# Length contraction and Time dilation calculation

## Summary:

Length contraction calculation

## Main Question or Discussion Point

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 realise 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.

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 travelled 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 travelled 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)
• 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.

Related Special and General Relativity News on Phys.org
BvU
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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 !

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
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2019 Award
That is indeed the general idea with the Lorentz transformation .

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.

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 travelling 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?

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?

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 travelled 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 travelled from one to the other and everyone measures it travelling 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.

Last edited:
• P J Strydom
jbriggs444
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If I see a vehicle moving relative to me, say I observe it travelling 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.

Last edited:
• P J Strydom, vanhees71 and Ibix
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 dont 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 travelling 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 dont know what it is.

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

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)
• 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
Homework Helper
2019 Award
I looked at it and feel like a sheep in a desert looking at a green light. I dont 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 travelling 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 dont 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.

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 dont like to hear something I can not explain.
Therefore this time around, I dont 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?

Oh, and thanks for listening.

jbriggs444
Homework Helper
2019 Award
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.

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 dont 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 dont 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 dont 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.

Last edited:
• vanhees71 and jbriggs444
Im all yours.
But, I am confident you will be able to get the theory through my thick scull.

##\Delta t## and ##\Delta x=c\Delta t## t ##\Delta x'=c\Delta t'##. ##v\ll c##.
As for the above, I dont have a flippen clue what it is.

jbriggs444
Homework Helper
2019 Award
As for the above, I dont 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.

Last edited:
• Ibix
Ibix
As for the above, I dont 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.

• jbriggs444
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 definately the problem.
I cant expect you guys to teach me the basics of this topic, if I dont 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. Is there any chance that you can perhaps link the aplicable mathematical formulas where I can go and learn and do some excersises on?