jun 6, 2009

### JesseM

Re: Confusion

v = dx'*(1 - v^2/c^2)/(dt' - v*dx'/c^2).

Dividing both top and bottom of the right side by dt' gives:

v = (dx'/dt')*(1 - v^2/c^2)/(1 - v*(dx'/dt')/c^2)

Multiply both sides by the denominator of the left side:

v - v^2*(dx'/dt')/c^2 = (dx'/dt')*(1 - v^2/c^2)

v = (dx'/dt')*[(v^2/c^2) + (1 - v^2/c^2)] = (dx'/dt')*1

So, this is one way of showing that v, the speed of the clock in your frame, is equal to (dx'/dt'), your speed as measured in the rest frame of the clocks. It's probably easier to just use the Lorentz transformation to show this though, you could pick two events on your worldline at the same position but different time coordinates in your frame, figure out the coordinates of these events in the other frame using the Lorentz transformation, and then show that in the other frame (difference in position)/(difference in time) is equal to v (the Lorentz transformation already assumes that the other frame has a speed of v in your frame). If you're not familiar with the Lorentz transformation I could show this as well if you like.

4. Jun 6, 2009

Re: Confusion

Thanks a lot I have to give it some thought. I am fairly familiar with the Lorenz Transformation there are just these little problems I have with reality :)

5. Jun 7, 2009

### DrGreg

Re: Confusion

3735928559,

The equations you quoted,

$$\mathrm{d}t' = \frac{\mathrm{d}t}{\gamma}$$.............(1)
$$\mathrm{d}x' = \gamma \mathrm{d}x$$..........(2)​

are valid only in special cases. As you say you know about the Lorentz transform, the equations that are always valid are

$$\mathrm{d}t' = \gamma (\mathrm{d}t - \frac{v}{c^2}\mathrm{d}x)$$...........(3)
$$\mathrm{d}x' = \gamma (\mathrm{d}x - v \mathrm{d}t)$$............(4)​

which can be inverted to give

$$\mathrm{d}t = \gamma (\mathrm{d}t' + \frac{v}{c^2}\mathrm{d}x')$$...........(5)
$$\mathrm{d}x = \gamma (\mathrm{d}x' + v \mathrm{d}t')$$............(6)​

These equations apply to two observers O and O' each measuring the same object P.

From (5), (1) is valid only in the special case when dx' = 0, i.e. you are measuring events that are at the same place according to O'. This includes the case P=O', when O is measuring O'.

From (4), (2) is valid only in the special case when dt = 0, i.e. you are measuring events that are simultaneous according to O. In your scenario, this is never true.