Question on simultaneous events.

[lky]

Suppose if I am in a ship traveling from points A to B (10 light years apart) at a relavistic speed of say 0.8c.

Then suppose if there is a very strong light bulb at both points A and B, and assuming that the light rays do not get weakened along the way.

Now if I am in the ship moving from A to B at 0.8c and I am in the midpoint of my journey, when I observe the 2 light bulbs turn on simultaneously.

Am I correct to conclude that the 2 light bulbs are indeed turned on simultaneously, as if viewed by a stationary observer, since that the speed of light is constant to all observers, regardless of their motion?

Or would the motion of the ship have any effect on this simultaneity?

Or I am incorrect to assume that the midpoint of my journey means light has to travel the same distance for both the cases of points A and B?

I would appreciate any help to clarify my doubts.

 

[Doc Al]

Originally posted by lky
Am I correct to conclude that the 2 light bulbs are indeed turned on simultaneously, as if viewed by a stationary observer, since that the speed of light is constant to all observers, regardless of their motion?

Everyone agrees that the light from each bulb arrived at the ship at the same time. But they disagree on whether the lights were switched on at the same time.

The fact that the two flashes reach the midpoint at the same time is evidence that they were turned on simultaneously according to observers in the rest-frame of the light bulbs. Observers in the ship will disagree that the lights were turned on at the same time. (Observers in the ship will conclude that light B must have turned on first, since it is moving towards the ship.)

Or would the motion of the ship have any effect on this simultaneity?

Simultaneity is relative to the observer's frame. Observers at rest with the bulbs and those in the ship will disagree on what is simultaneous.

Or I am incorrect to assume that the midpoint of my journey means light has to travel the same distance for both the cases of points A and B?

The light from each bulb is only seen to have traveled the same distance according to the observers at rest with the bulbs. Folks in the ship disagree.

 

[lky]

Just to be sure if I got you correct,

Because if both bulbs were turned on simultaneously, I (on the moving ship) would see the bulb at B turn on first.
Therefore if I am on the moving ship and I saw the 2 bulbs turn on simultaneously, it would mean that the bulb at A was turned on earlier than the bulb at B because I am moving from A to B.

And if C was instead a stationary observer at the mid point, he would see the bulb at A turn on earlier than B.

 

[gnome]

(Observers in the ship will conclude that light B must have turned on first, since it is moving towards the ship.)

This is very confusing to me.

Here's the way I see it:

Let's change the scenario, and say the lights were turned on simultaneously as viewed by an observer O who is at rest wrt points A and B and who is located at the midpoint between A and B, and at that same instant (before either observer sees the lights) observer O' in the spaceship passes directly by the same midpoint.

Edit: Just to clarify: here the lights are turned on at the instant that the spaceship passes the midpoint, so both observers will see the lights at some (different) later time.

In this situation, O' will first see the light from B at time T₁', and sometime later, at time T₂' will see the light from A. He measures the distance to B to be, say, d₁' and concludes that it was turned on at time T₁' - d₁'/c. Similarly, when O' sees the light arriving from A, he measures the distance to A to be d₂' and concludes that that light was turned on at time T₂' - d₂'/c. He then compares those two times he calculated in order to determine which light was switched on first.

But in the situation iky described, O' sees light coming at him from both points A and B, and each point is the same distance (call it d') away, and in each case the speed of the light is the same, c, so wouldn't he conclude that the time it took for the light to arrive at his location from each point was Δt' = d'/c and therefore the lights were turned on simultaneously? In this situation, stationary observer O, measuring the distances to A and B to be d, would be concluding that the time for the light to arrive at the same central location was Δt = d/c so he also concludes that the lights were turned on simultaneously. However, I think that because d' is smaller than d, observer O' concludes that the time that has elapsed since the lights were turned on is less than the elapsed time that would be determined by observer O.

If this is not correct, I'm in trouble. Please straighten me out.

 

[LURCH]

I'm going to disagree with you DocAl, which probably means I'm wrong, but maybe you could show me where I'm wrong. I would have said that if light from A and B meet at Mid-point M when M is occupied by the ship, then the observer on the ship will see the two come on simultaneously. And that an observer stationary relative to A and B (and M) would also agree. He would see that both lights were turned on simultaneously, but the ship was nowhere near M when this occured. In fact this observer would say that the ship was one year out from A when both lights came on, and the light from A took four years to catch up to the ship, arriving at the same moment as light from B reached the ship head-on.

This would be consistant with time-dilation, since to an observer inside the ship, the light from A closed the gap in one year, because the ship was one lightyear out from A at the time the bulb came on. Is this not so?

 

[gnome]

Yes, I think the stationary observer O finds that it took the spaceship (5ly)/(.8c)=6.25 years to travel from A to his position at the midpoint, whereas it took 5 years for the light from A to cover that distance, so as LURCH says, the ship was 1.25*.8 = 1ly from A when the light turned on.

But for observer O' in the spaceship, the trip from A to the midpoint took 6.25*√(1-.8²)=3.75 years, and he measures the distance from A as vΔt'=(.8c)*3.75 years = 3 ly. So he concludes that only 3 years have elapsed since the lights were turned on. Agree?

 

[lky]

LURCH, your explanation sounds plausible, but I'm assuming that the ship is indeed physically at the midpoint between points A and B.

Drawing a rough space-time diagram (see attached file), with really bad symmetry ( the gradient for both light from A and from B should be perpendicular instead).

From my not-so-good-understanding and space-time-diagram, it seems that the 2 light turns on at time t together. And both the ship and the stationary observer would see the light bulbs turn on together.

Or is it that the axis for the space time-daigrams are different for the moving ship and the stationary observer and the axis are instead skewed to each other?

Please help.

 

[Doc Al]

Originally posted by LURCH
I would have said that if light from A and B meet at Mid-point M when M is occupied by the ship, then the observer on the ship will see the two come on simultaneously.

All observers will agree that the light arrives simultaneously at the ship. So, the observer on the ship would see the the two light beams arrive simultaneously. This does not mean that observers would agree that the lights were turned on simultaneously; that is a deduction, not a direct observation.

And that an observer stationary relative to A and B (and M) would also agree.
He would see that both lights were turned on simultaneously, but the ship was nowhere near M when this occured.

See my comments above. No one "sees" the lights turned on simultaneously. But observers at rest with A-B will insist that they were turned on simultaneously. Observers in the ship will not.

 

[gnome]

Please tell me where I am wrong. If the observer on the ship doesn't think both lights were turned on 3.75 years ago, exactly when does he think each one was turned on?

 

[Doc Al]

Much of this discussion hinges on how this statement by lky is interpreted:

Originally posted by lky
Now if I am in the ship moving from A to B at 0.8c and I am in the midpoint of my journey, when I observe the 2 light bulbs turn on simultaneously.

I've been interpreting this to mean: The light from both bulbs arrives at the ship exactly as the ship is at the mid-point between the bulbs. (That's what "I observe the 2 light bulbs turn on simultaneously" means to me.)

Am I wrong about what you meant, iky?

 

[Doc Al]

Originally posted by gnome

But in the situation iky described, O' sees light coming at him from both points A and B, and each point is the same distance (call it d') away, and in each case the speed of the light is the same, c, so wouldn't he conclude that the time it took for the light to arrive at his location from each point was Δt' = d'/c and therefore the lights were turned on simultaneously?

No. The light bulbs are moving. So, when the light that reaches the ship first started out, the light bulbs were not equidistant from the ship.

In this situation, stationary observer O, measuring the distances to A and B to be d, would be concluding that the time for the light to arrive at the same central location was Δt = d/c so he also concludes that the lights were turned on simultaneously.

Right! Observer O will agree that each light beam traveled the same distance in getting to midpoint. So observer O thinks they were emitted simultaneously.

However, I think that because d' is smaller than d, observer O' concludes that the time that has elapsed since the lights were turned on is less than the elapsed time that would be determined by observer O.

Observer O' (ship) disagrees that the light beams were emitted when the bulbs were equidistant from the ship; so he thinks each beam traveled a different distance. Also, observer O' sees all the usual relativistic effects:
1) Moving clocks slow down
2) Moving clocks are no longer synchronized (if they are synched in their own frame)
3) Moving lengths shrink

If this is not correct, I'm in trouble. Please straighten me out.

I think we'll both survive!

 

[Doc Al]

When did the lights go on? Mysteries revealed?

Originally posted by gnome
Please tell me where I am wrong. If the observer on the ship doesn't think both lights were turned on 3.75 years ago, exactly when does he think each one was turned on?

Here's how to think of it.
Pretend that observers in O have three clocks: one at A, one at M (midpoint), and one at B.

According to O, these clocks are synchronized. When they all read T = 0, that's when lights at A & B were turned on. When the light reaches M, O thinks that his clocks all read T = 5 years.

What does O' think? First, he knows that the clocks in O are not in synch: they are way off. O' thinks that clock A is 4 years behind clock M! (Use Lorentz transformations to check this. \Delta T = \frac{vL}{c^2}) Thus, when the light reaches M, O' says that clock M reads T = 5, but clock A only reads T = 1. So O' thinks only 1 year has passed (on the clocks in O) since the light left A. But since the moving clocks are slow, that means that O' says that 1/.6 = 1.66 years have passed.

Similary, O' says that the clock at B is 4 years ahead of the clock at M. So, when the ship is at the midpoint, O' says that clock B reads T = 9. So O' thinks 9 years have passed (on the clocks in O) since the light left B. So O' says that 15 years (of his time) have passed.

So, according to the observers in the ship, the light:
- left A 1.66 years ago
- left B 15 years ago

(Assuming I didn't mess up my transformations; it's been a while )

 

[Eyesaw]

If my head was not simultaneous in time with my body,
I would expect it to fall off when I walk up a
flight of stairs. Relativity of simultaneity...time
dilation...lol.

 

[lky]

Originally posted by Doc Al
Much of this discussion hinges on how this statement by lky is interpreted:


I've been interpreting this to mean: The light from both bulbs arrives at the ship exactly as the ship is at the mid-point between the bulbs. (That's what "I observe the 2 light bulbs turn on simultaneously" means to me.)

Am I wrong about what you meant, iky?

Yup, that's what I meant.

 

[gnome]

Hurrah! I don't want to get too excited about this, but at least I did finally get the same result as you. Thanks Doc.

But I still don't see exactly what you are doing. Here's what I did.

First, I let O' claim that he is stationary & A and B are moving toward the left at .8c. Since the distance from A to B in the A-B frame is 10ly, it is 6ly in the O' frame, and when O' is at the midpoint, it is 3ly to each of A and B in the O' frame. We all seem to agree on that.

Next I let x' = the distance that A and B traveled since the lights went on, and t' = the time elapsed since that time. So, as to the light from B:
ct_B' = 3 + x_B'
And as to B itself:
.8ct_B' = x_B'
Solving those equations gives t_B' = 15 y

Similarly, as to the light from A:
ct_A' = x_A'
And as to A itself:
.8ct_A' = 3 - x_A'
Solving those equations gives t_A' = 1.67y

---------------------------------------------------

Alternatively, using the Lorentz transformations I find in my textbook:
t' = γ(t - vx/c²)
In frame O, x_A = x_B = 5y, and t_lightson = -5 (i.e. 5 yrs ago).
γ = 1/√(1-.64) = 1.67, so as to A:
t_A' = 1.67[-5 - (.8)(-5)] = -1.67y

and as to B:
t_B' = 1.67[-5 - (.8*5)] = -15y

----------------------------------------------------

Now, Doc, does my second calculation express EXACTLY the same approach as yours ('cause even though we got the same answers, it doesn't look the same to me [b(] )
You seem to be using a very abbreviated form of the transformation.
Am I still missing something that would make it easier for me to see?

 

[Janus]

I did a couple of animations some time back to illustrate this very thing. Now my illustration deals with a railway car and lightning strikes rather than spaceships and light bulbs, the the principle it illustrates is exactly the same.

The first one shows the sequence of events according to the observer stationary at the mid point.

http://home.teleport.com/~parvey/train1.gif

The exanding circles represent the leading edges of the light spheres.

Note that both lights turn on at the same time and that the light spheres reach both observers at the same time, so each of them "sees" the lights turn on at the same time.

The second one shows the sequence of events according to the observer on the Ship/train. Remember, from his position, he can consider himself as stationary and the lights as moving.

http://home.teleport.com/~parvey/train2.gif

First note that as each light turns on, the leading edge of the light still expands as a sphere from the initial point of emission according the viewpoint of the spaceship/train observer. But, the light sources move away from these points as time goes on. By the time the midpoint is reached, the lightsources and their intial points of emission are far removed from each other. Not also that the intial emission points always maintain the same distance form the train observer.

Now, since the train observer is not at the midpoint when either of the lights is initially emitted, this means he is closer to one of the emission points than the other at that time.

Thus in order for him to "see" the lights turn on at the same time, the light he is heading for has to start emitting first, followed later by the light he is departing from.

this means that as the midpoint passes by, the light spheres both arrive, and again both observers "see" the light come on at the same time.

Again, in order for order for both observers to "see" the lights turn on at the same time, the lights actually turn on at the same time for the observer with no relative velocity with respect to the lights, but for the observer with a relative velocity, they will have to turn on at different times.

IOW, events that are simultaneous in one frame are not simultaneous in another.

 

[Doc Al]

Originally posted by gnome

Now, Doc, does my second calculation express EXACTLY the same approach as yours ('cause even though we got the same answers, it doesn't look the same to me [b(] )

Yes! It's exactly the same.

You seem to be using a very abbreviated form of the transformation.
Am I still missing something that would make it easier for me to see?

For a complicated problem, I would just go right to the lorentz transformations and turn the crank. But for simple problems like this, I like to apply the "rules" for how clocks and metersticks behave. Of course, these rules are exactly equivalent to the lorentz transformations, but it makes me think I understand what's going on.

1) moving metersticks shrink (by a factor of γ)
2) moving clocks slow down (by a factor of γ)
3) moving clocks are out of synch (by a factor of ΔX V/c²)

Make sense? Get these rules in your bones and you can solve some problems quickly. (Derive them for yourself.)

 

[@Gents]

Lets assume the spaceman don't noe anything abt physic, and he is moving from A to B with a man stationed at M. If he reach M and saw both the light from A and B reached him at the same time, doesn't he think both are lighted simultaneously?

 

[lky]

Originally posted by @Gents
Lets assume the spaceman don't noe anything abt physic, and he is moving from A to B with a man stationed at M. If he reach M and saw both the light from A and B reached him at the same time, doesn't he think both are lighted simultaneously?

If the spaceman don't know anything about physics, not only would he think the explosions are simultaneous, he would also think that the Earth is flat.

 

[lky]

A big thanks to you guys for clarifying my doubts, especially Doc Al and his detailed explanation, and also Janus' animations.

 

[gnome]

Make sense? Get these rules in your bones and you can solve some problems quickly.

Solve problems: yes.
Make sense? NO!

[Janus: I'd like to learn how to write animations like those you posted. Would you mind showing me the code? Can you tell me what specifically to search for to find tutorials, etc.?]

I guess I'm able to use these transformations well enough, but play around with the numbers a little & it immediately becomes clear that even weirder things happen as soon as acceleration is involved. For example, in iky's problem O' travels from A to the midpoint in 3.75 years by his clock, or 6.25 years by the O clock. But suppose, instead, that he flew at .825c for [6 yrs O-time = 3.39 yrs O'-time] and .2c for [0.25 yrs O-time = 0.245 yrs O'-time], so now he arrives at the midpoint in the same 6.25 yrs O-time, but it's only been 3.64 yrs O'-time ("his" time). Now, he arrives at the midpoint, sees the light arriving from A and calculates that the light left A (1/√(1-.2^2) * (-5 + 5*.2) = 4.08 yrs ago, O'-time. But wait! The spaceship left A only 3.64 yrs ago (O'-time). The spaceship passed the light? Obviously not. So the acceleration must screw things up (unless I made a mistake in my conversion).

Anyway, if the space-travelers left A to fly to B, why would they claim that they (the space-travelers) are stationary and that A and B are moving? Isn't that just being ornery? After all, they know that they are the one's that are going to be accelerating and decelerating. Why not just keep O-time and make appropriate adjustments to the readings of their instruments?

EDIT: Spaceman says, "I'll fire my rockets & make that planet come to me."
HUH?
Yeah, yeah -- I'm just ranting.

 

[Doc Al]

Originally posted by lky
A big thanks to you guys for clarifying my doubts, especially Doc Al and his detailed explanation, and also Janus' animations.

I am glad that you found the discussion helpful. And, yes, Janus... cool animations!

 

[dlgoff]

And, yes, Janus... cool animations!

ditto. Thanks. I would like to use them to explain to my young daughter if that's okay.

 

[gnome]

Thanks, Doc. I very much appreciate your help, but I'm not quite "there" yet. I'm still wrestling with the question of why the space traveler moving from A to B, having begun in the AOB frame, and planning to end up in that frame, wouldn't choose to continue using AOB as his frame of reference, making appropriate conversions or even calibrating his instruments to measure time and distance as they would be measured in that frame. (As opposed to taking the - to me, at least - very strange position that he is the stationary "center of the universe" and everything else is moving relative to him.)
After all, if I can use the Lorentz transformations to determine how things look to him, he can also.

So, is he compelled to use his local frame of reference just to avoid making these adjustments, or is there something more fundamental that I'm missing?

 

[Doc Al]

Originally posted by gnome
Solve problems: yes.
Make sense? NO!

You're killin' me.

I guess I'm able to use these transformations well enough, but play around with the numbers a little & it immediately becomes clear that even weirder things happen as soon as acceleration is involved.

The lorentz transformations relate measurements between inertial frames. When things accelerate, you have to be careful in applying special relativity.

For example, in iky's problem O' travels from A to the midpoint in 3.75 years by his clock, or 6.25 years by the O clock. But suppose, instead, that he flew at .825c for [6 yrs O-time = 3.39 yrs O'-time] and .2c for [0.25 yrs O-time = 0.245 yrs O'-time], so now he arrives at the midpoint in the same 6.25 yrs O-time, but it's only been 3.64 yrs O'-time ("his" time). Now, he arrives at the midpoint, sees the light arriving from A and calculates that the light left A (1/√(1-.2^2) * (-5 + 5*.2) = 4.08 yrs ago, O'-time. But wait! The spaceship left A only 3.64 yrs ago (O'-time). The spaceship passed the light? Obviously not.

Bah! Careful. In lky's problem, we can't have the spaceship changing speeds. But we can fudge it OK by considering 3 frames of reference: O, O'(spaceship 1, v=.825c), O''(spaceship 2, v=.2c). After our astronaut travels for 6yrs (O time) he jumps to the new spaceship. (We assume that he and his pocketwatch makes the transistion in zero time with no problems.) Now he travels in the new ship (O'' frame) until he makes it to the midpoint. Yes, he claims that only 3.64 yrs have passed for him; and his new O'' friends say that the light left A 4.08 yrs ago. But they laugh and say, "Dude, your watch was wacky and your time means nothing to us! What you thought was ΔT (when you were on the other, faster ship) was really longer since you were moving compared to us."

 

[Doc Al]

Originally posted by gnome
Thanks, Doc. I very much appreciate your help, but I'm not quite "there" yet. I'm still wrestling with the question of why the space traveler moving from A to B, having begun in the AOB frame, and planning to end up in that frame, wouldn't choose to continue using AOB as his frame of reference, making appropriate conversions or even calibrating his instruments to measure time and distance as they would be measured in that frame.

I think you are thinking that he has a choice. But, no, the effect is real. His clocks (and his body) really do keep different time. Interesting thought about carrying a watch (purposely distorted to keep O time) but since he keeps changing frames (in our SR analysis) that won't work.

(As opposed to taking the - to me, at least - very strange position that he is the stationary "center of the universe" and everything else is moving relative to him.)

To treat himself as being at rest, he'd have to use general relativity. (Experts: help me out here.) We can find the answer using special relatvity, but we have to use the fudge of changing inertial frames. If you were that astronaut, I suspect you'd be very interested in how long the trip will take. (In your local "proper" time AND in the time "back home".)

After all, if I can use the Lorentz transformations to determine how things look to him, he can also.

Of course. As long as you stick to inertial frames. The astronaut does not: he really does accelerate.

 

[gnome]

I'm killing you?

I'll keep trying, waiting for my epiphany. In the meantime, hopefully, I'll get by just applying the rules.

Thanks again.

 

[Eyesaw]

If time is dilated for a moving object only in the
dimension of motion, what happens to the time variable
in the other two dimensions? Is SR suggesting
that every object has 3 different time dimensions? How?
What?

 

[gnome]

I think you're even more confused than I am. A given point in space-time has only one time coordinate. So there's only one dimension -- time -- that's dilated.

 

[Eyesaw]

In the Lorentz transformation, there are 6 coordinate
transformations: x -> x' , y' -> y', z-> z' for
position and tx -> tx' , ty -> ty' and tz -> tz' for time.
So why do they ignore the time coordinates in the y and
z axis in SR? Why is the space-time diagram only drawn
for one time dimension and one position dimension, t and x ?

Huh? What? What a boring nonsensical theory.

 

[gnome]

Where are you finding tx, ty and tz? Every reference to Lorentz transformations in Modern Physics (Serway, Moses & Moyer) and in Special Relativity (A.P. French) mention only
x -> x'
y -> y'
z -> z'
t -> t'

 

[Eyesaw]

Time is treated as a variable that is dependent on relative velocity
in SR. Since velocity is a vector with 3 components, there should be
3 transformation equations for time as well.

 

[gnome]

Relative velocity of one frame of reference compared to the other.

How you set up your coordinates is entirely arbitrary. The "easy" way is to set them up so the relative motion is along the x axis. Then you only need one transformation equation for time.

If you set up your coordinates in such a way that you have 3 components of velocity, you just make the calculations more complicated -- probably hopelessly complicated -- but the end result will be the same. There is still only 1 time axis each coordinate system.

 

[Eyesaw]

Originally posted by gnome
Relative velocity of one frame of reference compared to the other.

How you set up your coordinates is entirely arbitrary. The "easy" way is to set them up so the relative motion is along the x axis. Then you only need one transformation equation for time.

If you set up your coordinates in such a way that you have 3 components of velocity, you just make the calculations more complicated -- probably hopelessly complicated -- but the end result will be the same. There is still only 1 time axis each coordinate system.

You have no choice, things do not move in one dimensional space. It's implicit in the one dimensional Lorentz transformation that
the V is the x component of V where the y and z components are 0. If time is a function of Vx, it is also a function of Vy and Vz simultaneously. Or rather, not simultaneously since we believe
in SR.

 

[gnome]

What are you talking about?

The transformation is not concerned with various objects moving in various directions in 3 dimensions. It is merely translating the coordinates of an event in one frame of reference to the coordinates of the same event in another frame of reference. The motion of one frame in relation to another frame can always be described by a single vector, and you can always position your coordinate axes so that vector is parallel to the x-x' axes.

 

[Eyesaw]

"..it is merely translating the coordinates of an event"... in other words, something moving through space. You would be better of saying Special Relativity is only valid for one dimensional objects moving in one dimensional, field free space. In other words, never.

 

[gnome]

Dude, your invention of an x-component of time and a y-component of time makes as much sense as a y-component of x and a z-component of x. In other words, no sense at all.

As to
"there should be 3 transformation equations for time as well":
When are you publishing your book? Everyone else seems to manage just fine with only 1 transformation equation for time. The difficulty of the calculation is a function of how you set up the problem. As I said, you probably could set up a problem with one frame moving along some skewed line relative to the other, but you still wouldn't have three SEPARATE time transformation equations. You would have one hopelessly complicated time transformation equation dealing with three components of velocity, and then about ten hours later (relatively speaking ) you would end up with the same result for t' as you would have gotten with the common, simple time transformation equation had you set up the problem sensibly -- if you didn't screw up the calculations.

In other words, setting up a problem such that the relative velocity of the S' frame is parallel to the x-axis is NOT saying that this method is only useful for motion in 1-dimension. It's only saying, I don't want to do 10 hours of calculation when I can solve the same problem in 10 minutes instead. Note that within those two "well-behaved" reference frames you can have objects moving in many different, not necessarily parallel, directions (and speeds), and the same, simple translation equation works for all of them.

 

[Eyesaw]

Originally posted by gnome
Dude, your invention of an x-component of time and a y-component of time makes as much sense as a y-component of x and a z-component of x. In other words, no sense at all.

As to
"there should be 3 transformation equations for time as well":
When are you publishing your book? Everyone else seems to manage just fine with only 1 transformation equation for time. The difficulty of the calculation is a function of how you set up the problem. As I said, you probably could set up a problem with one frame moving along some skewed line relative to the other, but you still wouldn't have three SEPARATE time transformation equations. You would have one hopelessly complicated time transformation equation dealing with three components of velocity, and then about ten hours later (relatively speaking ) you would end up with the same result for t' as you would have gotten with the common, simple time transformation equation had you set up the problem sensibly -- if you didn't screw up the calculations.

In other words, setting up a problem such that the relative velocity of the S' frame is parallel to the x-axis is NOT saying that this method is only useful for motion in 1-dimension. It's only saying, I don't want to do 10 hours of calculation when I can solve the same problem in 10 minutes instead. Note that within those two "well-behaved" reference frames you can have objects moving in many different, not necessarily parallel, directions (and speeds), and the same, simple translation equation works for all of them.

Erm, no. Have you tried throwing a baseball at a 45 degree angle? If so, how did you make the reference frame of the ground parallel to the motion of the baseball? I suppose it's ok for SRists to move
the whole Earth in the direction of the baseball to accomdate the
one dimensional Lorentz Transformation.

Obviously velocity between two frames has 3 components,
and since time dilation and length contraction are a function of relative velocities in SR, time is also a vector with 3 components
so that there should be 3 transformation equations for time in
SR and not 1. Therefore, every object in SR has 3 time dimensions.

 

[Doc Al]

I can't decide whether this is hilarious or sad. No offense, Eyesaw, but have you even glanced at a relativity book before?

Originally posted by Eyesaw
Erm, no. Have you tried throwing a baseball at a 45 degree angle? If so, how did you make the reference frame of the ground parallel to the motion of the baseball? I suppose it's ok for SRists to move
the whole Earth in the direction of the baseball to accomdate the
one dimensional Lorentz Transformation.

Erm, yep. You do realize that this SR thing just happens to depend on the relative velocity, right? You also realize that "SRists" don't actually have to "move the whole earth", right? It's just a mathematical convenience.

Obviously velocity between two frames has 3 components,
and since time dilation and length contraction are a function of relative velocities in SR, time is also a vector with 3 components
so that there should be 3 transformation equations for time in
SR and not 1. Therefore, every object in SR has 3 time dimensions.

The relative velocity only has three components if you choose the wrong coordinate system. Unless you are a masochist (or, more likely, an idiot), you will wisely choose a coordinate system that aligns with the direction of velocity, making all but one component zero. (You do realize that velocity is an "arrow" with only one real dimension, right?) And if you don't, then your (single) time coordinate will merely be a messy function of four variables instead of just two. Have fun with that.

 

[Peterdevis]

Eyesaw,

I don't think that you understand SR.
In SR we have 4 dimensions (time and the tree space coördinates)
Then you define a frame (it must be inertial)(existing of four axes)
You can call them t,x,y,z. When you construct another frame with a relative velocity (0,v1,v2,v3) to the first frame. To transform the first frame to the second frame you need only four transformation rules.

If you choose (0,v1,0,0) as velocity, you become the traditionel time dilatation, and x contraction

 

[Eyesaw]

Originally posted by Doc Al
I can't decide whether this is hilarious or sad. No offense, Eyesaw, but have you even glanced at a relativity book before?

Erm, yep. You do realize that this SR thing just happens to depend on the relative velocity, right? You also realize that "SRists" don't actually have to "move the whole earth", right? It's just a mathematical convenience.
The relative velocity only has three components if you choose the wrong coordinate system. Unless you are a masochist (or, more likely, an idiot), you will wisely choose a coordinate system that aligns with the direction of velocity, making all but one component zero. (You do realize that velocity is an "arrow" with only one real dimension, right?) And if you don't, then your (single) time coordinate will merely be a messy function of four variables instead of just two. Have fun with that.

Two spaceships in space, one traveling at a 22 or 55 degree angle from the other. So, how do you set one reference frame parallel with respect to the other to perform a Lorentz transformation? Don't tell me we are just going to ignore the velocity components in the directions not parallel to the x direction for the sake of performing the Lorentz transformation? High school students had no problems working in 3 dimensional space-time prior to the advent of SR, but now we have PHDS who have problems believing cars can go up freeway onramps.

Time was made into a dependent variable in SR. Dependent on the the relative velocities between two frames in the x direction in 1 d space so that in 3d space, it becomes dependent on 3 velocity components, in contrast to Galilean transform where t is assumed to be an independent variable. Perhaps it would be easier to understand this in layman's terms: length contracts and time dilates in the direction of motion, thus with three different directions in space, length has the potential to contract in 3 spatial dimensions and time has the potential to dilate in 3 temporal dimensions.

 

[Doc Al]

Originally posted by Eyesaw
Two spaceships in space, one traveling at a 22 or 55 degree angle from the other.

Statements of angles are meaningless until you've defined a coordinate system. Nothing is preventing you from choosing a coordinate system that makes calculation easier. Nothing you do will change the physical fact that length along the direction of motion is frame dependent. If you wish to choose a coordinate system that makes the direction of motion 22 degrees from your x-axis, no problem; the results won't change! Length along that direction will still "contract", just like always.

So, how do you set one reference frame parallel with respect to the other to perform a Lorentz transformation?

By choosing a coordinate system that best matches the reality you are trying to model. But you don't have to choose a coordinate system that makes the equations easier.

Don't tell me we are just going to ignore the velocity components in the directions not parallel to the x direction for the sake of performing the Lorentz transformation?

If you (unwisely) choose a coordinate system in which the relative velocity is at 22 degrees with the x-axis, then of course you'll have to include the y and z components. But don't blame that on anyone but yourself!

You seem to think something special happens if you pick a coordinate system in which the velocity is not parallel to the x-axis. Nope. You still define an event by the same old four coordinates (t, x, y, z). All you've done is make the transformation equations connecting one frame to another more complex. The physics doesn't change. You don't magically get new x,y,z "components" of time---whatever that might mean.

If you want to get nuts, don't stop with the relative velocity at an angle. Why not have one coordinate system rotated and with a different origin as well? Let's make those equations really difficult!

High school students had no problems working in 3 dimensional space-time prior to the advent of SR, but now we have PHDS who have problems believing cars can go up freeway onramps.

You are in a dream world, my friend. Your approach is like trying to calculate the time it takes for a ball to fall, but insisting on using components that aren't parallel to the gravitational field. The ball doesn't care what coordinates you use. Do it the hard way, if you wish: you'll still get the same answer.

 

[Eyesaw]

You are in a dream world, my friend. Your approach is like trying to calculate the time it takes for a ball to fall, but insisting on using components that aren't parallel to the gravitational field. The ball doesn't care what coordinates you use. Do it the hard way, if you wish: you'll still get the same answer.

What kind of analogy is this? So there are no satellites orbiting in space?

Time was made into a dependent variable in SR- so what does it depend on? Velocity between two frames of reference. What is velocity? A vector with 3 components. I did not invent this ok? There are 3 dimensions in space so there has to be 3 velocity components that describes something in space- if you choose to ignore it so you can suit some dumb math formula, fine but don't drive a car near my town. You can have a 4 variable transform in Galilean transform because time is not a dependent variable but once you make it dependent, it's inexcusable for there not to be a separate time transformation for each spatial component.

Really, why do you think the Lorentz transformation was never written into a vector form like all other physical equations
prior to SR?

 

[Doc Al]

Originally posted by Eyesaw
What kind of analogy is this? So there are no satellites orbiting in space?

Huh?

Time was made into a dependent variable in SR- so what does it depend on? Velocity between two frames of reference. What is velocity? A vector with 3 components. I did not invent this ok? There are 3 dimensions in space so there has to be 3 velocity components that describes something in space- if you choose to ignore it so you can suit some math equation, fine but don't drive a car near my town.

What's your point? Time is dependent on velocity. No one is ignoring anything.

You can have a 4 variable transform in Galilean transform because time is not a dependent variable but once you make it dependent, it's inexcusable for there not to be a separate time transformation for each spatial component.

A meaningless statement. There are only four variables (Galilean or Einsteinian): t, x, y, z.

If you choose a dopey coordinate system, then t' = f(t, x, y, z). So what? Choose a better system, then t' = f(t, x). Your choice.

Rather than continue this silly discussion, why don't you derive your version of the lorentz transformations using an arbitrary velocity and whatever coordinates you like?

 

[selfAdjoint]

Two spaceships in space, one traveling at a 22 or 55 degree angle from the other. So, how do you set one reference frame parallel with respect to the other to perform a Lorentz transformation

The group of motions in special relativity is the Poincare group, consisting of the Lorentz transformations AND the 3-dimensional rotations. Since the ships are inertial (not accelerated) they keep a constant bearing toward each other. Rotate you coordinates until the x-axis points along that bearing and then do your Lorentz boost.

 

[Eyesaw]

Originally posted by Doc Al
Huh?

Huh, what?


What's your point? Time is dependent on velocity. No one is ignoring anything.

No it's not.



A meaningless statement. There are only four variables (Galilean or Einsteinian): t, x, y, z.

If you choose a dopey coordinate system, then t' = f(t, x, y, z). So what? Choose a better system, then t' = f(t, x). Your choice.

Rather than continue this silly discussion, why don't you derive your version of the lorentz transformations using an arbitrary velocity and whatever coordinates you like?

Just use the Lorentz transformations for x, and t but substitute
x and t for y and ty in the y direction and z and zt in the z direction. If you plugged in 0 for the velocity components, you get y = y', ty= ty', and z = z' tz = tz'. If they were non zero,
you get 3 sets of Lorentz transformations that's all.

Making time a variable leads to physical nonsense.

 

[gnome]

Just use the Lorentz transformations for x, and t but substitute x and t for y and ty in the y direction and z and zt in the z direction. If you plugged in 0 for the velocity components, you get y = y', ty= ty', and z = z' tz = tz'. If they were non zero,
you get 3 sets of Lorentz transformations that's all.

Making time a variable leads to physical nonsense.

Did you learn all this at the Ralph Cramden Institute of Physics, or is this your own discovery? You really have a profound misunderstanding of the concept of special relativity (or any relativity, for that matter). For example, the idea of a 4-dimensional universe has clearly escaped you entirely. Even worse, you seem to be suffering from a rather shaky concept of the meaning of "dependent variable", you can't seem to distinguish between the relative motions of frames of reference themselves and the motions of objects within those frames, and you appear to be laboring under the misconception that all motion must be described in relation to some set of UNIVERSAL COORDINATE AXES, aligned with the ether and with its origin presumably identified by the location of the GREAT PUMPKIN.

In short, you really ought to learn to crawl before you try to fly.

 

[Doc Al]

Originally posted by Eyesaw
Just use the Lorentz transformations for x, and t but substitute
x and t for y and ty in the y direction and z and zt in the z direction. If you plugged in 0 for the velocity components, you get y = y', ty= ty', and z = z' tz = tz'. If they were non zero,
you get 3 sets of Lorentz transformations that's all.

You can substitute the word "pizza" for t, and "banana" for x, if you want, but it's just nonsense.

How do I measure tx,ty,tz? Do I need a special watch (three special watches?) or do I just have to be facing the right direction?

 

[Doc Al]

Originally posted by gnome
Did you learn all this at the Ralph Cramden Institute of Physics...

Now that's funny.

 

[Eyesaw]

Originally posted by Doc Al
You can substitute the word "pizza" for t, and "banana" for x, if you want, but it's just nonsense.

How do I measure tx,ty,tz? Do I need a special watch (three special watches?) or do I just have to be facing the right direction?

Same way we currently measure time dilation and length contraction
in SR of course. That is to say, we assume it a priori. When did you ever try to measure the speed of light or any other speed in someone else's inertial frame? This whole notion of observing events in someone else's inertial frame is laughable. Where are these special SR telescopes on Ebay?