Length contraction and Time dilation calculation

[P J Strydom]

TL;DR Summary

Length contraction calculation

I was absent for a while due to personal constraints but I did keep myself busy with the Time dilation equation some member sent me a while back.

I decided to set a time limit for myself to learn and understand time Dilation and length contraction, which must be before December 2020, or I will just have to accept that I am unable to grasp the scientific principle of this topic.

I realize that my greatest shortcoming is definitely my rusted high school algebra which I am sharpening up again.

I therefore will humble myself to the fullest in anticipation that someone on this forum will take the time to direct me in the right direction as I wonder through this venture.

Well, I will start with the following.

I completely forgot who the member was that sent me the documents explaining the mathematical formulas to calculate length contraction and time dilation, furthermore, the printed copy I have is so badly deteriorated, that I am busy retyping the whole thing. Unfortunately, it contains a lot of mathematical inserts in Word, and it takes a while to go through the whole process. I finished the document, but found a few typos to fix.

Anyway, the document starts off explaining a simple Galilean transformation where a truck drives at 30Mph and I as a passenger throws a ball 60 feet, and the question is then asked, how far will the ball have traveled in 2 seconds.

Obviously, the document goes through the whole process to teach the calculation to arrive at the answer.

Then the document starts to teach the Special relativity transformation and works on a Linear function dependent on speed connection these 2 calculations that starts off with;

• S (frame at rest) and S1 (Moving relative to S at speed v)
• A beam of light is flashed along the Axis of motion of S1
• S calculates it’d distance as a function of time like this: Distance = rate X Time.
• x=ct. which is then re written as x-ct=0.
• If the beam is shot in the opposite direction, the equation will be;
• X=-ct rewritten as x+ct=0

Now we will look at the beam of light flashed in the S1 frame, and we get:

• X1-ct1=0
• And the flash in the opposite direction
• X1+ct1=0

This now gets connected, first for the equations where the light flashed with the axis of motion of S1. Then on the opposite direction. We get;

• X-ct=x1-ct1
• And x+ct=x1ct1

Then we insert the velocity of the S1 observer as α(v) and for the flash in the opposite direction, δ(v)
Now we have:

• X-ct= α(v) (x1-ct1)
• And x+ct= (δ(v)(x1+ct1)

Now these 2 functions will be added together and simplified.

Great stuff, I do understand the mathematical formula, because it is the old algebra function where 2 functions are expressed in Rate X Time = distance, added up, and deducted from the distance.

Simple example

Two cars started from the same point, at 5 am, traveling in opposite directions at 40 and 50 mph respectively. At what time will they be 450 miles apart?

	r​	t​	d​
Car A​	40	t​	40t
Car B​	50	t​	50t

40t+50t=450

90t=450

t=5 hours

Cool.

Now to get back to our equation.

We took the distance the light beam traveled to the axis of motion of S1 and added the distance of the S frame and inserted the different velocities into the equation. We will then simplify the equation (we will get to that later.)

My question,

• x-ct= α(v) (x1-ct1)
• x+ct= (δ(v)(x1+ct1)
• All added up
• 2x=x1[α(v)+ δ(v)]+ct1 [δ(v)- α(v)] divide and simplified gives
• X= x1[α(v)+ δ(v)]/2 +ct1 [δ(v)- α(v)]/2
• We then replace [α(v)+ δ(v)]/2 with γ(v)
• And [δ(v)- α(v)]/2 with β(v)

Which leaves

x= γ(v)x1 + β(v)ct1

Which is as indicated by the document as the result that we have the Distance transformation from S1 to S.

What are we attempting to achieve with this calculation?

I sat and drew pictures forever and still do not know what this formula means.

 

[BvU]

Hi,

What are we attempting to achieve with this calculation?

We are trying to express $x$ in the S frame in terms of $c$ and $x_1$, $t_1$ in the S1 frame.

It is the so-called Lorentz transformation, which google

My respect for your endeavour ! PF is a good place for help !

 

[P J Strydom]

Hi,

We are trying to express $x$ in the S frame in terms of $c$ and $x_1$, $t_1$ in the S1 frame.

Ha!
I did not even start and I am not sure what this means.
Does this mean, the observer in the S1 frame is using his x1 and t1 to determine where x is in the S timeframe?

 

[BvU]

That is indeed the general idea with the Lorentz transformation .

 

[P J Strydom]

So this does not have anything to do with Length contraction?
This is only to find where another observer is in relative position to you?

 

[Ibix]

So this does not have anything to do with Length contraction?
This is only to find where another observer is in relative position to you?

If I know where the ends of an object are as measured in my frame, I can deduce its length. If I know how to transform the end positions into the positions measured by some other frame then I can deduce its length as measured by that frame.

 

[P J Strydom]

If I know where the ends of an object are as measured in my frame, I can deduce its length. If I know how to transform the end positions into the positions measured by some other frame then I can deduce its length as measured by that frame.

OK, in my simplest understanding.
If I see a vehicle moving relative to me, say I observe it traveling to my right hand direction, and I measure its 2 end positions on my x coordinates, I can deduct how long it is.
The same with the observer who will measure my vehicles length. He will use his X1 co ordinates.
This I understand.
but I am now wondering why did we bring the light beams into this equation?

 

[P J Strydom]

Let me just recap.

We have S (who is in this situation stationary), and S1 who is moving relative to S,

Then we have a light flash in the direction of the motion of S1.

And we have a light flash in the opposite direction.

Does this mean that S1 will use the 2 positions of the “light flashes” on x1 after t1 expired, to determine where S is? or does he want to determine where S measures the light flashes' positions on his X co-ordinates?

Are we looking for S? or the positions of the light flashes on x after t passed?

 

[P J Strydom]

I will again be back tomorrow morning.
Greetings.

 

[Ibix]

This I understand.

Good (edit - but note jbriggs444's comment #11).

but I am now wondering why did we bring the light beams into this equation?

Because light beams have a nice property in relativity - they always have the same speed in all inertial frames. If I have some event $x_1,t_1$ and another event $x_2,t_2$ and I know light traveled from one to the other then $(x_2-x_1)=c(t_2-t_1)$. Then I'm saying that there's a transformation law that transforms $x_1,t_1$ to $x'_1,t'_1$ and $x_2,t_2$ to $x'_2,t'_2$. I don't know what the primed quantities are because I don't know the transformation law, but I do know that $(x'_2-x'_1)=c(t'_2-t'_1)$ because the light traveled from one to the other and everyone measures it traveling at $c$.

All the stuff with light rays going from A to B is just exploiting this fact in order to be able to deduce the transforms.

 

[jbriggs444]

If I see a vehicle moving relative to me, say I observe it traveling to my right hand direction, and I measure its 2 end positions on my x coordinates, I can deduct how long it is.

No. Not unless you measure those end positions at the same time. If you measure the end positions at different times you can get any answer you please.

In order to make sense of special relativity, one has to discard the idea that "at the same time" is meaningful by itself and accept that only "at the same time according to this frame" is meaningful.

Consider, that the car has a front bumper and a rear bumper. Both are moving rightward at some velocity v. We can plot the two positions versus time on a piece of graph paper. (green = front, red = back).

The length is the "difference between x coordinates for the two lines". But a line (i.e. a world line) does not have an x coordinate. A point on the line (i.e. an event) has an x coordinate. You have to be clear about exactly which two events you are picking out.

It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so.

 

[P J Strydom]

No. Not unless you measure those end positions at the same time. If you measure the end positions at different times you can get any answer you please.

In order to make sense of special relativity, one has to discard the idea that "at the same time" is meaningful by itself and accept that only "at the same time according to this frame" is meaningful.

Consider, that the car has a front bumper and a rear bumper. Both are moving rightward at some velocity v. We can plot the two positions versus time on a piece of graph paper. (green = front, red = back).

The length is the "difference between x coordinates for the two lines". But a line (i.e. a world line) does not have an x coordinate. A point on the line (i.e. an event) has an x coordinate. You have to be clear about exactly which two events you are picking out.

Tnx for this one and the graphic display.
I looked at it and feel like a sheep in a desert looking at a green light. I don't know what to do with it.
It shows X as a measurement of length between the 2 bumpers, that I understand.
then it shows t as the traveling distance of those 2 points.
I understand that.
Then it shows the true length on the left hand side as a measurement between the two x points.
I understand that too.
Then there is two diagonal lines comming from the left upward.
This I don't know what it is.

Shacks, I still have a long way to go.
Sorry pal.

 

[P J Strydom]

I am doing everything wrong to find out what I need to understand.

My error is that I am already thinking of the global calculation, whilst I skipped out on the step by step explanations of the calculation in question as I set out to do right in the beginning.

This is the calculation (for reference)

• x-ct= α(v) (x1-ct1)
• x+ct= (δ(v)(x1+ct1)
• All added up
• 2x=x1[α(v)+ δ(v)]+ct1 [δ(v)- α(v)] divide and simplified gives
• X= x1[α(v)+ δ(v)]/2 +ct1 [δ(v)- α(v)]/2
• We then replace [α(v)+ δ(v)]/2 with γ(v)
• And [δ(v)- α(v)]/2 with β(v)

Which leaves

x= γ(v)x1 + β(v)ct1

After careful thinking, I recon I should stay with the calculation and get the reasons behind every step.

Step 1

• x-ct= x1-ct1

Is this the meaning of the equation?.

• x-ct= x1-ct1 . The position of S determined by S1 as S1 took the position of the light beam on x1 after t1 expired. And the same for S who used the position of the light beam on x after t expired. (in this equation S and S1 are stationary relative to each other because there is no velocity inserted in the equation.)
• So, S measures S1 on his x co ordinate after t expired. And vice versa for S1 who measures S’s position using x1 after t1 expired.

Is this correct?

 

[jbriggs444]

I looked at it and feel like a sheep in a desert looking at a green light. I don't know what to do with it.
It shows X as a measurement of length between the 2 bumpers, that I understand.
then it shows t as the traveling distance of those 2 points.
I understand that.
Then it shows the true length on the left hand side as a measurement between the two x points.
I understand that too.
Then there is two diagonal lines comming from the left upward.
This I don't know what it is.

The green line is a graph of the position of the front bumper (on x, the vertical axis) versus time (on t, the horizontal axis). The red line is a graph of the position of the rear bumper in the same way.

Those lines depict the history of the bumpers. They are what we refer to as "world lines".

The problem of measuring the length of the car in real life is analogous to the problem of measuring the distance between the two lines on the graph. In both cases, the answer is ambiguous.

In the case of the real life car, you cannot determine the length of a moving car by subtracting the position of the rear bumper from the position of the front bumper. That is because the rear bumper has no fixed position. Similarly for the front bumper. You have to pick out one particular "event" in the history of the rear bumper and write down the position of that event. You have to pick out one particular "event" in the history of the front bumper and write down the position of that event. But which two events do you pick out? There is the ambiguity.

In the case of the graphs, you cannot determine the distance between the lines by subtracting the x coordinate of the red line from the x coordinate of the green line. That is because the red line has no fixed x coordinate. Similarly for the green line. You have to pick out one particular point on the red line and write down the x coordinate of that point. You have to pick out one particular point on the green line and write down the x coordinate of that point. But which two points do you pick out? There is the ambiguity.

Simultaneity is the answer. We pick out a point where the respective bumpers are at the same time.

Of course we know this for the case of the car. This is such an obvious fact that it seems not even worthy of mention. For the graph, the corresponding notion is picking out a pair of points with the same t coordinate.

The point is that "at the same time" and "with the same t coordinate" are very much analogous concepts. That's the key insight. That and the idea of rotating the graph paper.

 

[P J Strydom]

Except for the fact that I had to google a few English words (English being my third language), and had to read your post 3 times, I eventually understood exactly what you meant.
Thanks.
I yet still have to understand the steps of the calculations of the formula I posted, and I am sure I will then grasp the full concept of Time Dilation and Length contraction.

I am continiously derailed on my thinking about these concepts, so much that when someone tells me how we will be able to travel through space, and Time will slow down so much that we can travel to the end of the universe if we reach close to c, with time then being a fraction of the true time we know, I freak out because I just don't like to hear something I can not explain.
Therefore this time around, I don't want to make any assumptions on what will happen if..., but I want to understand the mathematical equations.

I am sure that if I work through the furmula step by step, I will eventually gain this knowledge.
What do you think?
Is this the way to go?

 

[P J Strydom]

Oh, and thanks for listening.

 

[jbriggs444]

I am sure that if I work through the furmula step by step, I will eventually gain this knowledge.
What do you think?
Is this the way to go?

Different people come to understand relativity differently. An explanation that works for one person will fail for the next person.

For me, the equations are not the way to understand it. For me, an equation without a mental picture of the thing being described is empty and meaningless. Boring. Geometry and graphs work for me. The metaphor of "rotating graph paper" is appealing.

Some others may willingly cast loose from intuition, embrace the mathematics and follow the equations where they lead. I salute those people. But I am not one of them.

If I follow a set of equations from start to finish and wind up at a startling conclusion then I will go back and ask "what implicit intuitive assumptions did I carry into this problem that these equations turn out not to respect"? I like to find the key assumption that is violated. If I can wrap my head around that, the rest of the theory will fall into place.

 

[P J Strydom]

For me, the equations are not the way to understand it. For me, an equation without a mental picture of the thing being described is empty and meaningless.

Thats what I am looking for.
The mental picture.
But so far I just don't get the mental picture in an understandable way.
It does not help if someone tells me and paint a mental picture about say...a light clock on the move will run slower. Its nice to see it mentally, but if there is no explainable evidence to support that picture, it remains a myth as a religion forced upon the ignorant.
Just as if one tells me, "Learn the maths", and I don't get the mental picture. Such clculations can just as well be furmulas given to someone suffering from dyslexia. It will mean nothing.
This is where I am.
The ignorant believer with dyslexia.
I need to learn the meaning of the formula, and to visualize the intent of what it calculates.

 

[Ibix]

But so far I just don't get the mental

My advice is to forget trying to derive the Lorentz transforms and accept them as given. Then show that they do perform as advertised - that if you transform two events that are separated by $\Delta t$ and $\Delta x=c\Delta t$ then, whatever the separation ofthe transformed events, $\Delta x'=c\Delta t'$. You also might like to check that they simplify to the Galilean transforms if $v\ll c$. That's not as thorough as deriving the transforms (it doesn't prove that the Lorentz transforms are the only transforms that can do this), but it's probably more achievable if your algebra is a bit shaky. It'll also help you get used to working with frames, which is a stumbling block for a lot of people.

Once you can do that, you can derive time dilation and length contraction formulae fairly easily. You can also look up Minkowski diagrams, which (to me, anyway) give the intuitive bit that the maths doesn't. But without the maths to support it, it isn't at all clear what they're doing.

 

[P J Strydom]

Im all yours.
I will follow your lead (even though I memorised the calculation I posted here hoping it will give the answer.)
But, I am confident you will be able to get the theory through my thick scull.

 

[P J Strydom]

$\Delta t$ and $\Delta x=c\Delta t$ t $\Delta x'=c\Delta t'$. $v\ll c$.

As for the above, I don't have a flippen clue what it is.

 

[jbriggs444]

As for the above, I don't have a flippen clue what it is.

What @Ibix wrote was:

if you transform two events that are separated by $\Delta t$ and $\Delta x=c\Delta t$ then, whatever the separation of the transformed events, $\Delta x'=c\Delta t'$

@Ibix is asking you to verify that the speed of light is the same in both frames.

If you have two events and you have a speed of light signal going straight from one to the other then $c \Delta t$ for the two events should equal $\Delta x$ for the two events. This should hold both for the $(t,x)$ coordinates for the transformed $(t',x')$ coordinates.

The Lorentz transforms are, of course, what you would use to take a $(t,x)$ coordinate as input and produce a $(t',x')$ coordinate as output.

The $v\ll c$ bit is part of a different test and is a bit more difficult to describe compactly. You would look at the Lorentz transform equations and the Galilean transformation equations side by side, assume that v is much less than c and see whether they produce nearly the same results.

 

[Ibix]

As for the above, I don't have a flippen clue what it is.

As @jbriggs444 says, I was suggesting you consider two events that are at $(x_1,t_1)$ and $(x_1+\Delta x,t_1+\Delta t)$, where those two events represent the emission and reception of a light pulse. Since you know that light travels a distance $c\Delta t$ in time $\Delta t$ you know that $\Delta x$ can be replaced by $c\Delta t$. You can then use the Lorentz transforms to transform the coordinates of these two events into another frame. But the events remain the start and end of a light pulse, so the difference in x' coordinates and t' coordinates must also have the ratio $c$ if light also travels at $c$ in this frame.

The point about $v\ll c$ is just that at low speeds (when $v$ is much smaller than $c$) you should recover the Galilean transforms that underlie Newtonian physics. This is usually called the Correspondence Principle, and stares that any new theory must approximate the old theory in cases where the old theory has been experimentally tested. If it doesn't, then the new theory has already failed experimental test and should be discarded. In this case, our old theory was the Galilean transforms, which have been tested in cases where $v$ is so small that $v/c\approx 0$, so it's worth checking that the Lorentz transform formulae look like the Galilean transform formulae when that's the case.

 

[P J Strydom]

Please give me the weekend to study what you guys just said.
I think it is a case that my algebra is much worse than I anticipated.
And this is definitely the problem.
I can't expect you guys to teach me the basics of this topic, if I don't even know what this Delta sign is that got connected to time and distance etc.
I hope you understand the predicament I am in here.
I want to calculate the ends of the world, but with the education of a preschool todler.

 

[P J Strydom]

Is there any chance that you can perhaps link the aplicable mathematical formulas where I can go and learn and do some excersises on?

 

[Ibix]

I don't even know what this Delta sign is that got connected to time and distance etc.

$\Delta$ is usually used to mean a difference. So for example if you have two things that happen at times $t_1$ and $t_2$ you'd define $\Delta t$ as $t_2-t_1$. That is, $\Delta t$ is common notation for a period of time, as distinct from a point in time. Since you often don't care about the ends of a time period, just its length, it's very common to see "one event happens at $t=t_1$ and the other at $t=t_1+\Delta t$".

Usually $\Delta t$ is a large time interval and $\delta t$ is a small one, but nit always.

Is there any chance that you can perhaps link the aplicable mathematical formulas where I can go and learn and do some excersises on?

Which formulae? The Lorentz transforms? They're easy to Google...

 

[P J Strydom]

Ho, no I have the Lorentz transformation.
And after telling me what major Delta is, I am fine.
I found another explanation by Shankar at open Yale courses whichg uses the Major Delta.
I will be back as soon as I went through his document.
https://oyc.yale.edu/sites/default/files/relativity_notes_2006_7.pdf

 

[P J Strydom]

OK, I think I understand the Mathematical equation and what it means after I had some time to think about it.
Remember the post.

Summary:: Length contraction calculation
My question,
x-ct= α(v) (x1-ct1)

• x+ct= (δ(v)(x1+ct1)
• All added up
• 2x=x1[α(v)+ δ(v)]+ct1 [δ(v)- α(v)] divide and simplified gives
• X= x1[α(v)+ δ(v)]/2 +ct1 [δ(v)- α(v)]/2
• We then replace [α(v)+ δ(v)]/2 with γ(v)
• And [δ(v)- α(v)]/2 with β(v)

Which leaves

The way the informative document posed the situation was where I got everything wrong.
It said:
Start with 2 frames of reference, one moving in respect to the other.
Assume the law of distance is
d=v*t
And speed of light is c
Frame 1 is S and is designated at rest.
The other frame is S_1 and designated to move at speed v, a beam of light is fired along the axis of motion of S_1
Timeframe S calculates its distance as a function of time like this if the beam is fired →
distance=rate*time
x=ct rewrite it as
x-ct=0 (S→)
If the beam is fired in the opposite direction, the equation would be:
x=-ct(S←)rewrite as
x+ct=0 (S←)
From the timeframe S_1 the beam is fired →
x_1=ct_1 (S_1→)
Rewrite as
x_1-ct_1=0 (S_1→)
If the beam is fired in the other direction
x_1=-ct_1 (S_1←)
Rewrite as
x_1+ct_1=0 (S_1←)
Now we must insert speed, v. But we do not know the speed, and will call it α(v)
We have:
x-ct=0 (S→)
and
x_1-ct_1=0 (S_1→)
Combining the equations for S^1→
x-ct=α(v)(x_1-ct_1 )
If the beam is shot in the opposite direction, we will use x= δ(v)
We have:
x+ct=0 (S←)
And
x_1+ct_1=0(S_1←)

And so on.
What I missed out on was the intention of how each observer moves in relation to the other, where the light beam is flasshed from, and what the velocities of each observer is.
I had this idea. (and forgive the non scientific wording)
I thought;
There is an "X axis" where both observer "A" and "B" is on zero.
On the "x axis's" zero point, a beam is fired toward the positive co ordinates.
and another beam is fired in the opposite direction to the "Negative " side.
Then the document say I should insert velocities for "A" and "B".

Everything in my mind saw the 2 obsevers and the 2 light beams taking off at X=Zero.

Now I have the picture as the following.
An "X-Axis" with a zero point where A and B meets and time is synchronised to zero.
AND the 2 light flashes ARE 2 EVENTS HAPPENING FROM THE "PAST" AND "FUTURE " POSITIONS ON THE X-AXIS, OR FROM THE NEGATIVE SIDE OF X AND THE OTHER FROM THE POSITIVE SIDE OF X.

Is this correct?