Lorentz Contraction in Supersonic Jets

AI Thread Summary
Supersonic jets traveling at speeds of approximately 3 * 10^-6 times the speed of light exhibit negligible length contraction due to their low velocity. The relevant formula for length contraction is gamma = L(proper)/L, where gamma approaches 1 at these speeds, indicating minimal contraction. Although the calculations initially suggest no observable contraction, it is clarified that any contraction is too small to be detected. Users discuss the limitations of standard calculators in providing precise values for gamma, suggesting the use of computer algebra programs or binomial approximations for more accurate results. Ultimately, the discussion emphasizes that while contraction exists, it is practically imperceptible at such low velocities.
Benzoate
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Homework Statement


Supersonic jets achieve maximum speeds of about (3 *10^-6)*c.

By what percentage would observe such a jet to be contracted in length?


Homework Equations



The equations for these this problem would be : 1/gamma= L(proper)/Length = 1/sqrt(1-v^2/c^2) ; beta=v/c

The Attempt at a Solution



1/sqrt(1-(v/c)^2)^.5 => 1/sqrt(1-(beta)^2)= 1/sqrt(1-(3e-6)^2)^.5=1=gamma; which leads to gamma=1=L(proper)/Length or L = (1/gamma)*L(proper) = > L=L(proper) leading me to the conclusion that the length of the jet did not contract at all .
 
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You must remember that you have approximated the value of gamma to 1, and that DOES NOT imply that no contraction occurs at all. A better answer would be that the contraction that does occur is too negligible to be observed.

Btw, your formula for length contrction: \frac{1}{\gamma} = \frac{L_o}{L}, is incorrect.
 
neutrino said:
You must remember that you have approximated the value of gamma to 1, and that DOES NOT imply that no contraction occurs at all. A better answer would be that the contraction that does occur is too negligible to be observed.

Btw, your formula for length contrction: \frac{1}{\gamma} = \frac{L_o}{L}, is incorrect.

Oh okay, thank you. the correct formula should be gamma=L(P)/L My TI 83 Plus will only make approximations to 1. is there any other computational tool I can used that will make more precise approximations?
 
Benzoate said:
is there any other computational tool I can used that will make more precise approximations?

I'm sure there must be one, but I'm afraid I don't know any. All calculators I have access to give me 1.
 
You can try a computer algebra program, but I can't guarantee it will work. Your best bet would probably be to use the binomial approximation on gamma.
Taking only the first two terms of the approximation should be exact enough here. If not, the approximation to three terms is:

(1+x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2
 
Last edited:
Benzoate said:
Oh okay, thank you. the correct formula should be gamma=L(P)/L My TI 83 Plus will only make approximations to 1. is there any other computational tool I can used that will make more precise approximations?

for this exp:
1/sqrt(1-(3e-6)^2)^.5?

I got
1.00000000000225
and more precisely
1.00000000000224997798170928355350144490508802983505104210845...:approve:
 
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