Special relativity: Calculating the relativity of simultaneity

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  • #1
billllib
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The question is in reference to calculating relativity of simultaneity. I am on the step where I take the time in Alice's frame from the front and from the rear clock and minus it to the get the total time. I end up with gamma squared etc (For more details see the picture below)

I have observer Bob who is stationary and who see Alice moving.

Now that I finished describing the question here is the question.

How does ## L' = (L_a) (v) ## etc become ## (L_b) (v) ## etc?

I start with the information L_moving because Alice is moving. I don't have Bob's frame. My point is I have L_moving not L_stationary. I am referencing to this formula ##L' = \frac L y ## In order to get L shouldn't I go ## L = L'Y ## but it looks like I go ## L' = \frac L y ##


picture below



When I say etc I am referring to ## \frac { (L_a) (V) (gamma^2) } { (c^2) } = \frac {(L_b) (V) (gamma^2) } {(gamma) (c^2)} . ##
 
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  • #2
kuruman
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Now that I finished describing the question here is the question.
I am sorry, but I don't think you finished describing the question. You did not specify what the physical picture is or what you are trying to calculate / understand / figure out. All we know is that Alice is moving relative to Bob. Please be more specific. Also, you have not defined the meaning of the symbols that you use. The equation ##L'=(L_a)(v)## does not make sense if the symbols have their conventional meanings.
 
  • #3
billllib
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i had an extra ## L' = ##. Just in case this wasn't clear I was referring to ## \frac { (L_a) (V) (gamma^2) } { (c^2) } = \frac {(L_b) (V) (gamma^2) } {(gamma) (c^2)} ## . Also I my question starts with the terms ## \frac { (L_a) (V) (gamma^2) } { (c^2) } = \frac {(L_b) (V) (gamma^2) } {(gamma) (c^2)} . ##

Now to the question. I start off ##L_a.##. I am never given ## L_b## How do I get ## L_b##?

I may have further question but here is where I will start. Thanks for the help.
 
  • #4
kuruman
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The relatoin you seek is in the equation you posted
$$\frac { (L_a) (V) (gamma^2) } { (c^2) } = \frac {(L_b) (V) (gamma^2) } {(gamma) (c^2)}$$,
Just cancel common factors that appear on both sides of the equation.
 
  • #5
billllib
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Okay got it thanks.
 
  • #6
billllib
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The relatoin you seek is in the equation you posted
$$\frac { (L_a) (V) (gamma^2) } { (c^2) } = \frac {(L_b) (V) (gamma^2) } {(gamma) (c^2)}$$,
Just cancel common factors that appear on both sides of the equation.


Okay but this brings up another question

I have T_total_A = t_A_rear - T_A_front (time to hit both clocks) In Alice's frame
T_A = seeing Bob move.
Can T_A also = Alice's stationary frame?
How do I differentiate between T_A = Alice's stationary frame and T_A = Alice seeing Bob moving? Or am I confused somewhere?
 
  • #7
kuruman
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Okay but this brings up another question

I have T_total_A = t_A_rear - T_A_front (time to hit both clocks) In Alice's frame
T_A = seeing Bob move.
Can T_A also = Alice's stationary frame?
How do I differentiate between T_A = Alice's stationary frame and T_A = Alice seeing Bob moving? Or am I confused somewhere?
At this point it will be helpful if you described the question clearly.
1. What is the physical situation above and beyond that Alice is moving relative to Bob?
2. What is given and in whose frame?
3. What are you trying to determine or figure out?
 
  • #8
Ibix
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@billllib - just to add to @kuruman's comment - special relativity isn't all that difficult, mathematically. The huge problem is learning to keep track of what's been measured or described by who, since a lot more of it is frame-dependent than in non-relativistic physics. Often, simply writing a clear statement of where and when each event happened and in which frame you are measuring is all you need. The way forward becomes clear then.

You haven't clearly stated anything about your scenario, so we really can't help.
 
  • #9
billllib
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Okay I did that and I think I got it. Might as well type it out here.

Bob is stationary. Alice is moving. I want to know where the clocks hit in Alice's frame.

Then in order to get Length contraction of Alice I go

L_Alice or L_moving = L_Bob or L_stationary /y.

Thanks.
 
  • #10
Ibix
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What clocks? Where are they and when are they there, and according to who? What velocity do they have, and according to who? What are they going to hit?
 
  • #11
billllib
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What clocks? Where are they and when are they there, and according to who? What velocity do they have, and according to who? What are they going to hit?



Well I am calculating relativity of simultaneity.

The velocity is v. It does not specify the amount in the video.

The velocity of Alice is going to the right. Alice is flying past Bob.

I have two light clocks at the end of Alice platform of length L. Both clocks are set to 0. The question is, how long does it take the light to hit the light clocks?

So I have a rear clock and a front clock. I calculate the time in both clocks. I minus the time in both the front and rear clock to get the total time of both clocks of Alice moving.

Then I apply length contraction because Alice is moving.

Then I get confused. Why Do I use Delta T B? Shouldn't I still have delta T A?

Please correct me if I made any mistakes with what I typed.



1592274914862.png
 
  • #12
kuruman
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It's not a matter of making mistakes, it's a matter of not explaining what's on your mind. This is what I understand so far.

This much is clear
Alice is on a platform that moves past Bob with some velocity ##v##.

Here it gets confusing.
I have two light clocks at the end of Alice platform of length L. Both clocks are set to 0. The question is, how long does it take the light to hit the light clocks?
You say that there are two light clocks at the end of the platform set to zero, presumably when the origins of the two frames coincide as is usually the case with such problems. Then you ask "how long does i take the light to hit the light clocks?" What light? Where does it come from?

Here it gets even more confusing.
So I have a rear clock and a front clock. I calculate the time in both clocks. I minus the time in both the front and rear clock to get the total time of both clocks of Alice moving.
No, you don't have a rear clock and a front clock. You have two clocks at the end of the platform, presumably next to each other. Each clock measures its own time. They are both on Alice's frame and synchronized by your own admission. So when you "minus" their times, you will get zero. You have not explained what you are trying to calculate.

If you are concerned with simultaneity, you need one clock on Alice's frame and one clock on Bob's frame and an event such as a flash of light that is observed by both Alice and Bob at times and locations as measured by them. The details of that is what's missing from what you ave aid so far.

Please read what you write before you post it and think of the reader who is totally unfamiliar with what's on your mind. Would the reader understand the picture you have in your mind only by reading what you wrote? We cannot help you if we don't understand what kind of help you need.
 
  • #13
Ibix
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Please correct me if I made any mistakes with what I typed.
Basically, we need to be able to do the experiment from your description. So you need to tell us what components there are and any special features of them, where they are, how you use them, and what you do with the results.

What you are doing is like saying: I want to measure ##g##. The time was 200s. I don't understand how to get ##g## from this.

What you need to say is: I want to measure ##g##. I take a mass on the end of a 1m long lightweight rod and hang it as a pendulum. I displace it slightly and measure the time for 100 swings, which is 200s. I am given a formula ##T=2\pi\sqrt{l/g}##, but I don't understand what numbers to put where.
 
  • #14
Ibix
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Addendum: I guess you are talking about Einstein's train thought experiment. But there are at least three variants of it, and this sounds like a fourth variant because it doesn't usually use clocks explicitly.
 

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