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Contradictory time dilation equations.

 
 
Reply Wed 16 Dec, 2020 01:24 pm
It is possible to derive 2 contradictory time dilation equations. Imagine that Sally is in a spaceship and moving to the right and is aiming a flashlight straight up and down so that Sally sees the light moving straight up and down and John is outside the spaceship and sees the light forming a triangle with the floor of the spaceship. Now imagine that Sally is aiming a flashlight upwards and towards the left while the spaceship moves to the right. Now the situation is exactly reversed. Sally sees the light forming a triangle with the floor and John sees the light bouncing straight up and down. Here's the details...

Sally is in a moving spaceship. John is outside the spaceship. Sally is moving to the right at .6c. The height of her spaceship is .8 light-seconds. If Sally has a light clock with the light bouncing straight up and down the light will make a 3-4-5 right triangle from the viewpoint of John. If the change in time for Sally is delta T_o and the change in time for John is delta T then the following equation can be derived:

delta T = delta T_o/((1-.6^2)^.5)

Now Sally has a light clock but this time she is holding a flashlight at an angle of 53.13 degrees above the horizontal and pointed to the left. Now the leftward movement of the light exactly matches the rightward movement of the spaceship from John's viewpoint. Now the light is bouncing straight up and down from the viewpoint of John and the light is making a 3-4-5 right triangle from viewpoint of Sally. If the change in time for Sally is delta T_o and the change in time for John is delta T then the following equation can be derived:

delta T_o = delta T/((1-.6^2)^.5)

The 2 equations are in direct contradiction to each other.
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Type: Discussion • Score: 0 • Views: 761 • Replies: 43
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maxdancona
 
  0  
Reply Wed 16 Dec, 2020 02:36 pm
@Fruityloop,
You aren't using the equations correctly which is why you are getting contradictory answers. I am not exactly sure what you are doing

What reference frame are you using for thesee calculations?
Fruityloop
 
  0  
Reply Wed 16 Dec, 2020 02:49 pm
@maxdancona,
If you are familiar with the famous situation with a person inside a spaceship that has a light clock aimed straight up and a person outside the spaceship seeing the light making a triangle with the floor the rest should make sense. The person inside the spaceship has a light clock by aiming the flashlight upward and to the left so that the situation is exactly reversed.
0 Replies
 
maxdancona
 
  1  
Reply Wed 16 Dec, 2020 06:59 pm
@Fruityloop,
If John and Sally are in two reference frames and each has a clock.

1. John will observe Sally's clock running slower than his.
2. Sally will observe John's clock running slower than hers.

Both of these will be equally valid under relativity.

Is this the "contradiction" you are talking about?
Fruityloop
 
  1  
Reply Wed 16 Dec, 2020 07:21 pm
@maxdancona,
No. It is possible to derive 2 contradictory time dilation equations simply based upon how Sally has her light clock set up, either straight up or at an angle up and to the left.
maxdancona
 
  1  
Reply Wed 16 Dec, 2020 07:29 pm
@Fruityloop,
I don't know what definition of "contradictory" you are using.

Relativity deals with reference frames with the understanding that all frames are equally valid. That is why I asked you about the reference frames.

There is no way that time dialiation leads to equations that are mathematically contradictory... if you understand what you are napping between two reference frames.
maxdancona
 
  1  
Reply Wed 16 Dec, 2020 07:30 pm
@maxdancona,
Do you understand what I mean when I say that all reference frames are equally valid?
Fruityloop
 
  1  
Reply Wed 16 Dec, 2020 08:29 pm
@maxdancona,
Let me try a specific example...
Sally has a flashlight inside her spaceship that is aimed straight up. She is moving to the right at 0.6c. The height of her spaceship is 0.8 light-seconds. Sally sees the light go 0.8 light-seconds in 0.8 seconds. John is outside the spaceship. John sees the light go 1 light-second along the hypotenuse of the triangle in 1 second.

So 0.8 seconds for Sally = 1 second for John.

Sally has a flashlight inside her spaceship that is aimed upward and to the left at an angle of 53.13 degrees from the floor of the spaceship. She is moving to the right at 0.6c. Sally sees the light go 1 light-second in 1 second along the hypotenuse of the triangle. John is outside the spaceship. John sees the light go straight up 0.8 light-seconds in 0.8 seconds.

So 1 second for Sally = 0.8 seconds for John.
maxdancona
 
  1  
Reply Wed 16 Dec, 2020 10:10 pm
@Fruityloop,
1. You are awkwardly expressing frames of reference. It would be more precise to say that When John observes his own clock go a second he observes Sally's clock go 0.8s. and when Sally observes he own clock go a second, she observes John's clock go 0.8s.

I am not exactly sure if what you are saying is the same as what I am saying. But it is correct.

There is no contradiction. That is how the physics works
Fruityloop
 
  1  
Reply Wed 16 Dec, 2020 11:09 pm
@maxdancona,
You obviously haven't grasped what I have written. Please read it again until you understand it. Maybe you do grasp it but don't realize the significance of this.
maxdancona
 
  1  
Reply Wed 16 Dec, 2020 11:43 pm
@Fruityloop,
I grasp the Physics just fine. I have a Physics degree and taught Physics.

I don't grasp what you are saying, which is why I am asking you to clarify it. Science has to be exact, you are expressing yourself in a way that isn't clear.

You are setting up two different reference frames. The time dilation measured in one frame of reference from another is pretty straightforward. I think you are confusing the reference frames. That is why I am trying to help you clarify what you are saying.
Fruityloop
 
  1  
Reply Thu 17 Dec, 2020 12:11 am
@maxdancona,
I've described this pretty clearly. I'm deriving a time dilation formula by having the light clock oriented differently. I have shown that moving clocks run fast instead of slow. A theory with contradictory equations must be discarded.

delta T = delta T_o/((1-.6^2)^.5

delta T_o = delta T/((1-.6^2)^.5)

These equations conflict with each other.
maxdancona
 
  1  
Reply Thu 17 Dec, 2020 12:58 am
@Fruityloop,
There is no contradiction in the physics. There seems to be a contradiction in your understanding. If you would like, I can try to help you learn

You keep repeating over and over that there is a contradiction. I think the problem is that you aren't understanding what a frame of reference is.

0 Replies
 
maxdancona
 
  1  
Reply Thu 17 Dec, 2020 01:17 am
@Fruityloop,
I think what you misunderstand is that what you are calling "delta_t" is dependent on the reference frame. Delta t in Bob's reference frame doesn't equal delta t in Sally's reference frame.

When you talk about time dilation, you need to explain who is measuring "t" and whose clock that are observing. If you confused the frames of reference you won't get the correct answer.
Fruityloop
 
  1  
Reply Thu 17 Dec, 2020 04:00 am
@maxdancona,
Let's try this again....
Do you know how the equation delta T = delta T_o/((1-v^2)^.5) is derived?
If you don't know, this is how...
Imagine 2 people, one of them is inside a spaceship, her name is Sally.
From floor to ceiling of her spaceship it is 0.8 light-seconds.
Imagine Sally is moving to the right at 0.6c.
The other person is named John, he is outside the spaceship.
Sally has light clock with the light source aimed straight up. Sally sees her light go straight up and bounce back and forth between the floor and the ceiling. According to John, because Sally is moving to the right the light forms the hypotenuse of a 3-4-5 right triangle with sides 0.6 light-seconds, height 0.8 light-seconds, and hypotenuse 1 light-second. By applying the Pythagorean theorem and some math we get delta T = delta T_o/((1-v^2)^.5) so in this specific case it is delta T = delta T_o/((1-.6^2)^.5).
Delta T is the change in time for John and delta T_o is the change in time for Sally.
So in this case 1 second for John = 0.8 seconds for Sally.

Now let's forget about that situation for a moment and concentrate on a different situation.
Imagine 2 people, one of them is inside a spaceship, her name is Sally.
From floor to ceiling of her spaceship it is 0.8 light-seconds.
Imagine Sally is moving to the right at 0.6c.
The other person is named John, he is outside the spaceship.
Sally has a flashlight inside her spaceship that is aimed so that it follows the same path upward and to the left as John saw in the first situation.
Sally sees the light form a 3-4-5 right triangle with the floor with sides 0.6 light seconds, 0.8 light seconds, and hypotenuse 1 light-second.
Because Sally is moving to the right at 0.6c and the leftward movement of the light is 0.6c, John sees the light bouncing straight up and down.
By applying the Pythagorean theorem and some math we get delta T_o = delta T/((1-v^2)^.5) so in this specific case it is delta T_o = delta T/((1-.6^2)^.5).
Delta T is the change in time for John and delta T_o is the change in time for Sally.

So 0.8 seconds for John = 1 second for Sally.
So why would moving clocks run fast or slow depending upon how Sally is holding her flashlight?
maxdancona
 
  1  
Reply Thu 17 Dec, 2020 07:38 am
@Fruityloop,
Quote:
Delta T is the change in time for John and delta T_o is the change in time for Sally.

So 0.8 seconds for John = 1 second for Sally.



This is what you are misunderstanding.

Any delta_t has to be measured in one frame of reference.

Measured from Sally's frame of reference delta_t_o (how Sally observes her own clock) is 1.0s. Measured from Sally's frame of reference delta_t (how Sally observes John's clock) is 0.8s.

Of course they are different. That is why it is called time dilation. There is no contradiction.

You are spiitjng out values without noting the frame of reference. That is why you are confusing the frames.
0 Replies
 
maxdancona
 
  1  
Reply Thu 17 Dec, 2020 07:51 am
@Fruityloop,
Humans have a very strong intuition that there is an absolute time... that some Universal clock ticks off absolute seconds that are perceived equally by everyone. That isn't true, but it is a very difficult intuition to shake.

If you take a beginning Physics class (something I recommend) you will be taught first to work with Gallileo's realtivity with objects on ships. Then you will work through the math of different frames of reference using Newton's laws. Here velocity is measured differently in different reference frames, but Newton believed incorrectly in absolute time. By thinking about Newton you will be able to think intuitively about reference frames. And you can then understand the problem with Newton's conception of time.

Generally students master Gallilean realtivity and Newton's laws before diving into Einstein's relativity.

You need to understand reference frames.
Fruityloop
 
  1  
Reply Thu 17 Dec, 2020 08:33 am
@maxdancona,
Quote:
Delta T is the change in time for John and delta T_o is the change in time for Sally.

So 0.8 seconds for John = 1 second for Sally.
That is what you are misunderstanding.

You're hung up on the variable names and that's why you're not getting it? Ok. I'll fix that.

Let's try this again....
Do you know how the equation delta A = delta B/((1-v^2)^.5) is derived?
If you don't know, this is how...
Imagine 2 people, one of them is inside a spaceship, her name is Sally.
From floor to ceiling of her spaceship it is 0.8 light-seconds.
Imagine Sally is moving to the right at 0.6c.
The other person is named John, he is outside the spaceship.
Sally has light clock with the light source aimed straight up. Sally sees her light go straight up and bounce back and forth between the floor and the ceiling. According to John, because Sally is moving to the right the light forms the hypotenuse of a 3-4-5 right triangle with sides 0.6 light-seconds, height 0.8 light-seconds, and hypotenuse 1 light-second. By applying the Pythagorean theorem and some math we get delta A = delta B/((1-v^2)^.5) so in this specific case it is delta A = delta B/((1-.6^2)^.5).
Delta A is the change in time for John and delta B is the change in time for Sally.
So in this case 1 second for John = 0.8 seconds for Sally.

Now let's forget about that situation for a moment and concentrate on a different situation.
Imagine 2 people, one of them is inside a spaceship, her name is Sally.
From floor to ceiling of her spaceship it is 0.8 light-seconds.
Imagine Sally is moving to the right at 0.6c.
The other person is named John, he is outside the spaceship.
Sally has a flashlight inside her spaceship that is aimed so that it follows the same path upward and to the left as John saw in the first situation.
Sally sees the light form a 3-4-5 right triangle with the floor with sides 0.6 light seconds, 0.8 light seconds, and hypotenuse 1 light-second.
Because Sally is moving to the right at 0.6c and the leftward movement of the light is 0.6c, John sees the light bouncing straight up and down.
By applying the Pythagorean theorem and some math we get delta C = delta D/((1-v^2)^.5) so in this specific case it is delta C = delta D/((1-.6^2)^.5).
Delta D is the change in time for John and delta C is the change in time for Sally.

So 0.8 seconds for John = 1 second for Sally.
So why would moving clocks run fast or slow depending upon how Sally is holding her flashlight?

So did changing the variable names help you to understand?
I can change the variables to w,x,y, and z if you like.
maxdancona
 
  1  
Reply Thu 17 Dec, 2020 09:18 am
@Fruityloop,
I would like you to define four variables

1. Time elapsed on John's clock as observed by John.
2. Time elapsed on Sally's clock as observed by John.
3. Time elapsed on John's clock as observed by Sally.
4. Time elapsed on.Sallys clock as observed by Sally.

If you keep these four variabless straight, you will see there is no contradiction.

I don't care what you variable namea you use, you need need to be clear on these four values.

So if you want me to understand simply tell me..

Value #1 is delta_t (or delta_o)
Value #2 is __________
Value #3 is __________
Value #4 is _________

I can fill in these blanks with the correct values and no contradiction. But please give me your answers and we can compare notes.
Fruityloop
 
  1  
Reply Thu 17 Dec, 2020 04:49 pm
@maxdancona,
I will take a specific example of 10 minutes passing for John and 15 minutes passing for Sally within their own frame of reference and Sally is passing John at 0.8c.

1. delta T which is 10 minutes for John.
2. delta T_o now we must apply 2 equations each of which are equally correct.

10 = delta T_o/((1-.8^2)^.5)
delta T_o = 10*0.6 = 6
So in John's 10 minutes he observes 6 minutes passing for Sally.

But the other equation is equally applicable as I have shown...
delta T_o = 10/((1-.8^2).5)
delta T_o = 16.6667
So in John's 10 minutes he observes 16.6667 minutes passing for Sally.

3. delta T_o now we must apply 2 equations each of which are equally correct.

15 = delta T_o/((1-.8^2)^.5)
delta T_o = 15*0.6 = 9
So in Sally's 15 minutes she observes 9 minutes passing for John.

delta T_o = 15/((1-.8^2)^.5)
delta T_o = 25

So in Sally's 15 minutes she observes 25 minutes passing for John.

4. delta T

So in conclusion all of the following statements are true:

John sees 10 minutes passing for himself.
John sees 6 minutes passing for Sally.
John sees 16.6667 minutes passing for Sally.

Sally sees 15 minutes passing for herself.
Sally sees 9 minutes passing for John.
Sally sees 25 minutes passing for John.

So how can John see both 6 minutes and also 16.6667 minutes passing for Sally?
So how can Sally see both 9 minutes and also 25 minutes passing for John?
 

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