Contradictory time dilation equations.

@Fruityloop,

Here is your mistake

1. You are setting the elapsed time on John's clock (#1) as 10min.
2. You are then correctly calculating the elapsed time on Sally's clock as measured by John (#2).

3. Your second equation is wrong. You seem to be setting the the elapsed time on Sally's clock as observed by John (#2) is 10min. Then you are calculating that the elapsed time John's clock as measured by John for this new situation is 16.6minutes.

You are confusing what you are calculating.

1 Reply

@maxdancona,

Quote:

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.

When you keep your variables straight what you calculated is.

1. In the case John observes 10min elapsed on his clock, he observes 6 minutes pass on Sally's clock
2. In the case John observes 10minutes pass on Sally's clock, 16.67 minutes he observes 16.67 minutes has passed on his own clock.
3. In the case where Sally observes 15min pass on her own clock, she observes 9 minutes pass on John's clock.
4. In the case where Sally observes 15 minutes passing on John's clock, she observes 25 minutes passing her own.

You have to use the function correctly.

T1 = T0 * √(1 - v^2/c^2)

T0 is always the time measured by X on X's clock (in the reference frame where X is stationary)

F1 is always the time measured by X on Y's clock (in the reference frame where Y is moving at speed v).

If you apply the equation correctly, there is no contradiction

1 Reply

@maxdancona,

You need to read this again:

Quote:

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)

You have completely ignored everything I wrote.
There are 2 equations that I have legitimately derived that are in conflict with each other.

Quote:

Your second equation is wrong. You seem to be setting the the elapsed time on Sally's clock as observed by John (#2) is 10min. Then you are calculating that the elapsed time John's clock as measured by John for this new situation is 16.6minutes.

Nope. My second equation is correct. I derived it in the same legitimate fashion as the first equation so it is just as correct.

Quote:

In the case John observes 10min elapsed on his clock, he observes 6 minutes pass on Sally's clock
2. In the case John observes 10minutes pass on Sally's clock, 16.67 minutes he observes 16.67 minutes has passed on his own clock.
3. In the case where Sally observes 15min pass on her own clock, she observes 9 minutes pass on John's clock.
4. In the case where Sally observes 15 minutes passing on John's clock, she observes 25 minutes passing her own.

Once again you have completely ignored the legitimate derivation of the second time dilation equation which shows that moving clocks run fast.

Quote:

You have to use the function correctly.

T1 = T0 * √(1 - v^2/c^2)

Once again you have completely ignored the legitimate derivation of the second time dilation equation which shows that moving clocks run fast.

So do you normally just ignore and dismiss what people write and don't even think about it?

2 Replies

@Fruityloop,

1. You derived you own equations.
2. You have found a contradiction in your own equations. This suggests that you made a mistake.
3. I used the proper equations and I don't get a contradiction.

I am trying to help you figure out what you are doing wrong.

0 Replies

@Fruityloop,

Not only did I read your posts, but I went through each of your calculations to see what you were doing. Then I listed why the results of your calculations were not what you said they were.

I am responding to you, and putting effort to see what you are doing, because you seem to be putting effort into actually doing the math (which is more than many people do here).

If you can see your own mistakes, then you will learn something. The math we are doing is pretty basic. Think of it this way, Einsteins theory of relativity is 100 years old (the concept of time dilation is actually a bit older). It has been studied by tens of thousands of people and is a core of basic Physics. The odds that you have found a simple mathematical flaw in basic Physics missed by students scientists and mathemeticians is pretty low.

The odds that you are misunderstanding basic Physics is considerably higher .. and I am helping you to find out how.

0 Replies

I was hoping for a productive discussion-- but here is the correct answer (where "correct" is defined as what is taught in second semester Physics courses ).

Time elapsed on Bob's clock observed by Bob - 10s.
Time elapsed on Sally's clock observed by Sally - 8s.
Time elapsed on Sally's clock observed by Sally - 10s.
Time elapsed on Bob's clock observed by Sally - 8s.

You get these values by using the time dilation equation shown above. The symmetry you see (values 2 and 4) above are a consequence of this equation.

The twin paradox is fun (where one twin loses time in a no -stmetretical way) is far more complex, and involves a round trip including acceleration and deceleration.

1 Reply

@maxdancona,

First situation Sally with flashlight aimed upward.
S is change in time for Sally.
J is change in time for John.

D is height of spaceship. L is hypotenuse.

S = D/c
L = (D^2 + (vJ)^2)^.5
J = L/c
D = Sc
L = Jc
J = ((D^2 + (vJ)^2)^.5)/c
J = (((Sc)^2 + (vJ)^2)^.5)/c
J^2 = ((S^2)*(c^2) + (v^2)*(J^2))/c^2
J^2 = S^2 + ((v^2)*(J^2))/c^2
J^2 - ((v^2)*(J^2))/c^2 = S^2
J^2*(1-(v^2/c^2)) = S^2
J^2 = S^2/(1-(v^2/c^2))

Finally....
J = S/((1-(v^2))^.5)

Second situation Sally with flashlight aimed at an angle of 53.13 degrees up and to the left.
It is very simple. Since the situation is exactly reversed replace every S with a J and every J with an S.

S is change in time for Sally.
J is change in time for John.

D is height of spaceship. L is hypotenuse.

J = D/c
L = (D^2 + (vS)^2)^.5
S = L/c
D = Jc
L = Sc
S = ((D^2 + (vS)^2)^.5)/c
S = (((Jc)^2 + (vS)^2)^.5)/c
S^2 = ((J^2)*(c^2) + (v^2)*(S^2))/c^2
S^2 = J^2 + ((v^2)*(S^2))/c^2
S^2 - ((v^2)*(S^2))/c^2 = J^2
S^2*(1-(v^2/c^2)) = J^2
S^2 = J^2/(1-(v^2/c^2))

Finally....
S = J/((1-(v^2))^.5)

Why is this wrong or incorrect?

1 Reply

@Fruityloop,

They are correct. In fact you are deriving the same equation twice.

In both cases you are calculating the elapsed time of the clock in the stationary frame of reference as observed from the stationary frame based on the elapsed time on the moving clock viewed from the stationary frame.

Assuming X is the stationary frame and Y is the moving frame you have derived the equation

T0 = T1 / ✓(1 - v^2)

Where T1 is the elapsed time on the moving clock as viewed from the stationary frame, and T0 is the elapsed time on the stationary clock as viewed from the stationary frame (yes I typed that correct, both are viewed from the stationary frame).

I think where you are getting confused is that you are switching which is the stationary frame. Once you do that, the values can't be considered equal.

Note that the usual expression of this equation is

T1 = T0 * ✓(1 - v^2)

This is the same equation as above (all you have done is muliplied both sides to solve for T0). You have derived only one equation.

In either form of the equation T1 (the elapsed time on the moving clock observed from the stationary frame) is lower than T0 (the elapsed time on the stationary clock as measured from the stationary frame).

Is you don't confuse the reference frames, the math is pretty straightforward.

1 Reply

@maxdancona,

Quote:

I think where you are getting confused is that you are switching which is the stationary frame. Once you do that, the values can't be considered equal.

Nope. The stationary frame is John in both derivations.

1 Reply

@Fruityloop,

Quote:

Since the situation is exactly reversed replace every S with a J and every J with an S.

When you do this, you change who is stationary and who is moving.

1 Reply

@maxdancona,

Quote:

Quote:
Since the situation is exactly reversed replace every S with a J and every J with an S.

When you do this, you change who is stationary and who is moving.

Nope. In the first situation Sally sees the light bouncing back and forth straight up and down. In the second situation John sees the light bouncing back and forth straight up and down. In the first situation John sees the light forming a triangle with the bottom of the spaceship. In the second situation Sally sees the light forming a triangle with the bottom of the spaceship. That's why all that needs to be done is to replace every S with a J and every J with an S. Nothing has changed about who is moving and who is stationary.

1 Reply

@Fruityloop,

You are getting what is clearly the same equation for what you claim are two different situations. You are obviously doing something wrong. But, instead of looking for your mistake and learning something, you are insisting that you are right and that math is wrong

I will explain to you again what you are doing wrong. You are confusing the two frames of reference.

There are four different values....

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.

You are only recognizing two of these. You calculate D/c in your second derivation which you ambiguosly call "time elapsed for John". This is actually the time elapsed on Sally's clock as observed by John. But you are using it as time elapsed on John's clock.

Your consistent mistake is confusing time elapsed on Sally's clock as observed by John with time elapsed on John's clock as observed by John.

For you to make this simple logic mistake is understandable. For you to persist in your error when your results are clearly wrong is not ... especially when someone is trying to patiently explain it to you.

1 Reply

@maxdancona,

In the first situation Sally sees the light bounce straight up and down and John sees the light form a triangle.
The math is applied and out pops the equation:
J = S/((1-(v^2))^.5)

In the second situation John sees the light bounce straight up and down and Sally sees the light form a triangle.
The math is applied and out pops the equation:
S = J/((1-(v^2))^.5)

These are 2 different equations that directly follow from 2 different scenarios. Since it is exactly reversed the change in time for each observer is switched. The second equation is just as valid and shows that moving clocks run fast. If all of the physics books were rewritten with the second scenario and second equation instead of the first scenario and first equation why would that be wrong or incorrect? Obviously it would be just as acceptable as the first scenario and first equation.

2 Replies

@Fruityloop,

I don't know why you think that I am confusing the 2 frames of reference. In Sally's frame of reference she sees the light bouncing up and down in the first scenario and she sees the light forming a triangle in the second scenario. These are both viewed from within her frame of reference. In John's frame of reference he sees Sally's light forming a triangle in the first scenario and he sees Sally's light bouncing straight up and down in the second scenario. These are both viewed from within John's frame of reference. Can you provide more detail about this confusion?

0 Replies

@Fruityloop,

This is incorrect. Are doing a calculation in Sally's frame of reference to calculate what would be straight up and down to John?

This is not at all symmetrical. The up and down case for the moving frame is a special case since the motion is tangential in the moving frame.

Think of it this way, the angle in the moving frame isn't measured the same way in the stationary frame.

1 Reply

@maxdancona,

Quote:

This is incorrect. Are doing a calculation in Sally's frame of reference to calculate what would be straight up and down to John?

Is a calculation done in Sally's frame of reference to calculate what would be an angle to John in the first example?

Quote:

This is not at all symmetrical. The up and down case for the moving frame is a special case since the motion is tangential in the moving frame.

Sally is moving to the right in both situations. So how is the up and down case special and and somehow different from the angle case?

Quote:

Think of it this way, the angle in the moving frame isn't measured the same way in the stationary frame.

I don't know what you mean or why this is a problem.

I honestly don't know what you are talking about.

1 Reply

@Fruityloop,

Let's try true or false statements.
In the first example the time for the light to travel a distance of 0.8 light-seconds from floor to ceiling took 0.8 seconds for Sally.
True or false?
In the first example the time for the light to travel a distance of 1 light-second along the hypotenuse from floor to ceiling took 1 second for John.
True or false?
Therefore, 0.8 seconds for Sally is equal to 1 second for John.
True or false?

In the second example the time for the light to travel a distance of 0.8 light-seconds from floor to ceiling took 0.8 seconds for John.
True or false?
In the second example the time for the light to travel a distance of 1 light-second along the hypotenuse from floor to ceiling took 1 second for Sally.
True or false?
Therefore, 0.8 seconds for John is equal to 1 second for Sally.
True or false?

So which statement is false and why?

2 Replies

@Fruityloop,

1..True
2..True
3..True

1..False
2..True
3..False

0 Replies

@Fruityloop,

The idea is to set up a "light clock" that can show a difference between a similar geometry is two frames of reference. We use light travelling perpendicular to the direction of V because it is "orthogonal" meaning the vertical direction doesn't interact with the effect being measured.

This isn't about playing games with a flashlight.

1 Reply

@maxdancona,

So if it didn't take 0.8 seconds for light to travel 0.8 light-seconds for John how long did it take and why would light not be traveling at c relative to John?

1 Reply

Contradictory time dilation equations.

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