Basic Special Relativity (Time Dilation)

AI Thread Summary
The discussion revolves around calculating the time difference between two atomic clocks due to time dilation experienced by the Concorde traveling at 375 m/s over 8000 km. The initial approach incorrectly assumed the Lorentz factor to be 1, leading to confusion about the expected time difference of 16.7 microseconds. Participants suggest using the binomial approximation to simplify the calculations and derive the correct time difference. The key step involves rearranging the equation T - T' = Tax, where T is the time in the rest frame and T' is the proper time. Ultimately, the correct application of these concepts resolves the confusion and leads to the accurate answer.
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Homework Statement



The Concorde traveled 8000km between two places with an average speed of 375 m/s. What is the time difference between two atomic clocks, one on the train and one at rest with respect to the train?

Homework Equations



T=AT' where A is the Lorentz gamma factor

The Attempt at a Solution



Seemed simple enough the distance divided by time would yield the T (the time taken from the rest frame) so (8x10^6 m/s)/(375) and then solve for T' (proper time which would be the time of a clock on the train correct?) but the speed of 375 m/s is so miniscule in comparison to the speed of light the lorentz factor comes out to be 1 (which isn't surprising) but in the back of the book it says the answer is 16.7 microseconds...what am I doing wrong? Thanks
 
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To a good approximation for your problem, you can approximate [1-(v/c)2]1/2 using the binomial approximation (1-x)a≈1-ax for x<<1.
 
bmb2009 said:

Homework Statement



The Concorde traveled 8000km between two places with an average speed of 375 m/s. What is the time difference between two atomic clocks, one on the train and one at rest with respect to the train?

Homework Equations



T=AT' where A is the Lorentz gamma factor

The Attempt at a Solution



Seemed simple enough the distance divided by time would yield the T (the time taken from the rest frame) so (8x10^6 m/s)/(375) and then solve for T' (proper time which would be the time of a clock on the train correct?) but the speed of 375 m/s is so miniscule in comparison to the speed of light the lorentz factor comes out to be 1 (which isn't surprising) but in the back of the book it says the answer is 16.7 microseconds...what am I doing wrong? Thanks

Not using a calculator with enough precision?
 
TSny said:
To a good approximation for your problem, you can approximate [1-(v/c)2]1/2 using the binomial approximation (1-x)a≈1-ax for x<<1.
How does this help though? that just makes 1-ax = 1 - (7.8125e-13) which still is very close to 1. Where/how do you produce the 16.7 nanoseconds?
 
Last edited:
Use the binomial approximation in the equation T = AT' without plugging any numbers in yet. Then see if you can rearrange for the quantity T-T'. Then plug in numbers.
 
TSny said:
Use the binomial approximation in the equation T = AT' without plugging any numbers in yet. Then see if you can rearrange for the quantity T-T'. Then plug in numbers.

I don't think you can )easily) solve for T-T' with the expansion... T=T'/1-ax ==> T-Tax=T'...unless I am missing some simply algebra i don't see how to manipulate to solve for T-T'
 
bmb2009 said:
I don't think you can )easily) solve for T-T' with the expansion... T=T'/1-ax ==> T-Tax=T'...unless I am missing some simply algebra i don't see how to manipulate to solve for T-T'

Rearrange your last equation as T - T' = Tax

As you said in your original post, you know how to get T for the right hand side.
 
TSny said:
Rearrange your last equation as T - T' = Tax

As you said in your original post, you know how to get T for the right hand side.

ahhh finally got it.. Thanks a bunch!
 

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