1. Dec 18, 2015

### Rahul Mohan P

Hi All;
I was trying to understand Lorentz Transformation equation and special theory of relativity, but as I compared the derivation with a thought experiment which I imagined I found the whole Lorentz Transformation Equation fails. The details of the problem is given below. I know I m wrong but I wish you help me to find where I went wrong.

Thought Experiment Details

Let us consider a spacecraft moving with a velocity v m/s, two observers are there, one inside the spacecraft (observer-1) and other watching spacecraft from earth (observer-2). Two light sources (L1, L2) are emitted inside the spacecraft from point ‘P’ and point ‘A’ simultaneously as shown in figure. After one second (time with respect to observer-2) the spacecraft moves ‘x’ m. Observer-1 measures light source L1 moved a distance ‘a’ and light source L2 a distance ‘b’, but observer-2 measures L1 moved a distance ‘a1’ and L2 a distance ‘b1’.

From ΔABC Lorentz Transformation Equation can be derived;

Ø x = v*t1....................................................................................(1)

Ø a = c*t.......................................................................................(2)

Ø a1 = c*t1...................................................................................(3)

where;

x is the distance travelled by spacecraft in time t1 (t1 is taken as 1 sec for simplicity)

t1 is time w.r.t observer-2

v is velocity of spacecraft

a is distance travelled by L1 in time t as per observer-1

c is speed of light

a1 is the distance travelled by L1 in time t1 as per observer-2

Applying Pythagoras Theorem to ΔABC;

a12 = a2+ x2

(v*t1)2 = (c*t)2 + (c*t1)2

─► (c*t)2 = (c*t1)2 - (v*t1)2

─►t2 = t12 (1- v2/c2)

t = t1√(1- v2/c2)....................................Lorentz Transformation Equation

Consider ΔPQR;

Since L1 and L2 are emitted at same time the distance travelled by both light will be same after a time interval ‘t’ as measured by observer-1, hence;

a = b

From figure the sides PR = AC

Hence from the ΔABC & ΔPQR it is clear that the distance travelled by L1 & L2 measured by observer-2 in time t1 is not equal.

But the distance travelled by L1 & L2 at time ‘t1’ from observer-2 should be equal since speed of light is constant from any frame of reference;

ie;

a1 = c*t1

b1 = c*t1

Since a1≠b1, the only thing which should vary is the speed of light measured from observer-2’s frame of reference.

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2. Dec 18, 2015

### Samy_A

I'll try again.

I suppose the problem starts here: "Two light sources (L1, L2) are emitted inside the spacecraft from point ‘P’ and point ‘A’ simultaneously as shown in figure." Simultaneous events in one frame of reference are not necessarily simultaneous in another frame of reference.

Last edited: Dec 18, 2015
3. Dec 18, 2015

### Rahul Mohan P

But how ?

4. Dec 18, 2015

### Staff: Mentor

Relative to what? I assume relative to Earth, but you should take care to explicitly specify these things, both for clarity and to make sure you yourself are clear on the problem you are posing.

Simultaneously according to whom? The observer on Earth, or the observer in the spacecraft?

As Samy_A has pointed out, in SR simultaneity is relative: if the emission of light from the two sources is simultaneous for one observer, it won't be for the other. But even if you don't yet understand how that can be, from a logical point of view, you can't simply assume that the emissions are simultaneous according to both observers, because doing so prevents you from even considering the possibility that they're not--and therefore prevents you from seeing how SR might explain why they're not. So for purposes of posing the problem, you need to pick: are the emissions simultaneous according to the observer on Earth, or according to the observer in the spacecraft? In other words, you can only assume one, not both.

In what length of time? You can't assume that observer-1's time is the same as observer-2's time, for the same reason you can't assume that the emissions are simultaneous for both observers. I assume you would pick observer-1's time here, but then you need to make sure you only draw deductions from it that are valid if observer-1's time and observer-2's time are not the same, since you can't assume they are the same.

5. Dec 18, 2015

### Rahul Mohan P

Ok then if observer-1 see the lights emit at the same time, then how will the observer-2 see the lights ? I mean for observer-2 which light is emitted first ?

6. Dec 18, 2015

### Staff: Mentor

Google for "Einstein train simultaneity" and also look for some of the many threads here about relativity of simultaneity. Most of the "paradoxes" of special relativity are the result of overlooking relativity of simultaneity, incorrectly assuming that because two things happened at the same time according to one observer they necessarily happened at the same time for another observer.

I also notice that in your post you said
That is NOT the Lorentz transform equation, it is the time dilation formula derived from the Lorentz transforms and you have to be very careful not to misapply it.

If observer one says that something happened at position $x$ and time $t$ and you want to know the position and time ($x'$ and $t'$) at which it happened according to observer 2, you don't use the time dilation formula, you use:
$t'=\gamma(t-vx)$
$x'=\gamma(x-vt$ where
$\gamma=1/\sqrt{1-v^2}$
These are the Lorentz transforms, and I'm measuring distances in light-seconds and time in seconds so that $c$ comes out equal to one and I don't have to worry about constantly dividing and multiplying it.

7. Dec 18, 2015

### Samy_A

I'll answer as a mathematician: apply the Lorentz transformation.
Let's take the two events: E1=lamp L1 emits light, E2=lamp L2 emits light.
Let's say that these event are simultaneous for the spacecraft, and occur at t=0.
The coordinates of these events in the frame of the spacecraft are: E1=$(0,x_1,y_1,z_1)$, E2=$(0,x_2,y_2,z_2)$.

In the frame of the earth, E1 will occur at $t_1'=\gamma (0-\frac{v_xx_1+v_yy_1+v_zz_1)}{c²}$) and E2 at $t_2'=\gamma (0-\frac{v_xx_2+v_yy_2+v_zz_2)}{c²}$) . You can see that there is no particular reason for the events to be simultaneous in the frame of the earth.
($\gamma$ being the Lorentz factor: $\gamma=\frac{1}{\sqrt{1-\frac{v²}{c²}}}$, c the speed of light.)

8. Dec 18, 2015

### Staff: Mentor

Let's say that the you are observer one inside the spaceship, and the according to you the two flashes are emitted at the same time.
One flash is emitted from point A: $x=0$, $t=0$.
One flash is emitted from point P: $x=-L$, $t=0$ where $L$ is the distance between points P and A according to you.

Then plug these values into the Lorentz transformation (don't forget that according to you, the earth and observer two are moving from right to left so $v$ is negative) to calculate the time and position at which these events happened according to observer two. Whichever $t'$ value comes out smaller.... That's the one that happened first.

9. Dec 18, 2015

### Rahul Mohan P

Ok now understand what went wrong. I should be more careful taking assumptions.Thanks for the replies.