# Pertinence of relativity of simultaneity

1. Aug 15, 2007

### ice109

i wanted to say relevance but i figured that would be too easy :rofl:

editing

there is something i'm not getting. yes if the speed of light is the same in all intertial frames simultaneity is relative but so what? can someone explain to me how this implies that time and space are relative? very very clearly please.

and can someone also tell me what is the interval in lorentz geometry?

if this is the equation in only 2 of 4 dimensions in minkowski space
$$\Delta t^2 - \Delta x^2 = \Delta t'^2 - \Delta x'^2=\tau^2$$( with primes being in moving frames.)

and it is an equation of a hyperbola then what is the $\tau$. it's obviously a constant but as far as the graph of a hyperbola i can't figure out what it represents.

Last edited: Aug 15, 2007
2. Aug 15, 2007

### pervect

Staff Emeritus
The fact that simultaneity is relative basically means when you say that you specify two events happen "at the same time" that you must specify which particular frame you are talking about. "At the same time" in frame S is not "at the same tine" in frame O, if O is moving relative to S.

The only reason this is a big deal is because people often omit this bit of information, and because it plays havock with some people's intuition.

3. Aug 15, 2007

### ice109

but this is supposedly albert's big eureka thing? this for him resolved a lot of paradoxes.

ok answer this question for me

we have two people, one moving at some velocity relative to the other. as one passes the other one of them turns on a light bulb. after any amount both are still at the center of the light cone. how do i recover anything like a lorentz transformation from this thought experiment? and how about the other question[edit: nm robphy explained this one in another thread]? and additionally how do you prove the interval in minkowski space is invariant?

Last edited: Aug 15, 2007
4. Aug 15, 2007

### robphy

The neatest way to proceed is to do a radar experiment... by having one observer send out another signal [a light bulb flash] a short time after the meeting event...then wait for its reflection to be received by the source.
Then analyze with the Bondi k-calculus.

Follow, e.g.,

With a little more work (not shown), you can obtain the Lorentz Transformations. [You can find this in [Dover] books by Bondi, by Bohm, and by Born... among other places.]

The beauty of this method is that all of the "relativistic factors" involved can be expressed in terms of k [an eigenvalue of the transformations you seek]. Of all of the "effects", the Doppler effect is emphasized... all others [time-dilation, length-contraction, relativity-of-simultaneity, velocity-composition, etc...] are of secondary importance.

I'm writing up a set of notes on the Bondi k-calculus... but I don't have time to work more on it at this time.

5. Aug 15, 2007

### JesseM

Yes, basically because if you assume that each observer defines the coordinates of events in terms of local readings on rulers and clocks at rest with respect to themselves (so if I see an explosion happen next to the 10-meter mark on my ruler and the clock at that position reads 5 seconds at the moment of the explosion, I'll assign it coordinates x=10 meters, t=5 seconds), with the clocks "synchronized" using the assumption that light signals all move at c in the observer's frame (which results in different observers defining simultaneity differently), then as long as each observer measures moving rulers to shrink by $$\sqrt{1 - v^2/c^2}$$ and moving clocks to slow down by the same amount, this will ensure that when Maxwell's laws (or any other fundamental laws of physics) are expressed in different inertial coordinate systems, they will always obey the same equations. This means there is nothing to differentiate one inertial coordinate system from another, no physical reason for thinking one observer's definitions of simultaneity or lengths or time intervals are more correct than any other's.
Well, suppose I'm on a moving ship and I want to synchronize clocks at the front and back in my frame, I can just turn on a light bulb at the center of the ship and set both clocks to read the same time when the light reaches them. But in a frame where the ship is moving, the front of the ship is moving away from the point where the light was turned on while the back is moving towards it, so if this frame also assumes light travels at the same speed in both directions, it must conclude that light hit the back before the front so the clocks are out-of-sync. This is enough to show relativity of simultaneity, then to show why each observer must see the other one's rulers shrink and the other one's clocks slow down you have to impose the condition that not only do they each measure the light move at the same speed in both directions, they both must also measure the light to move at c in their frame. With a little work you can derive the Lorentz transformation in this way.

6. Aug 16, 2007

### ice109

can you show me? i've attempted this before and gotten nowhere. additionally that doesn't address my specific light cone experiment

Last edited: Aug 16, 2007
7. Aug 16, 2007

### JesseM

A derivation of the Lorentz tranformation from these assumptions? I have a friend coming to visit for a few days tomorrow so I won't have much time for internet, but I'll try to get to it sometime next week.
I'm not quite sure what you're asking with that thought-experiment, but note that if both frames measure the speed of light beams coming from the bulb in all directions to have the same value of c, naturally that means they will both remain at the center of the light cone in their own coordinate system, because at any given time in that coordinate system each light beam will have traveled the same distance from the position (in that coordinate system) where the bulb was when it turned on.

8. Aug 16, 2007

### ice109

yes this is obvious the point is how is it possible? the answer of course is that space is skewed for each relative to the other but i want to know how to derive this.

9. Aug 16, 2007

### JesseM

Are you asking me to show that each will measure the speed of light to be the same if we assume that each measures the other's ruler to shrink by $$\sqrt{1 - v^2/c^2}$$, the other's clock ticks to be slowed down by the same amount, and each one synchronizes his clocks using the Einstein synchronization convention? Or are you asking for a derivation of the fact that rulers and clocks will be altered in this way if we assume that the laws of physics must be such that each measures light to move at c in all directions (i.e. a derivation of the Lorentz transformation from the two postulates of SR)? Or both?

10. Aug 16, 2007

### ice109

i want you to derive gamma without referring to anything except the two postulates of SR and and relativity of simultaneity and within the context of this thought experiment.

11. Aug 16, 2007

### robphy

Check out my page of animations at
http://physics.syr.edu/courses/modules/LIGHTCONE/LightClock/ [Broken]
(These animations illustrate a mechanism for proper time along a worldline using light clocks.)

There's more analytical detail in my paper at
http://arxiv.org/abs/physics/0505134

Note that the relativity of simultaneity is a consequence of the two postulates.

(The animations above doesn't use the k-calculus in the presentation. However, the paper makes reference to the k-calculus.)

Last edited by a moderator: May 3, 2017
12. Aug 16, 2007

### JesseM

I still don't understand what you mean by "within the context of this thought experiment". Your description doesn't even explicitly involve rulers or clocks, which are essential to any physical derivation--can I assume that when the light bulb is turned on, it coincides with the center of a ruler with clocks at either end, which is moving at velocity v in my frame? Then I can show that if the light moves at c in both directions in my frame, then in order for an observer moving along with the ruler-clock system to also measure the light from the bulb to have moved at c in both directions (so that we both measure the position the bulb was turned on to be the center of a light cone), in my frame the ruler must be shrunk by the gamma factor and the clocks slowed down by the same factor. Would that be acceptable?

13. Aug 16, 2007

### ice109

yes it would be more than acceptable; i would very much appreciate it

14. Aug 18, 2007

### JesseM

OK, so let's assume that in terms of my own coordinate system (which is defined in terms of rulers and clocks at rest in my frame, with the clocks synchronized using the Einstein synchronization convention so that light moves at the same speed in both directions in my coordinate system) the light bulb is turned on at position x=0 light years and time t=0 years. Assume a ruler is moving along my x-axis in the positive x direction with velocity v, with clocks at either end as well as at its center; at t=0 the center of the moving ruler is at position x=0, where the light is switched on, and the clocks on the ruler have been synchronized using the Einstein synchronization convention, so that they the clocks at either end will read the same time at the moment the light from the bulb being turned on reaches them.

Now we don't know how length contraction or time dilation works from the start, so let's say that if marks on the moving ruler are 1 meter apart in its own frame, in my frame they'll be L meters apart, and if the clocks on the moving ruler tick at a rate of 1 tick per second, in my frame they'll tick at T ticks per second. We need not assume that the L and T factors are equal, nor that their value is smaller than 1 as opposed to greater than 1 or equal to 1.

Assume without loss of generality that the ruler is 2 light years long in its own frame, so that in my frame its length is 2L light years. In this case, at t=0 the back end is at position x=-L ly, while the front end is at position x=L ly. Since the rod is moving at velocity v in the positive x direction, the back end's position as a function of time is given by x=vt - L, while the front end's position is given by x=vt + L. So, to figure out when the light from the bulb first reaches the back end in my coordinate system, take vt - L = -ct and solve for t, giving t = L/(c + v). Likewise, to figure out when the light from the bulb first reaches the front end in my coordinate system, take vt + L = ct and solve for t, giving t = L/(c - v). We can also plug these times back into the equations for position as a function of time to find the position of the front and back ends when the light first reaches them; the back end will be at position v*[L/(c + v)] - L = L*v/(c + v) - L*(c + v)/(c + v) = -Lc/(c + v), while the front end will be at position v*[L/(c - v)] + L = L*v/(c - v) + L*(c - v)/(c - v) = Lc/(c - v).

Now, suppose I have my own ruler which is at rest in my frame, and whose back end is permanently at position x = -Lc/(c + v) while its front end is permanently at position x = Lc/(c - v). In my frame the length of this ruler must just be [Lc/(c - v)] - [-Lc/(c + v)] = [Lc*(c + v) + Lc*(c - v)]/[(c - v)*(c + v)] = (2Lc^2)/(c^2 - v^2). The back end of my ruler will be at the same position as the back end of the moving ruler at the moment the light from the bulb reaches the back end of the moving ruler, while the front end of my ruler will be at the same position as the front end of the moving ruler at the moment the light from the bulb reaches the front end of the ruler. Remember that the clocks on the moving ruler were synchronized using the Einstein synchronization convention, so these events are simultaneous in the moving ruler's frame. Since each frame defines the "length" of a moving object as "the position of the front end at a given time minus the position of the back end at the same time", this means that in the moving ruler's frame, my ruler must be exactly the same length as the moving ruler, which we assumed earlier was 2 light years in its own frame.

Now we can use the fact that one of the postulates of SR says the laws of physics must be identical in different inertial frames, which means that if I see the moving ruler's length as different by a factor of L from its rest length, then in the moving frame *my* ruler must be different by a factor of L from its rest length in my frame (I guess to be really rigorous here, to justify this we'd also have to show that speeds are reciprocal, so that if I measure a moving ruler to be moving at speed v, then in the moving ruler's frame I must be moving at speed v as well; I'll give a little justification of this at the end of the post). So since my ruler has a length (2Lc^2)/(c^2 - v^2) in my frame, its length in the moving frame must be L times that, or (2L^2 * c^2)/(c^2 - v^2). But I also showed that its length in the moving frame must be equal to that of the moving ruler in its own frame, which was assumed to be 2 light years; this means that (2L^2 * c^2)/(c^2 - v^2) = 2, which if we solve for L gives L^2 = (c^2 - v^2)/c^2 = (1 - v^2/c^2), which shows that L is equal to the familiar lorentz contraction factor sqrt(1 - v^2/c^2).

Once you have the lorentz contraction factor I can think of two different ways you might go about finding the time dilation factor T. One way is to place a mirror at one end of the moving ruler, send a light signal from an emitter at the other end, and then impose the condition that the amount of time t for the light to go to the mirror and bounce back to where it was emitted, as measured on a clock next to the emitter, must satisfy d/t = c, where d is twice the ruler's length in its rest frame (the distance the light travels between leaving the emitter and returning to it). If the ruler's length in its rest frame is 2 light years and its length in my frame is 2L, then if the signal first travels in the same direction the ruler is moving in my frame (from back to front), the time for the signal to go from the emitter to the mirror in my frame will be given by the equation vt + 2L = ct, so t = 2L/(c - v). Then the time for the signal to return from the mirror to the emitter, this time traveling in the opposite direction as the ruler in my frame (front to back) will be given by vt = 2L - ct, so t = 2L/(c + v). So, the total time in my frame is 2L/(c - v) + 2L/(c + v) = [2L*(c + v) + 2L*(c - v)]/[(c + v)*(c - v)] = 4Lc/(c^2 - v^2) = 4L/[c*(1 - c^2/v^2)]. If we make use of the fact that we know L = sqrt(1 - v^2/c^2), then that means the time in my frame must be 4/[c*sqrt(1 - v^2/c^2)]. But we also know that whatever the time in my frame, the time as measured by a clock that's moving with the ruler must be T times that, so the time as measured by a clock next to the emitter must be 4T/[c*sqrt(1 - v^2/c^2)]. Meanwhile, the distance the light travels in the ruler's frame is twice the ruler's rest length, or 4 light years. So, d/t = c gives 4/(4T/[c*sqrt(1 - v^2/c^2)]) = c, or [c*sqrt(1 - v^2/c^2)]/T = c, which gives T = sqrt(1 - v^2/c^2), the familiar time dilation factor.

However, you may prefer to find the value of T using your original thought-experiment where, instead of light starting from one end and bouncing off a mirror on the other end, we instead have a light bulb which turns on when it is next to the center of the moving ruler, and clocks at either end of the ruler are synchronized so that they read the same time when the light from the bulb reaches them. As mentioned above in the third paragraph, if in my frame the bulb is turned on at time coordinate t=0, then the light reaches the back end at t = L/(c + v) in my frame, while the light reaches the front end at t=L/(c - v). So, the time interval between these two events in my frame is just [L/(c - v)] - [L/(c + v)] = [L*(c + v) - L*(c - v)]/[(c + v)*(c - v)] = 2Lv/(c^2 - v^2). Since both clocks are synchronized in their rest frame so that they read the same time at the moment the light hits them, in my frame the time on the clock at the back end will have advanced by T times this amount by the moment the light reaches the clock at the front end, so in my frame the two clocks will be out of sync by 2TLv/(c^2 - v^2). This means that if there is another clock at the center of the moving ruler, and it reads a time of 0 years at the moment it's next to the light bulb and the light bulb turns on, then in my frame the clock at the back end will read TLv/(c^2 - v^2) at the moment the bulb is turned on, while the clock at the front end will read -TLv/(c^2 - v^2) at the moment the bulb is turned on.

As mentioned earlier, in my frame the light reaches the back end at a time of L/(c + v) after the bulb was turned on. The clock at the back end will have advanced forward by T times this amount between the time the bulb turned on and the time the light reached it, and in my frame it started at TLv/(c^2 - v^2) when the bulb was turned on, so when the light reaches it the time on the clock at the back end will be TLv/(c^2 - v^2) + T*L/(c + v) = [TLv + TL*(c - v)]/(c^2 - v^2) = TLc/(c^2 - v^2). Since the clock next to the bulb read 0 at the moment it was turned on, this must be the actual amount of time for the light to travel from the bulb to the back end of the ruler as measured in the ruler's own rest frame. And we know the distance from the center of the ruler to either end is 1 light year in its own rest frame, so if distance/time must equal c in this frame, then we have [1]/[TLc/(c^2 - v^2)] = c, which gives (c^2 - v^2)/TLc = c which in turn gives TL = (c^2 - v^2)/c^2 = (1 - v^2/c^2).

If we know that L = sqrt(1 - v^2/c^2), then TL = (1 - v^2/c^2) is enough to prove that T = sqrt(1 - v^2/c^2). However, there's one missing step here, because as I said in my original derivation of L, we really need to prove that speeds are reciprocal, so that if the ruler's speed is v in my frame, then my speed is also v as measured in the ruler's frame. To show this, imagine I place a marker at the 0 mark of my own ruler, and the moving ruler notes the time according to its own clocks that the marker is next to its center and the time the marker is passing the back end of the ruler. We already assumed that the clock at the middle of the moving ruler reads 0 at the moment it coincides with position x=0 on my ruler. Then since the distance between the middle of the moving ruler and the back end is L in my frame, and the ruler's speed is v, the time in my frame for the back end of the moving ruler to reach the marker must just be L/v, so the clocks on the moving ruler must advance forward by T times this amount, or TL/v. But remember that in my frame the clocks at different points on the ruler are out of sync, so the clock at the back end didn't start out with a time of zero at the moment the clock in the middle read zero; instead the initial time on the clock at the back end was TLv/(c^2 - v^2) in my frame. So, by the time the back end passes my marker, its time must read TLv/(c^2 - v^2) + TL/v, which can be rewritten as [TLv^2 + TL*(c^2 - v^2)]/v*(c^2 - v^2) = (TLc^2)/[v*(c^2 - v^2)]. So, this is the time for the marker to move from the middle of the moving ruler to its back end as measured in the moving ruler's frame. Since the distance between the middle and the back end is 1 light year in the moving ruler's frame, the speed of the marker as measured in the moving ruler's frame must be distance/time = 1/[(TLc^2)/[v*(c^2 - v^2)]] = v*(c^2 - v^2)/TLc^2 = vc^2*(1 - v^2/c^2)/TLc^2 = v*(1 - v^2/c^2)/TL. And recall that in the previous paragraph I found TL = (1 - v^2/c^2), so plugging this in we find that the speed of the marker as measured by the moving ruler is v*(1 - v^2/c^2)/(1 - v^2/c^2) = v. So, this shows that my speed as measured in the moving ruler's frame must be equal to the moving ruler's speed in my frame. Note that we didn't need to know the value of L to get either this result or the earlier result that TL = (1 - v^2/c^2), so there isn't any circular reasoning in using this result to find L as I did in the fifth paragraph above.

Let me know if you have any questions about these derivations...if you have trouble following some of it, it may help to draw a diagram, or to plug in some specific numbers like v=0.6c.

15. Aug 20, 2007

### JDługosz

The book General Relativity from A to B by Robert Geroch does that, in very simple and lucid terms. I currently see used copies for \$3.37.

You can read the essay Nothing But Relativity on the web. It takes a different approach, in that the constant speed of light is not a postulate! Lorentz invariance is derived from 4 postulates involving the uniformity of space and time.

16. Aug 21, 2007

### morrobay

The spacetime interval constant in relation to the hyperbola is the ct intercept when x = zero, if interval is 9 then ct axis point equals 3.The hyperbola is generated by holding the spacetime interval constant while varying ct and x.