deda
Jan21-04, 04:03 AM
regarding my page Einstein vs him self (http://www.geocities.com/dr_physica/evhs.htm) I must admit I was wrong due to an optical illusion reading my primary book.
In "General physics 1" by dr.Zafir Stojanov Skopje 1995 I had this equations:
L'=L\sqrt{1-\frac{V^2}{c^2}} for contraction of length and
T=\frac{T'}{\sqrt{1-\frac{V^2}{c^2}}} for dilation of time
In "Special theory of relativity" by dr. Vida Zigman Belgrade 1997 I have this equations:
L'=L\sqrt{1-\frac{V^2}{c^2}} for contraction of length and
T'=T\sqrt{1-\frac{V^2}{c^2}} for dilation of time.
If one looks carefully he'll notice that those two sets of equations are actually identical but only expressed differently. I always tougth that the time measured in rest frame is the shortest but in fact it's the longest. when traveling at greater speeds the observer measures both shorter length and proportionally shorter time.
\frac{L'}{T'}=\frac{L}{T} this ratio is some speed that is constant and same in any frame. It's c. SR tries to measure every thing trhu c.
I'll sure have more questions as I make progress knowing SR. At this time I'll be just quite.
My appologies for all the incovinience.
In "General physics 1" by dr.Zafir Stojanov Skopje 1995 I had this equations:
L'=L\sqrt{1-\frac{V^2}{c^2}} for contraction of length and
T=\frac{T'}{\sqrt{1-\frac{V^2}{c^2}}} for dilation of time
In "Special theory of relativity" by dr. Vida Zigman Belgrade 1997 I have this equations:
L'=L\sqrt{1-\frac{V^2}{c^2}} for contraction of length and
T'=T\sqrt{1-\frac{V^2}{c^2}} for dilation of time.
If one looks carefully he'll notice that those two sets of equations are actually identical but only expressed differently. I always tougth that the time measured in rest frame is the shortest but in fact it's the longest. when traveling at greater speeds the observer measures both shorter length and proportionally shorter time.
\frac{L'}{T'}=\frac{L}{T} this ratio is some speed that is constant and same in any frame. It's c. SR tries to measure every thing trhu c.
I'll sure have more questions as I make progress knowing SR. At this time I'll be just quite.
My appologies for all the incovinience.