The Special Relativity Theory says that the moving clock is slower. It results from the the transformation equation for time that shows time dilation:

$$\Delta t' = \Delta t \gamma$$

where $\Delta t'$ is time measured in a reference frame considered stationary, and $\Delta t$ is measured in a reference frame considered moving (with respect to the stationary one) and $\gamma>0$*.

As you can see, whatever time period you choose as $\Delta t$, $\Delta t'$ will be always greater, because $\Delta t$ will be multiplied by $\gamma$. This means that the stationary clock will always measure larger number of seconds for a given number of seconds measured by the moving clock, and therefore the moving clock will always be the slower one according to SR.

*$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $

The Special Relativity Theory says that the moving clock is slower. It results from the the transformation equation for time that shows time dilatation:

$$\Delta t' = \Delta t \gamma$$

where $\Delta t'$ is time measured in a reference frame considered stationary, and $\Delta t$ is measured in a reference frame considered moving (with respect to the stationary one) and $\gamma>0$*.

As you can see, whatever time period you choose as $\Delta t$, $\Delta t'$ will be always greater, because $\Delta t$ will be multiplied by $\gamma$. This means that the stationary clock will always measure larger number of seconds for a given number of seconds measured by the moving clock, and therefore the moving clock will always be the slower one according to SR.

*$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $

The Special Relativity Theory says that the moving clock is slower. It results from the the transformation equation for time that shows time dilatation:

$$\Delta t' = \Delta t \gamma$$

where $\Delta t'$ is time measured in a reference frame considered stationary, and $\Delta t$ is measured in a reference frame considered moving (with respect to the stationary one) and $\gamma>0$*.

As you can see, whatever time period you choose as $\Delta t$, $\Delta t'$ will be always greater, because $\Delta t$ will be multiplied by $\gamma$. This means that the stationary clock will always measure larger number of seconds for a given number of seconds measured by the moving clock, and therefore the moving clock will always be the slower one according to SR.

*$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $

The Special Relativity Theory says that the moving clock is slower. It results from the the transformation equation for time that shows time dilation:

$$\Delta t' = \Delta t \gamma$$

where $\Delta t'$ is time measured in a reference frame considered stationary, and $\Delta t$ is measured in a reference frame considered moving (with respect to the stationary one) and $\gamma>0$*.

As you can see, whatever time period you choose as $\Delta t$, $\Delta t'$ will be always greater, because $\Delta t$ will be multiplied by $\gamma$. This means that the stationary clock will always measure larger number of seconds for a given number of seconds measured by the moving clock, and therefore the moving clock will always be the slower one according to SR.

*$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $