Relativity of Simultaneity Questions

[mucker]

Hi All,

I've been doing some reading on the above but having some problems understanding certain parts of it (maybe it's wrong from Wikipedia!) To simplify it for me I will first pose a simple scenario where we are not factoring in speed yet, then go from there.

Say we have two planets A and B and observer A. These planets are exactly 1ly from observer A and explode. Observer A will see and therefore conclude they exploded at the same time. Now let's say observer B is in a position so that he is 1ly away from planet A but 2ly away from planet B. He will observe that plant A exploded first.

My question is, if observer B is a physicist (and therefore is aware that light takes longer to reach him from planet B), and were to document these events, would he record them as how he observed them (not at the same time) or would he record them as happening at the same time (by factoring in the his distance from them)? In other words when we record/talk about simultaneous events (in the scientific sense) do we base it on what we observe, or what we conclude?

More questions to come soon when factoring in speed.

Thanks,

 

[Dale]

having some problems understanding certain parts of it

Yes, the relativity of simultaneity is the single most difficult concept in SR.

In other words when we record/talk about simultaneous events (in the scientific sense) do we base it on what we observe, or what we conclude?

It is about what we conclude. I.e. we are assuming intelligent observers that know the finite speed of light and correct for it.

 

[mucker]

Thanks Dale.

So from the above (when motion is not involved) observer A and observer B would be in agreement that the events happened at the same time. It seems that isn’t the case if motion is involved; take this diagram for example re simultaneity. Here are my questions:

From the blue man’s frame of reference I don’t get how he concludes (assume that’s what the diagram is implying) that the 2 lighting bolts happened at separate times. Is it because he is not aware that he is moving relevant to the lighting bolts? In other words, if he was aware he was moving, he would factor in the speed in and conclude they struck at the same time? Or is due to dime dialation? Or is it something else?

Another thing, I don't see why the above statements in my OP don't seem to be applicable here, it seems we could just use the same reasoning above to a moving object if you know how much you've moved in a set amount of time, would that work?

Just want to say that a fully understand how they observer them different, the bit I am struggling with is how they conclude it differently.

 

[Dale]

(when motion is not involved) observer A and observer B would be in agreement that the events happened at the same time.

Yes. Although B does not receive the light at the same time, B recognizes the finite speed of light, compensates, and concludes that they did happen at the same time.

It seems that isn’t the case if motion is involved;

Yes, even if B is co-located with A at the moment of receiving the light.

Is it because he is not aware that he is moving relevant to the lighting bolts?

The lightning bolts in this diagram are idealized and considered to be instantaneous. They do not have any duration, so they cannot have any speed. They are events, “points” in spacetime.

In other words, if he was aware he was moving, he would factor in the speed in and conclude they struck at the same time?

It is not about awareness. It is about choice. Because the laws of physics are the same in all inertial frames, he is allowed to use the frame where he is at rest if he chooses. Of course, he could also choose to treat himself as moving. Either way, it has nothing to do with unawareness. He is aware of all the motion in the scenario and he is allowed to use the frame where he is at rest.

it seems we could just use the same reasoning above to a moving object if you know how much you've moved in a set amount of time, would that work?

Certainly. If he chooses to treat himself as moving then he will correct for his motion and determine simultaneity according to the frame in which he is moving. But that would make the scenario much less instructive, so here he chooses to analyze things in the frame where he is at rest.

the bit I am struggling with is how they conclude it differently.

You should do the math by hand a couple of times, just to see how it works.

 

[mucker]

thanks Dale.

The lightning bolts in this diagram are idealized and considered to be instantaneous. They do not have any duration, so they cannot have any speed. They are events, “points” in spacetime.

I get that, I was referring to the light rays from the lightning bolts taking time to reach him

It is not about awareness. It is about choice. Because the laws of physics are the same in all inertial frames, he is allowed to use the frame where he is at rest if he chooses. Of course, he could also choose to treat himself as moving. Either way, it has nothing to do with unawareness. He is aware of all the motion in the scenario and he is allowed to use the frame where he is at rest.

I suppose what I am saying is that in the blue man's ref frame it looks like it's only describing what he observed - I mean it doesn't even use words like "conclude" like it does when talking about the green man's ref frame. It looks to me, from the lightning bolts at diff times (very botton frame), that he observed it that way, but says nothing about what he concluded of that observation.

 

[Dale]

I get that, I was referring to the light rays from the lightning bolts taking time to reach him

That is c in all inertial frames.

I suppose what I am saying is that in the blue man's ref frame it looks like it's only describing what he observed - I mean it doesn't even use words like "conclude" like it does when talking about the green man's ref frame. It looks to me, from the lightning bolts at diff times (very botton frame), that he observed it that way, but says nothing about what he concluded of that observation.

That is why I recommend that you go through the math yourself a few times.

 

[mucker]

One last question - does any of this have to do with time dilation at this point? Or can we understand it without having to bring that in?

 

[Ibix]

Time dilation comes in later. At this point we're just proving that "simultaneous" isn't something frames can agree on if the speed of light is invariant. We haven't introduced clocks yet so we can't compare their rates.

It's worth doing the maths to work out when the flashes arrive at the train guy and what ge must conclude about when the flashes happened, as @Dale advised. Post if you get stuck - it's why we're here.

 

[Dale]

One last question - does any of this have to do with time dilation at this point? Or can we understand it without having to bring that in?

All three, time dilation, length contraction, and relativity of simultaneity are important. The Lorentz transform encapsulates all three.

In working out the numbers, you should use the Lorentz transform. And 0.6 c is a good speed to use since relativistic effects are readily visualized and computed but things are still easy to draw.

 

[mucker]

Ok so let's give this a go. I assume you mean the lorentz transformation so I will use the equation from the wikipedia page here.

So before I start I need to know what metric to use. t is obvious but is v and x in metres? Or is v a percent to light speed like Dale used above?

 

[Dale]

is v and x in metres? Or is v a percent to light speed like Dale used above?

You can do it either way, as long as you are consistent. If $v$ is in m/s then $x$ is in m and $t$ is in s and $c$ is in m/s. If $v$ is in light years per year (fraction of $c$) then $x$ is in light years and $t$ is in years and $c=1$. Similarly with light-seconds.

 

[Nugatory]

@mucker Now that @Dale has straightened out the units…. If you’re going to work through an example I suggest that you choose $v=.6c$ or $v=.8c$, measure time in seconds and distances in light-seconds. You will find the arithmetic appreciably less messy.

 

[mucker]

ok I'll do it in m/s.

For the calc below I am using m/s. Ok so let's give this a go. Let t =3 and x=2 and v=1000. Let's just do the top half of t¹ first to make sure my math is correct:
t -vx/c²

= 3-(1000x2)/c2
= 3-(2000/c2)
= 3- 2.225300112107237e-14
=2.999999999999978

Is that right? (just the top half)

for botton half, which is the Lorentz factor I believe?

v2 / c2 = 1,000,000 / 8.987551787368176e+16 = 1.112650056053618e-11
so now we say 1- 1.112650056053618e-11 = 0.9999999999888735
then we square root this = 0.9999999999944367
then it's 1 / 0.9999999999944367
= 1.000000000005563

so t¹ = 2.999999999999978 / 1.000000000005563
= 2.999999999983289

Proably should have used bigger numbers lol. I realize (after doing the calc) I've used low crappy numbers (I was visualising the train example) that it would negilble but can someone just confirm the math is correct?

 

[mucker]

Then to work out x¹ we do top half first:

x-vt
= 2 - 1000*3
= -2998

then bottom half is the Lorentz factor again which we already know from above is 1.000000000005563.

so now 2998 / 1.000000000005563 =
-2997.99999998332000000000

Clearly something is wrong with how I worked out the x¹ location but I can't see where I went wrong?

 

[Nugatory]

but I can't see where I went wrong?

Stop.
Back up and retry your example with a few changes that will make it easier to see what is going on without obscuring the underlying physics:
1) Start with a much larger value of $v$, either .6c or .8c so that the gamma factor comes out to be a reasonable round number, either 5/4 or 5/3
2) Choose units in whic $c$ is equal to one. Seconds and light-seconds will be the closest to something that you can visualize: we all know what a second is, and a light-second is about seven times around the world.
3) Do an Google image search (that's an image search, not a regular one) for "Einstein train simultaneity". Look at some of the graphs that finds.

(Also, be aware that you cannot trust a calculator when working out to 12 or 14 decimal places)

 

[mucker]

Ok so if i change v to 0.6c I get a Lorentz factor of 1.25

So doing it again...
t¹ = top half of equation = t -vx/c2
= 3 - 0.6*2/1²
=3- 1.2
= 1.8

so t¹ = 1.8/1/25 = 1.44 ly
Correct?

NOTE: I had no idea it would get so messy using m/s, which is why I went ahead with it. m/s seemed simpler at the time...

 

[mucker]

Would someone be so kind as to confirm if I have calculated it correctly above?

Also, I've simualated the Einstein thought experiment now in a spreadsheet and it does the Lorentz transformation to show what obervers B should conclude; I'll share a screenshot of it next after I know I've got the formula correct first!

thanks

 

[Ibix]

You've correctly calculated $\gamma=1.25$ but you then seem to have divided by it instead of multiplying.

 

[mucker]

Thanks Ibix.

So the big line separating the top formula from the bottom formula here doesn't mean divide by?

I worked out the top to be 1.8 and bottom as 1.25 (which you said is correct) so from that I see it as top formula outcome divided by bottom formula outcome, therefore 1.8/1.25.

 

[Doc Al]

So the big line separating the top formula from the bottom formula here doesn't mean divide by?

Of course it does, it's a fraction. But the Lorentz factor is one over the bottom formula:
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

 

[Ibix]

Thanks Ibix.

So the big line separating the top formula from the bottom formula here doesn't mean divide by?

View attachment 287388
I worked out the top to be 1.8 and bottom as 1.25 (which you said is correct) so from that I see it as top formula outcome divided by bottom formula outcome, therefore 1.8/1.25.

$\sqrt{1-v^2/c^2}=\sqrt{1-0.6^2}=0.8$, but the Lorentz factor is one over that. You can multiply by 5/4 or divide by 4/5.

I can confidently predict that this is not the only time you will make that mistake...

 

[mucker]

Hold on, I think i see your point now Ibex. What you are saying is that I came to the correct answer to work out Lorentz factor but by a different (wrong?) way?

 

[mucker]

Ibix because you didn't mention t and my answer to Lorentz factor was correcton I thought your comment about dividing rather than multiplying was in ref to me calculating t, no?
Thinking about it now, I think you may have misread the post. When I said 1.8/1.25 I was referring to t, I think you thought I was still working out the Lorentz factor at that point?? Just for some context I did use the 1 over formula to work out the Lorentz factor btw - I just didnt show the calculation (well I did in the m/s post), I just shown the outcome (first line). The rest of the post was working out t (which I think you thought was still me working out Lorentz factor??)

 

[Doc Al]

Even though you calculate the Lorentz factor, you need to realize that your equations are not written in terms of that factor. (You are adding it mentally, and making an error.) Try writing your Lorentz transformation equations in terms of $\gamma$, like this:

$$t' = \gamma( t - v x/c^2)$$

 

[mucker]

Thanks Doc, I will give that a go. But just to clarify, that equation I quoted is not mine but from Wikpedia. Just want to check - so is the equation (I provided) wrong, or how I interpreted it?

If I use your formula I get 1.25*1.8 = 2.25 correct?

If correct:
Either the original formula I used is wrong or I worked it wrong. If I am wrong, I can't see how:

doesn't equal:
t -vx/c²
/ (divided by)
γ

If I'm wrong, can you please explain where I am getting confused? thanks buddy

 

[Doc Al]

Thanks Doc, I will give that a go. But just to clarify, that equation I quoted is not mine but from Wikpedia. Just want to check - so is the equation (I provided) wrong, or how I interpreted it?

There's nothing wrong with the equations you quoted (interpreted properly). They are just the standard Lorentz transformations. But note that they don't mention or involve the "Lorentz factor" -- you added that bit.

 

[Doc Al]

Thanks Doc, I will give that a go. But just to clarify, that equation I quoted is not mine but from Wikpedia. Just want to check - so is the equation (I provided) wrong, or how I interpreted it?

If I use your formula I get 1.25*1.8 = 2.25 correct?

If correct:
Either the original formula I used is wrong or I worked it wrong. If I am wrong, I can't see how:

View attachment 287397
doesn't equal:
t -vx/c²
/ (divided by)
γ

If I'm wrong, can you please explain where I am getting confused? thanks buddy

In that quoted equation, you are dividing by $\sqrt{1 - v^2/c^2}$, which is NOT $\gamma$. (It's 1/$\gamma$.)

 

[mucker]

But note that they don't mention or involve the "Lorentz factor" -- you added that bit.

I think I see now. These equations are all new to me, I thought
that equation (you just quoted (how do I do that btw?)) was the lorentz factor as it looks so similar to it. I thought there was some kind of typo so I replaced it with the actual Lorentz factor... I am learning...thanks for clearing it up!

 

[mucker]

Ibix thanks Ibex, sorry for being dumb!

 

[Doc Al]

(how do I do that btw?)

I assume you are asking how I wrote the equations that I showed. I used "Latex", a tool for writing math stuff. It's much easier than it looks! You can get started here: https://www.physicsforums.com/help/latexhelp/

 

[Dale]

Ibix thanks Ibex, sorry for being dumb!

As @Ibix said, we have all made that same mistake multiple times on this road.

 

[Ibix]

Ibix thanks Ibex, sorry for being dumb!

It's easy done. I still do it from time to time. To be clear, you correctly calculated $\gamma=1/\sqrt{1-v^2/c^2}$ and correctly calculated $t-vx/c^2$ but then incorrectly combined them as $(t-vx/c^2)/\gamma$ instead of $\gamma(t-vx/c^2)$. As a general policy I'd recommend trying not to mix expressions containing gammas with expressions containing the square root form because if you do it's easy to substitute the wrong value as you did.

 

[mucker]

Yep, so now I've got that out the way and get the math (finally ) let's get back on track to my OP. I am trying to apply it to the train thought experiment. Let's just focus on one lighning bolt for a second.

from observer 1 (green man, stationary) a lighning bolt strikes 2ls away (he knows the distance for this example). He becomes aware of it at t2 obviously. If we were to plot this event on a spacetime diagram from his perspective do we say t=0, x=2ls, or is it actually t=-2, x=2ls? (because the current frame of reference of an observer is always rooted in t=0, so therefore this event lies 2 seconds in his past). I was at first thinking the former but after thinking through it for a while I think you would record it as the latter...

 

[Ibix]

because the current frame of reference of an observer is always rooted in t=0, so therefore this event lies 2 seconds in his past

You're always at t=0? So your watch always reads the same time? That's odd - I might stay in the same place but my watch always advances.

 

[Nugatory]

If we were to plot this event on a spacetime diagram from his perspective do we say t=0, x=2ls, or is it actually t=-2, x=2ls? (because the current frame of reference of an observer is always rooted in t=0, so therefore this event lies 2 seconds in his past). I was at first thinking the former but after thinking through it for a while I think you would record it as the latter...

It is indeed the latter, but not for the reason you say. Negative values of $t$ aren’t necessarily in the past, they’re just before whatever moment we decided to call $t=0$.