# Why do physicists multiply the time component of a space-time vector by $c$ to define the conventional four-vector?

When I think of a space-time vector, I think of it as a geometrical object in Minkowski space with the physical properties of magnitude and orientation wrt other space-time vectors. I can use four numbers to represent it uniquely using measurements from rulers and clocks that are physically different to one another. From this, I can define the differential space-time vector as:

$$dx_i = (d\vec r, dt)$$

Instead, physicists decide to convert the time component of the above into a space interval measurement through multiplication by $c$, giving the differential four-vector as:

$$dx_i = (d\vec r, cdt)$$

To me, it now looks like a vector with four space components which isn't consistent with the original space-time vector where the time component is physically different, and hence can't be combined with the other three space components: Why is this conversion done?

• I would say that physicists don't actually do what you're saying. What they really do is employ a system of units in which $c=1$.
– user4552
Feb 18, 2018 at 1:36
• I think this question deserves more credit than it currently gets and is more fundamental than only pertaining to differential four-vectors. Why is it "allowed" to just multiply time by c? We wouldn't (couldn't?) do that in non-relativistic space-time diagrams/problems, would we? Often it is introduced to students simply as a conversion factor that "makes light move at 45°" but it obviously does more: it changes time into distance (even in natural units c is a velocity: length per time). So do I assume correctly that this is reasonable because c is what fundamentally "connects" space and time? Feb 3 at 12:36

First, we want all four components to have the same units. This is done to make the Lorentz transformations unitless, like rotations are. Second, it requires less writing than dividing the spatial components by $c$.

• Space-like and time-like vectors can't be rotated into one another entirely, so I don't see how they can be made to have the same units Feb 18, 2018 at 0:42
• @Physikslover They can be mixed with one another and if you’re allowing such mixing they better have the same units! Feb 18, 2018 at 0:57

It's important that all components of a vector have the same units if we want to think about transforming coordinates in a way that is consistent. That is, we want the Lorentz transforms $x^{\mu}\to\Lambda^{\mu}_{\,\,\nu}x^{\nu}$ to be dimensionless (corresponding to complexified rotations in four-dimensions).

More conceptually, if we wish to treat space and time on the same footing, then it is important that we use the same units for them. Typically theorists do this by picking units in which one lightyear and one year are the same thing ($c=1$).

I hope this helps!

I'll throw my hat in here as well, but include the all important graphic: If you scale your time axis by $c$ (or set $c=1$), then light rays traveling at $c$ automatically travel at $45^\circ$ (a line with slope $c/c = 1$). See the left figure. Without the scaling, light-like paths get scrunched down by the spatial directions. See the right figure. Clearly, the left figure is more pleasing than the right one. This is what is meant when we put the time axis "on equal footing" with the spatial ones.

The separate time and space units from Euclidean space+time are combined into a single space-time unit in Minkowski space-time. This now makes $c$ dimensionless whereas before it had units of $ms^{-1}$

Therefore, multiplying the time-like component in a four-vector by $c$ doesn't do any sort of unit conversion and merely scales the relative magnitudes of time-like vectors to their corresponding space-like vectors.