# Not completely getting Lorentz transformations

1. Jul 5, 2010

### calebhoilday

Iv just been reading a physics text book and i feel iv completely missed something. It may help to draw a diagram and to read the thread slowly. Sorry if it is a little thick.

My understanding of Special Relativity is that it allows two seemingly conflicting principles to co-exist, these being the principle of relativity and the principle of invariant light speed.

If one considers the hypothetical where there exists two points, point A and point B, an observer that is stationary relative to these points and an observer that moves from point A to point B at 0.5 C. If a ray of light is emitted from point A to point B then according to classical relativity the relative speed between that ray and the moving observer is 0.5 C. As this is relative speed is not permitted by the principle of invariant light speed. Special relativity states that Lorentz trasformations affect any moving body to ensure that the speed of light is perceived to be C relative to the observer themselves. Lorentz transformations affect the moving observer so that instead of observing 0.5 C as the light ray speed, it is perceived as C. Lorentz trasformations speed up the perceived speed of the light ray by slowing the rate of time relative the the stationary observer and contracting length.

Now here is my problem. If one considers the same situation but adds in a light ray that moves in the opposing direction, that being from point B to point A, the relative speed between this ray and the moving observer is 1.5 C; to maintain that this light ray as C, Lorentz transformations would have to increase the rate of time and increase length of the moving observer, so the light ray is slowed. As these rays require different Lorentz transformations, one of these rays is not going to be perceived to be C; i can not see how it would be possible to apply two different Lorentz transformations. The outcome seems paradoxical.

2. Jul 5, 2010

### Ich

The speed is not only perceived as c, it is c. Lorentz transformations don't affect moving bodies, they transform between equally valid coordinate systems.
No. They transform between coordinate systems in a specific way. The best way is to actually use the transformations, as you learn a whole lot of SR by doing so.
If you insist on using time dilation and legth contraction (these are not the Lorentz transformations!), be sure to add the relativity of simultaneity. Without it, nothing makes sense.

3. Jul 5, 2010

### BruceG

The magic formula for adding velocities (following from Lorentz transformation) is

s = (v + u) / (1 + vu/c^2)

In the case that v = c (light)

s = (c + u) / (1 + u/c^2) = c [multiply top and bottom by c]

If we reverse the direction of the observer of light ( u -> -u) we get

s = (c - u) / (1 - u/c^2) = c again

Does this help at all?

4. Jul 6, 2010

### Rasalhague

Probably your confusion comes from the perfectly natural assumption that when two events are simultaneous in one spacetime reference frame (=coordinate system), they ought to be simultaneous in all reference frames. This turns out not to be the case!

Have a look at the attached diagram. The white lines are the axes of one reference frame (vertical is time, horizontal one dimension of space), the red lines the axes of another reference frame, moving from left to right along the space axis of the first reference frame at some velocity relative to the first. The time axis of the red reference frame is tilted slightly relative to the time axis of the white. Likewise the space axis of the red is tilted slightly from the space axis of the white. The time axis of the red reference frame is the trajectory (also called, in the jargon, the "world line") of an object at rest in that frame. Any other object at rest in the red frame will follow a trajectory parallel to the red time axis. Similarly any object at rest in the white reference frame has a trajectory (world line) parallel to the vertical white time axis.

The blue lines at 45 degrees from the white axes represent the paths of light rays. They're at the same angle as each other from the red axes. It's impossible to represent this as 45 degrees too, but this is just an artefact of the diagram. The sameness of the angles represents the fact that light travels at the same speed in both frames.

The space axis of either frame is a line of simultaneity, a snapshot of the universe at one instant: all events on the white space axis happen at the same time as each other according to the white frame. All events on the red space axis happen at the same time as each other according to the red frame. Notice that--very counterintuitively--the red and white frames disagree about which sets of events are simultaneous to each other. I've drawn in a few more red lines parallel to the red space axis. I've drawn a white line parallel to the white space axis. All events on any given red line happen at the same time as each other, in the red frame. All events on that other white line happen at the same time as each other.

In particular, if rays of light emitted from the mutual origin of these reference frames are measured by the white frame as arriving at two equidistant locations in space at the same time (the upper corners of that triangle), those two events won't happen at the same time as each other when measured by the red frame. According to the red frame, which is moving to the right with respect to the white frame, the light ray that's going right arrives at the right corner sooner, before the ray that's going left arrives at the left corner. But of course, thinks someone at rest in the red frame, the light has less far to travel from the spatial origin of the red frame to the right corner, so "natually" it's going to get there sooner. And in the red frame, the light has further to travel to reach the left corner, so it takes longer, providing it travels at the same constant speed.

It's bizarre, but consistent (with logic and experiment). Without taking the relativity of simultaneity into acccount, you do end up with paradoxes such as the one you describe.

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5. Jul 6, 2010

### Staff: Mentor

I agree. Here's a simple setup based on Caleb's post, for Caleb to work through (or anyone else who's interested). To simplify the math, I use light-years (ly) for distance and years (y) for time. This makes c = 1 ly/y, and the Lorentz transformation equations become:

$$x^\prime = \gamma (x - vt)$$

$$t^\prime = \gamma (t - vx)$$

where

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

Let there be two spaceships A and B at rest in frame S, located at x = 0 and x = 10 ly respectively. They carry clocks that are synchronized in frame S. At t = 0 they fire light pulses towards each other. In frame S, each pulse arrives at the other ship at t = 10 y.

So we have four events with the following coordinates in frame S.

1. Ship A emits pulse A at x = 0, t = 0.
2. Ship B emits pulse B at x = 10 ly, t = 0.
3. Ship A receives pulse B at x = 0, t = 10 y.
4. Ship B receives pulse A at x = 10 ly, t = 10 y.

In frame S, we can calculate the velocity of pulse A from the event coordinates as follows:

$$\frac {\Delta x}{\Delta t} = \frac {x_{reception} - x_{emission}}{t_{reception} - t_{emission}} = \frac {10 - 0}{10 - 0} = 1$$

which is as it should be. Using the same method, we calculate the velocity of pulse B as -1, which is negative because it's moving in the opposite direction.

Let there be a third spaceship, C, traveling from A to B at a speed of v = 0.6 ly/y. It is at rest at x' = 0 in frame S', which has velocity v relative to frame S. C passes A at t = 0 in frame S, at which time it sets its clock to t' = 0. So x' = 0 and t' = 0 correspond to x = 0 and t = 0 and we can use the standard Lorentz transformation equations.

Exercise: Use the Lorentz transformation to calculate x' and t' for each of the four events listed above. Then calculate the velocity of each pulse in frame S', as $\Delta x^\prime / \Delta t^\prime$.

6. Jul 6, 2010

### calebhoilday

Thank you all for your replies,

It's going to take me a while to digest your posts and hopefully i will get it once i have. Im just going to post math i did, as it's different to the velocity addition formula BruceG has used. If you could look at it and find the particular flaw in my logic, it may help me not misinterpret what you have written.

The Formula

As my issue concerns the velocity of a light ray relative to the moving observer, the formula i have derived is:

M= (S-V)*(1/(1-V/C))

M is the velocity of an object relative to the moving observer
S is the velocity of the object relative to the stationary observer
V is the velocity of the moving observer relative to the stationary observer
C is the speed of light in a vacuum

The (1/(1-V/C)) part of the formula is the simplification of Lorentz transformations that alter the relative speed.

The First Situation

The observer with motion is moving from point A to point B at half the speed of light and a pulse of light is also moving from point A to point B

M= (C-1/2C)*(1/(1-1/2))
= 1/2C*2
= C

Lorentz transformations have maintained the Speed of the pulse of light to be C, relative to the moving observer.

The Second Situation

The observer with motion is moving from point A to point B at half the speed of light and a pulse of light is moving from point B to point A

M= (-C-1/2C)*(1/(1-1/2))
= -1.5C*2
= -3C

Lorentz transformations have increased the speed of the pulse of light to 3C relative to the moving observer.

Thank you once again.

7. Jul 6, 2010

### Staff: Mentor

No, this is incorrect. In your notation, the correct formula is

$$M = \frac {S - V} {1 - SV/c^2}$$

See for example

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

in which his u' is your M, and his u is your S.

I'm sorry, I don't understand what you mean by this.

Last edited: Jul 7, 2010
8. Jul 7, 2010

### Rasalhague

If, for simplicity, we set c = 1, for example by using units of years and light years, we can derive the velocity addition formula from the Lorentz transformation equations like this:

$$S=\frac{dx}{dt}, \enspace M=\frac{dx'}{dt'}$$

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

$$M=\frac{d x'}{d t'}=\frac{\gamma(dx-V dt)}{\gamma(dt-V dx)}=\frac{\frac{dx}{dt}-V}{1-V\frac{dx}{dt}}=\frac{S-V}{1-V S}$$

Notice that the gamma cancels as we divide an interval of space by an interval of time. This is the formula to use if object and "moving observer" are going in the same direction. If they're going in opposite directions, the minus becomes a plus. Try it for S = c = 1.

If you want to switch to other units, just divide the velocity squared term in the denominator by c^2. (When in doubt, just use dimensional analyis to balance the units: saves having to memorise where all those pesky c's go!)

9. Jul 7, 2010

### Rasalhague

Small typo here: the U from the Hyperphysics page has crept into the denominator in place of your S.

10. Jul 7, 2010

### Staff: Mentor

Thanks for catching that. I missed one substitution, even after writing the correct version on a piece of paper which is still right here next to my computer. I've fixed it.

11. Jul 7, 2010

### Rasalhague

No problem. There's always something, isn't there!

12. Jul 7, 2010

### Rasalhague

Here's a 2d Minkowski diagram of your two situations. To fit them into the same diagram, I've supposed the pulses are emitted and received simultaneously with respect to the yellow ("stationary" observer's) frame.

The vertical yellow lines are lines of constant x (same space coordinate) in your "stationary" frame, for example the worldlines (trajectories through spacetime) of the two points in space A and B. Note that points in space are curves in spacetime, in this case straight lines. Points in spacetime are represented by points in this diagram. Points in spacetime are also called events.

The horizontal yellow lines are lines of constant t (same time coordinate) in your "stationary" frame.

The red line labelled R is the worldline of your "moving observer". (R for Red or Rocket.) The red lines parallel to this are lines of constant x' (same space coordinate); your "moving" observer, by definition, stays in the same place in what's called their own rest frame, their position is always denoted by the same value of x', just as the "stationary" observer is always located at the same value of x.

The other red lines, those not parallel to R, are lines of constant t' (same time coordinate) according to the red frame. Events on one of these red lines happen at the same time as each other according to the "moving" observer.

I haven't shown the "moving observer's" worldline all the way up to the event of its intersection with the worldline of the spatial point B, but you can imagine it carrying on all the way; it doesn't really affect the calculation (unless we're considering the practical matter of whether the "moving" observer has to go there to check their instruments or something like that).

Anyway, the blue lines that cross stand for the pulses of light, one from A up to B, one from B up to A. The diagram is slightly wonky, as I just draw it by eye in Photoshop, but it's meant to be to scale for geometric units, such as years and light years, so that the pulses travel one unit of space per unit of time, and their worldlines have a slope of c = 1.

The diagram is from the perspective of the yellow frame, that of the "stationary" observer. To fit the red frame into the same picture, we have to take liberties with Euclidean geometry and show its orthogonal lines as skewed.

It might be hard to see what's going on with all those lines, but the idea is that the worldlines of the light pulses have a slope of 1 in both yellow and red frames. Even in the skewed-looking red frame, where squares look like parallelograms, each light pulse goes one unit along the x' axis for every unit it goes alone the t' axis. The pulse from A to B travels a shorter distance, and does so in less time. The pulse from B to A travels a longer distance, and does so in more time: it sets off earlier and arrives later, due to the relativity of simultaneity.

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13. Jul 7, 2010

### calebhoilday

When i considered qualitatively, that there appears to be an inconstancy i decided to model M based on the assumption that C = (C-V)*x which results in x= C/(C-V)=1/(1-(C/V)).

The reason i modeled M instead of S, is because it seems more likely that S would be known considering that the stationary observer was able to determine V.

If i was to model S my approach would to have been to manipulate M by multiplying by the function X as it has to be transformed to the context of the stationary observer.

The fact that in the velocity addition formula provided, that M and V are being manipulated by dividing it by a function, it leads me to think that my V and the V in the formula are different; that being that the V in the formula is the velocity of the moving observer relative to itself, which will always equal 0. If this is true then this would be consistant with my model that when V=0 that M=-C if S=-C

Then again im no physics professor and im probably wrong.

Thank you for your replies once again. If you could explain how my model assumptions were incorrect, i think i might finally get it.

14. Jul 8, 2010

### Rasalhague

Could you explain how you got this first equation?

I assume C is c = 299 792 458 m/s, and x is the position the light pulse has reached in the rest frame of your "stationary" observer, which is also the rest frame of spatial points A and B; and V, as you've said, is the velocity of the "moving" observer relative to the "stationary" observer.

One problem with this formula is that the units are inconsistent. In conventional units, C is a speed, with units of length divided by time, whereas (C-V)*x has units of length squared divided by time. In geometric units, where c = 1, you'd have a dimensionless quantity on the left equal to a quantity with the dimension length on the right.

For something moving at constant velocity,

$$\Delta x=\frac{\Delta x}{\Delta t} \Delta t$$

and when that something is a pulse of light in a vacuum, in spacetime with no significant gravitational curvature, $\Delta x / \Delta t$ = c. In geometric units, c = 1, so $\Delta x = \Delta t$. The triangular capital delta symbols denotes a difference in x (an interval of x, some amount of x), a difference in t. In curved spacetime, we'd have to replace them with the derivative, dx/dt, the instantaneous velocity, which is the limit of this ratio as delta t approaches zero. Finding the correct formula for transforming this velocity from one inertial reference frame to another is then just a matter of individually transforming the space interval in the numerator and the time interval in the denominator according to the usual Lorentz transformation equations that jtbell gave.

There's nothing wrong with your basic plan: to derive M from S. But yes, you could do it the other way around. Which calculation you'd have make would depend on the question being asked. A simple way to measure the speed of light in both directions in the "moving" observer's rest frame (red) would be to test how long it takes to reach two equidistant spatial points in in that frame (red), rather than A and B (as defined by yellow). In that case, the situation would be exactly equivalent to this one: you'd essentially just have changed the names of the observers. Nothing wrong with that, as "moving" and "stationary" are relative and purely arbitrary names. Alternatively, you could start with the knowledge of how long the light takes, according to red, to reach points A and B defined by yellow, which are moving in the red frame, and work back from there. That's a little bit more complicated, but once you've accepted that the speed is c in the red frame, this too is really the same problem as your original one.

I used V the way you defined it: the velocity of the "moving" observer relative to the "stationary" observer. The Hyperphysics site jtbell linked to used v for your V, u for your S, and u' for your M. None of these formulas has a variable representing "the velocity of the moving observer relative to itself", which is always equal to 0, as you rightly state. S=-C implies M=-C not only when V=0, but for all possible values of V. In geometric units, where C=1:

$$\frac{S-V}{1-SV}=\frac{-1-V}{1-(-1)V}=\frac{-(1+V)}{1+V}=-1$$

15. Jul 8, 2010

### calebhoilday

My statement on nature of the velocity addition formula was just a quick hunch. After looking at the formula, i can see that my assumption that Lorentz transformations alter the length and time of an observer, to maintain C based alone on on the velocity of the observer are incorrect. This is based on the fact that the manipulator for the velocity subtraction formula is C^2/(C^2-VS); for S is within it. If based on my assumption and using the Lorentz transformations provided (my length contraction formula worked out to be different, due to my manipulator) i arrived at C^2/(C^2-V^2), which doesn't maintain C at all.

Im going to do some more reading and try to get my head around the new perspective i need to take on. If i have any more issues, ill start a new thread.

16. Jul 8, 2010

### calebhoilday

quickly is ((1-(V^2/C^2))^0.5)*((1-(S^2/C^2))^0.5) = 1-(VS/C^2) ?

my formula has a typo, its 1/(1-(V/C)) rather than 1/(1-(C/V))

C = (C-0.5C)*1/(1-(0.5C/C))
= 0.5C*(1/0.5)
= C

17. Jul 8, 2010

### Naty1

In all probability, you have missed a lot more than than you realize.....we all do.

Go back and read the same material in a few weeks or months after hanging out here and reading in the Relativity Forum and you'll probably find a LOT of insights that did not occur to you the first time around. Leonard Susskind says there is a lot of physics we "are not wired for" meaning they do not comport with everyday experience and are tough to get in perspective.....

And then there is quantum mechanics waiting to "fool" you..........

18. Jul 8, 2010

### Dickfore

what happens when $u = -c$?

19. Jul 9, 2010

### Rasalhague

In BruceG's way of labelling the variables, u is the velocity of the new frame with respect to the old, and v is the velocity of something else wrt to old frame, in this case a pulse of light. There's no such things as an inertial reference frame with a speed of c relative to another inertial reference frame, so the situation of u = c or u = -c doesn't arise. If v = -c, then s = -c, where s is the velocity of light wrt the new frame.

20. Jul 9, 2010

### Dickfore

How does the formula:

$$V = \frac{v + u}{1 + \frac{v \, u}{c^{2}}}$$

tell me about the restrictions on $u$?