# Question from I. KINEMATICAL PART § 1. Definition of Simultaneity

1. Feb 9, 2005

### arrell

Suppose we have two frames (frame 1 and frame 2) which are passing each other infinitesimally close at a relative velocity v with respect to x axes in both frames..
We have two points, a and b, in frame 1 with separation L along the x axis.
We have two other points, c and d also in frame 1 where point c is infinitesimally above (y axis) point a and point d is infinitesimally above point b (so c and d are also with separation L along the x axis).

We fire lasers simultaneously from a, b, c, and d (all in frame 1) orthogonally to vector v towards frame 2 .
In frame two, there is a fully reflective surface opposite to point a and b
and a laser detecting sensitive surface opposite to c and d.

Let g and h be the points where the laser beams (which were fired from c and d) strike and are detected in frame 2.

Let e and f be the points in frame 1 where the laser beams (which were fired from a and b) strike in frame 1 after being reflected from frame 2.

Questions:
What is the separation in frame 1 of e and f in terms of L?
What is the separation in frame 2 of points g and h in terms of L (with respect to a coordinate system in frame 2 that is congruent to that of frame 1 when the two frames are at rest)?

2. Feb 9, 2005

### JesseM

When you say "towards frame 2", what does that mean exactly? Frames don't have a particular location in space, they're more like coordinate systems that assign coordinates to every point in space. Do you mean "towards the origin of frame 2"? If so, where are the origins of frame 1 and 2 located at the moment the lasers are fired?

3. Feb 9, 2005

### arrell

For purposes of this question, we are considering a plane surface that is in frame 2 (ie moving at the same v as frame 2) and is parallel to the axis of the points in frame 1 and infinitesimally close to the axis of the points in frame 1. Also at the time of the lasers firing, the plane surface extends in
the +x and -x directions sufficiently that the laser pulses all strike somewhere on the plane. I should have also made clear that for purposes of the question, the lasers are emitting an infinitesimally small pulse (ie a photon).

4. Feb 9, 2005

### JesseM

If the surface is infinitesimally close to the point where the photons from the lasers are emitted, then they will be detected and reflected infinitesimally close to the point where they were emitted.

5. Feb 10, 2005

### arrell

Do you mean that when the distance from e to f in frame 1 is measured by a frame 1 measuring stick that the separation = L ?

Do you mean that when the distance from g to h in frame 2 is measure by a frame 2 measuring stick that the separation = L ?

6. Feb 10, 2005

### JesseM

If the distance from g to h is L in frame 1 where they are at rest, it must be $$\sqrt{1 - v^2/c^2}L$$ in frame 2 where they are moving with velocity v.

7. Feb 12, 2005

### reilly

arrel -- You have posited the basic train expt. of Einstein, but in a somewhat more complex arrangement. So, what's to worry? The solution to your problem is pretty much explained in most basic books on relativity. Read, and you will find the answers to your problem. It's all about the LT.
Regards,
Reilly Atkinson

8. Feb 15, 2005

### Physicsguru

Reilly, why do you place so much faith in the Lorentz transformations. Are you positive that simultaneity is relative?

Regards,

Guru

PS: One other thing, what is the speed of a charge density wave relative to the center of inertia of that which supports it? Greater or less than c.

9. Feb 15, 2005

### JesseM

What does that question mean? In a philosophical sense, it's possible there could be some single "true" definition of simultaneity, relativity just says there's no physical experiment we can do that will pick out a preferred frame--this is equivalent to saying that all the fundamental laws of physics will turn out to be Lorentz-invariant. And besides Maxwell's laws, all the new laws of physics which have been discovered in the twentieth century have either been Lorentz-invariant, or are understood to be approximations to a more fundamantal Lorentz-invariant theory (like how nonrelativistic QM is understood as an approximation to relativistic quantum theories).

10. Feb 15, 2005

### Physicsguru

The meaning of simultaneity is physical, not philosophical. Maybe the meaning of the word 'simultaneity' is philosophical, but the concept at issue here is physical.

Next, the laws of physics must be formulated so as to be invariant under a Galilean transformation, if simultaneity is absolute.

Take Newton's second law for example:

$$\vec F = m \vec a$$

Let S denote an inertial reference frame. Let the statement $$\vec F = m \vec a$$ be necessarily true in this frame.

Now, let the origin of frame S be moving in reference frame S. We wish to inquire as to whether or not the second law of Newton is also true in frame S.

Suppose that the position vector of a particle in frame S is given by:

$$\vec r = x \hat i+y \hat j+z \hat k$$

and that the position vector of the particle in frame S is given by:

$$\vec r^\prime$$

Let the position vector of the origin of S in frame S be given by:

$$\vec R$$

The following statement about the vector triangle in S is true at moment in time t in frame S:

$$\vec R + \vec r^\prime = \vec r$$

For a Galilean transformation we have to have t=t, therefore dt=dt hence:

$$\vec V + \vec v^\prime = \vec v$$

Where V is the velocity of the origin of frame S in frame S, and v is the velocity of the particle in frame S, and v is the velocity of the particle in frame S.

Now, suppose that at moment in time t=0, force F is exerted on the particle of inertia m, so that in frame S we have:

$$\vec F = m \frac{dv}{dt}$$

The external force acting upon the particle is not a frame dependent quantity. So let F denote the force acting upon the particle in the primed frame. The condition for S to be an inertial reference frame is:

F-F=0

So that we have:

$$\vec F - \vec F^\prime = m \frac{dv}{dt} - m \frac{dv^\prime}{dt}$$

$$\vec F - \vec F^\prime = m \frac{d}{dt}( \vec v - \vec v^\prime)$$

Therefore:

$$\vec F - \vec F^\prime = m \frac{d\vec V}{dt}$$

So the condition for S` to be inertial is for that V above be constant as measured in frame S.

Regards

Guru

Last edited: Feb 15, 2005
11. Feb 15, 2005

### JesseM

OK, so do you agree that all of the most fundamental laws discovered in the twentieth century were invariant under the Lorentz transformation but not under the Galilei transformation? So if the real fundamental laws were Galilei-invariant that would require a pretty radical revamping of all of modern physics, yet the new laws would have to reproduce all the predictions made by the old laws which have been verified experimentally to a high degree of accuracy. The unlikeliness of this is why most physicists "place so much faith in the Lorentz transformations".

12. Feb 15, 2005

### Physicsguru

Welp they're wrong about simultaneity, hence the fundamental laws of physics should not be made to be invariant under a Lorentz transformation, so yeah some re-vamping needs to be done. I don't think the laws of physics care whether or not we know them.

Regards,

Guru

13. Feb 15, 2005

### JesseM

We don't "make" the laws of physics, we discover them. Physicists may use Lorentz-invariance as a criterion when postulating new equations, but it's experiment that decides which postulated equations describe nature and which don't. And Lorentz-invariant theories are the ones that have consistenly been favored by experiment in the twentieth century--do you deny this?

14. Feb 22, 2005

### arrell

Just a simple answer to my questions would be of real help to me and I would be very appreciative.

I'm just simply trying to follow Einstein's paper line by line. My 'posit' was to try to find out if what I thought had to be a basis for one step in his derivation is what I think it is.

If the length in both cases (e to f and g to h) is not L, I'm in a whole lot of trouble early on at almost the start of the derivation he's making in
I. KINEMATICAL PART § 2. On the Relativity of Lengths and Times

In terms of the physical world, e to f in frame 1 seems like it has to be L.

If g to h in frame 2 were to be:
$$\sqrt{1 - v^2/c^2}L$$

then when the photons are reflected from frame 2 they are now originating in frame 2 and symmetry would seem to indicate that e to f in frame 1 would be:
$$\sqrt{1 - v^2/c^2}(\sqrt{1 - v^2/c^2}L)$$ which isn't exactly the same as L. As I say, I get into a lot of trouble about there if g to h is not equal to L.

You really do have to carefully follow the experiment and the points exactly as I layed it out to see what I am trying to figure out.

15. Feb 23, 2005

### JesseM

Why would symmetry indicate that? If an object at rest in one frame has length X in that frame, its length in another frame moving at v relative to the first one will be $$X\sqrt{1 - v^2/c^2}$$. But if an object moving in one frame has length X in that frame, its length in the second frame will not be $$X\sqrt{1 - v^2/c^2}$$.
Well, as I said before, your language is confusing, since photons don't "originate" in one frame vs. another, I guess what you're saying is that the source of the photons is at rest in a given frame. You also have a profusion of points which are infinitesimally close to each other--if a and c and g and e are all infinitesimally close to one another, what is the point of giving them different labels, for example? Both frames will assigning a and c and g and e the same coordinates anyway.

Could you explain what it is you're trying to figure out, exactly?

Last edited: Feb 23, 2005
16. Mar 2, 2005

### arrell

My understanding is that
frame 2 is moving at v with respect to frame 1 (and points stationary in frame 1)
and symmetrically
frame 1 is moving at v with respect to frame 2 (and points stationary in frame 2)

Photons have to emit from somewhere. In this case, I chose that photons emit originally from points a, b, c, and d which are stationary in frame 1.

Only
a and c
b and d
are infinitesimally close.
While
a and b
c and d
are separated by significant distance L.

When the photons originally from a and b are reflected from frame 2, the points at which they are reflected from frame 2 are stationary in frame 2
so now
the photons travel back towards frame 1 which is moving at v with respect to the reflection points and strike at points e and f in frame 1.

Last edited: Mar 2, 2005
17. Mar 2, 2005

### JesseM

OK, I think I see the problem. It's true that if the two points on the source which shoot the photons have a separation of L in frame 1, then they will have a separation of $$\sqrt{1 - v^2/c^2}L$$ in frame 2. But what you have to realize is that this does not mean frame 2 will see events c and d (or g and h) happen at a distance of $$\sqrt{1 - v^2/c^2}L$$ apart; the reason is that if events c and d (or g and h) happen at the same time in frame 1, then they happen at different times in frame 2, so in the time between them the source will have moved somewhat in frame 2. If g happens at coordinates x=0, t=0 in frame 1, and h happens at coordinates x=L, t=0 in frame 1, then using the Lorentz transform we can show that in frame 2, g happens at x'=0, t'=0 while h happens at x'=$$\frac{L}{\sqrt{1 - v^2/c^2}}$$, t'=$$\frac{-vL}{c^2 \sqrt{1 - v^2/c^2}}$$.

Note that in frame 2, at time t'=0 the first laser gun was at position x'=0 and the second laser gun was at position x'=$$\sqrt{1 - v^2/c^2}L$$, so since both guns move a distance vt' in any time interval t', at the earlier time t'=$$\frac{-vL}{c^2 \sqrt{1 - v^2/c^2}}$$ both laser guns must have been a distance $$\frac{v^2 L}{c^2 \sqrt{1 - v^2/c^2}}$$ to the right, meaning the first laser gun was at position x'=$$\frac{v^2 L}{c^2 \sqrt{1 - v^2/c^2}}$$ and the second was at position x'=$$\sqrt{1 - v^2/c^2}L \, + \, \frac{v^2 L}{c^2 \sqrt{1 - v^2/c^2}} = L(\frac{1 - v^2/c^2}{\sqrt{1 - v^2/c^2}} \, + \, \frac{v^2/c^2}{\sqrt{1 - v^2/c^2}}) = \frac{L}{\sqrt{1 - v^2/c^2}}$$, just where h must have happened in frame 2 according to the Lorentz transformation.
I understand that, I'm just saying it initially confused me when you said that the photons "originate from frame 1" when what you really meant was that they originate from sources which are stationary in frame 1; this is not a typical shorthand used by physicists.
Yes, I understand that--but I am correct that a, c, g, and e are all infinitesimally close, right? Likewise, aren't b, d, f, and h infinitesimally close? If so, you have eight different labels but only two distinct points in spacetime. What do all these points add to the thought-experiment? Why not just say something like "one photon is emitted at a, another is emitted at b at the same time in their rest frame, the distance between a and b is L in their rest frame, what is the distance in another frame moving at v relative to their rest frame?"
Again, you're talking about "frames" like they're things which can emit and reflect photons, and which occupy some particular position in space, while the standard meaning of a "frame" is just a coordinate system which fills all of space. I think what you really mean is something like "When the photons originally from a and b are reflected from a mirror at rest in frame 2, the points on the mirror where they are reflected are stationary in frame 2 so now the photons travel back towards the source which is at rest in frame 1 and which is moving at v with respect to the reflection points and strike at points e and f on the source which is at rest in frame 1". But again, I don't see what the mirror-reflection adds to the thought-experiment. For one thing, since you're placing the mirror infinitesimally close to the source, then each photon will be emitted, reflected and reabsorbed by the source at the same spacetime coordinates, regardless of which frame you're using. Second of all, even if the mirror wasn't infinitesimally close so these events happened at different locations, the question of how the mirror is moving is irrelevant--the photons will only make contact with the mirror for a single moment, so it doesn't make any difference whether the mirror is at rest in frame 1, at rest in frame 2, or moving in both frames; this will have no effect whatsoever on the coordinates of the three events as seen in either frame.

Last edited: Mar 2, 2005
18. Mar 4, 2005

### arrell

The distance from the plane of frame 1 to frame 2 (orthogonal to the direction of v) is constant. No matter where a, b, c, and d are located on the x axis of frame 1, when the photons are fired simulaneously towards the plane of frame 2, the photons will travel the same distance and at the same velocity c. If they leave frame 1 simultaneously, there is no way in a physical world where they can arrive at the plane of frame 2 other than simultaneously or the reflected photons arrive back in frame 1 other than simultaneously. The photons are travelling in the y direction in the framework of a Lorentz transform. There is no preference between them. There are no observers or relational photons involved in the x direction during the travel of the photons in this situation. There are just after the fact 'measurers' who measure the separation of e and f, g and h after those points have all been created.

19. Mar 4, 2005

### JesseM

I think the problem here is that you're unaware of one of the foundational concepts of the theory of relativity, the "relativity of simultaneity" (edit: never mind, I just realized the thread title refers to it so you're obviously aware of it, but maybe you don't understand the reasoning behind it). Relativity of simultaneity says that different reference frames have different definitions of what it means for two spatially separated events to happen simultaneously, so that if one frame says two such events happened at the same time, that guarantees that every other inertial reference frame will say the two events happened at different times. To understand this, you have to understand the physical procedure that each observer would use to synchronize two clocks at different locations which are at rest in his own frame. Einstein's suggestion was that each observer would synchronize his clocks using the assumption that light travels at the same speed in all directions in his own frame; so, if I set off a flash at the midpoint of two clocks, and each clock reads the same time at the moment the light from the flash reaches them, then I define them to be "synchronized" in my frame. But you can see that if different observers use this procedure, then clocks which are synchronized in one observer's frame will be out-of-sync in another's frame--if I see you flying by on a spaceship, and you set off a flash at the midpoint of the ship, then from my point of view the clock at the back of the ship is moving towards the point where the flash occurred while the clock at the front of the ship is moving away from that point, so if I assume the light travels at the same speed in both directions in my own frame, that means in my frame the light from the flash will reach the back clock before it reaches the front clock.

The importance of the relativity of simultaneity in SR can be seen from the fact that the first section of Einstein's 1905 to this topic of clock synchronization:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Last edited: Mar 5, 2005
20. Mar 6, 2005

### arrell

I'm trying to ascertain the answers posed, if possible, with clear constructs in the physical world without resorting to 'observers'. As I said, I would like to use 'measurers' rather than 'observers'. As you so aptly point out, 'measurers' are only meaningful if photon detection points are created simultaneously. (and, yes - I have been over and over the first couple sections of the 1905 paper)

As an aside, I like photons. The emission and detection of photons have to be among the 'pristine' events in the universe. And we are in a fascinating day and age where we can actually create the emission and detection of single photons.

If, in frame 2, the arrival of photons at g and h are not simultaneous, then the arrival at one point has to occur first and the arrival at the other point after that. If there is no intrinsic preferentiallity that can be ascribed to the photons emitted from c and d, what physical law(s) do we use to determine at which point g or h arrival/detection occurs first ??

As I say, I would really like to avoid observers for the moment and just try to deal first with this question of simulataneity from physical laws/properties.
--------------------------------

But, if you want to only consider answers based on observers, then the first thing I would want to consider would be the set of ALL observers in frame 2. I haven't spent time thinking about what all of those observers would 'observe', but if and only if we have to deal with observers, I would be wondering if there is a subset A of all observer(s) that would indeed see the arrival at g and h as simultaneous. And of the observers outside of subset A, would exactly half of them see the detection at g as occuring first and exactly the other half see the detection at h as occuring first ? If this were to be the case, then it would seem as though the only 'neutral' conclusion would be that the detections at g and h are intrinsically simultaneous in frame 2.

Of course, the whole theory in I. KINEMATICAL PART stands on the crux of the postulate: