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Alfred Centauri
Alfred Centauri
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My professorIn my lecture notes, there are two frames S and S'. Now the isThe prime frame moves with uniform velocity with respect to the unprimed frame. In this frame now, she drivesderives the time dilation equation in the following way he:

She assigns $t$ as time on thein S as t,and distance xas $x$. Now from inverse transformations $t_1= \gamma (t_1' + \frac{vx_1'}{c^2})$ ;;$ t_2 = \gamma (t_2' + \frac{vx_1'}{c^2})$ then:

$$t_1= \gamma (t_1' + \frac{vx_1'}{c^2})$$

$$t_2 = \gamma (t_2' + \frac{vx_1'}{c^2})$$

Now, if betweenthe both the positions are the same, how S' is S' moving?

My professor lecture notes there are two frames S and S'. Now the is prime moves with uniform velocity with respect to this frame now she drives the time dilation equation in the following way he assigns time on the S as t, distance x. Now from inverse transformations $t_1= \gamma (t_1' + \frac{vx_1'}{c^2})$ ;;$ t_2 = \gamma (t_2' + \frac{vx_1'}{c^2})$ then if between both the positions are same how S' is moving?

In my lecture notes, there are two frames S and S'. The prime frame moves with uniform velocity with respect to the unprimed frame. In this frame, she derives the time dilation equation in the following way:

She assigns $t$ as time in S and distance as $x$. Now from inverse transformations:

$$t_1= \gamma (t_1' + \frac{vx_1'}{c^2})$$

$$t_2 = \gamma (t_2' + \frac{vx_1'}{c^2})$$

Now, if the both the positions are the same, how is S' moving?

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Nobody recognizeable
Nobody recognizeable
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Problem on derivation of time dilation

My professor lecture notes there are two frames S and S'. Now the is prime moves with uniform velocity with respect to this frame now she drives the time dilation equation in the following way he assigns time on the S as t, distance x. Now from inverse transformations $t_1= \gamma (t_1' + \frac{vx_1'}{c^2})$ ;;$ t_2 = \gamma (t_2' + \frac{vx_1'}{c^2})$ then if between both the positions are same how S' is moving?