Lorentz Transformations Question

1. Aug 13, 2010

aguycalledwil

I've spent a large portion of my day trying to figure this out and I figured my best answer is likely to come from here. Forgive me if I'm wildly wrong about anything, I'm somewhat basic with physics, largely due to the fact that I'm 15 and my maths is limited to a GCSE level.

My dilemma is this. It is to my understanding that I can use the Lorentz factor to determine the magnitude of spacial contraction or time dilation. For example, if I wanted to find out how many of 'my' seconds a clock took to tick whilst hurtling along at half the speed of light, I could just use the Lorentz Factor as follows:

t'=1/√1-(1.5*10^8)/(3*10^8)

So then what are the full Lorentz transformations used for? If I can calculate the extent of a time dilation or a length contraction with the Lorentz Factor, what are the full transformations used to calculate, and how do I use them?

As aforementioned, I am very unexperienced and it could be that everything I have written is wrong. :P

2. Aug 13, 2010

zhermes

I'm not entirely clear on what you're looking for, but here's a general reply:

The lorentz transformations are used to translate between two reference frames, one 'at rest' and the other with some velocity. This is important for predicting what a 'moving' observer would see or measure in any situation in which they have a velocity comparable to that of light.

Does that help?

3. Aug 13, 2010

aguycalledwil

Well that's what I thought, but could I not just use the Lorentz factor in the way described above to reach this value? I read Einstein's 'relativity' and it lists the Lorentz transformations but then goes on to use the Lorentz factor to describe the time dilation of a train.

4. Aug 13, 2010

Staff: Mentor

The time dilation and length contraction formulas apply only in very specific cases (e.g. the events must be co-located in one frame for the time dilation formula). The Lorentz transform is the general equation and must be used in any case where the requirements are not satisfied for the more specialized formulas.

Personally, I would prefer that the length contraction and time dilation formulas never be taught as such. The source of most "paradoxes" stems directly from novice students misusing them when they should be using the Lorentz transform instead. The number of times that they save any effort is vastly outweighed by the number of times that they cause problems. For myself, I always use the transform and never the simplified expressions.

5. Aug 13, 2010

bcrowell

Staff Emeritus
This boils down to a question of the meaning of the off-diagonal terms. That is, the Lorentz transformations have the form
$$t'=At+Bx$$
$$x'=Ct+Dx$$ ,
where A and D are interpreted as the time dilation and length contraction. Why do we need B and C? C is present even in Newtonian mechanics, and it's simply what makes the frames move relative to one another; it shifts the origin of one relative to the origin of the other with the passing of time. B describes the fact that simultaneity is observer-dependent. Without B, all observers would have to agree that events occurring at t=0 were simultaneous. In Newtonian mechanics, B=0.

6. Aug 13, 2010

Staff: Mentor

To make bcrowell's key point in other words:

Length contraction and time dilation do not give a complete picture of relativistic effects. You also have to take into account relativity of simultaneity, which most popular-level or elementary treatments either ignore or gloss over.

Suppose you have two identical clocks at rest with respect to each other, separated by distance $\Delta x_0$ in their rest frame, and synchronized so their time readings always agree in their rest frame. In another reference frame, moving at speed v along the line joining the two clocks:

1. The distance between the (now moving) clocks is smaller: $\Delta x = \Delta x_0 \sqrt{1 - v^2 / c^2}$ (length contraction).

2. The (now moving) clocks tick at a slower rate, i.e. the time between ticks is now $\Delta t = \Delta t_0 / \sqrt{1 - v^2 / c^2}$ (time dilation).

3. The (now moving) clocks are no longer synchronized; their readings differ by the amount $\Delta t = v \Delta x_0 / c^2$ (relativity of simultaneity).

All three of these effects can be derived from the Lorentz transformation. In general, you can solve a problem using either the two Lorentz transformation equations or the trinity of length contraction, time dilation and relativity of simultaneity.

7. Aug 13, 2010

jason12345

As Dalespam has pointed out, the time dilation and Lorentz contraction formulas are special cases. I would go further and say they are the most misused formulas complete novices use in trying to solve problems in relativity. Don't even look at them until you've used the LT equations to get a feel for transforming the time and space values from one frame to another. When you've done that, do it for the special cases:

1. Transform the space time coordinates for 2 events at the same time, but different locations - Lorentz contraction formula.

2. Transform 2 events at the same location, but different times - time dilation formula.

By all means, stick with rudimentary, popular texts on relativity if you want to casually discuss it with friends. To really understand it, buy a book, work through the exercises, and expect to spend years gaining experience.

8. Aug 14, 2010

aguycalledwil

Okay, thanks for the help guys, it's hugely appreciated. I do have two other queries on the matter, and again forgive me if they're elementary. In the time dilation transformation, does the 'x' refer to the distance traveled along the x-axis, or the length of the subject moving (after taking contraction into account)? And secondly, what are the conditions of these 'special cases' wherein using the simple formulas is acceptable?

9. Aug 14, 2010

Staff: Mentor

The time dilation formula only applies for a clock which is at rest in an inertial frame. The length contraction applies only when both ends of a rod are at rest in an inertial frame and the measurements of the positions of the ends are made simultaneously in each frame.

10. Aug 17, 2010

DaTario

Expressing in more direct terms, length contraction applied to a bar allows you to compare its lengths as measured by two observers, but says nothing about where are the ends of this bar in each observer's frame of reference. To do this we use LT's.

Best wishes

DaTario

11. Aug 18, 2010

Dickfore

Best derivation for time dilation formula I've seen is through the use of the light clock.

It is depicted in the following picture:

A flash shines lines and starts a counter at the same time. The light ray reflects off of a miror and is detected in a detector. Neglecting the horizontal distance between the emission and reception point and assuming the distance to the mirror is $L_{0}$, we deduce that the light signal travels a total distance $2 L_{0}$ during this round trip. Because the speed of light is equal to $c$, the time interval that between emission and reception is:

$$\Delta t_{0} = \frac{2 L_{0}}{c}$$

Next, consider the same situation according to an observer with respect to whom the clock is moving with a velocity $v$ in the direction perpendicular to the axis of the clock (the direction FD - M). The three important instants of the new situation are depicted on the following picture:

There are 2 important things:

1) Distances in the direction perpendicular to the relative velocity do not change (this can be proven by a thought experiment involving a solid body with a piece cut from it and by considering what happens when the cut out piece is moving relative to the rest of the body). Therefore the distance from the flash/detector (F/D) to the mirror (M) is still $L_{0}$.

2) Once emitted, a light pulse travels independently of the source as a spherical front with the same speed $c$ (in vacuum) according to all observers.

The events of emission, reflection and detection at a particular physical spot on the clock, however, are an objective reality independent of what frame of reference we use to describe the situation. Thus the signal must be emitted at the (instantaneous) position of F, reflected at M and detected at D. That is why the light signal's path is actually the dashed inverted V-line on the picture. The times $t_{1}$ and $t_{2}$ indicate the times necessary for light to travel from F to M and from M to D, respectively.

The displacements of the clock in the horizontal direction are also denoted. Using Pythagoras' theorem, we have:

$$\left\{\begin{array}{l} (c \, t_{1})^{2} = L^{2}_{0} + (v \, t_{1})^{2} \\ (c \, t_{2})^{2} = L^{2}_{0} + (v \, t_{2})^{2} \end{array}\right. \Rightarrow t_{1} = t_{2} = \frac{L_{0}}{\sqrt{c^{2} - v^{2}}}$$

The total round-trip time for the signal is:

$$\Delta t = t_{1} + t_{2} = \frac{2 L_{0}}{\sqrt{c^{2} - v^{2}}}$$

Expressing $2 L_{0} = c \, \Delta t_{0}$ from the discussion of the first situation and doing some algebra, one arrives at:

$$\Delta t = \frac{\Delta t_{0}}{\sqrt{1 - v^{2}/c^{2}}}$$

This is the time dilatation formula. It shows that the same clock shows different time intervals relative to different observers (reference frames). The shortest time interval is measured by the observer relative to which the clock is stationary. This time interval is called proper time and, according to our discussion it is actually $\Delta t_{0}$. According to all other observers, the time interval between the same events is longer (dilated) because:

$$1 - \frac{v^{2}}{c^{2}} < 1 \Rightarrow \gamma \equiv \frac{1}{\sqrt{1 - v^{2}/c^{2}}} > 1$$

Similarly, you can deduce the formula for length contraction. Just impart a velocity to the light clock in the direction of its axis and consider the situation according to observers where the clock is stationary and where it is moving with velocity v.

You can also derive the velocity addition formulas by using a modified apparatus where a stream of particles is used in one stage of the round trip and a light is flashed as soon as a particle is detected on the return trip. The details of these derivations may be found in University level textbooks.