Flashlight Interval

Classical Mechanics Level 1

Now consider tossing the flashlight across the same room. It travels the same distance $\Delta x$ , but at a velocity less than $c$ .

Is the interval $\Delta s^{2}$ between tossing the flashlight and the flashlight hitting the wall greater than, less than, or equal to 0?

0 < 0

0

1 solution

Adam Strandberg
Feb 11, 2016

Let the average velocity of the flashlight be $v$ , where $v < c$ . Then $\Delta t = \frac{\Delta x}{v}$ and the interval is $\Delta s^2 = \Delta x^{2} - (c \frac{\Delta x}{v})^{2} = \Delta x^{2} (1 - \frac{c^{2}}{v^{2}})$

since $v < c$ , $\frac{c^{2}}{v^{2}} > 1$ , so $\Delta s^2$ is negative.

In the example where it is shown that (delta s)^2=0 :

v = dist covered/time difference

c=distance covered by light / time difference

then how can delta t=c / delta x ??

c/delta x = 1/delta t . no?

Ananya Aaniya - 5 years, 4 months ago

here, in this explanation, s is the time interval.

So, how can s=(delta x)^2 - .... ? how can the relation be taken as linear without the parameter time?

Ananya Aaniya - 5 years, 4 months ago

s isn't the time interval but the total amount of distance travelled trough spacetime, i think.

Francisco Martins - 2 years, 1 month ago

doesn't the length contraction have the formula: x'^2=x^2(1-v^2/c^2)?

Borcea Sara - 4 months, 1 week ago

Flashlight Interval

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