Lorentz transformation And Einstein's Laws.

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To=(time of moving object.)
T=(Time of standing object.)
^=power of number.
sqrt=square of number.(x^1/2)


Law of einstein say: T= To / ( sqrt( 1 - u^2/c^2 ) )
Lorentz law say: T= (To + u*Xo/c^2) / ( sqrt( 1 - u^2/c^2 ) )


Why there is difference??
Is for other things?? I thought is for the same reasons...
 
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  • #2
Nugatory
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The time dilation formula that you're calling "the law of Einstein" is derived from the Lorentz transformation.

There's only one way to understand the different meaning of the two formulas, and that's to do the derivation yourself. Start with the Lorentz transforms, and use them to answer the questions:

When my clock advances from time ##t## to ##t+\Delta{t}##, how much does the moving clock advance? What does the moving clock read when my clock reads ##t##? What does the moving clock read when my clock reads ##t+\Delta{t}##? What is the difference between those values?
 
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So,for example we have two humans. One in earth and one in spaceship.
To count the difference of time that they count,which law is the correct?

Both,counts the time of standing,relative to the time of moving person.
 
  • #4
Nugatory
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So,for example we have two humans. One in earth and one in spaceship.
To count the difference of time that they count,which law is the correct?

Both,counts the time of standing,relative to the time of moving person.
Clock on earth advances from ##t## to ##t+\Delta{t}##. What did the spaceship clock read when the earth clock read ##t##? What did the spaceship clock read when the earth clock read ##t+\Delta{t}##? Use the Lorentz transform to answer these questions.

The difference between these values will give you the amount of time that passed on the spaceship while an interval ##\Delta{t}## passed on earth. What does that formula look like?
 
  • #5
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The einstein law,will not give me correct answers??
 
  • #6
Orodruin
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What you call the Einstein law is determining the relation between time intervals for stationary and moving objects. What you call the Lorentz law is just the time component of the Lorentz transformation, which relates space-time coordinates of different inertial frames.

If you start out with both objects at x = 0 at t = 0 (which according to the standard setup Lorentz transformation is also t' = 0 and x' = 0) and the moving object moves with velocity v, then at time t = T the moving object will be at x = vT. The t' coordinate (i.e., the time that a clock moving with the moving object will show) is then (according to the Lorentz transformation)

t' = T0 = (t - vx/c^2)/sqrt(1-v^2/c^2) = (T - v^2 T/c^2)/sqrt(1-v^2/c^2) = T sqrt(1 - v^2/c^2)

resulting in the time dilation

T = T0/sqrt(1-v^2/c^2)
 
  • #7
stevendaryl
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To=(time of moving object.)
T=(Time of standing object.)
^=power of number.
sqrt=square of number.(x^1/2)


Law of einstein say: T= To / ( sqrt( 1 - u^2/c^2 ) )
Lorentz law say: T= (To + u*Xo/c^2) / ( sqrt( 1 - u^2/c^2 ) )


Why there is difference??
Is for other things?? I thought is for the same reasons...
If you have two events [itex]e_1[/itex] and [itex]e_2[/itex] that take place at different times and locations, and you let [itex]\delta t[/itex] be the time between the events in one frame, frame [itex]F[/itex], and [itex]\delta x[/itex] be the distance between them in that frame, then in the other frame, frame [itex]F'[/itex], you have:

[itex]\delta t' = \dfrac{1}{\sqrt{1-u^2/c^2}} (\delta t - u\ \delta x/c^2)[/itex]
[itex]\delta x' = \dfrac{1}{\sqrt{1-u^2/c^2}} (\delta x - u\ \delta t)[/itex]

Now, let's look at two special cases:

Case 1: the two events take place at the same location, according to frame [itex]F[/itex].

In this case, [itex]\delta x = 0[/itex]. So the formula for [itex]\delta t'[/itex] becomes:

[itex]\delta t' = \dfrac{1}{\sqrt{1-u^2/c^2}} \delta t[/itex]

Case 2: the two events take place at the same location, according to frame [itex]F'[/itex].

In this case, [itex]\delta x' = 0[/itex]. This means that [itex]\delta x = u\ \delta t[/itex].

In this case, the formula for [itex]\delta t'[/itex] becomes:

[itex]\delta t' = \dfrac{1}{\sqrt{1-u^2/c^2}} (\delta t - u\ (u\ \delta t)/c^2)[/itex]

which simplifies to:

[itex]\delta t' = \sqrt{1-u^2/c^2} \delta t[/itex]
 
  • #8
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If you have two events [itex]e_1[/itex] and [itex]e_2[/itex] that take place at different times and locations, and you let [itex]\delta t[/itex] be the time between the events in one frame, frame [itex]F[/itex], and [itex]\delta x[/itex] be the distance between them in that frame, then in the other frame, frame [itex]F'[/itex], you have:

[itex]\delta t' = \dfrac{1}{\sqrt{1-u^2/c^2}} (\delta t - u\ \delta x/c^2)[/itex]
[itex]\delta x' = \dfrac{1}{\sqrt{1-u^2/c^2}} (\delta x - u\ \delta t)[/itex]

Now, let's look at two special cases:

Case 1: the two events take place at the same location, according to frame [itex]F[/itex].

In this case, [itex]\delta x = 0[/itex]. So the formula for [itex]\delta t'[/itex] becomes:

[itex]\delta t' = \dfrac{1}{\sqrt{1-u^2/c^2}} \delta t[/itex]

Case 2: the two events take place at the same location, according to frame [itex]F'[/itex].

In this case, [itex]\delta x' = 0[/itex]. This means that [itex]\delta x = u\ \delta t[/itex].

In this case, the formula for [itex]\delta t'[/itex] becomes:

[itex]\delta t' = \dfrac{1}{\sqrt{1-u^2/c^2}} (\delta t - u\ (u\ \delta t)/c^2)[/itex]

which simplifies to:

[itex]\delta t' = \sqrt{1-u^2/c^2} \delta t[/itex]

So,that what I call eistein's law,is when two phenomena,are located at the same location(dx=0) ??

Thank you by the way!!!
 

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