# How far does a photon move in the 4th dimension when it travels one light second? [closed]

In one second a photon moves 3x10^8 meters through the three spatial dimensions.

Light's velocity is 3x10^8 m/s.

If the photon moved at all in the fourth dimension, it's velocity would no longer be 3x10^8 m/s, but something different.

Ergo a photon must remain stationary in the fourth dimension.

QED

Edit:
Einstein and Minkowski defined the fourth dimension (I'm rather surprised I have to remind all the experts here of this). Do you agree? Please use their definition. How far does a photon move in the 4th dimension when it travels one light second?

## closed as unclear what you're asking by garyp, CuriousOne, AccidentalFourierTransform, ACuriousMind♦, user36790 Apr 23 '16 at 4:48

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• What is the 4th dimension? – CuriousOne Apr 22 '16 at 0:38
• Presumably related to physics.stackexchange.com/q/250303 and therefore to physics.stackexchange.com/q/33840 and the various duplicate thereof. – dmckee Apr 22 '16 at 0:45
• Comments are not for extended discussion; this conversation has been moved to chat. – dmckee Apr 22 '16 at 1:36
• See the comment above about extended discussion. Hold it in the chat room or not at all. – dmckee Apr 22 '16 at 2:23

Any two points in spacetime are linked by a four-vector that physicists conventionally write as $(x^0, x^1, x^2, x^3)$, where $x^0$ is normally the timelike dimension and the other three components are spatial. If we use the usual Cartesian coordinates in flat spacetime we'd generally write the four-vector as $(t, x, y, z)$.

In this case suppose the light starts at the origin and is directed along the $x$ axis then after one second its position is $(1, c, 0, 0)$. Assuming the fourth dimension is taken to mean time then the distance the light has moved in the fourth/time dimension in one second is, well, one second.

The question is presumably based on the idea that the distance moved by the light would be given by some kind of Pythagorean expression:

$$s^2 = t^2 + x^2 + y^2 + z^2$$

and if $t \ne 0$ this contradicts our observation that the distance moved is:

$$d^2 = x^2 + y^2 + z^2$$

That's because the quantities $s$ and $d$ are different things. $d$ is the coordinate length of the three-vector $(x, y, z)$ i.e. we are ignoring the change in the time when calculating it. This is entirely reasonable under normal circumstances, and if for example you are an engineer trying to construct a bridge please continue to calculate distances in this way! However $d$ is not a relativistic invariant and will have a different value for different observers.

The quantity $s$ is the proper length or alternatively the norm of the four-vector. Unlike the coordinate length mentioned above, the proper length is a relativistic invariant (technically a Lorentz scalar) and has the same value for all observers. We calculate the proper length using an equation called the metric, and for flat spacetime the equation turns out to be:

$$s^2 = -c^2t^2 + x^2 + y^2 + z^2 \tag{1}$$

This is called the Minkowski metric. Note that the fourth/time dimension has a negative sign. This sign change is what distinguishes the Lorentzian geometry used in relativity from the Euclidean geometry used by Pythagoras, and that negative sign is responsible for all the weird effects like time dilation and length contraction.

If you use equation (1) to calculate the proper distance the light moves one second you'll find that it turns out to be zero. The proper distance moved by anything travelling at the speed of light is always zero. Such trajectories are called null geodesics.