# Why is light affected by time dilations in space-time curvatures

My previous question on this site gave me an answer to what affects light - space warping or time warping. The answer is- both. But what now doesn't make sense to me is why light is affected by time dilations. I read that the time component of the $$4$$-velocity of light is $$0$$, so then how does time dilation affect light when it can't interact with it?

Also, can someone provide a qualitative way to express how the individual contributions of space-warping and time-warping have and effect on light's speed/extent of light diverging from its original path?

Any help is appreciated!

• light follows geodesics, not paths. there are no interactions, if the geodesic changes it follows it. – anna v May 14 at 14:39
• @annav What I meant to ask was how much does the path of the geodesic of light change because of space-curving and how much does it change because of time-curving (so what are their individual contributions)? – Apekshik Panigrahi May 14 at 15:19
• It will depend on the energy momentum tensor in the region of interest. In a simplistic analogy it is like asking :water flowing down the mountain what paths it will take? will it depend on the x, y, or z direction? it will depend on the topology of the mountain. In the case of geodesics time is one of the variables and it will depend on the solutions of the general relativity equations for those masses existing there. – anna v May 14 at 16:16
• "the time component of the $4$-velocity of light is $0$" - This is incorrect.The 4-velocity of light is a light-like vector whose magnitude is $0$, because the squared magnitudes of its time and space components are equal: $(u^0)^2 - \vec u ^2 = 0$ – safesphere May 16 at 8:03

You are correct, light moves in the time dimension with speed 0 and in the space dimension with speed c when:

1. measured locally

2. in vacuum

Now, you are asking why light is then affected by time dilations.

Now the answer is local measurement. When you do a measurement from far away, you might get a speed different from c (yes, even more then c is possible), because:

1. time dilation is caused by the difference between the stress energy (it is a misconception that gravity is caused by mass, in reality it is caused by stress-energy) at two points in space

2. when you do a measurement from a far away point, the point where you are (the observer) might have a different strength (stress-energy) gravitational zone, then the space point where you actually measure the speed of light

Now, you need the learn about the four vector. You have to accept that the four vector is set up so, and the universe is set up so, that the magnitude of the four vector needs to be c always. Now light moves in the time dimension with speed 0 as you say. And it moves in the spatial dimentisons with speed c.

But this is only true for local measurements in vacuum.

Here I have to add a note based on the correct comments, because the four velocity vector of photons (with no rest mass):

The four-velocity defined here using the proper time of an object does not exist for world lines for objects such as photons travelling at the speed of light

Now to make the explanation a little bit easier, I am using an example for photons with four velocity vectors and changing time components in gravitational fields, that would make it seem like the photon could behave as if its four velocity would be changing only when viewed from far away (Earth), where the stress-energy is different then the place where the photon is actually passing by the Sun. But in reality you have to use an affine parameter for a photon's four velocity vector instead of proper time. This is because dτ=0 for a photon. The affine parameter is a scalar and is invariant under Lorentz transformations.

AS soon as you make a measurement from far away, gravity (the difference of the gravitational field between where light passes and where you make the measurement from) will cause time dilation, and that will mean that the gravitational zone will cause time to seem to pass slower next to the Sun where light passes (relatively compared to a clock on Earth). Now the four vector's magnitude needs to be a constant always, and light seems (in reality it does not) to have started to move in the time dimension (it seems to start to experience time as we do). In reality light does not move in the time dimension, but because of the difference between the stress-energy at the Sun and Earth, time dilation will cause clocks at the Sun to seem to tick slower then clocks at Earth. Thus, light will have to seem to slow down in the spatial dimensions, the spatial dimensions will have to compensate to keep the magnitude of the four velocity vector constant.

• "the magnitude of the four vector needs to be $c$ always" - This is incorrect for light. The magnitude of the 4-velocity of light is always zero (see my comment above). "light seems to have started to move in the time dimension (it seems to start to experience time as we do)" - This is also incorrect. The (always) non-zero time component of the 4-velocity of light does NOT mean that light moves in time. To calculate the 4-velocity of light, we have to use a different affine parameter instead of proper time, because the proper time of light is always zero. – safesphere May 16 at 8:17