special relativity - Does travelling towards something at relativistic speed cause it to appear to speed up?

If one were to travel towards a giant TV floating in space, at a fraction of the speed of light, would a video playing on the TV appear to play faster?

I'm guessing the answer is no. From what I understand, because the speed of light in a vacuum is the same for all observers, the video will not speed up. And because the light from the edges of the screen have a longer distance to travel, the image at the edges will be lagged behind the image in the center.

I'm expecting that we'll get some weird lensing of the screen, which I haven't wrapped my head around.

Question is inspired by this video and a bunch of other videos attempting to explain relativity to the layman.

Answer

Actually, the received transmission does speed up as per Doppler blue shift.

If you'd look directly at the TV screen with a telescope, then the image would also be blue shifted, for the same reason. If you'd look at a digital transmission of the image, you would retrieve the correct pixels, but the rate at which the frames are received would still speed up. The proof parallels closely that for the Doppler blue shift.

Let the TV and the receiver be Einstein synchronized to the moment when they pass each other, $t = t' = 0$, and let the TV come toward the receiver at relative velocity $v = \beta c >0$ in the positive $x$ direction. Say the TV, at $x' =0$ in its rest frame, emits consecutive frames at times $-t'<0$, $-t'+T'<0$. If $\gamma = 1/\sqrt{1-\beta^2}$ is the factor of time dilation, the receiver sees these frames emitted at $$ {\bar x} = -\gamma v t' = -\beta \gamma ct'\\ {\bar t} = - \gamma t' $$ and $$ {\bar x}_1 = -\gamma v t' + \gamma vT' = -\beta \gamma ct' + \beta \gamma cT'\\ {\bar t}_1 = - \gamma t' + \gamma T' $$ and so he receives them at his location $x=0$ at times $$ t = {\bar t} + \frac{|{\bar x}|}{c} = - (1-\beta) \gamma t'\\ t_1 = {\bar t}_1 + \frac{|{\bar x}_1|}{c} = -(1-\beta) \gamma t' + (1-\beta)\gamma T' $$ From this the period at which the frames are received is $$ T \equiv t_1 - t = (1-\beta)\gamma T' = \sqrt{\frac{1-\beta}{1+\beta}}< T' $$ and therefore the frequency is $$ f \equiv \frac{1}{T} = \sqrt{\frac{1+\beta}{1-\beta}}\;f' > f' $$

Similarly, a transmission from a TV moving away from you would appear slowed down due to red shift ($\beta \rightarrow -\beta$).