In conclusion

The previous transformation law cannot be true if the speed of light is a constant in every frame. Consider measuring the speed of light $v_B$ in frame $B$ via our previous definition $v_B=\Delta x_B/\Delta t_B$ . We have

$v_B= \frac{\Delta x_B}{\Delta t_B}=\frac{\Delta x_A-v\Delta t_A}{\Delta t_A}\neq \frac{\Delta x_A}{\Delta t_A}=v_A$

In other words, no velocity remains unchanged so this transformation law can't be right if the speed of light is the same in every reference frame. Therefore, to agree with observation we need to properly define how the coordinates (which, after all, are the values of physical clocks and rulers) of relatively moving reference frames relate. This is the domain of special relativity, which we will get to in the next set.

#Physics

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Comments

Why do we take it for granted that the velocity of light is a constant?

Agnishom Chattopadhyay - 6 years, 12 months ago

we don't. We experimemt a observe the speed of light is a constant.

David Mattingly Staff - 6 years, 12 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$