# Who will see the light first O or O’?

1. Dec 24, 2013

### powermind

Hi,
Suppose O’ is moving forward with respect to O at speed v. A light behind him is switched on at t’ and x’ according to him (O’). x' is less than zero.

By using Lorentz transformation, we get that O will see the light before O’. Am I right?

Regards,

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2. Dec 24, 2013

### WannabeNewton

You're doing more "dirty" work and less conceptual work if you're explicitly using Lorentz transformations for this. Draw a space-time diagram using the $t'-x'$ axes of the inertial frame of $O'$ and draw in this diagram the worldlines of $O'$, $O$, and the light signal that originates at the desired time and spatial location relative to $O'$. Where are the intersections between the worldline of the light signal and the worldlines of $O$ and $O'$? Are the intersections time-like separated from one another? Why do we care if they're time-like separated or not?

3. Dec 24, 2013

### ghwellsjr

Depending on when and where the light turned on, there are cases where either one could see the light first or both at the same time in all frames and in other cases it could be frame dependent. If you want to have a single answer to your question, you need to ask a more specific question.

I would also suggest that you keep your scenario along one line, that is make sure that your two observers and the light source are all along the x-axis (with y=0 and z=0).

And if you have already worked out the solution using the Lorentz Transformation so that you think that O will always see the light before O', then it would be immensely helpful for you to show your work so that we can understand specifically where you are coming from and where you are going.

4. Dec 25, 2013

### powermind

Thank you all.

I will explain my point in different way because I found my first post is not clear.

The time in Lorentz transformation equations refers to the time when the event occurred and not when the signal was received by the observer.

When the event occurred it possible occurred in the frame of observer O before the frame of observer O’ who is moving forward along x-axis with respect to O. This could be happen when the event occurred behind of O’.

Here I am confused!

Suppose frame O itself is moving forward with respect to another frame let me name it O2. Then that event should occur in this frame O2 before frame O and O’ because what applied on O’ and O it could apply the same with O an O2. And so on.

This loop means that the event not occurred! But it occurred! What is wrong then?

I hope you understand my point!

Last edited: Dec 25, 2013
5. Dec 25, 2013

### A.T.

The term "before" only makes sense, when comparing times of two events in a single frame. Comparing coordinates between frames is meaningless, because it depends on an arbitrary choice of the coordinate origin.

6. Dec 25, 2013

### Staff: Mentor

You cannot use the word "before" this way. Given two events, A and B, and a coordinate system O you say that A occured before B iff $t_A<t_B$ in O. Given one event, A, and two coordinate systems, O and O', you cannot say that A in O came before A in O' even if $t_A<t'_A$. It simply isn't a word that is used to make comparisons between frames.

7. Dec 26, 2013

### ghwellsjr

Your last post is also not clear. Here are some suggestions:

Use only one frame.

Keep everything in line (y=0 and z=0).

Describe the locations and speeds of each observer with coordinates in that frame. For example, you could say that each observer passes through the origin but one (O) is stationary and the other one (O') is traveling at some speed v.

Provide the coordinates of the light flash. Where and when did it occur?

Ask more questions about what happens when we transform to a different frame.

If you have already worked out some answers, show them to us.

Last edited: Dec 26, 2013
8. Dec 26, 2013

### powermind

1- The frame O’ moves forward with velocity (v = 0.5c m/s^2) along x-axis with respect to the fixed reference frame O which is a stationary.
2- When two frames O and O’ are coincide, t and t’ are set to 0.
3- A light is located in the frame O’ at x’ = -10m, y’ = 0 and z’ = 0;
4- The light is switched on at 10s according to the frame O’.

I hope it is very clear now and nothing is missing.

So, λ = 1.154700538, v = 0.5c m/s, x’ of the light = -10m and (t’ = 10s when the light is switched on).

If the time takes 10s according to the frame O’ after the coincidence, the time according to the frame O will take λt' = 11.54700538s.

Now, when is the light switched on according to the frame O?
The answer is: $t = λ(t' + \frac{vx'}{c^2})$ = 11.54700536s < 11.54700538s which means that the event occurs first in the frame O then it will occur in the frame O'.

Per the equation $t = λt' + λ\frac{vx'}{c^2}$, when x’ < 0, t’ >= 0, t is the time of event and λt' is the actual time or current time according to the frame O, then t < λt'. No need to think too much to explore that!

Let us agree to continue …

Last edited: Dec 26, 2013
9. Dec 26, 2013

### Staff: Mentor

This statement literally doesn't make any sense. It is like asking "which color is the quietest: squishy or sour?".

You cannot assign a temporal ordering for one event in two different frames. Words like "first", "second", "before", "later", and so on simply cannot be applied in this context.

Please see my previous post on this topic.

EDIT: Here is another way to think of why this is a nonsense statement. Consider two reference frames, frame O where t is defined by the standard Gregorian calendar and frame O' where t' is defined by the Hebrew calendar. Pearl Harbor was attacked at t = 1941 in O and at t' = 5702 in O'. Would you therefore say that the attack on Pearl Harbor occured first in O and then in O'? Or since the current time in O is <5702 would you say that Pearl Harbor has not yet even been attacked for Hebrews? Hopefully this helps you understand why your statements above are so problematic. The time coordinates assigned by different frames simply cannot be used to make any temporal ordering statements, particularly not about a single event.

Last edited: Dec 26, 2013
10. Dec 26, 2013

### powermind

I understood your point before, DaleSpam. But I cannot accept it!
1- Can you tell me what is the benefit from the time derived by Lorentz transformation then?
2- Why are the different times t and t' mentioned in the equation if there is no relationship between them?
3- If O' see two events occurred simultaneously, O will not. So there is delay and relationship for sure!

Can you clarify your point deeply? Do you want to say that the comparison between two frames should be determined based on different frame other than O and O', but this frame is unthinkable so no comparison so what about the three questions above???

11. Dec 26, 2013

### Staff: Mentor

That is fine. These concepts are difficult and often need further explaining. However, the proper response is to ask follow-up questions as you have in this post, not continue to post mistakes that have already been pointed out to you. We are trying to provide an educational benefit to you and also to other people who are following the conversation without participating.

The benefit of the Lorentz transform is simply that the laws of physics are invariant under it. This means that we can always solve a problem by Lorentz transforming it to a more convenient frame, solving it there, and transforming it back. We don't have to worry about adapting the laws of physics to the new frame since the same laws apply in both.

There is a relationship between them, it just isn't a temporal ordering relationship.

Certainly, but here you are talking about two different events and examining the temporal ordering of the two events. This is not problematic at all.

The problem is trying to establish a temporal ordering for one event in two different frames. That simply cannot be done, temporal ordering relationships are between pairs of events, not pairs of frames. What you are trying to do is similar to sorting scents by their pitch. It simply is not an applicable ordering for the things being sorted.

No, I am saying that this kind of comparison can only be made between different events. The comparison cannot be made between two frames at all.

Last edited: Dec 26, 2013
12. Dec 27, 2013

### A.T.

So you can compare the $\Delta$t of two events within one frame to their $\Delta$t in another frame.

The equation is the relationship.

Yes, see 1).

Your argument why the event never happens has nothing to do with relativity or even physics. It is a very basic fallacy about coordiante systems that could be paraphrased for simple geometry:

Let P=(1,0) be a point described in the coordinate system S. Let S' be another coordinate system shifted along the X-axis such that P'=(2,0). In system S' the point is further to the right than in system S. Suppose frame S' itself is to the left with respect to another system let me name it S''. Then that point is even further to the right in system S'' than in S and S'' because what applied on S’ and S it could apply the same with S' an S''. And so on. This loop means that the point cannot be located so it doesn't exist! But is exists!

Do you see why this is nonsense?

13. Dec 27, 2013

### ghwellsjr

It's clear enough that I think I can figure out what you want but first let me point out some problems.

1) You should not specify motion to the frame in which you are specifying your scenario. Since you defined the coordinates of the light in frame O', you should make that your single frame.

2) When you specify a speed as you did under your first bullet, it is sufficient to say v=0.5c. The units of m/s^2 has a couple problems. First, you have seconds squared which isn't appropriate for speed (that's what we need for acceleration). Second, since you're using units of seconds, you probably want to use units for distance of light-seconds. This is also the case for your third bullet.

3) You should say that observer O' is stationary in the frame and observer O is traveling at -0.5c and that they both passed through the origin.

More problems here. You are using the wrong symbol for gamma. You are using Lambda. The symbol you want is γ.

Instead of x' = -10m you should use units of light-seconds and it's obvious that's what you were really doing.

Here's a spacetime diagram depicting everything as I think you want it:

Observer O' is shown in blue and observer O in green. The light is shown in red. Note that O sees the light before O' sees it. Note the times according to their own clocks (the dots represent 1-second intervals of time). O sees it at his own time of about 11.6 seconds and O' sees it at his own time of 20 seconds.

This isn't correct. You neglected to use the x' value of -10 light-seconds in your calculation. Here is the original spacetime diagram transformed to a speed of -0.5c, making it the rest frame of observer O:

As you can see the time the light turns on in the O frame is around 5.8 seconds and no where near 11 seconds.

Again, it appears that you used a value of x'=0 in your calculation so it is wrong.

Let us wait until you get everything right.

But, in the mean time, you should notice that the answer to your question of which observer will see the light first is the same in all frames. In both of the above frames, Observer O' sees the light at his time of 20 seconds and observer O sees it at his time of about 11.6 seconds.

Here is another frame where both observers are traveling at the same speed in opposite directions:

And another frame moving at -0.3c with respect to the original frame:

Note that each observer sees the light according to his own clock the same in all frames. Note that the light propagates at c in all frames. Note that the Coordinate Times and Distances are different in each frame.

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14. Dec 28, 2013

### powermind

ghwellsjr, I would like to thank you for your effort. It is really appreciated.

I updated my first question in my post #4. So, no need to get when the signal was received by the observers since the time in Lorentz transformation refers to the time when the event occurred and not when the signal was received by the observer.

I neglected to use the x' value of -10 light-seconds because the case does not need it. In details, the time spent on frame O’ is equivalent to the same time multiple by γ. So, when O’ spent 10s, the O spent 10γ second.

Anyway, your last post proves that I am right. The observer O will see the light before O’ sees it.

Regards,

15. Dec 28, 2013

### ghwellsjr

OK, so when you say "see", you don't mean when the light reaches an observer, you mean when the event of the light flash occurs according to an observer's frame. Then we only need to look at the first two diagrams that I provided you in post #13.

The first diagram is the rest frame of O' and the event (shown as a red dot) occurs at t'=10 (because you defined the problem that way) and you agree with this:

The second diagram is the rest frame of O and the event (still shown as a red dot) occurs at t=5.8s but you claim it occurs at 10γ which is 11.547s:

Since you "neglected to use the x' value of -10 light-seconds" then you didn't transform the event shown by the red dot. Instead you are transforming the event shown by the black dot in this diagram:

And it transforms to the rest frame for O as shown by the black dot in this diagram:

The black dot does have a Time Coordinate of t = 10γ = 11.547 seconds, but I don't see why you think this event fulfills your requirement.

I also don't see why if the time for O is 11.547 s and the time for O' is 10 s, you conclude that "observer O will see the light before O’ sees it". Isn't that backwards?

But even if it weren't backwards, I don't understand why you think you can neglect to use the x'-coordinate. Where'd you get that idea?

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16. Dec 31, 2013

### powermind

I will explain what my problem is.
Same example, let the event occur at t’ = 5s instead of 10s according to O'. If we apply the Lorentz transformation, the event will occur at t = 0s and x = -8.66 light-seconds according to O.

Now,
1- When the observer O and O’ are coincide, they reset their clock to zero.
2- So, the time (t) derived by Lorentz transformation should be read from the watch of the observer O after the reset. So how t=0s and may be less than zero although the event occurred after the reset of the clock of both observers!
What do these values mean or represent?

If two events occurred in frame O’, the Lorentz transformation can derive the time and location of each event according to O. what do the values represent?

See what I thought in more details:

Suppose two events A and B occur at the same time according to O’ after coinciding at t=t’=0. Let x’_a = -10 light-seconds, x’_b = 10 light-seconds, v = 0.5c. Then at t’=11s I will get t_a = 6.9s, x_a = -5.2 light-seconds, t_b = 18.48s and x_b = 18 light-seconds.

I thought that the observer O is reading clock from his watch starting from t=0s. When his clock points to 6.9s, the event A will occur and the distance between him and A is -5.2 light-seconds. Then he waits few seconds until his clock points to 18.48s. At this time, the event B will occur and the distance between him and B will be 18 light-seconds. Is this correct? Or am I wrong?

Last edited: Dec 31, 2013
17. Dec 31, 2013

### Staff: Mentor

OK, let's call the event A. The "four-vector" spacetime coordinates of A in O' are $(t',x',y',z')=(5,-10,0,0)$.

Yes, the spacetime coordinates of A in O are $(t,x,y,z)=(0,-8.66,0,0)$.

The Lorentz transform already assumes that origins of O and O' coincide at the same event in space and time. There is no need for a "reset", it is already included in the Lorentz transform you did above.

There is no reset, so I am not sure what you are talking about.

Coordinates don't necessarily represent anything. They are just arbitrary labels assigned to events in any of an infinite multitude of ways.

The inertial coordinates are a little special as they represent distances and times measured by a hypothetical lattice of inertially co-moving rods and synchronized clocks. However, physics doesn't constrain you to use such coordinates, and you can use coordinates that represent angles (polar, spherical, etc.) other such non-standard labels. You can even use coordinates that don't have a specific coordinate for time, such as null coordinates. It is not useful or important in general to assign meaning to the coordinates themselves.

18. Dec 31, 2013

### powermind

Look any example in the internet, it should mention that two observers set their clock to zero. So the initial time is zero!
How do you say there is no reset?

19. Dec 31, 2013

### Staff: Mentor

Sometimes people repeat the same thing more than once in order to emphasize a point. They may even use different words to repeat the same information adding emphasis on something they wish to communicate.

That said, do you have a particular internet link that is confusing you?

20. Jan 1, 2014

### powermind

Thanks DaleSpam,
I am here to learn. I have tried to find the answer from several sites before writing my question here but I failed.

My question is:
You said this is wrong. Then you said:
You increase my confusion! in Galilean transformation, everything is logical and clear and its coordinates represent the expected in that time. But when it is replaced by the Lorentz transformation, it needs more clarification.

Regards,

Last edited: Jan 1, 2014